help-23

[stoic dune]

I believe the first part is:
f(λx) = λf(x)

The way you'd "check" that is to simplify f(λx) and λf(x), and show they are actually the same thing

[hasty mountain]

f(x1+x2)=f(x1)+f(x2)

[lean otter]

does the Ax = a + d matter at all then?

that's what i'm most confused on

[stoic dune]

What's A?
I see a T in the question, but no A

[lean otter]

Tx sorry

[stoic dune]

You may be trying to picture this linear operator as a matrix multiplication, but that isn't necessary here

[lean otter]

yee

something like that

its not necessary?

[stoic dune]

T simply takes (a b, c d) and returns the number a + d

[lean otter]

ah

[stoic dune]

(Note it is possible to view this as a matrix multiplication! But you definitely don't want to do it here)

[lean otter]

so as long as it satisfies these 2 its a linear transformation right

[stoic dune]

Exactly! In other words, T should be able to move over scalars, and split over +

[lean otter]

icic tyvm

the matrix multiplication theorem is what got me confused

and theres not much information online about it 😅

appreciate it a lot

[stoic dune]

The matrix multiplication is:
[a]
[b] * [1 0 0 1]
[c]
[d]

[lean otter]

ye chatgpt told me that

[stoic dune]

So note the form of the input matrix is not preserved

[lean otter]

but why can 2x2 be represented as 1x4

also v sry for taking ur time

[stoic dune]

They're "isomorphic vector spaces" which is a fancy way to say these two are structured identically

[lean otter]

i see

[stoic dune]

Basically adding 2×2s together is really the same as adding 1×4s together

[lean otter]

do u think the professor would prefer to prove by this

since we havent learned abt that idt

[stoic dune]

Yeah definitely

[lean otter]

betbet

thanks muchh

[stoic dune]

I'm only saying this for fun. It doesn't relate to the question as your professor wants it done

[lean otter]

ah

[manic bane]

did u solve that one nest

[lean otter]

no

lmfao

[manic bane]

isn't it just the derivative

[lean otter]

yea the book tells me to do that

[manic bane]

u can show it with calculus properties

[lean otter]

would u have time to explain

if not is ok

[manic bane]

so f'(x+y) = f'(x) + f'(y)
and f'(k * x) = k* f'(x)

[lean otter]

and since differation is linear, scalar addition and multiplication holds right

[manic bane]

Yeah your calculus rules give those two properties

so D is the function that takes the derivative

[lean otter]

do uk the specific name for the rule mbmb

[manic bane]

uh

i believe addition is called the sum rule

[lean otter]

ohh

wait but product rule doesnt hold

for multiplying derivs no?

[manic bane]

it's just for scalars

[lean otter]

in proving linear transformation right

[lean otter]

damn

alr ty i wont take any more of ur time

appreciate it i understand a little more now

[safe radish]

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[left sapphire]

i need help im a dutch guy so idk how to type all of it in english but ig here we go, so i have finals tommorow and im currently studying for math its about square roots, power roots, and stuff like that but im stuck if with wierd formulas and stuff help is needed

[fathom adder]

Sure sends the questions where you are stuck

[twin panther]

!noask

[left sapphire]

ok

[twin panther]

!ask

what’s the command

!donotask

[fathom adder]

Idk

[twin panther]

!ask

!noquestion

[left sapphire]

this is the test of a friend

[fathom adder]

Guys stop xd

[left sapphire]

and i have final of it tomorow

but idk what to do

[twin panther]

This is easy?

Lowk

[left sapphire]

not for mw

i missed alot of classes bcs of some reasons

and didnt get this test nor information

[twin panther]

What do u need help with?

With what

Equation

[left sapphire]

kindof everything

in particular

[fathom adder]

Don't judge on ppls level

[left sapphire]

the 2nd picture

[twin panther]

There’s no need to explain it, you mustc watch a video

[left sapphire]

i know its easy if i get the information about i

[twin panther]

https://youtu.be/B4zejSI8zho?si=DR-q4_mXewC3RlhZ

[left sapphire]

is it that?

[twin panther]

,w frac(1)(sqrt(c)

[flat frigate]

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Requested by joninavaltingojer

[twin panther]

I honestly don’t undertake how you could get this wrong

,w 1/sqrt(c)

[flat frigate]

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Requested by joninavaltingojer

[twin panther]

8:2
4:2
2:2
1

$2\sqrt(2)$

[flat frigate]

Jónína Valtingojer

[left sapphire]

so thats the solution

[twin panther]

Yes….

[left sapphire]

of

[twin panther]

$\sqrt{8}$

[left sapphire]

1/sqrtc

i mean

[flat frigate]

Jónína Valtingojer

[left sapphire]

the teacher corrected it on his page

and the answer to aquare root 8 was 8 1/2

[twin panther]

That’s also correct

2 * 2 * 2 = 8

[left sapphire]

and to 1/sqrt (C)

it was

c-1/2

[twin panther]

1sqrt(c)
1sqrt(c)

You eliminate 1 and you have sqrt(c)

[left sapphire]

so square root c was correct?

[twin panther]

@left sapphire

[twin panther]

No

[left sapphire]

teacher said to him it was wrong

that was his original answer

[left sapphire]

as u can see here

[twin panther]

It’s 1 my dude

[left sapphire]

so the answer is 1

[twin panther]

$1\sqrt{c} / c$

Answer

[flat frigate]

Jónína Valtingojer

[left sapphire]

this is chinese for me

[twin panther]

Yeah

[safe radish]

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

[left sapphire]

i dont get this shi

this this

wait

[twin panther]

I have to go.

