help-23

[runic pine]

?

[frigid locust]

here u go , (idk why its pink)

[timid scroll]

its breat cancer awareness

day

Ok I got it too

[frigid locust]

,rccw

[timid scroll]

And then I need to multiply the original by 1/2 to get rid of the 2

🤑

[runic pine]

What

[frigid locust]

eh?

[timid scroll]

Yeah

[runic pine]

No?

[timid scroll]

Yes

[frigid locust]

what u mean by original?

like divide the entire question by 1/2?

[timid scroll]

The (2+ cot x)^-2

So the answer is

1/2 (2+ cot x)^-2

[frigid locust]

ah ok , thats cuz of the integral power rule that i wrote up

.

[timid scroll]

Ye its like a bit of a jumpy method I do

impossible to follow

frl

I would get lost if someone explaind this the exact way im explaiing it

[runic pine]

Your way of writing is completely out of mind ngl

[timid scroll]

[frigid locust]

btw dont forget the +c part

very imp

[timid scroll]

Yes ! I no joke always forget

I just updated it rn on the paper too

[frigid locust]

btw in ur exams do u always write like this

[timid scroll]

Worse

https://tenor.com/view/awarding-randall-mc-holland-chicago-fire-decorating-honoring-gif-21688821

https://tenor.com/view/congrats-gold-trophy-good-job-good-job-award-gif-3050735871148737880

https://tenor.com/view/win-emma-johansson-olympics-medal-celebrate-gif-22491251

Ok thank you for my ted talk

I appreciate the fun and help

We had

Which was good

Thank you

.close

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[safe radish]

@mortal radish Has your question been resolved?

[safe radish]

@mortal radish Has your question been resolved?

[safe radish]

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[hollow flicker]

Hello. I am having trouble understanding the proof of theorem 1.10 in Functional Analysis by Rudin (p. 10-11). The statement is suppose $K$ and $C$ are subsets of a topological vector space. $K$ is compact, $C$ is closed, and $K \cap C = \varnothing$. Then $0$ has a neighborhood $V$ such that $$(K+V)\cap (C+V) = \varnothing$$. The proposition makes sense and I can follow the proof after the statement that $$(x + V_x + V_x + V_x) \cap C = \varnothing$$; however, it is this statement that I am not sure how Rudin derived.

[flat frigate]

Buttermilk Pancake

[hollow flicker]

My assumption is that because V is clearly an open neighborhood of 0 we can apply the proposition to V to generate the symmetric neighborhood V_x

semi-relevant SE post: https://math.stackexchange.com/questions/4753083/rudin-functional-analysis-theorem-1-10 except they're asking about a different part of the theorem

[light shoal]

x is in the complement of C and C is closed right?
C^c is a neighborhood of x
shift everything so that x is at the origin
then C^c - x is a neighborhood of the origin

now use the proposition to build yourself a V_x which is a symmetric neighborhood of the origin such that V_x + V_x + V_x lies within C^c - x, then shift everything back: x + V_x + V_x + V_x lies within C^c

[safe radish]

@hollow flicker Has your question been resolved?

[hollow flicker]

yeah, that makes sense, thanks @light shoal

[safe radish]

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[shrewd topaz]

how come they multiplied both sides by 25

and not just 5

[shrewd solar]

the fractions are squared so their denominators are 25

they skipped that part it seems

[shrewd topaz]

ohh right

i forgot about the square

thanks

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[cold brook]

hey can anyone explain spherical segment more to the detail for me

[cold brook]

if i got something like this

for some reason r1 - v1 should equal the v

and i cant wrap my head around it why

[fossil forum]

Well if it's a circle then they're both radii

The diagram looks stretched tho

[cold brook]

ive made it just quickly in paint to demonstrate

[fossil forum]

Right then they're both radiuses

r1 = v1 + v

[cold brook]

yeah that makes sense

yeah ok

i see it now

its simple

thanks

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[lone arch]

If $A_1$ and $A_2$ are countably infinite sets, then show that the cartesian product $A_1 \times A_2$ is also countably infinite.

[lone arch]

So, we need to find a bijection $f: A_1 \to A_1 \times A_2$.

Correct?

[obtuse jackal]

do you know that N² is countable ?

[lone arch]

Yes

Ok, then N x N -> A_1 x A_2 might be better

[obtuse jackal]

I'm sure you can find that one easily

[lone arch]

Yep

Thanks

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[lone arch]

.reopen

[safe radish]

✅

[lone arch]

Wait, how would we define that

$(A, B) \mapsto$ uhh

(A n A_1, B n A_2)

Would be possible

But

That wouldn't be injective, would it

[obtuse jackal]

why set operations ?

you're trynig to find a map from N² to A x B ?

[lone arch]

Yes

[lone arch]

Well we can't just do (Q, P) i-> (Q, P)

[obtuse jackal]

well almost

[lone arch]

Because A and B don't include all elements from N

[obtuse jackal]

or do they ?

[lone arch]

Wait, they're countably infinite

[obtuse jackal]

doesn't A contain N since it contains phi(N), which is a copy of N ?

