help-36

[final saddle]

@rain sentinel Has your question been resolved?

[ripe jewel]

<@&268886789983436800>

[tulip coyote]

What was the ping for, may I ask? Something that already got deleted?

[ripe jewel]

yup

[tulip coyote]

Fairs I guess

[agile nacelle]

scambot, banned for activity elsewhere

[final saddle]

@rain sentinel Has your question been resolved?

Channel closed

Closed by @rain sentinel

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[spare plover]

<@&286206848099549185>

[spare plover]

I want to show l^2 is strictly convex

[shell condor]

You need to send your question first, and then if no one responds for 15 minutes then ping helpers mate

[late rose]

!15m

[final saddle]

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.

[late rose]

!show

[final saddle]

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

[spare plover]

I actually showed $ | | x + y | |_2 \leq 2$

But I can't figure out how to show the equality doesn't hold

[shell condor]

Can you do differentiation?

[spare plover]

Yes I can

[shell condor]

Well apologies, but i am not qualified for this question, I could possibly try and decipher it in some time and then do this, but then I'd be wasting your time

[spare plover]

Don't worry. I appreciate your time as well

Thanks anyway

I'll keep channel open

[shell condor]

Sure man, apologies again

[final saddle]

@spare plover Has your question been resolved?

[spare plover]

<@&286206848099549185>

I wanted to argue by contradiction

So, as $l_2$ norm is $| |x | | = \sum_{n=0}^{\infty} \ |a_n|$

[soft zealot]

Gol D Roger

[spare plover]

I apply it at $| | x + y | | = | | x | | + | | y | | = 2$

[soft zealot]

Gol D Roger

[spare plover]

But I still can't conclude a contradiction

[spare plover]

<@&286206848099549185>

Please, I just need to deal with that sum

I tried squaring both sides

I tried substracting both sides

<@&286206848099549185>

[gray radish]

triangle inequality?

[gray radish]

think about the unit ball

is it open or closed

@spare plover

[spare plover]

It is closed

I mean, I'm gonna send the full process till now/

[gray radish]

is ur space finite dimensional

oh ye it is

[spare plover]

Let $x,y \in l_2$ be s.t. $ \Vert x \Vert_2 = \Vert y \Vert_2 = 1$. Using the triangle inequality (which we know we can use since $l_2$ is a Banach space), we have

\begin{align*}
\Vert x + y \Vert_2 \leq \Vert x \Vert_2 + \Vert y \Vert_2.
\end{align*}

By hypothesis, we also know that $ \Vert x \Vert_2 = \Vert y \Vert_2 = 1 $, so

\begin{align*}
\Vert x + y \Vert_2 \leq 1 + 1 = 2.
\end{align*}

Now, suppose that there exist some $x,y \in l_2$ (with $ \Vert x \Vert_2 = \Vert y \Vert_2 = 1$) such that

\begin{align*}
\Vert x + y \Vert_2 = 2.
\end{align*}

Then, by definition of norm in $l_2$ , we have that $ \Vert x \Vert_2 = \sum_{n=0}^{\infty} \ |a_n| $ and similarly, $ \Vert y \Vert_2 = \sum_{n=0}^{\infty} \ |b_n| $.

Then,

\begin{align*}
\sum_{n=0}^{\infty} \ |a_n + b_n| = \Vert x + y \Vert = \Vert x \Vert_2 + \Vert y \Vert_2 = \sum_{n=0}^{\infty} \ |a_n| + \sum_{n=0}^{\infty} \ |b_n|.
\end{align*}

Where $|x|$ denotes the usual metric in $\mathbb{R}$ (i.e. $ |x| = \sqrt{x^2}$).

[spare plover]

Yes it is

But how can I conclude a contradiction there?

I sense is some arithmetical trick

But I can't find it online ):

[soft zealot]

Gol D Roger

[spare plover]

I tried squaring it, idk if I can use GM AM inequality

Any ideas?

<@&286206848099549185>

[gray radish]

idk apart from showing the line segments for any $x,y \in l_2$ is in the interior of the space

[soft zealot]

frisysics

[gray radish]

since ur working in euclidean space the contradiction is kind of easy to do

[spare plover]

I feel the same but I am just too dumb to figure out lmao

[gray radish]

the unit ball being closed makes the way ur doing it so weird to do

[spare plover]

Yeah... I just tried using definition and algebra, not geometric arguments

[gray radish]

you can have a line connecting x to y, and do a contradiction through that

[spare plover]

But if ball was not closed, I don't think statement is still correct though

[gray radish]

if the ball was open then if

[spare plover]

Is that the same as saying $ \Vert x - y \Vert$ and using inverse triangle inequality?

[gray radish]

$\Vert x + y \Vert_2 = 2$ then that would be a contradiction since the open ball must have d(x,y) < r where r is 2

[soft zealot]

frisysics

[gray radish]

instead of $\leq$

[soft zealot]

frisysics

[spare plover]

I tried it, but substracting an infinity amount of numbers works awful for me

[spare plover]

Oh I see, so if ball was open, result may be trivial?

Hmm I see, but then again, ball is closed

[gray radish]

yeah

😔

[spare plover]

I need to figure out how to use that $ \Vert x \Vert_2 = \Vert y \Vert_2 = 2$

Since norm is an infinite series, then it must be that at some point, entries are zero, isnt it?

Or is there some series that converges to 2 and are "nice" in the sense that it has a pretty finite sum

[gray radish]

i think there is

[spare plover]

I mean, the argument for showing $l_1$ and $l_{\infty}$ are not strictly convex is using the sequences $(1,0,0, ...)$ and $(0,1,0, ...)$.

Maybe I can use them too here?

[soft zealot]

Gol D Roger

[spare plover]

Or use some Cauchy Criteria for convergence to show it can't exist such sequence s.t. $ \Vert x \Vert_2 = 1$

[gray radish]

this is probably an insanely dumb way to do it but

take z = 1

$$
\sum_{k=0}^{\infty} \frac{1}{k!} \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = 2
$$

[soft zealot]

frisysics

[gray radish]

wait no you'd take z to be the first series im getting sleep deprived

[spare plover]

lol

Get some rest bro, thanks for your help till now

[gray radish]

$$
\sum_{k=0}^{\infty} \frac{z^k}{k!} = 2
\ \text{where} \ z = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}
$$

[soft zealot]

frisysics

[spare plover]

I'll keep chat open until I solve it, help may be apreciated

[spare plover]

Hmm, and what about 1?

