help-13

[cedar kiln]

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[shut wadi]

a

[shut wadi]

$5 thru $2000 is range

and 1 thru 1000 cumin seeds in domain

and ofc its real numbers

because what if you wanan do like

1.5 kilograms for $7.5

this correct?

[crystal raptor]

<@&268886789983436800>

[half stratus]

o shoot i messed up

will i get banned

i didnt relaize i copied the link

lemme remove it

well i removed it

[shut wadi]

aaa

<@&286206848099549185>

aint no way pinging 1000 people still does nothing

thats wild

frick it

im just going to risk it its a 25% its right

[twilit knot]

I think it is

Sir have patience

I will review it after dinner

[kindred cobalt]

Hola

@shut wadi

You are right

[shut wadi]

ok

[shut wadi]

but this guy thats verified

says its integers

[kindred cobalt]

Hmm

Trust him

[lunar lynx]

No. Lol.

It should be real numbers.

[kindred cobalt]

Hey it’s you again

[shut wadi]

so this verified guy is wron

[lunar lynx]

Hey. You didn't sleep ?

[shut wadi]

and i was right

?

[kindred cobalt]

I got a nap in

I’ll try again tonight

[lunar lynx]

He/she is.

[kindred cobalt]

Definitely

I should have more confidence in myself

[shut wadi]

bro had a 116 thanks on his reply 💀

[shut wadi]

same

[kindred cobalt]

Usually when I hear the word verified I immediately 2nd guess myself

Anyway is the answer correct?

[shut wadi]

same

yea

before you guys leave

[lunar lynx]

Yes

[shut wadi]

this is correct maximum

-1, -1 and

[kindred cobalt]

It said absolute

[shut wadi]

OH

sos

so only the -5, 5

not the -1, -1

[kindred cobalt]

I would go with that

[kindred cobalt]

Yep

[shut wadi]

yeah mb i make alot of mistakes

[kindred cobalt]

It’s fine

[shut wadi]

from reading too fast

i see "maximum point"

[lunar lynx]

Yup. A big old cross does it.

[shut wadi]

and im like yup

[lunar lynx]

Lol

[kindred cobalt]

Earlier today I was useless in solving a modular arithmetic problem because I didn’t sleep

[shut wadi]

what

i dont even know what that is lol

[kindred cobalt]

It’s basically

Fancy arithmetic with a lot of extra steps

[shut wadi]

ah

👍 👍 👍

[kindred cobalt]

Yeah that’s all you need to know, (until college)

[shut wadi]

real

[kindred cobalt]

?

Wdym

[shut wadi]

yes

[kindred cobalt]

Oh

Sorry idk a lot of slang 💀💀

[shut wadi]

btw is invertable function just reversed

[shut wadi]

ha your good

me neither

i know some

[kindred cobalt]

I learned what TMI was like a week ago

[shut wadi]

but tiktok goes wild

i dont go on tiktok

[kindred cobalt]

Exactly

[shut wadi]

never have

[kindred cobalt]

I go on it to just hear bible quotes and some stem stuff

And memes

[shut wadi]

^

[shut wadi]

yeah

stem stuff seems fun

[kindred cobalt]

Use le Google

[lunar lynx]

You mean switch x and y ?

[shut wadi]

yes

[kindred cobalt]

Yeah

[shut wadi]

so if

[lunar lynx]

No.

[shut wadi]

ahhh

so its just

[kindred cobalt]

Switching x and y is a rotation

[shut wadi]

if x = 8

then

[kindred cobalt]

Well in plane geometry

Yeah

[shut wadi]

g^-1(8) = -8

[muted bear]

Its a reflection

[muted bear]

Not always

[lunar lynx]

Why am i confused ?

[kindred cobalt]

A rotation by 180 degrees in the Cartesian plane is (x,y) —> (-x,-y)

[shut wadi]

yeah

ik those

learned them in pre-algebra

[kindred cobalt]

Or (x,y) —> (-y,-x)

Idk remember the exact formula I think it’s the first one I wrote

[shut wadi]

270?

[muted bear]

Draw the line y=-8 percy

[shut wadi]

ok

[kindred cobalt]

No I think that’s wrong

[shut wadi]

idk

[muted bear]

This is a reflection across y=-x

[shut wadi]

oh you said -8

i was confused

[kindred cobalt]

Lemme check my notes

[lunar lynx]

Btw, @muted bear In general, idea of switching x and y should work to get inverse. Right? Except for domain and range issues in some cases.

[shut wadi]

notes is a word i rarely use

i've only recently started taking seriously notes

[muted bear]

Yes, the function should only have one x avlue at the y value

[shut wadi]

as ive been trying to learn python

[kindred cobalt]

What he said

[muted bear]

Oh my god we're not getting anywhere

[kindred cobalt]

I’ve made ai’s and a bunch of stuff in that

[kindred cobalt]

Agreed

[shut wadi]

bro this is becoming discussion lo

[muted bear]

The line y=-8 should be a straight horizontal line

[shut wadi]

nice

[shut wadi]

oh I was just reflecting y=8

[shut wadi]

🍄

[muted bear]

Max can you stop posting memes please

[kindred cobalt]

Can relate

[kindred cobalt]

Rip

Alr

[shut wadi]

relatable

[lunar lynx]

Percy, focus on solving first. Someone is trying to help you. Be sincere.

[shut wadi]

Yes

[kindred cobalt]

https://tenor.com/view/gigachad-chad-gif-20773266

[shut wadi]

max is very distracting

@

[kindred cobalt]

Hey I’m saying that @lunar lynx said a very true thing

[shut wadi]

@muted bear

[kindred cobalt]

Actually focus, I’ll stop

[muted bear]

Yes

Now that line intersects the function at what x value?