[lone arch]

I shouldn't imagine them as {1, 2, 3}, lol

[lone arch]

What's phi(N)

[obtuse jackal]

until you're applying countability, in which case it's either to apply a theorem or to make a sequence out of it

[obtuse jackal]

the bijection N -> A

[lone arch]

There is a bijection from A_1 to N

So yes, A_1 must contain N

[lone arch]

So can we just do (Q, P) -> (Q, P)

[obtuse jackal]

it's just a bit of renaming

[lone arch]

Why renaming? $f: \mathbb N \times \mathbb N \to A_1 \times A_2$ with $(A, B) \mapsto (A, B)$

[obtuse jackal]

because you don't literally have N subset A

[lone arch]

Ah

How do we rename it

Wdym

Ah

g: N -> A_1

so g(A)

g(B)

[obtuse jackal]

let $\phi, \varphi$ be the bijections with A and B

Then just consider $(n, m) \to (\phi(n), \varphi(m))$

[flat frigate]

Bezier

[lone arch]

Yes

Thank you

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[patent flame]

6/2 = 3

Sqrt(3) remains as normal

That’s why it’s 3 sqrt(3)

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[lunar lagoon]

can someone help me with this please

[obtuse plover]

@lunar lagoon whatve u tried

[lunar lagoon]

This is what I drew

Is this correct?

@obtuse plover

[obtuse plover]

nah its not

its projected horizontally

[lunar lagoon]

oh

so its not going up

how is it supposed to look like then?

cus im kinda confused my the question

[obtuse plover]

yea ill show u

1 sec

[lunar lagoon]

cus its saying its been projected 2m from the edge

not from the edge

kk

[lunar lagoon]

oh

ok

so its like rolls for the first 2m then its goes off the edge?

[obtuse plover]

yep

[lunar lagoon]

so what do i do after that part then?

do i add the time it rolls

[obtuse plover]

yea its a 2 part problem. u have to consider: 1. the rolling time and final velocity on the table and 2. the time it took for falling off the table and hitting the floor

@lunar lagoon did u get it

[lunar lagoon]

sorry i had to go do something

i try it now

for the rolling time i got 1/10

[obtuse plover]

yea thats right

[lunar lagoon]

what values would be for the horizontal?

would s = -1.2m

and a = -9.8

[obtuse plover]

for the falling off the table u mean?

yea

[lunar lagoon]

yes

but i dont have any other values to use suvat

unless im missing something

[obtuse plover]

i think i know what ur missing

but lets see

which formula are u trying to use

ultimately we wanna find time of falling

so something with t preferably

[lunar lagoon]

one of the ones that have t^2

these ones?

[obtuse plover]

yea same thing

[lunar lagoon]

how can i use these tho since i only got the horizontal speed

velocity

[obtuse plover]

exactly, so if we're looking in the vertical direction, whats the intitial velocity in the y direction?

[lunar lagoon]

0?

[obtuse plover]

correct

[lunar lagoon]

ah ok i see it now

makes sense

so t = 0.606

then i add them together

so my answer is 0.706

[obtuse plover]

no i didnt get 0.606

[lunar lagoon]

😦

0.4948?

[obtuse plover]

d = 0.5at^2

yea

[lunar lagoon]

i see so the fiinal answer is 0.5948

[obtuse plover]

yea

[lunar lagoon]

ok

thank you for the help

[obtuse plover]

np

[lunar lagoon]

have a good day

.close

[obtuse plover]

u2

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[split steppe]

i need help pls

[last heath]

use the formula

[split steppe]

hmm?

[last heath]

A=bh

[split steppe]

yes but wouldnt that be wrong?

[empty gyro]

Why

[split steppe]

cuz it will find area of this?

nvm

i oh ty

[rustic goblet]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[split steppe]

how would i find this?

b = 9 + x

[empty gyro]

You don't need to. Unless you were asked to

You could use Pythagoraean theorem

[split steppe]

hm

so the answer would be 63?

[last heath]

ye

[split steppe]

9 x 7

[empty gyro]

No

[split steppe]

i mean the other dude said 54

[empty gyro]

No it would not

[split steppe]

hmm

[last heath]

who dude

[empty gyro]

Base x height

[last heath]

the one got deleted?

[split steppe]

idk a random dude whos message got deleted

[split steppe]

yh

[last heath]

oh i see

[split steppe]

yh cuz now im extra confused cuz both the questions are on

[split steppe]

which is 9 x 7?

=63

[empty gyro]

The height is not 7

[split steppe]

how so?

oh

6?

[empty gyro]

Yes

[split steppe]

alr i get it

ty for the help c:

!close

uhm

[last heath]

damn sorry i didn't see this 6 coming lol

but it's .close

[split steppe]

.close

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[quasi meadow]

I have this question and I do not even know where do start

[obtuse plover]

so just take a random point

for example

(1,1)

plug in the values of x and y into the formula for dy/dx

then u should get 1(1 + 1) which is equal to 2

therefore the slope at (1,1) is 2

so plot that with a steep line accordingly

[quasi meadow]

i see, and which points do I do that for? would I go from y=-2 to y=2 and same for x?

