help-0

[keen fossil]

ok wait

<@&286206848099549185>

[tacit arch]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[keen fossil]

i cant

im not on phone

i can talk it out

but i cant screenshot anything

[tacit arch]

...

You can clearly type numbers

[keen fossil]

ok so

i went from

7√45 - 2√80 right

7√9 √5 - 2√16 √5

and then i did this thing where you simplify 16 into 4√5

so now we have 7√9 √5 - 2 times 4√5

which basically equals

7√9√5 - 8√5

wait

so then

7 times 3√5 - 8√5

equals 21 - 8√5

which technically equals uhh

13√5

wait so did i do it wrong

probably

[tacit arch]

Use a calculator to check

[keen fossil]

yea that was correc

thanks riemann

.close

[lone heart]

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[serene marlin]

im not sure what im doing wrong

or where to start to fix it

[tepid sundial]

thats your difference quotient formula

ill tell you your final answer 8x is correct but you didnt setup the limit correctly

[serene marlin]

okie!

.close

[lone heart]

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[lone heart]

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[vague briar]

how would i start off this indentity? (prove the identity is real)
tanx + cotx = 2csc2x

[tardy stag]

turn everything into sin and cos

[vague briar]

my teacher insists that i chose one side and stick to it throughout the equation, Should i start off with double angle identity on the right?

[tardy stag]

a neat trick is that you can go down one side and up the other

but yeah sure double angle seems like a good idea

[vague briar]

would i convert that to 1/sin2x and then expand out with it in the denominator? or should i times both sides then factor it out

[tardy stag]

should i times both sides
i thought you weren't allowed to do that

would i convert that to 1/sin2x
yes that seems like a good idea unless you have an identity for csc2x memorized lol

[vague briar]

no im not alowed to, also okay ill try that.

figured it out, thanks!

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[echo goblet]

Can someone help me with another related rates problem?

[stark crater]

Differentiate both sides with respect to time

[echo goblet]

Ok so I got

[stark crater]

You need to raise the 8+x^3 to the -1/2 power

But other than that it looks good

You want to find how fast x is changing, so solve for dx/dt

[echo goblet]

oohhh thats the part i missed

ok thanks

.close

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[vapid shuttle]

I'm working on this, but not getting very far.

So, what I've gathered is that each face has atleast 3 edges

the total degree of the faces is 2 times the edges

and

[ocean seal]

Austin

[vapid shuttle]

also eulers formula applies because it is a 3D convex polyhedra

so v-e+f=2

so I have all of these tools at my disposal

but really not seeing anything to do with them

I want to create some sort of contradiction I'd assume. Suppose P does not have a triangular face or a vertex of degree 3, then........

but I don't see at all what this tells me about any of the stuff I know has to be true for my polyhedra

any ideas or comments?

@rose sigil

@bright hedge

[bright hedge]

Have you found an example where it has a triangular face and no vertex of degreeb3

[vapid shuttle]

I have not, I don't really know how to draw 3D convex polyhedra at all

so

there's also that

[bright hedge]

Tbh I don’t think you can

[vapid shuttle]

so how could I find an example?

[bright hedge]

Because if Hou have a triangular face and it’s 3D one of the edges need to be connected to some other face

Which gives you a third degree

Which is weird

[vapid shuttle]

if you have a triangular face

it is degree 3

one of the edges does need to be connected to some other face yes

what do you mean after that though?

Chad there is also this

[bright hedge]

Like if this were true why does it mention triangular face

[vapid shuttle]

if there are no triangular faces

then all of the faces have degree >= 4

so this becomes

$2e=\sum_{F}d(F)\geq 4F$

[ocean seal]

Austin

[vapid shuttle]

if there are no triangular faces

could we work with that to show there is a vertex of degree 3?

[bright hedge]

Pigeonhole almost solves this from here

[vapid shuttle]

yeah that would be better

that is a better method

not pigeonhole, idk what you mean yet

but

Suppose there is no triangular face show this implies a vertex of degree 3

and additionally proving

suppoose there is no vertex of degree 3 show this implies a triangular face

instead of contradicting both at once

[bright hedge]

This is false

[vapid shuttle]

that is nicer aprroach

[bright hedge]

Cuz we showed it would have to have a vertex of degree 3 to be 3D

[vapid shuttle]

?

how did we show that

[bright hedge]

Start with the triangular face. If none of the vertexes had degree 3, they all have degree 2. But then we just have the triangular face which is not 3D

[vapid shuttle]

that seems like too trivial, why would the part about a vertex of degree 3 be included then?

