help-33

[slim turtle]

Bro I agree but I have to use these cause they tell us to

[red nimbus]

2k/n² -> 0

[slim turtle]

Ohh a IIT aspirant , eh?

[still temple]

not like i have a choice

[red nimbus]

,rcw

[still temple]

just cause you dont understand,doesnt make it wrong

[slim turtle]

💀 pooki

[red nimbus]

yea i get what dx is but technically it should bee dx = (n-1)/n

[still temple]

why💀

[slim turtle]

I'm also one

[red nimbus]

because delta X = (b-a)/n

[still temple]

thats not small bro

[still temple]

🐥

[red nimbus]

so you used anything that is small lmao

[still temple]

did bro just say dx and delta are same

[still temple]

no bro it has to be infinitecimally small

thats the point of dx

[red nimbus]

why didnt you then take 1/n³

or 1/n²

[still temple]

cause you need x also whats x then

[red nimbus]

no i didnt but you said you applied the limit definition of integral

[still temple]

yes

[red nimbus]

,, \lim_{n \to \infty} \Delta x = \lim_{n \to \infty} \frac{b-a}{n} = \dd x

[elfin berry]

bacc

[slim turtle]

Bro don't argue, I don't think it's his fault

[red nimbus]

oh i am not arguing i am just trying to understnad

maybe i am missing out something

i dont think it's coincidence she got 1/3

[slim turtle]

Idk maybe it is

[still temple]

yes thats right

ill just write it its hard to type

[red nimbus]

.>

[red nimbus]

If we integrate from 1 to n then dx would be here 1

i will just wait

[red nimbus]

I just saw it LMAO

[slim turtle]

😅

[slim turtle]

I'mma try the limit in a nice way

[red nimbus]

oh

[still temple]

no its not

[slim turtle]

Maybe.

[red nimbus]

,, \frac{\frac{1}{n} \left ( \frac{k^2}{n^2}+\frac{2k}{n^2} \right )}{1+\frac{k^2}{n^3}}

[elfin berry]

bacc

[red nimbus]

thats what you meant with equal powers

[still temple]

ykw sure i just guessed it

[still temple]

yup

[red nimbus]

(k/n)²

[still temple]

yup

[red nimbus]

can be substituted with u = k/n

[still temple]

yup

[red nimbus]

it seems intuitive

to skip the non equal terms like that

thats the least to say

[red nimbus]

,rcw

[elfin berry]

Couldn't find an attached image in the last 10 messages.

Sep 10, 2024, 5:53 PM

[still temple]

no bro theres a reason

[red nimbus]

ok bro

enlighten me

[still temple]

sigh im fucked for tommorow but ill explain it anyways

i cant finish my hw

[red nimbus]

you remind me of my old friend

[still temple]

give me 2 min\

ill do it

[red nimbus]

@coral cloak

two JEE fighters ⚔️

I forgot @supple hazel

that's my indian gang

I get it

[red nimbus]

pookie i get it

i saw what you did

i was just confused about your algebra

i notice how very fast thinking yall are

this should be published

[still temple]

ok what

is

happening

?

i just came back from eating

[red nimbus]

i saw what you did with that substitution

bro

[still temple]

i had such a good meal

[red nimbus]

you wont recover

[still temple]

and i came back with looking at integrals

[red nimbus]

if you see what happened

[red nimbus]

that aint for the weak

[red nimbus]

go sleep bro

[still temple]

its 1/3

how

that

s

[still temple]

wait

my brain is enlarging help :c

.

[still temple]

thats good

[red nimbus]

if everyone had 1/3 of your brain we would be at mars

conquering the universe

[still temple]

fr

[red nimbus]

my indian gang is off

they training for JEE

[still temple]

im not following your writing

[still temple]

ok i hope you understood but i have to go study more cause i have no choice

[red nimbus]

gogogog

[still temple]

wow

[red nimbus]

i just was confused about the beginning but i got it

[still temple]

ill re write it

[still temple]

i could see

[red nimbus]

[red nimbus]

hell yeah

[still temple]

what now

i cant see it

[red nimbus]

now there is this Indian theorem to ignore all non-equal terms

[still temple]

what

[red nimbus]

and you can factor 1/n

[still temple]

[red nimbus]

it's very small so consider it as a differential conveniently

tbh it's as good as Rishabh's approach

[slim turtle]

💀 The Fundamental Theorem of India.

[red nimbus]

with the k²/n³

[red nimbus]

yea

[still temple]

am i wrong

[red nimbus]

inf/inf 💀

[still temple]

this is the neatest i can write

[red nimbus]

😭

[still temple]

yes

[red nimbus]

wasnt you last time asking me something about derivatives

[still temple]

huh i did?

[red nimbus]

you said somethign like does it work or i dont think it works like that

[still temple]

where did inf/inf and 0 /inf came from

WHY IS THERE AN INTEGRAL THERE

??????????????????????????

