help-28

[torn jolt]

Precisely

[stiff axle]

yes

[torn jolt]

Nah im making is even more confusjng anyways

There cannot be a cube w/ 4 shaded sides.

Because to have 4 shaded sides, it MUST be the cube itself

It cannot be covered

So there can be a maximum of 3 shaded sides.

[stiff axle]

i see

aight ty

i think i get it now

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[primal trail]

I need help. So let's say P= -½ + 2 × |⅓-⅚| and Q= 2,5 × 4,8 : (-1,5).
What is P/Q?

[primal trail]

The correct answer is supposed to be -16, but I got -1/16

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[dusky canopy]

idk where to start with this question. Would appreciate some help 🙏

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[gritty rose]

try finding a pattern for n=2,3,4

[dusky canopy]

oh yea, forgot to mention. The teacher did say "attempt" to find the general solution to this without finding a pattern

*he said this for the whole assignment overall, not just this one question. So, I assume he doesn't want us to solve it by finding a pattern ._.

[gritty rose]

what a stupid thing to suggest

do it anyway

[dusky canopy]

yea i don't really see a pattern ngl

plus the expansion even for n = 2 is stupidly long

[gritty rose]

show it for n=2 and 3

[dusky canopy]

the only reference he gave was this:
which i didn't really figure out how to manipulate it into this form

[gritty rose]

wot

this is all you need for n=2 case

[dusky canopy]

ye

but

it's damn long

[gritty rose]

[dusky canopy]

I ain’t expanding this man -_-

I ain’t seeing a pattern

And ma teacher says this is one of the normal difficulty questions on the mid term exam, I’m like so done for wth

And we’re just three weeks in into year 1 linear algebra

yea let's just call it here. ima wake up later and try to do this thing again from scratch by looking at his ppt lecture

what a cancer unit 1 introduction to linear algebra

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[lethal aurora]

can someone teach me how to graph in 3d cordiniate system? for calc 3?

[lethal aurora]

<@&286206848099549185> DIRE NEED OF HELP PLZ

on paper, not on geogebra or desmos

Like i can graph (0,0,1)
Or (0,0,6)

[stiff summit]

are you required to draw a 3d graph on paper in homework or in an exam?

[tame cradle]

What

[lethal aurora]

but idk how to graph the point (5,3,2)

Yeee

Like look

[stiff summit]

isometric paper?

[lethal aurora]

ill give u an exampple

[tame cradle]

First draw a 3d coordinate system

[lethal aurora]

Nono just regular paper

[tame cradle]

Represent x y and z clearly

Then we proceed

[lethal aurora]

Like this

See the verticies

I needa be able to point those and connect dots to draw a tetrahedra

[stiff summit]

you don’t need to be able to plot them, just visualise them, right?

[lethal aurora]

I need to be able to plot them

and draw the tetrahedra

like a rough sketch

Like (0,0,0) = super EZ

(3,0,0) = Ezzz

(0,4,0) = ezz

(0,0,5) = ezz

But sumtimes he be giving sum hard points like (6,5,3)

how would I graph that point?

on a 3d coordinate?

[stiff summit]

why? does the question explicitly require it

[stiff summit]

because this one doesn’t

[lethal aurora]

During exams he requires it

this one doesnt require it because its homework

he asks us to use geogebra

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[manic heart]

doing either composites or functions and just lost how to find the domain

[jolly horizon]

Alright have you found the composite function yet

[manic heart]

I don't even know what that means 😭 My class started 3 days ago and its been confusing

Like I’ve been trying to do it off of the help video but I’m lost because of the +3 next to the x and idk why the all real numbers on the help is 0

because if I do the times x on everything idk if its still 2/3 -1x or if I combine the 2 and 3?

[dusk dust]

what r u having difficulty with exactly

[dusk dust]

a composite function means

[dusk dust]

the input of a function

is another function

like for example f(x) is a function

[manic heart]

Ohh

[dusk dust]

here input is x right

now take another g(x) = x^2

[manic heart]

Then I think I know them? The teacher legit just started teacher about them yesterday

[dusk dust]

so basically a composite function is f(x^2)

like before inputting x in the function

u first substitute in g(x) and use That output as an inpur for f(x)

[dusk dust]

can u show the full working

[manic heart]

Uhm I might just take a pic of my paper cuz I’m doing this on paper but the assignment is online

[dusk dust]

anything is fine

i just want to see ur method

[dusk dust]

what did u do in the 3rd line?

[manic heart]

the one that has the 1 on top or the x times by the stuff?

[dusk dust]

yes

the standard way to simplify it is take x+3 as LCM and make the denominator as one fraction

and then bring it up

but when finding domain u usually should not simplify

take the 2nd line for example

u have just written f(g(x))

[manic heart]

thats how it was written in the video and how to put them together so I've been just going it like that

[dusk dust]

yeah that si correct

now we just have to check the points where f(g(x))

is undefined

take that small 2/x+3

for a function to be defined every single term has to be defined right

[manic heart]

ok

[dusk dust]

so for this to be defined

x+3 not equal to 0

x not equal to -3

we have found one point where x not defined

now for the other point

we have 1/(2/x+3) -1

the denominator of this fraction must be defined right

so we can say

(2/x+3) -1 not equal to 0

= 2 is not equal to x + 3

x is not equal to -1

so domain will be all real numbers except -1 and 3

[manic heart]

ok lemme try that

omfg the question online changed

wait i think im slowly getting them tho

[dusk dust]

mhm

tell where ur getting stuck

[manic heart]

its just hard to do it because i suck with word descriptions 😭

[dusk dust]

ah

ur going in the write direction

[manic heart]

klojhds

[dusk dust]

what u ve written in the second line dont simplify it

u start analysing from there

[manic heart]

it still says I've been getting them wrong

[dusk dust]

oh i see

the domain ir R-{3,-1} not {3,-1}

[manic heart]

it gives me a answer key that shows me the right answer but it remakes me do the question with new numbers but it doesnt give the thought process

[dusk dust]

ohhhh

i see

[manic heart]

idk what i did wrong this time 😭

this is what it gives when I press for answer

[dusk dust]

oh im so sorry

it is -3

i didnt see the question properly

because if x + 3 needs to b 0

x must be -3

[manic heart]

ok

[dusk dust]

mb

[manic heart]

so for this one x must equal 4?