[shut wadi]

0

oh

no

oops

my brain

7

[muted bear]

Yes

that means g^-1(-8)=7

Remember that g(g^-1(x))=x

if x=-8

g^(g^-1(-8))=-8

g^-1(-8)=7

[shut wadi]

so this is just 7

[muted bear]

And we know g(7)=-8

[shut wadi]

i thought it would be 8

because

we were just switching

[muted bear]

Switching x and y values is a reflection on the line y=x

[shut wadi]

ok

so we aren't doing reflection

we are inverting

[muted bear]

Inverting is a reflection

[shut wadi]

so we are reflecting

y=-8

i mean

y-8

[muted bear]

In a way

[shut wadi]

y=8

into y=-8

so the answer is just -8

[muted bear]

No

[shut wadi]

but if inverting is reflection

and we were doing inverting

hen

then

[muted bear]

I think its easier to think of inversion like this:

You would find a normal function value like f(4)=6 by drawing a vertical line a x=4 and then drawing a horizontal line where it hits the graph

to find the inverse, you work backwards

f^-1(6)=4 because you draw a horizontal line at y=6 and draw a vertical line where it hits the graph

[shut wadi]

So

[muted bear]

y=6 not x=6, pretty bad typo there

[shut wadi]

So

[muted bear]

So to find g^-1(-8), you draw a horizontal line at y=-8 and then draw a vertical line where it hits the graph

[shut wadi]

But

[muted bear]

The value it hits on the x axis is your amswer

[shut wadi]

The line hits y=8

[muted bear]

sorry i keep forgetting the -

[shut wadi]

How do I find the answer tho u h

[muted bear]

You draw the horizontal line at y=-8

[shut wadi]

Yeah and it hit 7

[muted bear]

Yeah

Your answer is 7

[shut wadi]

But

[shut wadi]

You said we couldn’t reflect

You said we can’t change from y=8 to y=-8

[muted bear]

No you cannot

[shut wadi]

Then why did we go down to -8. From 8

[muted bear]

When

[shut wadi]

Look

[muted bear]

That was an example

[shut wadi]

Look

We were at y=8

In what magical method did you go down to y=-8

[muted bear]

we are looking for g^-1(-8)

[shut wadi]

Yes

[shut wadi]

But like

The “(-8)”

Is x

So how is that y = -8

[muted bear]

When you are doing inverses, you are looking for "what value of x is there such that g(x)=-8"

if g(x)=y, then y=-8

[shut wadi]

Yes

I still don’t get how we went down from 8 to -8

[muted bear]

Where did the +8 come from

[shut wadi]

Y = 8

The original question

[muted bear]

the original question is g^-1(-8)

[shut wadi]

Ok

[cedar kiln]

@shut wadi Has your question been resolved?

Channel closed

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Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cinder phoenix]

How I'm a supposed to find the period of sin5x?

I need to determine if the fonction is periodic and find his period

But how can I find a period of these fonctions?

[deep oriole]

isnt the period how often the function repeats itself?

[cinder phoenix]

Hum kind of

The correct answer is 2π/5

But how haaha?

[deep oriole]

in the usual sin curve, the function repeats every 2 pi

sin5x repeats 5 times as quickly

so 2pi/5

[cinder phoenix]

Okay

2pi/5 on 5 for what reason?

[deep oriole]

?

[cinder phoenix]

Why 2π/5

And not just 2pi

[slate lintel]

,tex .transformation rules

[deep oriole]

because sin5x is stretched by 1/5 from sinx in the x direction

[wraith dagger]

Hayley

[deep oriole]

so its compressed by 5, hence any repetitions in the sin5x curve happen at 1/5 times the distance

[cinder phoenix]

Ah yes

You're right

Thx bud

[deep oriole]

np

[cinder phoenix]

Sinh?

Honestly I'm in engeeniring and never seen what is sin hyperbolic

[dire geode]

where are you getting sinh from

[deep oriole]

kind of looks like this

[cinder phoenix]

just need to find the period of sinh2x

But seems 1

It's 1*

[deep oriole]

sinhx isnt a periodic function though

[cinder phoenix]

That's why it's one yes I know 🤣

Lmao

[deep oriole]

lol

[cinder phoenix]

So sin(πx/L) Wich sin repeat it self for 2pi it's gonna be what 2L because you expend by L the ondulation?

The waves*

[deep oriole]

the period of sin(πx/L) will be 2L yeah

[cinder phoenix]

Ah yes because he obviously the π included in sin

All right then now let's go on Fourier series fuck that shit hahahahahah

[deep oriole]

lmao

[cinder phoenix]

.close

[cedar kiln]

Channel closed

Closed by @cinder phoenix

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[cedar kiln]

Please don't occupy multiple help channels.

[jolly sable]

I didn't understand the formula how it is actual works

[lunar lynx]

Do you know what's shortest distance between any two lines?

I mean... What line would you draw between two lines if you were asked to draw the shortest line joining two lines?

[jolly sable]

Yes. I know a formula

[lunar lynx]

Not formula.

[lunar lynx]

.

[cedar kiln]

@jolly sable Has your question been resolved?

[jolly sable]

Perpendicular

[lunar lynx]

Yes.

Now, are you familiar with cross product of vectors ?

[jolly sable]

Yes of course

(a.b=a.bcosthita)

[lunar lynx]

No.

That's dot product.

