help-39

[light helm]

if you were to combine
$$\frac{C}{x+1} + \frac{D}{(x+1)^2}$$
you'd get
$$\frac{Cx + (C + D)}{(x+1)^2}$$
which is of the form
$$\frac{Mx + N}{(x+1)^2}$$

[jolly parrot]

ℝαμΩℕωⅤ

[daring wedge]

If you look at the last step, the degree of above is one less than the below one

That's what I imagined. In this too

[visual hazel]

Example: (x³+3)/(x²+1) = x + (3-x)/(x²+1)

Here u go
U can take out x and have the degree one less now

[daring wedge]

What niw

I'm lost

Som1 help pls

[pearl pond]

@daring wedge Has your question been resolved?

[daring wedge]

<@&286206848099549185>

[azure plume]

what exactly is your question?

@daring wedge

[daring wedge]

Pls check

Am I doing right

[azure plume]

1 sec let me look it over

2-1/2+0=4?

?

[daring wedge]

It's D lmao

Sorry

[azure plume]

oop

[daring wedge]

Not 0

Hehe

[azure plume]

yep ur good @daring wedge

[daring wedge]

Thank you sire

Very very much

[azure plume]

.close

[daring wedge]

I needed this

[azure plume]

wait can u do .cose

*.close

[daring wedge]

Ofc

Dotclose

Dot close

[azure plume]

no like

[daring wedge]

.calose

.shut

[azure plume]

close

use .close

[daring wedge]

Use .close

Whaaaaaat

Close

Dorclose

[azure plume]

like this

.close

[daring wedge]

Door close

TarryTheCow
. close

[azure plume]

😭

[daring wedge]

.C lose

.cl ose

[azure plume]

leave it its fine

[daring wedge]

Lmao

I'm playing

Hehehehe

.close

[pearl pond]

Channel closed

Closed by @daring wedge

Use .reopen if this was a mistake.

[daring wedge]

Have a good day

Or night

[azure plume]

k bye

[daring wedge]

Thanks

[azure plume]

.solved

frick

@daring wedge

[daring wedge]

What?

.solved

[azure plume]

im acc gon jump out a window lmfao

[daring wedge]

But I closed it

Hehe

.close

.reopen

[pearl pond]

✅

[azure plume]

nm

[daring wedge]

Bwahahahahahha

[azure plume]

yep

[daring wedge]

. Close

Dot close

Door close

Door 🚪

.close

[pearl pond]

Channel closed

Closed by @daring wedge

Use .reopen if this was a mistake.

[daring wedge]

Bye bye cow

I'm not a vegan so be scared of me

[pearl pond]

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[autumn fossil]

correct

mhm

,w 3*(2/3)^3 + 4*(2/3)^2 - (2/3) - 2

hm

oh wait lol

[jolly parrot]

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[autumn fossil]

Q5 is correct

[inland ivy]

Question 6 also seems correct to me

56 can’t be a multiple of 3 so a and b must be wrong

[autumn fossil]

oh i thought he answered b for some reason lol

yeah, its correct

[willow marsh]

Alright thank you guys

[pearl pond]

@willow marsh Has your question been resolved?

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[wintry bobcat]

i'm lost

[crude echo]

!show

[pearl pond]

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

[crude echo]

Have you sketched f(x)

What is domain here?

[robust oar]

do you get the f^(-1) and domain ?

[pearl pond]

@wintry bobcat Has your question been resolved?

[wintry bobcat]

no i dont understand where to start

how would i sketch it

[crude echo]

Okay

[wintry bobcat]

okay i think its a v shape

right?

and plus one is right one?

[crude echo]

What are the possible values of x here?

[wintry bobcat]

all real numbers, and the y is always positive

[crude echo]

I suggest you read the que. Again with bit of a care

[wintry bobcat]

oh okay

only 0 and above

[crude echo]

Yea

[wintry bobcat]

sorry

[crude echo]

Yeah

[crude echo]

Nw

So does that tell you anything

[wintry bobcat]

well

i know the positive of x in absolute value

is this area

on the right

and the plus one would move it to the right i think

[crude echo]

Yea

So your function would just be?

[wintry bobcat]

{1, infinity)

?

[crude echo]

Yeah

But in terms of x?

I mwan f(x) = ..

[wintry bobcat]

i'm confused

[crude echo]

I mean

[wintry bobcat]

this is the domain right

[crude echo]

No

Domain is x>=0

[wintry bobcat]

oh then that's the range i mean

wait no

[crude echo]

Yea

[wintry bobcat]

okay

[crude echo]

Yep that is the range

[wintry bobcat]

so now i have to find the inverse

[crude echo]

Yea

Can you write the function in terms of x?

[crude echo]

I mean you know how to do this right

[wintry bobcat]

yeah switch f(x) and x

[crude echo]

Yea

[wintry bobcat]

f(x)= |x| +1

[crude echo]

Can we drop the mod ?

