help-39

[stuck vapor]

i mean

tje (0.065/4) by 400000

[short sparrow]

the interest per quarter is the 4th root of the interest per year, because it gets stacked 4 times during the year

the same principle applies to each month, but with 12

[stuck vapor]

wait but when i do (0.065/4) x 400,000 i get 6500 which is considered correct according to the answer sheet

same for 12

[stuck vapor]

representing per month

is that rlly right? bc like sorry i was just checking if the method was correct

[short sparrow]

if the interest goes up as a smooth exponential function, then my method should be correct.
But if it gets recalculated once every year or if it is a fixed rate forever, then your method is correct.
I don't know enough about banking to know for sure which one of the two is how it should be calculated.

[stuck vapor]

yeah me neither 😭 ur method makes more sense to me but ig it gets recalculated if thats the answer

bc i tried ur method first but didnt get the right answers according to the answer sheet

but anyways thanks a lot for helping! i appreciate it!

.close

[pearl pond]

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[pearl pond]

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[solid siren]

How do I solve with u sub

[solid siren]

supposedly meant to use u² so even if there’s an easier way please show both

@ when u reply

What I’ve tried

[sharp smelt]

let x^3=t

:D

isn't that overkill

[prime bramble]

I'm fairly sure you can get away with u = x^3 + 1

[solid siren]

js do regular u sub bruh

Don’t do wack stuff

[coarse harbor]

Perhaps

[urban briar]

bruh what are u doing

have u tried the substituion

[coarse harbor]

Tbh it's the first thing that came to mind and it works so yeah

[urban briar]

u = x^3

[solid siren]

Why would I leave out the 1

if u look at the photo u can see what I tried

[urban briar]

k

[solid siren]

didn’t know where to go or if I should do sum else

[prime bramble]

I will quietly reiterate this

[solid siren]

I tried that dk where to go

[sharp smelt]

,rccw

[jolly parrot]

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

May 20, 2025, 4:02 PM

[solid siren]

.

[urban briar]

i legit

dont understand

what the issue is

[sharp smelt]

,rccw

[prime bramble]

you used u^2 for some reason

don't do that

[solid siren]

Read the message

[prime bramble]

there isn't a need to

I did read it

[solid siren]

^

[prime bramble]

you don't need it

[solid siren]

Read it again cs u clearly didn’t understand it

[prime bramble]

you have to use u^2?

is this a rule set by the problem itself?

[solid siren]

teacher said I have to

[prime bramble]

[urban briar]

are u sure u understood correctly

[prime bramble]

okay, in that case

[urban briar]

just substitute

x^3 + 1

[solid siren]

Yes gng

[prime bramble]

set u = sqrt(x^3 + 1)

[solid siren]

how will I get rid of the 4x^5 after

[urban briar]

just set u = x^3

• 1\

$u = x^3 + 1$

[jolly parrot]

bagelguy3

[urban briar]

$du = 3x^2 dx$

[solid siren]

Bagel

Read the message I sent

[jolly parrot]

bagelguy3

[urban briar]

You've sent a lot of messages

which one

[solid siren]

have to use u²

[urban briar]

BRUH

have to?

as if in

no other choice?

Why tf would your teacher give you some assignment like that?

[solid siren]

idk

[urban briar]

alright then

to this

first substitue

u = x^2

then do

[solid siren]

bruh what

[urban briar]

$v = u^{1.5} + 1$

[jolly parrot]

bagelguy3

[urban briar]

that way you get $x^3 + 1$

[jolly parrot]

bagelguy3

[prime bramble]

I think if you set $u^2 = x^3 + 1$, then we get $2u du = 3x^2 dx$, which we can rearrange for $x^2 dx = \frac{2}{3}u du$

[solid siren]

I initially js wanted to do u=x³+1 and then u²=x^6+2x³+1 so i have a power of 5 in my derivative but idk how to put it in

[prime bramble]

combine that with $x^3 = u^2 - 1$

and you can rewrite the entire integral in terms of u

[jolly parrot]

higher!

[urban briar]

bruh

look

[prime bramble]

then it becomes a standard integral

[urban briar]

$\frac{du}{3x^2} = dx$

[jolly parrot]

bagelguy3

[urban briar]

take the constant term out

and use that to divide the x^5

[solid siren]

3x²

[urban briar]

and then u get x^3 = u-1

[solid siren]

Isn’t it

[urban briar]

yes

bruh where are u going wrong

i dont get it

[solid siren]

What am I subbing it

U guys are legit doing 2 dif problems atp

[urban briar]

wait

just show me the problem

[solid siren]

2 dif solutions

[urban briar]

please

[solid siren]

Bro scroll up

[prime bramble]

yes, my bad

[solid siren]

it was the first thing I sent

[urban briar]

i just wnana know i, talking about

YES

THAT

[prime bramble]

I can stand aside and let bagel do his thing

[urban briar]

IS WHAT IM TALKING ABOUT

just do

u = x^3 + 1

[solid siren]

I’d rather you 😭

[prime bramble]

well, you can try my sub

[urban briar]

look u get

$\frac{4}{3} \int (u-1)\sqrt{u} du$

[jolly parrot]

higher!

bagelguy3

[urban briar]

$\frac{4}{3} \int u^{1.5} - u^{0.5} du$

[jolly parrot]

bagelguy3

[urban briar]

this is pretty easy to evaluate

thats it

I dont see wheat your problem is

[solid siren]

So if I have 2udu/3x²=dx and then I sub that in and sub the u in u just get 8x³u²/3 don’t I

[urban briar]

I never try to get powers in my substitution'

[prime bramble]

our integral is $4 \int x^5 \sqrt{x^3 + 1} dx$, yeah?

[jolly parrot]

higher!

