help-33

[stable path]

ill go reread, thank u again

"we perform the operation opposite to what we see"

interesting

is there a rule for this

name*

[ornate ginkgo]

ive no idea what the name of the rule is atm

something about equality of operations

[stable path]

hmmm ok ill go from there

[ornate ginkgo]

$a = b + 2 \implies a - 2 = b$

[elfin berry]

blanket

[ornate ginkgo]

$a = bc \implies \frac ac = b$

[elfin berry]

blanket

[stable path]

ok so in scenerio 1

if i were to move b instead

would it be

a - b = 2

i get the multiplication/division ones no problem

[ornate ginkgo]

yes

[stable path]

so in this example

how come its not just all addition on the top

because of the parenthesis?

nvm i get it

.close

[marsh citrus]

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[green rover]

i have no idea how to solve this

[marsh citrus]

@green rover Has your question been resolved?

[green rover]

<@&286206848099549185>

[odd crest]

you have to represent it as a sin or cosine

so you have y

= some sine function

what is y in this case?

[green rover]

idrk

@odd crest i thought y was randomly assigned

[odd crest]

No, y is the dependent variable in this case

X is the independent

[green rover]

i think the dependent variable is 3.5

and independent is 2.25

that right? @odd crest

[odd crest]

No thos are constants

Variables change

If you read the question again what are things you read that are changing in the question

That are not single values

[green rover]

the months and the precipitation

[odd crest]

Brilliant!

And now which one of thos depends on the other

[green rover]

precipitation depends on the month

[odd crest]

Exacly so in this case your independent variable is months

And dependent variable is prefipitation

C

So number of months is x

And the amount of precipitation is y

[green rover]

got u

so Y = Sin X ?

[odd crest]

Yes but its not quite that simple

Unfortunately i have to take a shower

So im gonna be gone for a bit

Btw you answered the first question alread

[green rover]

<@&286206848099549185> can someone else help till they come back

[odd crest]

Im back

[marsh citrus]

@green rover Has your question been resolved?

[odd crest]

@green rover do you know the general formula for sin functions?

Actually nevermind that it turns out you dont need it

Since what you need to find is the amplitude and the period

Do you know how to get an amplitude of a sin function?

[marsh citrus]

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[sleek crater]

can you do y1 - y2 also?

[glacial grotto]

wdy

wdym

[still temple]

replace them by actual numbers and try to verify if your logic is right

[sleek crater]

thats too much effort

[still temple]

it takes 2 seconds

is y2 - y1 the same as y1 - y2 the same thing as asking

is 5 - 1 the same as 1 - 5

[marsh citrus]

@sleek crater Has your question been resolved?

[marsh citrus]

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[untold badger]

Hello :) I have a rather simple question on how to do this problem

[untold badger]

It’s in a digital Logic class ^

I’m confused on how to get started mainly and if those problems would be the same as

5X = 1 mod 7

[runic temple]

Rewrite it as 5x=7y+1

But recall that it can also be rewritten as 5x=7(5y+2)+1

Since the choice of y is somewhat arbitrary

Basically keep adding 7 until you get a multiple of 5

[untold badger]

Oooooh okay lol

Ty Ty :)

.close

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[fathom thicket]

Angle of elevation is something I just haven't quite gotten down yet. Would someone know the necessary formula to work on this question and questions adjacent?

[fathom thicket]

However thorough an explanation would be appreciated.

[silver pawn]

you use the cosine rule

[fathom thicket]

I see, and Cosine would yield two seperate angles?

[merry terrace]

You could solve for x

And then use cosine rule

[merry terrace]

Cause the satalite seperates the angles

[silver pawn]

you can just use the cosine rule to find the angles directly

[merry terrace]

Oh it is angle of elevation

Not depression

Then use cosine law

[fathom thicket]

Alright so Cosine law , I wasn't sure how to approach it seeing as my knowledge concerning Trig isn't one to be described as vast so thanks for the help. That's all concerning the question. 👍

[merry terrace]

Watch a yt video for a visual explanation

[fathom thicket]

.close

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[glass silo]

Oh this one

Definitions time

[fierce belfry]

yes

lemme send

i have like 0 clue how to continue

[glass silo]

Well you have a good start [missing some bits but I can see what you're doing!]