CROSS PRODUCT.

[deep oriole]

does this strictly have to be done with cross product?

[lunar lynx]

They want to understand the formula which uses the cross product.

[deep oriole]

true i didnt see the other picture

[lunar lynx]

Also, for 3-d lines, i think that's the most efficient way anyway.

[jolly sable]

Then you are talking about sin thita sorry cross ×

[lunar lynx]

Yes

Also, that's just magnitude. Cross product gives a vector quantity.

[jolly sable]

A×B = |A||B|sin thita n cap

[lunar lynx]

|A×B| = |A||B|sin theta

So, do you know how's the direction of a cross-product found ?

[jolly sable]

N cap = A×B/|A||B|

Here sin thita 90°

[lunar lynx]

What? You are getting a vector by doing division operation on two scalar quantities. 😐

You need to learn right hand curling method to find direction of cross-product.

[jolly sable]

I don't think left side will be mod

[lunar lynx]

The index finger points in the direction of the velocity vector v. The middle finger points in the direction of the magnetic field vector B. The thumb points in the direction of the cross product F.

[jolly sable]

But if you give mod then direction magnitude is 1

[lunar lynx]

So, a vector is equal to a scalar ?

[jolly sable]

Bro you are making things confusing

Just wait and let me write

[lunar lynx]

Thumb will give you cross product.

[lunar lynx]

You're writing half the things in wrong notation. What can i do?

[lunar lynx]

Yeah. That's good.

I'm asking if you know how to find direction of that n(cap) at end ?

[jolly sable]

So here if we take cross product of these lines then we can find perpendicular direction which will be perpendicular to both of them

But we want our line of perpendicular to each other

Maybe it is in 3d so it can be happen

By flaming law of left hand rule

[lunar lynx]

If a line is perpendicular to line A and line B, isn't it perpendicular to both ?

[lunar lynx]

And Alexander Fleming's dad rolls in his grave. Lol

[jolly sable]

Maybe right hand rule

[jolly sable]

Then both are parallel

It's taking too much time. Could you make it fast? I need to go

[lunar lynx]

Alright. But that still means that each line is perpendicular to both lines in our question.

[lunar lynx]

Sure. You do the cross product to get the perpendicular direction to both line. Now, since distance will be shortest along this - we'll find any distance between two random points on both lines(one on line A and other on line B) and just take a component of it across the perpendicular direction unit vector.

That's what they did.

[jolly sable]

Then surely it is 3d

If it's in 3d then they must be perpendicular to each other right?

I mean line A and line B are at 90°?

Or they can happen on any angle

[lunar lynx]

They can be in any angle.

[jolly sable]

But they will not intersect

[lunar lynx]

I mean. Angle between lines in 3-d isn't exactly defined unless they intersect.

[lunar lynx]

Yeah. If they intersect, shortest distance will be just zero.

[jolly sable]

So they took any random points over these lines

[lunar lynx]

Yes. a1 and a2.

[jolly sable]

I didn't understand it actually

How it works

Could you make it more clear?

Sorry for deep dive

[lunar lynx]

You said that you had to go. Lol. So, i just fast forwarded.

[jolly sable]

They multiply it with unit vector of perpendicular line

[lunar lynx]

Wait for a minute. Let me draw a diagram.

[jolly sable]

I understand the terms so it's more interesting than going out

[lunar lynx]

Damn it. I have to write again.

@jolly sable See this.

I'm not good at drawing or such so it took me so much time.

Hopefully you'll get it now.

Our lines are DC and AB.

TO find distance between them we took a random point on DC as E. Another random point on AB as G.

We find EG.

But we know that that's not the shortest distance.

So, we want its component across the perpendicular to both line which i have drawn as EF.

Since that's a right angle triangle, component across perpendicular is EG cos(theta).

Now, we know that
$$\vec{EG} \cdot \vec{EF} = |\vec{EG}|\cdot|\vec{EF}|\cdot \cos(\theta)$.

[wraith dagger]

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

[lunar lynx]

So, we just divide by $|\vec{EF}|$ to find $EG \cdot \cos(\theta)$ that we need.

[wraith dagger]

Enemagneto

[jolly sable]

Awesome drawing

Why were we doing cross product (just to show how perpendicular works?

[lunar lynx]

Yeah.

I was ascertaining if you understood direction or not.

[jolly sable]

Can we not divide it by |EG| cos

We are finding |EF| no??

EG =EFcos thita no?

[lunar lynx]

No. EG is hypotenuse.

We always take component of hypotenuse.

[jolly sable]

I see

Thank you

.close

[cedar kiln]

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[cedar kiln]

Available help channel!

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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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[gloomy wing]

if my friend and i are playing an amongus game of 10 people and 2 impostors, what is the chance we both get impostor?

[gloomy wing]

i have a feeling it's 1/45 cuz 2/10 for first impostor times 1/9 for second

but idk probability lol

[south tundra]

Should be 2/45

[gloomy wing]

oh why

[south tundra]

Actually wait

Oh, yeah, 2/45

So, you start by calculating the amount of all possible pairs from the 10 players

If we assume the order doesn't matter in the algorithm that among us uses, there will be 10 choose 2 pairs

[upper abyss]

I like your strategy of (2/10)(1/9), but remember there's two ways to select the players

[south tundra]

And 10!/(8! * 2!) = 9 * 10/2 = 45

[upper abyss]

(P1 then P2) vs (P2 then P1)

[south tundra]

Yeah we don't know if the algorithm chooses two of the players simultaneously or separately

[upper abyss]

Without how exactly the algo works, we don't know for sure haha

I'm going to guess there's a "pity" system where players who haven't been imp become more likely

But maybe not

[south tundra]

Either way, the answer's 1/45 or 2/45 depending on the algorithm I guess

Assuming it gives equal chances

[gloomy wing]

ah

[upper abyss]

But I would assume the algo is basic. 2/45 imo

[cedar kiln]

@gloomy wing Has your question been resolved?