I mean, if some series converges to 1 then argument brokes, but if some series converges to 2, then there is not problem, since splitting the sum does not split the result (in a nice way at least)

So yeah, series should converge to 1, not to 2

But idk if that's possible

[gray radish]

i mean z has alternating terms no?

like +,-,+,-

[spare plover]

Yeah, but idk how to argue that inserting parenthesis or reordering the series still has the same result

Since it does not converge absolutely

[gray radish]

yeah. .

wait

are x and y distinct? @spare plover

[spare plover]

Yes

By assumption

If not, result is trivial again, isnt it?

Bc it's in the unit ball

[gray radish]

and they're both unit vectors?

can't you have the contradiction be x = y

[spare plover]

Yeh

[spare plover]

Yes I can

But... how?

Are you trying to say $ | | x - y | |_2 = 0$?

[gray radish]

yeah

[spare plover]

Hmm, why?

[gray radish]

well

i feel like it has something to do with

either

showing the series for $ | | x | |_{2}$, same for y is equal to each other and thus x,y must be non distinct i.e the same

or the distance between them is 0 which means they're the same point

that is d(x,y) = 0

[spare plover]

Hmm

So, do I have to compare the nth term of the sum both sides?

And apply some "uniform convergence" argument?

To conclude that the distance is zero?

[gray radish]

yeah something like that

theres probably an easier way than that

[spare plover]

But is that possible?

[spare plover]

Hmm

Maybe do smt like this?

[gray radish]

yeah

[spare plover]

Hmm

Let me write it

[gray radish]

https://math.stackexchange.com/questions/1356842/is-l-2-norm-a-strictly-convex-function

ok i sleep now

[spare plover]

There's just one little problem with that answer. I'm supposed to show it without the use of inner products

[gray radish]

bro

😭

😭 😭 😭 😭 😭 😭 😭 😭 😭 😭

[spare plover]

We still haven't properly studied Hilbert Spaces, so yeh

[gray radish]

ok that makes sense

[spare plover]

Dont worry man, go to sleep

[gray radish]

just do the boundary unit ball definition

like the line segment thing

[spare plover]

There may be more ppl here

[gray radish]

ye

[spare plover]

I'll try

[spare plover]

You have done more than enough, so thank you for that

[gray radish]

i feel like i didn't help that much ngl

but was fun

[spare plover]

Perhaps you gave me more intuition behind, which is greately appreciated

[spare plover]

So <@&286206848099549185> can any other helper give me a hand?

[wet steeple]

Bro who studies Hilbert space in this age

😭 😭

[spare plover]

Idk man, I need to graduate soon, so perhaps, that's why?

[wet steeple]

I even dropped complex analysis i can't help u...I'll go help out in important doubts like application of Pythagoras theorm

[spare plover]

Truly based, go ahead

Chat still will be open

Got it! I finally solved it

Thanks @gray radish 🙂

.close

[final saddle]

Channel closed

Closed by @spare plover

Use .reopen if this was a mistake.

[gray radish]

im watching a dota 2 vid how did you end up doing it

[spare plover]

I'll send it in just a sec

But basically, show inverse triangle inequality using normal one, and since the equality holds in one iff in the other, then I got what you said, $| | x - y | | = 0$

[soft zealot]

Gol D Roger

[spare plover]

But it's a norm so hell yes

[smoky egret]

@spare plover for future questions i recommend that you ask in the topic channels

since there you get faster responses

the usual level of questions asked in the helped channels is up to hs and the very beginning of uni math more or less

so people who know how to answer questions of more involved topics rarely visit help channels

[final saddle]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• 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.

[unborn elk]

Im recreating a video in minecraft and want it to be exact. The player in the video is in adventure mode and can't see block outlines. Can someone measure this hallway in blocks?

[unborn elk]

Would the measurements for each doorway help? I can figure those out. The video is "Matchbox Round 1"

Here's some lines if that helps

Wair it didnt send righht

Hey I drew new lines they wont send

[quiet flume]

they sent for me

[unborn elk]

Oh ok

There

[harsh musk]

i love this

[late rose]

math in real world application ahh

[harsh musk]

"were never gonna use this irl"

[unborn elk]

Ok but this looks calculatable

How do I do this

Should I consult artists instead

Cuz perspective

[harsh musk]

i think artists would know better

[quiet flume]

the fov would make a difference here no?

looks like 13 to me

[unborn elk]

Is that a calculated or is that a guess

[quiet flume]

i mean its not any math done but thats what i counted

[unborn elk]

Also the fov is shown by the nearby blocks so compare the blocks to the size of the first one on the far left

[late rose]

im just gonna guess that the ratio of the pixel length of the left side of the stripped oak log to the right side of the stripped oak log, is the same as the ratio of the length of right side of the stripped oak log to the right side of the next orange concrete block.

lemme test this

[final saddle]

@unborn elk Has your question been resolved?

[unborn elk]

No

Shoot its a bot

Why did I reply

[unborn elk]

Did you do it yet

[late rose]

not yet

[unborn elk]

Maybe 8? This is just a guess btw

[unborn elk]

Have you started it or not yet

[late rose]

ye im doing it rn

[unborn elk]

Alr thanks

Just remember the perspective doesnt have the corner in the middle so the sizes might be different between the left and right of the corner

ChatGPT says 8 but idk how much I trust that

More lines

[late rose]

i think it might be solvable by considering the pixel length of the initial stripped oak block, and the pixel length of the final orange concrete block, gonna need more time to think about it

[final saddle]

@unborn elk Has your question been resolved?

[unborn elk]

Tomorrow if it hasn't been resolved I might just create different sizes in game and see which one looks right

[late rose]

i dont really know too much about 3d projection and honestly you might have better luck asking in #geometry-and-trigonometry

If it helps anyone though, i did capture a few example values on my test world

[unborn elk]

Alr thanks

.close

[final saddle]

Channel closed

Closed by @unborn elk

Use .reopen if this was a mistake.