[fierce belfry]

[glass silo]

One good idea is to note that
[
\abs{g(x) - c} < \varepsilon \implies -\varepsilon + c < g(x)< \varepsilon + c
]

[elfin berry]

@glass silo

[fierce belfry]

hmm

how does that halp XD

[glass silo]

Well, the idea is that we want to show that $f + g$ diverges to infinity - so, like the definition of that would be that for any $K > 0$, there is some positive real number $m_{K}$ such that whenever $0 < \abs{x - a} < m_{K}$, we have that $f(x) + g(x) > K$, right?

[elfin berry]

@glass silo

[fierce belfry]

sorry why diverge lol

u mean tends to?

do they mean the same thing

but yes i agree with the sentence XD

[glass silo]

Yea tends to infinity, different phrasing same meaning

[fierce belfry]

ahh i see

i agree

[glass silo]

But note how we have that basically:
\begin{enumerate}
\item For any choice of $\epsilon >0$, there is a $\delta_{\epsilon, g}>0$ such that whenever $0 < \abs{x - a} < \delta_{\epsilon, g}$, we have that $-\epsilon + c < g(x)< \epsilon + c$, and
\item For any choice of $M>0$, there is a $\delta_{\epsilon, f}>0$ such that whenever $0 < \abs{x - a} < \delta_{\epsilon, f}$, we have that $f(x) > M$
\end{enumerate}
Those are the from the definitions and my rewriting

And green tea

I see you 🫶

[fierce belfry]

wait what

wait

oh cra

[glass silo]

[fierce belfry]

wait lemme read

[glass silo]

Give me one second to edit that!

[elfin berry]

@glass silo

[fierce belfry]

[glass silo]

[fierce belfry]

hmmmmm

wait

okay

so like

ok but yes

sorry had to do something

i understand

ok hmm

so like

[fierce belfry]

wait is this supposed to be sufficient

[glass silo]

Cool, but then if we took those two up there and added them together, and let $\delta_{\epsilon} = \min{\delta_{\epsilon, g}, \delta_{\epsilon, f}}$, then we then have that whenever $0 < \abs{x - a} < \delta_{\epsilon}$, we have that $-\epsilon + c + M < f(x) + g(x)$, right?

[elfin berry]

@glass silo

@glass silo

[fierce belfry]

wait sorry shouldnt the arrow be the other way

[glass silo]

[Alternatively I could've written that $f(x) + g(x) > -\epsilon + c + M$]

[elfin berry]

@glass silo

[fierce belfry]

wait reee where did that come from

oh wait

oh

IM STUPID

okay

i get it

okay

wait

is that sufficient

[glass silo]

Cool cool, now this is almost what we want for the definition of divergence...

[fierce belfry]

[glass silo]

https://tenor.com/view/well-yes-but-actually-no-well-yes-no-yes-yes-no-gif-13736934

[glass silo]

See how it's been written here, it would be much nicer if we could have this here(!)

But we can pick our $\epsilon$ and $M$ such that we can get that, based off a given $K$

[elfin berry]

@glass silo

[fierce belfry]

wait did we just introduce the K

[glass silo]

[I used the letter K rather deliberately so that it doesn't get confused and it's clear what we're doing here]

[fierce belfry]

ohh

okay

[glass silo]

Me mind reader 🔮

[fierce belfry]

[glass silo]

Anyways, based on that $K$, I would quite like to choose an $\epsilon$ and $M$ such that $-\epsilon + c + M$ cancel down nicely to $K$

[elfin berry]

@glass silo

[fierce belfry]

[glass silo]

Do you have any thoughts on what choice you'd like to make for epsilon, and for M?

[glass silo]

The sad cat?

[fierce belfry]

[fierce belfry]

hmm

can i choose the bigger one?

[glass silo]

Hmm, what do you mean by that?

[fierce belfry]

max{episilon, M}

i thinK

[glass silo]

We're given a K, and want to choose an epsilon, and an M

[fierce belfry]

oh wait

i thought u meant choose K from episolon and M

hmm

could we do K + epislon + M?

wait then thats similar to the previous one hmm

[glass silo]

Welllll, that isn't exactly an answer to the question really

[fierce belfry]

okay lemme write everything down from scratch and think again

[glass silo]

This one is a bit of a think

[nocturne fiber]

ooh analysis question

how's this going

[fierce belfry]

i cant think

[glass silo]

We have quite a bit of the work done for it

[fierce belfry]

lemme think

wait its analysis?

im doing calc 1

[glass silo]

Welll limits are part of analysis tbf

[fierce belfry]

do u have any hints

XD

[nocturne fiber]

I mean the hint is to look at the definition you know

[fierce belfry]

hmm

ok just to be clear

we have arbitrary K

and i want to find espilon and M from K

K is anything right

[glass silo]

Anything positive yes

[fierce belfry]

hmm

[glass silo]

[though with that said, negatives should be fine without affecting the definition I think? Do you have the definition from like your course that you're working with?]

