help-10

[weak belfry]

I think i know

is it just tthe trig graphs
a

[pine sail]

Yes, those.

[weak belfry]

nd how they only go up to 1 and down to -1

[solemn ravine]

yes, the important takeaway here is not complex numbers, it's the fact that sin(x) is between -1 and 1

[weak belfry]

then what about like

3sin(x)

doesn't that go up to 3

[solemn ravine]

yes

[weak belfry]

oh but if i rearrange

it will just be a normal sine graph

[solemn ravine]

what do you mean

[weak belfry]

like

if 3sin(x) = 2

sin (x) = 2/3

[pine sail]

Yes.

And this definitely has solutions.

[weak belfry]

do yo uguys enjoy this stuff

[solemn ravine]

yeah i guess that's a good way to think about the algebra you are doing, you are manipulating unknown trig graphs into forms which you know things about

yeah man more than anything in the world

[weak belfry]

makes sense

[weak belfry]

https://tenor.com/view/kermit-death-stare-stare-look-car-gif-17449959

[pine sail]

Define enjoy.

[weak belfry]

do u have fun

doing math

[turbid wing]

maths

[weak belfry]

are u having a dopamine release

[pine sail]

Lol I could continue this by asking define fun and, stuff. But I was kidding, haha of course maths is fun!

[solemn ravine]

solving hard maths problems is 99% suffering and 1% dopamine release

but that 1% hits like a motherfucker

[turbid wing]

no it's actually the best way to release dopamine 😭

I was stuck on a problem yesterday for like 3 hours

and got the solution after that. was so happy

[weak belfry]

at that point I would just skip the question

[solemn ravine]

yeeeeee, makes you feel like a god, right?

[weak belfry]

mindset

[turbid wing]

hard questions are your greatest asset

[pine sail]

There are no hard questions.

[weak belfry]

[great]est [ass]et

[solemn ravine]

questions are hard until you solve them, then they become trivial

[weak belfry]

much wisdom

[turbid wing]

I mean the questions from which you learn something new

[turbid wing]

I actually took it as the Greatest Integer Function

bruh

[fickle turret]

meth

[royal basin]

is there an actual problem being discussed here

if not then may i please direct all of y'all to #discussion or #math-discussion

[weak belfry]

thanks for the help guys :))

,close

.close

[obtuse pebble]

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[plucky patio]

is it supposed to be (n+2)/3?

[lofty jay]

no, thats the notation for binomial coefficient

[solemn ravine]

are you aware of this formula?

[obtuse pebble]

@plucky patio Has your question been resolved?

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[outer locust]

1 peasant changed rabbits to hens and 2 rabbits were equal to 3 hens, each hen produced the eggs that were equal to ⅓ of the total number of hens, the peasant sold the eggs, 9 eggs will cost as many pennies as each hen that produced the eggs and totally he got 72 pennies so how many hens and how many rabbits he had?

[obtuse pebble]

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[proven zephyr]

is that all the information you have?

[obtuse pebble]

@outer locust Has your question been resolved?

[outer locust]

yuh

<@&286206848099549185>

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[fluid snow]

Can anyone give me an example of an even function that doesn't go through the origin?

[fluid snow]

Apparently only odd functions have to go through the origin

[kind hawk]

x^2+1

or even simpler just 1

[fluid snow]

oh, and thats even?

[kind hawk]

yes

check the definition of even

what i mean here is the constant function f(x)=1 everywhere

[fluid snow]

odd: $f(x) = -f(-x)$

[warm shale]

Kepe

[fluid snow]

even: $f(x) = f(-x)$

[warm shale]

Kepe

[fluid snow]

right?

[kind hawk]

yes

[fluid snow]

oh

thx!

.close

[obtuse pebble]

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[fluid snow]

$f(x) = x$ is odd because it's point symmetrical at point (0, 0), it's basically flipped by 180°, right?

[warm shale]

Kepe

Kepe

Kepe

[fluid snow]

So it's neither odd nor even

[kind hawk]

well it's not point symmetrical around (0,0)

[fluid snow]

but it's point symmetrical at (0, 1)

[fickle turret]

f(-x) = (-x) + 1

[fluid snow]

yeah

[fickle turret]

Sorry, I did actually missread it

[fluid snow]

"Functions which map to themselves when rotated 180° around the origin are odd."

The origin in the case of $f(x) = x + 1$ should be (0, 1), no?

[warm shale]

Kepe

[fickle turret]

With the Origin (0|0) is meant

[fluid snow]

oh

[fickle turret]

So f(x) is not symmetrical about the origin

[fluid snow]

thx!

.close

[obtuse pebble]

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[echo thistle]

how can i know if this enters or not

i think they're called conversion or diversion ?

yea prob

but how can i tell

[obtuse pebble]

@echo thistle Has your question been resolved?

[obtuse pebble]

@echo thistle Has your question been resolved?

[short zodiac]

What do you mean by "enters"?

Do you want a solution to this integral? Is that your question?

[echo thistle]

i know the solution its just that idk how it got solved

it is Convergence

[short zodiac]

You mean the steps to evaluate it?

[echo thistle]

yes

[short zodiac]

You know the final answer tho right?

Only the steps is what you want

[echo thistle]

yeye

[obtuse pebble]

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[rain coral]

How can I approach this question?

do I just do it normally

[devout solar]

why wouldnt you do it normally?

[rain coral]

when I mean normally like

just switching sides

[devout solar]

just solve it

[rain coral]

alright

Is there like

no tricks

[devout solar]

well if the answer is imaginary we throw it out ig

[rain coral]

.close

[obtuse pebble]

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[dire otter]

how do I find the surface area?