Goodbye!

<@&286206848099549185> Help them.

[left sapphire]

damn thats harsh ngl

that

[digital sparrow]

<@&268886789983436800> pls do something about jonina

lmaaooo

[left sapphire]

just needed help with something im stuck on bcs i missed class and stuff and then she starting to be rude #

[twin panther]

It’s isqrt(81)

[left sapphire]

ok

[twin panther]

,w minus square

[flat frigate]

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Jónína Valtingojer

[twin panther]

@left sapphire

[digital sparrow]

hey bro is your question how to do root(-81)

[left sapphire]

ye?

no

[twin panther]

I already answered them.

[left sapphire]

4th root of -81

idk the english word for that

[twin panther]

Square root*

[digital sparrow]

ohh

[twin panther]

@left sapphire

[left sapphire]

no

[twilit furnace]

It's not a square root lol

[left sapphire]

nth-power roots

[digital sparrow]

${-81}^{1/4}$

[twin panther]

https://images.app.goo.gl/upXY54y1BXB3AzXL8

[flat frigate]

Wumpus Man

[twin panther]

Incorrect.

[twilit furnace]

You're not being helpful, please stop @twin panther or I'll time you out for now

[left sapphire]

ur incorrect ur saying its square roots but its different

[digital sparrow]

$\sqrt[4]{-81}$

[twin panther]

No comment…

[flat frigate]

Wumpus Man

[left sapphire]

y

e

[digital sparrow]

lmao

[left sapphire]

thats what the quation it

i need to reform it into womthing with x/x

and then find the solution ot it

to it

or find the solution imediatly

so thats

sqrt -81 1/4

or isnt it

so the answer is -3

[rugged patio]

Unsolvable

[rugged patio]

this is unsolvable

[digital sparrow]

you should try writing in polar coords

[left sapphire]

wich means/

[digital sparrow]

i am not sure myself how to do this

i recommend pinging helpers

[left sapphire]

im not good in math

[safe radish]

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.

[rugged patio]

no, there isn't root of negative numbers

[left sapphire]

oh alr

[rugged patio]

Its literally unsolvable

[left sapphire]

so what do i write here bcs she said i have to solve them and stuff

[twin panther]

^^^^

[digital sparrow]

i think its best if we ignore jonina

[twin panther]

Be for real

[twin panther]

what did i do?

[empty gyro]

You said this correctly. Are you trying to write $\sqrt[4]{-81}$ in exponent form, or are you trying to simplify it?

[flat frigate]

SWR

[left sapphire]

uhm

the question is

calculate without a calculator, so write down the necessary intermediate steps, a result is often not sufficient to show that you have taken the right steps

[rugged patio]

Maybe you have did something wrong in the solution and you got this while you shouldn't have

[left sapphire]

and they gave me

$\sqrt[4]{-81}$

[flat frigate]

afrikaan

[rugged patio]

If you have taken right steps

[empty gyro]

In the set of real numbers, yes. But with complex numbers, this is perfectly legitimate

[left sapphire]

idk its all i knoq

idk what to do

[twilit furnace]

It doesn't look like OP is working with complex numbers at all

[digital sparrow]

acc to your profile you are 15yo or in 9th/10th
in 11th you will learn about complex numbers
pls stop messaging now

[twilit furnace]

What does the teacher expect of your methodology/results? What are you trying to perform?

[empty gyro]

It may be a bit difficult because it will involve solving $\sqrt{i}$, which is annoying if you don't immediately know it

[flat frigate]

SWR

[left sapphire]

yuh i know but ye like i said i missed the test and some classes bcs of some reasons and yea now i have finals without any knowledge in this

[rugged patio]

I study 11th grade mathematics 😭

I know you put i

[left sapphire]

i know how to do some thing like 3th power root of 27 to the power of 2 is 9

and stuff

[rugged patio]

for that purpose if you have the thing

[left sapphire]

but some things are pretty diffevult

[empty gyro]

It's hard to help you, sadly. While $\sqrt[4]{-81}$ is "reducable" it is hard to say what you need to do because we don't have the context here on what your teacher is expecting. As ghost said, i see no other complex numbers in your assignments, so this problem is very suspicious

[flat frigate]

SWR

[left sapphire]

this was the question

[rugged patio]

quastion

[left sapphire]

calculate without a calculator, so write down the necessary intermediate steps, a result is often not sufficient to show that you have taken the right steps

[empty gyro]

With this context, best i can think of is $3\sqrt{i}$, or $\frac{3\sqrt{2}}{2}+i\frac{3\sqrt{2}}{2}$ in standard complex form, or $3e^{i\pi/4}$ in polar form, or $3i^{1/2}$ in exponent form. It's really hard to say what your teacher wants right now

[left sapphire]

ye idk brother

but then still

the other thing she asked us is to make a aquare root of 4 to the power of -0,5

[flat frigate]

SWR

[empty gyro]

That makes more sense at least. I wish i could help but i must leave now. Good luck to you

[left sapphire]

thanks

if anyone else can help me ping me pls

[safe radish]

@left sapphire Has your question been resolved?