[solid siren]

4x⁵

yea

4 in front

Yea

It

That is it

[prime bramble]

so if we write $2u du = 3x^2 dx$, we have $x^2 dx = \frac{2}{3} u du$

[jolly parrot]

higher!

[solid siren]

Yea

[prime bramble]

our substitution was $u^2 = x^3 + 1$, so $x^3 = u^2 - 1$

[jolly parrot]

higher!

[prime bramble]

now write the integral as $4 \int x^3 \cdot x^2 \sqrt{x^3 + 1} dx$

[jolly parrot]

higher!

[prime bramble]

we have an x^3, and an x^2 dx

[solid siren]

I gtg but can u explain a bit more later

Like in DMs

[prime bramble]

we can sub our expressions for those into both

[solid siren]

My next class abt to start

[prime bramble]

and the sqrt is just u

[prime bramble]

ahh, I'd rather not go to DMs, apologies

I like having the TeXit bot here

you can come back to this channel when you're free again

[solid siren]

K

[prime bramble]

or open a new one and ping me

[solid siren]

It’ll be like 4 hrs or sum probably

[prime bramble]

that's okay too

[pearl pond]

@solid siren Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

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[split depot]

hi

[split depot]

looking for an answer to a simple question

but the answer probably isn't as simple as the question so if there's anyone who's experienced in math that could help me out

[prisma elk]

ask

[toxic lichen]

yeah just ask your question lmfao

[split depot]

if we're asked to prove a certain relationship between some variables in a question, if we build up an equation solely consisting of whatever variables we need for the final proof can we always arrive at the given answer in more discrete math problems?

[toxic lichen]

feels vague

[robust kraken]

yh give an example

[sonic hound]

yes, that would help

[autumn fossil]

stated like this, if you're supposed to prove e.g. a = b and you managed to form an equation a + b = 10, then it's of no use

[split depot]

uh alright lemme see

[prisma elk]

Equations are often useful, but discrete mathematics frequently requires domain-specific reasoning

[split depot]

'If a particle moves under uniform acceleration describes successive equal distances in times t1, t2, t3 respectively. Prove image

[versed mica]

is this like google translate

[sonic hound]

1/t1 - 1/t2 + 1/t3 is not symmetric

[split depot]

if we build an equation with the given constraints consisting of only t1 t2 t3 and no constant terms can we always arrive to that answer?

[autumn fossil]

it doesnt have to be symmetric, does it?

[toxic lichen]

rhs is symmetric but lhs isnt

[split depot]

nah nah i just sent the thing we have to prove as an img cause it might be confusing if I type it

[prisma elk]

you need extra information

because that equation isnt provable unless there are certain constraints on t_i

[split depot]

could you explain a bit more?

[prisma elk]

drop more context

[split depot]

the only other information (we're sort of meant to assume this) is that there's an initial velocity (unknown)

[feral sedge]

t1<t2<t3?

[split depot]

that's the whole question though

[prisma elk]

where is it from?

[split depot]

wait let me re check if I typed it from

[sonic hound]

does this mean that the initial velocity is non-zero?

[split depot]

it wrong*

[split depot]

yeah yeah

[split depot]

my math class tute

[prisma elk]

what book

[split depot]

it's just a part of simple linear motion equations

[split depot]

it's not really from a book, its a tute from my tuition

the question's correct as far as I'm aware

I just gave that as an example

hey I'm not rushing or anything but just to confirm, this help request is still active right

[prisma elk]

yes

[split depot]

got it

[proven pawn]

I can help with this but idk what your original question was

[split depot]

do you mean you cant see it or that my explanation is unclear

[proven pawn]

I mean, I can help with the question in the image, if thats what you want help in

[split depot]

oh, no no, I gave that as an example to make my actual question more clear

[proven pawn]

ah ok

[prisma elk]

how do you solve that though?

[split depot]

I was asking if for discrete math problems, if we're asked to prove a certain relationship between some variables

[proven pawn]

I would find the times in terms of s and a

[split depot]

if building up a relationship with just those terms, can definitely lead you to arranging it in the desired way

[prisma elk]

yes but still it doesnt simplify nicely though

[split depot]

try cancelling out the s and the initial velocity

by either substitution or division

[proven pawn]

time for 0 to s=t1, s to 2s = t2, 2s to 3s = t3
so t1 = sqrt(2s/a), t1+t2 = sqrt(4s/a), t1+t2+t3 = sqrt(6s/a)

im assuming u=0

then calculate both lhs and rhs in terms of s and a and see if they come out to be equal

[split depot]

u isnt 0

[proven pawn]

oh

[split depot]

unless you didn't do that intentionally

did*

[proven pawn]

same method probably works thuogh

[split depot]

yeah probably

anyways aside from that question, does anybody know the answer to my initial question

would be great if there's someone who could tell me

is there some general or formal proof of things like that? I've had this question forever but I don't for sure know the answer

[pearl pond]

@split depot Has your question been resolved?

[split depot]

still active?

[pearl pond]

@split depot Has your question been resolved?

[pearl pond]

@split depot Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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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[cold pollen]

How do I factor quadratic equations?

<@&286206848099549185>

[azure ether]

Some numbers times to give u c, and plus to gives u b

Any specific examples?

[cold pollen]

omg

that helps so much

lets go

now i can factor the equation x^2+7x+12

with your help i figure out that 3x4 is 12

which also adds up to 7

so the answer it

is

[azure ether]

Yep

[cold pollen]

(x+3)(x+4)

W teacher

12/10 rating

[azure ether]

Bingo

[cold pollen]

ok

now

how do i factor a binomial

[pastel umbra]

...Isn't a binomial already a factored expression?

