help-23

[faint seal]

for 3 specifically we could have something like (1,1,0) and (1,omega,0) where omega=(-1+sqrt(3)i)/2 so that (2,1+omega,0) doesn't lie in that set

[obtuse jackal]

it's more standard to call it j afaik

1, j, j²

[faint seal]

i have never seen that notation

[obtuse jackal]

where did you see omega?

[faint seal]

maybe it's more common in higher math

[faint seal]

every place i've seen the cube root of unity it's always been omega lol

[obtuse jackal]

omega is used for the root of unity of a random power

[faint seal]

almost everywhere

[obtuse jackal]

for cube we often use j

[faint seal]

interesting

[obtuse jackal]

at least in my experience

[faint seal]

maybe it's an industry thing

[obtuse jackal]

nah

just physics folks aren't gonna use j

cause j is i over there

[faint seal]

we hate physicists

[obtuse jackal]

good

[obtuse jackal]

what about CS majors?

[faint seal]

if we have something like $U={(x,x)\in \mathbb{R}^2}\cup {(x,2x)\in \mathbb{R}^2}$

it's the two lines x=y and 2x=y

[obtuse jackal]

the intersection is just {(0,0)}, are you sure about that?

[faint seal]

whoops

[flat frigate]

kheerii

[obtuse jackal]

go on

[faint seal]

if (x,y) lies in U then (-x,-y) lies in U

because either y=2x or y=x, and in both cases we can find (-x,-y) to satisfy the same equation

but (1,1)+(1,2) = (2,3) is not in this set

[obtuse jackal]

so?

[faint seal]

so it's not a subspace

[obtuse jackal]

it's not closed under addition

[faint seal]

oh wait

shit'

this still feels like it's false though

[obtuse jackal]

what feels lacking?

[faint seal]

scalar multiplication

[obtuse jackal]

it does look lacking

[faint seal]

if we just take the integers

[obtuse jackal]

yes

[obtuse jackal]

this is called a subgroup

[faint seal]

like U={(x,0):x\in Z}

then it's not closed under scalar multiplication

[obtuse jackal]

a group is a nonempty set with some operation called addition that must be associative, and have additive inverses
Must have an additive identity (i.e. 0)
no multiplication operation

[faint seal]

oh so it's just a vector space without scalar multiplication

[obtuse jackal]

pretty much

[faint seal]

but without commutativity either

[obtuse jackal]

those are specifically called abelian groups

[faint seal]

the ones which are commutative or the ones which aren't?

[obtuse jackal]

(after Niels Henrik Abel)

[obtuse jackal]

abelian is a fancy word for groups to mean commutative

[faint seal]

i've heard that name before

Abel summations

[obtuse jackal]

see any intro abstract algebra ressource for further information

[obtuse jackal]

yeah he did a few things in analysis

[faint seal]

where is it?

[obtuse jackal]

because groups are the most elementary structure in abstract algebra

I'm not going to recommend you an AA book myself

[faint seal]

oh wait I read "see my intro abstract algebra course"

[obtuse jackal]

yeah no I'm not a prof
And even if I were, I def wouldn't be teaching AA

[faint seal]

lol why not?

you don't enjoy it?

[obtuse jackal]

not my thing

[faint seal]

gotcha

[obtuse jackal]

maybe if I took the time, but atm I'm much more interesting in studying analysis and proba/stats

as far as math is concerned

[faint seal]

yeah prob/stats is definitely much more interesting

imo

[obtuse jackal]

for me it'd be either stochastic processes/martingale theory, or more specifically markovian decision processes

a generalization of Markov chains

[faint seal]

sounds super cool!

hopefully one day I understand those words haha

[obtuse jackal]

if you see a grad class in proba theory or stochastic processes, it probably covers it

(martingales, not MDPs)

MDPs lean a lot more into CS

[faint seal]

that's a long way away for now, I'm in high school right now

i'm just studying linalg for fun right now

[obtuse jackal]

well you're doing good

there's no such thing as knowing too much linalg anyway
not in math nor in CS

[faint seal]

thank you!

i do have a bad habit of rushing things but i think i'm going at a decent pace right now

[faint seal]

yep

so essentially I need a subset which is closed under scalar multiplication but not under addition

[obtuse jackal]

||short term memory||

[faint seal]

xy=1?

oh xy=0

[obtuse jackal]

yeah, a union of 2 lines works

the U you gave earlier works

[faint seal]

oh yeah huh it does

[faint seal]

this does not though

any two lines should work yeah

okay

it isn't

because sin(x)+frac(x) is not periodic

[obtuse jackal]

this classic always comes up every now and then

[faint seal]

oo

[obtuse jackal]

this is actually surprisingly hard to prove

[faint seal]

so I have heard

but I couldn't think of a better example

so

¯_(ツ)_/¯

[obtuse jackal]

I looked it up in the past, finding a provable exemple is not exactly simple

[faint seal]

i have also looked up a variant of this kind of thing before

i dont remember what they did tho

[obtuse jackal]

https://math.stackexchange.com/questions/1079/sum-of-two-periodic-functions

[faint seal]

oh yeah I saw this exact post before

[safe radish]

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[viscid crane]

I think this is impossible

[viscid crane]

I would deeply appreciate a step by step way, this is so hard, how am I supposed to take the biggest exponent

There’s like so many things inside a single question

[fickle monolith]

,rccw

[safe radish]

@viscid crane Has your question been resolved?