[fierce belfry]

all the definitions i know are the first two lines haha

any more hints lol

[glass silo]

Wellllll, if we assume that c is positive non-negative for the time being...

[nocturne fiber]

are we allowed algebra on limits btw

[fierce belfry]

no eyedea

[nocturne fiber]

if we are

[fierce belfry]

HMMMMM

[nocturne fiber]

it immediately follows

[fierce belfry]

i think im just stupid tbh

why am i not getting this

[nocturne fiber]

dont worry lo l

it's always like that

ur not stupid

[glass silo]

...then we would have that this here, we can say that
[
f(x) + g(x) > -\epsilon + c + M \geq M - \epsilon
]

[fierce belfry]

[elfin berry]

@glass silo

[fierce belfry]

yes

[glass silo]

Given a K, we would like to choose a M and an epsilon such that the M - epsilon becomes K

There's a rather nice choice we could do for that, right?

[fierce belfry]

wait i get the idea but like

we just birthed the K out of nowhere

are we trying to use K as a lower bound (sorta)

for the infinity

[glass silo]

That's the idea - we are given a positive K, and we want to then get that we can force f + g to be greater than that K over a certain interval

That's the explicit definition of diverging to infinity

[fierce belfry]

okay that makes more sense now

cant we just equate K to M - e tho

[fierce belfry]

sorry am i being dumb D:

[glass silo]

You can't equate K to anything

You can equate epsilon and M to whatever you want though...

[nocturne fiber]

here's an idea, why not we try letting M-eps = K

[fierce belfry]

Wait but i thought we cant do that

[nocturne fiber]

we're not touching K

[fierce belfry]

Me lost

[nocturne fiber]

let K = M-eps not o-K
let M-eps = K this is ok

the first one says we choose K based on M-eps which is not fine

the second one says we choose M and eps based on K, which is okay

so be specific about what

that's why the word "let" exists

[glass silo]

[and remember that we said that we want it such that M - epsilon "turns into" K based on the choice of M and epsilon]

2 - 1 = 1, quik maffs

[nocturne fiber]

woah woah woah

,w 2-1

[elfin berry]

Results provided by WolframAlpha

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

Requested by DerpZ#9141

[glass silo]

[fierce belfry]

WAIt

Something just snapped

Ok sorry i kinda need to go

But

I got the idea

Ill attempt it again

Without referencing

@glass silo @nocturne fiber

Tysm gyts

Guys

[marsh citrus]

@fierce belfry Has your question been resolved?

[marsh citrus]

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[brave cloak]

I don t understand how to calculate this i need help
i have to find which is correct here

[brave cloak]

the answers is sqrt(1/4b) but i don t know why

[light frost]

$\sqrt{\frac{1}{4}b}=\sqrt{\frac{1}{4}}\sqrt{b}$

[elfin berry]

jay.

[brave cloak]

how?

@light frost

[light frost]

this rule

[brave cloak]

there is a minus in front

[marsh citrus]

@brave cloak Has your question been resolved?

[marsh citrus]

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[orchid sedge]

Hi! I could use some help with the following question:

[orchid sedge]

What i think i have to do first, is write up a polynomial with the p(4) and p(-4) = 0, so:

(t-4)*(t+4)

But i get unsure when i progress here, am i supposed to try to copy the polynomial below? If so i need to multiply with another (t-4)

[still temple]

yes, this is right

[worn nimbus]

every element of your base should have this factor correct

[still temple]

you want to write the poly as a multiple of (t-4)(t+4)

[orchid sedge]

Alright, So will I write the poly like: a2xt^2 x (t-4)(t-4), a1 x t x (t-4)(t+4), a0 x (t-4)(t+4)

Or have i missunderstood?

[still temple]

sure, thats fine (what you have written is equal to (a2*t^2+a1*t+a0)(t-4)(t+4))

[orchid sedge]

ah ok, so that should be the polynomial right?

[still temple]

yea

find the coefficients a2, a1, a0

[orchid sedge]

Alright!

So find the coefficients so that it relates to the polynomial in the question? t^3-4t^2-16t+64?