[red nimbus]

it was about a limit where I suggested to use the definition of derivative

[still temple]

scroll up i cant doit again

[red nimbus]

it's very grenzwertig

[still temple]

i dont remember but i probably did

[red nimbus]

but basically substitute x = k/n

[still temple]

i haveto go do organic chem hahahahahah illl cry

bye

who is this girl

is she einstein

??????????

[red nimbus]

then you get bounds n/n and 1/n which is like 0 if n goes to inf

[still temple]

fr bro

[red nimbus]

no

[still temple]

im generally curious how much you thought about this problem

[red nimbus]

but she prob knows what she done

[still temple]

like how do you get to this level

[red nimbus]

you dont

[still temple]

into factoring 1/n randomly

[red nimbus]

well if you know such theorem exists it becomes intuitive

[still temple]

indian coaching institutes,i study like 12 hours a day

i use this discord as a break

are you indian

[red nimbus]

so little?

[still temple]

???????

[still temple]

yeah

u are the math demon

what is pre university math

[red nimbus]

we need indian math battles here

[still temple]

yes

i still dont understand

pls

[still temple]

?

[red nimbus]

before university

[slim turtle]

+1

[still temple]

bro there are kids better than me in my class

[red nimbus]

average indian reply if they get complimented

[still temple]

im like rank #5 in my class

[still temple]

there is no way

sorry i cannot believe that

i cant

[still temple]

no im not joking

[red nimbus]

some people achieve this playing fortnite solo

[red nimbus]

that's like every reply i got from indian friends i have I complimented on their math

[still temple]

im #2 in my class at maths

[red nimbus]

but i fw gang frfr

[still temple]

there is this kid who has straight A+

[still temple]

i mean youre doing better than the american highschoolers they dont even learn this shit

[red nimbus]

there is a spectrum on math skills to say the least

[still temple]

you are in school right

[still temple]

my professor haven't even taught us limits or matrices

i just learn them on my own

[red nimbus]

damn

[still temple]

i mean i was forced to learn them,but they are easy so idc

yeah i have been self learning for ab 2.5 years

[red nimbus]

why you dont care

[still temple]

good job

[red nimbus]

i can tell you care the most

[still temple]

fr

[red nimbus]

calling strangers poookie

and putting them before your homework

[still temple]

i mean even if i dont want to i am good at it so its ok

[red nimbus]

it's ok bro

i almost got tears stop

[still temple]

its only cause its chem homework i just dont want to

[still temple]

have you tried to solve milenium problems

[still temple]

whats that

OH NO

guys dont tell her

PLS

dont

i do pathfinder and blackbook

just dont tell her

idk what mellenium is

[red nimbus]

you get much money

[still temple]

dont

[still temple]

ooh noice

[red nimbus]

mathematical problems that have been unsolved for over thousand years or some

[still temple]

theres no way i could do that

YEA

there IS

im not that smart

[red nimbus]

yes pookie you can

:)

[still temple]

ty deer

[red nimbus]

https://tenor.com/view/deer-buck-stag-shikanoko-shikanoko-nokonoko-koshitantan-gif-17301713205920170507

[still temple]

eer

[still temple]

how are your college entraces like

enterances

how to s[pell

[red nimbus]

there are no enter races

we good

entrances

[still temple]

entrances

[still temple]

what

is that

[red nimbus]

i wish i had the ability to torture myself 12 hours

[still temple]

i dont i have adhd once my meds run out im over

[red nimbus]

now you also take meds

[still temple]

me too

[red nimbus]

thats where i cross the line

[still temple]

and if i forget to take it then also over

[still temple]

yes meth

im in high school

[red nimbus]

ah now i understood

[still temple]

like

WTF

[still temple]

same im 17

same

[red nimbus]

you do methmatics

[still temple]

bro jkjk

[still temple]

fr

[red nimbus]

no i almost believes your ass

thanks for confirming

[still temple]

i think i have to go aldol now

stupid chemistry

okay

ew

[red nimbus]

https://tenor.com/view/breaking-bad-walter-white-gif-8086630669522815021

[still temple]

im actually #1 at chemistry

surprise

[red nimbus]

https://tenor.com/view/youre-joddamn-right-jod-joddamn-walter-white-breaking-bad-gif-24853055

[still temple]

i am the worst in my class i just cant remember

[still temple]

you dont have to

if you try to remember IN CHEMISTRY you will never be good

bro what do you even have in your sylabus

i dont remember even 10% of the things my teacher taught me

but i can do any exercise anytime

we have over 40 chapters in chem and i need to know everything

chill

i also have same

exeptions mechanisms etcetc

yes

mechanisms are the best

really cute things

[still temple]

ill give you a question ill see if you know

points at TNT mechanism

[still temple]

go ahead

no wait

nah u gotta chill

i will give you a easier one

i dont know if i can understand chemistry in other language

its just so weird

i learn it in my natal language(hint: its not english)

the names of the elements are same in every language

not really

ph is benzene substituent

i is iodine

c is carbon

naoh is sodium hydroxide

o is oxygen

there you go

[still temple]

now can you understand?

yeah

D

i think

im not sure

i havent done chem in like 1 year

bcz summer break

what is that *

i never seen that in my entire life

i did it hehehehhehe

its haloform test

what is halofoorm test

i have no idea what heat does tho i just eliminated options

[still temple]

google

is it D

?

noh

its c

ok im actually going now

bye

see ya

.close

[marsh citrus]

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[stark field]

guys does someone knbow

why sometimes without adj(A)

swear i saw without the adjunct matrix

did i see wrong?

or this is formula for non 2x2 matrices?