[versed cave]

of what

[dire otter]

rectangle

[versed cave]

base lenght * height lenght

[dire otter]

thanks

[versed cave]

[dire otter]

.close

[obtuse pebble]

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[grim dragon]

lim n--> 0 SINn - COSn - Xn + A/n^2 - n^3 = 1/2
find X , A

[fierce lagoon]

What the hell

Dude your syntax

[grim dragon]

umm should i write it

[fierce lagoon]

There's like 3 different interpretations of that

Hold on

[grim dragon]

1sec i'll draw it

[fierce lagoon]

Is it, and Gimmie a sec

Just latex it

$$\lim_{n\to\infty}\left(\sin{(n)}-\cos{(n)}-xn + \frac{A}{n^2-n^3}\right) = \frac{1}{2}$$

[warm shale]

Umbraleviathan

[fierce lagoon]

@grim dragon this?

[grim dragon]

this one

[grim dragon]

this

[fierce lagoon]

Okay

[grim dragon]

actually i knew the answer but i want to know how to solve it

[fierce lagoon]

$$\lim_{n\to 0}\frac{\sin{(n)}-\cos{(n)}-xn + A}{n^2-n^3} = \frac{1}{2}$$

So you have that

[warm shale]

Umbraleviathan

[grim dragon]

yeaaah

[fierce lagoon]

If you plug in 0

So you can try a couple of things

Plug in 0, and make it so it becomes 0/0

In that sense, A has to be 1

$$\lim_{n\to 0}\frac{\sin{(n)}-\cos{(n)}-xn + 1}{n^2-n^3} = \frac{1}{2}$$

[grim dragon]

yeah that's true

[warm shale]

Umbraleviathan

[grim dragon]

but howw

[fierce lagoon]

Then you can use L'Hopitals rule

[grim dragon]

hmm how you know that a is 1

[fierce lagoon]

_ _

[grim dragon]

it would become 1+0 -x*0 + A = 0

[fierce lagoon]

Yeah

[grim dragon]

A well be -1

hmm weird

[fierce lagoon]

Well no

Sin(0) = 0

It's $$0-1-0x+A=0$$
$$A = 1$$

[grim dragon]

1-A = 0 A =1

[warm shale]

Umbraleviathan

[grim dragon]

oh true i change sin with cos

my bad

[fierce lagoon]

So that's how A = 1

Then use L'Hopitals rule

[grim dragon]

thx then L hopitals

yeah i find X

its 1

[fierce lagoon]

$$\lim_{n\to 0}\frac{\cos{(n)}+\sin{(n)}-x}{2n-3n^2} = \frac{1}{2}$$

[grim dragon]

cosn +sin n - x =/2n - 3n^2 = 1/2

[fierce lagoon]

Frick

[warm shale]

Umbraleviathan

[fierce lagoon]

There

Again, make it a L'Hopital case

[grim dragon]

1+0-X

[fierce lagoon]

x = 1

[grim dragon]

true wow

thx dude

[fierce lagoon]

And now check if that works

Is that the answer lol?

Sometimes this way doesn't always work

[grim dragon]

yeahhh hhhhhhh actually i start learn math

[grim dragon]

actually i solve it

but i want to check the asnwer

i have question

harder than this one

[fierce lagoon]

Uh

Well let's see

@grim dragon

[grim dragon]

this one

d is connected

idk what the symbols u use instead of d

[fierce lagoon]

Is d(n) a function?

Wdym by connected

[grim dragon]

yebb

[fierce lagoon]

Well you can assume d(2) = 8 since 2-2≠1

[grim dragon]

maybe in english is Continuous

[fierce lagoon]

Yeah okay that makes sense

So what you can try is this

If you plug in 2 for the top limit

[grim dragon]

my sensei do this one idk why

[fierce lagoon]

$$\lim_{n\to 2} \frac{n-2}{d(n)-4n}=\frac{1}{3}$$
$$\lim_{n\to 2} \frac{0}{d(n)-8}=\frac{1}{3}

But 0 ≠ 1

We can make this a L'Hopital case by making $d(2) = 8$

[grim dragon]

He reversed the first equation

[warm shale]

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

[fierce lagoon]

He raised both sides by -1

What you get it this:

[fierce lagoon]

So you need to do this

So that

[grim dragon]

is that easier or the sensei

[dusky drum]

I would say Sensei way

[fierce lagoon]

Well hold on

[fierce lagoon]

Is it supposed to@be 2d(n) or 2d(2)

[grim dragon]

d(n) = 8 by using L law ? are u sure it works

[grim dragon]

2d(2)

[fierce lagoon]

This is how I would do it

[grim dragon]

i think we should reversed then make it equal 0/0

[grim dragon]

leem see

[fierce lagoon]

Gimmie a sec I gotta type this out

$$\lim_{n\to 2}\frac{n-2}{d(n)-4n}$$

Let $d(2) = 8$, and let $d'(2) = 7$

$$\lim_{n\to 2}\frac{n-2}{d(n)-4n} \rightarrow \frac{0}{0} \text{ (L'Hopital's applies)}$$

$$\lim_{n\to 2}\frac{1}{d'(n)-4}$$
$$\lim_{n\to 2}\frac{1}{7-4}$$
$$\frac{1}{3}$$

So you know what d(2) and d'(2) could be

[grim dragon]

yeah Derivation of the first

[warm shale]

Umbraleviathan

[grim dragon]

but why d`(2) = 7

not 8

[fierce lagoon]

7-4 = 3

@tardy epoch lurker alert

[grim dragon]

oh i see

[fierce lagoon]

🤡 guys they added Riemann as an emote

[fierce lagoon]

So just plug it into your limit here

[grim dragon]

but stil didnt find

[fierce lagoon]

Plug it in

[grim dragon]

oh thats it

dude u make math easy

hhhhhhhhhhhhhh

[fierce lagoon]

L'hhhhhhhhhhhhopital

[grim dragon]

i'll be glad to have friend like you

[fierce lagoon]

Lol

Np

[grim dragon]

😂 thx so much dude

[fierce lagoon]