[wintry bobcat]

like

this?

f(x)= |x|

[crude echo]

No

[wintry bobcat]

whats mod

[crude echo]

Absolute valu

Value

[wintry bobcat]

why would we remove it

f(x)= x+1

[crude echo]

Yeah

[wintry bobcat]

okay so

[crude echo]

We can write this

But why ?

[wintry bobcat]

idk why we removed the absolute value

let me think

[crude echo]

Okay

[wintry bobcat]

because there's no negative x

[crude echo]

Righttt

[wintry bobcat]

okok

so

x= y + 1
x - 1 = y

[crude echo]

Yep

What

f(x)=x+1

What is the inverse function

[wintry bobcat]

x=f(x)+1

[crude echo]

How did you got this

[wintry bobcat]

i switched

them

isnt that how you get inverse

[pearl pond]

Channel closed

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[crude echo]

.reopen

[wintry bobcat]

.reopen

[pearl pond]

✅

[wintry bobcat]

okay go on

[crude echo]

I mean

You have to let x be on one side of equation and everything else on the other side

Can't just switch f(x) and x

[crude echo]

Can you explain what did you do here?

Wiz are you following?

[pearl pond]

@wintry bobcat Has your question been resolved?

[warm sentinel]

An equation for f^-1 would mean a piecewise function

[pearl pond]

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[timid isle]

basically, you change f(x) to y

y = x + 1

now rearrange for x
x = y - 1

now rewrite x as f^-1(x) to say the result of the equation is the inverse function

and change y to x

f^-1(x) = x - 1

[timid isle]

mb bot its not my help channel

[crude echo]

I was helping him

He left

[crude echo]

Close the channel :3

[timid isle]

wait fuck

I thought giraffe asked for help

[crude echo]

Noep

[timid isle]

I scrolled more up

!close

[crude echo]

Dot close

[timid isle]

.close

[pearl pond]

Channel closed

Closed by @timid isle

Use .reopen if this was a mistake.

[wintry bobcat]

sorry i was doing some more

[wintry bobcat]

well after switching x and y

i tried solving for y

[crude echo]

Ohh

So what is the inverse function now?

You need help right?

[wintry bobcat]

yeah

[wintry bobcat]

f(x)^-1 = x - 1

that's my answer

[crude echo]

Ywah

Rightt

I mean correct

[wintry bobcat]

okay so

domain is the same

and the range is what

[timid isle]

f^-1(x) ≥ 0

f^-1(x) = x - 1

so d oma in is x ≥ 1

the domain of f(x) is the range of f^-1(x)

the range of f(x) is the domain of f^-1(x)

[wintry bobcat]

ohh

i didn't know you could do that

like this

[timid isle]

yea u gotta remember it but it's easy to remember

[wintry bobcat]

alright ill keep that in mind from now on

[pearl pond]

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[chilly slate]

Hi, I have a question about parametrising a sphere into cilindrical coordinats the general question is not that important I just want to know if I'm doing it correctly:
Let a Sphere be defined by the equation: $$x^2 + y^2 +z^2 = 2z$$

[jolly parrot]

Mephisto

[chilly slate]

so... I let $$x^2 + y^2 = u^2$ because x = ucos(t) and y = usin(t) which results in $u^2 + z^2 =2z$$\

[jolly parrot]

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

[chilly slate]

but I'm not sure what to do next? do I solve for z or...

[grim fractal]

Yes

[chilly slate]

ah alright, so this will give me the parametrisation?

or for the z component atleast

[grim fractal]

Yes

[chilly slate]

okay, and let's say I'd want to take a volume integral of the volume between a code and this sphere... how do I do that?

I mean how do I choose my boundries?

the question specifices me to use a triple integral, but I have only two variables to work with

$$\integral{}{}\integral{V}{}\integral{}{} dudt$$

[jolly parrot]

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

[fathom oasis]

that does not form a sphere

[chilly slate]

It should, It's a sphere with z=2 as center point

[fathom oasis]

plot it and see

,w plot x^2+y^2+z^2-2z=0

[jolly parrot]

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Requested by stipendi

[fathom oasis]

ah crap you're right lol sorry

[chilly slate]

I'm still clueless on the integral part, so if you have any ideas I'm all ears

I'll try to plot the problem hold on

I have to calculate the area between the cone and the sphere using cilindrical coordinates and a triple integral

the cone is a basic x^2 + y^2 =z^2

[fathom oasis]

cylindrical coordinates contain three different parameters

there are two angle parameters and one radius parameter

and the absolute value of the determinant of the jacobian is r^2 * sin(the north pole angle)

[chilly slate]

ah okay

so lets say I have
$$\begin{cases}
{x = ucos(\theta)}\{y=usin(\theta)}\{z=\text{(still have to solve for it)}}
\end{cases}$$

[jolly parrot]

Mephisto

[chilly slate]

there isn't anything for the north angle

same goes for the cone

$\iiint_V r^2sin(\phi)d\theta d\phi du$

[jolly parrot]

Mephisto

[chilly slate]

is it simply this? with 0 and 2pi as bounds for phi?

or do I have to calculate the bounds based on the intesection points?

in this case it looks like pi/4

[pearl pond]

@chilly slate Has your question been resolved?