[cedar kiln]

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

[lone ferry]

not sure how to approach

[flint plinth]

the triangle you drew looks useful

you should be able to get cos(theta) from that

[lone ferry]

-4/5?

[flint plinth]

side note, it's odd that they gave you both sin(theta) and csc(theta) since the latter is just the reciprocal of the former

[flint plinth]

yep

[lone ferry]

okay tysm!

[lone ferry]

yeah 😭

.close

[cedar kiln]

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[slate lintel]

,calc 1/2 * 12 * 5^2

[wraith dagger]

Result:

150

[slate lintel]

👍

yes, $s(t) = \frac12at^2 + v_0t + s_0$

[wraith dagger]

Hayley

[slate lintel]

[umbral dew]

wait

can we ask physics ques also?

[hybrid wren]

So I believe so?

[umbral dew]

oh

[hybrid wren]

Isn’t it?

[umbral dew]

ig

n1u1=n2u2

n1 = 7.86

let n2 = x

[hybrid wren]

ty bro

[umbral dew]

bro

wait

lmao

how many kg in 1 g?

bro what?

yes, but better write it as 1/1000

okay now, how many m^3 in one cm^3?

no

1 cm = 10^-2 m

write?

right?

[hybrid wren]

yes

[umbral dew]

so cubing both side

gives us, 1cm^3= 10^-6 m^3

nope

10^-6 m

[hybrid wren]

sorry

[umbral dew]

no

bro

meter is bigger than cm

how it can be 100000

it is

[hybrid wren]

oooh

[umbral dew]

1/100,000

[hybrid wren]

sorry i thought the other way

[umbral dew]

okay so now we have -
1g = 1/1000 kg
and 1 cm^3 = 1/100,000 m^3

yeah but, its the other way

were not finding, how many cm cubes in one m cubbe

we are finding

how many m cube in one cm cube

theyre not the same

[umbral dew]

okay with this now?

now put the values in the expression

yes, im sorry

youre right

idk

[umbral dew]

solve this

write 1/1000 as - 10^-3 and 1/1,000,000 as 10^-6

and that would make it much easier to solve

okay?

yes

this this right

this is

ty

[hybrid wren]

so my answer was correct

work was kinda funky

[umbral dew]

yes, but when i asked the comversion ques

you were not able to ans

youre ans was right

but idk how did you get it lmao

[hybrid wren]

i seen a post lol

could i show you?

[umbral dew]

not wrong tho, but they didnt show you the process behind how they actually got this

they shouldnt have written 7.86g/cm^3, it should it be just 7.86

[cedar kiln]

@hybrid wren Has your question been resolved?

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[cedar kiln]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[crimson sedge]

Let $a_n$ denote the number of strings of length $n\ge 1$ using the alphabet [
\set{x,y,z,2,3,4,5,6,7,8,9}
]
that do not contain any consecutive digits that divide each other.

\vs{3 mm}
Find a recurrence relation for $a_n$

[tropic oxide]

are x, y and z actually J, Q and K

[muted bear]

methinks its just
a_n=C(a_{n-1})

All you have to do is discount the the cases when one divides the next in the sequence

[crimson sedge]

yeah sure but like

how to 'discount' something in recurrence relations sounds a bit odd to me

like,

[crimson sedge]

oop i just realised i messed up

okay

[muted bear]

Oh crap i think im wrong too

When it comes to 3 characters, e.g. 248 is discounted twice

Does xx count as valid or invalid

[crimson sedge]

valid

its a character

[muted bear]

Best i can come up with is a silly graph for now

[crimson sedge]

so like

not like

[muted bear]

x|x lmao

x has a unicode value

I have learnt nothing, but the diagram exists now

[crimson sedge]

Wow that's a lot of lines

[muted bear]

It sure is

[royal loom]

it's missing a line

[muted bear]

No friggin way

I dont see the missing line

[royal loom]

look harder

[muted bear]

Do tell

[royal loom]

well so basically you see

the thing is

that basically when you are drawing the lines

so the lines intersect yes and then as you continue to draw them

you end up missing one of them

so basicall

[muted bear]

[royal loom]

I just wanted you to look for lines for awhile

[royal loom]

[muted bear]

What

[royal loom]

[muted bear]

Oh god not again

conswcutive digits that do not divide each other
Any consecutive numbwr doesnt divide, silly

I maybe tired or just not qualified enough, but im out of ideas

[cunning forum]

mods member syndrome

[muted bear]

Lmao

Actually it might be helpful to think about the graphs complement

Use PIE and somethjng realted to this maybe

Idk

[cunning forum]

7 is basically another letter

wait no since 77 isnt allowed

[muted bear]

Oh right i have to include the selfloops

I couldnt think of a pretty way to make it using the software

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

hmmmm

this feels too confusing 😅

[crimson sedge]

this actually seems like a fun question too tho

"prove that there doesn't exist any two consecustive digits that divide each other"

[muted bear]

Four worda: Euclidean Algorithm

Worda 💀

[crimson sedge]

four

[muted bear]

I need to go to bed

Euclidean Algorithm states gcd(a,a+1)=gcd(a,a+1-a)=gcd(a,1)=1

[crimson sedge]

i see

anyways i will keep thinking for the original question

[muted bear]

This feels like a computer question

Feels very similar to finding the number of solutions to a latin square

[brittle lynx]

Yeah could easily code a dp for this and then plug it into oeis

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[summer igloo]

How does a calculator determine what the correct angle is for a given cosine or sine since there are potentially 2 different answers when asking for arc sine or cosine?