[desert dirge]

thats for all invertible matrices

[stark field]

hmmm

oh w8 adj(A) would then have to be " matrix of algebraic cofactors" not the adjunct?

denoted as A* or C

[desert dirge]

thats included in the adjunct

[stark field]

[marsh citrus]

@stark field Has your question been resolved?

[marsh citrus]

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[crude valve]

is my professor punking me

is it just false because the initial point p on line L is not within hte half plane

ik its reversed, can i just say the contrapositive is true or is that wrong

[marsh citrus]

@crude valve Has your question been resolved?

[marsh citrus]

@crude valve Has your question been resolved?

[main idol]

is what false?

[marsh citrus]

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[terse brook]

anyone got advice with this

[terse brook]

just not really sure how I should show that it repsresents a circle

[quartz plank]

what relationships do you know with sin(theta) and cos(theta)

[terse brook]

first thing I think of is sin^2(theta) + cos^2(theta) = 1

or tangent = sin/cos

[quartz plank]

yeah exactly

is there any way you can use that in this context

[terse brook]

uh

should I try to like isolate sin and cos?

[quartz plank]

yeah see what happens

[terse brook]

sounds good, will try that and check back

I think I am getting it but let me work it out fully before I say its solved

This look right?

[quartz plank]

yep

[terse brook]

perfect, I'll leave this open while I try 50 but if I get then ill mark it solved

thank you

yeah seems all good, thank you so much and have a nice day

.close

[marsh citrus]

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[willow bloom]

what would be the most precise definition for f(x) = |x-1| + |x-2|?

[willow bloom]

or

They're both equal numerically, but the first one seems weird to me

formally

"se" is translated as "if"

[trim quest]

Neither one is more precise, they're exactly the same

Are you referring to the < or <= signs?

[willow bloom]

Yes

[trim quest]

Either one is fine, in both cases f(1)=1

[willow bloom]

But wouldn't that be a coincidence?

In this func at least

[trim quest]

I guess. But there isn't really any reason to prefer one over the other

Neither one is more typical than the other

[willow bloom]

That can be personal bias, but I've seen more resolutions using this structure
|x| =
x, if x >= 0
-x, if x < 0

[willow bloom]

I only needed to show that f(x) = |x-1| + |x-2| equals this

[willow bloom]

But I've reached that first instead

Took me alot of time to realize that they're both the same

[trim quest]

What exactly is the prompt? Just to express |x-1| + |x-2| as a piecewise function?

[trim quest]

or did they tell you specifically to prove that it is equal to this exact function

[willow bloom]

express f(x) as

[trim quest]

That's actually how the problem is given?
Or is that just the solution they give?

[willow bloom]

It gave me the resolution. I just needed to show that It's true

[trim quest]

Yeah I really don't think either one is that much more common than the other

You can just show that f(1)=1 and f(2)=2

and that pretty much shows that both definitions are equivalent

since at every other value of x they are explicitly equal

[willow bloom]

The only thing that confused me there, was this different way of defining f(x)

Just need to get used to this representation too, I guess

[trim quest]

Yeah, honestly the most common and convenient way to define this would just be |x-1| + |x-2|

They're just trying to get you used to piecewise functions, I'd guess

in practice, there's really no reason not to just do |x-1| + |x-2|

[marsh citrus]

@willow bloom Has your question been resolved?

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[still temple]

not sure I understand y=3x so m is 3 and both the x and y intercept are 0 so how can I have two ordered pairs to graph

[marsh citrus]

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[still temple]

no its no closed

.closed

.close

[marsh citrus]

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[sinful atlas]

can someone help guide me on how to work this out 🙏

[lyric bay]

Hmmm fractional decomposition won't work here

So maybe let's try long division 🙂

[sinful atlas]

NOOO

[lyric bay]

Or do you wanna learn Newton's method of optimization

[sinful atlas]

um

long division will work

hold on

[lyric bay]

That's what I thought

[sinful atlas]

smarter baby smarter 😞

OMG

THNAK YOU SO MUCh

i guess i was trying too hard to find a simpler solution by like

idk finding a way to factor the num and denom

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[old girder]

Plug in the intervals for X but for 8+h h will replace X

[devout pebble]

Average rate of change is basically the slope from (8, f(8)) to (8+h, f(h+8))

[timber knot]

Hmmm got it

[patent sundial]

you should close this channel

[patent sundial]

Do it by typing: .close

[marsh citrus]

@timber knot Has your question been resolved?