[chilly slate]

<@&286206848099549185>

[chilly slate]

.close

[pearl pond]

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[pearl pond]

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[fierce kestrel]

was trying to figure out if this was the right direction to go in

i know that if it was just sqrt(4n^3), then i can leave it as 2n(sqrt(n))

but this is sqrt(4n^3-1), so it must not work, so im not sure what exactly is supposed to happen, if what ive been doing isnt working

i did try this (on the left), but i dont think thats the right direction either (evident by the fact i kinda just paused and didnt bother to finish where i was at)

[sterile tusk]

the left is good

you’re comparing the series to the series of $\sum_{n=1}^{\infty}\frac{1}{\sqrt{4n^3}}$

[jolly parrot]

y0shi

[sterile tusk]

which we know converges by p-series

[fierce kestrel]

yeah, i know that it converges

[sterile tusk]

mhm so then just carry out the limit

if it equals a positive finite number

then both series must converge or both diverge

by the limit comparison test ofc

[fierce kestrel]

oookaay... uh, lemme see how far i get it then

... this is gonna be a whole bunch of L'Hopital rules...

[sterile tusk]

well not if you know the general rule for rational functions

generally for the limit as n approaches infinity of some rational function

if the degree of the numerator is greater than the degree of the denominator, it’ll approach infinity

if it’s less, then it’ll approach 0

[fierce kestrel]

...hm, okay, so with, that in mind

[sterile tusk]

if they’re equal, then it’ll approach the ratio of the highest order coefficients

[warm sentinel]

take the 1/2 out!!!!

[sterile tusk]

mhm so this limit would equal?

[fierce kestrel]

O_o

[sterile tusk]

it matches with the other series we’re comparing it to

less confusing to keep it inside

[fierce kestrel]

okay so if the limit is just sqrt(4n^3)/sqrt(4n^3-1)
whatever value n would be, the denominator is still less then the numerator

[sterile tusk]

mhm

but would that -1 matter

as n gets really really big

[fierce kestrel]

it shouldnt

[sterile tusk]

yep so we can ignore it

[fierce kestrel]

its barely a difference, its just 1

[sterile tusk]

as n gets really big

so really it’s just the same as $\lim_{n\to\infty}\frac{\sqrt{4n^3}}{\sqrt{4n^3}}$

[warm sentinel]

engineers be like

[jolly parrot]

y0shi

[sterile tusk]

and see how they cancel out

[fierce kestrel]

so it'd be equal to 1, they cancel out

[sterile tusk]

yep that’s it

[fierce kestrel]

and since 1 isnt 0 it converges by LCT

[sterile tusk]

yep

[fierce kestrel]

[sterile tusk]

more so that it’s a finite positive number but

that should work

[fierce kestrel]

i sure the heckerdoodle hope so

hopefully the ratio test stuff i gotta do wont go nearly as bad as this did

thank you

.close

[pearl pond]

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[pearl pond]

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[unborn token]

I am having trouble evaluating a limit. lim as r→9 of √r over (r − 9)^4. I have asked chat gpt and it just subsitutes random values to make it make sense and watched a video where they simply subsituted 1 into the top but that didnt make any sense to me. can someone help me please?

[eternal tulip]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[eternal tulip]

you're trying to find $\lim_{r \to 9} \frac{\sqrt{r}}{(r-9)^{4}}$?

[jolly parrot]

🫎 Moosey 🫎

[unborn token]

yes

[deft lynx]

I mean it's not that convenient, but multiple applications of L'Hospital could help

[eternal tulip]

You should be able to note something immediately if you plug in 9, but if you want to be more rigorous, you'll need to find $\lim_{r \to 9^+} \frac{\sqrt{r}}{(r-9)^{4}}$ and $\lim_{r \to 9^-} \frac{\sqrt{r}}{(r-9)^{4}}$

[jolly parrot]

🫎 Moosey 🫎

[eternal tulip]

this is not of the form of lhopitals rule.

[deft lynx]

Oh, sure, that's right, sorry 💀

My bad

[unborn token]

does the limit not exist then?

[deft lynx]

3:30 am here, i think I'm gonna hit the sack

[eternal tulip]

if infinity or -infinity are not valid answers, then yes, the limit does not exist

[unborn token]

haha gn thanks for the advice

[unborn token]

but if they are?

[eternal tulip]

then you would look at both of the cases listed up here

[unborn token]

ah ok I was being stupid. thanks for the help

[eternal tulip]

you understand what lim r->9^+ and r->9^- means yes?