[cold pollen]

i meant to say trinomial aura goat

[pastel umbra]

(also not the Bennet thumbs up lmao)

[cold pollen]

shut up

bu

bum

jk

ur very nice

@azure ether

[pastel umbra]

Poe's Law, just be careful 🤣

[cold pollen]

how to factor trinomials

[cold pollen]

whats that

[pastel umbra]

"Something you mean as a joke may, and often will, be vague enough to be taken seriously by someone"

[cold pollen]

jk ur a nice guy

[pastel umbra]

But trinomials aren't necessarily something generally factorisable

All that defines a trinomial is that it's a three-term polynomial

[azure ether]

There are multiple ways to do

[cold pollen]

ok sigma

[cobalt hinge]

I presume that OP is talking about a quadratic?

[cold pollen]

no shit sherlock

[azure ether]

Yeah

[rough forge]

it's complex 🥀

[cold pollen]

hang it up unc

[azure ether]

I can find a helpful video that explains better than me

[pastel umbra]

I'd've thought so, but that was OP's first comment, so...

[cold pollen]

ok bet

[cobalt hinge]

Well, not all trinomials are quadratic

[cold pollen]

thank you beta

[cold pollen]

very sorry sir

https://tenor.com/view/happy-tuesday-tuesday-atleast-not-monday-tuesday-happy-tuesday-morning-happy-tuesday-good-morning-gif-11632253283275856846

[pastel umbra]

x^100 + x + 1 is a trinomial, for example

[cold pollen]

skibidi indeed

[cobalt hinge]

Dude, you can stop being weird kek

[pastel umbra]

That doesn't make it something that can be factorised

Yh it's just getting in the way

[cold pollen]

ok im actually sorry

[cold pollen]

im sorry

i thought i was funny

[pastel umbra]

There's a time and a place

[azure ether]

https://youtu.be/-4jANGlJRSY?feature=shared

[cold pollen]

let me use my ai for the transcript to sumarize

[pastel umbra]

though every now and again is okay (just don't constantly spam jokes)

[cold pollen]

ok srry

[azure ether]

It’s only 12 min

[cold pollen]

my final exam is in 8

[rough forge]

attention span is less than 12 sec

[cold pollen]

💀

[cold pollen]

btw im not joking @azure ether

[azure ether]

https://tenor.com/view/baffling-gif-24471655

[cold pollen]

https://tenor.com/view/spongebob-squarepants-gay-rainbow-lgbt-pride-month-gif-6443547037069682625

[azure ether]

Just look some of the example not all

[cold pollen]

ok

[azure ether]

https://tenor.com/view/pride-month-lgbtq-rights-gay-allies-lgbtq-allies-lgbtq-ally-gif-998048842246359674

Bro lol

[cobalt hinge]

Can you guys not

[azure ether]

Go study and close the chat

[cold pollen]

oh i forgot

im sorry

:closechat

/closechat

[azure ether]

.close

[pastel umbra]

!done

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[cold pollen]

.cose

.close

[pearl pond]

Channel closed

Closed by @cold pollen

Use .reopen if this was a mistake.

[rough forge]

,calc cos(e)

[cobalt hinge]

What are these .close permutations

[jolly parrot]

Result:

-0.91173391478697

[cobalt hinge]

Free bingo

[cold pollen]

,clac sigma

,calc cos(e)

[jolly parrot]

Result:

-0.91173391478697

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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!

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

[balmy lily]

I need help with my 3d model for my math exploration. Im doing it in geogebra
I have problem with the lettuce... like idk how to do the form with functions

[balmy lily]

and with the cheese i tried make it a square but i only get circles

[pearl pond]

@balmy lily Has your question been resolved?

[balmy lily]

<@&286206848099549185>

[pearl pond]

@balmy lily Has your question been resolved?

[balmy lily]

.close

[pearl pond]

Channel closed

Closed by @balmy lily

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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!

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

[willow veldt]

is it true that There is no scalar equation of a line in three space?

[sharp vigil]

it's true that you have to either give it as a parametric equation, or as a system of two scalar equations

[willow veldt]

thank youuuuu

.close

[pearl pond]

Channel closed

Closed by @willow veldt

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[pearl pond]

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Send your question here to claim the channel.

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

[cosmic crag]

Hello

Is the Maclaurin's series a generalised version of the Taylor's series?

[karmic pier]

pretty much

mclaurin’s centers around x=0

taylor centers around x=a, a being real

[sharp vigil]

the opposite, taylor series are a generalized version of maclaurin series

[karmic pier]

https://tenor.com/view/theyre-the-same-picture-the-office-pam-the-office-us-gif-20757621

[cosmic crag]

So, I've been doing Mclaurin's series.

[karmic pier]

if you can do mclaurin’s you can do taylor

it’s just a bit more work

you “choose” an a where you want your series to be centered around, or “accurate”

[cosmic crag]

Do you have a video for Taylor's series?

[karmic pier]

usually you’ll be asked to find the taylor series around a specific number

https://youtu.be/LDBnS4c7YbA

[cosmic crag]

Can I ping one of the helpers

[karmic pier]

what for

[cosmic crag]

I need more videos

i need a large group of videos to learn Taylor's series

[karmic pier]

is that necessary?

try giving this one a watch

if you’d like

[cosmic crag]

It is a bit superfluous

[karmic pier]

i see

[cosmic crag]

But Taylor's series is necessary for my work

[karmic pier]

you want to understand what it’s for

how it works

https://youtu.be/3d6DsjIBzJ4

this won’t teach you how to work with them but it’ll show you what they’re for and how they work

[cosmic crag]

I watched this video

it is a great video

[karmic pier]

good

[cosmic crag]

But he talks about the Mclaurin's series unfortunately

[karmic pier]

it’s the same thing

all you do with taylor is change where it’s centered

[cosmic crag]

I gotta do the same drill for Taylor's now

[karmic pier]

mclaurin’s series are only accurate around 0

taylor’s series are accurate around whichever “a” you use

that a can also be 0

[cosmic crag]

All the approximations are different, aren't they?

when we change the a, the approximation changes accordingly

[karmic pier]

what do you mean

[cosmic crag]

The same approximation doesn't work for two different points

[karmic pier]

ohh

you mean if you can express the series as a function of x and a

[cosmic crag]

Let's say I found an approximation for cos x near 0.