[bright hedge]

That’s what I’m saying it’s weird

[vapid shuttle]

and why does it say "Or"

[bright hedge]

Oh wait

Does it mean exactly degree 3

[vapid shuttle]

yeah it does man

[bright hedge]

My b

[vapid shuttle]

okay so

Suppose there is no triangular face show this implies a vertex of degree 3
and additionally proving
suppose there is no vertex's of degree 3 show this implies a triangular face

this is our approach

seems decent enough right?

[bright hedge]

Yeah

[vapid shuttle]

okay that's better

and if there is no triangular face

so working on the first statement

then we have this

and we have to think about if this would be useful to imply there is a vertex of exactly degree 3

[bright hedge]

Can we find an example where the polyhedra has no vertexes of degree 3

[vapid shuttle]

like I said earlier

I don't know any examples of 3D convex polyhedra

[bright hedge]

Yes octahedron

And it has triangular faces

[vapid shuttle]

https://tenor.com/view/platonic-solid-gif-26298583

this?

how tf am I supposed to imagine that

[bright hedge]

Be better idk

[vapid shuttle]

skill issue

pain

okay

[bright hedge]

Yeah

[vapid shuttle]

so an example

the thing we are trying to prove seems likely it is true

rejoice

but no progress

[bright hedge]

https://tenor.com/view/we-are-so-back-we're-back-cat-happy-gif-960412808382061232

[vapid shuttle]

https://tenor.com/view/stickydacat-gif-26502121

[bright hedge]

What goes wrong if it has all square faces

[vapid shuttle]

like a cube

nothing goes wrong

but the vertices have degree 3

[bright hedge]

Suppose they’d I didn’t

[vapid shuttle]

..

then our cube is not a cube

and is just like empty strings floating attached to eachother in my mind

[bright hedge]

V - E + F = 2

[vapid shuttle]

square faces

degree 4

4C-E+V=2

why does this help

[bright hedge]

E is multiple of 4 too

[vapid shuttle]

no

it could be a multiple of 2

[bright hedge]

Yes dumbass all the faces are squares

[vapid shuttle]

but in like a more abstract sense

or are we just working only with the cube

[bright hedge]

Ik we dealing with this first

[vapid shuttle]

ok cube

[bright hedge]

Anyways this says the # of vertices is 2 mod 4 but it should be a multiple of 4 a contradiction VM

[vapid shuttle]

4(C-E)+V=2

[bright hedge]

Wait

Nm

[vapid shuttle]

lamo

[vapid shuttle]

@lone heart ban him please

[bright hedge]

Yeah this is tough

Lmso

[vapid shuttle]

@ocean seal

[bright hedge]

@vapid shuttle

[vapid shuttle]

@fierce herald

bruhhh

I went to office hours for this one and he gave me a hint

but it isn't helping

[bright hedge]

What was it

[vapid shuttle]

it was basically the stuff I told you

eulers formula should apply

and

2e= total degree of all the faces >=3f

basically the sketch was

since 3d convex polyhedra -> planar -> eulers formula -> contradictons

that is what Ricky said

and he did as an example proving that K5 is not planar

[bright hedge]

Because V - E + F is not 2?

[vapid shuttle]

no

because 2e = 20 and 20 is not >= 21

[bright hedge]

Ah

[vapid shuttle]

2e=total degree of all the faces >= 3f

20 >= 21

[bright hedge]

Do you have an equivalent classification for triangular face

Other than face of degree 3

[vapid shuttle]

no

[bright hedge]

I think the triangular faces come in pairs

[vapid shuttle]

idk if that is true or not

idk how to prove if that is true or not

idk if that helps us if it is true or not

[vapid shuttle]

@fallen verge

[fallen verge]

bink

[vapid shuttle]

bonk

[fallen verge]

oh

this problem

the one i was lowkey scared of

[vapid shuttle]

I got hints in office hours

[fallen verge]

tis fine i will power through

[vapid shuttle]

maybe they will help you

[fallen verge]

v-e+f=2

[vapid shuttle]

so basically what I got was

yes eulers formula

but also

2e=(total degree of faces) >= 3f

[fallen verge]

assume the opposite and prove by contradiction?

ok yeah

[vapid shuttle]

no

I don't think that is a good appraoch

a better one would be

suppose there is no triangular face

show this implies a vertex of degree 3

and then prove the opposite statement yknow

[fallen verge]

its time to

[vapid shuttle]

instead of trying to contradict both

[fallen verge]

d-d-d-d-d-dual

you can use duals since they are convex polyehedra

[vapid shuttle]

what are duals?

also Garlic

another potentially good hint for you

https://mathoverflow.net/questions/106835/does-eulers-formula-imply-bounds-on-the-degree-of-vertices-in-a-3-polytopal-gra

this

[fallen verge]

duals are sort of the inverted polyhedron

[vapid shuttle]

which I couldn't quite follow

[fallen verge]

hold on lemme think

[vapid shuttle]

I also recieved a hint for #4 if you wanted to talk about that one

[fallen verge]

couldnt think of anything quick, checking the mathoverflow problem

actually just consider one vertex

[vapid shuttle]

consider any one vertex of our 3D convex polyhedra

[fallen verge]

if it cannot have triangles and has to have at least 4 shapes, what must be around the vertex?