[unborn token]

yes

so for 9+ it would be positive infinity and then for 9- negative infinity?

[eternal tulip]

no

[unborn token]

damn

lol

wouldnt 9+ make the denominator obsolete making it infinity?

[eternal tulip]

you are right in that limit as r approaches 9 from the right (9^+) is positive infinity

and you are right in thinking 9^- -9is a negative number

[unborn token]

but because it is ^4 it is positive?

[eternal tulip]

mhm

so it becomes positive infinity

[unborn token]

i see

thank you very much for the help and helping work through it

[eternal tulip]

so you could say the limit is positive infinity, if allowed, but if not, then DNE is fine

[unborn token]

alright, thanks once again

do i have to do anything to close this?

[eternal tulip]

yeah you can just do .close

[unborn token]

.close

[pearl pond]

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[pearl pond]

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[cinder ledge]

help

[pearl pond]

@cinder ledge Has your question been resolved?

[pearl pond]

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[tulip shale]

i need help

<@&286206848099549185>

[warm current]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[tulip shale]

Sorry

I’m just in a hurry

.close

[pearl pond]

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[vague fiber]

How do you find critical values with a significance level 0.01

[vague fiber]

Does this mean that alpha is .01?

[cosmic charm]

yeah

though usually it should be in 1 - alpha

[vague fiber]

So if alpha is 0.01 and my alternate is >

Then i just look at 0.01 without dividing

[cosmic charm]

yep

[tulip shale]

I have already submitted it

[vague fiber]

Do i use t or z test

Table

[cosmic charm]

you know the population var/sd?

if yes use z

if not use t

[vague fiber]

I know the pooled variance

[cosmic charm]

is this two samples?

[vague fiber]

Its a paired t test

Wait no i dont think thats pooled variance actually

I have SD for the bottom of the equation to solve for the test statistic

To be safe ill use t table

Is it just 0.01 and sample size -1

[cosmic charm]

yeah

[vague fiber]

Thank you

[pearl pond]

@vague fiber Has your question been resolved?

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[cunning vault]

can someone tell me how to do this question?
x^x=3^(x+9)

[tulip ore]

you here?

[cunning vault]

yeah

[tulip ore]

are you in the mood to guess?

theres no way you can directly solve for x in x^x = 3^(x+9)

the best you can do is to prove that there is only up to 1 solution (which is bigger than 1)

[cunning vault]

ohh i see thank you

[tulip ore]

if you want to guess, you should aim for a power of 3

[cunning vault]

I do until x(ln(x/3))=9ln3 and then get stuck

[tulip ore]

with x ln(x/3) = 9 ln 3,

can you guess x as a particular power of 3 which would satisfy the equation

[cunning vault]

yeah ， thank you

[pearl pond]

@cunning vault Has your question been resolved?

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[rose sentinel]

What did I do wrong

[hallow raven]

check your anti derivatives

[rose sentinel]

Anything in particular?

I don’t see it

[hallow raven]

the anti derivative of (e^2x)/4 is (e^2x)/8

also the 1/2 part isn’t integrated

[rose sentinel]

??

Huh

Oh

Wait

Yeah I just took the derivative 💀

[hallow raven]

dw it happens a lot lol

at least to me

usually i just differentiate my anti derivative to check

[rose sentinel]

Me too lol

I forgot

Ok I got it right thanks

😁

.close

[pearl pond]

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[daring valley]

i dont understand square roots, what's the square root of 144

[rose sentinel]

What times what is 144

[daring valley]

idk

i dont go to school

[rose sentinel]

Well there’s lots of resources online to help you

[rose sentinel]

Do you understand the concept of squaring a number?

[west sapphire]

if nothing else you can brute force it

[daring valley]

nvm ill just use a calculator

[rose sentinel]

ie $2^2$ =?

[jolly parrot]

Nathan

[rose sentinel]

Don’t cheat your way out of arithmetic it’s good to know these basic calculations

[daring valley]

nvm i got the answer its 12, thanks guys

.close

[pearl pond]

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[pearl pond]

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[vapid bough]

Hey

[vapid bough]

Is the tangent defined for the point A

[muted remnant]

no

How do you define a tangent, too?

[vapid bough]

You take a point at the right and left of the point at which you wish to have your tangent

draw a secant to both those points

[muted remnant]

That's the relevant concept

[vapid bough]

and move the closer and closer

Yeah not much of a defination

[muted remnant]

I am asking more formally, if you know about the limit definition

[vapid bough]

Nope

Could you tell me?

[muted remnant]

Do you need to know?

[vapid bough]

yes

[muted remnant]

I'll copy from wikipedia

[vapid bough]

oki lol np

But before that

[muted remnant]

[L = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}]

[vapid bough]

The doubt was not this, I knew the answer to this

[jolly parrot]

fishwhale

[vapid bough]

This is just the derivative

[muted remnant]

If this exists at a the function exists at a

Yes the slope is the derivative

[vapid bough]

ah

[muted remnant]

Tangents at points are derivatives at points

[vapid bough]

oh ok

One curve can only have one tangent at a point?