[vagrant elbow]

no there aren't

[tropic oxide]

it doesn't

arcsin returns an output between -90° and 90° [or -pi/2 and pi/2]

arccos returns an output between 0 and 180° [or 0 and pi]

in each case there is one and only one output

[summer igloo]

If I ask for the inverse -.86 how do you know if that's angle 150 or 210?

[tropic oxide]

you accidentally a word.

did you mean arcsin(-0.86) or arccos(-0.86)?

also you should not drop the leading zero like this. -.86 just looks so ugly.

[summer igloo]

arccos

[tropic oxide]

ok

arccos returns an output between 0 and 180° [or 0 and pi]

which of 150 and 210 lies in that range?

there's only one.

(btw cos(150°) is -sqrt(3)/2, not -86/100)

[summer igloo]

-0.86 I view as just a ratio, I didn't put it over 100

[tropic oxide]

you understand that $-0.86$ is equal to $-\frac{86}{100}$ but ISN'T equal to $-\frac{\sqrt{3}}{2}$, yes?

[wraith dagger]

Ann

[summer igloo]

Well, true, I guess I was just assuming that we we're talking about sqrt(3)/2 ratio since I gave the angles of 150 and 210.

But I can be more precise then

[tropic oxide]

if you mean sqrt(3)/2 then write sqrt(3)/2 lol

anyway,

arccos returns an output between 0 and 180° [or 0 and pi]
which of 150 and 210 lies in that range?
there's only one.

[summer igloo]

But I did state the exact angles so it's not hard to figure out from there

But you can get a negative cosine from any angle between 91 and 269, no?

[tropic oxide]

am i not expressing myself clearly, or am i getting unheard on purpose?

arccos returns an output between 0 and 180° [or 0 and pi]

it is for PRECISELY the ambiguity of which you speak
that the range of arccos is restricted this way

also not all angles are an integer number of degrees. you meant "between 90° and 270°"; this includes angles such as 90.01°

[summer igloo]

I guess what I am asking if I only have the x-coordinate then, from that, I could say to myself, (Okay, it's either this or that, i.e., angle 150 or 210) even if I don't know the sine value.

[tropic oxide]

should i repeat myself

like 17 more times

arccos(t)

LIES BETWEEN 0 AND 180

IT LIES BETWEEN 0 AND 180. it lies between 0 and 180 !!! it LIES BETWEEN 0 AND 180 !!!!!!!!

[summer igloo]

Right, repeating is unnecessary since this your words are still visible. I don't think you're reading my question. I tried to phrase it better last post.

[tropic oxide]

i don't understand your question.

[kindred violet]

The previous post is confusing

[tropic oxide]

but ok yeah fine

[kindred violet]

Can you try to rephrase or simplify?

[tropic oxide]

im clearly the mean bitch here CLEARLY

[summer igloo]

I'm saying that if you only knew the cosine value, you could say, okay, then I know it's either one of these 2 angles depending on sine value which is unknown?

[tropic oxide]

this is vague and poorly worded/notated in a way i don't like.

but yes.

[summer igloo]

Okay, that's all I was asking, I was trying to get a feel for getting knowledge in a quick manner, which is done a lot in math without getting exact answers, so that you can be guided by your next steps.

[tropic oxide]

if you have an angle θ and you know only the value of cos(θ) and that θ lies somewhere on the unit circle,
then yes, there are 2 options for what θ itself could be, depending on whether sin(θ) is positive or negative (which you cannot find out as is)

no, you were asking how it is determined which value to output for arccos.
and i THOUGHT i answered it several times, but clearly i am too much of an idiot to communicate that in a 100% misunderstanding-proof manner.

[summer igloo]

I stated from the beginning that -.86 (or if you wanna get really precise sqrt(3)/2) tells me the angle is either 150 or 210 depending ...

But none the less, thank you for the input. I'm here to learn, not to argue. 🙂

.close

[cedar kiln]

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[tropic oxide]

there really is no need to thank me for verbally abusing you for 3 pages worth of discord convos.

i said NOTHING of value.

[summer igloo]

But you did still help by providing truth, so it's all good.

[cedar kiln]

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[tidal wasp]

can anyone explain the central limit theorem to me in this context?

[cedar kiln]

@tidal wasp Has your question been resolved?

[prisma gull]

@tidal wasp

[tidal wasp]

im here

[prisma gull]

Essentially C.L.T says suppose you have n independent, identically distributed random variables

Then their sum will be asymptomatically normaly as n goes to infinity

you get that right?

[tidal wasp]

yup

[prisma gull]

The fourth step

just makes it into a standard normal variate

E(Y)= nmhu

[tidal wasp]

nmhu?

[prisma gull]

V(x)= n sigma^2

where mhu is mean

[tidal wasp]

oh i see

[prisma gull]

ie according to C.L.T, sum of iids ~ N(n mhu, nsimga^2)

Y follows $N(n \mu, n \sigma^2)$

[wraith dagger]

.doc

[prisma gull]

so definitely we can standardise any normal variable to standard normal variable

by substracting it’s mean and diving the standard deviation

That’s what’s done in Normal (0,1) step

[tidal wasp]

but how is the variance n simga^2?

i get that the mean is n mu

should the var not be (nmu)^2 ?

cus var(nX) = n^2*var(X)

nvm i get it

thx

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[thin roost]

Is this property valid for sum?