[vapid shuttle]

something larger than a triangle?

[fallen verge]

yes

[vapid shuttle]

more than 2 edges

more than 3 edges actually

[fallen verge]

at least 4 edges

[vapid shuttle]

yes

I did get that earlier

[fallen verge]

oh wait they need not be regular

[vapid shuttle]

our formula becomes

[fallen verge]

my dumbass

[vapid shuttle]

2e=(total degree of faces)>=4f

[fallen verge]

if theyre regular, its a lot easier

[vapid shuttle]

if there is no triangular faces

we have that

[fallen verge]

are you allowed to use the fact that K_5 is not planar?

[vapid shuttle]

OHOHOOHOOHO

IWNDIWNIDNWDow

Garlic

we proved that as an example

as a hint

so

yes

I imagine you are onto somethin

because he showed us that as an example hint for this question

[fallen verge]

hmm one sec

im trying to find a way to "simplify" the graph to get a K5

[vapid shuttle]

Garlic

I do already have the eulers formula applies for convex 3D polyhedra

btw

if that is what you are looking for

[fallen verge]

yes i know

[vapid shuttle]

okay just making sure

I'm gonna be reading my textbook to see if there is any helpful examples

tag me if u think of something pls

[fallen verge]

alright

[lone heart]

@vapid shuttle Has your question been resolved?

[fallen verge]

@vapid shuttle i think i have a way

it might be a little finicky

but it works

[vapid shuttle]

no way!??!!

https://tenor.com/view/customer-service-do-you-know-the-way-du-yu-kno-de-wey-gif-13342854

[fallen verge]

so remember how i wanted to prove that a K5 must exist in the graph somewhere?

so we look at a vertex in the graph

[vapid shuttle]

sure

[fallen verge]

we know that there must be at least 4 polygons connected to it

that creates a graph with 8 out of the 10 edges

after simplification

[vapid shuttle]

Garlic

can I have 10 minutes and then talk?

[fallen verge]

sure

[vapid shuttle]

@fallen verge okay I am ready when you are

so we look at a vertex in the graph

this vertex has at least 4 edges connected to it

if it does not have 3 edges connected to it

indeed this is true

when you say, "with 8 out of the 10 edges" what do you mean?

don't we only have 4 edges from our vertex? not 8? and the 10 edges of what?

[maiden glen]

I think I have a solution if you are interested

[vapid shuttle]

Yes please!

[maiden glen]

Suppose every face has $>3$ edges. Then, $4f\leq 2e$ (at least $4$ edges per face, but $2$ faces per edge), so $2f<e$. If every vertex has $>3$ incident edges, then $2e\geq4v$, or $e\geq 2v$ (similar double counting argument).
By Euler's formula,
\begin{align*}
v-e+f&=2\
2v-2e+2f&=4\
2v-2e+e&\geq4\
2v-e&\geq4
\end{align*}
so $e\leq 2v-4$, violating the bound on the second inequality with vertex/edges.

[vapid shuttle]

why does 4f<=2e => 2f>e ?

this seems the opposite of true

[maiden glen]

probably mistyped the inequality

hold on

[vapid shuttle]

also what are incident edges?

[maiden glen]

edges that touch the vertex

[vapid shuttle]

ah ok

[maiden glen]

an edge (a,b) is said to be incident to vertices a and b

[vapid shuttle]

I've never actually worked with them in a coordinate system

[ocean seal]

Desync

[maiden glen]

ah, that's not a coordinate

it's an ordered pair

[fallen verge]

ah thats better than my idea

[maiden glen]

formally, edges are ordered pairs (or tuples for hypergraphs) of vertices

[vapid shuttle]

formally ordered?

but anyways

ah

okay so

if every face has >3 edges

then 2e=sum of all the degrees of the faces

2e>=4f

yes

I had gotten this far earlier

okay and then

2f<e

[maiden glen]

that should be nonstrict

I'm used to typing with macros but they don't activate on discord 😭

[vapid shuttle]

okay so 2e>4f

why?

every edge has atleast 4 faces

but what if they all just have 4

is that not allowed someway

oh nonstrict

the other one

I see sorry

[maiden glen]

yeah

[vapid shuttle]

and should it be flipped aswell

nvm

not flipped

geez

[maiden glen]

inequalities

in terms of graphs it's basically this theorem

my proof is just probably scuffed

[vapid shuttle]

suppose every face has >3 edges (so no triangular faces) then this impies 2f<=e yes. if every vertex has strictly greater than 3 edges, (so no vertex with 3 edges) then 2e>=4v or e>=2v