[muted remnant]

For it to be well-defined yes

[vapid bough]

ok and if it has two, then we say that, that tangent at that perticular point it not well defined?

[muted remnant]

At A it depends a little on what you mean by tangents

[muted remnant]

Yes, 2 or more = not a single one = not well defined, I think this is good equating for now

[vapid bough]

Alright thank you for clearing that up

Just one more thing

[muted remnant]

I want to expand on this because the left-side tangent is likely infinity, that's actually still considered existing and well-defined enough, since infinity is a solidly defined concept
Additionally the right side is just some downward slope so some finite negative number.
We know (\infty \neq \text{some finite negative number})

[vapid bough]

How do you know for sure that this point A has two tangents

You could say that there is a gap in the function

but the gap is a point

[jolly parrot]

fishwhale

[muted remnant]

What is the formal definition of continuity?

[vapid bough]

I haven't got there yet, I was just on tangents

[muted remnant]

ah I see

[vapid bough]

I was following the method where you draw a secant to two poinits from left and right and make them appoach the center point

so the line that those two secants approaches is the tangent is what I tought

[muted remnant]

Ok I need to back track a bit

It is correct to say that a function is not differentiable at a point if the left and right derivatives do not agree.
Here it is not as meaningful because a function not defined at a point means it cannot be differentiable at that point

Oh wait

Ahhhhhh I'm being slow

[vapid bough]

np

🙂

[muted remnant]

Ok

1. function not defined at a point -> function differentiability at that point undefinable
2. function whose limits do not agree with that point -> function not differentiable (you may be asked to prove this)

So this is case 2 of that

[vapid bough]

oh ok

[muted remnant]

And the formal way of saying it is, a function that is not continuous at a point cannot be differentiable at that point

Basically continuity is agreement of limits at points with functions values at those points

[vapid bough]

ahh, so if the limit is same from both the sides, we say that the function is continues at that point

?

wait say there was a physical gap between the functiono

Like this

[muted remnant]

yes

[vapid bough]

then there would be some angle theta between those two tangents right?

[muted remnant]

This is continuous (at every point, including A)

Just not differentiable (at A, and the sudden angle switching points)

[vapid bough]

But if you approach the limit from left at A and from right, you will get differrent values

because I am saying that there is a physical gap between those two curves

[muted remnant]

No, with positive angle theta

I agree with you if theta is exactly 0

[vapid bough]

wait, those two lines that you see coming down are tangents that I drew

[muted remnant]

Yes so if your theta is strictly positive

[vapid bough]

They are not part of the function

[muted remnant]

Then the limit at A is exactly A

Ok sorry that's not coming through in the drawing

But do you get what I mean?

[vapid bough]

Yes my bad

[vapid bough]

yeah I got that

[muted remnant]

Niceee, I think that should give more intuitive sense as to what is going on too

And continuity/diff in 1D should be intuitive

[vapid bough]

Yes but that was not by doubt lol

lets call those two tangents t1 and t2 (again the function is just the curve, those lines coming down are tangents)

[muted remnant]

yes

[vapid bough]

in that case, the tangent would be not defined as you would have two tangents obviously as you can see

[muted remnant]

Yes let's call them left and right tangents (fact that we are working in 1D lets us do this, we are literally exhausting all cases)

[vapid bough]

but as theta goes to 0 or in other words, the gap between those two curve decreases, the tangents two would come closer?

[muted remnant]

And yes they need to agree exactly

[muted remnant]

Aaaaaaaaa

I see your doubt

[vapid bough]

Is this picture

[muted remnant]

Ok no, that's not how we would define a tangent in that case

[vapid bough]

the gap is infinitly small

because we have only take out one poinit

right?

[muted remnant]

yes

I know what you are seeing now

[vapid bough]

yeah

[muted remnant]

I don't know what to say about that intuition other than it goes against the definitions

The 'tangent' not definable at A in the case of theta=0, would be a flat tangent

[vapid bough]

So the teacher I was seeing this from just said, the tangent is undefined and moved on, but ideally we would have to rigorously prove that the tangent is not defined because this way, it seems that it is

right?

[muted remnant]

Well actually that's probably not true, I don't know. I think it's safer to say the tangent is not definable at A if you get theta=0

Yeahhhh I'm giving up like your teacher

Because the limit in the case of theta=0 is exactly the meeting point, and NOT equal to A. So we know the function is not continuous, and non-continuity -> non-diff

[vapid bough]

makes sense

Thank you for your help

🙂

[muted remnant]

I think

For your 'wrong intuition'

There's always value in wrong intuition btw

It's just a matter of like, making it work

[vapid bough]

oh yeaah, that is how you realize the need of rigour

[muted remnant]

I think you just need to try weaker definitions and see if any of the tangents stick around

But for that you need a lot higher math :)

[vapid bough]

yeah

I can't wait to start real analysis

lol

[muted remnant]

So yeah, for now stick to simpler math and say the tangent is not defined

[vapid bough]

k

good talk

[pearl pond]

@vapid bough Has your question been resolved?