[south aspen]

?

this is so ez

afrikaan

u still want the answer

show me the question

@left sapphire

[left sapphire]

no im good

[safe radish]

@left sapphire Has your question been resolved?

[safe radish]

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[west hedge]

What is the “a” called in mathematical term

A

Sass

Aasaa

Help!

SOS

Power

As

[hard crest]

argument ig

[west hedge]

What

[west hedge]

I cannot

I cannot find such meaning in Cambridge dictionary

Do not lie to the fellow of lord Mickey

Please

Lord Mickey, guide me

[honest perch]

no one calls it anything in particular

[west hedge]

What

How professor describes

Then

[honest perch]

it's just a, nothing more

[west hedge]

How professor, professionals describe it then

What

Unbelievable

[peak estuary]

argument is fine

[west hedge]

Unbelievable

[honest perch]

why do you want a name

[peak estuary]

or colloquially you can just say input

[west hedge]

I would always trust a man with default avatar

Thank you

The question is solved

[honest perch]

what

[peak estuary]

or do you mean that for say b^x we call b specifically the base?

do you mean some variant for that specifically for log?

I dont think that exists

[west hedge]

I believe you, man with a default avatar

You looks credible and reliable

Thank you

.close

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[west hedge]

I luv you

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[smoky canopy]

i'm struggling to do this one:
Inside an isosceles trapezoid with sharp angles of 60 degrees, and with the shorter base being 1, a circle is inscribed. Find the length of the trapezoid's diagonal.

[cunning pasture]

What's the use of the circle being inscribed?

[smoky canopy]

if i knew i would probably solve the problem

[topaz tree]

It touches all of the sides

I think that is the meaning here

[cunning pasture]

What does it have to do with the question?

[topaz tree]

Idk prolly a misdirection or something

[cunning pasture]

Or maybe it's given as a hint, that may help solve the question.

[smoky canopy]

wouldn't call it a misdirection these are practice for university application exam

[topaz tree]

four angles and one side uniquely identifies a tetragon(or quadrilateral)

[cunning pasture]

Oh great

[smoky canopy]

we are not constructing i need to find the length of the diagonal

we have top side the smaller one being 1

[ornate skiff]

https://en.wikipedia.org/wiki/Tangential_quadrilateral might be relevant. The circle touches all 4 sides of the trapezoid, so it tells something of the trapezoid... (it doesn't happen to all trapezoids)

[smoky canopy]

yeah i know that i just don't know how to use that

[cunning pasture]

wait

The circle thing has to do with the question.

Not all quadrilateral are tangential.

[smoky canopy]

well yeah

[cunning pasture]

Also, in a tangential quadrilateral, the sum of opposite sides is equal.

You should draw the altitude from the shorter base to the longer base and find the longer base in terms of the non-parallel sides.

[cunning pasture]

Now use this fact.

[safe radish]

@smoky canopy Has your question been resolved?

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[wary dagger]

Could someone help me with this

[topaz tree]

domain

[wary dagger]

I got it nvm

Thank u thi

.close

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[twin tree]

i think the ans to a) is 200degree but im not sure

i also dont know the ans to any other questions

[ornate skiff]

Need to use the relation between the measure of an angle with the tip on the center of circle and the measure of the arc it describes on the circle...

[thin bridge]

can you try marking which angle you think they want you to find for part a)

[ornate skiff]

aka...for example...m(POR)=m(PR) (the arc)

[twin tree]

how do i do that?

[twin tree]

humm wait

[cunning pasture]

Central?

[thin bridge]

no, that's incorrect

[twin tree]

oh

[cunning pasture]

The central angle is basically the reflex angle POR.

[thin bridge]

first mark the arcPSR (on the circumference)

[twin tree]

ohk

Now

[thin bridge]

what you indicated was <PSR which is in part b)
show what arc you've highlighted

[twin tree]

okk

i have a doubt

[twin tree]

By this, should i just draw a line from S to Q

[thin bridge]

no

[twin tree]

oh

[thin bridge]

sry also made a typo there,

[twin tree]

Np

[thin bridge]

do you know how arc naming works

[twin tree]

sryy i dont

is it jst a line on the circle like a curve

[thin bridge]

yes

[twin tree]

oh

[thin bridge]

with two letters, it'll be the minor arc, (the shorter part from one point to the other)
with three letters, you start from the first letter, go through the middle letter to get to the third

[twin tree]

ohh

[twin tree]

so PS will be a minor arc right

[thin bridge]

yes

[twin tree]

so how can i solve a)

uhmmm sir??

[thin bridge]

try attempting to mark arcPSR again

[twin tree]

okii

[thin bridge]

yes

now can you identify the location of the central angle of that

[twin tree]

yeeeeeee

is it 260degree

[thin bridge]

yes

[twin tree]

oh

for b) PSR is 130 right

[thin bridge]

no

[twin tree]

then how can i solve it??

[thin bridge]

identify which arc subtends <PSR

and then the central angle of that arc

[twin tree]

uhmmm sry i didn't understand

[thin bridge]

try first looking up inscribed angle theorem

[twin tree]

okiii

so divide by 2

sry im a little dumb

[thin bridge]

divide what by 2

[twin tree]

idk central angle ig

[thin bridge]

and what's the corresponding central angle for <PSR

[twin tree]

260

[thin bridge]

no

[twin tree]

100?