[obtuse plover]

yea

idk why they said [-2,2,1] though

instead of just [-2,2]

[quasi meadow]

yeah no idea

and for b is there a formula for me to follow?

[obtuse plover]

no formula, just get all the terms with y to one side, and all the terms with x to the other side, and then integrate both sides

[quasi meadow]

so I would get y by itself and just normal integration with no bounds?

[obtuse plover]

idk what u mean by y by itself, but yea no bounds

o are u saying the final solution is y = ..something.. ?

[quasi meadow]

no I was just wondering if there were bounds but worded the question weird. My bad

and then for C, how would I find that solution

[obtuse plover]

u just use the solution to the diffeq u found in part b, and then for the +C part, plug in (1,-1) to find it

[quasi meadow]

im sorry, I am not sure I understand what I plug in for C

[obtuse plover]

just find the answer to b

then ill help u out

tag me when ur done

[quasi meadow]

alright

@obtuse plover

[obtuse plover]

ok lemme take a look

ok that looks good i think

now just plug in (1,-1) into (x,y) in ur equation

and solve for c

[quasi meadow]

alright

I got 0

which can't be right

[obtuse plover]

ok yea somethings fishy

lemme look again

yea hmm

ive tried it a couple a different ways now

-1 = e^(0.5 +c) - 1

which becomes

0 = e^(0.5 +c)

which doesnt make sense

unless we make it

[quasi meadow]

cause you take the ln of both sides right?

[obtuse plover]

ce^0.5 = 0

[obtuse plover]

yea i mean

we do that, but the ln shouldnt affect us if we have a valid initial point

it doesnt seem like (1,-1) is

[quasi meadow]

so is it a trick question and there is no specific solution

[obtuse plover]

i doubt it, maybe im just not seeing something, just close this channel, and open a new one, and give ur solution to part b, and ask for guidance on part c. maybe someone else can see if im missing something

[quasi meadow]

sounds good, thank you for all your help

.close

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[dusty pier]

How can I find I in this situation?

Context: I am working on a 3D engine and I have screen space coordinates that have been calculated using a reverse engineered index (I) function. I then am unprojecting, rotating, and translating the screen space coordinates to get the world coordinates. I then am rotating, translating and reprojecting them to the lights position. (I am doing a technique called shadow mapping) now that I have the screen space coordinates of the light, I now need the index number of that position in the list. Using the original x and y calculation functions, how can I reverse engineer then to get I? I have tried this 3 times to no avail.

[dusty pier]

Either I am stupid, or I am overlooking something obvious; I need some fresh eyes.

by the way, "u" is a placeholder for a binomial (y+yb) y is not equal to the y in the equation, it is its own variable, and yb is also its own variable standing for ybound

[obtuse plover]

u want to solve the equations for I?

[dusty pier]

yes, I want to find I using the equations

[obtuse plover]

ok so first u should recognize that ur first equation is just

$x = RL(I-u) - 240$

[flat frigate]

Stephen

[obtuse plover]

no need for all the extra parentheses

and then what do u think we should do

if we want to get I by itself, whats the first term we should move to the side of the x

[dusty pier]

so my thinking right is that i would move 240 to the x side like any other algebra problem

$x+240 = RL(I-u)$

[obtuse plover]

yes we move the 240

but dont forget

if its a -240 on the right side

what do we do to both sides?

[flat frigate]

V1ewsh0t

[obtuse plover]

ah there u go

ok good

[dusty pier]

sry mistype

[obtuse plover]

ok now what do u think

[dusty pier]

then we divide rl from both sides

[obtuse plover]

yes

[dusty pier]

$(x+240)/RL = I-u$

[flat frigate]

V1ewsh0t

[obtuse plover]

yes

now what?

[dusty pier]

then add u

[obtuse plover]

$u + \frac {x+240}{RL} = I$

there u go

[flat frigate]

Stephen

[dusty pier]

but because we don't know u, we need to find u using the y equation

[obtuse plover]

$u = y _1+ y_B$

right?

but this isnt the y of the 2nd equation right

so

lets call it y_1

[flat frigate]

Stephen

[obtuse plover]

right?

[dusty pier]

yes

[obtuse plover]

ok so now for the second equation, we want to solve for u?

[dusty pier]

yes

[obtuse plover]

ok so we have

$y = uR - 180$

[flat frigate]

Stephen

[obtuse plover]

solve for u

tell me what u think we shd do

[dusty pier]

add 180, divide R

[obtuse plover]

yes

$u = \frac {y+180}{R}$

[flat frigate]

Stephen

[dusty pier]

$\frac {y+180}{R}+ \frac {x+240}{RL} = I$

[flat frigate]

V1ewsh0t

[dusty pier]

is that really it?

[obtuse plover]

does this equation come into play?

[dusty pier]

no

[obtuse plover]

ok

[dusty pier]

because we don't know those variables

[obtuse plover]

so u wanted to solve for I right?

isnt that what u wanted?