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[halcyon whale]

ABCD is a parallelogram
AE, BE, CG, and DG bisect the internal angles
Prove that EFGH Is a rectangle

[halcyon whale]

Anyone have an idea how to solve this?

I think I need to figure out that BFC = 90 Degrees And AHD = 90 Degrees

[pearl pond]

@halcyon whale Has your question been resolved?

[halcyon whale]

<@&286206848099549185>

[pearl pond]

@halcyon whale Has your question been resolved?

[tulip ore]

thats correct

now as a hint, consider that what BAD + ADC is, knowing ABCD is a parallelogram

[halcyon whale]

okay

90?

[tulip ore]

do you see angle BAD?

[halcyon whale]

yes

[tulip ore]

its an obtuse angle, right?

[halcyon whale]

yeah

[tulip ore]

that means its bigger than 90 degrees, right?

[halcyon whale]

yeah

[tulip ore]

so whats BAD + ADC?

[halcyon whale]

180?

[tulip ore]

do you know why BAD + ADC is 180?

[halcyon whale]

because its an parallelogram

[tulip ore]

you said 90 last time you know

are you sure you know why?

[halcyon whale]

oh

no

oh i know

because a + d = 180

i don't know how its called

this theorem

[tulip ore]

dont bother with the name

[halcyon whale]

alright

[tulip ore]

it sounds like you just memorized that its true

thats not good

but itll work for now

since you know BAD + ADC is 180,

[halcyon whale]

yeah

[tulip ore]

split this sum up into BAH + HAD + ADH + HDC

now is BAH = HAD?

[halcyon whale]

i think

but why

oh its because AE, BE, CG, and DG bisect the internal angles?

[tulip ore]

thats the reason

AE is the same thing as AH

so AH splits BAD in half

BAH and HAD are those halves

[halcyon whale]

how do you know that AE is the same thing as AH

[tulip ore]

what do you think

[halcyon whale]

mhm

you mean that AH = HE?

how AE is the same thing as AH

[tulip ore]

AE is the same line as AH

[halcyon whale]

oh

right

[tulip ore]

you know we're talking about angles

we're not talking about sides

so AH is the same line as AE

[halcyon whale]

ohh

[tulip ore]

AH is not the same segment as AE

[halcyon whale]

alright

[tulip ore]

now you know HAD is half of BAD

similarly ADH is half of ADC, right?

[halcyon whale]

Because of his slope?

[tulip ore]

no

[halcyon whale]

I don't really get what you mean

[tulip ore]

🤔

when you place three letters in a row like ADH,

what does that mean?

[halcyon whale]

They are equal

Oh

[tulip ore]

youre saying that when I type ADH, it means "They are equal"?

[halcyon whale]

The same angel

[tulip ore]

ADH

I am only saying

ADH

[halcyon whale]

You mean the angel right?

[tulip ore]

what does the three-letter ADH mean by itself?

[tulip ore]

yes

[halcyon whale]

Like the angel

yeah

i know

[tulip ore]

its spelled angle

not angel

you didnt

now I say ADH is half of ADC

[tulip ore]

you implied you knew what this meant

[tulip ore]

use what you know to see that this is true

[halcyon whale]

mhm

I don't know

[tulip ore]

do you know what this sentence means

AE, BE, CG, and DG bisect the internal angles?

[halcyon whale]

I think

LIke H1 = H2?

And F1 = F2

[tulip ore]

please stop making up names

without labelling a corresponding picture with those names

be professional

you didnt even say D1 = D2

we're talking about angle D

cmon

[halcyon whale]

AHD = GHE

Oh

[tulip ore]

also no

those are vertical angles

they are equal for a different reason

[halcyon whale]

So its means that AHD = CDH

[tulip ore]

you misspelled ADH

[halcyon whale]

Oh mb

ADH

[tulip ore]

so since ADH = CDH

[halcyon whale]

Yeah

So they are equal

[tulip ore]

and ADH and CDH are both part of ADC

[halcyon whale]

right

[tulip ore]

what does that mean about ADH?

[halcyon whale]

so ADH = ADC:2

[tulip ore]

yes

[halcyon whale]

nice

[tulip ore]

and similarly what about HAD and BAD?

[halcyon whale]

you mean BAH?

[tulip ore]

no

[halcyon whale]

mhm

Like BAD:2 = HAD

[tulip ore]

also correct

now lets consider HAD + ADH

at first it looks like we dont know this sum

but using the facts we just mentioned, we can calculate it

[halcyon whale]

Like D + A = 180

But its D:2 + A:2

so HAD + ADH = 90

[tulip ore]

correct

now you know those two angles add up to 90

[halcyon whale]

yes

[tulip ore]

youre very close to proving that AHD is 90 degrees

what is the final step to get from HAD + ADH = 90 to AHD = 90?