[final saddle]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• 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 mango]

Does this look okay? I have no clue how to do 1.7 so it is not included

[coarse copper]

Still need help

[balmy mango]

Yip, you are very quick, I just sent this message😅

[coarse copper]

Wait

Determining when the new growth function, (g(t)=150e^{2t}), exceeds the initial function requires the formula for the initial function. The image provided does not include this formula, making it impossible to solve for the time when the new function becomes higher than the initial one [1.1].

[soft zealot]

Celestial

[coarse copper]

Is it possible

[balmy mango]

What the helly

The information I sent was everything I was given plus the questions I answered

[proper dagger]

did you just pass this into gpt

[viral oriole]

Seems like it

[proper dagger]

!noai btw, if so

[final saddle]

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).

[coarse copper]

.

[proper dagger]

1.1 is incorrect for the green leaf beetles

you subbed t = 1 for g(0)

1.3 overspecified: you are asked for the point in time (aka just the time value)

1.4 consider describing what equals infinity here and what it means for the population of the green leaf beetles

1.5: while quadratic is no doubt correct, I think they may be looking for another word (this one I'm not sure about)
1.6: inaccurate explanation for the domain being all of R. the domain is all of R because there is no real number that makes a polynomial undefined (or, polynomials are defined for all real numbers)

1.7 is basically asking you to compare the two functions to see when the new function produces values exceeding that of the old one (for the red bark beetles)

so basically, the range of values of $t$ where $r_{new}(t) > r_{old}(t)$

[soft zealot]

Hanako(x, y); ∂(fox)/∂x

[proper dagger]

I think that's all I have to comment on for now

[balmy mango]

Thank you so much for checking it!

[proper dagger]

nps. and I hope you could get 1.7 too

that being said

[balmy mango]

Yeah I'm going to sit and try to do it

[proper dagger]

is this a live test, or a past year paper, or an assignment?

[balmy mango]

This is for an assignment, would never paste a test in here. The lecturers just aren't very helpful and I do not have friends to ask if they can check or compare work so I ask here. I do the work myself so I hope it does not go against any rules in the server

[proper dagger]

if it is an assignment it's generally fine. summary assessments/tests constitute academic dishonesty though

but yeah, all the best!

[balmy mango]

No it's not a summary assessment or a test. Thank you! May we meet again comrade.

[proper dagger]

aye. anything else?

[balmy mango]

That is it for now, I will ask again once I finish question 2.

.close

[final saddle]

Channel closed

Closed by @balmy mango

Use .reopen if this was a mistake.

[proper dagger]

aight

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
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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.

[strong flax]

hel

[plucky rover]

Well, ask your questions

[strong flax]

idk

look how im doing in my class

just struggling with school man

it sucks

[quartz flame]

its okay, just try your best

what class is it

[final saddle]

@strong flax Has your question been resolved?

[final saddle]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• 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.

[stoic temple]

Hello, I don't understand why this is so. 😦

[stoic temple]

!status

[final saddle]

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

[stoic temple]

1

[desert mantle]

do you know that $\int_a^b f(x) dx = -\int_b^a f(x) dx$ ?

[soft zealot]

Denascite

[stoic temple]

Yes

[desert mantle]

its the exact same idea

going on the line from a to b and going on the line from b to a cancels each other out

[stoic temple]

Wait wdym cancels out

$\int_a^b f(x) dx + \int_b^a f(x) dx=0$ ?

[soft zealot]

David🥛😋

[stoic temple]

Like this or something else

[desert mantle]

like that

[stoic temple]

Oh okay

Um I'll try this again after class

.close

[final saddle]

Channel closed

Closed by @stoic temple

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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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.

[blazing fable]

Help me how to solve this

🙁

[silver maple]

!show

[final saddle]

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

[blazing fable]

I’m stuck at finding the..

This part

[dreamy warren]

the big circle’s area is 4pi, you subtract the three semicircles (middle one fully inside = pi/2, sides overlap giving total removed area (5pi/3 − 4 + √3)

so shaded area = 4π − (5pi/3 − 4 + √3) = (7pi/3) − √3 + 4

so a+b+C+d=17

[final saddle]

@blazing fable Has your question been resolved?

[final saddle]

@blazing fable Has your question been resolved?

[final saddle]

@blazing fable Has your question been resolved?

[blazing fable]

I dont’ get it (after removing 3 semi-circles)

[dreamy warren]

oh so after removing the 3 semicircles we’re just calculating how much of the big circle’s area is left

one small semicircle (the middle one) is completely inside, so its whole area gets subtracted

the other two on the sides only overlap partly

[blazing fable]

Yes so,

I stuck at finding the part

At both sides, outside the shaded circle of the 2 semi circles

[dreamy warren]

you have to find where each semicircle meets the big circle (the intersection point is at x = 3 − √3), then calculate the overlap area between them

[blazing fable]

Yeah so basically I know what to find but idk how to do

[dreamy warren]

that overlap for one side comes out to (−2 + 7π/12 + √3/2)

[blazing fable]

?

[dreamy warren]

ahh okay

[blazing fable]

I still don’t understand(sr for bad English)

[dreamy warren]

yeah so basically to find that part, take the equations of the circles the small semicircle (center at (1,0), radius 1) and the big circle (center at (3,1), radius 2)

[blazing fable]

Aha

So we’re trying to get the area by coordinating

[dreamy warren]

yeah then you have to find the intersection x-coordinate between the left small semicircle top and the bottom of the big circle

[blazing fable]

The small one is

(X-1)^2+Y^2=1

And the bigger one is

[dreamy warren]

you take the small semicircle’s top curve y = √(1 − (x − 1)²) and the bottom of the big circle y = 1 − √(4 − (x − 3)²)

[blazing fable]

(X-3)^2+(Y-1)=4

[dreamy warren]

yso ou find where they intersect by solving √(1 − (x − 1)²) = 1 − √(4 − (x − 3)²)

[blazing fable]

Yeap

[dreamy warren]

so you should get x=3 − √3

[blazing fable]

How do I know the intersection of 2 circles?