[dusk dust]

4 and u have to find out for which value of x 1/(x-4) + 2 is 0

because the denomintor of the f(x) must also not be 0 right

[manic heart]

would I -2?

Finally got it right 😭

this isn't the same thing but idk why this got wrong when they look the same

[void nova]

What's 3x2? 😉

[manic heart]

oh wait nvm I forgot the make the 2 on 3x2 a square symbol nvm

😭

[manic heart]

I just realized that thanks 😭

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[clever dawn]

i need help on all of this

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[clever dawn]

thx chat, alot of help

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[sly parrot]

i dont even know where to start

[echo marsh]

what do you have to do

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[sly parrot]

?

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[torn jolt]

so basically they're just integrating the differential of u(x, y) if I understand correctly

so they're integrating wrt x and y

[leaden ermine]

you cannot integrate wrt two variables simultaneously

if you look closely you can notice this is the total differential

[hasty violet]

partial integration ahh

[leaden ermine]

so you can see u_x and u_y

[hasty violet]

jk

[torn jolt]

yeah

i don'

[leaden ermine]

If you integrate one wrt to one variable you got u(x,y) + constant function

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[clear rampart]

For B

Our teacher want us to compute this with geometric series, anyone know how to start?..

[naive monolith]

First, find your rate and write the first 3 terms

[clear rampart]

0,005

[naive monolith]

This is an investment, so the rate would be 1+0.005

But ye

[clear rampart]

Yes

[naive monolith]

So what would your first term be

[clear rampart]

I understood the first question, we solve for future value

[naive monolith]

Oh lmao sorry

[clear rampart]

Np, its the b question that is mindf me

[naive monolith]

The investment continues, but every month we draw out a certain amount

[clear rampart]

yes

[naive monolith]

So from out last term, imma just call it Tn

We create a new Gp

With Tn as the principle

There will be a certain amount withdrawn each month, so use a pronumeral to represent it

[clear rampart]

This will be our first term right

The investment continuous

[naive monolith]

Yep

But were withdrowing an amount

[clear rampart]

We want to withdraw something a

[naive monolith]

We assume it’s after it compounds

[naive monolith]

Uhhh

Sum of gp terms?

[clear rampart]

?

Gp?

[naive monolith]

Geometric progression

Its the uhhh

Hang on a sec

$S_n = a \frac{1 - r^n}{1 - r}$

[glossy valve]

Sho

[naive monolith]

this thing

do you know it?

[clear rampart]

Yes

[naive monolith]

yep

[clear rampart]

I have solved for A but it’s not correct

[naive monolith]

ok

[clear rampart]

Already tested it

[naive monolith]

did you use this for a?

can I see your A?

[clear rampart]

in the first question or?

[naive monolith]

Wait

Uh

Yea

Part A

[clear rampart]

its 200

This is part a

[naive monolith]

O

This is correct?

[clear rampart]

yes

[naive monolith]

Ah

Icic

Wait a sec

[clear rampart]

We need our yearly 6% interest rate to be monthly

[naive monolith]

You used sigma here

[naive monolith]

But why not use this?

[clear rampart]

Well

I dont know

[naive monolith]

Yea

If youre confused

Write the first 3 terms

[clear rampart]

of what?

[naive monolith]

Of your part B

[naive monolith]

This -a would be your first term

[naive monolith]

Oh

Yea

That

[clear rampart]

If I were to solve for A now

we multyple by the denominator

right

to the future value

and then devide by the top to get A alone

[naive monolith]

I would just put the thing in the calculator but yea sure

[clear rampart]

its wrong method

[naive monolith]

Hang on a sec

Oh lmao

Wait

Your rate is still goong

Its not just your tn

Tn

[clear rampart]

Tn?

It is

[naive monolith]

Your ans from A

Tn(1.005)^180

[clear rampart]

Well now its 15 years not 40 years as in part a

[naive monolith]

You forgot it still compunds

[clear rampart]

hmm

I dont understand

[naive monolith]

So

I have X amount still in the bank

[clear rampart]

The 398298 still compounds?

[naive monolith]

Yea

[clear rampart]

Yes i know

but the thing is that the

If i were to use the formula u said

Ohhhhhhhhhhh

Like that???

[naive monolith]

Thats a minus there

[clear rampart]

ye plus i mean

in the left side right?

[naive monolith]

Yep

Should work

[clear rampart]

really

then i multyple by the denominator and devide?

although im wondering if it should not be 179 in the right hand side as N

[naive monolith]

The final withdraw is still a term

So 180

[clear rampart]

so this should be correct?

[naive monolith]

If its not, idk what happened

[clear rampart]

Ok i will check, thanks so much

if i multyple the (1-(6/1200) to the (1+(6/1200))^180

it will be (1+(6/1200)^179

[naive monolith]

Different symbols, doesnt work

Is the ans 3361 or smth

Around?