Lets say I found another approximation for cos x near pi.

The approximations are not the same

[karmic pier]

then change a arbitrarily to “move” it around

[cosmic crag]

The same approximation does not work

[cosmic crag]

I haven't got around to that part yet. I have only watched that one video on Mclaurin's series and Taylor's series

[blissful cloak]

for cosine the radius of convergence is infinity, so in that case the approximation does work in either the x=0 or x=pi case, although they do generate different terms yes

[cosmic crag]

Can you elaborate on the former part?

An approximation exists both for x=0 and for x=pi.

Have I correctly interpreted your thoughts?

[blissful cloak]

[
\begin{cases}
a=0&\quad\cos(x)=\sum_{n\ge 0}\frac{(-1)^nx^{2n}}{(2n)!}\
a=\pi&\quad\cos(x)=\sum_{n\ge 0}-\frac{i^n((-1)^n+1)(x-\pi)^n}{2n!}
\end{cases}
]

[jolly parrot]

PajamaMamaLlama

[blissful cloak]

both of these series converge on the true value of cos(x) for any value of x

[cosmic crag]

What are both of these series?

Are they the formulae for Taylor series and Mclaurin's series?

[blissful cloak]

they are both Taylor series, although the MacLaurin is only the a=0 series

MacLaurin series are a "subset" of Taylor series with a=0

[cosmic crag]

Can you explain the formula for the second Taylor series?

Particularly the (((-1)^n)+1)

How did Taylor arrive at this formula for the coefficient of the variables?

@blissful cloak

[blissful cloak]

Well Taylor didn't, WolframAlpha did lmao but Taylor's result asserts [f(x)=\sum_{n\ge 0}\frac{f^{(n)}(a)(x-a)^n}{n!}] WolframAlpha jsut determined (f^{(n)}(a)), which in this case just happened to include ((-1)^n+1)

[jolly parrot]

PajamaMamaLlama

[blissful cloak]

,w series cos(x) at x=pi

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xxmrfancyu2xx

[cosmic crag]

Wait

Why are the coefficients of the generalised Taylor series, the same as the coefficients of the Mclaurin's series?

at pi and at 0, the coefficients of the series are the same.

<@&286206848099549185>

[blissful cloak]

no they're not, not quite

[
\begin{cases}
\mathrm{Taylor Series}&\quad f(x)=\sum_{n\ge 0}\frac{f^{(n)}(a)(x-a)^n}{n!}\
\mathrm{MacLaurin Series}&\quad f(x)=\sum_{n\ge 0}\frac{f^{(n)}(0)x^n}{n!}
\end{cases}
]

[jolly parrot]

PajamaMamaLlama

[blissful cloak]

observe, the maclaurin series is obtained when setting a=0 in the taylor series

[cosmic crag]

Right, because it is 1, -1/2

and the Taylors series for cos x near pi has -1 and 1/2

[pearl pond]

@cosmic crag Has your question been resolved?

[cosmic crag]

Done

<@&286206848099549185> done

[crystal hill]

do .close

[cosmic crag]

.close

[pearl pond]

Channel closed

Closed by @cosmic crag

Use .reopen if this was a mistake.

[cosmic crag]

<@&286206848099549185> this channel is being occupied

.close

.reopen

[pearl pond]

✅

[cosmic crag]

please close this channel from your side

[autumn fossil]

It stays in the occupied section for a while, it only moves after like 5 mins usually

.close

[crystal hill]

huh what

[pearl pond]

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[rugged lava]

you should probably clarify which specific problems you need help with

[autumn fossil]

is that past exam or current one?

[high garden]

.close

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[pearl pond]

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[desert cliff]

can you guys explain to me why this is wrong?

Thank you :))

[sour iris]

What is the questions

[edgy wren]

rewriting
-|x-1|+2 simply the c is 2 and there is -1 for a

[desert cliff]

what i did was multiply the -1 by 2 as they were next to each other

[edgy wren]

that's not how it works

[desert cliff]

the question was sketch y=2-|x-1|

[edgy wren]

it is seperated by the minus sign

[sour iris]

Heres how you do it:

[desert cliff]

yea but that is same as -1 right

[sour iris]

imagine the line without the absolute value sign

and it's just a line

that is flippe

from the x-b or where

that's it

[edgy wren]

yep but still the expression is still like this

[desert cliff]

oh because its same thing as y = +2 -1 |x-1|

[edgy wren]

yep

[desert cliff]

so if i rewrite it it would be y=-1|x-1| + 2

?

[edgy wren]

yep

[desert cliff]

.done

[edgy wren]

.close...

[pearl pond]

@desert cliff Has your question been resolved?

[pearl pond]

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[last walrus]

how would i find the answer to f and g

i know f = 155 and g = 15

but idk how i would work it out

[last walrus]

<@&286206848099549185>

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[gilded portal]

can someone explain the signed test? like i know how to do it when alternate hypothesis is not equal to median
am i supposed to use Z score? we were taught with binomial distribution
how do i manage the case where alternate hypothesis is lesser than or greater than median

[gusty pollen]

Do you only work with the binomial distribution? Or also with normal distribution or another one because the way to do it depends on which distribution you are using

@gilded portal

[gilded portal]

we havent really been taught with normal

is normal just approximation of binomial? like mean = np, sd = root npq and then continuity correction

[gusty pollen]

no the normal distribution is for quanitative variables while the binomial is for qualitative variables

can you show me an excersise or theory of your book that I can see how they teach it to you @gilded portal

[gilded portal]

we learnt it on rstudio

so just pbinom(minimum of + and -,n,0.5)

[gusty pollen]

so H0 = x, and H1 =/ x

and with the critical values

right?