[thin bridge]

yes

[twin tree]

oki

[thin bridge]

(look at the arc the angle is facing)

[twin tree]

100?

arc PQR

[thin bridge]

<PSR faces the minor arcPR
which the (non-reflex) central angle POR is facing

[twin tree]

oh

i get it

uhh sir

am i wasting your time rn???

Im sry

.close

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[cyan quail]

What percentage of this circle is coloured

And all the dots are equally spread

Please help 😭😭😭

<@&286206848099549185>

[safe radish]

@cyan quail Has your question been resolved?

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[cosmic lily]

The figure below shows a sphere suspended from a wire with 1
meter long, fixed at point 𝑂.
The center of the sphere oscillates between points 𝐴 and 𝐵, which are symmetric
relative to the vertical line 𝑟.
Line 𝑟 passes through point 𝑂 and is perpendicular to line 𝑂𝑆.
At the initial instant, the center of the sphere coincides with point 𝐴.
Assume that, 𝑡 seconds after this initial instant, the center of the sphere is at a
point 𝑃 ​​such that amplitude, in radians, of angle 𝑆𝑂𝑃 is given
(approximately) by s(t) = Pi/2 -Pi/6 x cos ( square root 9,8 x t )
In the following two paragraphs, do not use the calculator, except for possible numerical calculations.
2.1. Determine the distance from the center of the sphere to the straight line 𝑂𝑆, at the initial instant.
2.2. Determine the instant at which the center of the sphere first passes line 𝑟.
Displays the result in seconds, rounded to the nearest tenth.

[cosmic lily]

im so cooked

[topaz tree]

!status

[safe radish]

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

[cosmic lily]

1

[topaz tree]

Ok, so from the beginning

draw a triangle OPD, where D intersects OS at a right angles

[cosmic lily]

D?

[topaz tree]

Like this

[cosmic lily]

oh ok mb( my english kinda bad) I put a line to create a 90 angle

K

[topaz tree]

so, <SOP is already given right?

[cosmic lily]

yup

[topaz tree]

at the initial moment t = 0

and we have to find the distance DP at the initial moment

are you getting me?

[cosmic lily]

yeah

im just taking notes at the same time sry

[topaz tree]

PD = sin(<SOP)

trigonometry

sin(<SOP) = PD/OP = PD

now, <SOP = pi/2 - pi/6 cos(0), because t = 0

so, <SOP = pi/3

[cosmic lily]

oh ok

[topaz tree]

sin(pi/3) = PD = sqrt{3}/2

this is the first part

so, height at initial moment = PD = AD = sqrt{3}/2

now when the sphere is passing through r, what is the angle <SOP?

[cosmic lily]

90

[topaz tree]

yes

pi/2 = <SOP = pi/2 - pi/6 cos(sqrt{9.8}t)

when we simplify this we get cos(sqrt{9.8}t) = 0

so, $\sqrt{9.8}t = \pi/2$

[flat frigate]

Pro_Hecker

[topaz tree]

$\therefore t =\frac{\pi}{2\sqrt{9.8}}$

[flat frigate]

Pro_Hecker

[safe radish]

@cosmic lily Has your question been resolved?

[safe radish]

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[runic rampart]

I don't see what's wrong with my sine series

I found the formula like this

then plugged in n=1,5,7,11

but it's telling me my sequence is wrong?

<@&286206848099549185>

[safe radish]

@runic rampart Has your question been resolved?

[runic rampart]

<@&286206848099549185>

[runic rampart]

.close

[safe radish]

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[safe radish]

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[wind shoal]

i dont get what they mean by both directions is it this

[deep lily]

No

Your question is in the form of "A if and only if B"
In both directions means you need to prove that "If A then B" and also "If B then A"

[wind shoal]

hmm

can i show the proof of determinant

not equal to 0

to show both A and B are invertible

[fathom jewel]

E.g. You wanna assume (AB) is invertible and then prove that A and B must be invertible.
Now the other direction is, you assume A and B is invertible, and you wanna show that their product is also now invertible.

[wind shoal]

to prove that A and B must be invertible. u have to do this no

wait gimme a sec

[wind shoal]

like this

assume (AB) is invertible and then prove that A and B must be invertible.

[wind shoal]

and the other way is this no

[wind shoal]

by their product is invertible

[fathom jewel]

[wind shoal]

isnt this what i did in the image assume A and B is invertible, and you wanna show that their product is also now invertible.

[fathom jewel]

Yea, I think it's fine

You assumed A and B are invertible, therefore used their inverses to somehow prove that AB = BA = I

[wind shoal]

kk ty

.close

[safe radish]

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[safe radish]

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[fair goblet]

Where is my mistake for problem 13? I can’t identify it

[drowsy moss]

it's not done, but other than that I think its fine so far

[fair goblet]

Ah I thought I made a mistake

I guess I didn’t

How would I find C?

If I plug in the values

It would be 0 and 0

And I get 0 = ln(C)

[drowsy moss]

yep

oh. actually. there is a small mistake.

[fair goblet]

Hm?

[drowsy moss]

ln(e^-y) = ln(-x^2/2 + C)
You have the - outside the ln it should be inside

[fair goblet]

I’m having trouble visualizing

Can you explain

More

Sorry

[drowsy moss]

$-e^{-y} = \frac{x^2}2+c \implies e^{-y} = -\frac{x^2}{2}+C\implies \ln(e^{-y}) = \ln(-\frac{x^2}2+C)$

[flat frigate]

Zybikron

[drowsy moss]

you have $\ln(e^{-y}) = -\ln(\frac{x^2}2+C)$

[flat frigate]

Zybikron

[fair goblet]

Ic I get it

Thank u

[safe radish]

@fair goblet Has your question been resolved?