Is it like this and find the values

(The second one is -6x not -6y mb)

[dreamy warren]

okay so to find where two circles intersect, you just set their y-values equal because at the intersection point both circles have the same (x, y)

[dreamy warren]

yep exactly that’s the right idea

[blazing fable]

(3/2, sqr(3)/2)

I think it’s the intersection point

[dreamy warren]

yep that’s right!

now that you’ve got the intersection point (3/2, √3/2), you use it to find the overlap area between the small semicircle and the big circle

[blazing fable]

Am I on the right track

[dreamy warren]

yeah

[blazing fable]

But the line between (3/2, sqr(3)/2) and (1,0) is a curve

[dreamy warren]

yeah it’s the arc of the small semicircle

so the overlap region is bounded by two curves (the small semicircle arc above and the big circle’s lower arc below)

[blazing fable]

Yes..?

I have no idea about the 2 curved

Curves*

[dreamy warren]

yeah that’s right the two curves are just the arcs of both circles that intersect

[blazing fable]

Can you explain me how to get the area

[dreamy warren]

yes, im just calculating one sec

[final saddle]

Channel closed

The owner is missing!

[dreamy warren]

from x=1 to x=3/2the top boundary is the small semicircle y = √(1 − (x − 1)²) and the bottom boundary is the big circle’s lower branch y = 1 − √(4 − (x − 3)²) so integrate the difference

oh

[blazing fable]

.reopen

Wait integrate?

[dreamy warren]

from x=3/2 to x=2 the big circle is below the x-axis so the overlap is just the small semicircle part so integrate y=y = √(1 − (x − 1)²)

[dreamy warren]

yes you do an integral

[blazing fable]

Um

I’m so sorry but I’m year9 now

I only learned just a bit of integral

[dreamy warren]

ahh okay lemme find another method

[dreamy warren]

sorry 😭

[blazing fable]

Thx!

Np

It’s actually AMC10

[dreamy warren]

you can find the overlap using circular segment formulas instead of integrals

do you know those?

[blazing fable]

?

Circular segment?

Oh I know that

I just searched it cuz I learned it in another language 😩

Hmm but like

[dreamy warren]

hmm but i think it may be a bit complicated later on

[blazing fable]

How?

Yeap

[dreamy warren]

you can maybe break the big circle into smaller easier shapes

[blazing fable]

I’m thinking about this shape

But I don’t know the angle😨

[dreamy warren]

120?

no wait sorry

[blazing fable]

Sorry I need to go to bed 🙁

Thx..!

[dreamy warren]

im sorry, i'll figure out the solution and send it to you

you can view it later

[blazing fable]

oh no

I tell my mom that I'm learning math and she accepted me

[dreamy warren]

haha

[blazing fable]

lol

[dreamy warren]

don't stay up if you're too tired tho

[blazing fable]

It's fine

I actually need to lock in

[dreamy warren]

so you can divide the large circle into four easier parts

[blazing fable]

😮

how?

[dreamy warren]

the top semicircle of the big circle (above the center line), the bottom sector with a 120° central angle, the small triangle inside that sector and the little leftover area under the semicircles but above the diameter

if you can visualise it

[blazing fable]

You meant the top semicircle by the semi-circle in the middle of those 3 semi-circles

[dreamy warren]

nah nah the top semicircle here means the upper half of the big circle (the large one with radius 2)

[blazing fable]

I don't get the afterall things

[dreamy warren]

the bottom sector is a slice of the lower half of the big circle

[blazing fable]

under the line of the bottom of 3 semi-circles?

[dreamy warren]

the central angle of that slice is 120°, because if you draw lines from the circle’s center f to where the small semicircles touch, it forms a 30°, 60°, 90° triangle

[dreamy warren]

yep exactly

[blazing fable]

I think i get it

[dreamy warren]

that line basically forms the flat base of the three small semicircles, and below it is where the big circle continues that curved portion under that line is the 120° sector we’re talking about

[blazing fable]

aha

[dreamy warren]

now inside that same sector there’s a triangle made by those two radii and the chord that connects their ends

[blazing fable]

with 120 degree?

ohhhh wait

[dreamy warren]

the angle between those two radii at the circle’s center is 120

[blazing fable]

then if we make that line,

the semi-circle at one side is divded into 2 which are both

um what should I say

a part of the circles

nah no

[dreamy warren]

?

[blazing fable]

nothing

[dreamy warren]

did you get the first 3?

[blazing fable]

yes

( I think so )

[dreamy warren]

then the remaining region is the strip above the three small semicircles but below the line that divides the big circle

that’s the leftover space you still see shaded between the small semicircles and the diameter

[blazing fable]

yes

[dreamy warren]

can you form the diagram now

[blazing fable]

You just meant by the region between the diameter line and the line of the bottom of the semicirlces but except the region of semicircles.. isn't it?

[dreamy warren]

kind of like this

[blazing fable]

ah yeah yeah

120degree is the top angle of the triangle isn't it

[dreamy warren]

yeah

[blazing fable]

and the next..?

[dreamy warren]

now you calculate the area

the area of part a would be 2 pi

the sector which was the second part would be 4/3 pi

triangle area would be root 3

[blazing fable]

wait wth

[dreamy warren]

for calculating the area of the fourth part which is the leftover

[blazing fable]

we are basically finding the area of the white region to get the leftover

[dreamy warren]

yes

[dreamy warren]

fourth part is kinda tricky but if you imagine 2x2 square whose side length is 2, the area of the square is 4

[blazing fable]

yes

[dreamy warren]

now place a quarter of the big circle (a quarter of radius 2) inside one corner of that square the quarter-circle has area 1/4 pi * 2^2 = pi

[blazing fable]

wait can you describe it on the picture I cannot send you a picture

[dreamy warren]

so the part of the square not covered by that quarter circle (the four little corner regions if you tile four such squares around the circle) has area 4- pi

yeah sure one sec

okay here's an easier version

you want to calculate the area in blue shaded region so you would subtract the areas of the semi circles from the area of the rectangle

[blazing fable]

ohhhhhh

[dreamy warren]

so that would be 4-pi

[blazing fable]

bro but

the middle part of each side of semi-circle

it should be curve

[dreamy warren]

we are imagining the rectangle here

but you are right it is curved but were making a construction here if you get what i mean

like how we did with the triangle

[blazing fable]

so are we gonna subbtract the part

omg how do I say this

[dreamy warren]

yeah its not gonna include in our answer because we only need the blue part

[blazing fable]

oh my god

!!!!!