[dusty pier]

yes

[obtuse plover]

do u need anything else?

[dusty pier]

well I am about to test it real quick and I will let you know the results

[obtuse plover]

ok

[dusty pier]

and this is what I was worried about: original Index: 3374, new Index: 35.1458333333

and I did this with the x and y being the products from the original equations

L and R are also the same ones used in the original equations

[obtuse plover]

im unsure as to what that error means

[dusty pier]

me too.

and the original equations are correct. I have triple checked them multiple times

i have to be right back

[safe radish]

@dusty pier Has your question been resolved?

[dusty pier]

Hi, yes I am back

<@&286206848099549185>

[safe radish]

@dusty pier Has your question been resolved?

[safe radish]

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[soft chasm]

Does anyone know how to use rguroo

[safe radish]

@soft chasm Has your question been resolved?

[soft chasm]

No

[safe radish]

@soft chasm Has your question been resolved?

[soft chasm]

No

[safe radish]

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[slate thorn]

is there a quick way to do this?

[terse lichen]

what does the sampling distribution want

the mean and variance/standard deviation?

[safe radish]

@slate thorn Has your question been resolved?

[lean otter]

that is just what i was wondering

[terse lichen]

theres formulas for that

mean of sample = mean of population

σ(sample) = σ/sqrt(n) * √(N-n / N-1)

[safe radish]

@slate thorn Has your question been resolved?

[safe radish]

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[devout echo]

Suppose we have a bag with $10$ slips of paper in it. Eight slips have a $3$ on them and the other two have an $8$ on them.

How many $8$'s do we have to add before the expected value is at least $7$?

What I have tried so far:

Let N be equal to the number of 8 slips we must put in

8/(10 + n) * 3 + [(n+2)/(10+n)] * 8 >= 6
24/(10 + n) + (8n + 16)/(10 + n) >= 6
24 + (8n + 16) >= 6 * (10 + n)
24 + (8n + 16) >= 60 + 6n
24 + 8n + 16 >= 60 + 6n
24 + 16 >= 60 + 6n- 8n
40 >= 60 -2n

Therefore, N = 10

[flat frigate]

ChocolateFudge

[devout echo]

Pls ping me if you come

[devout shale]

@hot thistle

[quiet juniper]

@devout echo why >=6? Dont you want the expected value to be at least 7

[devout echo]

wait

why

did i do that

Dammmit

i spent

so long

Thanks very much lol

[eternal carbon]

u solved it correctly for "6" though so as long as u can repeat ur process ur good

[quiet juniper]

np lol

yeah

[devout echo]

.close

[safe radish]

Channel closed

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[solar hazel]

you look so cute now with the pink diamond

[quiet juniper]

uwu

[devout shale]

@safe radish Ban please

[split fulcrum]

I’m gonna need an animal factoid here to control the situation

[solar hazel]

channel is closed

it's a free for all

[quiet juniper]

uhhh hallucigenia are an extinct animal from over 500million years ago

[tall bough]

Spaghetti balls

[devout shale]

we do not yet have a !redir image

tragic

[devout echo]

.open

8/(10 + n) * 3 + [(n+2)/(10+n)] * 8 >= 7
24/(10 + n) + (8n + 16)/(10 + n) >= 7
24 + (8n + 16) >= 7 * (10 + n)
24 + (8n + 16) >= 70 + 6n
24 + 8n + 16 >= 70 + 6n
24 + 16 >= 60 + 7n- 8n
40 >= 70 -2n

Therefore, N = 15

[eternal carbon]

true

[split fulcrum]

It’s time you make one

[solar hazel]

.reopen

[devout echo]

but wrong for sum reson

[safe radish]

✅

[devout shale]

.open

/open

[eternal carbon]

hmmm

[split fulcrum]

Sir do you have a doubt?

[solar hazel]

CHANNEL IS REOPENED SERIOUS MODE NOW EVERYONE

[devout echo]

lol

[eternal carbon]

last line 2n becomes n

[devout echo]

oh carp

crap

[eternal carbon]

"oh carp" is correct

[devout shale]

https://tenor.com/view/carp-fish-holy-carp-holy-crap-fishy-gif-26978677

[devout echo]

i adjusted my equation

but now that part lol

[eternal carbon]

we do a li'l bit of adjusting of our equations

[devout echo]

YESSSSSSSSSS

Thank you all

[devout shale]

Glad I could clear your doubt

[devout echo]

I believe in math supremacy

yes

[split fulcrum]

You’ve gotta be squidding me

[devout echo]

thx everyone

[devout shale]

⚔️

[solar hazel]

close the channel so we can disengage serious mode

[split fulcrum]

CHARTTTTT

[devout echo]

.close

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Channel closed

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Use .reopen if this was a mistake.

[solar hazel]

thx

[split fulcrum]

@junior smelt

[solar hazel]

@junior smelt

[devout shale]

Please don't ping chartbit for your doubts

@junior smelt

[solar hazel]

water beam has no doubts atm austin...

[split fulcrum]

Sir I’m going to have to relocate you to channel #❓how-to-get-help

[devout shale]

Is this true?