[still temple]

equal to that poly, yes

[orchid sedge]

Ok, then i get a2=0, a1=1, a0=-4

[still temple]

$\polylongdiv{x^3-4x^2-16x+64}{x^2-16}$

[elfin berry]

Toby

[still temple]

yup correct

[orchid sedge]

How do I go forward now? I have the coefficients, how do i determine a base for H and the coordinates?

[still temple]

you already found the basis to be (t-4)(t+4)

the coefficient is a1=1, a0=-4. ie (t-4)

[orchid sedge]

Ohh, I see

Yea I get that

But that is not the coordinates right? The answer is supposed to be a vector

[still temple]

there is only one generator for H

so the vector is 1d

(ie just (t-4))

oh wait

does P3 mean polynomials of degree less than or equal to 3?

[orchid sedge]

I believe equal

[still temple]

it wont be a subspace if its only equal

[still temple]

this is wrong

[orchid sedge]

Then it is less, I translated it to english myself so might have been wrong there

[still temple]

the coords are a1=1, a0=-4

ie the vector (1,-4)

[still temple]

and the basis is what you had here

ie t(t-4)(t+4) and (t-4)(t+4)

sorry for the confusion

[orchid sedge]

No thats okay, I am just a bit confused in how I am supposed to write the answer. It wants them as the Polynomials in my choosen base

[still temple]

maybe write $t^3-4t^2-16t+64 = t(t-4)(t+4) -4(t-4)(t+4)$?

[elfin berry]

Toby

[orchid sedge]

In an example, my professor answer likes this

[still temple]

ah

$H=\mathrm{Span}{t(t-4)(t+4),(t-4)(t+4)}$

[elfin berry]

Toby

[orchid sedge]

That right there did it!

Thank you so much!

[still temple]

basically, the things you had here without the 'a's

and not a2 since that has degree 4 which is bigger than 3

[still temple]

np :D

[orchid sedge]

Yea that i understood, if we were to use another polynomial we would get to high of a degree

Curiosity question, why dont we write a last polynomial as just t? Since we have t^3 in the first, t^2 in the second

[still temple]

because no degree one polynomial has two roots

ie, non zero line goes through both (-4,0) and (4,0)

[orchid sedge]

Ohhh right!!! Since our base needs to contain (t-4)(t+4)

[still temple]

yup

[orchid sedge]

Alright! That makes sense

Thank you, and have a great day!

[still temple]

https://cdn.discordapp.com/emojis/690630274907897887.gif?size=32&quality=lossless

you too!

[orchid sedge]

Is it !close?

!close

[still temple]

.close

[orchid sedge]

.close

[marsh citrus]

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[sick sedge]

I'm not sure how to deal with the part after I factor r

[sick sedge]

I got to $r(cos\theta^2 - rsin\theta^2) + 14rsin\theta = 0

[still temple]

(put the eq in dollar signs)

[sick sedge]

how

[elfin berry]

Nim

[sick sedge]

am I suppose to factor inside the bracket?

like (x^2-y^2)(x^2+y^2) kinda stuff?

nvm i solved it

.close

[marsh citrus]

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[short wadi]

How did we get rid of the denumerator?

[still temple]

rationalise by multiplying by the conjugate

[short wadi]

yeah I get that

[light frost]

(a-b)(a+b)=a^2-b^2

[short wadi]

yes I know this

I’m here now

[still temple]

expand

and cancel with the t on top

[light frost]

simplify the denominator

[short wadi]

But it’s subtraction not addition in the denumerator

I mean it’s not multiplication

[still temple]

(4+t)-(4-t) = ?

[short wadi]

Yes you can see in this

[still temple]

what are you stuck on then?

[short wadi]

I’m just stupid

Thanks

.close

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[still crypt]

does anyone know how to solve this,

[still crypt]

i think my answer is wrong

[obsidian bone]

Looks right

[marsh citrus]

@still crypt Has your question been resolved?

[marsh citrus]

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[slim quartz]

Does this just mean to replace x using -2 or 3 and have 2 answers

[main idol]

yes

[pallid pine]

Yes, 1 answer associated with p(-2) and one answer associated with p(3)

[cobalt sentinel]

Wow insane camera

[marsh citrus]

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[still temple]

prove A-(B int. C) = (A-B) U (A-C) analytically

[still temple]

<@&286206848099549185>

[vast shoal]

Hi is int. ∩ here

Intersection

[still temple]

yeah

[vast shoal]

i hate these proof questions tbh

[still temple]

me too

[stoic saddle]

"analytically"?