[dreamy warren]

im just dropping perpendiculars from point e and g

[blazing fable]

why am I so naughty

that's actually tru

[dreamy warren]

and join them to form a rectangle

[blazing fable]

we only need the blue part and subtract it from the whole circle

[dreamy warren]

yes!!

[blazing fable]

the blue part is 4-pie

?

[dreamy warren]

yess

now the rest is easy

[blazing fable]

wait I just got confused that the answer is blue part

[dreamy warren]

huh are you still confused?

[blazing fable]

4-pie + 2pie+

[blazing fable]

nope I just got it I think

[dreamy warren]

okay

[blazing fable]

so it's 4-pie+2pie+4/3pie-

the triangle was

[dreamy warren]

so total shaded area would be (area of the top semicircle (1) + area of sector (2) - area of triangle (3) + the area of the remaining parts (the blue shaded region)

[blazing fable]

yes

so the top part is 2pie,

the middle part is 4-pie

and the bottom part is 4/3 pie - 2

[dreamy warren]

why -2?

triangle area is root 3 if you calculate

[blazing fable]

cuz the traignel

isn't the base

bro i thought the base is 4

so it's like

[dreamy warren]

you don't know the base

[blazing fable]

2x2xsin120?/2

[dreamy warren]

yeah

[blazing fable]

sqr(3)

[dreamy warren]

yep, now you get the answer

[blazing fable]

so the answer is 7/3 pie + 4 - sqr(3)

[dreamy warren]

yeah

[blazing fable]

wow

[dreamy warren]

compare it with a , b , c and d values which adds up to 17

[blazing fable]

wow

I now love this question

[dreamy warren]

me too haha

[blazing fable]

hehe

[dreamy warren]

but integration is the easiest way to solve it

[blazing fable]

anyway really appreciate you

um

[dreamy warren]

ofcc also im sorry for stealing so much of your time 😭

[blazing fable]

nono

I fell bad for that

[dreamy warren]

noo don’t feel bad 😭

[blazing fable]

I ithink i need to develop how to think creatively to find out the easiest way when doing this competition

[blazing fable]

haha..

I meant i feel bad for stealing "your" time

[dreamy warren]

yeah the best way would be to always sketch or mentally divide the figure into familiar shapes

[blazing fable]

Thank you so much!!!

.close

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[spare merlin]

Hey i have to approximate a sample using linear regression.
for a straight (ax+b) we had the formula:
$a= s_{xy}/s^2_x$ \
$b= \bar{y} - a* \bar{x}$
So i can approximate the sample pretty easily with that.
My problem is that the next assignment says to aproximate the same sample not with ax+b but with
$x^2 +c$
I have no idea how to approach this tbh. The only thing i have is the sample and the covariance matrix for the samples i used for the linear apporach

[soft zealot]

VZL

[final saddle]

@spare merlin Has your question been resolved?

[vital crag]

!original

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[spare merlin]

i have thought about it a bit more and the only way i can think of is optimizing:

$\sum_{i=1}^n (y_i -(x_i^2 +2c))$

[soft zealot]

VZL

[spare merlin]

but then i still wouldnt know what c is or am i tripping?

btw c translates to:
Approximate the Sample with the function/ model $f_c(x)=x^2+2c$

[soft zealot]

VZL

[final saddle]

@spare merlin Has your question been resolved?

[spare merlin]

<@&286206848099549185> :(

[amber valley]

Hello

[spare merlin]

hi

[amber valley]

Can you translate the full question in English?

My Dèutsch requires practice.

[spare merlin]

Given a bivariat sample (x1,y1) .....
a) is just finding the covariance matrix, i did that

[spare merlin]

and b is basically just that but with the linear formula

idk if sample is the correct word tho

[spare merlin]

also this should be
$\sum_{i=1}^n (y_i -(x_i^2 +2c))^2$

[soft zealot]

VZL

[spare merlin]

.close

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[late trellis]

Hello, doing some Galois Theory work now :3

Currently looking at the 4th polynomial here. We know α = 2cos(2pi/9) is a root, but what we don't know is that Q(α) generates the other two roots too. I'm not seeing a helpful factorisation - i feel like there should be something really obvious im missing since the other 4 polynomials are all pretty much trivial.

[late trellis]

specifically, im trying to find the splitting field of $\mathbb{Q}$ for the polynomial $f(X) = X^3 - 3X + 1$.

[soft zealot]

BoggledToggo

[final saddle]

@late trellis Has your question been resolved?

[late trellis]

hi <@&286206848099549185> it has been a bit :3

[vital crag]

can also try #groups-rings-fields

[open leaf]

I’m ngl I’m a lil stupid so <@&286206848099549185> what’s. 9X2X7.192 divided by 92*20

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[final saddle]

@late trellis Has your question been resolved?

[late trellis]

I'll end it since it's being answered elsewhere o7

Oops I pressed wrong one

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[twilit fern]

Question and answer

[twilit fern]

I'm getting -36 for angle DAE, which is not even in options.

Can anyone cross check. Also is there any better way to do this instead of finding all same angles, mark them x, y, z, etc. and try to find relationships between them.

[final saddle]

@twilit fern Has your question been resolved?

[odd seal]

your eqn 1 which is 2a + z - 2x = 180 should become 3a - 2x = 180 after subbing in a = z, but you instead continued with a = 180 + 2x

<@&268886789983436800>

[twilit fern]

Thank you. I subtracted instead of adding 😅

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[waxen basin]

I am having trouble understanding manipulating factorials.

Or how to get the number of terms using this:

$\abs{R_n} = \abs{S - S_n} \le a_{n+1}$

[soft zealot]

NullSquare

[waxen basin]

Would I just substitute the (n+1) term for k?

So the above would be something like:

[soft zealot]

NullSquare

[waxen basin]

But I am not sure where to go from here?