[south tundra]

No

[thin roost]

Alr

.close

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[cedar kiln]

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[sudden yoke]

And I also gotta sleep

I'll do this tomorrow then

.close

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[solar peak]

can someone help me calculate the eigenvalues and eigenvectors of this matrix?

[solar peak]

i have 0 clue

[crystal raptor]

same as ever, solve $\mathrm{det}(M-\lambda I) = 0$

[wraith dagger]

ΣΑC

[solar peak]

ok i have 0 and 2, is that correct?

0 twice tho

[brittle lynx]

,wolf eigenvalues {{1,i,0}, {-i,1,0},{0,0,0}}

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by _daili

[brittle lynx]

Yes it's correct

[solar peak]

now how do i get the eigenvectors?

i just get 2 rows of 0's

[brittle lynx]

find the null space

of $A - \lambda I$

[wraith dagger]

DAILI

[solar peak]

you mean {{1,i,0,}, {0,0,0}, {0,0,0}} ?

[cedar kiln]

@solar peak Has your question been resolved?

[cedar kiln]

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[sleek storm]

Why does this equation look like this in 3D?Shouldn't it be a regular parabola on the xz plane?

[vagrant elbow]

y needs to be constant for that

[sleek storm]

lemme see

[vagrant elbow]

There are multiple xz planes

[sleek storm]

How should the equation look then?

[sleek storm]

Along the y axis.

[vagrant elbow]

just something like z = x^2, y = 1 or something

[sleek storm]

Let me try..

So the calculator assumes any y?

[vagrant elbow]

all y

[sleek storm]

Ahh, I see.... Thanks!

.close

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[gentle yacht]

The following image displays a card house that is built to three levels. The first image displays F(1), the second shows F(2), and the third is F(3).

The amount of cards in the card house of x floors is defined by the formula: F(x) = (3x^2 + x)/2. Explain how this formula was built.

[vestal bear]

what's n

[gentle yacht]

my aplogies, it's actually x

[fallen moat]

try list out the first few cases 🙂

[gentle yacht]

That would be 2, 7, 15

[fallen moat]

in terms of 2, 2+5, 2+5+8

[gentle yacht]

I see, the next one should then be 11, as it is an increaase of 3 for each time, but what you come to such observation? I wouldn't have been able to spot such pattern

[fallen moat]

it's because we can see that the pattern is like putting the F(1) on the base of F(2) to become F(2)

so we split it into F(2)=F(1)+something
for easier observation

[gentle yacht]

It's an interesting observation. Have never actually used such method before.

[fallen moat]

it will become more and more common

[gentle yacht]

Okay. While we now know that f(x) is defined by x-numbers added together, where x_n = 3n - 1, how can be continue?

[fallen moat]

3n-1 is a good observation

do you recall the formula for sum of integers

[gentle yacht]

I am afraid I do not. This topic is quite new to me, and you should have already spotted that by looking at how I explained how f(x) is defined using words, but I should have enough mathematical background information to understand this anyway

Shouldn't the sigma-sign be related of some sort to the formula, though?

[fallen moat]

hmmmm

1+2+3+...+n = n(n+1)/2

[fallen moat]

this?

[gentle yacht]

That is unfamiliar

I'm unsure of how the formula is n(n+1)/2

[fallen moat]

hmmmm

gimme some time

(if other helpers wanna help, feel free)

ah, there we go

took less time then i thought

[gentle yacht]

It seems like I might need to have some addintional mathetmatical knowledge to understand this, anyway. I am bet that n(n+1)/2 should be cruical to understand before continuing

[fallen moat]

https://m.youtube.com/watch?v=bWZwF1H9YbU

[gentle yacht]

How lovely to see that the sigma sign was indeed something that could be used here

Finished the video. Seems intuitive to me. Thanks for sharing the video!

[fallen moat]

now that we have
sum from i=1 to n (i) = n(n+1)/2

we can start our work

we can rewrite the F(n) into

$\sum_{i=1}^{n}(3i-1)$

[wraith dagger]

Biscuity

[fallen moat]

so many typos lol

[gentle yacht]

Take your time

[fallen moat]

now that we can write it in this way, we need to get use to the summation operations

I'll first type out the next step to see if you understand

$3\sum_{i=1}^{n}(i)-\sum_{i=1}^{n}(1)$

[gentle yacht]

Could you grant me a couple of minutes to grasp this equation, i just need to wrap my head around it as I have not brushed up the way of using the sigma-symbol, which I have implied that I do know how to use

[fallen moat]

sure, take your time

[wraith dagger]

Biscuity

[fallen moat]

typo again lol

sorry

[gentle yacht]

Thank you. I will just input the current video that I am watching in this chat for the purpose of logging information, as I might want to refer back to these videoes if I were to forget any of this in the future

https://www.youtube.com/watch?v=5jwXThH6fg4

[fallen moat]

sure!

[gentle yacht]

The "i" at the bottom of the sigma notation tends to have an increase of one the entire time

$\sum_{i=1}^{n}(i)$

[wraith dagger]

Seed

[fallen moat]

true

and then stop when it reaches the top number, which is n

[gentle yacht]

we first have 1*i, then 2 times i, then three 3 times i, until we indeed reach n times i

[fallen moat]

if you don't like i, you can use k or other variables as long as it didn't obstruct the calculation

[gentle yacht]

What you have essentially done is converted the observation that we had made into the folloiwng equation that you had written

$\sum_{i=1}^{n}(3i-1)$

[upper ruin]

No, that's not how it works

[wraith dagger]

Seed

[gentle yacht]

Would you mind clarifying?