[marsh citrus]

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[stoic tiger]

I've got the following form:

[stoic tiger]

(b^4 - a^4)/(b^3 - a^3)

and i'm trying to figure out the behavior as b approachess a

im pretty sure this is a limit problem, on which imm a bit rusty

intuitively, due to the context of the problem, I expect this to go to zero

but i cant articulate why this is

i mean obviously either term invidivually goes to zero

but how do I know the top goes the zero faster than the bottom? if it were the other way around, this woud surely blow up yeah?

[fervent rampart]

you can factor the numerator & denominator using difference of squares/difference of cubes

[cobalt sedge]

you can factorize both the numerator and the denominator and see if it does go to 0

[stoic tiger]

hmm okay let me think about that

[split comet]

Did you progress?

[marsh citrus]

@stoic tiger Has your question been resolved?

[stoic tiger]

I was able to decompose the top into (b^2 - a^2)(b^2 + a^2), and then that into (b-a)(b+a)*(b^2+a^2)

and then after that

the denominator into (b-a)(b^2 +ab+a^2)

then we lose a factor of b-a which is nice

but not quite clear to me where to go after that

[split comet]

You're going correct

The whole becomes

(b+a)(b²+a²)/(b²+ ab +a²)

Now put the limit

[stoic tiger]

well if I plug b = a, Im pretty sure a lot of that reduces, ending p with just 2a

oh

hold on

i make error

[split comet]

You want help?

[stoic tiger]

i think it actually reduces to 4a/3

[split comet]

Absolutely

That's your answer

[stoic tiger]

huh alright

thanks a lot to all of you

i appreciate it

[split comet]

no problem!

[stoic tiger]

.close

[marsh citrus]

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[raw quest]

how do i do

part c ?

[patent sundial]

Hi

I am back

[raw quest]

oh my

hello !

[patent sundial]

for the rescue

[raw quest]

thank u sir

[patent sundial]

Hii

So...

[raw quest]

yeah part c

[patent sundial]

Do you know whats a double root?

[raw quest]

wait its

not a double root

its 1 root is double the other

[patent sundial]

Oh sorry

sorry sorry

Ok, so one root is x and the other is 2x

[raw quest]

mhm

pretty much

[patent sundial]

How can we start?

(I think the quadratic formula might come in handy again)

no?

[raw quest]

hm

i just dont know how to

use it

for this scenairo

[patent sundial]

Right!

[raw quest]

oh my god

wait

i think ur right

[patent sundial]

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[elfin berry]

moaforlife

[raw quest]

wait

i think

i see where ur going

lemme give it a go

[patent sundial]

Yeah right!

[raw quest]

oh my

yeah i got it

hahaha

thank u 🙏

.close

[marsh citrus]

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[patent sundial]

[marsh citrus]

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[balmy root]

Help

[balmy root]

Need some clearance of doubts in integration

Wait nvm

.close

[marsh citrus]

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[patent stirrup]

hi is this correct?

[marsh citrus]

@patent stirrup Has your question been resolved?

[viscid peak]

total standard deviation should be 0.7 * √18

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[swift olive]

I know how to do the others

but trig function i have no idea how to do algebraically

[night mica]

just sub theta =0

[swift olive]

I wasn't given a chart of their values so idk what they are

0/1

ig it's fine to just use my calculator

¯_(ツ)_/¯

pretty sure ill have on my test

ty

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[wheat rover]

the limit can be simplified further

you cannot put that yet

[swift olive]

.reopen

[marsh citrus]

✅

[swift olive]

what do you mean?

[wheat rover]

put sin(2theta) = 2sin(theta)cos(theta)

[night mica]

I don t understand ...if the function is continuous at x=0 then the limit if f(0)

[wheat rover]

cos theta cancels out in numerator and denominator

[swift olive]

no

it's a function

[night mica]

I don t see the reason to add another step to the computation

[swift olive]

f(x) / g(x) != f / g
you can't just get rid of the input to a function

[wheat rover]

bro

[night mica]

bri

[wheat rover]

i said simplify

i m not saying to delete the input

search double angle identity

[swift olive]

the answer key just put's the answer as 0

[wheat rover]

that's fine as well but the question says to simplify it algebraically

[swift olive]

find the limit

not simplify necessarily

don't need to simplify cause continuous 🤔

[wheat rover]

yeah ig

[swift olive]

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[wheat rover]

that works this way as well

[swift olive]

i think it's fine

[marsh citrus]

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[still temple]

Graph the following function...

How could I have known that the answer is C and not D.