Np

[grim dragon]

i wanna ask u how i could plug d` instead of d

so at first i should make the d be d`

2d(2)

i should say nd`(n) - 2d``(2)/1

[fierce lagoon]

$$\lim_{n\to 2}\frac{nd(n)-2d(2)}{n-2}$$
$$\= \lim_{n\to 2}\frac{16-2(8)}{2-2}\rightarrow\frac{0}{0}$$
$$\Rightarrow \lim_{n\to 2} [nd'(n)+d(n)]$$

@grim dragon

[warm shale]

Umbraleviathan

[grim dragon]

d2 is 8 ?

cuz we said d(n) - 4n = 0

d(2) - 4*2= 0

d(2) = 8

why u dont said n * d`(n) + d(n) * 1

@fierce lagoon

and i know that d(any number ) is = 0

when u take the d` if it

[fierce lagoon]

I just did L hopitals

[fierce lagoon]

thats the same as $nd'(n) + d(n)$

[warm shale]

Umbraleviathan

[grim dragon]

hmm but u forget the 2d(2) oh now this one well be 0

[fierce lagoon]

2d(2) goes to 0

[grim dragon]

yeaah

thx

.close

[obtuse pebble]

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[candid jungle]

A chess player moves a knight from the location (3, 2) to (5, 1) on a chessboard. If the bottom-left square is labeled (1, 1), the translation made is ? If the player moves the knight from (5, 1) to (6, 3), the translation made is?

[candid jungle]

i tried to graph the point of geogebra

but the question options make no sense

2 squares up, 1 right
1 square down, 2 squares left
1 square down, 2 squares right
2 squares down, 1 square right

1 square up, 1 square right
2 squares up, 1 square right
4 squares up, 2 squares right
2 squares up, 1 square left

yeah i got that

i plotted the points

OH WAIR

2 up 1 right

bro 💀 i spent 15 mins on that

thanks

[obtuse pebble]

@candid jungle Has your question been resolved?

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[obtuse pebble]

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[timid silo]

Struggling with Percentiles,

specifically on this question:

Find the 85th percentile score in the following test results.

{95, 88, 70, 75, 83, 70, 66, 91, 68, 76, 82} (11 in total)

{66, 68, 70, 70, 75, 76, 82, 83, 88, 91, 95} ordered

To solve it, all you do is divide the percentile(85p) by 100 and then times that by the amount of numbers there are (11), which you then get 9.35 as your final answer. And looking at how it's ordered 88 which is the 9th number — it should be our 85th percentile, but somehow this conclusion was wrong. Apparently, the answer was 91, but I don't understand how that is. Perhaps I'm using the wrong formula?

Can anyone help me understand where I went wrong in the math?

[timid silo]

I'm not in school or anything just trying to get a grasp on statistics since I'm taking a machine learning course for python

[boreal dew]

70 appears twice

[timid silo]

I've gotten on YT and seen all the different ways to do it, but the formulas are all different.

Percentile rank = p / 100 x (n + 1)

or

Percentile rank = p / 100 x (n + 0.5)

or

Percentile rank = p / 100 x (n )

[timid silo]

yeah I don't know why either

[boreal dew]

maybe the person asking the question expects you to treat the ordered numbers as a set (not counting replicating numbers)

then you'll have index 8.5 and it all depends on how you treat it

if you floor or roof it

[timid silo]

yeah, it can go .85 x 12 or 11.5 or 11 which can all lead to different answers, on the multiple-choice answers I could get either depending on what formula I go with. And don't know which formula is the right one for certain problems so I'm left confused when I get it wrong

[boreal dew]

https://en.wikipedia.org/wiki/Percentile_rank applying the formula listed on the wikipedia article seems to lead to the correct answer

[obtuse pebble]

@timid silo Has your question been resolved?

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[wanton dagger]

So I want to draw this information up on a triangle

[wanton dagger]

Would it be this? The arcos is confusing me here

[prime scarab]

soh cah toa

[nocturne minnow]

Not quite, recall that $\cos \theta = \frac{a}{h}$

[warm shale]

dldh06

[wanton dagger]

Ok

Oh yh ok makes sense

So this

Thanks

.close

[obtuse pebble]

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[fervent cradle]

what is the distribution of the difference of two d20s?

[royal basin]

https://anydice.com/program/29ac8

[fervent cradle]

so the difference between two iid. discrete uniform random variables

hmm

oh i can parameterise that

neat

[royal basin]

same as 2d20 - 21 as it seems

[fervent cradle]

why the hell should that be true lmao

hrmmmmmm

it looks so nice but i don't understand why lmao

oh, wait

nope

wait

maybe

probability generating functions to the rescue?

actually i can't be bothered, i'll finish this program and then maybe come back to try and work out why the maths works

.close

[obtuse pebble]

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[fervent cradle]

.reopen

[obtuse pebble]

✅

[fervent cradle]

actually what even is the cumulative distribution function here, my brain is fried

ok so if we added 20 to x to get x', we would have that, for x' =< 20, y = 0.0025x'

so for x' =< 20, we can can just sum from 0 to 20 over x'

so that'll be 0.00125x'(x'+1)

so y = 0.00125(x-20)(x-19) for x =< 0, and 1 - 0.00125(x+20)(x+19) for x > 0??

no

y = 0.00125(x+20)(x+21) for x =< 0, and 1 - 0.00125(20-x)(21-x) for x > 0

nope

y = 0.00125(x+20)(x+21) for x =< 0, and 1 - 0.00125(20-x)(19-x) for x > 0

ok

[obtuse pebble]

@fervent cradle Has your question been resolved?

[obtuse pebble]

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[peak charm]

does this question look sus or can it be solved

[lethal sand]

it indeed can be solved, what have you tried

[obtuse pebble]

@peak charm Has your question been resolved?

[high arrow]

That means it’s a vector

[peak charm]

i found theta but theta is a decimal value

so idk

[lethal sand]

why should it matters if theta is a decimal value

anyway, you have found the angle between vectors a and b, can you find the magnitude of their sum?

[peak charm]

we cant use calculators

[peak charm]

yes

[lethal sand]

you actually dont need to use a calculator

good

what would the magnitude of the sum be then?