@gilded portal

[gilded portal]

yeah

[gusty pollen]

you know how to calculate the critical values?

@gilded portal

[gilded portal]

critical value with Z yea

ok wait

[gusty pollen]

@gilded portal do you perhaps want a powerpoint from my school where there are examples

[gilded portal]

yea that would help

here there are 25 + signs

[gilded portal]

so in z calculation do i take 24.5-mean or 25.5-mean

[gusty pollen]

ahhhh you learn it this way

okay now i understand

here first you form you hypothetis H0=50

H1 =/ 50

then you take you values and each time the value is above 50 you write a +

[gilded portal]

H1>50 right?

[gusty pollen]

yeah sorry my bad didn't read it to good

if the value is below 50 you wright -

if it is 50 you ignore it

here if you do it with you values you get 24 +'s and 11 -'s

so n= 24 + 11

Under H0 the number of + signs follows a Binomial Distribution with n=35 and p=0.5 (from you H0 hypothethis)

Mean: 𝜇 = n*p = 17.5

σ = the root of [n*p(1-p)] = 2.96

Z-value is= (24- 0.5- 17.5)/ 2.96

= 2.03

then look up P(Z >= 2.03) = 0.0212

significane level= 0.05

so 0.0212 < 0.05

=> we reject H0

Conclusion: There is significant evidence at the 5% level to conclude that the median weight is greater than 50 kg.

@gilded portal do you understand it?

@gilded portal you still here?

@gilded portal hello

do you understand it

[pearl pond]

@gilded portal Has your question been resolved?

[pearl pond]

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[torn sky]

In these tasks, we're given the task to determine the original function, the first derivative, and the second. Now I know that when the y value of the first derivative is 0, that it's extremal point, if the second is 0, it means there is a turning point. Yet I find it quite hard to determine which one is which. Is there some trick to this?

[finite jolt]

Well blue cant be either original function or f’ because its increasing while the other 2 are zero

[wooden merlin]

"...when the y value of the first derivative is 0, that it's extremal point...", speaking of the original function.

[finite jolt]

so blue is f’’

[wooden merlin]

Where do you mean by this?

[safe prairie]

you should be able to see that when green curve is flat, red curve hits 0

[finite jolt]

At x=0

[torn sky]

Im not sure if my reasoning here is correct

but

wouldn't it be 1. blue 2. red and 3. green

mainly suspecting this as the blue function has 3 points where it hits the x axis

red has 2, green has 1

[wooden merlin]

All three functions bypass 0, including the blue one.

[wooden merlin]

Focus on red and blue in the interval [-2, -1] ish., look their relation.

[safe prairie]

that dont matter

[pearl pond]

@torn sky Has your question been resolved?

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[sonic hound]

Find any three positive integers A, B and C so that (C^2 - AB)/(A+B - 2C) is a perfect square.

[sonic hound]

A condition is that A is not equal to B and (C^2 - AB)/(A+B - 2C) is not 0.

[sonic hound]

I got that root(AB) <= C <= (A+B)/2

[toxic lichen]

hm

and how'd you get that?

i think i managed to cook up an entire family of solutions to this btw, though not exhaustive by any means

[sonic hound]

If C > root(AB), then we also want A + B - 2C > 0 to make the whole expression positive.
=> (A+B)/2 > C > root(AB)
If C < root(AB), then we want A + B < 2C to make the whole expression positive.
This gives (A+B)/2 < C < root(AB), which is absurd.

[toxic lichen]

ok yeah i guess that's true

here is my idea

start with $\frac{C^2 - AB}{A+B-2C} = n^2$ and do some algebraic fuckery to it

[jolly parrot]

Ann

[sonic hound]

Hi.

I am back

There is another condition:
n < A < 2n + 1 and the same for B.

[pearl pond]

@sonic hound Has your question been resolved?

[sonic hound]

<@&286206848099549185>

[ruby cargo]

So we have to show that $\exists a \neq b, c \in \bN. \sqrt{\frac{c^2 - ab}{a + b - 2c}} \in \bN$?

[sonic hound]

yes

i mean that's a condition

[jolly parrot]

@ruby cargo

[sonic hound]

What does E represent?

[ruby cargo]

Such exists

[sonic hound]

Also, with n < A,B < 2n + 1

Prove the inexistence or prove the existence of at least one.

[ruby cargo]

I'd start by trying, $a + b - 2c = c^2 - ab$ because then, you can just get 1 as the result of expression and 1 is in $\bN$ so yea

[jolly parrot]

@ruby cargo

[sonic hound]

1?

[ruby cargo]

well you just have to show that there exist such numbers right?

[sonic hound]

yes, integers

positive integers

[ruby cargo]

Yea

[sonic hound]

can you give the set?

[ruby cargo]

?

[sonic hound]

Like {A,B,C,n} in order

an example

?

[ruby cargo]

Actually Ann's way is more robus here

have you tried using it yet?

[ruby cargo]

Sorry

!noans

[pearl pond]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[sonic hound]

Ann told her answer before I gave the condition n < A,B < 2n + 1

[ruby cargo]

Fair, is the n in, n < a, b < 2n + 1 the same one in ann's equation

[sonic hound]

yeah, of course

we can make a bundle of inequalities purely in A, B and C.

[cinder thistle]

what is n?

[sonic hound]

See above.
n^2 = (C^2 - AB)/(A + B - 2C)

n is positive

[cinder thistle]

ok, you can prove there is no solution, use modulo n^2+n is a hint

use anns method

[sonic hound]

i will

gtg

thanks y'all

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[little terrace]

how did my teacher find j?