[plucky elk]

Could start by multiplying top and bottom by the radical conjugate

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[wet pendant]

Anybody can answer my question ? Why in each equivalence, we indicate x' and y' are prime among them ?

[wet pendant]

write it in one time, isn't enough ?

[tardy mango]

Lowers the chance of someone forgetting/missing the condition

That’s rlly it

[wet pendant]

Ok thanks

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[sly seal]

can some one please help with this

$

Find all the integer pairs (m, n) such that:
[
m^3 + n^3 + 99mn = 33^3
]

\textbf{Solution:}

Starting with the given equation:
[
m^3 + n^3 + 99mn = 33^3
]

Rewriting:
[
m^3 + n^3 = 33^3 - 99mn
]

Using the sum of cubes factorization:
[
m^3 + n^3 = (m+n)(m^2 - mn + n^2)
]

Substituting, we have:
[
(m+n)(m^2 + n^2 - mn) = 33^3 - 99mn
]

Simplifying further:
[
(m+n)(m^2 + n^2 - mn) = 99(363 - mn)
]

Assuming (m+n = 99), we use the identity (m^2 + n^2 = (m+n)^2 - 2mn). Then:
[
m^2 + n^2 = 99^2 - 2mn = 9801 - 2mn
]

Plugging this back, we get:
[
99(9801 - 2mn - mn) = 99(363 - mn)
]

From here, we attempt to solve for (m) and (n). Testing (m+n = 99) and (m^2 + n^2 = 363), we find:
[
mn = 4719
]

The quadratic equation for (m) and (n) becomes:
[
x^2 - 99x + 4719 = 0
]

However, solving this equation does not yield integer solutions, suggesting an error in the assumptions or steps. It is unclear where the mistake lies.

$

[flat frigate]

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

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[worthy plover]

in part a i got 3/2 as the answer, what does part b mean when it asks for the constant term?

[neat kiln]

X^0

Ie the term with no x

[worthy plover]

ohhhhh

ok thank you

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[sick ermine]

can we talk about horizontal stretching and condensing of logarithmic functions or do i need to provide a specific example?

[mortal sandal]

Ik what you mean without an example if that's what you're asking

[sick ermine]

my problem is actually very broad, so i'm going to try a tutoring site

thank you!

.close

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[novel nebula]

Need help

Solvin

Huh

Meant to send this lmao

<@&286206848099549185>

[vernal solar]

,rotate

[vernal solar]

Oh it's the same image

My bad

[novel nebula]

Bwahhh

[vernal solar]

Are you sure your factorisation on the LHS bottom fraction is correct?

Actually

[novel nebula]

Itsss uhm

4x^2+5x+1

[vernal solar]

nvm

[novel nebula]

So

4

[vernal solar]

What are you trying to do?

[novel nebula]

Simplify it :p

[vernal solar]

Right okay

[novel nebula]

~〰︎~

:3

[vernal solar]

I think (5x-4)/((4x+1)(x+1)) is reasonably the most simplified form

[quick crater]

whats the original problem

[quick crater]

which one are you trying to do?

[novel nebula]

I gotta simplify to find the

Other stuff right?

[drowsy moss]

you can find all of that information from the simplified form given.

[novel nebula]

Do I still gotta simplify the denominator on the 1st one?

The 4x^2+5x+1

[vernal solar]

It has done that for you

[drowsy moss]

they're the same. the factored denominator is given on the right

[novel nebula]

So I'm only looking at this?

[drowsy moss]

sure

[novel nebula]

@.@

Well

Yes or no?

[drowsy moss]

you can just look at that one, you don't have to. but you can.

[novel nebula]

I feel like-

That's wrong...

[drowsy moss]

ok, then just look at that one 🤷

[novel nebula]

Well no cuz it's polynomials so...

Gotta solve both sides...

[vernal solar]

They are the exact same thing

It's the same as saying 5(x+2) = 5x+10 all you have done is rearranged the term

If you solve it or simplify it correctly you would just get 0 = 0

[drowsy moss]

if you factor the left side, you get the right side. They're the same. The problem has given you the simplified form to make the problem easier for you to answer all those questions.

[novel nebula]

I doubted you guys hard, sorry

You're right tho

[safe radish]

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[tribal needle]

hey i need help

[agile kindle]

post ur question

[tribal needle]

pk

ok

A circular park of radius 20metres is situated in a colony. three boys adam,sam and david are siting at equal distance on its boundary each having a toy telephone in his hands to talk with each other. Find the length of the string of each telephone

[unique crypt]

As a rule of thumb, particularly for this server, don't "ask to ask", just ask. It's faster that way.

[unique crypt]

Okay so

can you draw this for us?

i unserstand it, I just want to make sure you understand this as well

[tribal needle]

yeah

[unique crypt]

okay it's fine i don't actually need you to draw it i just want to make sure you did

so you understand how we essentially have an equal triangle

inside of this circle

where each boy sits at each vertice of the triangle

[tribal needle]

ok

[unique crypt]

what is the angle of an equilateral triangle?