[clear rampart]

no

[naive monolith]

LmAo wtf

Can I see the answers?

[clear rampart]

3344,34

[naive monolith]

Bro

Im ngl

[clear rampart]

This is how she did it

I’m just trying to understand

[naive monolith]

This is the craziest way ive seen it done

Like

[clear rampart]

Me too I don’t understand a shit

[naive monolith]

Ill

Do it

Hang on a sec

Ok

I see

I honestly hate this q cause its not specific enough

Part A is weird cause this basically says that you compund before you put 200 in

This is saying it withdraws and then you compound it

[clear rampart]

Yes i know

its really weird

im starting to wonder if she have done something wrong

[naive monolith]

The only thing you need to do here is multiply a by 1.005

[naive monolith]

In part a, she compounds nothing at first and then adds 200 dollars

Doesnt make sense to me but meh

Part b, i thought shed keep it that way

[clear rampart]

waaaait

the compounding

in the left hand side

should be 179

then it should be right?

[naive monolith]

Let me write it

This

I disagree with it, but its the answer

Technically this is the correct one

[naive monolith]

But since you can divide both terms by a, it can be 179

[clear rampart]

i dont understand man

[naive monolith]

T1 is the first month

T2 is second month

We keep going until at T180, the balance is 0

[clear rampart]

yes

I understand that

We will then + over the aR

and then revide everything by R(R^180-1)/R-1

so what i forgot was a 1,005

[naive monolith]

Yea

But thats not your fault

Its the teachers ngl

[clear rampart]

okeyy

but do you know why?

[naive monolith]

Cause

I thought it would compound

And then you withdraw

But it withdraws and then compounds

You should ask your teacher to clarify these things so you know which is which

[naive monolith]

If it was my way, it would be 398298.15R-a for first term

Not aR

[clear rampart]

Okok

Me too

That whas why i kept getting wrong

[naive monolith]

Nice

Anything else

?

[clear rampart]

Not really

for C

or wait

for c

its the same thing?

[naive monolith]

.

Man

The wording

Or like

English

Of this q is bad

[clear rampart]

Yes I KNOW

Its so fucking weird

[naive monolith]

Im guessing it just wants the yearly deposits if it was 20 years?

Yea same thing

[clear rampart]

Okay

Thank you so much

Weird question huh

whole class is i denial

in

[naive monolith]

Yea

Just maybe ask your teacher to clarify when it compounds

Then youll be fine

[clear rampart]

Yes

Thank you so much

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[woeful anvil]

Can someone explain to me how the “Objective Function” for d was found?

[woeful anvil]

This is a optimization problem

[umbral dome]

pythagorean theorem

[woeful anvil]

Oh I forgot

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[woeful anvil]

Thanks

[leaden ermine]

cereal killer

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[glad hound]

Working with limits... I'm not entirely sure what to do with some of these symbols

[glad hound]

I understand ε to be just "a really small number", per https://www.rapidtables.com/math/symbols/Basic_Math_Symbols.html ...

But for δ it's simply defined as the "delta function" - that's... not helpful.

[umbral dome]

well it's not the delta function in this instance anyway

basically ε represents a small distance between the output of a function and the limit, and δ represents a small distance between the input of a function and the point it is approaching

[glad hound]

... but if the input is a variable...

I mean... x doesn't approach anything, really..

[umbral dome]

we read [ \lim_{(x,y) \to (0,0)} ] as ``the limit as $(x,y)$ approaches $(0,0)$''

[glossy valve]

cloud

[glad hound]

Sure

So it doesn't matter if the function is undefined at that point?

[umbral dome]

as long as it is defined in the neighborhood of that point then it's fine

[glad hound]

Ok, so...

... what can I actually do with that information?

Like, in the example..

δ is essentially > 0, but < then..... sqrt(ε/5) is essentially 0, so...

[umbral dome]

basically we want to establish that, if (x,y) is very close to (0,0) (we call that distance δ) then f(x,y) will be very close to 0 (we call that distance ε). we want to show that for any desired output distance ε, we can choose an input distance δ that will make it true

i will note that the ε-δ definition of two-variable limits is a bit more complicated but has a very similar meaning to the ε-δ definition of one-variable limits. so you may want to look into that first

[glad hound]

The explanations there are just as devoid of meaning

"Here's a symbol.. we're not going to define it or explain it... moving on."

English is not my instructor's first language, either...

So... where does |f(x,y) - 0| come from? Why are we considering it?

[gritty rose]

Comes from the first sentence

[glad hound]

along any line y = mx, the lim is 0?

So.. why "consider" it, and then in the very next step, drop it out of the equation?

[gritty rose]

Because y=mx is not sufficient for a proof

[glad hound]

Right, because the limit would have to be 0 everywhere

Do you know of a better resource that explains why the chosen steps are taken? My text and my instructor don't explain why things happen in these examples

There's too much I don't understand in the text.. I'd be asking "why this, why that" for hours

[gritty rose]

just a technique you memorize and repeat elsewhere

or work backwards from epsilon to find what delta has to be in sqrt(x^2 + y^2) < delta

[glad hound]

Ok, but if I don't know "why" I'm doing something, that makes it extraordinarily difficult to apply it to an equation that takes a different form

[gritty rose]

then work backwards

should be easier once you understand the one-dimensional case first

read those two https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

[glad hound]

thanks

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[frigid phoenix]

if i have z= xy and i want to take the partials i would get
dz/dx = y
dz/dy = x
but here x and y becomes constants.
i'm not sure how to use this result. what does it means graphically?