[dusty flame]

Trigonometry…

Try drawing a right triangle for ur vector T

This is an example of vector decomposition if u wanna look into it

[sour iris]

The horizontal component of the force vector T has the same magnitude as the force vector A.

[little terrace]

this is so confusing

how do i do it

[sour iris]

Hi. Could you send an image of your work?

I could walk you through the steps.

[little terrace]

well i've basically just done the first step of j

and sketched the graph

then i checked the answers and got confused

[sour iris]

So you can rewrite vector T as having vertical and horizontal components.

The vertical component of vector T should equal the objects weight. This is because the object is not moving at all in the vertical direction.

The horizontal component of vector T should have the same magnitude as vector A because the object is not moving in the horizontal plane.

[little terrace]

which is sin60 and cos60?

for i and j parts

[little terrace]

okay

[sour iris]

In this case, the hypotenuse is vector T.

[little terrace]

oh so you just do sohcahtoa

but you don't know the adjacent, do you?

or is it equivalent to a

[sour iris]

Yes it's equivalent to vector A

[little terrace]

okay

[sour iris]

Because the object is not moving horizontally at all.

[little terrace]

yes but how do you know what A is?

[sour iris]

So the forces need to balance out.

You need to find the vertical force first.

You know from the problem that the vertical component of vector T is equal to the weight of the object.

[sour iris]

The opposite side's length is equal to the weight of the object.

[little terrace]

okay

i understand these individual parts, how do i put it together?

so T = 10, sin 60 = A/10?

if T is the hypotenuse?

[sour iris]

The tension is not 100 N. The vertical component of the tension is 100 N.

You can imagine the tension force with two components: one vertical and one horizontal.

[sour iris]

Which is represented here.

[little terrace]

isn't the gravity 100N?

[sour iris]

The weight of the object is 100 N.

[little terrace]

the weight of the object is 10kg

or is that equivalent?

[sour iris]

W = m * g.

[hazy bear]

What is the force that's acting downward?

[sour iris]

W = 10 kg * 10 m/s^2.

W = 100 N.

[little terrace]

oh okay

[verbal whale]

Weight ≠ mass

[little terrace]

gravity

[hazy bear]

I mean the value of that force according to the figure

[little terrace]

100N

[verbal whale]

Weight is a force (due to gravity), whereas mass is a scalar quantity, not a vector

[hazy bear]

Now what the value of force acting upward

[little terrace]

no idea

-100N?

[hazy bear]

It is the vertical component of T

[little terrace]

how do i find that

[hazy bear]

What is formula for vertical and horizontal components of T if the angle is given? Do you know?

[little terrace]

no i dont

[sour iris]

The vertical component of vector T is equal to the weight of the object. They both balance each other out.

[little terrace]

this is what my teacher did

[hazy bear]

Consider hypotenuse is T and angle it makes with base is theta, now tell me what's sin theta?

Opposite side represent the vertical component and adjacent side represent the horizontal component of T

Now tell me sin theta

[little terrace]

can someone just help me do it please i dont know what you're talking about

[verbal whale]

Well if you don't know this honestly you won't be able to do anything related to inclined vectors

[hazy bear]

If you don't know basics of vectors, idk why you even doing this.

Ok let me tell you a trick

[little terrace]

so vertical T is 100N?

[hazy bear]

If any vector is given (say T) and the angle it makes with base is given (say theta) then
$$Horizontal Component of T = T sin{\theta}$$

[jolly parrot]

Shubham1029

[sour iris]

Correct. The vertical part of vector T is 100 N. Equal to the weight, and it cancels it out.

[little terrace]

uhh it's sin(60) *100?

[sour iris]

The object is stationary so they must cancel out

[hazy bear]

Yes

[little terrace]

ohh okay

[hazy bear]

No wait not 100*sin(60)

It's T sin 60

[little terrace]

so i got for sin(60)*100 = -30.48

[little terrace]

oh okay

um

[hazy bear]

Ok i am back

If upward force is T sin60 and downward force is 100 N and I tell you that net force is 0 then make an equation.

[little terrace]

j: Tsin(60) - 100 = 0

[hazy bear]

Yes

So find the value of T from here

[little terrace]

Tsin(60) = 100

T = arcsin(60) + 100?

[hazy bear]

No

arcsin gets multiplied

[little terrace]

oh okay

T = arcsin(60) * 100?

[hazy bear]

Yes

So T is

[little terrace]

is my calc supposed to be in radians or degrees?

[hazy bear]

Degrees

[little terrace]

okay

it wont let me do it

its a "non-real calculation"

[hazy bear]

Don't you know sin 60°

[little terrace]

what does that mean

[hazy bear]

What is your age?

[little terrace]

17

[hazy bear]

Huh alright i am getting this done fast now fr

[little terrace]

=100/srt3/2?

[hazy bear]

Yes

Now what's T

[little terrace]

200/sqrt3

[jolly parrot]

The following error occured while calculating:
Error: Syntax error in part "√3" (char 7)

[little terrace]

u have to do swrt

[hazy bear]

Yes

[little terrace]

okay

im kms officially

[jolly parrot]

The following error occured while calculating:
Error: Syntax error in part "\sqrt {3}}" (char 6)

[little terrace]

thank you so much!!

[hazy bear]

Whatever just use calculator

[little terrace]

my test is in 30mins 😵

okay thanks

.close

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[sour iris]

$\frac{200}{\sqrt{3}}$

[jolly parrot]

hello

[pearl pond]

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[lost umbra]

What exactly Is wrong with this

[lost umbra]

Dual enrollment stats

[pearl pond]

@lost umbra Has your question been resolved?

[lost umbra]

<@&286206848099549185>

[rough forge]

I don't think just because the donut hole is smaller it changes something, the ratios would still stay the same

[pearl pond]

@lost umbra Has your question been resolved?