[peak charm]

A^2 + B^2 + 2ABCos theta

[lethal sand]

well almost

the square root of that

[peak charm]

yes

[lethal sand]

and you have got that theta=arctan(3/4) correct?

[peak charm]

whats arctan

[lethal sand]

the inverse tangent

[warm shale]

waler

[lethal sand]

anyway, was that the angle theta you got?

[peak charm]

ye i got tan inverse

[lethal sand]

of 3/4

great

ok so basically in order to find cos(theta), you dont need a calculator

you just need geometry

here, how would you visualise tan(theta)=3/4?

[peak charm]

gimme 2 min

[peak charm]

hmmm thats what

there is no ratio for 3/4

[lethal sand]

3/4 is a ratio itself

lets say if we have a right triangle, where one its angle is theta

what would the ratio between the opposite side and the adjacent side to theta be?

[peak charm]

tan theta = 3/4

[lethal sand]

yes

so basically theta is just an angle in a right triangle where the opposite side to theta is equal to 3, and the adjacent side to theta is equal to 4

can you find cos(theta) now?

here's the visualisation

[peak charm]

cos theta= 4/5?

[lethal sand]

yes

so what is the magnitude of the sum of vectors a and b?

[peak charm]

ohh

$\sqrt{A^2 +B^2 + 2AB \frac{4}{5}}$

[lethal sand]

again, missing the square root and cos(theta)=4/5 not theta=4/5

[peak charm]

hmmmm yes

[warm shale]

Invisible

[lethal sand]

what you are writing implies that theta=4/5 not cos(theta)=4/5

[peak charm]

my bad

i forgot to remove the cos

$\sqrt{A^2 +B^2 + 2AB \frac{4}{5}}$

[warm shale]

Invisible

[lethal sand]

yes, much better

[peak charm]

so it must be A then

but where does 2AB disappear

[lethal sand]

huh?

what do you mean by that?

[peak charm]

option A

wheres AB

[lethal sand]

you should ask your teacher about it then

[peak charm]

hmmmmm

[lethal sand]

but the correct answer is what we got

[peak charm]

alright thanks

.close

[obtuse pebble]

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[timid silo]

1. A student taking a 100-question multiple choice exam has an 80% chance of answering any given question correctly, independent of all other questions.

(a) What is the exact probability that the student scores an A on the exam; that is, they answer between 90 and 100 (inclusive) questions correctly? Write your answer as a sum; you need not simplify it.

(b) Approximate the probability in part (a) using the normal distribu- tion. You may leave your answer in the form Φ(b) − Φ(a) for two numbers a and b and need not use the normal table to evaluate this.

[timid silo]

can someone verify this, im not sure if i did this correctly especially since online calculators doesnt seem to be properly calculating part a

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[pure walrus]

Anybody good at desmos? I have a list of things which are sometimes defined, and sometimes undefined. I want to take the minimum of the list of things, and if all of them are undefined, output 0.

[pure walrus]

I have a super hacky solution but it would be nice to have something cleaner

[obtuse pebble]

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[midnight summit]

Problem 21

[midnight summit]

I factored f(x) and got to |x-4| = epsilon

From here, I'm unsure where to continue or if this is correct

[wind crypt]

Remember that epsilon is a positive number that's given to you. You're only allowed to choose a positive number delta (which may depend on epsilon) such that you can use the definition to prove the limit. I assume You're doing q21 here. What you want to show is that |f(x)-6|<epsilon and it looks like you've simplified that expression to get the |x-4|<epsilon expression. Now, you have to use the assumption that |x-4| < delta to show this inequality.

[midnight summit]

.close

[obtuse pebble]

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[rotund vector]

Hey could someone help me out with this question?

[rotund vector]

The line segment CD is parallel to the y-axis. If C is (3,3) and D is (q,5) find the value of q

[wind crypt]

Have you tried drawing a picture?

[rotund vector]

No however I know that the only type of line parallel to the y axis is a vertical line, and lines parallel to the y axis do not have a gradient so I’m not sure how to approach this question.

[wind crypt]

That's a good start. Now the question would be: what choice of q would make the line segment from point C to D vertical?

You might want to think about the equation of a vertical line

[rotund vector]

x=a?

[wind crypt]

Yes

And that has to be for both of the points if they're on the vertical line

You should write both of the equations that follow from that

[rotund vector]

Alright I’ll try that

Still didn’t manage to figure it out

What values do I plug into the equation?

[wind crypt]

Remember the first coordinate is the x coordinate, that is the offset from (0,0) in the direction x. And the second is the y-coordinate.

So, if you apply x=a to the point (5,10) for instance, you'd have 5=a

since x =5 and y = 10

[rotund vector]

Oh I see

So I would get 3=a?

[wind crypt]

yes, that's right

[rotund vector]

So now how would I find the value of q?

[wind crypt]

You can do the same thing you did with (3,3) with (q,5) to get another equation

[rotund vector]

So for x=a I can use either x or y coordinate or do I have to use a specific one?

[wind crypt]

It has to be x, since the x in the equation is referring to the x-coordinate

[rotund vector]

But I get q=a

[wind crypt]

Yes, that's right

Remember that you also had the equation 3=a

Can you combine them?

a is just a constant. It's the same for any point on the line.

You can think of it as the x-position of the line

[rotund vector]

Oh ok

[wind crypt]

Well, a is the same in both of the equations. So if two things are equal to the same thing, then they're also equal to...

https://www.desmos.com/calculator/9rk5mfkz9d

This might be helpful to look at, too. You can use the slider to change the a-value and see the line we're talking about.

That's not quite it

Think of 3 and q as both being x-coordinates of some points on the line

What x-coordinates are possible for a point on that line?

e.g. the one with equation x=2

[rotund vector]

Just one

[wind crypt]

Yes

And since 3 and q are both x-coordinates that happen on the line, what does that say about them?