[sand fable]

you can start by graphing log(2 - x) then using absolute value properties to continue

(in truth, neither C nor D is a very good match for what this graph looks like, but C is slightly better i guess…)

[marsh citrus]

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[frosty thicket]

i have answered what i can. all i need is for someone to help me whether my answer is correct

[frosty thicket]

1.) AB = 0 -> A = 0 \/ B = 0

2.) 2(A+B) = AB -3

kin hell yk what ill just take a picture of my paper

14.) is neg(P) -> neg(R)

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[proven ridge]

the question states Use the symmertry of each quadratic function to din the maxiumum or miniumum points sketch each graph showing all axes crossing points . Am trying to solve for 1A I just need to know how to find the range for my graph

[proven ridge]

How do i know the range for my quadratic equation\

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[proven ridge]

<@&286206848099549185>

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[elfin berry]

bacc

[red nimbus]

I would like to know if my proof is correct or if there is some fallacy

[lucid zenith]

how did you get this

that term looks to be between 0 and 1

which would mean that this inequality should be reversed

[red nimbus]

.close

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[red nimbus]

.reopen

bruh

[marsh citrus]

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[lyric kelp]

idk how to find if true or false

[fervent rampart]

consider the dimensions

[lyric kelp]

its when they said multiply A

like how do i mulktiply a whole matrix by a row

[knotty trellis]

row vector, so it's just matrix-vector multiplication

[fervent rampart]

treat the row as a matrix with 1 row and however many columns

[lyric kelp]

do you eman column row expansion

like this?

[fervent rampart]

yes but notice the difference between this and the question you are asked

[lyric kelp]

uhh thing is

idk whats matrix vector multiplication

[fervent rampart]

matrix vector multiplication is multiplying a matrix by a vector (where a vector is a matrix with only 1 row or only 1 column)

[lyric kelp]

ah

i found this

[lyric kelp]

which im not sure how its different from this

[fervent rampart]

the size of the matrices determines whether they can be multiplied together and the size of the result

so if i have a matrix A with size m×n and a matrix B with size r×s then:

• for AB to make sense we need n = r
• AB will have size m×s
• for BA to make sense we need s = m
• BA will have size r×n

[lyric kelp]

ah then from this can i assume that

[lyric kelp]

a here

would have 1 coloum

[fervent rampart]

you can assume that A has the appropriate size to be multiplied with B, but you should not make any further assumptions

[lyric kelp]

then

[lyric kelp]

its false beause it would have to be multiplied by the jth row of A column

to get ith row of matrix AB

cuz otherwise it might not be ith row

if taken the whole matrix

oh it was false okay i got it tysm

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[still temple]

I need research help in discrete math / computer science

[main idol]

help channels are for specific problems/questions. do you have something

[still temple]

yeah, basically I need to choose a good research problem

but is specific in discrete math

[main idol]

"picking a research problem" isn't really specific, and probably out of scope to most helpers

try cs server in #old-network

or ask in #discrete-math

[devout mauve]

also you cant just say research and expect that people know what you mean. do you mean actual research? I doubt it

[still temple]

sorry lol, ok I will be more specific

I found a wiki page about the partition problem, and there were a lot of problems there.
so I am thinking what is a good problem to try in there

or a generalization variation etc

[main idol]

...

we don't know what you're interested in

[still temple]

without being too hard for undergraduate lol

[main idol]

try whatever looks like you can do on your own or watch youtube videos on the problems

[still temple]

I am interested in how can we apply the partition problem in other areas of mathematics

[main idol]

still too vague

[main idol]

do this

[still temple]

ok I will look on youtube. or something
can giveme some general ideas for thinking in a research project?
I am trying to do something new (without being too much professional, I am just at undergrad lol)

I mean, still good, but maybe something short in extension

[main idol]

no, again nobody knows what you know. and again, help channels are for specific math problems

[still temple]

oh lol , sorry
I will check then

[still temple]

where could I ask this when I find something interesting?

[main idol]

#discrete-math or #math-discussion

[still temple]

thanks

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[signal dove]

I have problems progressing in my calculations in this question. The question states:
'Given three linear independent vecotors A, B and C, construct a vector X who builds the same angle with all three vectors. Where X can be written as a linear combination as i have done in my calculation. Then i am to decide the scalars, alpha, beta and gamma only using the vectors A, B and C.

Thanks for any help!

[marsh citrus]

@signal dove Has your question been resolved?

[signal dove]

<@&286206848099549185>

[mellow crag]

whilst you can do it like this

we have problems with the fact that A.X has A.B and A.C terms

we'd like all of that junk to cancel out

||instead, consider what happens if we write X in terms of AxB, BxC and CxA||

[craggy totem]

Is in your case A•X the standard inner product?

[signal dove]

Yeah it should be

[craggy totem]

Well i guess it has to be

[signal dove]

yeah

[mellow crag]

did this make sense @signal dove ?

[signal dove]

Yeah i get what you meant but i don't see how that would simplify anything

[mellow crag]

lemme upload a photo

[craggy totem]

The only thing I could add if X is a linear combination of the basis then A•X is equal to A•(a*A)
(With a being the coefficient for the part of A in the combinations of X)
Since you can write X as the combination, apply linearity for each vector component and it will all turn to zero except if you look at the A component (since they all create a basis)

[signal dove]

You mean if A =(a,0,0)?