I understand up until here

[maiden glen]

so substitute 2f<=e into euler's formula

and we get e<=2v-4

which can't hold at the same time as e>=2v

[vapid shuttle]

so contradiction

yet eulers formula should hold

for convex 3D polyhedra

so it must be the case that one of our assumptions

no triangular faces or no vertex with 3 edges

is true

[maiden glen]

I think it's slightly clearer phrased as a direct proof

assume no triangular faces, so 2f<=e

[vapid shuttle]

but that is the gist indeed?

[maiden glen]

by euler, e<=2v-4

[vapid shuttle]

I think I do understand that

[maiden glen]

so the >3 degree for all vertices can't hold, so no triangular faces implies degree 3 vertex

[maiden glen]

yeah

[vapid shuttle]

ty Desync, do you mind if I take some time to try to sort through this and then tag you once I have ?

maybe 10-15 mins

[maiden glen]

sure

[vapid shuttle]

okay awesome

@maiden glen hopefully I’ve adequately summarized your points?

I think it makes good sense

[maiden glen]

yeah

technically, both assumptions could be false, so you should handle that separate case if you're doing it as a contrdiction

[vapid shuttle]

if both assumptions are false

then we have a triangular face and a vertex with 3 edges

this is fine

[maiden glen]

sure yeah

[vapid shuttle]

I should write

"atleast one.."

but it is fine

[maiden glen]

yeah

[vapid shuttle]

nonetheless

alright ty very much desync

that one had me stumped

goodnight!

.close

[lone heart]

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[maiden glen]

time for lectures for me

[vapid shuttle]

rip

[lone heart]

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[quick thorn]

Hello, I was wondering if anyone knows how to do this step by step. Any help would be appreciated, thanks.

[quick thorn]

for context, its a high school maths investigation for calculating the area of k trapeziums under this sine curve:

I know I can use integration but I am trying to find other methods

[lone heart]

@quick thorn Has your question been resolved?

[lone heart]

@quick thorn Has your question been resolved?

[lone heart]

@quick thorn Has your question been resolved?

[quick thorn]

ok

does anyone know why this works

[quick thorn]

at least for natural numbers

[lone heart]

@quick thorn Has your question been resolved?

[quick thorn]

Yeah ill try that

[lone heart]

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[alpine sable]

To get Mass can you do M=F(g+a)?

[dim oasis]

Seems like more of a physics question, we have a server in #old-network

[alpine sable]

I’ve asked plenty of physics questions here before

[tardy stag]

F = ma

[worthy oar]

Yeah thats correct.

It's F = ma

[dim oasis]

If it has mass in it and you know the other variables, I don't see why you can't find mass with it

[alpine sable]

Ok thanks

[tardy stag]

keep units attached

it's an easy check

[alpine sable]

,rotate

[alpine sable]

How would I do 12?

[tardy stag]

figure out how fast you were accelerating the cart

[alpine sable]

Would final velocity be 0?

[worthy oar]

Initial will be 0 I think.

[alpine sable]

Oh yea

To get vf can I use gravity for acceleration in this case)

?

Or would it not work like that

[worthy oar]

Nope.

As it's not vertical.

[tardy stag]

you are accelerating the cart, gravity isn't doing anything here

[alpine sable]

👍

Can’t I get Vf from doing x/t

[tardy stag]

no because you're not traveling at a constant velocity throughout that time

[alpine sable]

[tardy stag]

hi you should get your own channel, refer to #❓how-to-get-help

[alpine sable]

oh, i apologize. thank you

I’m actually not sure where to start can I get some help

[tardy stag]

you should have an eqn for motion under acceleration

but here I'll give u one

$s_f = s_0 + v_0t + \f12a_0t^2$

[ocean seal]

hayley

[alpine sable]

👍

[tardy stag]

oh that should just be at² bc this only works for constant acceleration

[alpine sable]

Is that right

It seems too small

[tardy stag]

it is too small

you elected to divide by 1/2 which confused you instead of multiplying by 2

[alpine sable]

Yeah sorry I just noticed

I got 9.122kg

Thanks I think I got it

👍

[worthy oar]

You got it. Good job.

@tardy stag Thanks for the help 👍

[alpine sable]

@tardy stag want me to make you anything in return for you helping me?

[worthy oar]

Wow looks nice.

[tardy stag]

they look so pretty! but those look like they take so much time, I'd feel bad asking you to make something like that for the like two minutes I spent on you

[worthy oar]

2 minutes is still time lol

[alpine sable]

.close

[lone heart]

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[alpine sable]

You can decide next time if you want anything

[lone heart]

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[strange fractal]

um..