[umbral dome]

x and y are not constants, we just "treat them as constants" for the purposes of differentiation. so dz/dx is a function of x and y (although it depends only on y)

[frigid phoenix]

ok this was my next question indeed. ok thanks, i'll think more about it

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[slow ermine]

Here is what I'm trying to achieve

[slow ermine]

Here's my current progross:

Im really close to replicating this effect but my calculations are still off, I'm not sure if this is the type of math I can get help with here but I'm giving it a shot.

My goal is to apply rotation to texture Gui's on each individual face so the green side of the gradient is always facing the dot.

This is done in two ways:

1. Get the direction of the cube relative to the reference point (or the dot in this scenario

2. Get the angle to rotate each face's texture based off their local direction relative to the reference point

3. Calculate how much the face is facing the direction of the reference point within a value of -1 to 1 ("1" meaning the face is facing directly at the point, "0" meaning the face is directly perpendicular, and "-1" where the face is facing the direct opposite direction)

4. Using the face direction calculation, offset of the gradient texture taking the angle the texture is rotating into account.

What's wrong with my calculations?

[full forum]

@slow ermine Has your question been resolved?

[full forum]

@slow ermine Has your question been resolved?

[full forum]

@slow ermine Has your question been resolved?

[full forum]

@slow ermine Has your question been resolved?

[full forum]

@slow ermine Has your question been resolved?

[buoyant raptor]

Are you just trying to illuminate the surfaces?

Like the dot is a light source and the faces / portion of faces that face the source are coloured green

[slow ermine]

No each face has a color gradient, I am basically adjusting the rotation of the gradient and the offset of each face individually relative to the cubes orientation

[buoyant raptor]

No but what is the final result you’re trying to achieve?

[buoyant raptor]

From here at least, when you say the green side always faces the dot, it should be the same as what I said

Essentially the green parts are illuminated whereas the red isn’t

[slow ermine]

Sort of, but instead of thinking of it as a light source, think of it as a direction. So basically the threshold is just a line in 3D space. Once the cube reaches this line, any face crossing it will change to red

[frigid phoenix]

what software are you using?

i know i'm not helping solving your problem, but the way i would do this would be in a shader, if i had access to the position of each point on the surface, would be to map the gradient onto the distance from surface point to the light.

[buoyant raptor]

Maybe if you explained how the coordinate system is setup, someone will be able to help

I'm not entirely familiar with roblox scripting

What's CFrame?

[full forum]

@slow ermine Has your question been resolved?

[slow ermine]

I was thinking about using trig formulas to measure the side of each face and place a point where gradient gradient line ends, then simply change the offset and the rotation of each adjacent face so the points line up with each other but idk

Something likes this?

[slow ermine]

What points would you use?

[slow ermine]

CFrame is just the combination of two vectors in a 3D space. The first tells its position and the 2nd tells its orientation relative to the world axis

[frigid phoenix]

the shaded point position on the surface of your object. I'm suggesting to just remap a distance field.

[slow ermine]

So like, use a plane as the threshold, and use the intersecting points on the edges of each cubes face to map each gradient?

[frigid phoenix]

No, in a shader you would just have access to the world position of each point on the surface that is being shaded, means you get the coordinate of the intersection between camera ray and surface.
Once you have that, for every point on screen, calculate the distance from your object and map a gradient on it. This normally is parallelized so you can do this with many many points. It's a computer job.

It works with any object of any polygon count.

What you are trying to do is certainly interesting anyway, but I can't help you there.

[slow ermine]

Hmm that is interesting, I changed my approach however by intersecting a 2D plane with the cube, and created showing this intersection on the edges exclusively. Here’s what I came up with:

The idea now is for each face with at least 2 points, get the midpoint between them to calculate the gradient offset

But I need to figure out how to choose the right point for each plane to get the right rotation, I haven't figured this part out yet unfortunately.

From these calculations do you have any idea how I can get the 2 points on each plane? https://pastebin.com/Jfqh9WY8

[frigid phoenix]

@slow ermine do you know blender?

[slow ermine]

Yes

[frigid phoenix]

I can show you how how to do the gradiend thing in shader if you want. but it's not your method.

[full forum]

@slow ermine Has your question been resolved?

[slow ermine]

Yeah please

Who knows it might help

[eternal stone]

So basically you have a 2d plane of a 3d space and you want to assign one color to all points in one semi-space, and another color to all points in the opposite semi-space. Is that right?

Are you looking for a solution for cube specifically, or do you want the most generalised version?

[eternal stone]

@slow ermine Anyways, here's an idea that might help with rotation of Gui. Let's say the normal of a given face is $q \in \mathbb{R}^3$. Let's say that the normal of separating plane is $p \in \mathbb{R}^3$. In order to calculate the vector parallel to the intersection of the face and the plane, consider the vector product of $p$ and $q$: $v=[p,q]$. The angle between $v$ and an appropriate edge of the face is the angle of rotation of your GUI.

[glossy valve]

EQUENOS

[eternal stone]

So yeah the angle between v and w will be the angle of rotation of your GUI

Offsetting the gui should be easy

[slow ermine]

That makes sense, but how do I know which edge to measure as W?

[eternal stone]

It depends on which edge the GUI sees as the bottom line

Attach a surface GUI, add a black frame with scale coords {0, 0} and you'll see the "upper left corner of the face"

The lower edge would be the w vector of that face

[slow ermine]

Alright and after figuring out which edge is the bottom, how would I define it exactly?

[eternal stone]

Assuming you know the CFrame and size of your cube, it's easy to calculate the coordinates of all 8 vertices: v_1, ..., v_8

After that the dege between say vertices 1 and 2 would be v_2 - v_1

[slow ermine]

I guess what I mean to say is how do I get this point so I can calculate the angle?