[tribal needle]

60

[unique crypt]

good

so we've got this right

we know the red length, correct?

[tribal needle]

yes

[unique crypt]

okay now

how would you solve this?

i want to know how much geometry you know

have you worked with radians...?

[tribal needle]

RADIUS

[unique crypt]

okay

hm

you've worked with sin, cos, tan though, correct?

[tribal needle]

no

[unique crypt]

oh

okay uh

well have you been given a formula for these situations?

[tribal needle]

i know the basics

[unique crypt]

okay

well

you could break it down into two right angled triangles

let me pull up geogebra it'll look nicer that way

[tribal needle]

ok

[unique crypt]

so we've got this situation

yeah?

[tribal needle]

yes

[unique crypt]

can you tell me what the angle CBO will be?

[quick crater]

you can use 30-60-90 triangle here

[tribal needle]

30

[unique crypt]

well they're already hinting at it above

good

perfect

now, the problem is, we can;t use trigonometry right now

because we have no 90 degree angles

[tribal needle]

ok

[unique crypt]

however looking at what was sent above, can you think of a way we could use trigonometry?

maybe draw something that'll give us a right angle

[tribal needle]

by constructing OE perpendicular to BC

[unique crypt]

good!

so

[tribal needle]

dc is not equal to 20

20 is radius

[unique crypt]

oh radius 20

that's fine

OC = 20

good catch

anyways now we've got all we need

we just need to find BE

and then multiply that times 2

does that make sense?

[tribal needle]

yes

[unique crypt]

perfect

which side do we know the length of?

[tribal needle]

BO AND CO

[unique crypt]

good

so now

you are familiar with sohcahtoa?

[tribal needle]

can u tell me what it is

[unique crypt]

it's just a nice way of summarizing the following:

[tribal needle]

ok

[unique crypt]

you've looked at this?

[tribal needle]

yes

[unique crypt]

perfect

so, knowing the angle 30

we want to find its adjacent angle

so we'd use "CAH"

which tells us Cos(30) = Adjacent/Hypotenuse

What's the hypotenuse?

then you can do Cos(30) * Hypotenuse = Adjacent

and remember whe want the sum of two adjacents

so

2 * Cos(30) * Hypotenuse = 2 Adjacent

which is your answer

[tribal needle]

okay

thank you

[safe radish]

@tribal needle Has your question been resolved?

[safe radish]

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[unique crypt]

b) find all vectors v = (x,y,z) in S such that x+z=0

so we take the linear combination of S

which leaves us with:

$(\alpha + \beta + \gamma, 2 \alpha - \beta - 7 \gamma, \alpha + 2 \beta + 4 \gamma)$

[flat frigate]

ransik (gmdn)

[unique crypt]

and we know x+z = 0

so:

$2 \alpha + 3 \beta + 5 \gamma = 0$

[stiff bobcat]

3\beta ?

[unique crypt]

yep

[flat frigate]

ransik (gmdn)

[unique crypt]

anyways

not sure where to go from here

i could solve for one of the variables but

[stiff bobcat]

this defines a plane in (\alpha,\beta,\gamma) coordinates, so its generated by two vectors

[unique crypt]

yes

[stiff bobcat]

and these should give the coordinates of two vectors in S

[unique crypt]

okay so

$2 \alpha = - 3 \beta - 5 \gamma$

[flat frigate]

ransik (gmdn)

[unique crypt]

$\alpha = - \tfrac{3}{2} \beta - \tfrac{5}{2} \gamma$

[flat frigate]

ransik (gmdn)

[stiff bobcat]

for this kinda stuff you can like, first assume \gamma=0 and get some vector (\alpha,\beta,0)

then assume \beta=0 and get (\alpha',0,\gamma')

and then show that these are linearly independent, and thus a basis for the subspace of solutions

[unique crypt]

they are clearly LI

[stiff bobcat]

right these two vectors in the right hand side happen to be these basis vectors yes

[unique crypt]

so then i apply each of these to the basis of the subspace?

[stiff bobcat]

the coordinates of these vectors are coordinates in S remember

[unique crypt]

i don't think i'm following sorry

[stiff bobcat]

so e.g. $(\alpha,\beta,\gamma)=(-\frac 32, 1, 0)$ being a solution means that [\alpha(1,2,1)+\beta(1,-1,2)+\gamma(1,-7,4)] is a multiple of $(1,0,-1)$, i.e. the generator of the subspace ${(x,y,z): \ x+z=0}$

[flat frigate]

derivada.schwarziana

[unique crypt]

okay

[stiff bobcat]

or equivalently that [(x,y,z)=\alpha(1,2,1)+\beta(1,-1,2)+\gamma(1,-7,4)]satisfies $x+z=0$

[flat frigate]

derivada.schwarziana

[unique crypt]

shouldn't it be (1,0,1)?