[strange fractal]

why is it -1/2 at the start

[tardy stag]

so that the y intercept matches

[round geyser]

when x = 0, what is y

[strange fractal]

do i have to do that every time to check my answer

[round geyser]

yeah

[strange fractal]

🥲🥲🥲🥲🥲🥲

why cant it be normal 😭😭😭😭

[strange fractal]

-6

so to make it pos 3

u need -1/2...?

[tardy stag]

yeah

[strange fractal]

this is devastating

.close

[lone heart]

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[potent plank]

In this video of 3blue 1 brown}
https://www.youtube.com/watch?v=e50Bj7jn9IQ&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=15

He is equating the mean of eigen values and mean of diagonals

Why is it equal ? Am I missing something?

[glacial patrol]

Trace of a matrix is sum of its eigenvalues

[potent plank]

Oh ok got it, I will check for the proof on the internet thanks.

.close

[lone heart]

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[strange fractal]

I thought everything to the power of 0 is 1

[strange fractal]

But if its -2^0

Is it -1?

no

t

[alpine sable]

(-2)^0 not -(2)^0

[strange fractal]

oh

i c ic

ok thx

.close

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[slender granite]

help would be greatly appreciated. This shouldn't be hard I just don't understand what could be wrong here

❤️

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[acoustic ether]

tf

[keen pasture]

<@&268886789983436800>

If you are NF fan: Great music taste

Don't think it is possible as the server has to many members

[pliant oracle]

Please don't troll / be inappropriate like this @alpine sable

[acoustic ether]

indeed but not for your bichazz ass

and please stop this nonesence

[tardy stag]

I'll give u 2 weeks

[acoustic ether]

2 weeks is kinda much i can do same in 2 days with yo mom

but

am a normal man without any autism or dawn sindrom so

and please stop this. this is chat for normal ppl not with retardations like u

be more serious

[tardy stag]

.close

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[mellow linden]

how come the direction vector if a line is parallel to the norm of a plane? or is that false

[mortal trellis]

what is the relationship between the line and the plane

[mellow linden]

not sure

[vale wigeon]

!original

[lone heart]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[mellow linden]

Ex. 12. Give an equation of the plane that is orthogonal to the line (x, y, z) = (1, 0, 1) + (1, -1, 2)t andpasses through the origin.

Why can n = (1,-1,2)

[mortal trellis]

thats what it means for a line to be orthogonal to a plane

[vale wigeon]

^

[mortal trellis]

basically by definition

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[real gazelle]

(Please don't use the words "retardation" or "autism" etc as insults, this is unacceptable in this server)

[acoustic ether]

i wont but that guy was really messy (sorry)

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[sly tundra]

Guys I found the lamest username of all time

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[stone breach]

.claim

claim

[stone breach]

Can somone solve the Integral for me? and also explain their steps? please

i have no idea how to solve the integral and it seems complicated

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@stone breach Has your question been resolved?

[noble frost]

L = int_{a}^{b}sqrt(1+f'(t)²)dt

[stone breach]

.close

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[alpine sable]

this question right here

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[alpine sable]

<@&286206848099549185>

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[alpine sable]

.close

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[scarlet ridge]

45-The image representes a center circumference 0.

It is known that:

• [RST] is an equilateral triangle inscribed on the circumference;

• [SU] is the side of a square inscribed on the circumference.

Determines, in radians:

45.2- UST

[scarlet ridge]

idk why its not pi/6

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[woven sphinx]

how would i workout which ones are subsets

[woven sphinx]

and how would i test whether they are closed under addition or multiplication

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[thorn niche]

help

ok

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[thorn niche]

help

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[forest heron]

i have to use the product rule to differentiate the function, but I'm lost on where to start and the process

[coral thorn]

do you know what the product rule is?

[limpid turret]

Do you know product rule?

[forest heron]

yes i do but what mean to differentiate the function

[limpid turret]

calculate the derivative

[coral thorn]

how’d you learn the product rule without learning what differentiate means

[forest heron]

i was having a brain fart thanks for the help

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[nocturne iris]

hello again

[nocturne iris]

Show that the subgroups of ( (\mathbb{Z}, +) ) are the ( a\mathbb{Z} ) where ( a \in \mathbb{N} ).

[ocean seal]

lilisworld

[nocturne iris]

i need a hint

<@&286206848099549185>

it's my section

[prime anchor]

Sorry, it's my fault

[nocturne iris]

np

<@&286206848099549185>

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[fair musk]

what is |a + b|^2 and why

[mortal trellis]

what are a and b

[fair musk]

any real numbers

[mortal trellis]

then the same as (a+b)^2

cause it doesnt matter whether a+b is negative or positive, the square is the same anyway

[fair musk]

If we want to prove that for all real numbers a and b, |a + b| <= |a| + |b| is true, can I do cases?