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[iron basin]

im not sure on how to finish subdivision (g) in this question

[iron basin]

im locked on the first few steps, but got lost pretty quickly with the remaining steps

[pearl pond]

@iron basin Has your question been resolved?

[vital crescent]

i think 10 might help if you swapped x and y

-zQy ->-zQx or -xQy

[iron basin]

oh yeahh

it works

thanks

.solved

[pearl pond]

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[vital crescent]

np

[pearl pond]

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[sharp smelt]

I was thinking (x^3/3,0,0,0),(0,x^2/2,0,0),(0,0,x,0),(0,0,0,1)

[solid pier]

you want to find elements of P_3, so your idea is fine but the way you're writing the basis is wrong

[sharp smelt]

hmm

[solid pier]

like x^3/3 is a vector in itself

for example

[sharp smelt]

I can define an ordered basis

right

[sharp smelt]

yeah, I get that

[solid pier]

great okay. so an ordered basis is just going to be one list

[sharp smelt]

so (x^3/3,x^2/2,x,1)

[solid pier]

yes great. you also want a basis of P_2

as mentioned in the problem

[sharp smelt]

ah yes

I didn't read that bit

I think x^2/2,x,1 works

[solid pier]

yeah and it seems like you already had this in mind when you chose your basis of P_3

wait

not quite

i misread your answer lol it's sort of late

when you take D of the first basis element of P_3 (in your basis) what do you get

[sharp smelt]

3x^2

oops

x^2

[solid pier]

okay

and that's what the first element of your basis should be in P_2

because the first column of D is [1;0;0]

[sharp smelt]

hmm

right

x^2,x,0

so just x^2,x

[solid pier]

that's not a basis

you can do the same thing for the other ones. like when you take D of the second basis element what do you get?

and similarly for the third and fourth, what do you get?

[sharp smelt]

x,1,0

[solid pier]

right.

so the second basis element in P3 is going to be (0;1;0;0) and when you take D of that you will get (0;1;0)

which will be the second basis element of P_2

[sharp smelt]

yes

thanks

[solid pier]

ok great, so you know that x is the second basis element

and then for the third basis element, you know that (0;0;1;0) has to map to (0;0;1)

[sharp smelt]

yes

[solid pier]

okay so what is D of your third basis element

[sharp smelt]

0

[solid pier]

D(x)

[sharp smelt]

1

[solid pier]

yep

and that's a basis for P_2

[sharp smelt]

Cool

Thanks!

.close

[pearl pond]

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[sharp smelt]

Can I have a hint

[sharp smelt]

I was thiinking of starting off with a basis of V , then defining the basis of W as an extension of T(e_i)

or is that the wrong. idea

[solid pier]

the problem with this is that if you have a set of vectors that are linearly independent in the domain, they don't have to map to a set of linearly independent vectors in the codomain.

[sharp smelt]

yea, I'l have to deal with that

[solid pier]

like i'm sure that an idea like this can work but you can modify the idea a little bit to get something cleaner.

[sharp smelt]

I mean I think I need to start with a basis of V

right

[solid pier]

that is, ||start with a basis for range(T)||

[sharp smelt]

rank nullity I guess

will help here?

[solid pier]

yeah this is another way to modify the basis of V idea

like if you extend a basis for the kernel of T to the rest of V then the rest of the vectors in the basis will map to something nonzero in range(T)

[sharp smelt]

Let dim(V)=n, dim(range T) =m. Then $dim(Ker(T)) = n=m$. We start off with a basis of $ket(T)$ $e_1,e_2,\dots e_{n-m}$. We extend this to a basis of $V$. $e_1,e_2,\dots, e_{n-m} , e_{n-m+1}, \dots, e_n$.

[jolly parrot]

wai

[toxic lichen]

Let dim(V)=n, dim(range T) =m. Then dim(Ker(T)) = n=m.

[solid pier]

n-m

[toxic lichen]

[sharp smelt]

TYPO 😭

[toxic lichen]

typos, typos, typos...

[solid pier]

okay, this is a good start. now what do you think will happen when we push these basis vectors through T?

[sharp smelt]

I can get a basis of Im(T) from this and then extend that

[solid pier]

right, so you first have to show that these guys form a basis for Im(T). you already know that the image of your basis spans Im(T) because the basis itself spans V. so all you have to do is show that the vectors are linearly independent

[sharp smelt]

What I was thinking of doing is define a basis by constructing it recursively ( not sure if it's the right word here)

[solid pier]

i don't really understand which basis you're trying to construct

[sharp smelt]

that of W

[solid pier]

you mean the extension from the one that you got from Im(T)

?

i mean i have no doubt that you're going to be able to do that without a problem. and yes, you can do that inductively

please use a different help channel, this one is occupied.

[solid pier]

i just want to make sure that you understand how to use rank-nullity to show that pushing the non-kernel vectors through T gives you linearly independent vectors in W

[sharp smelt]

hmm

wait, so what exactly should I do

[sharp smelt]

pushing the non-kernel vectors via T doesn't HAVE to give LI vectors

does it

[solid pier]

they do by rank-nullity and the way we constructed our basis of V

[sharp smelt]

like say V is 10 dimensional, withe kernel 5, and the rest of the vectors mapping to a 1D vector space

hmm, but by rank nullity that isn't possible

is it

[solid pier]

right rank-nullity says that the dimension of the image would have to be 5

[sharp smelt]

yes

but 6 vectors could map to those 5 vectors

right

[solid pier]

yes, okay so hear me out

[sharp smelt]

okay

[solid pier]

you have $e_{n-m},...,e_n$ mapping to 0, and $e_1,\hdots, e_n$ mapping to nonzero vectors in $W$. but since the dimension of Im(T) is n, and the $T(e_1),\hdots, T(e_n)$ span Im(T), then we know that they must be linearly independent

[jolly parrot]

smay

[solid pier]

sorry for the terrible tex

we have n vectors that span an n-dimensional subpsace of W

[sharp smelt]

yes

[solid pier]

so they must be linearly independent

by the definition of dimension.

or whatever other lemma that you have for dimension.