Or would it be more like:

[soft zealot]

NullSquare

[final saddle]

@waxen basin Has your question been resolved?

[loud sundial]

$\frac{(n+1)!}{2} \geq 10^4$

[soft zealot]

Civil Service Pigeon

[loud sundial]

$(n+1)! \geq 2 \cdot 10^4$

[soft zealot]

Civil Service Pigeon

[loud sundial]

@waxen basin the left hand side is strictly increasing for n__>__0, so trial and error from here

[waxen basin]

Ok! But how do I get rid of the factorial? Factorials are still not intuitive to me. 😛

I mean sure at infinity k! is effectively equal to (k+1)! But what do I do here?

[loud sundial]

the left hand side is strictly increasing for n__>__0, so trial and error from here

Factorials of grow very quickly so you won’t be testing for long

[waxen basin]

Ok. Is it okay that I still don't like it? 😆

(Thank you!)

[loud sundial]

To each their own

[waxen basin]

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[craggy atlas]

Find the number of ways in which 8 different flowers can be strung to form a garland so that four particular flowers are never separated.

[craggy atlas]

Why is it not 5! 4!

[formal trail]

that would be true if (for example) you only put the four inseparable flowers on the left side of the garland and put the other 5 on the right

[versed crater]

How many ways are there to string 3 different flowers together?

[formal trail]

but you have the freedom to choose where in the garland the group of four can go

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[craggy atlas]

so how do I do it?

[final saddle]

@craggy atlas Has your question been resolved?

[craggy atlas]

Find the number of ways in which 8 different flowers can be strung to form a garland so that four particular flowers are never separated.

[final saddle]

@craggy atlas Has your question been resolved?

[craggy atlas]

<@&286206848099549185>

[craggy atlas]

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[lethal drum]

hi

[lethal drum]

can somebody help me

with the determinent

supposed to solve for x

[zealous storm]

i mean can’t u just compute the determinant

of both things and equate them

so like that option always exists

[lethal drum]

thats what am having trouble with

[zealous storm]

now whether you can do it quicker is a secondary call ig

[lethal drum]

the first determinent will be this

but the second one

[zealous storm]

do you know laplace expansion?

[main mirage]

^ Additionally, ||which column/row looks easiest to expand with on the right hand side?||

[zealous storm]

or cofactor or whatever algorithm you use to compute the determinant

[lethal drum]

what i know is theat u can choose the colum on row with the most zeros

then discard that and start multiplying right?>

[zealous storm]

sounds like laplace expansion

[lethal drum]

yea idk its name

[zealous storm]

yeah so which row or column

do you think is optimal?

[lethal drum]

the 010

[zealous storm]

yes

this is the "sign chart"

assuming you're familiar with it

[lethal drum]

i remember the teacher saying if the position is even its positive or if its odd then negative

or something like that

right?

as in a11 or a21

[zealous storm]

yeah well (1,1)

what position is even?

maybe it isn't so clear

[zealous storm]

u can just remember it with this

if u like

it starts at +

in a_11

then u keep swapping between + and -

or rather every adjacent element must have a different sign

with reference to whichever "node element" you're fixated on

if that makes sense

[lethal drum]

alright so going back to the question how would i use this?

[zealous storm]

do you remember this?

if we're looking at a_13

we can imagine deleting the row and column containing a_13

that gets us the

a_21 a_22
a_31 a_32

matrix

the same logic extends to finding the corresponding "smaller determinant" for a_23 and a_33

[lethal drum]

but we dont need any other matrixes right?

the the first one?

[zealous storm]

wdym?

[lethal drum]

just this one

[zealous storm]

u go from a large 3x3 matrix

to a sum of scaled

2x2 matrices

which u can more easily compute

[lethal drum]

so i gotta calculate 3 matrixes?

[zealous storm]

in general yes

[lethal drum]

3 2x2

[zealous storm]

but in your case

only 1, right?

since the other two matrices

will be scaled by 0

[lethal drum]

ah so since their outside element will be 0 so we disregard them thats what u mean?

cause the zero will be multiplied by the whole thing

[zealous storm]

0 \begin{vmatrix}
3 &-2  \\
5 & 1
\end{vmatrix} - 1\begin{vmatrix}
x &-2  \\
1 & -1
\end{vmatrix} 

+ 0\begin{vmatrix}
x &3  \\
1 & 5
\end{vmatrix}```

so yeah the ones w the coefficient of 0 can just be omitted

notice that i "retrieved" the signs of the matrices

[zealous storm]

using this

[lethal drum]

so it should look like this

and then we just multiply the matrix like usual and find x right?

[zealous storm]

yes

[zealous storm]

maybe u should write all of this down

but i guess that's more of a stylistic choice

[zealous storm]

also there's no multiplication of matrices going on

but if u meant computing the determinant on the right

then sure

after that it turns into an elementary algebra problem
namely, solving a quadratic

[lethal drum]

yea thats it

1 more thing

when i asked my friend about this he sent me this

supposedly

this is what the professor gave him

am wondering where the extra 1 came from

or is that just something i should disregard?

[zealous storm]

(-1) corresponds to the sign from this chart

[lethal drum]

yes

[zealous storm]

1 * (the 2x2 matrix) is the actual laplace expansion

[zealous storm]

^

let's assume our a_23 is indeed 1

then yeah 1 * determinant of that 2x2 matrix

[lethal drum]

so hes not just turning it into a negative, hes multiplying it by negative 1?

[zealous storm]

we append a - in front of because of the sign chart

[zealous storm]

they're the same(?)

[lethal drum]

yea but the way its written makes it seem like it came out of nowhere

but thats what he did right?

[zealous storm]

not really

the general formula is something like $\sum_{i = 1}^n b_{ij} {\color{red}(-1)^{i + j}} M_{ij}$

[soft zealot]

mmmm7

[zealous storm]

if u look at the red part

u just end up raising (-1) to some power

depending on whether i + j is even or odd

it's similar to what u mentioned earlier

"position is even"

but that's ambiguous

in particular, if we denote an element as a_{ij} where i is the row it belongs to

and j corresponds to the column it belongs to

then i + j must be even for the sign to be positive

otherwise it must be negative

[lethal drum]

alright so if in the chart if its a negative position then multiply the coefficient by -1. thats it right?