[eternal stone]

You don't need that point

The angle between v and w is equal to $arccos\frac{(w, v)}{|w|\cdot|v|}$

[slow ermine]

Oh I must've misunderstood then sorry

[glossy valve]

EQUENOS

[eternal stone]

Or any other formula you find convenient

[slow ermine]

Would v be the angle of the 2 intersecting points on the face?

[eternal stone]

v would be the vector such that if you draw it at one intersection point on the edge, it will point to another such intersection point

[slow ermine]

Oh okay I understand what you mean now

[slow ermine]

What calculation would you use to determine which face a point is on?

The function I use doesn't seem to work as I intended:

function getPointsOnFace(cube, point)
local cubeCFrame = cube.CFrame
local cubeSize = cube.Size
local localPoint = cubeCFrame:PointToObjectSpace(point)

local halfSizeX = cubeSize.X / 2
local halfSizeY = cubeSize.Y / 2
local halfSizeZ = cubeSize.Z / 2

if localPoint.Z >= halfSizeZ then
    return "Front"
elseif localPoint.Z <= -halfSizeZ then
    return "Back"
elseif localPoint.X <= -halfSizeX then
    return "Left"
elseif localPoint.X >= halfSizeX then
    return "Right"
elseif localPoint.Y >= halfSizeY then
    return "Top"
elseif localPoint.Y <= -halfSizeY then
    return "Bottom"
else
    return nil
end

end

[eternal stone]

You don't need that function as far as I can tell
v = p × q, where × denotes the vector product
guiAngle = arccos((v, w)/(|v| |w|)), where (•,•) denotes the scalar product

In other words, v = (p.y q.z - p.z q.y, q.z p.x - q.x p.z, p.x q.y - p.y q.x)

[slow ermine]

The math is a a bit advanced for my current understanding

[eternal stone]

Essentially the vector product p×q is an operation that constructs a vector perpendicular to both p and q

So if p and q are normal vectors of 2 planes, then their vector product is a vector parallel to the intersection line of those planes

[hybrid jay]

hey guys!

don't wanna disturb, but just wanna say something

[eternal stone]

Okay

[hybrid jay]

currently in the 7th grade, but according to standardized tests my math level was that of a college student

so can i try to tackle one of your problems?

in extremely good at algebra

[eternal stone]

Of course, you can select any channel with a problem that you find hard enough

[hybrid jay]

i know ur more knowledgeable than me

sorry to disturb

bye

[eternal stone]

Feel free to help other people here, there're plenty of channels :)

[hybrid jay]

but can i help him too, or only one person can?

[slow ermine]

The wording is a bit confusing to me, im more of a visual leaner, can I get a bit more detail that way by chance?

[eternal stone]

Yes, you can help Burdy as well

[hybrid jay]

ah okay

what is you working on exactly lol

i just wanna know

[eternal stone]

#help-28 message

[eternal stone]

Sure, here's a pic

Blue vector and teal vector are normals to the given planes

Orange vector is their × product

It happens to perfectly align with the red line (intersection of planes)

A face of the cube is a plane. The other plane in our case is the separator plane

(The plane that cuts the cube into red and green parts)

[hybrid jay]

yo here is a more modified version of your function, given that you already have cube's dimensions and its position, you can just simply check local coordinates of point relative to cube function getPointsOnFace(cube, point)
local localPoint = cube.CFrame:pointToObjectSpace(point)

local halfSizeX = cube.Size.X / 2
local halfSizeY = cube.Size.Y / 2
local halfSizeZ = cube.Size.Z / 2

if math.abs(localPoint.Z) >= halfSizeZ then
    return localPoint.Z > 0 and "Front" or "Back"
elseif math.abs(localPoint.X) >= halfSizeX then
    return localPoint.X > 0 and "Right" or "Left"
elseif math.abs(localPoint.Y) >= halfSizeY then
    return localPoint.Y > 0 and "Top" or "Bottom"
end

return nil -- Point is inside the cube

end

seems like ur working with a function to determine which face of a cube a point is on?

or am i tripping

im so clueless 😭

you guys there?

[slow ermine]

I think I get what you mean generally, but the way you're explaining the normals for the plains is a bit confusing

Its possibly because I only understand normals in a specific programming language

Its not pretty, but I was able to add the intersection points and separate them into tables to define which faces the intersection shares. take a look:
https://pastebin.com/jQcdQRzx

It was able to successfully separate the points from any face I decide to select. The blue shows the points touching the bottom face.

[slow ermine]

@eternal stone

local function convertPointVectorsToRotation(faceName, point1, point2)

local halfSize = Cube.Size / 2

-- Define the bottom edge vector based on the face name
local bottomEdgeVector

-- Determine the bottom edge vector based on the specified face
if faceName == "Front" then
    bottomEdgeVector = Vector3.new(halfSize.X * 2, 0, 0)
elseif faceName == "Back" then
    bottomEdgeVector = Vector3.new(halfSize.X * 2, 0, 0)
elseif faceName == "Left" then
    bottomEdgeVector = Vector3.new(0, 0, halfSize.Z * 2)
elseif faceName == "Right" then
    bottomEdgeVector = Vector3.new(0, 0, halfSize.Z * 2)
elseif faceName == "Top" then
    bottomEdgeVector = Vector3.new(halfSize.X * 2, 0, 0)
elseif faceName == "Bottom" then
    bottomEdgeVector = Vector3.new(halfSize.X * 2, 0, 0)
end

-- Calculate the vector from point1 to point2
local vectorAB = point2 - point1

-- Calculate the angle between the two vectors using the dot product
local dotProduct = vectorAB.Unit:Dot(bottomEdgeVector.Unit)
local angleRadians = math.acos(dotProduct) -- Get the angle in radians
local angleDegrees = math.deg(angleRadians) -- Convert to degrees

return angleDegrees

end

[eternal stone]

Nice, it almost worked

It probably misses some checks but that's it

[crude prawn]

can somwone. help me

[slow ermine]

Do you see something wrong with how I calculated the rotation?