[stiff bobcat]

(x,y,z)=(1,0,1) doesn't satisfy x+z=0

[unique crypt]

doesn't (1,0,-1) satisfy x-z=0?

ohh

wait

im an idiot

it satisfies z = -x

which satisifes z + x = 0

or x + z = 0

[stiff bobcat]

yup

[unique crypt]

okay

so there's two solutions

[stiff bobcat]

two linearly independent solutions

there's infinite solutions, but they form a subspace of dimension 2

[unique crypt]

okay

[stiff bobcat]

generated by these two

i.e. any linear combination of these two is a solution

[unique crypt]

i don't know why i'm having such a hard time understanding this. this is usually my strong part in linear algebra.

the vectors i've found

correspond to (alpha,beta,gamma)

and alpha beta gamma are the coefficients by which we multiply the basis of S to generate S

[stiff bobcat]

yes exactly

[unique crypt]

okay

so if i were to grab the basis of S

[stiff bobcat]

in other words they're coordinates in the basis S

[unique crypt]

and multiply it by these vectors

i'd end up with a subspace of dimension two

so

a plane

[stiff bobcat]

yes

[unique crypt]

ok

[stiff bobcat]

idk if this helps but these coefficients $\alpha,\beta,\gamma$ can also be obtained by solving [
\begin{pmatrix}1&1&1 \ 2&-1&-7 \ 1&2&4\end{pmatrix}\begin{pmatrix}\alpha \ \beta \ \gamma\end{pmatrix}=\begin{pmatrix}1 \ 0 \ -1\end{pmatrix}
]

[flat frigate]

derivada.schwarziana

[unique crypt]

so the solution then are all the vectors in S generated by the two vectors found

[stiff bobcat]

by noting, as in here that (1,0,-1) generates the space {x+z=0}

[unique crypt]

okay that could be useful

but I usually have an easier time just operating through equations

but yeah i can see that

okay so my answer then would be all the vectors v in S spanned by the two vectors provided

[stiff bobcat]

yes, the ones you get by replacing \alpha, \beta, \gamma here basically

[unique crypt]

ok

so i have to do that

then i can come up with a solution

a "proper" solution if you will

[stiff bobcat]

yeah, but basically that's the idea

like, if one of the solutions you got is $\alpha=-3/2$, $\beta=1$ you get that [(x,y,z)=-\frac 32 (1,2,1) + 1(1,-1,2)] and any multiple of it verifies $x+z=0$

[flat frigate]

derivada.schwarziana

[stiff bobcat]

so that's one of the generators of the space of solutions

[unique crypt]

yep

[stiff bobcat]

(and \gamma=0 ofc)

[unique crypt]

i'm solving that right now

yeah

wait a darn

they're LD

so it's unidimensional?

[unique crypt]

sorry for the ping i just wanna make sure

it's not a plane is it

it's dim 1

a line

[stiff bobcat]

hmm

[unique crypt]

yeah looks like it

this is the solution from another student

and chatgpt seems to agree as well

interesting

also their approach looks

far more elegant than mine

i could have just used a simoultaneous equation lol

oh well

that's it then

thanks!

.close

[safe radish]

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[stiff bobcat]

np

[stiff bobcat]

I think the vectors inside the span were LD to begin, in fact S is two dimensional. The intersection of two planes can be a line

so it makes sense

[safe radish]

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[supple skiff]

help

[austere cypress]

ask

[supple skiff]

i have no idea why this is true

i can give the context of the problem if needed.

this is what gpt gave me and other than that i think i understand the equation

it has to do with like reference frames and stuff so idrk

hello? @austere cypress ?

[empty gyro]

!nogpt

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[empty gyro]

!original

[safe radish]

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

[supple skiff]

Ok

this is for part a. basically just asking for the derivative and second derivative, but later in part b i need my stuff expressed in terms of theta so i can pliug that in but i don't know the polar link stuff

@empty gyro @austere cypress u guys still there

[empty gyro]

big text

[supple skiff]

K just checking 😊

[empty gyro]

$\hat{e}r(t)=\cos \theta(t)\hat{I}+\sin\theta(t)\hat{J}$ and $\hat{e}{\theta}(t)=-\sin\theta(t)\hat{I}+\cos\theta(t)\hat{J}$. So $$\frac{\dd}{\dd t}\left[\hat{e}r(t)\right]=\frac{\dd}{\dd t}\left[\cos \theta(t)\hat{I}+\sin\theta(t)\hat{J}\right]=-\dot{\theta}(t)\sin\theta(t)\hat{I}+\dot{\theta}(t)\cos\theta(t)\hat{J}=\dot{\theta}(t)\hat{e}{\theta}(t)$$

[flat frigate]

SWR

[empty gyro]

@supple skiff

[supple skiff]

yeah im readin it

so like

how do yk that the ehat theta = -sintheta(t)I + Costheta(t) J

is this just like a polar conversion fact or something

[empty gyro]

Rotation matrices

[supple skiff]

idk im taking an online class and i miss a lot of fundamentals

oh

[empty gyro]

you need some linear algebra knowledge

[supple skiff]

:/

okay well thanks i guess ill cite you on my project

[empty gyro]

[supple skiff]

i might be back if this gets tricky again im on the last problem but

actually

since youve read it all

for part d

can you help me understand that bc i don't entirely understand this reference frame stuff

my extend of knowledge on like matrices and reference frame conversions is extrodinarily limited

i can find the rest of this stuff through like online notes and such but i don't get how these are different

[empty gyro]

You're just using $\hat{e}r(t)=\cos \theta(t)\hat{I}+\sin\theta(t)\hat{J}$ and $\hat{e}{\theta}(t)=-\sin\theta(t)\hat{I}+\cos\theta(t)\hat{J}$ again

[flat frigate]

SWR

[supple skiff]

so if i were to graph the two resulting vectors even after i convert it to polar it would have the same vector value?