[fresh wasp]

if your question has been answered you can close the chat with .close goodnight ❤️

[safe radish]

Channel closed

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[fair goblet]

Haha sorry I’m new here

Thanks

[fresh wasp]

its ok

[safe radish]

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[manic arch]

it more of an open ended question but
anyone here recognize any function that looks like this ?
im trying to find a function that will fit this curve
its in logarithmic view just in case you havent noticed

[left gyro]

try switching away from a logarithmic view then

[manic arch]

nahh, i need the precision in the small value
it's looks like a normal function in linear

[left gyro]

it looks like at the ends its straight, to represent that you can draw in those portions as logarithms

have you confirmed what function is looks like when its linear?

[left gyro]

what function did you come up with

[manic arch]

some kind of sum of 4 "normal distirbution"

[left gyro]

you are aware that adding four normal distributions results in another normal distribution?

those are normal distributions

add them

[left gyro]

also, have you confirmed that this is correctly the result of adding the normal distributions?

[manic arch]

i mean, i modify the normal formula so
it simply a^x^2 * b and play with the variable

[left gyro]

are you getting this data from adding the results of 4 normally distributed data?

[manic arch]

im getting this curve by guessing

the data was taken from neuralink comression challenge actually

[left gyro]

goddammit

can you be more specific on what the data is

[manic arch]

so
i ahve this frequency of signal fluctuation between -255 and 255

[covert yoke]

I think the user is saying it the PDFs of a few normal distributions added together

Not the actual random variables

[left gyro]

I have a feeling that is not correct and need more context

maybe theyre thinking "to add two normal distributions you add their PDFs"

[manic arch]

im not well versed in statictic btw, so im kinda clueless on what to say

[covert yoke]

True

[left gyro]

yes or no?

[manic arch]

but the point is im trying to find a function that will fit the curve

[covert yoke]

Where did this set of points come from?

Directly from the data given?

Or did you mess with them somehow?

[manic arch]

the difference between 1 data and the next

[covert yoke]

And these first differences are binned in a histogram?

[manic arch]

yea

[covert yoke]

They do not seem to be a direct time series.

[left gyro]

(when youre done, Im seeing https://news.ycombinator.com/item?id=40531157 which may not be a good sign)

thinktechthurs

The Neuralink compression challenge seems impossible

Having a job means that you have to do two main tasks. The first is doing whatever your job description says. If you're an engineer, you have to build widgets; if you're in sales, you have to sell widgets; if you're a designer, you have to design better widgets. The second is doing whatever your manager says. Unfortunately, it's not rare that yo...

Hacker News · May 31, 2024, 3:18 AM

[covert yoke]

Oh ueah

[manic arch]

im just having fun lol

[covert yoke]

The neuralink challenge is real dumb, but it's a fun data set

[left gyro]

oh I see

so youre trying to add 4 normalish PDFs to get the data to match is all?

[manic arch]

yea, its just a decent guess i ahve rn

[covert yoke]

You could probably manage a good fit with only 2 based on the logarithmic display

Why did you settle on 4?

[left gyro]

to me, was it that you just tried to put in more until it looked good?

[manic arch]

yea lol

this what each of these 4 funtion i add up looks like

[covert yoke]

What if you use just the 3rd and 4th one?

[covert yoke]

Interesting

Anyway, to answer your direct question, no I'm not aware of a function that behaves in this manner.

[left gyro]

got an idea

[left gyro]

try stretching the data horizontally and see if it appears normal

[manic arch]

in which direction ?

[left gyro]

stretching presumably means to make the data horizontally wider

[manic arch]

oh wait horiznotal lol

it loks normal if you see it close enough

[left gyro]

it sort of looks too weird not to be

the issue here is that the center and the edges are comparatively thin

normal would be thick in the center compared to the edges
theres other distributions which have differing amounts which could work for your guess

[manic arch]

what is it ?

[left gyro]

well I dont know if theyre worth guessing, but you can try searching up t-distribution

theres also 1/(1+x^2) and 1 / cosh x which you can try scaling around (they may not entirely be distributions)

[manic arch]

it seems like a good start

aight ty so much

[left gyro]

what you could consider trying out is finding a good picture for the far ends in the logarithmic view

it looks like it goes to lines, a^x and scaled copies will draw lines in the logarithmic view

[manic arch]

wait sorry its not histogram, but frequency

.close

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[cinder halo]

I can't read that. Can you make it darker?

[safe radish]

@zinc salmon Has your question been resolved?

[safe radish]

@zinc salmon Has your question been resolved?

[safe radish]

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[odd folio]

Can someone see if I did it right pls

[severe pond]

try submitting it

and see what it says

green means correct

red means incorrect

[odd folio]

The homework is a one try thing so I’m trying to get someone to review my answers 😭

[severe pond]

show the full screen

[odd folio]

Ok

[severe pond]

you good?