[solar hazel]

we're just saying hi to our friend chartbit

[devout shale]

Ah I see

[split fulcrum]

I have no doubts

[devout shale]

I had a doubt

[tall bough]

Please be serious, this is a serieus server, i am asking it seriously.

[solar hazel]

channel is closed it's a free for all now

[devout shale]

Sir yes sir

[split fulcrum]

Sir do you have a doubt?

[quiet juniper]

[devout shale]

I have a doubt about what serieus means though... then I can comply

[quiet juniper]

i need to know what im being theeatened with before i comply

[tall bough]

Serieus means serious, but for posh people, which you aren't! 😝😝

[devout shale]

math war

[quiet juniper]

ok i comply i comply

[devout shale]

good we will win

⚔️

[split fulcrum]

sir do you have a doubt

[devout shale]

I do

#help-15

[tall bough]

Sustained

[split fulcrum]

Okay, I will come check it out

[tall bough]

What's a light cone

Is it the cone in Penrose diagrams

[devout shale]

it's part of my doubt aswell

[quiet juniper]

https://giphy.com/gifs/QBRwq5uxTCMbAQS5Ic

[split fulcrum]

all I know is that light comes point to the future

so stay bright my friend

[hot thistle]

woah spoilers man

im only on season 2

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[ionic zenith]

doubt any1 would know but how would i regress this on a calc

[ionic zenith]

2nd pic is answer

this is my calc

but also which regression

and do i just include x values , 5, 25, 45 and 75 without 70?

@me if anyone responds

^^

[ionic zenith]

how do i get the equation

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[brazen cargo]

How do you show this?

[frozen marlin]

,rccw

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[stone crypt]

Can someone pls check my work for part b and c

and also please help me with part a and b

I don't at all understand how to do part A and D

<@&286206848099549185>

[ember bronze]

how to integrate this?

[stone crypt]

use the u substitution method

also, do u mind going to another channel?

.close

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[winter flume]

First of all use definite integral property and put

Cos(pi-x)cos(pi-2x)
Property is lim0-a f(x)dx = lim0-a f(a-x)dx

[tribal coral]

?

u still need us? (u closed the channel)

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[hollow silo]

hi guys i have a simple question concerning the cosine trigonometric function, i noticed some contradiction on it, we all now the general formula but not how its really works, lets suppose we have a right triangle of radius of 1 and an angle of 9 degrees, of course to find the ratio we do this, x = r * cos(9) , but cos function itself is equal to adjacent side/hypotenuse, but the problem is we don't know the adjacent side length ! so how can we find the ratio of adjacent side to the hypotenuse ?

[quiet plume]

That's the point of cos. You give it an angle and it spits out the ratio between the adjacent side and hypothenuse to this angle in any right triangle that has that angle. The definition is similar for other trigonometric functions.

You seem to feel like it's circular because we end up using the definitino in the computation (i.e. we find the ratio using the ratio)

[hollow silo]

thanks but how its really works? how the algorithm of the cos function can find the ratio by knowing the angle ?

[quiet plume]

As to how it is computed in the first place, historically, it was computed manually in tables, originally in a table of chords, and we used known values of common angles (originally in increments of 15 degrees) and would interpolate for value in between. There have been many ways of approximating such functions in the past.

Nowadays, calculators calculate these values numerically using Taylor series or something called the CORDIC algorithms, which you can probably look up if you want to know more.

[hollow silo]

ah okeeeeyyyy haha got it, i noticed that many people know HOW to use the trigo funtion but NOT able to explain how its really works

[quiet plume]

Yeah I guess the confusion stems from how we say that cos(t) = adj / hyp.

[hollow silo]

yes

[quiet plume]

So it looks as though we need the ratio to get cos(t) in the first place.

[hollow silo]

exactely

[quiet plume]

But we don't really need it since we can approximate the actual value to however precise we want it to be, and that's good enough.

[hollow silo]

im like that i always try to have a deep understanding of math i mean thinking in the same place as Pythagor or fibonacci or other math scientists

and those functions intrigues me a lot

especially in cos(theta) equal to adj/hyp, we have to know the numerator and the denominator

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[acoustic crescent]

can somoene help me to solve this. i dont know how to approach it

[marsh walrus]

have you ever heard difference of squares?

[acoustic crescent]

yes

[marsh walrus]

there you go

[acoustic crescent]

!done

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[acoustic crescent]

.close

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[sharp cave]

i dont know if this is the right place to ask but should where should that 'implies' go : where it is now, in the next line or both ? (this is strictly page arrangement)

[peak estuary]

well it looks shit either way. so I wouldnt worry too much about it

way too cramped

that said, I would put it in the next line

[sharp cave]

ty

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[dense tinsel]

hello

[dense tinsel]

how do i solve?

[brazen ravine]

alternate angles

So those arrows on the line mean they’re parallel

so it’s 53

[dense tinsel]

ohh

then do 180-53

oops sory nevermind

i just realize

[brazen ravine]

do 180-66 to find the angle inside the quadrilateral

since angles on a straight line equal 180

[dense tinsel]

ohh

then 360 -

[brazen ravine]

yeah

[dense tinsel]

what about this?