1. if a < 0 and b < 0, they're equal
2. if a >= 0 and b >= 0, they're equal
3. if a < 0 and b >= 0, the left side is less than OR EQUAL TO the right side.

◻️

[mortal trellis]

is it?

the last case?

what if a is so much negative that |a+b| is huge again

you can do it by cases but it needs to be slightly more involved

[fair musk]

i dont follow

doesnt my case work for that too?

[mortal trellis]

you just state the result

where is your argument

[fair musk]

I argue the same thing as the problem says

[mortal trellis]

you basically just write "case 3: works out"

[fair musk]

case 3: if a > b and b != 0, then |a + b| < |a| + |b|. If a < b and b != 0, then |a + b| < |a| + |b|.
if b = 0, then they are equal.

[mortal trellis]

you just state what you want

you dont state why

[fair musk]

we want that to be true for all cases so the entire thing is true

[mortal trellis]

well we want it to be true

doesnt mean that it automatically is

we still have to argue that it is true

[fair musk]

Oh so we just add "assume that this is true" at the beginning?

Then prove that assumption?

ok so then i add "assume that this is true." and then do my cases.

[mortal trellis]

no you cant just assume what you want to show

so one possible case is that b>0 and a<0, but 0<-a < b. then 0< a+b<=b and therefore |a+b| = a+b <= b = |b| <= |a|+|b|

I have to go now

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[silver tangle]

a

[silver tangle]

can someone explain the last term

[mortal trellis]

if you start at a vertex v and have two neighbors u,w, then a path of length 4 could be v,w,v,u,v. but you dont want to include those

[silver tangle]

does it also count u->v-w->v-u because its deg*(deg-1)*2 to count his case and the one u mentioned?

since deg edges can come in and only deg-1 can go out

@mortal trellis

[mortal trellis]

well you can either go to u or to w first

but if you start at u and then go over v to w and back thats just the same path

[silver tangle]

so 4 ways to count it then?

[mortal trellis]

why 4

[silver tangle]

start at v or u = 2, start at w go left or right = 2

and need to choose 2 points adjacent to v

[mortal trellis]

you cant go right at w. I am not sure what you mean

ok sry its too late for me to be awake and trying to help

good luck

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[foggy mica]

How do I simplify this to that? Can someone break it down into steps? Thank you.

[wild umbra]

square both side

[foggy mica]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

ok that makes sense

thanks

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[thin sedge]

what quadrant is -5pi/3 on the unit circle

[thin sedge]

first quadrant?

,rotate

[thin sedge]

Nick

finish typing

brb

yep

is it the first quadrant

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[high zealot]

z table related

[upbeat gorge]

You might want to draw this so you know what part of the z-table to look at

[high zealot]

yeah this is when im stuck at lol

[upbeat gorge]

If x exceeds 150, are we concerned with values to the left or to the right of 150?

[high zealot]

right?

@upbeat gorge

[upbeat gorge]

Yes

[upbeat gorge]

So probability is area under the curve, right?

So that means the area to the right of 150 is 0.9

[upbeat gorge]

So both of these are wrong

[high zealot]

huh then what should this look like?

[upbeat gorge]

Sketch this

[high zealot]

?

[upbeat gorge]

Yes (though my that is not a good-looking bell curve lol)

[high zealot]

lol i agree i drew that with 2400 dpi mouse

so the probability for this is z score + table value?

@upbeat gorge

0.5*

[upbeat gorge]

I mean it depends on your z-table

[high zealot]

pls explain

[upbeat gorge]

you’re looking for the mean tho

[high zealot]

right

[upbeat gorge]

Well first you want the z-score of 150

[high zealot]

my z table only goes to 3.09

[ocean seal]

CST (please ping when replying)

[high zealot]

isnt 150 the x ?

[upbeat gorge]

It is

[high zealot]

i dont get how the z score could be found in this case

looked online and its 1.28 but idk why

@upbeat gorge

[upbeat gorge]

Can I see your z-table

[high zealot]

its this but 1st decimal only goes to 3.0

wait hold on no

[upbeat gorge]

Oh ic

Yeah you want to look for 0.90 inside the table itself

[high zealot]

This is the one

[upbeat gorge]

ah

So it calculates the area from the mean to the z-score

in our case, the area from z-score to mean plus the entire second half is 0.9

[high zealot]

Ok

[upbeat gorge]

So we’re looking for what value

[high zealot]

The z score of x > 90

?

Sorry I m lost at this point

@upbeat gorge

[upbeat gorge]

Find ?