Given there's only one possible x-coordinate

Yes, they're both x-coordinates of points on the line. But they certainly could be equal, couldn't they? Say they aren't equal. What would that look like in our picture?

Say I gave you three points, all on the y-axis. They're (a, 1), (b, 10), and (c, 20). What are a, b, and c?

Remember what it means for a point to be on the y-axis

Yes, so their x-coordinate must be 0

What are the x-coordinates

No worries

[rotund vector]

So how would I proceed

I know that q and 3 are both x coordinates

[wind crypt]

Right

And you also know that they're both a, right?

[rotund vector]

Yes

[wind crypt]

And a is a constant that never changes

[rotund vector]

Yes

[wind crypt]

If two things are equal to the same thing, aren't they also equal to each other?

[rotund vector]

Yes

So 3=q

[wind crypt]

Hence...

Yes

[rotund vector]

Haha yes

Alright what’s next?

[wind crypt]

Well, that answers your

question

I don't know if you have any others

[rotund vector]

Oh yes hahahaha I forgot

Sorry i stayed up till around 4am doing math so I only got around 3 hours sleep

So probably best if I go to sleep

Thanks for helping me out

[wind crypt]

Yes, that's probably best

Half of the learning happens in your sleep

I'm glad I could help

[rotund vector]

100% right

[wind crypt]

To free this channel, you have to type .close I think

[rotund vector]

Oh ok

.close

[obtuse pebble]

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[obtuse pebble]

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[past solar]

How do you prove that a^2 + b^2 = 1234567, isn't possible for any integral solution?

[timid silo]

integral solution?

[past solar]

integer solutions.

[timid silo]

oh

[royal basin]

consider this equation mod 4

squares can only ever be 0 or 1 mod 4, while right hand side is 3

[timid silo]

(a+b)=sqrt(1234567+2ab)

[obtuse pebble]

@past solar Has your question been resolved?

[past solar]

sorry, What does it meant when the rhs is 3?

[kind hawk]

the RHS is 3 mod 4

[past solar]

Oh I see I see.

so a^2+ b^2 can't have a remainder of 3 when divided by 4.

.close

[obtuse pebble]

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[timid silo]

is anybody good at chemistry?

[timid silo]

for so far I know

mol = 6,022*10²³

mol is not even a unit it is just a number

and the avogadro constant is

N0 = 6,022*10²³/mol

so if I replace the mol with the number i get

N0 = 6,022*10²³/(6,022 x 10²³) = 1

avogadros constant being equal to one?

[fickle turret]

No

We have n = N / N0

If N = 6.022•10^23 => n = 6.022•10^23 / (6.022•10^23•1/ mol) = 1 mol

[timid silo]

google says Na = 6,022*10²³/mol

[fickle turret]

That's what I wrote

[timid silo]

lets ignore everything and just think about Na

then what if we replace mol with the number

so it comes out as 1

[fickle turret]

Solving for N0 we get N/n. No unit / mol

[timid silo]

so if N isnt equal to n

wait

i never heard about the N tho i need to research

[fickle turret]

N is just the number of molecules or atoms

[timid silo]

yes i see

[fickle turret]

You can't just set N = n as on the left hand side we have a constant, which implies N ~ n.

The only value where It'd be true is N = n = 0

[timid silo]

I thought the n is the amount of molecules

not N

for example if we have 36g of water

[fickle turret]

Du bist Deutsch, oder?

[timid silo]

ja

[fickle turret]

n ist die Stoffmenge in mol

[timid silo]

ja

die stoffmenge gibt doch an wieviel wir haben oder nicht

zb bei 36g wasser haben wir

n(H2O) = 2 mol

das heißt wir haben 2 mal 6,022*10²³ wasser moleküle

[fickle turret]

Die Stoffmenge n gibt an, wie viele Teilchenpakete wir haben => 2. Mit der Avogadrokonstante N0 kannst du die Anzahl der Pakete in die Anzahl der Teilchen N umrechnen. Also 2•6,022*19^23

[timid silo]

aso da wird eig mathematisch gesehen nur

das mol in die zahl umgesetzt

wir haben gesagt

wir haben 2 mol wasser moleküle

jetzt haben wir

2 mal 6,022*10²³ Teilchen (wassermoleküle)

[fickle turret]

Stimmt

Das 1/mol brauchen wir nur, damit sich das mol mit dem 1/mol rauskürzt und die Einheit rausfällt

[timid silo]

ja eben ich hatte es schon vermutet es geht eig nur um einheiten umformen

also verändert sich ja der zahlenwert mathematisch gesehen nicht

nur die einheit

[fickle turret]

Aber das ist falsch

[timid silo]

hmmm

[timid silo]

das ist genauso als hätten wir eine zeit t=2 minuten und wir multiplizieren sie mit einer konstante 60s/min

dann haben wir t in sekunden

[fickle turret]

N0 = N/n

[fickle turret]

Genau. Ist im Prinzip das gleiche 😀

[timid silo]

eben aber

die konstante 60s/min ist ja im endeffekt eine 1

[fickle turret]

?

[timid silo]

weil min = 60s und wenn wir einsetzen kommt eine 1 heraus

und bei N0 wäre es ja genauso

[fickle turret]

Wenn du teilst bekommst du 60s/min

Und das als 1 definiert

[timid silo]

eben

[fickle turret]

Aber was sagt das über die Konstante aus?

[timid silo]

in diesem fall für diese zeitberechnung garnichts

aber bei der berechnung in mol in teilchen oder teilchen in mol

ich war nur verwirrt weil

wenn wir mit N0 umrechnen

ist es ja dasselbe prinzip

und N0 wäre eine 1 im endeffekt

es rechnet nur um aber verändert keinen echten physikalischen wert

[fickle turret]

Jetzt verstehe ich dein Problem

[timid silo]

warte warte

1 mol IST 6,022*10²³ oder 1 mol ENTSPRICHT 6,022 x 10²³ ?