[mellow crag]

and now note that A.BxC = [A,B,C]

so the problem is easy now

(i.e. set alpha = |A|, beta = |B|, gamma = |C|)

[signal dove]

Ah yes i see what you mean, but wouldn't then A,B and C have to be orthogonal, which it never states in the problem description that they have to be? or am i wrong?

[mellow crag]

wdym?

CxA is perpendicular to A so dotting it with A always cancels out etc

[signal dove]

Yeah that i get, but don't you consider BxC=A which isn't necesarilly true?

[mellow crag]

no

the challenge of this problem is basically to make A.X /|A| =B.X/|B| = C.X/|C|

[mellow crag]

now if you look at my expression here, dot X with A to get

alpha(A . B x C)

now A.BxC is the scalar triple product

so if we set alpha = |A|, beta = |B| and gamma = |C|

we get that A.X/|A| = [A,B,C] (i.e. the scalar triple product) so all of them are equal

[craggy totem]

What the fuck, im in my 6th semester of studying math and never ever have I heard of the scalar triple product

[mellow crag]

huh

it was first term for us

[signal dove]

lol that i have at least

Alright I get what you're saying

[craggy totem]

Im gonna check my scripts of my firsr courses but I have never seen this occuring

But that was a smart way to solve it

[signal dove]

Just intuivly this seems like 'to simple' of an anwser,

but i guess it should work

thanks for the help, gonna do som testing with it to see if it holds

[mellow crag]

nw!

[mellow crag]

ngl i didn't come up with it myself lol, it appeared in a past exam paper i did

[signal dove]

haha thats nice!

[craggy totem]

I see but still, its something I have never heard in my life. Particularly vector product was done in one really short subchapter about the concept of orthogonality

[signal dove]

.close

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[signal dove]

.reopen

[marsh citrus]

✅

[signal dove]

When i did some testing i didn't add upp correctly.

it*

[mellow crag]

huh interesting

i might have done my algebra incorrectly somewhere but that's the idea

[craggy totem]

It should work though, since the triple scalar product is commutative

[signal dove]

Isn't this the problem, or am i wrong, but that since BxC not necessarily = A then we change the equation?

[craggy totem]

Why should that be an issue

It doesnt matter what basis we used to contruct as X, as long as the coefficient were picked dependant on A,B and C

[signal dove]

But lets say here we put alpha=|A|, we are then left with A*(BxC)
And from the next one we will get: B*(CxA) right?, but what says that these are equal?

[craggy totem]

Well A(BxC) is equal to B(CxA)

The triple scalar product is commutative

[signal dove]

isn't it anti-commutative

[craggy totem]

I just read on it and it says its commutative

LY could prolly say it with 100% guarantee

@mellow crag (sorry)

[signal dove]

Yeah you are right!

BUT WHY DOESNT IT WOOOORK

im to tired for thiss

wait one minute

i'm stupid

brb

[craggy totem]

OH MY FUCKING GOD I JUST REALISED WE DID TALK ABOUT IT

But I was getting hammered by college level math back then, that I was unable to understand it

OH RIGHT IT ISNT COMMUTATIVE BUT IT RETAINS ITS VALUE WHEN YOU "ROTATE" THE VECTORS IM THERE

Ups Caps lock

[signal dove]

yeah right but it actually holds in this case

[craggy totem]

Yeah thanks to how we constructed the new basis

[signal dove]

yeah

I checked now and it holds, thans for the help 🙂

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[marsh citrus]

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[still temple]

Midsegments of triangles

i have no idea where to start

[limber marsh]

Well, can you identify any similar triangles here?

[still temple]

i have no idea

t r and s make one

[limber marsh]

And that's similar to which other triangle?

(there's only so many triangles here)

[still temple]

p q s

[limber marsh]

Yep

[still temple]

my brain doesnt comprehend all of this

[limber marsh]

And can you tell me how long QS is compared to RS?

[still temple]

alot more

um

rs is like half of qs

[limber marsh]

Exactly

[still temple]

is it 20

[limber marsh]

Because pqs and trs are similar,
That means if QS is twice the length of RS, PQ is twice the length of TR

[limber marsh]

Bingo 🎉

[still temple]

my oh my

well i guess i cannot say that word

there

[still temple]

i finished the assignment thank u

[limber marsh]

You're welcome!

[still temple]

.close

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[still temple]

wait

can u help me with this

[limber marsh]

?

[still temple]

i think thats it

after

[limber marsh]

Well, consider EPC and ENA ( the triangles)

It looks awfully similar to the last question you had, no?