[brazen ravine]

uhh

[dense tinsel]

where u go

help plz

[brazen ravine]

i think u have to imagine another triangle that extends in the bottom right corner

so 180 - 153 is one angle in that triangle

[dense tinsel]

ohh kk

its ok ill just open anew ticket

tysm

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[lone arch]

#1195800467880558672 message

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[lone arch]

#1195800467880558672 message

[brazen cargo]

How do you solve this?

.rotate

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[dim meteor]

How many automorphisms $Z_2 \cross Z_2 \cross Z_2$ have?

[flat frigate]

The Great D

[dim meteor]

The given hint is "Pick a set of three generators of this group. An automorphism turns each generator into another element of the group. What the three generators turn into completely determines the automorphism."

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[lone belfry]

This is my work so far. Anyone able to help?

[safe radish]

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[lone belfry]

<@&286206848099549185>

[uncut gate]

uh can you re-take the pic on a different angle

[lone belfry]

Yeah np

[uncut gate]

i'm on pc and i don't really want to tilt my head like a wild animal

[lone belfry]

Makes sense lmao

[uncut gate]

okay I would suggest writing sqrt6 + sqrt3 on the equation given x^2 - 6 times sqrt x -3 is 0 but it's harder than I thought

so you get an outlandish shit like this

okay I got it

@lone belfry

the equation should be like this

and if we place x with sqrt6 + sqrt 3 we should end up with this

-sqrt3 - sqrt3 is NOT -sqrt6

[lone belfry]

Ok lemme get the paper

[lone belfry]

How did you get this

[uncut gate]

we just did the equality wrong lol

[lone belfry]

Elaborate

[lone belfry]

Ok

Lemme try to break this down

Where did the x -3 come from after the -2rt3

Oh nvm

What do I do from here

Quadratic formula?

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[rigid sapphire]

.close

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[true violet]

Hi, can anyone help me with this? I need to see which comparison is valid

I know the second one is valid because n^2 > n^2 - sin(n)

are there any other valid ones?

@junior smelt Is something wrong with my rationale?

dang

[junior smelt]

It's quite sus, after all you'd be implying that sin(n) > 0 with that

[true violet]

because n oscilates between -1 and 1

[junior smelt]

sin(n) does sure, n is an integer

but e.g.

,calc sin(1)

[flat frigate]

Result:

0.8414709848079

[junior smelt]

...is positive, while...

,calc sin(5)

[flat frigate]

Result:

-0.95892427466314

[junior smelt]

...is negative

[true violet]

so would 1/n^2-sin(n) be wrong?

in comparison to 1/n^2

[junior smelt]

You're doing direct comparison right? (rather than e.g. limit comparison)

[true violet]

yes

0.5n² < n² + sin(n)

0.5n² > -sin(n) true

[junior smelt]

If so then you'd have to be careful

[true violet]

So the comparison fails
n² > n² + sin(n)

0 > sin(n) which is not for all n true
Even for very big n
You can take 0.5n² no??

[junior smelt]

You could sure, and that would be valid to do (noting that you can "pull constants out" of the summation), hence my comment that "you'd have to be careful"

Of course, something like 2n^2 > n^2 + sin(n) is equivalent to n^2 > sin(n), which is true for positive integer n (noting n^2 >= n > sin(n) and all)

[true violet]

so would 1/n^2-sin(n) be wrong to compare with 1/n^2? Im kinda confused

[junior smelt]

Well, erm, yes and no-

[true violet]

haha

the question is"For each pair of series below, the convergence of the first series is unknown, while the convergence of the second series is known. Decide which options are valid applications of the comparison test."

[junior smelt]

Actually wait-

[true violet]

this could be more than one correct ans for this question too

[junior smelt]

For the second one, you'd want to have the second series (or some scalar multiple of it, maybe(?)), the 1/n^2, as the "greater" one

Don't think there's a way to do that? but then I might be slow

[true violet]

well it doesn't specify the greater one so yeah you'd be right

i think even n > n -1 comparison could be valid too

for the otehr option

[junior smelt]

Generally you want either the convergent one being the greater one, or the divergent one being the smaller one, else you don't really get information

[true violet]

fair enough, guess the question could've been more specfic eh lol

[junior smelt]

And yep, cause then 1/n < 1/(n - 1), the series of the latter diverges so the one you're thinking about must diverge too

[true violet]

alright im gonna take your word for it haha.

.close

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[leaden pendant]

Someone had 1/3 of a pizza, someone had 1/4th of a pizza and someone had 1/5 of the pizza. The remaining pizza was cut in 3 even slices and had a diameter of 50.
Whats teh Diameter of the whole pizza?

[buoyant shadow]

lmao what

[stoic dune]

How do the remaining three slices have a diameter

[buoyant shadow]

you're not reading it right

the pizza that was cut had a diameter

the 3 slices do nothing

[stoic dune]

Uh, okay...
How does the remaining pizza have a diameter?