[fickle turret]

:= ist ein mathematisches Zeichen

[timid silo]

gehe an eine technische schule wir machen da einen großen unterschied

zwischen ist und entspricht

[fickle turret]

Es heißt sowas wie „ist definiert als“

[timid silo]

also kann ich mathematisch sagen 1 mol = 6,022*10²³ ?

mol ist soweit ich weiß nur eine zahl und keine einheit

(wird als einheit betrachtet, das ist mir klar, aber ist mathematisch wie eine zahl)

[fickle turret]

mol ist an sich eine Einheit. Mit der Avogadrokonstante ist klar, dass dahinter eigentlich eine Zahl steckt, ja

[fickle turret]

Was ich damit sagen will. Du nimmst eine Definition

Teilst

Ziehst die selbe Definition wieder heran

Kein Wunder, dass dann 1 = 1 herauskommt

[timid silo]

mathematisch war das nichts falsches oder?

ich wollte mir nur bewusst machen dass N0 nur verwendet wird um umzurechnen aber als zahl selber nur eine 1 ist und nichts physikalisch verändert

[fickle turret]

Im Gegenteil. Wenn da jetzt 1 = 0 herauskommt, wäre die Definition an sich schon sinnlos

[fickle turret]

n ist plump gesagt nur eine Möglichkeit Teilchenpakete aus Teilchen zu bilden. Und N0 ist der Umrechnungsfaktor. Man könnte auch N0 = 3 machen. Das hätte dann aber nicht so hilfreiche Eigenschaften

[timid silo]

ja genau teilchenpakete weil n in mol angegeben ist und mol gibt ein "paket" von 6,022*10²³ teilchen an

[fickle turret]

Es ist aber keine Naturkonstante wie c, Epsilon 0 oder die planksche Konstante

[fickle turret]

Ja

[timid silo]

ja des ist mir klar

ist jetzt nicht so wichtig aber

N0 wäre doch dann 1 tho oder nicht

😂

falls nein lass ich es sein ich hab sowieso alles verstanden was ich brauche

[fickle turret]

Wie ich versucht habe zu erklären. Die 1 ergibt sich dadurch, dass du die selbe Definition in sich selbst einsetzen kannst, sodass es sich rauskürzt. Es ist auch keine simple 1, sondern 1 = 1, also eine Wahre Aussage.

[fickle turret]

.

[timid silo]

ja genau so

habs jetzt wirklich verstanden

danke tobias

[fickle turret]

Bitte

[timid silo]

schönen tag

.close

[obtuse pebble]

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[fickle turret]

Dir auch

PS. Einheiten ohne Wert rüberzubringen macht keinen Sinn

[timid silo]

ja

[obtuse pebble]

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[barren finch]

hey friends!

hopefully a quick question here

currently looking to understand why the lower limit is pi/2 in this case

let me know if you need to see my calculations, but i dont exactly see why we'd rewrite 0 as pi/2?

i got 2 alright for the upper limit, which i am super happy with, but i have no idea why Wolfram put that there

also seriously considering signing up to see the step-by-step solutions, any thoughts on that? gonna be doing maths for a good while lol

[cerulean tulip]

Howd u do it

[barren finch]

its a bit messy because im still trying to get used to doing integrals by substitution

also i believe i dont need to substitute the lower limit if its 0, so im unsure if i need to still do that using the Newton-Leibniz formula fully

[obtuse pebble]

@barren finch Has your question been resolved?

[cerulean tulip]

Sorry im back

[barren finch]

dw about it

[cerulean tulip]

So here u can do a u substitution

[barren finch]

t = u

here

[cerulean tulip]

U = sqrt x

[barren finch]

directly below the original excercise, separated by dotty line

[cerulean tulip]

Then u shpuld get t^2/(t^2 + 1)

And then you should do some algebraic manipulation

So you can integrate it

[empty monolith]

i think j know how my dudes

[cerulean tulip]

Also t /(1+t2) is not 1/(1+t)

I think thats why

[barren finch]

ah damn i totally forgot that we'll have 2*t

i have no idea what my tutor thinks that i would be able to solve this but ok

[empty monolith]

you dont always have to limit your substitution with algebraic expressions

trig identities are also useful in integration

[barren finch]

right

[empty monolith]

if you know the differential equations for each functions

[cerulean tulip]

U integrated 1/(1+t)

[obtuse pebble]

@barren finch Has your question been resolved?

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[barren finch]

ty guys

[obtuse pebble]

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[dire umbra]

3x * e^4x + x^3 * (e^4x * 4)

[dire umbra]

how does it become this x^2 * e^4x [ 3 + 4x)

[night isle]

Hello can anybody help me. With indices?

[dire umbra]

chat is taken

[night isle]

Ooh sorry

[dire umbra]

How does this 3x * e^4x + x^3 * (e^4x * 4) become this x^2 * e^4x [ 3 + 4x]

[teal prawn]

ig it should be x * e^4x[ 3 + 4x^2]

[dire umbra]

No?

[teal prawn]

what?

[dire umbra]

you typed

x * e

it is x^2 * e

[teal prawn]

(3x)(e^(4x))[ 3 + 4x^(2)]

[teal prawn]

where?

$3x × e^{4x} + x^{3} × e^{4x} × 4$

[warm shale]

kinglacto

[teal prawn]

is this what your question looks like?

[dire umbra]

no

the 3x is 3x^2

$3x^2 × e^{4x} + x^{3} × (e^{4x} × 4)$

[warm shale]

Sai

[sacred oak]

common factor

[teal prawn]

that isnt what you typed here but ok in that case, it is, indeed x^2 * e^4x[3 + 4x]

[dire umbra]

my mistake

[dire umbra]

what do you mean?

[teal prawn]

factor out x^2 * e^4x

[sacred oak]

e^4x is common

also x^2

[dire umbra]

so the e^4x + e^4x should be 2e^4x?

[sacred oak]

nop

ir u have a something that repeats in every term

its a common factor

like this

[dire umbra]

yes

so e * 2x + 2x = 2x e?