[still temple]

yeh

different shape tho

and alot more numbers

letters

[limber marsh]

There's a lot of irrelevant information

And I'd hardly call it a different shape

It's just presented a bit differently

Whoever wrote this question was trying to be tricky
Don't let them trick you

[still temple]

yeah

thats what i was thinking

i got it

.close

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[round gorge]

for this question, I am confused a little. I understand that the question is trying to trick you by saying that these values are part of the increasing techincally. But aren't there 2 non increasing points? It isn't increasing at -2 or 2 since those are exclusive brackets

[placid prawn]

I mean just x^2+4 seems like a possible equation for f(x)

[round gorge]

oh wait im dumb

hol up

that means the vertex is at 0, 4 right

[placid prawn]

yes

[round gorge]

but the function isnt increasing or decreasing at -2

or 2

because the bracket is exclusive

if it was [2, infinity) that would make more sense for example

[placid prawn]

does the question state that it must not be increasing at 2 or just that it is increasing from 2 to inf

[round gorge]

dont the () state exclusive

[] is inclusive

[placid prawn]

yeah the given intervals are exclusive but it doesnt necessarily mean something else must be true at those points

[round gorge]

okay so its liek a trick question kindas

[placid prawn]

maybe

[round gorge]

how would i algebraicly get x^2 + 4

its a thinking question which means not a regular question

[placid prawn]

quadratic implies that the highest exponent of x in the equation will be 2, the info it gives about the intervals shows you that the parabola will be open upwards, and the given y-intercept shows that f(0)=4; so from the 1st and 2nd facts you can start with x^2 and with the 3rd fact you get x^2+4

[round gorge]

yeah okay, so just logic basically. i get it, i was just wondering if there was an algebraic way. thanks tho

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[stark glacier]

Hello. Is this a good place to ask about "significant figures" in physics?

[stark glacier]

nvm

.close

[marsh citrus]

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[stark glacier]

no actually i'm fuming i dont know what to do.

.reopen

[marsh citrus]

✅

[tired lava]

i think so

the digits right

[old talon]

Ik that it’s - cos

[stark glacier]

.occupied

!occupied

[marsh citrus]

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

[old talon]

?

[stark glacier]

So I have this question:
"
The time for 30 pendulum swings is measured 10 times.
53.1s, 53.3s, 52.8s, 53.4s, 52.8s
52.8s, 53.4s, 53.4s, 53.2s, 53.2s.
a) Find the average value of the measurements and the uncertainty.

b) Determine the period (the time for one pendulum swing) and the uncertainty in this.
"
And another question:
"
The acceleration of a body with initial velocity (1.8 ± 0.3)m/s is given to be (3.4±0.3)m/s^2.
a) What is the speed of the body (3.5±0.1)s after start

b) How far has the body moved during this time?
"

===================
For 1a) the first bit, I simply added all the measurements together and /10 to get the "mean". Then I found the uncertainty by recognizing the highest and lowest measurement, and took mean - (highest OR lowest) = uncertainty value.
so for 1a) I wrote 53.1 +- 0.3 as my answer.

For 1b) Since the mean time was calculated for entire 30 swings, i took (mean time)/30 to get the Period.
I got 1.77, and I know that uncertainty doesn't change in value when dividing by a constant. But in the answer, I wrote "2 +- 0.3" because (mean time) was divided by a number with one significiant figure.

When I sent those answers written, I only got 1/2 points and I don't know what I did wrong.

[chilly latch]

Would you like a quick anwser?

[stark glacier]

To compare with mine yes.

[chilly latch]

I don't want to type all the explaining

[stark glacier]

I simply need to see what I did wrong in terms of sig figures.

It's driving me nuts, because it's not easy to ask

[chilly latch]

1a) Average time: 53.14 ± 0.28 s
1b) Period: 1.77 ± 0.01 s
2a) Speed after start: (13.7 ± 0.5) m/s
2b) Distance moved: (31.5 ± 1.8) m

[stark glacier]

And you considered significant figures?

[chilly latch]

I have

I have a friend on a call with me

He believes what I said is right

I might be wrong

[stark glacier]

I don't really know why 0.28s is an accepted answer, because it has two sig figures but the question/values in calculation contains 3 sig figures.

many small details, im lost lol

[chilly latch]

Well

Wait

OH

Its 0.3 not 0.28

I think

That's what my friend said to tell you

Now I think about it he's right

[stark glacier]

how about 53.14? Isn't it supposed to be 53.1 because the question contains 3 sig figures?

[chilly latch]

Because when you calculate an average or any derived quantity, the result should be rounded to reflect the precision of the measurements involved.

The rounding rules

Don't forgot

[stark glacier]

53.1 +- 0.3 is the correct answer for 1a)? 0.28 rounds to 0.3 and 53.14 rounds to 53.1

[chilly latch]

Yeah

What are you not understanding

[stark glacier]

And for period, how did you end up +- 0.01?

[chilly latch]

Step 1

[stark glacier]

do uncertainties change when you divide it by a constant?

[chilly latch]

Period = Average Time over 30 = 53.1 over 30 = 1.77

Step 2

[stark glacier]

what

ah, time/30

but yeah 1.77, got that.