[fallen moat]

lemme clarify

to see, I'll choose some expamples

$\sum_{k=1}^{10}(k)$

[wraith dagger]

Biscuity

[fallen moat]

this would be equal to

1+2+3+4+5+6+7+8+9+10

[gentle yacht]

Begin at one, ends at 10, that's clear, but the K sign besides the symbol is what is confusing

[fallen moat]

sure

I'll continue

[gentle yacht]

i apologise if I inturrupted

[fallen moat]

$\sum_{k=1}^{5}(k+1)$

[wraith dagger]

Biscuity

[fallen moat]

it's equal to
(1+1)+(2+1)+(3+1)+(4+1)+(5+1)

[gentle yacht]

That made this clearer
I might need to show my compehension by confirming that I have understood the formula that you had pasted above
If we would assume that n = 3, which is that we are looking for f(x) in the card house (which we earlier saw is equal to 15 by counting each card), this equation should solve it like this:

(3(1) - 1) + (3(2)-1) + (3(3)-1), and this perfectly suits my earlier simplified explanation that f(x) is defined by x-numbers added together, where x_n = 3n - 1

of course, adding them up would be 15

[fallen moat]

correct, but one thing

since we defined f(x), we used x this variable.
then if we wanna talk about 3n-1
which is like
f(n)=f(n-1)+(3n-1)
we may wanna use another variable

let's say we can use d as in difference between terms
or t as in terms

e.g.

f(1)=t_1

f(2)=t_1+t_2

f(n)=t_1+t_2+...+t_n

or

$f(n)=\sum_{k=1}^nt_k$

[wraith dagger]

Biscuity

[fallen moat]

anyways that's kinda abstract lol, but that's the way

[gentle yacht]

You may want me to take a short moment to grasp the abstractness of what you have just said

[fallen moat]

hehe sorry

[fallen moat]

anyways we'll come back to this

[gentle yacht]

Yeah I struggle a bit with that last bit that you gave me, but I do understand the formula that you had written for the equation

Now it's essentially about converting it to an actual equation

I might want to use the same way that the bloke in the youtube video that you have shared with me did, expect that we're dealing with some slight differences

[fallen moat]

,align
f(1)&=\sum_{i=1}^{1}(3i-1)&=(3(1)-1)&=2\
f(2)&=\sum_{i=1}^{2}(3i-1)&=(3(1)-1)+(3(2)-1)&=2+5\
f(3)&=\sum_{i=1}^{3}(3i-1)&=(3(1)-1)+(3(2)-1)+(3(3)-1)&=2+5+8

[wraith dagger]

Biscuity

[fallen moat]

lol

anyways

[gentle yacht]

While TeXit may have had some issues, I am able to comprehend what you're trying to write

[fallen moat]

yey

[gentle yacht]

It is perfectly intuitive

[fallen moat]

now for the hard part

consider this

$\sum_{k=1}^{n}(3k)$

[wraith dagger]

Biscuity

[gentle yacht]

3 + 6 + 9, ... n?

[fallen moat]

which is equal to
3(1)+3(2)+3(3)+...+3(n)

now we can factor the 3 out

and become

3(1+2+3+...+n)

and hence

$3\sum_{k=1}^{n}k$

[wraith dagger]

Biscuity

[fallen moat]

that's how we factor out the 3 out of the summation sign

[gentle yacht]

another side for the sigma notation...

so it simply means 3 times anything that k is equal to?

[fallen moat]

$3 \text{times}\left(\sum_{k=1}^{n}k\right)$

[wraith dagger]

Biscuity

[fallen moat]

it means this

[long arrow]

$$\times$$

[wraith dagger]

Modus

[gentle yacht]

$\sum_{k=1}^{n}(3k+3)$

[wraith dagger]

Seed

[gentle yacht]

3k+3 can be simplified to 3(k+1)

[fallen moat]

yes

[gentle yacht]

can we therefore place the 3 on the left side of the sigma sign and write k+1 instead?

is this what it essentially means?

[fallen moat]

yep

and that also works for any constant

[gentle yacht]

Great, then we are currently on the same track

[fallen moat]

good, now addition

I'll use f(n) for simplicity

f(1)=(3(1)-1)

f(2)=(3(1)-1)+(3(2)-1)
= 3(1+2) - (1+1)

f(3)=(3(1)-1)+(3(2)-1)+(3(3)-1)
=3(1+2+3) - (1+1+1)

f(n)=(3(1)-1)+(3(2)-1)+(3(3)-1)+...+(3(n)-1)
=3(1+2+3+...+n) - (1+1+1+...+1)

[fallen moat]

with n number of 1's

all good till here?

[gentle yacht]

sure, but as far as I am concerned, should not we be subtracting ones, rather than adding them?
f(1) = (3(1)-1)?