[sacred oak]

$4x^2+5x^3$

[warm shale]

NicoA

[sacred oak]

$x^2(4+5x)$

[warm shale]

NicoA

[teal prawn]

2x * (e + 1) is what it would be

[dire umbra]

but common factor?

[teal prawn]

is 2x

[dire umbra]

Yes?

[sacred oak]

https://youtu.be/U6FndtdgpcA

[dire umbra]

So 2x is the common factor

[dire umbra]

Thank you I am gonna check it out

[sacred oak]

np

[dire umbra]

.close

[obtuse pebble]

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[obtuse pebble]

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[slim cove]

Wait so the matrix reads 0 0 0.1, 0.4 0.3 0.3?

And what does the vector on the right read

[timid silo]

a and b probably

[slim cove]

Interesting, I feel like this is not a standard notation, is this from a textbook or something or just your lecture notes

[obtuse pebble]

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[pallid acorn]

Did I work out everything correctly

<@&286206848099549185>

Hehe

[uncut robin]

yes

[pallid acorn]

Thanks ^_^

.close

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[timid silo]

what is "topology
of uniform convergence on compact"

[timid silo]

i mean, topology is a group of subsets satisfied some properties

aahh i got

is a topology of all uniform convergences?

[obtuse pebble]

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[drifting pier]

I know of a polynomial equation of fourth degree that it touches the x- axis at points x1 = -1 x2 = 1 and that it has a low point at ( 0 | -( 1,2)). What is the equation to the function?

[quiet saffron]

What is (0 I -(1,2)) i have never seen a notation like that before

[drifting pier]

(0,-(1/2))

[quiet saffron]

Oh i understood now

[drifting pier]

I can try writing it in latex

[quiet saffron]

It only touches x axis at points x1=-1 and x2=1

Or we have to find other twi

Two

[drifting pier]

Thx but somebody is exhaling it to me in PM

[quiet saffron]

Ok

[drifting pier]

. close

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[timid silo]

can I post vector calculus questions here, or where?

[timid silo]

Im so confused how this works

[royal basin]

you can post questions of any level in these help channels

[timid silo]

ohh!

I though this was my schools discord haha!

[royal basin]

if you wanted to post in a topic-specific channel, #calculus or perhaps #multivariable-calculus would be a fitting place. but here is fine too, so you can just post your question(s) here.

[timid silo]

I am typing my question: it sa doozy.

so anyway, I require a bit of help--I am studying math methods by Arfen and Weber. And it starts off with vector analysis, nothing too bad.

It does go over the rotation of a coordinate axes, which does not make any sense to me whatsoever, could be the summation notation, which honestly never made all that much sense to me... I have gotten better at it? It does say a lot of things like
x'=xcosΦ+ysinΦ
y'=-xsinΦ+ycosΦ

Any advice on learning how to rotate a coordinate axes? like what sort of resources should I be aware of?

The next area that does not make sense to me is the triple scalar product, triple vector product,
triple scalar: A* BxC=BCxA =CAxB and also with the reverse of the cross product, reversing order multiply by negative one, ok

I guess this one is not too bad.

Triple vector product
Ax(BxC). so a vector orthogonal to both A and BxC... in 3D does that make it 0? what purpose does this serve??

Next section is gradient and I understand what that is at a basic level
<d/dx, d/dy,d/dz>

If I want the gradient of a function, f(r), it just the <fx,fy,fz> of said function?
and if f(r) is a function f(r(x,y,z)) in order to find the gradient, I have to use the chain rule? so dr/dr * dr/dx and etc?

Next is divergence: Grad* r.
if r=xx^+ yy^+zz^
Just becomes dx/dx+dy/dy+dz/dz or 3?

My question is when they try to generalize this formula:
they do
Grad*(r f(r)) where r is a vector and f(r) is a function,
d/dx ( x f(r) ) + d/dy ( y f(r) ) + d/dz ( z(f(r) )
r is the vector given above. what I dont get is how and why does this give
3f(r) + x^2/r df/dr + y^2/r df/dr+ z^2/r df/dr ?
There was no x^2 in the previous example, so why would this one have it?

So next was curl, which is just Grad x V or Grad x r and then the subject of the lapacian and successive applications of the gradient? Which I guess would make sense if I understood how to actually work with these vectors a bit better...

Like how does one learn these vector identities?

sorry for the long essay

Thx for any help!

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[timid silo]

Hello goodevening in identifying if its polynomial or not?
Is the key point for this is that the exponent is non negative integers?

[kind hawk]

yes. polynomials are sums of terms c*x^n with n a non-negative integer and c some constant

[timid silo]

what is your question exactly

Im just wondering how do i identify if its poly or not

How do you determine if an equation is a polynomial or not

[kind hawk]

an equation is something else. an equation somehow involves an equality sign = while a polynomial never does that

a polynomial is something that looks like this: $a_0+a_1x+a_2x^2+a_3x^3+\ldots+a_nx^n$ for a non-negative integer $n$ and some constants $a_0, a_1, a_2, \ldots, a_n$

[warm shale]

Denascite

[timid silo]

Wait so if the equation has an equal sign its not classified as a polynomial cuz it never does that

?

[kind hawk]

you can classify it as a "polynomial equation" but not as a polynomial

[timid silo]

Oh

Yeah im talking about how do i identify if its a polynomial equation or not

[kind hawk]

a polynomial equation has polynomials on both sides of the equality sign

[kind hawk]

and polynomials look like this

[timid silo]

Okay

So wait
If the example is like k³ + 3k² + 9k - 3 = 0

Its not a polynomial equation cuz there's no polynomials?

On the other side

[kind hawk]

k³ + 3k² + 9k - 3 is a polynomial

0 is also a polynomial

[timid silo]

Ohg

Ohhh

Okay okay

[kind hawk]

k³ + 3k² + 9k - 3 = 0 is a polynomial equation

the constant functions are polynomials

they are of the form $a_0$ with just some constant $a_0$

[warm shale]

Denascite

[timid silo]

okay

So if the equation is like 3k³+...+....