[empty gyro]

it should

it better

[supple skiff]

alright

for part e which acceleration vector do i use or does it matter since they are both the same

or is a different one entirelyt becuase its on the bead or something

actually thats a stupid question if they are oth the same vector

okay last question and then ill let you go my math king

do you know any resources i can find about how to convert to i j reference frames

ive done it like twice and both times ive done it wrong

or like how do i do it in general

[empty gyro]

You just need some basic study in rotation matrices

[supple skiff]

should i just search it up on like khan academy?

the function i gotta rotate isn't hard

is what i would say if i knew what i was doing but comparitively ig it looks slight

[safe radish]

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[unique crypt]

In P2 consider the subspace S

b) Find a base B' of P2 which contains the basis B found in item (a)

hold on a sec

green is the basis i found for a)

and now i'm supposing i need a basis B' which contains an additional vector v3 which is LI to the other two vectors

but i'm not sure how i'd solve this

i suppose just set each vector multiplied by a scalar

expand and solve?

so something like this? maybe?

i suppose an obvious choice for v3 is x^2

or just 1

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[ruby tulip]

Do elements of a set have to be of the same type, or can they be of different types? {1, (2 , 3), {4, 5, 6}}

[peak estuary]

thats a valid set

[unique crypt]

sets are just a bag

you can put many things in a bag

you can even put in other bags

[ruby tulip]

Ok, so it is possible to mix different mathematical object types in a set. Thank you.

[unique crypt]

you can have nothing in a set even

which doesn’t mean the set doesn’t exist

it just means the set contains the set nothing

[ruby tulip]

I'm not asking about cardinality. I'm asking about types

[unique crypt]

i believe that’s the way it works

yeah you can have different types of objects in a set

[ruby tulip]

thank you

I was just double checking, because I'm reading a book on linear algebra that I regard highly, but the author made a small mistake.

[burnt notch]

I also knew that had to be the same type, actually

But maybe it's allowed in certain contexts (?)

[peak estuary]

well its just generally not that useful to have different types in a set

it basically never occurs

[ruby tulip]

Should vs could

[peak estuary]

you generally define sets by taking the set of all elements which specify some condition

and usually then the elements have to be of the same type anyway if they are supposed to satisfy the same condition

[ruby tulip]

ok, thanks all. I appreciate the answers.

.close

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[halcyon light]

How dyu integrate this

haha

[simple granite]

Misread as differentiate

dont ask how

[halcyon light]

yeah no my brain kinda just

automatically says quotient too

[simple granite]

have you tried u sub

[halcyon light]

Has to be u sub right?

yeah I'll get that fraction

but then multiplied by 2 root 5-x^2 du

oh shit yeah

i can just sub in u

[simple granite]

You dont even need u sub tbh you can see that -(5-x^2)^1/2 just solves it

if Im not being braindead

[halcyon light]

oh yeah that looks like it works

[simple granite]

remember this rule

[halcyon light]

ahhh

I kept thinking ln

[simple granite]

wait lemme get a pic for it

[halcyon light]

🤦‍♂️

[simple granite]

hld up

If you have a function in this form

[junior smelt]

(I think it’s -[that], but the overall point still stands, can be done by inspection )

[simple granite]

you can use reverse chain rule

[halcyon light]

yeah thx

[junior smelt]

Oh wait nvm, the line breaking made that look strange

[halcyon light]

I've gone over that it j keeps slipping out my brain

idk when i try u sub

it doesnt look like it works

idk if u can read that XD

oh shit i got my shadow on it too

[simple granite]

Lets try decipher that

[halcyon light]

wait

i differentiated wrong too

omg

[simple granite]

it happens to the best of us

[halcyon light]

XD gna redo it quickly

dunno if this is at all legible

[simple granite]

I feel bad for your teachers

[halcyon light]

hahahahaha

[simple granite]

i swear thats an 11 not an x

read the first thing on the second line

du/du

[halcyon light]

oh no thats meant to say dx

prob should try write neater huh

[simple granite]

did you get the right answer with u sub

if not can you just redo with legible handwriting so i can tell you where you went wrong

[halcyon light]

this parts right isnt it?

If I make u = to the denominator

forgot du at the end

[simple granite]

Im sorry but bro I cant not understand a thing

Ill just walk you through the steps

[halcyon light]

OMG I thought u was equal to square root 5-x^2

AHHHH

that makes so much more sense

thanks

lol

[simple granite]

try that now

[halcyon light]

yeah I got it thanks

.close

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[shrewd hazel]

ME

[charred saffron]

you

[shrewd hazel]

5=a•2squared +3

does that

equals

0.5

and if yes

[digital sparrow]

nope

[shrewd hazel]

how

OH

it’s 5/7 right…

[digital sparrow]

do you mean $5 = a\cdot2^2 + 3$

[shrewd hazel]

no

a TIMES

[simple granite]

Yes its 0.5

[shrewd hazel]

how

pls i feel so dumb

[simple granite]