[halcyon whale]

oh so its means AHD = 90

right

because 90 +90 = 180

[tulip ore]

thats it

now you do the same thing for the other side of the parallelogram

[halcyon whale]

so its the same thing

so BFC = 90

[tulip ore]

yep

[halcyon whale]

and AEB = 90

And DGC = 90

Oh so now thats a rectangle

[tulip ore]

yea

[halcyon whale]

cooool

thanks man @tulip ore

[tulip ore]

np

[halcyon whale]

.close

[pearl pond]

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[pulsar stump]

why is this correct?

[pulsar stump]

this should be wrong the above

i myself proved

oh wait

i get it

.close

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[icy badger]

...

[pulsar stump]

given A is invertible and is of structure n x n (square matrix), how do i prove there exists a k natural number such that:

$A^k = I_\mathbb{F}$

[jolly parrot]

Ayanokoji

[pulsar stump]

also given $\mathbb{F}$ is a closed Field

[jolly parrot]

Ayanokoji

[pulsar stump]

idk how that helps

[pearl pond]

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[pulsar stump]

<@&286206848099549185>

[pearl pond]

@pulsar stump Has your question been resolved?

[pulsar stump]

.close

[pearl pond]

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[willow chasm]

my teacher specifically wants this in point-slope form, but I don't know how much to include:

y-1=1(x-0)

I want to simplify it but I don't know if its more correct to leave everything in there

[tulip ore]

this is usually enough

[willow chasm]

I don't know if it should be that or y-1=x

[tulip ore]

you should do y - 1 = 1(x - 0)

[willow chasm]

ok ty

.close

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[midnight haven]

I had a doubt
What does series expansion actually mean
For example if I take sin(x) series so what does that series mean?

[wet osprey]

You can take it as a definition of the sine function

[sharp vigil]

we can write that [ \sin x = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 + \cdots ] which means that in the limit as the number of terms in the series that we add approaches infinity, the polynomial we get approaches the sine function

[jolly parrot]

cloud

[autumn fossil]

don't look for some deep hideen meaning

there are similar series for lots of other functions

e^x = 1 + x + x^2/2! + x^3/3! + ...

[midnight haven]

But my question is
Will it ever reach sine function?
If not then what's the point

[wet osprey]

When you take the limit it is equal

[sharp vigil]

not in a finite number of terms, but it can get arbitrarily close to the sine function, which is why it's equal in the limit

[midnight haven]

Can u show?

[wet osprey]

We say it converges to the function

[autumn fossil]

basically it means that if you add enough terms, you can get arbitrary precision

[wet osprey]

What I meant is the definition of a new function named “sin” can be defined by that infinite sum (or series)

[midnight haven]

So are series even useful then?

[wet osprey]

Very

[sharp vigil]

similarly we can say that the function $f(x) = \frac 1x \ne 0$ for any given value of x, but regardless we say that [ \lim_{x \to \infty} \frac 1x = 0 ]

[jolly parrot]

cloud

[autumn fossil]

ofc, in the limit to infinitely many terms, it's actually equal to sin(x)

[sharp vigil]

you can get a very good approximation for sin(x) by adding the first few terms

[midnight haven]

Ooo real real

[wet osprey]

Have you ever heard of sinx = x for sufficiently small x

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

Yep
A lot

In physics it was used in so so so many derivationa

Derivations

[pearl pond]

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[pearl pond]

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[brittle onyx]

is the answer D for this?

apparently the answer's C and i dont get how

[odd robin]

Yeah I am getting d too

[brittle onyx]

ok thank you

.close

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[quasi fern]

when does the object start moving?

and does it move at varying speed or constant speed throughout?

[covert pivot]

Wait @brittle onyx didn't you just send this above?

A few hours ago

[odd robin]

U can see this above in the graph

It starts at 1 sec

[covert pivot]

.

[covert pivot]

u sure

[odd robin]

Oops

2*

But pretty sure ans is d

[brittle onyx]

yea i did and Nemesis gave me an answer but i asked it in the physics discord a while ago and they said it was C but didnt give me any reasoning

[covert pivot]

I feel like it would be D tho

Since its displacement north is decreasing

[brittle onyx]

but my doubt is "instantanous" velocity

[covert pivot]

Which means its moving south no?

[covert pivot]

How so

[brittle onyx]

its asking for that specific instant

[covert pivot]

Yeah

[brittle onyx]

wait nvm

[covert pivot]

The velocity is the same from between 2 to 8 seconds

[odd robin]

Yes

5

South

[pearl pond]

@brittle onyx Has your question been resolved?