With no equal sign

Its not an equation?

[kind hawk]

yes

an equation has to include an equality sign

[timid silo]

Okay thanks

Could someone check this for me

So i could see if im right?

[kind hawk]

what are your answers?

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[timid silo]

Poly
Poly
Poly
Poly
Poly
Non Poly
Poly
Poly

[timid silo]

@kind hawk ( ╹▽╹ )

[kind hawk]

no

take a closer look at 2, 4, 5

[timid silo]

Its not polynomial equation because they only have one variable?

In all polynomials

[kind hawk]

each of those 8 only have one variable. that's fine

[timid silo]

Hmm

OH WAIT

There's negatives

So it cant be a polynomial equation cuz there's negative integers?

[kind hawk]

in the exponents, yes

[timid silo]

Oh wait

If the denominator has a variable but no exponents its already negative?

[kind hawk]

we have 1/x=x^(-1)

or in general $\frac{1}{x^n} = x^{-n}$

[warm shale]

Denascite

[timid silo]

Ohhhh

Okay okay thanks alot

Synthetic Division
2a³ - 3a² - 4a - 17 ÷ a - 3

My answer that it is not factorable

[kind hawk]

the question doesn't ask anything about being factorable

it doesn't matter whether those terms are factorable or not

[obtuse pebble]

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[obtuse pebble]

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[lyric pawn]

how does Euler's tau(a) function work in this case?

nvm, it's actually easy

in this case it's tau(a) = (3 + 1) * (3 + 1) * (1 + 1) * (5 + 1)

.close

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[rustic wraith]

is the arrangement of numbers on the real number line like rational then irrational and rational
for example in the study of limits is pi+ rational....saw this theory somewhere...

[grave halo]

There are more irrational numbers than rational numbers so no I don't think so

[kindred oasis]

Between every two numbers there are infinite rational numbers and infinite irrational numbers, so it doesn't mean much talking about which number is between which

[grave halo]

I don't remember the name of the proof but between two rational numbers there are always more irrational numbers than there are rational numbers between two irrational numbers

[obtuse pebble]

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[full finch]

Is this true or false?

[brave bramble]

It's not really meaningful

"The axe is a good idea when you find you need an axe"

[full finch]

so it is true

[brave bramble]

Define "helpful".

[full finch]

of use

[brave bramble]

Like, let's say it is somehow "quicker to consider the elements in the complement than it is to consider the number of items in A". Do you actually need this count?

Similarly, are you suggesting taking the complement just to get this count? Maybe you don't need to.

This statement is trying to bypass a course in set theory by being vague.

[timid silo]

just answer the kids question

[full finch]

calm down

[timid silo]

?

im helping you

lmao

[full finch]

@brave bramble so its true but its a useless statement kinda thing

[brave bramble]

No. I would ask "What does this really mean"? And then reply with true/false once it has been cleared up.

I am worried the statement is really saying "You should always calculate the complement explicitly to get its cardinality" which is very false

[timid silo]

@full finch its true idk why kaynex is trolling you

[brave bramble]

But if the statement is saying "Calculate the complement when calculating the complement is the best choice" well then duh.

True, but not usefully true

[full finch]

@timid silo thanks bro. Idk why @brave bramble is trying to deceive me.

.close

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[timid silo]

yeah no problem man

hmu anytime you need help

just block kaynex idk

i been getting a lot of complaints about him

[slim cove]

Kaynex is just trying to point out that the question is a tautology

"It is useful to consider the complement when it is easier to consider the complement"

[obtuse pebble]

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[rough epoch]

im getting a bunch of these in my SAT practice:

[rough epoch]

is there a faster way/trick to do them other than using elimination or substitution?

[hexed agate]

I don't think so no

[rough epoch]

ok thanks. just making sure - in case it could save me some time

.close

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[errant oracle]

|f(x) - f(a)| = |sqrt|x+2| - sqrt 2|

also if i set delta = 1 we have |x| <1

|f(x) - f(a)| = |sqrt|x+2| - sqrt 2| <=|sqrt 3 - sqrt 2|

[long peak]

this might simplify with the difference of two squares
as in |f(x)-f(a)|*|f(x)+f(a)|=|f(x)^2-f(a)^2|

[errant oracle]

how does that help here

i need to show |f(x) - f(a)|<epsilon

wait i think i understand

[long peak]

the |f(x)^2-f(a)^2| is easier to work with, and if you show it's <epsilon you could divide both sides by |f(x)+f(a)| to get
|f(x) - f(a)|<epsilon/|f(x)+f(a)|<epsilon (since |f(x)+f(a)|>sqrt(1)+sqrt(2)>1)

[errant oracle]

let me send my work 1 sec

[errant oracle]

i understand it now tysm

actually 1 thing

|f(x)+f(a)|>sqrt(1)+sqrt(2)>1
is this a result of -1 < |x| < 1

[long peak]

I misread it but to do that you assume that delta<1

[errant oracle]

yes

[long peak]

by misread I mean I thought this said delta less than 1

but yes

because |x|<1 you have sqrt(2-1)<=f(x)<=sqrt(2+1)

that's a messy explanation but it's essentially recognizing that f(x)+f(a) is always positive and then bounding it below

[obtuse pebble]

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[high snow]

Wait is this a typo and an actuall error

[faint prism]

The answer is 15. Did you thought $x/3 = 12$?

[warm shale]

Potet

[high snow]

huh can you explain

?

[faint prism]

How did you came to the answer?

Sorry, there’s something extremely wrong with my English today.

[high snow]

c^12 * c^3 = c^x

12*3 = 36

x = 36

[high lily]

which exponent law are you applying

and/or are attempting to apply

[high snow]

oh wait nvm

a^m * a^n = a^m+n

thats the correct law

i was doing (a^m)*n = a^m+n

.close

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[exotic oak]

Hello.