5=4a+3, 4a=2, a=2/4=0.5

[shrewd hazel]

HUH

[digital sparrow]

wait

do you mean

[shrewd hazel]

put it in this thing

[flat frigate]

George (Wumpus Man)

[shrewd hazel]

yes

[simple granite]

what part do you not get

[shrewd hazel]

WAIT CUS

it’s

a times 4 + 3

can’t i do then

a times 7

and then 5 divided by 7

[simple granite]

its 4a+3, not 4a+3a

[shrewd hazel]

okay 4a + 3 and then what u do minus 3 and then divided by 4

right

[simple granite]

yes

[shrewd hazel]

HELP IM SO DUMB BUE THAT WAS SO EADU

thanks yall love u all

[digital sparrow]

[shrewd hazel]

.close

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[simple granite]

what does the textbook solution get

a=b is trivial

isnt that just a=b

[main mural]

if a and b = 0, then a=b

[simple granite]

a=b provides an infinite number of solutions

a,b=0 is one of them

[main mural]

hi btw

i should start doing that ☠️

idk if you need determinants here tho

[spiral saddle]

placing the row vectors as columnsin a matrix and compute det != 0 is possible?

did you tried that ?

[main mural]

if they are linearly dependant then
$\exists s_1, s_2, s_3 s.t. $\
$s_1(a,b,b) + s_2 (b,a,b) + s_3(b,b,a) = 0$

[flat frigate]

artemetra

[spiral saddle]

you were talking about determinant tho

[main mural]

from here you get an augmented matrix and you solve and find for which values a,b there are non zero solutions for s_1, s_2 and s_3

fair enough

maybe if you can assume that $a=\lambda b$ and substitute and solve for $\lambda$

[flat frigate]

artemetra

[main mural]

but otherwise no clue tbh

[safe radish]

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[main mural]

if you first notice that a=b is a solution you can try polynomial long division too

this is how it factors btw

.reopen

[safe radish]

✅

[main mural]

https://www.youtube.com/watch?v=9EpU7TeIUC8

first time seeing this myself too lol

but it works

[simple granite]

interesting never thought of this before

now that i see it, it is seems like a common sense extension of normal polynomial division

i wonder if there a quick way to do this like synthetic division

perhaps that works the same way too

[safe radish]

@vapid flint Has your question been resolved?

[main mural]

notice that a=b works again for the second polynomial

[simple granite]

How on earth did you see that so quickly

[main mural]

yep

[main mural]

idk lol

[safe radish]

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[glossy wave]

wich of the 2 formulas i have to apply here?

[glossy wave]

and what happens to -1

[slim lion]

brutha what the fuck is this

on the left

[stoic saddle]

the first image is $(1+\frac{-1}{n^{2}+n+2})^{n^2+1}$

isnt it

i cant read

[flat frigate]

CherryMan

[glossy wave]

yes

yes

[glossy wave]

its the one

[glossy wave]

my phone is dead

[stoic saddle]

is it the limit as n goes to infinity or 0

[glossy wave]

it just says calculate the limit of sequnce

[glossy wave]

infinity

[stoic saddle]

guessed that much

sorry i have no idea u need someone else to help you

[glossy wave]

<@&286206848099549185>

[proper moth]

Got it.

[glossy wave]

?

[proper moth]

Generating latex wait

$\lim_{n \to \infty} \left(1 - \frac{1}{n^2 + n + 2}\right)^{n^2 + 1} = \lim_{n \to \infty} \left[\left(1 - \frac{1}{n^2 + n + 2}\right)^{-(n^2 + n + 2)}\right]^{\frac{(n^2 + 1)}{-(n^2 + n + 2)}}$

[flat frigate]

viiviiiix

[proper moth]

Now, it'd be
$e^\lim{n \to \infty} \frac{n^2(1+\frac{1}{n^2})}{-n^2(1+\frac{1}{n}+\frac{2}{n^2})}$

[flat frigate]

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

[glossy wave]

its e^-1?

the result

[proper moth]

Yeah, I got 1/e

[glossy wave]

eh same thing

thx

[proper moth]

np
You should check your answer with the book as well.

[glossy wave]

it doesn t have answers

its just an exercise sheet our teacher sent

[safe radish]

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[split ether]

sin^2x is symmetric around x = pi/2

[lean otter]

you could even solve this

and check

that they are equal

[safe radish]

@modern bloom Has your question been resolved?

[split ether]

Do you really need to know that those are equal?

[fathom jewel]

Why ≠ How

[split ether]

What's the original problem?

[safe radish]

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[fathom jewel]

rage quit

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[inland ibex]

What statistical test should I use for the following:
The propability Y that the predictions are correct is the same across the binary group X

P(Y|X1) =µ1
P(Y|X2) =µ2
H0: µ1=µ2

[safe radish]

@inland ibex Has your question been resolved?

[safe radish]

@inland ibex Has your question been resolved?

[safe radish]

@inland ibex Has your question been resolved?

[severe scarab]

depends

[inland ibex]

on what

[severe scarab]

Are you familiar with the two proportion z test

[inland ibex]

yes, that's what I was thinking about

but I was wondering if there was another oen

[severe scarab]

nah thats your best choice if Y is binary and want to directly compare the two probabilities

[inland ibex]

okk, my data is categorical so not a lot to choose from I think

ok thank you

[severe scarab]

all goods

[inland ibex]

.close

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[orchid horizon]

kurze frage weiß jmd was 5.12 bedeutet

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@orchid horizon Has your question been resolved?