[odd folio]

Yeah wait I was on another problem

[severe pond]

please don’t waste time

[severe pond]

brother

what’s the title of the assignment

you’re still covering that

do you need help with an exam

[odd folio]

Bro that’s literally top and bottom of my screen

[severe pond]

no it’s not

[odd folio]

Ok I’ll send another one

[severe pond]

😴

[odd folio]

Bro my wifi sucks damn hold on😭😭

See

[severe pond]

yea we got you bro

just wait a second

trust

[plucky elk]

Lmao why did you delete part of the url

[odd folio]

Bro what are u talking abt the url is right there

[severe pond]

time in top right

😭😭🙏🏻

[odd folio]

Yah it’s timed

[severe pond]

timed homework

🤔

[odd folio]

Yeah u never heard of it

[severe pond]

100 points possible

[odd folio]

Yup

[severe pond]

brother you clearly spent time trying to hide this

it took you 5 minutes

[odd folio]

?

My WiFi just atusllt sucks

I’ll just ask chat instead

[versed wave]

heres a joke, what do you call a homework that is graded, timed and only allow one submission attempt?

a god damn test

[severe pond]

how is this guy not gone

[stiff bobcat]

I thought smay was handling it

[severe pond]

yea they’re not typing anymore

[stiff bobcat]

this looks like a (timed) test and cheating is against the rules here.

[gleaming hawk]

Not his first time doing it either

[severe pond]

really?

[lusty breach]

Dang

[severe pond]

gigagoat

[gleaming hawk]

Look at previous messages

[lusty breach]

Blud really thought 😭😭

43 mins in the top right

[gleaming hawk]

Mf sent full blown video of the test question 💀

[lusty breach]

Fr

[lusty breach]

Hey knief

[gleaming hawk]

#help-23 message

[lusty breach]

How’s your math class going

[gleaming hawk]

Look at this crap

[severe pond]

bruh

[lusty breach]

Are you still doing topology or whatever

[severe pond]

yea i just saw his message history

[severe pond]

no that was one homework

just an intro shit

[lusty breach]

So what are you doing now

[severe pond]

i finished my last homework for stats class today and i just have finals now

[severe pond]

this dude really got help lol

crazy

[lusty breach]

No way 💀

[severe pond]

look at his message history

he sent videos and screenshots of his exam

[lusty breach]

Alr

Damn

So this isn’t the first time 😭

[severe pond]

yep

[safe radish]

@odd folio Has your question been resolved?

[gleaming hawk]

.close

[safe radish]

Channel closed

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[lean otter]

i dont understand

[versed wave]

understand what?

[open junco]

Take r^n common

[gleaming hawk]

They factored out r^n

[lean otter]

oh lol

[gleaming hawk]

What exactly is there to understand?

[lean otter]

silly me had the same question a while ago

thanks

[open junco]

Probably miss read

[lean otter]

.close

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[safe radish]

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[open junco]

Which one

[nimble blaze]

I wanna see if my answers are correct

for all of them

its part of the same question

is 2 a correct? Chatgpt says otherwise

[trim swan]

chatgpt is still really bad at math

[nimble blaze]

yeah ik it sucks, I just wanna make sure I am doing it right

cuz my final is tomorrow

in like 8 hours

and Im doing the practice paper rn

[trim swan]

You are right

[nimble blaze]

Thanks bro, what about the rest

[trim swan]

This is obviously wrong according to the graph

Everything else so far is right

but what about b?

[nimble blaze]

I was confused on that, how do I approach the problem?

[trim swan]

Have you learned about derivatives?

[nimble blaze]

yes

[trim swan]

the expression in part b is a derivative

[nimble blaze]

right

so do I need to integrate that?

[trim swan]

no, you just need to interpret it as a derivative

[nimble blaze]

okay, so factor and replace x with 0?

[trim swan]

wait wait

[nimble blaze]

okay, by the way thanks a ton for helping out

[trim swan]

remember $f'(x) = \li{h}{0} \frac{f(x+h)-f(x)}{h}$

[flat frigate]

tatpoj

[trim swan]

np

[nimble blaze]

oh yes!

so how do I proceed with the next step?

[open junco]

So f'(-4)

[trim swan]

$$\li{h}{0} \frac{f(x+h)-f(x)}{h} = f'(x)$$
so
$$f'(-4) = \li{h}{0} \frac{f(-4+h)-f(-4)}{h}$$

[nimble blaze]

which is 1.5!!?

[trim swan]

it's f'(-4), not f'(4)

[flat frigate]

tatpoj

[nimble blaze]

So 2?

or like 1.9999

[trim swan]

First of all, do you see why it's f'(-4)?

[nimble blaze]

Yes because the x value is replaced by -4

its the f prime

[open junco]

Yah sorry

[trim swan]

Yeah, the expression in part b is the definition of f'(x) evaluated at x=4

[nimble blaze]

your good bro, thanks for helping out

[open junco]

That basically means slope at that point

[trim swan]

all good

[nimble blaze]

I see

[trim swan]

yeah, f(-4)=2, but not f'(-4)

[open junco]

So I would guess dne

[nimble blaze]

I see, so its DNE

[open junco]

You understand why dne

[nimble blaze]

Yes because we are finding f prime of -4 not just f(4)

also guys, is there a good AI tool for checking your answers?

that you guys recommend

[trim swan]

not AI, but for certain things WolframAlpha is really good. But not if you'd have to upload a picture

[trim swan]

The reason f'(-4) is undefined, is because the slope of the function at x=-4 is undefined, because of the sharp point there.