[high zealot]

@upbeat gorge oh the .40!

my table says ?= .1554

[upbeat gorge]

0.4 is not the z-score

0.4 is the probability in the inner rows/columns of the table

[high zealot]

right i mean the z score for .40 = .1554

@upbeat gorge

or is it the z score closest to .40?

.4015?

[upbeat gorge]

I hope btw you know probability of a z-score is not the z-score

[high zealot]

oh mb i confused those 2 together

let me try again, the z score corresponding to the probability on the table closest to .40 is 1.28

right @upbeat gorge

[upbeat gorge]

Yes

That is best

Although

Recall 150 is under the mean

So the z-score isn’t 1.28

[high zealot]

could you show what that situation looks like graphically?

so its actuallly -1.28?

@upbeat gorge

[upbeat gorge]

Yes

[upbeat gorge]

Refer to my image a while ago

[high zealot]

the red part you drew is the 150 right

is the mean the center line? or is it a dot or value within the red shade? could you explain that?

@upbeat gorge

[upbeat gorge]

The red part is the probability lmao

[high zealot]

😅 but i don't get what "under the mean" means

[upbeat gorge]

basically if it’s below the mean the z-score is negative

And if it’s above the mean it’s positive

[high zealot]

where is the mean

or is the mean just 0

[upbeat gorge]

We’re looking for the mean

(The z-score of the mean is 0, but the mean itself isn’t 0)

[high zealot]

hm

ok

[upbeat gorge]

But yeah, you have the z-score, you have X, and you have the SD

you can now solve for the mean

[high zealot]

yeah i just wanted to make sure i don't confuse anything

thanks for your help man really appreciate

[upbeat gorge]

np

[high zealot]

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[hardy aspen]

how come for this when I move 20 to the right, it subtracts it and I eventually end up with x = -1 right

[hardy aspen]

so

i solved this

did the sam ething

and ended up with

d=7

checked the answer sheet and its apparently 5?

and the only way to get that

is to do the opposite and keep it the sanme when i move 3

to the right

[ocean seal]

QuasiStar 超新星

[hardy aspen]

ah so the answer sheet was wrong

alr ty i knew sum was up

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[sweet basin]

Wouldn’t both sides of the equation be the same?

If 2ln(e) is the same as (ln(e))^2

[weak field]

Urs would have been true if it was ln (e^2)

But it's (ln e)^2

[sweet basin]

oh

so how would that be expanded

you can’t expand it right

[weak field]

Solve ln e
And then square it

Follow the brackets

It asks for squaring on ln e

[sweet basin]

Thanks sorry for the ping

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[rustic nebula]

I need to show that if 5 divides n with a remainder of 4, than 10 divides 2n^2 + 2n + 3 with a remainder of 1.
I dont know what I did wrong but I found that, if 5 divides n with remainder of 4, than 2n^2 + 2n +3 has a remainder of 7

[rustic nebula]

<@&286206848099549185>

found the error

I wrote down the problem wrong to begin with. my math is right

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[cloud token]

Can I get some help on part b?

[cloud token]

I was thinking of something like this but obviously it includes integrals

also pretty unfamiliar with indicator random variables so not sure how to apply that here

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[lone heart]

@cloud token Has your question been resolved?

[cloud token]

<@&286206848099549185> ;-;

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[harsh sparrow]

I’m horrible at math and could really use some help with a couple of problems, starting with this one. :(

[formal shard]

So for this, your domain would be your range of x-values. From the graph you can see that the line goes from x = -9 to x = 8. So from this, you could say that your domain would be between -9 and 8, which in interval notation would be Domain = [-9, 8]

I'm assuming the next question is for range, in which you would do the exact same thing but for the y-values instead

[harsh sparrow]

Ohhhh okay thank you

For some reason I kept getting [8, -9]

I could also really use help on this one too

[formal shard]

Like thats what the website was saying is the answer?

[formal shard]

while technically not wrong, its more common to go with the smallest value first

[harsh sparrow]

No it’s what I kept getting myself but I had a feeling it was wrong :(

[formal shard]

It wasn't completely wrong, you just need to make sure you're putting the smallest value, in that case -9, first

you had the right numbers, just need to put them in the right order

[harsh sparrow]

Ohhh okay that actually helps a lot I’ll have to fix some of my past answers then

[formal shard]

For this you would do the same, but since you arent given any values on the axis you would just need to count the number of lines

[harsh sparrow]

Huh ?