[dusk prawn]

it's over ☠☠😭😭🙏🙏

[edgy isle]

verstehst du (1)? Oder ist der aufbau der matrix unklar?

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[shrewd timber]

How do I like combine functions?
Like if I have
y = f(x) and y = -f(x)
how do I make it so I have just one term without changing any proporties of these functions?

[main mural]

define what you mean by "combine"

[lime dust]

https://tenor.com/view/dragon-ball-z-fusion-dbz-goten-trunks-gif-8263700045051414172

[shrewd timber]

having like an easy example of two half circles y = sqrt(r^2 - x^2) and y = -sqrt(r^2 - x^2) to have them combined would mean to just have y^2 = r^2 - x^2

these two functions y = sqrt(r^2 - x^2) and y = -sqrt(r^2 - x^2) just combined into y^2 = r^2 - x^2, and still have the same proporties, just they're now combined

[main mural]

aha i see

problem is though that y^2 = r^2 - x^2 isn't an expression anymore

but that is still okay

the key here is multiplication

you want to rewrite y=f(x) as y-f(x) = 0 and y=g(x) as y-g(x) = 0

then the combined graph would be (y-f(x))(y-g(x)) = 0

[shrewd timber]

I think it would be easier if I say exactly what I'm doing

[main mural]

so in your case you would do (y-f(x))(y+f(x))=0 which is the same thing as y^2 - (f(x))^2 = 0

[shrewd timber]

this is my case

[main mural]

what the fuck

[shrewd timber]

expanded equation for an ellipse

https://www.desmos.com/calculator/2qvmaosobb

[main mural]

lmao

well

[main mural]

it's this

[shrewd timber]

and if I square it on both sides, the equation doesn't hold the change of variable k

it only works if they're separated

[main mural]

wdym

it does precisely the same thing as the two things separately

i'm playing around with it rn

what is your goal?

[neat kiln]

The longest function I have ever seen

[shrewd timber]

let me just

it broke again

damn

[main mural]

what graph are you expecting to get?

[shrewd timber]

wait

[shrewd timber]

I'm gonna send fixed one

https://www.desmos.com/calculator/nv6h6gfcon

now

[main mural]

https://www.desmos.com/calculator/z5cqhu19n5

it works

i just took (y-blah blah)(y-blah blah2) = 0

[shrewd timber]

THX!!!!

.close

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[earnest rain]

how to handle that different radius?

[earnest rain]

.close

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[eager sable]

Can somone help me with algebra geometry

[eager sable]

im revising for exam

[safe radish]

@eager sable Has your question been resolved?

[agile kindle]

what is your question

[safe radish]

@eager sable Has your question been resolved?

[eager sable]

ill show

wait its jsut algebra

[visual swift]

when you have fractions, multiply both sides by common denominator

[eager sable]

???

can you walk me through it

[vapid panther]

he means like for the first quesiton multiply the leftmost fraction by 3/3 and the rightmost by 2/2 so that you can "merge" them (now same denominator)

then calculate from there

[eager sable]

.close

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[golden granite]

do I have to divide 441 on both sides

but then what would that result in

.close

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[smoky pier]

what series tests do i use for these problems to determine if they converge/diverge?

[tardy mango]

3. As $n \to \infty$, what is the summand "approximately" equal to? What test allows you to use this approximation? \ \
4. Recall that $\frac{a^c}{b^c}=\left(\frac{a}{b} \right)^c$

[flat frigate]

Civil Service Pigeon

[broken scroll]

5 should be easier try that one first

As n goes to infinity

It’ll be either -1 or 1

For numerator

And the denominator will get bigger and bigger

Meaning that it’s oscillating with waves getting smaller and smaller each time

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[jade berry]

What does it mean by "the scale drawing of the space shuttle is shown opposite."?
Am I supposed to do this with physical textbook to solve rest of the problems?

[sleek sentinel]

i dont see any other information given here

[jade berry]

exactly

[sleek sentinel]

so you'd have to measure yeah

interesting problem though i've never had to do this

[jade berry]

I am currently doing standard math 2

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[late violet]

guys how do i do this

[azure delta]

draw a picture

[late violet]

i did

what nexxt

[azure delta]

find the radius

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[ionic lark]

You there?

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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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[rough matrix]

hi

[rough matrix]

how to do 7b plz

[lean otter]

Plug in that point into your answer from Part A, then use the point-slope formula to get your tangent line.

[rough matrix]

working on it rn ill lyk how it goes

k i got this for y prime

now idk how to do next step

[lean otter]

That looks good. Now plug in 1/8 for x and (3 sqrt 3)/8 for y in order to get your slope.

[rough matrix]

ok thas the problem

idk how to solve for b

wait ill write it

and send where im stuck

are u able to go therough the steps with me?

this isnt even hw just review for a test and I have been stuck for like an hour

@lean otter

[lean otter]

Show where you are stuck

All you have to do is plug in your point into your y' formula

Then use y - y1 = m (x - x1) and you're done

[rough matrix]

this whole next part of finding b in y=mx+b

i like y equals mx plus b

[lean otter]

What did you get for m?