[eternal stone]

It's probably the direction of vectorAB that messes things up

[azure plank]

Please may someone help me

[sour marsh]

burdy is the original

!original

[full forum]

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.

[sour marsh]

!occupied

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[slow ermine]

I've tried testing each face separately by changing the order of the vectorAB but it doesn't really fix the problem, maybe there's something else wrong?

[slow ermine]

Thank you

[eternal stone]

Hm, one of the issues is that acos(x) returns an angle between 0 and pi, however, in reality we may need the angle between -pi and 0, depending on the direction of rotation

So perhaps atan2 would be a more suitable function

Actually, roblox studio provides the :Angle() method for Vector3 objects

So I guess we want bottomEdgeVector:Angle(vectorAB, q), where q is the normal of the face we're working with

[slow ermine]

Alright, I implemented it. And the speed at which the angle is adjusting seems equal to how much I rotate the cube which is good. However, there's 2 problems:

1. The angle does a 180 degree rotation when one of the points of VectorAB reaches the bottom right point of the bottomEdgeVector (the purple line)

2. The rotation of the gradient was not quite lined up with the rotation of VectorAB still.

I was, however able to keep the rotation perpendicular to VectorAB by changing:

"bottomEdgeVector:Angle(vectorAB, q)" calculation so that the result is negative:

"-(bottomEdgeVector:Angle(vectorAB, q))"
Which gave me this result,

I think its very close, but here is the new function. It defines the bottom edge vector differently by first getting the bottom left point and the bottom right point. Idk if there's a better way?

https://pastebin.com/i7jKeKAC

I think the rotation flipping has to do with the vector calculation

Because flipping VectorAB
From: (point2 - point1)
To: (point1 - point2)
flips the gradient as well, as you see from the video

So basically flipping the points changes the direction of VectorAB

[eternal stone]

Yes, of course
If the vector is flipped then the rotation is off by 180 degress

Are you sure point1 and point2 are always in correct order?

[slow ermine]

Hmm they do seem to swap places here.

The red is point1 and the blue is point2

But why?

[eternal stone]

It depends on your implementation

[slow ermine]

Well the script is 400 lines long currently, so its a bit difficult to decipher where exactly it does wrong. But I debugged to make sure the points being sent were always the same instances and it was true. So I believe the problem is this block of code where my intersect calculations is for some reason swapping the existing positions of the points when the cube orientation reaches a certain point
https://pastebin.com/XQVVTpc5

[eternal stone]

oof

I assume pairs(edges) messes up the order?

Maybe ipairs(edges) could do the trick

[slow ermine]

Didn't fix :/

[eternal stone]

[full forum]

@slow ermine Has your question been resolved?

[slow ermine]

I figured out why they were flipping. I had set the function to loop through the edges of the face in a specific order, which was causing the index of the points to change positions making them swap orders

Just need to figure out why the vector angle calculation is off now... any ideas?

[torn jolt]

I don't see the issue

What do you mean the vector angle calculation is wrong?

[slow ermine]

Its all good I figured it out

[torn jolt]

Right, anything else?
If not, you can close this channel by typing .close

[slow ermine]

.close

[full forum]

Channel closed

Closed by @slow ermine

Use .reopen if this was a mistake.

[eternal stone]

Oh, nice, congratulations!

[slow ermine]

And the rotation was off because the canvas size of my surface Gui was 800 by 600 😂

Did so much worrying about that for nothing

Anyways I appreciate the help 👍

[eternal stone]

Lol

[boreal briar]

it is 10 question

4.5

is the answer

[empty viper]

I need help

I need help with no. 1

[humble gulch]

Cross multiply to get quadratic

And apply quadratic formula

[full forum]

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@silk swift Has your question been resolved?

[full forum]

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[grizzled basin]

3-96(b)

[grizzled basin]

I do not understand the notation of Z sub 6 or [n] sub 6 and cant find it anywhere else in the book

[rare dock]

Z_6 is the set {[0], [1], …, [5]}

[k] is the set of numbers of form 6a + k for some integer a (they have k remainder when divided by 6)

i.e. the numbers that are k mod 6

[grizzled basin]

oh so [0] = {0, 6, 12, 18, ...}?

[rare dock]

yea

[grizzled basin]

ok what is [n]_6

is it just some element of Z_6

[rare dock]

oh by [k] i meant [k]_6

[grizzled basin]

ok

is [n]_6 just an element of Z_6

[rare dock]

yea

[grizzled basin]

i dont really understand what the functio nis doing, then

what would [4]_6 [n]_6 look like?

or does it not matter for the sake of the problem to show it isn't injective

[rare dock]

well it’s just [4n]_6

[grizzled basin]

i see

sorry im still stuck on how to show a situation where [n]_6 is not equal to [m]_6 but f([n]_6) = f([m]_6)

ohh [0]_6 is not equal to [3]_6 but f([0]_6) = f([3]_6) 😁

.close

[full forum]

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[azure marlin]

Hello could I get some help with this question?