[sharp smelt]

let me try to write a proof?

[solid pier]

if you have n vectors that span an n-dimensional space, they have to be linearly independent. sure you can go ahead and prove this fact if you want to.

[sharp smelt]

I prove it earlier, it's fine

[solid pier]

okay great

so now what we have is that T(e_1), ... T(e_m) are a set of linearly independent vectors in W

and they span Im(T), and furthermore, the other basis vectors that we have map to 0

[sharp smelt]

wait, one minute

I'm kind of confused, so the rest of the n-m vectors map to

[solid pier]

my indexing is wrong

sorry

but

[solid pier]

the other n-m vectors were in the kernel so they map to 0

[sharp smelt]

n-m vectors map to a m D vector space

[solid pier]

i am thinking of e_1,...,e_m as the non-kernel basis vectors

no

[solid pier]

m vectors do

n-m vectors mapped to 0

[sharp smelt]

ah yes

Let dim(V)=n, dim(range T) =m. Then $dim(Ker(T)) = n-m$. We start off with a basis of $ket(T)$ $e_1,e_2,\dots e_{n-m}$. We extend this to a basis of $V$. $e_1,e_2,\dots, e_{n-m} , e_{n-m+1}, \dots, e_n$. We the define a basis. We then define $T(e_{n-m+1})=f_1, T(e_{n-m+2})=f_2,\dots, T(e_m) = f_m$. We then extend this to form a basis of $W$.

[solid pier]

n-m

[jolly parrot]

wai

[solid pier]

this looks good

so now you have to show that the matrix does what you want

and then reorder your basis so that it's right

[sharp smelt]

How do I show this gives the desired matrix, I just explicitly write it out?

[solid pier]

well, you know what your basis vectors map to

like what does e_1 map to

that will be the first column of T in the basis that you chose

similarly for all the other guys

[sharp smelt]

yea

[solid pier]

in your definition of the bases e_1 maps to 0 right

so the first column would be (0;...;0)

since e_n-m+1 maps to f_1 then the n-m+1 column will be (1;0;...;0)

[sharp smelt]

yes

got it

thanks!

:D

ngl axler's matrx problems are quite fun

compared to other books

[solid pier]

matrices are pretty cool

people don't like them for some reason

i really think they're neat

[sharp smelt]

We aren't introduced to them properly

[solid pier]

how would you change the way that we are introduced to them

[sharp smelt]

They should first be introduced in the context of linear maps

not in terms of a way to solve equations by guassian elimination

[sharp smelt]

with heavy motivation

[solid pier]

ah, yeah. i agree

[sharp smelt]

Like I hated , absolutely HATED that part of LA

[solid pier]

they are sort of different pictures of the same object in a sense

[sharp smelt]

yea

[solid pier]

i mean gaussian elimination takes on a whole different picture when you think about it as modifying a map

[sharp smelt]

yea, but it isn't taught that way

[solid pier]

i think my linear algebra class was pretty good in the sense that we covered both sort of at the same time.

i am pretty fortunate.

[sharp smelt]

maybe we will in LA 2

which I happen to be taking a year early ! ( this coming sem(

[solid pier]

nice :0

[sharp smelt]

I'm on my summer break rn

[solid pier]

yeah LA courses that are not a first course really emphasize the linear transformation part.

it's good

[sharp smelt]

I was fortunate enough to personally follow axler for my first course though we used another book in class

was pretty nice

[tropic saddle]

exercise: prove the same claim by gaussian elimination

the reason gaussian elimination is often taught first is because its so immensely powerful

[solid pier]

i love gaussian elimination

[tropic saddle]

you can even apply it to questions like this

[sharp smelt]

hmm

[sharp smelt]

for that don't I first need a matrix

[tropic saddle]

yes

just pick any bases for V and W

you need some knowledge on elementary matrices for this approach. and that invertible matrices and basis changes are the same thing

[sharp smelt]

the latter part I do not know

or atleast do not recall

[tropic saddle]

given any invertible matrix A and any basis B, you can see A as a basis change from B to some other basis or from some other basis to B

prove that first

[sharp smelt]

Can I do this in like 2 days, once I warm up to the idea of matrices a bit more, I've never really personally used them conceptually before

I'll be sure to do this then, I'll make a note

[tropic saddle]

do it whenever

[sharp smelt]

Thanks

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[pearl pond]

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[gilded portal]

yes yes

[gilded portal]

sorry i had to go

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

How to do q7

[autumn fossil]

calculate the straight lenght and curved length separately

the curved length is just certain fraction of the circumfence of whole circle

and the straight length can be calculated with some trigonometry

[midnight haven]

The chord and the arc length?

For wht?arc length?

I calculated fhe arc length and chord and then added it but im not getting the correct answer

[oak ivy]

you got the line length?

for perimeter add the line + 10

[midnight haven]

What i did

[oak ivy]

perimeter of this segment meaning the perimeter of the entire circle shown?

ok

[midnight haven]

Ig so

The whole perimeter i need to get

[oak ivy]

The angle would be the other one right

cause by 120 we find the other segment

[midnight haven]

So take it as 240?

[oak ivy]

That's what I think

[midnight haven]

I did that also

[oak ivy]

And add that line

[midnight haven]

I took the arc length as 240 but i got a bigger answer for arc length

Answer is supposed to be 29.6 cm

Maybe i got the chord wrong?

For arc length

[oak ivy]

Also

don't u need to add

that line

[midnight haven]

Which line?

[blissful salmon]

Arc length at the bottom is correct. It’s probably the chord length that’s wrong.

[oak ivy]

yea