[nimble blaze]

Yeah I tried that for the image I sent

[nimble blaze]

I see

so if it was a straight line then we could find it?\

[trim swan]

a straight line, or even just a smooth curve

just not a jagged corner

[open junco]

Yes

[trim swan]

or a broken curve like at x=5

[nimble blaze]

I see, thank you so much guys, yall doing lords work here

is there a limit on number of questions I can get answered on this server?

[trim swan]

nah, just close this channel before you open another one

[nimble blaze]

Alright, I'll be back if I got another question. Have a good one

[trim swan]

Same to you

[nimble blaze]

.close

[safe radish]

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[hasty wagon]

I need help

[burnt notch]

Make a substitution, n = k-1

And you'll see you obtain something well known

[hasty wagon]

?

,close

.close

[safe radish]

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[azure plover]

Where’s the height?

[azure plover]

Need to find area.

<@&286206848099549185>

[dawn kite]

You can use pythagoras to find all the sides

And then find put the area of the two triangles and add them together

[azure plover]

Okok

[dawn kite]

Is that all or do you have any further questions?

if no you can close the session with .close

[azure plover]

Did I do the area of half circle correctly? @dawn kite

[dawn kite]

i think you used the diameter of the circle not the radius

so you used 11 but 11 is the diameter of the circle

so to get radius you just divide the diameter by half

so instead of 11^2 x π x 1/2 it is 5.5^2 x π x 1/2

[safe radish]

@azure plover Has your question been resolved?

[azure plover]

@dawn kite
Area and circumference of halfsicrlce

Pi * r^2 * 1/2
Pi * r

[dawn kite]

no the circumfrence is
d * Pi
and Area is simplified
d/2 * Pi * 1/2

and d is diameter

[azure plover]

WHAT!

I mean a half circle

@dawn kite

<@&286206848099549185>

[dawn kite]

area of a half circle?

[azure plover]

I did it right

Pi * r^2 * 1/2

[dawn kite]

Yes

but you calculated with the diameter and not the radius

[azure plover]

Yh I fixed that

@dawn kite

@dawn kite when I was finding cirumfetence of semicircle

I did pi * r was I correct?

[dawn kite]

No it is pi * r^2 / 2

because r^2 is not the same as 2r

[azure plover]

I did this right

And circumference of a semi circle

Pi * r

[hoary compass]

No

hey

[dawn kite]

yes

[azure plover]

Ur trolling me

[dawn kite]

So you cant just remove r^2 by diving by 2 because r^2 means r * r

[hoary compass]

I'm korea man

[dawn kite]

But you can revove 2r by diving by 2 because 2r is just 2 * r

[hoary compass]

Do you guys know the formula that you've never seen before?

[dawn kite]

so r^2 is not the same as 2r

[azure plover]

I did it right

Pi * 11^2 * 1/2

And

2pir circumference circle
Semi is pi * r

[dawn kite]

what did you put as r?

did you use the red one as r or the green one?

[azure plover]

I did firstly 11 cm

But then I fixed it

To 5,5 cm

[dawn kite]

okay with 5.5 that is correct then

[azure plover]

I know I should use radius

I just forgot that it was diameter

[dawn kite]

yes that is correct

[azure plover]

Ok then I’m good on the exam

Im do happy

I thought I was gonna do everything wrong

[dawn kite]

well then gl then

[azure plover]

Ty

[dawn kite]

any further questions?

If no you can close the session

with .close

[azure plover]

No

[safe radish]

@azure plover Has your question been resolved?

[safe radish]

@azure plover Has your question been resolved?

[azure plover]

.close

[safe radish]

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[hard wigeon]

.help

[safe radish]

Commands:

• clopen: .close, .reopen
• consensus: .poll
• factoids: .tag
• help: .help
• version: .version

Type .help <command name> for more info on a command.

[hard wigeon]

.open

[civic parcel]

hiii

[safe radish]

@civic parcel Has your question been resolved?

[safe radish]

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[knotty jungle]

any help please

The number N = 1000 can be written as the product of four numbers, e.g.
N = 4 x 5 x 5 x 10.
Of course, this is not unique, because there is also true
N = 2 x 2 x 5 x 50.
In this exercise we want to check in how many ways 1000 can be written as the product of four natural numbers, i.e.
N = a x b x c x d, (1)
with 2 ≤ a ≤ b ≤ c ≤ d and a, b, c, d ∈ N. Note: We do not want permutations to count,
so therefore the choice a ≤ . . . ≤ d.

(a) Check in how many ways 1000 can be written as a product of four numbers. (1)

[knotty jungle]

@buoyant shadow if u could help thatd be nice bc ur good at these problems

[safe radish]

@knotty jungle Has your question been resolved?

[simple coral]

I am thinking about writing a script to solve this problem

[knotty jungle]

yeah u can

I’m going to have to do this in matlab

But I’m trying to find a formula first but I can’t

@simple coral if u find something lmk

its due friday

[simple coral]

For sure I am on my way creating the script

[safe radish]

Channel closed

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[knotty jungle]

.reopen

weird

[safe radish]

✅

[safe radish]

@knotty jungle Has your question been resolved?

[safe radish]

@knotty jungle Has your question been resolved?