[zealous storm]

yeah the chart is easier to look at

and u can draw it out

it's just like checkerboard

if the sign chart shows + then you multiply by 1

if it shows a - then you multiply by -1

[zealous storm]

this formula here already incorporated the sign thing btw

i'd be focused more on the

element * (determinant of 2x2 thing) * (sign)

[lethal drum]

yea i got the chart figured out i was just confused cause hes actually wiriting (-1)(1) instead of just saying -1

[zealous storm]

yeah it's a perspective thing

if u look at it from a general formula pov

[lethal drum]

should be good now.

[zealous storm]

then he's doing (-1)^(1 + 2) = (-1)^3 = (-1)

but if u use the "sign chart" then it makes sense to just append a negative in front

well obviously they're all equivalent

so this ends up being a more philosophical discussion

[lethal drum]

alright thx for the help

am done here

.close

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[plain rain]

hello. could someone remind me what we do when we wanna solve for x in this situation? : x³ = y

[worthy wren]

cube root both sides

[plain rain]

yeah, but what it y is negative?

if*

[worthy wren]

,w cube root of -13

[soft zealot]

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[plain rain]

oh i see

oke thanks

.solved

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[left trail]

@tardy agate sorry for the ping. you asked how I got my left over area for 14. I had the volume of the sphere minus the cyldiner but it doesn't seem to have a way to simplify it

[tardy agate]

oh yeah np, so u made 2 mistakes

1. You took both the spheres radius and cylinders radius as same
2. The shape of the part that is removed from the sphere is actually not a cylinder it is something like a capsule

[left trail]

how can the radius of the two not be the same? and why is it a capsule and not a pure cylinder?

I am thinking of the problem in this sense:

[tardy agate]

the hole is drilled through it so passes through from the top and bottom leaving behind a cylindrical hole. So in ur figure the top and bottom domes on the cylinder will also be removed. If their radius is same then the entire sphere will be gone. In the figure u can see what would happen if they were the same.

It is called a napkin ring

[left trail]

wait I am mis interpreting so when he says it is drilled from 0 to 2h that means the whole sphere but how can we know the sphere is of 2h?

how can we not deduce anything of our radii for our shapes?

[tardy agate]

it is not, it will look like this

u will get final answer in terms of r,R,h but ig u will be able to eliminate using this triangle

[left trail]

ok so I see that r is the radius of the cyclinder

also in your image the cycldiner goes to the end?

[left trail]

this would mean r is one of the radii for one of the circles that we use to find the volume of the sphere

[tardy agate]

no

how can it go to the end, h is also smaller than R

[tardy agate]

yeah r is smaller than R

[left trail]

I am having trouble seeing where it stops then

[left trail]

but for the volume of the sphere we need to take the sum of all the circles that make it and then we wouldn't be able to use either r or R because they are specific radii and we need a general r when we set up the intergrall

[tardy agate]

use something else to represent it then, or just use spheres equation x^2+y^2+z^2 = R^2, for a circle at height z from centre ull get x^2 + y^2 = R^2-z^2, so the radius is root(R^2 - z^2) integrate this wrt z

[tardy agate]

it stops at h from centre or R-h from top

[left trail]

ins't your R a speicfic radii though

or you mean that R is the radius of any circle that makes the spehre

[tardy agate]

yes, R is the radius of the sphere

read the second part

that is any circle that makes the sphere

its radius is root(R^2 - z^2) where z is its distance from center

sorry ive gtg now but click these hints if u understood the above figure and dont know how to move forward. you can ask other helpers if u are still stuck

you can do it in 2 ways:
||1. Volume of Sphere - Cylinder - Domes, find dome volume using integration||
||2. Integrate to find volume of remaining shape, (from -h to h) dz * cross section area|| ||cross section u can find as bigger circle(the eq i gave above) - pi*r^2(smaller circle of cylinder)||

[left trail]

ok. Thanks for taking the time

[final saddle]

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[tired glen]

...

[warm python]

As $a$ is relatively prime to $n$, $a$ is a generator of $Z_n$.
Let $x \mod n =k$. Then $\phi(x)= \phi(ka) = ka \mod(n)$

[soft zealot]

wai

[desert mantle]

so 2 mod n = 2, so phi(2)=phi(2a) ?

and for k=1, phi(1a)=1a mod n = a ?

[warm python]

<@&268886789983436800>

[warm python]

hmm?

[desert mantle]

well thats what you wrote

[warm python]

ioops

should be k b mod n

[desert mantle]

when working in Z_n you should forget the entire mod n notation

note that they wrote phi(a)=b, not phi(a) = b mod n or something

[warm python]

so should just be kb

[desert mantle]

what exactly

write it again

fix the mistakes

[warm python]

As $s$ is relatively prime to $n$. $a$ is a generator of $Z_n$. Let $x mod n =k$. Then $\phi(x)= \phi(ka) = kb$

[soft zealot]

wai

[desert mantle]

so 2 mod n = 2, so phi(2)=phi(2a)

[warm python]

how did you get phi(2)=phi(2a)

okay

[warm python]

yea

[desert mantle]

but if a isnt 1 then now clearly phi isnt injective

[warm python]

but a is clearly a generator.

so phi(a)= b would mean phi(ka) = kb; where ka=x

[desert mantle]

then fucking write that

[warm python]

Isn't that exactly what I've written

[desert mantle]

are you reading what you are writing

the letter a is not even included in the defining equation linking k and x

[warm python]

oh right

got muddled up

tq

.close

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[meager relic]

Eine dreistellige natürliche Zahl hat die Quersumme 14. Liest man die Zahl von hinten nach vorn und subtrahiert 22, so erhält man eine doppelt so große Zahl. Die mittlere Ziffer ist die Summe der beiden äußeren Ziffern. Wie heißt die Zahl?
Pls speed i need this

[meager relic]

Gauss algorhytm is needed

[faint locust]

Man nenne die Ziffern a, b und c.

<@&268886789983436800>

Kannst du die Zahl mithilfe von a, b und c ausdrücken?