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[midnight haven]

not sure how to approach this

[wet osprey]

I’d think about factorising 2010

[pearl pond]

@midnight haven Has your question been resolved?

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[covert pivot]

Can you tell me what angles/sides you used to get that it is ASA?

[short blade]

2 angles and the included side of one triangle must be congruent to two anfles and the incleded side of the other triangle

[covert pivot]

Are you sure that is ASA?

Is the congruent side between the two angles?

Question 1 btw sorry that I didn't specify

[short blade]

2 angles and the side in
between are congruent is asa

wait is it aas then?

@covert pivot ?

[covert pivot]

yeah

[short blade]

how can u tell the diffrence beteween aas and asa

in that one

[covert pivot]

Because the side isn't between the angles

ASA the side is between the angle

AAS the side touches one of the angles but not both

[short blade]

ok i see

then for the second one?

for the second one is it the second option

[covert pivot]

I think maybe the first?

[short blade]

how do we get another person to look at it to see what they think

any idea on the third one?

[covert pivot]

third is the first choice

you have two sides congruent

then the third side is reflexive

so it is also equal

[short blade]

ok thank you

what do you think about the second one

[covert pivot]

I feel like its the first one also

Because we have the reflexive side

then because its a perpendicular bisector we know that ACB = BCD

and AC = CD

[short blade]

i am just going to do the first one

[pearl pond]

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[rose sentinel]

How do I do 27

[pearl pond]

@rose sentinel Has your question been resolved?

[rose sentinel]

<@&286206848099549185>

[vocal lark]

Working on it...

[rose sentinel]

Thanks

I calculated the surface area I think but I don’t know what to do from there

[vocal lark]

Multiply by 0.5mL 👀

[rose sentinel]

Yeah I did

Got it wrong

Let me see what I got again

[vocal lark]

Is this your integral?

[rose sentinel]

No you forgot to add 1 on the inside of the square root

It simplifies it

[vocal lark]

Oh I messed up

Forgot the root

[rose sentinel]

Let me write my root

@vocal lark

[vocal lark]

Yes

Multiply that by .5 ml

[rose sentinel]

Okay well we’re both wrong

[vocal lark]

Lol

Oh...

Times 5000

[rose sentinel]

The correct answer is 226.2 L

[vocal lark]

Convert to L

[rose sentinel]

Tried that

[vocal lark]

Times 5000 woks?

[rose sentinel]

Wait

There’s two enamels

An interior and exterior

So it’s 144+144

[vocal lark]

Ah, there ya go

[rose sentinel]

Still wrong

[vocal lark]

Multiply the pi

[rose sentinel]

Not a terrible idea

Still wrong

Wait

I got

2261.9

That’s too much of a coincidence

Uh

I don’t see what I’m doing wrong

[vocal lark]

🤷

Maybe the branch isn't what I think it is

[rose sentinel]

Wait let’s walk through the calculation

[vocal lark]

Lemme demos it

[rose sentinel]

The surface area of the wok is 288

Wait

It’s 0.5 mm

Not cm

We had to transfer unit

I got it 😎

Wait

Okay so the enamel is 0.05 cm

[vocal lark]

Oh...shoot

[rose sentinel]

Yeah no I got the right answer

He now I’m cknfused

[vocal lark]

The conversion was wrong?

[rose sentinel]

It’s 0.05

That’s why

[vocal lark]

1mL = 1cm^3

[rose sentinel]

But still

[vocal lark]

But we were doing 0.5mm

[rose sentinel]

Isn’t there two enamels

[vocal lark]

And calling that a mL

[rose sentinel]

Or is the sum of them 0.05

[vocal lark]

Half a

Hang on one sec

[rose sentinel]

Okay

It says EACH enamel is 0.5 mm thick

And there’s two enamels

So the total thickness is 1

mm

So the volume of enamel is 288 *0.1

28.8

Times pi

90.5

Times 5000

Okay so it’s exactly double

The correct answer

Yeah so maybe it’s worded really poorly

It suggests there is double the thickness but in order to get the right answer I must use 0.5mm not 1mm

[vocal lark]

Idunno I'm still getting 144pi L

,rotate

[rose sentinel]

$2880.12*\pi5000\frac{1}{1000} = 2 * 226.2$

[jolly parrot]

Nathan

[rose sentinel]

@vocal lark

It’s twice the right answer

[vocal lark]

How did you get .1x2

[rose sentinel]

0.05

Wait

Okay ignore the 2

Just 0.1

[vocal lark]

Yeah that's one side's enamel

[rose sentinel]

0.1 is the combined thickness

[vocal lark]

So you are doubling TWICE?

[rose sentinel]

No no lol it was a typo

[vocal lark]

So why do you need the 2?

[rose sentinel]

It was a typo

I don’t

it’s just 0.1

No 2

[vocal lark]

Well it should be same as mine then?

[rose sentinel]

well I got 452

So

Liters

Okay well

I think the bottom line is they must have worded it poorly right?