[stable nest]

find the parts that look like parts of circles

[green merlin]

break the shape into semicircle arcs

[azure marlin]

and then what

[stable nest]

find the arclength

[green merlin]

do you know how to find the perimeter of an arc

[azure marlin]

no clue

[green merlin]

oh

do you know how to find the perimeter of a circle

[azure marlin]

yes

wait u mean circumference

[green merlin]

yes

[azure marlin]

yeah

I do

kinda

[green merlin]

do you know how to find the perimeter of a semicircle then

[azure marlin]

no

[green merlin]

like the arc part (not including the diameter)

[azure marlin]

I have no clue 😭

[green merlin]

It's the circumference / 2

because the semicircle is half of a circle

[azure marlin]

ok

[torn jolt]

What are the measures of all those smol squares tho? Is it mentioned?

[azure marlin]

No

thats the entire question

@torn jolt

[torn jolt]

I have solved a similar question in the past so i thought i could solve this one too but the length is not mentioned so ig it's gonna be too hard for me

Like this, if measures were given it would have been possible for me 🫠

[azure marlin]

What grade are you in?

@torn jolt

[torn jolt]

12th 🫠

[azure marlin]

So abt to gradute highschool?

[azure marlin]

Do you take math?

[torn jolt]

Umm just a guess but can't we just "assume" each square is 1 unit tho like cm or m 🫠 and then solve?

[azure marlin]

yes

[torn jolt]

Like we do in a unit circle thing

[azure marlin]

Ig so

ig

[torn jolt]

Yup i have got maths even tho i didn't want to take it 😭 i was studying all night lol

[torn jolt]

Lemme try, gimme a min ig

[azure marlin]

Did you find out?

[torn jolt]

Yeah i was just drawing the damn figure, i solved 1/4 i would say

[azure marlin]

alr

Its ok

[azure marlin]

You don't have to do it

[torn jolt]

Ig that's the way atleast

[azure marlin]

but

this question is easier

but too hard for me

Any clue?

[torn jolt]

Lemme try, just a min

[azure marlin]

ok

[torn jolt]

Done

It should be 14π

[torn jolt]

(atleast acc to me)

[azure marlin]

ok

ITS CORREcT

UR A GENIUS

oML

[azure marlin]

what abt this hehe

[torn jolt]

See there are 4 "semi circles" in it of 12 units, 6 units, 2 units and 8 units. So you just use the circumference formula on each of them "separately" and divide each of them with 2 as well (because they are all semicircles yk)

[azure marlin]

ok

[torn jolt]

Wait what grade are you in btw? Coz this is really easy, lemme do it in one sec

[azure marlin]

Ehm

Middleschool

[torn jolt]

See in the first row, x value is given i.e. 0

[azure marlin]

mhm

[torn jolt]

So put that in our main equation i.e. y = 1.5x + 2

So that gives us
Y = 1.5 * 0 + 2
Y = 0 + 2 = 2

[torn jolt]

So Y = 2 is the answer for the first row

In the second row we got Y's value instead of X. So just put Y's value in our main equation

[azure marlin]

wait so

for

this what do I put

[torn jolt]

5 = 1.5x + 2
Subtract 2 from both sides
3 = 1.5x
Divide both sides by 1.5
2 = x

[torn jolt]

2

[azure marlin]

what tbat the next one

5 y

[torn jolt]

(i make silly mistakes sometimes so i can be wrong but nah i am sure this is correct

[azure marlin]

wait so what are the next 3?

[torn jolt]

For the next one

[torn jolt]

This is for the 2nd one

[azure marlin]

what

Im so conufsed

whats the number
💀

[torn jolt]

2

[azure marlin]

ther both 2?

@torn jolt

[torn jolt]

Yup

First one is 2
Second one is 2
Third is 6.5
4th is 1

[azure marlin]

ty

does this look correct

[torn jolt]

Yup (if i didn't make some silly mistake 😔)

[azure marlin]

except for the bottom one

I changed it

I mixed up the y and x

right?

[torn jolt]

Nope the sequence of the answer is
2
2
6.5
1

[azure marlin]

Thats legit what I put

💀

[torn jolt]

Yeah it should be correct 🙂

[azure marlin]

YEAHHH

W

I CAN SLEEP NOW

[torn jolt]

Oh it was

[torn jolt]

Yup gn mate

[azure marlin]

ty

gn

[full forum]

@azure marlin Has your question been resolved?

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[split heath]

Please help how does this make sense

[stable nest]

what's average rate of change formula?

[split heath]

bro what

all i need to do is find how many points the slope is zero

and the solution isnt explaining how that came to fruition

it draws a line with 3 points on it and it says the answer is two?

[stable nest]

yes

the interval is [b, 1]

we know the point b must lie on the same y as the point (1, f(x))

so just draw a line

[split heath]

I don't know how that answers the question though

nevermind

.close

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[stable nest]

... just look at the formula and the interval and you'll understand

[split heath]

.close

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[quiet jungle]

help someone has chegg premium? cause i really need the answer for this one or If you can solve this that would be great

[low sail]

First of all physics

Second of all use kirchoffs loop rule

[quiet jungle]

i tried and i get -277.5 its not close to the actual final asnwer

[quiet jungle]

could u try and solve it?

[low sail]

Show your working

[quiet jungle]

heyy

[low sail]

Which loop did you use?

[hybrid zephyr]

He said he doesn't want current sources

Use source transformation

Uk how to do that?

@quiet jungle

[quiet jungle]

i dont unless u show me

plss

solve it

im so desperate right now

[hybrid zephyr]

We should be helping not solving

Ok si

So*

Source transformation

Let me explain

If u can find a current source parallel to resistor

Then u can use source transformation

To convert it into voltage source with resistor in series

In here u see

5 uA and 2Mohm in parallel

Use this

v=ir

[quiet jungle]

10 v

[hybrid zephyr]

Yes

So now

Remove those two

And put 10V and 5Mohm in series

[quiet jungle]

hmm??