[meager relic]

Ne

[meager relic]

Das ist alles was gegeben ist

[faint locust]

Das ist schon klar, damit ergeben sich auch gleich 3 Gleichungen. Ich meinte eher folgendes:

Die natürliche Zahl können wir jetzt ausdrücken als a * 100 + b * 10 + c

Macht das Sinn?

Weil a, b und c die Ziffern sind

[meager relic]

Kannst du es erklären bitte

Ja aber woher die 10

[faint locust]

Weil das die Zehnerdarstellung ist. Wir haben eine dreistellige natürliche Zahl, ich nenne die Ziffern a, b und c

(die Zahl ist also abc)

und man kann das auch schreiben als a * 100 + b * 10 + c. Ein Beispiel:

Sagen wir, wir haben 684

Das können wir ja auch schreiben als 6 * 100 + 8 * 10 + 4

Oder 432. Das ist dann 4 * 100 + 3 * 10 + 2

[meager relic]

Ah ok

Wie geh,es weiter

[faint locust]

Und unsere Zahl ist jetzt halt abc, also a * 100 + b * 10 + c

Quersumme 14 bedeutet a + b + c = 14

[meager relic]

Ja

[faint locust]

Wenn man die Zahl von hinten nach vorn liest, hat man cba

Also ist der Wert davon c * 100 + b * 10 + a

Liest man die Zahl von hinten nach vorn und subtrahiert 22, so erhält man eine doppelt so große Zahl

Kannst du das also als Gleichung aufschreiben?

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[sonic sky]

so I need someone to help explain what the x1x2<1 means here

I understand the first part that 4^2=2 and thats the radius and it includes the inner and the circle

but how do you know where to draw those parabolas

and how do you know what quadrant

the asymptotes

[storm haven]

Those are not parabolas

You can rearrange x1x2 <= 1 into x2 <= 1/x1

[sonic sky]

yeah I got that

but how do u know what quadrant to put them in

[storm haven]

Wdym

[sonic sky]

like

if the sign was the other way around

would the asymptotes be in the other two quadrants

[tired walrus]

only when x1 is known >0

[sonic sky]

this is just the last thing I got left for my revision so

[storm haven]

Can you clarify

[slim turtle]

If it's positive then 1 and 3

[sonic sky]

and is it not convex when the asymptotes cross through that region

[sonic sky]

wdym

like if the sign was > that way

[storm haven]

Top right and bottom left quadrant

[sonic sky]

yea ik

[storm haven]

Oh, then it would be the area above the 1/x curve for x>0

And below the curve for x<0

[sonic sky]

and whats with that green line

he drew

ahh okay

[storm haven]

Do you still need help?

[storm haven]

It's quite hard to know what he meant to do by drawing that green line without any context

[sonic sky]

aight

so just one last thing

by looking at any question they might have like that in the exam

whats the procedure to drawing the graph

I understand the x1^2+x2^2<4 shit

[storm haven]

If it has 2 or more equations, find the areas that satisfy the two then find their intersection

[storm haven]

If youre struggling with graphing functions then you might have to review your previous lessons

[sonic sky]

no lessons sort of has this sort of stuff

first ive seen of it

[storm haven]

What 💀

Graphing functions is literally one of the most basic things in alg

[sonic sky]

nah

like this sort of question

I mean

I just dont get where the asymptotes are meant to be drawn

and what quadrant and where they start etc especially on this example

[storm haven]

I think organic chemistry tutor might have a video on graphing functions

That might help

[sonic sky]

aight thanks bro

.close

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[final saddle]

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[tranquil pine]

.

i don’t understand this symbol, also i don’t understand how to interpret it as arrows in a graph

[tranquil pine]

heres an example of a partial order

[plucky rover]

You just draw an arrow from x to y if x ≤ y

That's all

[tranquil pine]

can you please explain

i don’t understand what that means or how it’s interpreted

[plucky rover]

I can't actually, I gotta run unfortunately

If this is still open when I'm back I'll catch you

[tranquil pine]

no problem

[tired walrus]

btw jsyk if your message starts with a full stop it PREVENTS channel opening

[tranquil pine]

full stop?

[tired walrus]

the . character

period. dot. whatever you wanna call it

[tired walrus]

this order relation is not the same as the one from your earlier pic

$\preceq$ is not a symbol with any fixed meaning. rather its job is to play the role of a partial order relation and look similar to, but distinct from, the ``regular'' order symbol $\leq$.

[soft zealot]

Ann

[tranquil pine]

same goes for strict orders? With the < sign

[tired walrus]

$\preceq, \prec, \succeq, \succ$ are entirely analogous to $\leq, <, \geq, >$ yes

[soft zealot]

Ann

[tranquil pine]

I see

btw is every relation a subset of a cartesian product

I forgot the name

and if we say R is a relation of A does that imply R ⊆ A x A?

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@tranquil pine Has your question been resolved?

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[stoic marsh]

,claim

[drowsy epoch]

claimed

[stoic marsh]

Hi

[tired walrus]

claiming happens by sending any message btw.

[stoic marsh]

ohhh thanks

[tired walrus]

and the first one is normally supposed to be your question

which, yeah, go ahead and post that now.

[stoic marsh]

So basically i got a test tomrorrow which i have to prepare for and the theme is relative frequency cumulative frequency avreage median kvartilbredde variasjonsbredde.

histo diagram

Quartiles and Interquartile Range (IQR / Quartilbredde)

@drowsy epoch

and Range

i know this one

@rustic sequoia

[tired walrus]

aight yeah you're better off asking in #study-discussion

[brisk charm]

yeah the help channels are better for getting help w specific math questions, whereas #study-discussion is better for advice about studying

[final saddle]

@stoic marsh Has your question been resolved?

[final saddle]

@stoic marsh Has your question been resolved?

[final saddle]

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[indigo swan]

Have i taken then second derivative correctly?

[loud sundial]

$\frac{\cos \alpha}{\sin \alpha}$ is $\cot \alpha$, not $\tan \alpha$.

[soft zealot]

Civil Service Pigeon

[indigo swan]

🫠

Well if I had put cot

What I did after

Is that right

@loud sundial

[loud sundial]

yes

[indigo swan]

Okay thanks

.close

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[loud sundial]

[final saddle]

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[final saddle]

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