[knotty jungle]

i got 26

@simple coral

now 8

[simple coral]

I think the script will be overkilling, but to put it simple

from the description of the problem a, b , c and d can equal each others and they can be as small as 2, so if we give all of them 2 then we will get 2 * 2 * 2 * 2 = 16 which not what we want we need to get 1000 so must one of them be 1000 / 2 / 2 / 2 which equal to 125

it can't be a, b or c because none of those can be greater than d so the value of d = 125 it can't be greater, but it can be less 125 as the values of a, b, c increase

now to get the next value of d and c (we selected c because it's the next greater number after d we can't select a or b because that will contradict the problem conditions), d is going to be 1000 / whatever we will give to c / 2 / 2

so we can say that d = 1000/c/2/2, the value of c should be a number that doesn't make d none integer

integer = N

also it doesn't make the whole number greater than 1000

[knotty jungle]

okay interesting

[simple coral]

we need to keep following the condition so we can't increase a be greater than b or b to be greater than c or c to be greater than d

[simple coral]

I don't know if I made it more complicated 😅

the next value for c is 5 now let's see if we can increase b

d = 1000/5/2/2 = 50

[simple coral]

we can also use 10 for c

but assuming you got the pattern then you can work your way to get the right answer

[knotty jungle]

ur supposed to make a function for

The next question

Using the same principe

And plugging in 1million

Should give u 224

For the number of possibilities

So I’m trying to make one

I have a few ways

[simple coral]

oh, ok then I think a script will really come in handy in this case

[knotty jungle]

But the most efficient and fastest one gives u a higher grade

[simple coral]

you do programming as well?

[knotty jungle]

matlab

not that good in programming in general lol

[simple coral]

I didn't use this, but on the internet I saw about it, it looks like for studying programming

[simple coral]

it's easy

basically all you need to break down the problem and make a loop so the computer can repeat that code in that loop over and over again this loop should work on extracting the solutions

but first of you need to break down the problem as I was doing early

i am just rewriting the script

[lean otter]

Could anyone tell me well Pythagoras thereom formulas

[simple coral]

please open another channel for your question

[lean otter]

Which one

[simple coral]

#help-18

@lean otter #help-8

[knotty jungle]

@simple coral any progress?

[simple coral]

yes, but I am afraid it's going to be a little bit complex (the code) I am using Javascript by the way

[knotty jungle]

yeah its def not the easiest either way

im trying diff methods but the less loops the better

[buoyant shadow]

no idea how matlab works

[knotty jungle]

do u know an efficient way to calculate it ?

i can try get it in matlab myself its just theres so many ways

[buoyant shadow]

no, it's just "ask computer"

[simple coral]

I will finish and send it to you, in Javascript but converting to any another programming language is so easy

the hardest part is to put the code together

[knotty jungle]

oki ty

[buoyant shadow]

@knotty jungle

does 8 sound right lol?

[knotty jungle]

for 1000?

[buoyant shadow]

yes

[knotty jungle]

yes thats correct

for 1000000 it should be 224

[buoyant shadow]

yeah

[knotty jungle]

crazy

[buoyant shadow]

good news is matlab looks normal

it should be translatable

[knotty jungle]

what does prev do? @buoyant shadow

apparently the same command exists in matlab

[buoyant shadow]

it's not a command it's a variable name

it could be anything instead of prev

[knotty jungle]

ah

[safe radish]

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[dire terrace]

Yo

[dire terrace]

Help

i need it

Sigrid brings a 1.2 m long tube, open at one end, down into a lake. The tube is held vertically with the open end facing downward. When this happens, the water level in the tube rises. Determine how far below the water surface the open end of the tube is if the water level has risen to a quarter of the tube's length.

its a physics question

btw

the answer is 3.7m idk how they got that

when the tube itself is 1.2m long

[safe radish]

@dire terrace Has your question been resolved?

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[obsidian pivot]

Show that every natural number that bigger than 6 can be written as addition of two coprime numbers that bigger than 1.

[obsidian pivot]

We can write odd numbers with form of n = 2 + (n - 2)

But i do not know how to write even numbers as addition of two coprime numbers that bigger than 1 😞

[empty gyro]

This may be where the "bigger than 6" qualifer might be handy. Not saying I know the answer, but that would be the first thing I look into

why would 2, 4, and 6 fail when the rest succeed?

[obsidian pivot]

Hmm

Let me think

A bit

In fact 5 is succeed, whose smaller than 6

But i cant say some thing for 4

[empty gyro]

Yeah. 5 still works, but we're considering evens now

(even though 3 does not work)

we know one thing, the two numbers must either be both even or both odd. I imagine you can rule one of those cases out quickly

[obsidian pivot]

1 + 1 + 1 + 1 = 2 + 2 = 3 + 1 = 4

[empty gyro]

I don't follow here what you are trying to do

[obsidian pivot]

Okayy

You saying that both of that coprimes must be even or odd

I will come back with examples

[peak siren]

both of the coprimes can’t be even

they wouldn’t be coprime

[obsidian pivot]

They both gonna be odd then

Great

Or one even one odd

[peak siren]

sounds good

[empty gyro]

do even and odd ever sum to even?

[obsidian pivot]

Nah

That is a olympic level question and im not very well in it

So forgive my ignorance

Both of coprimes must be odd btw

2m = 2n-1 + 2n-1

2m = 2q-1 + 2p-1

[empty gyro]

Oh yeah? I see.

I'm never good at competition math, but #competition-math knows about competition math

[obsidian pivot]

That what i was looking for then

Also

Do you know any sources for studying number theory ?

I also find out why they limited that natural set at 6

2q-1 + 2p-1 = 6