[chilly latch]

Wait

When dividing by a constant, the relative uncertainty remains the same.
The relative uncertainty in the average time is:
0.3/53.1 apply this relative uncertainty to the period

[stark glacier]

but 1.77, how many signficiant figures do you include here? You divide 3sig/1sig, so aren't you supposed to round this value to 1 signfigcant figure?

[chilly latch]

Uncertainty in peirod = (0.3/53.1) x 1.77

Step 3

Perform the Calculation

[stark glacier]

are you using chatgpt?

[chilly latch]

No

[stark glacier]

ok.

[chilly latch]

I have a friend whos helping me

[stark glacier]

saying "step 1", "step 2", it's not really normal.

but ok.

[chilly latch]

0.3/53.1 = 0.00565

0.00565 x 1.77 = 0.01

[chilly latch]

That's how I talk/write

[stark glacier]

wait.

[chilly latch]

What

[stark glacier]

(0.3/53.1)*1.77, where did this come from?

[chilly latch]

Here

Not there

Wait

[chilly latch]

Here

[stark glacier]

I think I understand.

[chilly latch]

Good

I'm leaving

Bye

Sorry for the attitude

I'm just sick

cya

[stark glacier]

oki

.close

[marsh citrus]

Channel closed

Closed by @stark glacier

Use .reopen if this was a mistake.

[stark glacier]

.reopen

[marsh citrus]

✅

[stark glacier]

I need help understanding why 1.77 is the answer and not "2" because of significant figures.

[chilly latch]

I would love to help you more

But I can't

I'm tried af

[stark glacier]

alright. i'll be asking others who may be passing by this chat.

""
The time for 30 pendulum swings is measured 10 times.
53.1s, 53.3s, 52.8s, 53.4s, 52.8s
52.8s, 53.4s, 53.4s, 53.2s, 53.2s.
a) Find the average value of the measurements and the uncertainty.

b) Determine the period (the time for one pendulum swing) and the uncertainty in this.
"
And another question:
"
The acceleration of a body with initial velocity (1.8 ± 0.3)m/s is given to be (3.4±0.3)m/s^2.
a) What is the speed of the body (3.5±0.1)s after start

b) How far has the body moved during this time?
"

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[marsh citrus]

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[lime heron]

The constant term is 168. Find k.

[lime heron]

Honestly, I am so clueless...

Pascal's Triangle and combinations really aren't helping.

[desert dirge]

you can find which term it is
itll happen when [x^4]^(7-n) * [x^{-3}]^(n)=x^(-7)
n starting at 0, if thats of any help

[lime heron]

I fear I understand nothing

[desert dirge]

consider the bottom, im saying you can use it to find the term where you have a x^(-7) in the expansion, which when you distribute the x^7 from outside of it will leave you with the constant term

so if you find n, you also know what the 7Cr value will be since youll have r
(the n in my original message)

or you could just do the expansion wholly, 7 isnt that much

[lime heron]

WHAT-

I HAVE NEVER SEEN THAT

IN MY LIFE

Im so sorry but i have no clue what that is

Ive done sigma notation but we havent learnt that exactly

but we learnt sigma notation just this week

[desert dirge]

eh? i assumed this is what you meant by pascals triangle

[lime heron]

No

1
1 1
1 2 1
1 3 3 1

etc.

[desert dirge]

yeah i know

[lime heron]

& also the other option of calculating the term

[desert dirge]

the nCr follows that

[lime heron]

the

[lime heron]

yes that

the nCr

i just leearnt it today

[desert dirge]

here the thing that look like vectors are nCr

n is the top value r is the bottom one

[lime heron]

what are vectors?

[desert dirge]

ah

this means nCk

[lime heron]

the nk represent

n and r?

Oh we learnt that format

but not with sigma notation-

[desert dirge]

im not entirely sure how to explain it in a way that really adds anything to the picture

[lime heron]

Oh that's fine

but would u admit my teacher is insane

literally refreshed ourselves on pascal's triangle and learnt this nCr today

AND WE DIDNT EVEN DO QUESTIONS LIKE THAT

AND HE EXPECTS US TO DO ALL THEESE QUESTIONS

[desert dirge]

if theyre asking you to do that without yet teaching binomial expansions, then yeah

its probably in your next few lessons though

[lime heron]

binomial expansions?

[lime heron]

no he said our unit is done

we're having a quiz next monday

[desert dirge]

odd

[lime heron]

s in expanding binomials?

as in

[desert dirge]

this is a binomial expansion

[lime heron]

well ive seen it in

[desert dirge]

for natural n

[lime heron]

(x+2)^7 or so

and ive expanded those

but i dont see how that can possibly help w/ figuring out k

The constant term would be

7C7 wouldn't it?

Which is just 1

Since the constant term is last?

[desert dirge]

the last term in your expansion would be k/x^21

[lime heron]

Wouldn't it be

[desert dirge]

but even when you distribute the x^7 its still not a constant

[lime heron]

k^7/x&21

k^7/x^21

(k/x^3)^7

[desert dirge]

you are correct

mb