[fallen moat]

ah

i hate my typos

fixed

see if that makes sense now

[gentle yacht]

To an extent it seems good
if i have observed this correctly, the last equation should be equal to:

=3(1+2+3+...+n) - (n)

oh, wait

[fallen moat]

yep

[gentle yacht]

i can also observe a similar pattern in the first one, too: 3(1+2+3+...+n) = 3 * ((n-1)(n))/2 , as we had earlier talked exactly about that

[fallen moat]

n(n+1)/2

[gentle yacht]

Yes, in addition to multiplying it by 3

i have done a typo. Should've been n+1, as you had written, and not the n-1

[fallen moat]

no worries

[gentle yacht]

but otherwise it should then be:

3 * (n(n+1)/2) - n

[fallen moat]

correct

[gentle yacht]

We're slowly progressing to our sweet (3n^2 + n)/2

let's see
n(n+1)/2 = (n^2 + n)/2

so it's
3 * (n^2 + n)/2 - n (you might want to take your time reading this)

= (3n^2 + 3n)/2 - n

then we can add the n to the fraction by turning it into 2n/2

[fallen moat]

perfect

[gentle yacht]

(3n^2 + 3n - 2n)/2

exact identical to our initial equation

except the varaible is x, haha!

[fallen moat]

haha

[gentle yacht]

I thank you a lot for the assistance with this, Bisuity, I appreciate every minute of your time. The information that you have given me is gold

[fallen moat]

Glad that I can help

to be honest, i was about the take a break, but suddenly I saw your question. it reminds me of my highschool years
stacking cards as high as possible lol

[gentle yacht]

I might need to try to recognise patterns such as the 2 + 5 + 8, because that might be the difficult thing to do, since the rest of the process of finding the correct formula after having found this pattern is a piece of cake, with your help

[fallen moat]

you can try doing this splitting for further exercises, it's usually helpful even if it's just a simple case

[gentle yacht]

It is sweet that you dedicated some additional time to help me. I cannot thank you enough, and I sincerely apologise for taking the time off your break. I won't take more of it, as you do really deserve it

[fallen moat]

no worries lol

[gentle yacht]

Again, thank you. Now that I got this clear, I can close this chat. I wish you a great day further

[fallen moat]

have a nice day!

[gentle yacht]

.close

[cedar kiln]

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[cedar kiln]

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[hot python]

"Classify the shape based on the following description.
Opposite sides are parallel and congruent, and all vertices are right angles"

[hot python]

this is like

a square

right

[lunar lynx]

No.

[hot python]

oh

rectangle..?

wait yeah

[lunar lynx]

Yes

[hot python]

tyyyyy

.close

[cedar kiln]

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[unique dew]

I need help with this problem

[unique dew]

I found that u_1 = 2

[long arrow]

is S_n a specific kind of series? anything else?

[unique dew]

but i tried to set Sn = (n/2)(u_1 + u_n)

it didn't really work

[long arrow]

so

this is arithmetic series?

[unique dew]

i doubt it

the difference between the terms is not the same

[long arrow]

so

[unique dew]

Ah

i think i got it from here

but stay jic

[long arrow]

$u_n=S_{n}-S_{n-1}$

[wraith dagger]

Modus

[unique dew]

got it

.close

[cedar kiln]

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[astral spruce]

How does (n-1)! + (n-1)(p-1)(n-2)! simplify to p(n-1)! ?

[silent finch]

(n-1)!+(p-1)(n-1)(n-2)!=(n-1)!+(p-1)(n-1)!=p(n-1)!

[astral spruce]

ohh

ic

thanks!

.close

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[cedar kiln]

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[carmine moss]

i have a question

[carmine moss]

not really question in math

but i am grade 7

so u can guys help me in anyquestion or examplle it for me?

[crimson sedge]

?

What question

[carmine moss]

[crimson sedge]

What’s the question

[carmine moss]

my school not start

srry i anoyying u

[cedar kiln]

@carmine moss Has your question been resolved?

[rain drift]

lol but like what's your question? Do you have a specific example?

[carmine moss]

nah just having some ideas about server

[cedar kiln]

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[cedar kiln]

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[warped fractal]

the unit is functions, part b of this question asks you to prove that DF = 4-x and that CE = x. in part a, you had to express GA and BH in terms of x. i calculated that

GA = 8-x, BH = 4-x

however i’m not really sure how to prove that DF and CE are the above values. the only way i can kinda see is it by making a triangle APF, but even so, you would have to introduce another variable for DF as you don’t know what the length of AF is.

[silent finch]

DF+CE=4

Because they are radiii

[warped fractal]

isnt the radius of the circle 8 tho since i think you can say P is the origin?

okay wait so i’ve drawn them as circles right, and i’m pretty sure here DK = PB and AP = CK

but i’m not sure how to prove that

[silent finch]

DF=DK=4-x. Do the same with CK=CE

[warped fractal]

but how do i know that DK = 4-x

wait nvm i got it

okay so from here, it asks you to find the area where the dog is free to roam in terms of X. i assume this would just be the solid curve. how could i find the area since the shape is kinda wack

[silent finch]

Split it into sectors of circles

[warped fractal]

how would you split it?

[silent finch]

Semi Circle with diameter GH + quarter circle with radius AG + quarter circle with radius BH + quarter circle with radius DF + quarter circle with radius CE

[warped fractal]

thank u so much bro 🙏 literal lifesaver

[silent finch]

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[hard ember]

Help with partc

[cedar kiln]

@hard ember Has your question been resolved?

[hard ember]

<@&286206848099549185>

[grand plume]

So just find the point where average slope = instantaneous velocity?

[cedar kiln]

@hard ember Has your question been resolved?

[cedar kiln]

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[shrewd owl]

Hello

[shrewd owl]

I need help

[tropic oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[shrewd owl]

2

I already got started

[tropic oxide]

how far'd you get

[shrewd owl]

Ik C1=0 and C2=2

but I don't know what to do with (pi/2)

the problems that I am use to is y(0)= some number

ex: y(0)=2

[cedar kiln]

@shrewd owl Has your question been resolved?