[leaden pendant]

Someone had 1/3 of a pizza, someone had 1/4th of a pizza and someone had 1/5 of the pizza. The remaining pizza was cut in 3 even slices and had a diameter of 50.

[buoyant shadow]

Whats teh Diameter of the whole pizza?

[leaden pendant]

i think its the diameter of the original pizza

you have to find out

[buoyant shadow]

just skip it

think of it as not encouraging bad problems to be presented

[leaden pendant]

ok

.close

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[spiral fable]

So this is a method to find the shortest distance between a line and a point (vectors) using the cross product

My question is why did he divide U with its magnitude before taking it's cross product with AP (U is the direction vector of the line)

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[hot thistle]

it says it on the orange box to its right

to determine the unit vector in the direction b

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[brazen parrot]

how come this is wrong

[brazen parrot]

i was just copying this method

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[buoyant shadow]

3c2 × 6 × 2

or 3c2 × 24 / 2

[brazen parrot]

why?

[buoyant shadow]

4c2 = 6, there's 6 ways to place those 2 empty boxes in case 3

[brazen parrot]

Ohhhhhhhhhhh

Ohhhhhhh

Wait why do we multiply by 2

I forgot why i did that

[buoyant shadow]

if you chose where the empty boxes are and you chose which 2 letters to put there, you can put them backwards or not, there's still 2 options

[brazen parrot]

Oh 2!

[buoyant shadow]

you can do 3c2 × 4! / 2!

[brazen parrot]

How come we do it different to my second one

So the second one i did like 3C1 to fill in the empty slot from the 4 distinct letters excluding R so 3

And then i multiplied by 4

Because the letter can go in 4 boxes

[buoyant shadow]

right, that actually works too

the first letter goes in 4 boxes, and the second in 3

4×3 is again 12

[brazen parrot]

Huhh

Is it ok if u draw something 🥲🥲

[buoyant shadow]

it's the same thing as with the museum

when there was 6 time slots

[brazen parrot]

what museum

[buoyant shadow]

that wasn't you

my bad

[brazen parrot]

Haha

Is it ok if u just draw like the box diagram

[buoyant shadow]

i don't mind drawing, i just don't know what to draw

[brazen parrot]

oh 🥲🥲

[brazen parrot]

what do u mean by this?

the first letter goes in 4 boxes?

but we have RR in the 4 boxes

[buoyant shadow]

there's 6 ways to choose where the R go
then for each of those 6, you have 6 ways to fill the empty spots
PQ, PS, QS, QP, SP, SQ

so 36

[brazen parrot]

i think what i did wrong here is that its supposed to be 3c2 x 4 x 3

because

oh wait

[buoyant shadow]

and there are tons of ways to get there, they are all correct

[brazen parrot]

same thing u said?

since theres 1 spot gone theres 3 places

left for the letter to go

[buoyant shadow]

yes

[brazen parrot]

do we count RR as 1?

like dis?

[buoyant shadow]

yes, there's no second way to put Rs, they jsut go everywhere

[brazen parrot]

wait..

thats wrong

oh

tyhis makes a lot mroe sense

more

[buoyant shadow]

yes

[brazen parrot]

first one is for the case with 2 R's

and 2nd one is for case with 3 R's

[buoyant shadow]

you write the letters you chose, so RRPQ, and shuffle the entire password

[brazen parrot]

wat

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[nocturne blaze]

can some1 help please

isnt the answer basically A?

it decreases from j to l

but answer key says am wrong

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[deep talon]

\int \frac{sec^2x}{tan^2x-1}

[tepid walrus]

You have to wrap your LaTeX in $$ for it to be picked up by the bot. So $\ \int \frac{sec^2x}{tan^2x-1}$

[flat frigate]

ℑμΤ𝛄𝛗θ

[deep talon]

ty

[deep talon]

this is my first time here ehe

[tepid walrus]

No worries :)

Also, for this integral I suggest using substitution, and then either hyperbolic trig functions if you know them, or partial fractions

[deep talon]

i got the ans just noww

yup i used them

ty

[tepid walrus]

Oh welp, good job!

[deep talon]

thank you

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[lethal quartz]

how would i go about differentiating this?

[obtuse plover]

Whayve u tried

whatve

@lethal quartz

[lethal quartz]

i was taught that ln() and e cancel eachother out or something like that

[obtuse plover]

well yea, but theres an x there too

use properties of logs to separate the log

[lethal quartz]

i tried differentiating ln into 1/xe^7x but didn't know what to do there

log rules?

ohhh

ok i think i got it

[obtuse plover]

cool

[lethal quartz]

ok got it thanks

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[lean otter]

The "answer" here is surely incorrect no?

[hard crest]

depends on how they're numbering but... yes I think you understood the concept if nothing else

they seem to start at 0

[lean otter]

I see, so they see the first term x^8 as term 0

[hard crest]

yeah

which I guess is easier in some sense

since that way the term number is the power of y

[lean otter]

Is that a standard 'notation'?

[hard crest]

i don't think so

[lean otter]

I'll have to ask one of the tutors

But thanks though!