[formal shard]

So its similar to the first question where yoiu want to find the range of x-values in the graph

[harsh sparrow]

Ohhh okay

[formal shard]

Since there are two lines, you would just go off of the furthest left point and furthest right, which in this case is the whole graph

[harsh sparrow]

Last question,

[formal shard]

and since there is no numbers labelling how far it goes you would just need to count

[harsh sparrow]

I have no clue how to do this what do ever :’)

[harsh sparrow]

Okay that’s easy tyty

[formal shard]

I'm not exactly sure what it's asking, since theres two questions technically

Ignoring the top line, you would just substitue the x-values into the equations, so for example in the first one a) f(x) = x - 4 find x = 7, you would just put 7 where the x is for f(x) = 7 - 4 and so on for the other questions

[harsh sparrow]

OMG okay that makes things a lot simpler

Thank you :)

[formal shard]

happy to help

[harsh sparrow]

.close

[lone heart]

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[plucky dagger]

help

[plucky dagger]

To be more specific. The last two parameters
I have an equation already that fits the first three
r(x)=(x-2)(x-1)/(x+3)^2(x-2)

Please someone help

[formal shard]

This looks like it also fits the fourth

just looking at it analytically

[plucky dagger]

it touches I think. but it doesnt intersect or have an x intercept

thats why i am confused

[formal shard]

touching it still counts as an x-intercept technically

[plucky dagger]

I know it bounces due to multiplicity but idk how to do that

[formal shard]

since it also asks for a turning point at the same point for the last condition

[plucky dagger]

would I have to change something in the numerator to still have it touch and have an intercept at (1,0)?

[formal shard]

Just graphed it, and looks like there is still an intercept

[plucky dagger]

ohh ok. I thought the bend would be more drastic. Like a parabola alsmot

almost

so does my equation seem to fit all parameters?

[formal shard]

It definitely fits the first 4, i'm just very stumped by the last one sorry

[plucky dagger]

No worries. Hopefully I get a 9/10. Thank you for the help. I don’t have a great professor so I normally teach myself and I just had trouble with this one.

[formal shard]

It seems like it will also fit the last condition

I'm not extremely familiar with how to check for it, but the concavity is positive at and after 1

[formal shard]

I understand that, currently taking a break from studying for a course under a similar professor

[plucky dagger]

The multiplicity is odd I think if you take the two factored pairs. So that would mean it touches and bends

If I’m not mistaken

I think I got it. Thank you again!

.close

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[tribal valve]

WIAT

[naive valley]

ow my eyes

[tribal valve]

i think i know

how

[long axle]

Or if u wanna get that result in a more convoluted way, do some division

[tribal valve]

yeah thats what i was thinking

so just long division x^4/x^4-16

[long axle]

Nah just do this

Same thing just shorter

[tribal valve]

$

$\frac{x^4}{x^4 - 16} + 16$

[ocean seal]

dabbingpotato

[long axle]

$\frac {x^4 -16+16}{x^4 -16}$

[ocean seal]

Stephen

[long axle]

This^

[tribal valve]

oh

can we just do that

[long axle]

Now u can just break the fraction apart

[long axle]

It’s just adding 0 so

[tribal valve]

oh

ah

so you get

$1 + \frac{16}{x^4 - 16}$

[ocean seal]

dabbingpotato

[long axle]

Yes

[tribal valve]

so just

wait

right hand side integral is hard

[long axle]

Why?

[tribal valve]

nvrmind

power rule

bruh

[long axle]

Wat

[tribal valve]

no

not power rule

is it trig sub

[long axle]

Rewrite it as difference of squares

[tribal valve]

i'm ending myself

right

$\frac{A}{x-2} + \frac{B}{x+2} + \frac{CX + D}{x^2 + 4}$

[ocean seal]

dabbingpotato

[long axle]

Ew

Nick u got this bruh

LOL

[tribal valve]

bro

[long axle]

gg

Org chem tutor go hard

[tribal valve]

i know how to do it

its just

gonna take

lots

of

time

and

i

hate

clamaculus

fuck

[long axle]

Well it’s a good thing u hate clamaculus and not calculus cuz calculus can be pretty fun once u get the hang of it

[tribal valve]

wait

why didn't we just use partial fraction decomposition in the first place

oh wait

nvrmind

i know why

okay i can do this by myself from now

but basically

you would just use ln on a/x-2 and b/x+2

and just use u sub on cx + d/x^2 +4

right

ok i got it

thank you guys so much for the help

.close

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[junior warren]

could anyone explain how that 2x3 maxtrix was formed here

is it just this?

aka applying the transformation to the basis elements of P2, then representing them in columns of a matrix

[lone heart]

@junior warren Has your question been resolved?

[lone heart]

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[raven field]

how is f not differentiable at one it has a defined point and no discontinuity

[tardy stag]

it's a sharp turn though

[fallen verge]

if you take the derivative, you will notice it is not continuous

[raven field]

o

i forgot about that one

😭

tysm