[rough matrix]

m is y prime

[lean otter]

Yes, what numerical value did you get for m?

[rough matrix]

o shoot i forgot to substitue all x and y values

one sec

ok ill send the pic

this is where i am at currently

[lean otter]

Ok, that's fine. When you simply that m value in a calculator, it actually comes out to -√3

Now you're left with (3√3) / 8 = (-√3)(1/8) + b

And you can simply solve that for b.

[rough matrix]

how do u simplify that m value to that

wait i simplified a bit more i will send again

one moment

is this good so far @lean otter

[lean otter]

Like this

[rough matrix]

this is needed?

for the question

is the last line I wrote valid?

[lean otter]

Yes. In the end, it simplifies down to -sqrt 3

[rough matrix]

my last line is valid?

[lean otter]

Yes

[rough matrix]

photomath is saying it would simplify to like root 3 over 4

its saying the b value would be root 3 over 2 i believe

so that means my equation is wrong somewhere

[lean otter]

I'm saying the final m value is -√3.

After you solve all of that for b, you should get √3 / 2 as the b value

[rough matrix]

ok so i messed up somewhere

doesnt that turn into this:

[lean otter]

Yes, thats all correct.

But 3√3 raised to the 1/3 power is equal to √3

So you should then have 3√3 / 8 = (-√3 / 8 ) + b

So you get b = 4 √3 / 8, which simplifies to √3 / 2.

[rough matrix]

But 3√3 raised to the 1/3 power is equal to √3 that confuses me

how does that work

[lean otter]

3 * √3 = (3^1) * (3^(1/2))

= 3^(3/2)

The cube root of 3^(3/2) is 3^(1/2), which is the same as √3

It seems like most of the difficulty of this problem is just wrangling radicals and fractions, rather than the actual calculus itself

[rough matrix]

yes

i agree

if possible

can u write that more clearly

[lean otter]

Are you good up to this point ?

[rough matrix]

good upto this point atm

[rough matrix]

ok now i understand this pic

[rough matrix]

got it

so far

do we eventually get here?

wait what should i get for b

o ok

i got it

just the b value

now I have to get equation

[lean otter]

Yes. But but to make it easier on yourself, just write 3^(3/2) as 3√3 and 3^(1/2) as √3.

[rough matrix]

wait what is my m again?

is it that?

[lean otter]

Yes. But that top part simplifies to -√3.

Your final equation is (3√3)/8 = (-√3)/8 + b

[rough matrix]

no i gtta get this:

so if i times everything by 8 i dont get it

i think something is off

[lean otter]

Multiply everything by 2, then move everything to the left side, and you should arrive at that solution.

[rough matrix]

is this right?

before simplification

[lean otter]

It is y = -√3 x + (√3) / 2

The 1/8 was part of your x-value, not your m-value

[rough matrix]

ohhh

wow very much appreciated

i thought this question would be way easier cause never had smth where we had to manipulate radicals and stff.

is this the easiest way to do this?

[lean otter]

It's easier if you're allowed to use a calculator

Unfortunately the problem just had some obnoxious numbers and that just made the arithmetic trickier than necessary

[rough matrix]

ok ya im surprised my teacher assigned this

i have one more question

if u dont mind

its the last one

in this worksheet that I confused on

if u dont mind

[lean otter]

Sure, whats the question ?

[rough matrix]

number 9

rlly confusing

for me

[lean otter]

It might be easier to think about if you rewrite it as xy^3 + x^2 y = 3 and you have the point (2, 1)

Solve for y', and then plug in (2, 1)

[rough matrix]

point 1,2?

[lean otter]

Sorry (2, 1)

[rough matrix]

how do u know we have this point

[lean otter]

The problem says that f(2) = 1.

2 is the input, the x-value.

And 1 is the output, the y-value

[rough matrix]

wait how do u recognize that implict differentiation is in this question

[lean otter]

f(x) and y are sometimes used interchangeably.

And honestly not sure how else you'd solve it besides implicit differentiation

[rough matrix]

And honestly not sure how else you'd solve it besides implicit differentiation. o rlly?

ok

ok

ill lyk how it goes

when do we consider x to be its own function

is it like if it has an exponent

or is x its own function in this equatino

because technically a number is a functino and so i would use product rile

rule

[lean otter]

You're finding dy/dx, so you have to treat x like a function. So yes, use the product rule here.

[rough matrix]

my teacher said something about how x is a variable and y is the function in implicit diferentiation

trying to understand y he said this

[lean otter]

Ok yeah I misspoke x is a variable, and y is a function of x. That's why you also multiply by y' whenever you differentiate y with the power rule or whatever

Just use implicit differentiation like normal

[rough matrix]

but if u use the product rule that means x is a function

when would x be consider a finction

if x is exponent that would be a function right?

isnt any number considered a function also?

since it passes vertical line test

[lean otter]

Yeah

[rough matrix]

so then i should always treat y as a function and x even in implicit differentiation?

[lean otter]

Yes. Differentiating xy^3 for example is (1)(y^3) + (x)(3y^2)(y')

[rough matrix]

okay but in my notes it is not like this