[hybrid zephyr]

Remove the 2Mohm resistor and 5uA current source

Add them in the circuit in series

[quiet jungle]

10 x 5?

sorry

[hybrid zephyr]

10*5?

Wait a sec

I'll draw and send

[quiet jungle]

okok thank you

[hybrid zephyr]

Mhm

@quiet jungle

[quiet jungle]

ok so the 5 becomes 10 after transformation

[hybrid zephyr]

?

Yes

[quiet jungle]

next?

[hybrid zephyr]

Now it's simple enough right

Simple circuit

Add all voltages

Add all resistors

Solve

[quiet jungle]

-5/18?

=

[hybrid zephyr]

Idk I didn't solve

[quiet jungle]

-277.7

real

wrong

[hybrid zephyr]

Adding doesn't mean summing up

U have to check

Sign convention

U missed there maybe

[quiet jungle]

like in the final answer here is -577.5

[quiet jungle]

i didnt miss anything i extracted the other side to get i

then divide both sides

[hybrid zephyr]

Send process once

[quiet jungle]

well i might miss the signs

[hybrid zephyr]

Well no

Sorry I messed up

I sent the wrong figure

Like this sry

@quiet jungle

[quiet jungle]

add all v and r?

[hybrid zephyr]

No

They not in series

Kirchoff laws

[quiet jungle]

KVL?

[hybrid zephyr]

Yes

KCL also works

[quiet jungle]

well well

look at this bwahah

[hybrid zephyr]

What happnd

[quiet jungle]

did i do it right?

sorry

@hybrid zephyr

[hybrid zephyr]

I don't understand anything u wrote

Cuz I don't see the equations

[quiet jungle]

its KVL

[hybrid zephyr]

U need 3 loop equations

How did u just have numbers only

[quiet jungle]

all the 12 7 10 is V and the rest are i

[hybrid zephyr]

U cannot add like that

I told you

[quiet jungle]

sorry

[hybrid zephyr]

Voltages are not in series

Not the resistors

[quiet jungle]

lets do a white board thing

pleasee

😭

[hybrid zephyr]

Dude

Bruhh

Wait

[quiet jungle]

ok ok

[hybrid zephyr]

I'm not able to find the thing u need to solve

U are not able to solve cuz u r not familiar with basics

Idk how to explain

[quiet jungle]

.close

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[white smelt]

You are given two coordinate system in three dimensional space. You are also given the general matrix of transformation of them (a column of a1, a2 and a3, then with b and with c). Under which conditions the second coordinate system will also be Cartesian system, if we know that the first one is the Cartesian system?
I know how to do that for the 2d space, by I can't prove it for the 3d space

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[white smelt]

.reopen

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✅

[white smelt]

Lines are perpendicular and so on

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[fossil wedge]

how to compute this quickly?

[fossil wedge]

it says that we can ignore the terms of degree >5, but why?

[left bone]

its a good enough approx without them

[fossil wedge]

is that because of the o(x^5) that we choose to ignore the terms of degree >5 ?

[austere cove]

Essentially the o(x^5) captures the rest of the terms in the series.

[fossil wedge]

yeah

[austere cove]

Because the actual series is infinite

[fossil wedge]

so o(x^5) represents a polynomial that for x-->0 is "smaller" than x^5

so any polynomial with degree >5

i think i understand but i need the steps...

[tender tapir]

You actually know exactly what the polynomial is, since it's actually just the Taylor series of sin(x) at x=0 inside ()^3

[fossil wedge]

but if we actually expanded this, then it would be like (a+b+c+d)^3 and 3a^2c = 3x^2 * o(x^5) = o(x^7) would be the little o with the lowest degree

so we'd have to ignore the terms with degree >7, not >5...

[tender tapir]

I think if you move the little o out of the power, the lowest degree that doesn't get fully accounted for is x^9, since the lowest degree that doesn't get included from the Taylor series is x^7. And the lowest combination with it would be x*x*x^7=x^9

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[fossil wedge]

sorry i dont understand

[tender tapir]

Tbh. I'm not really familiar with the little O notation, only big O, but if I understand it correctly, x^9 being not fully accounted for would make o(x^7) as you suggested.

[fossil wedge]

okay thank you!

[tender tapir]

The difference is only that you expanded o(x^5), while I expanded the lowest degree of the residual series of o(x^5)

[torn jolt]

it is good also to fisrst put x^3 in factor of all

[fossil wedge]

so this would just become x^3 - x^5/2 + o(x^7)

that becomes x^3 - x^5/2 + o(x^5)

[fossil wedge]

i did a^3 + 3a^2b + 3a^2c

[tender tapir]

Which is the lowest degree that you want fully accounted for?

x^5?

[fossil wedge]

i was trying to solve this actually

i dont know how far i have to expand it

[tender tapir]

This screams L'Hopital, doesn't it? 😅

[fossil wedge]

idk, i was trying to use taylor series...

anyway i think i solved it

[fossil wedge]

thanks very much 🙂

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[red frost]

Please help

So

I've been kinda stuck on this question for a bit

[torn jolt]

Ping a helper not me

[red frost]

I thought you helped some else

oh ok

[olive saffron]

What have you done so far

where have you got

[red frost]

I haven't got anywhere

@olive saffron

[olive saffron]

ok

[red frost]

so i've done 3700x1.05

= 3885

2x12=24

24+11 = 35

and it says it's compound intrest each month