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[vernal yoke]

Anyone can help with how to get the answer

[vernal yoke]

I know it’s there but I wanna know how to evaluate the limit

[mortal arrow]

Think about the tan graph and reflect it across y=x

however there aren’t multiple y values so it’s like just one of the lines of the tan graph

That’s how I imagine it

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[fervent girder]

Will it not just be the area under the parabola that is translated in x and y direction?

[fervent girder]

a(x-h)^2+k integrate this equation from 0 to 0.66?

[potent seal]

Why parabola?

It's given it's a sin curve

[fervent girder]

so its a sin equation?

[potent seal]

Yup

But not just sin(x)

It would need a few modifications

[fervent girder]

so its need like amplitude and stuff

[potent seal]

Yeah

[fervent girder]

ohhh

im dumb

ty

[potent seal]

We all are

Np

[fervent girder]

LMFAO

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[lean otter]

Let n⩾2 be a natural number. Solve the following system of linear equations, are arbitrary real numbers.

[lean otter]

Any idea where to start?

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@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

anyone?

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@lean otter Has your question been resolved?

[lean otter]

can someone help me where to begin with this?

<@&286206848099549185>

[lean otter]

<@&286206848099549185>

[lean otter]

<@&286206848099549185>

[lean otter]

<@&286206848099549185>

anyone can help with my problem?

[grizzled seal]

I don't think I'm able to help (new to LA myself), but I think your question needs edited? are you just supposed to solve for any n?

[lean otter]

yeah, so the n should be => than 2

I have given these instructions to solve it, but i don't even know where to start it

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@lean otter Has your question been resolved?

[magic pivot]

Hey Brk, I'm not entirely sure what the purpose of this problem is. It looks more like a format for a bigger idea- like the beginning of a chapter type of deal. The subscript n just means "number", which is reinforced by the "......", which just means it does on and on. What you should look to is the "a". What this problem is saying, is that even though a bunch of different numbers are being added, the variable "a" makes the sum of the system 1- like how you set something to the power of 0.

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[hot thistle]

@lean otter do you have an image of the original question

[eternal carbon]

aside from what can be interpreted as a grammatical error

that pretty much is the original question

[eternal carbon]

try solving via elimination for n = 2, n = 3 and see what happens

[hot thistle]

well we can guess that, sure. but i would also like to see the original question just in case

[eternal carbon]

"we can guess" i believe i've seen this problem before so i wouldn't say i'm guessing

also i just solved the problem as posted above and i believe it's a well-designed problem

[eternal carbon]

i mean to say it seems very familiar, i haven't been able to locate its source

[hot thistle]

you are certainly guessing since we don't know where brk got this problem from. i understand you are probably right, but often helpees withhold information or ask the wrong thing

i don't think there's any harm in asking for and looking at the original problem

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[dapper mirage]

Im confused why my intervals dont match any of the answer choices

[dapper mirage]

My intervals are on the bottom right

Wait let me post a better pic of the question

[chilly atlas]

You say it's decreasing from -inf to -3?

[dapper mirage]

yeah

[chilly atlas]

Well, look at your answer

x = 3 or -3

doesn't one of those cause problems?

check the domain of the original function

[dapper mirage]

the question doesnt give us a domain for the og function

[chilly atlas]

any function has a domain by nature of existing

what made you choose x = -3 instead of x = 3?

[dapper mirage]

I didnt choose either of them

arent those both critical points

[chilly atlas]

One of them is

You have to check if both possible critical points are within the domain of your original function

Just try plugging in x = -3 to the original function and see what you get

I would also recommend looking at a graph of the original function, if your instructor allows it

[dapper mirage]

we have been taught to find it by using the first derivative

so could I just plug those points into the derivative?

-3 and 3

[chilly atlas]

Try plugging them both into the original function

You will notice that one of them turns out to be impossible

Which means it's outside of the domain of the original function

Remember, you're trying to find where the original function is increasing and decreasing, so it's domain is still relevant

[dapper mirage]

idk how to plug it into the OG function without putting it into a calculator

[chilly atlas]

What is the log of any negative number?

[dapper mirage]

undefined

[chilly atlas]

Thus, -3 is not a valid critical point

It must also mean that the original function's domain starts at 0

And since there is only one critical point at x = 3, the first interval of increasing or decreasing must be from (0, 3)

Based on this information, I would change your number line

Then use the derivative to determine whether or not the function is decreasing or increasing over (0, 3)

That's half of the problem already. Then you just have to determine whether the function is increasing or decreasing on the other side of the critical point. The other side of the critical point should have a domain of (3, ...)

You'd have to determine whether or not it goes to infinity, or some number

If you get stuck again I'd be glad to help

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[near basin]

You want to determine the height of the screen at a drive-in movie theater. You use a cardboard square to line up the top and bottom of the screen structure. The vertical distance from the ground to your eye is 5.5 ft and the horizontal distance from you to the screen is 14 ft. The bottom of the screen is 6 feet from the ground. Approximate the height of the screen to the nearest tenth.

[near basin]

Thats the question i dont know the process

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