[still temple]

that just means mathematically other than using venn diagrams @stoic saddle

[stoic saddle]

so, by element-chasing...?

[wary yew]

To show both are equal, we can show both left and right are subsets

Let x ∈ A-(B∩C); then x∈A and x∉B∩C

x∈A and x∉B or x∈A and x∉C

x∈(A-B) or x∈(A-C) or it's union

[still temple]

can you explain this step a bit

[vast shoal]

If $x \in A - (B \cap C) \
x \in A$ and $x \notin (B \cap C) \ x \in A$ and $x \notin B$ or $x \in A$ and $x \notin C
\ x \in (A-B)$ or $x \in (A - C) \ \ x \in (A -B) \cup (A - C)$

[still temple]

3rd line can you elaborate a bit

[vast shoal]

A - (B int C) means

Set of elements that belongs to A but not to (B int C)

So either it belongs to A and doesnt to B or belongs to A or doesnt to C

[elfin berry]

ColdTee

[still temple]

thank you

[vast shoal]

But you also have to prove the reverse is true

That's what sucks

[still temple]

why though that

[vast shoal]

I don't know but it is required i guess

[still temple]

does these arrows have a specific name and significance in maths that you used what do they stand for

[vast shoal]

Oh them

It's the implies sign

[still temple]

ok

.close

[marsh citrus]

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[still temple]

is this right?

[glass silo]

,calc (4 * 6) + (2 * -2)

Linearity of integrals

[elfin berry]

Result:

20

[hollow veldt]

nope check for option E

[still temple]

ok

[glass silo]

All but one should be “trivial” to check, I’ll say…

[still temple]

k so A is true

[hollow veldt]

B is true

[still temple]

yeah b is true

[glass silo]

Really? Why?

[still temple]

wait im confuzzled

let me refer back to my notes

rq

[glass silo]

Come back to c later, do the others

[hollow veldt]

what about E

[still temple]

Is this testing if I know the properties?

of the fundamental theorem

ad rules

and

well 6*-2 is -12 but that's not what the answer is

it's not necessarily the answe4r

the only thing secured is additivity and subtraction

not multiplication

when you have the same integral

definite integral

not multiplication

that why?

[glass silo]

That’s a good point, have you checked D and E yet?

[still temple]

nien

well

[glass silo]

[of course you have for D, I answered you already on that, but forget that I did]

[still temple]

I know e is true

because that rule where you can flip

the signs

if you do that for both of them

it stays the same

I mean

It functions

properly

but yea

Thx for the help g

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[long cape]

What exactly is meant here by Rationalize the Numerator? Is it a typo with Symbolab?

or is Rationalize the correct term, when you combine two fractions into one?

I always thought Rationalize meant to remove sqrts

[mystic minnow]

when you combine fractions with a common denominators, you get an irrational numerator. They suggest you rationalize the numerator

[long cape]

Why is it irrational to combine two fractions with common denominators?

[mystic minnow]

when you combine the fractions, the numerator has a radical in it. making the numerator irrational.

meaning it has a radical in it

[long cape]

Hmmm, I have to think about this..

[mystic minnow]

'rationalize' just means get rid of the radicals

[long cape]

1/2 + 2/2 = 3/2. These fractions have common denominators. Would it suggest 3/2 is irrational? Or am I misunderstanding?

[light frost]

The numerator when you combine the two fractions becomes 1-sqrt(t-1)
They wanted to get rid of the square root (aka rationalize) so they multiplied the conjugate (1+sqrt(t-1)) and then canceled t

[mystic minnow]

no, it doesn't have a radical in it

[long cape]

Oh, I didn't pick up that part from this explanation

how would you word this to mention radicals?

is the idea "roots" are irrational? or powers too

like ^2 and ^1/2. only ^1/2 (sqrt) would be irrational?

[long cape]

yeah thats what I thought.. only when a radical can be drawn, is it considered irrational

[mystic minnow]

this function has irrational numerator and denominator because it has radicals in the top and bottom

[long cape]

yeah

I often hear "rationalize the denominator", I think it depends on the question if you need to bother or not.. don't think professors are too picky with that

my line of thought, is that if it starts with a radical in the denominator i think it's OK to answer with that too

otherwise I will rationalize

but for this on Symbolab

it says "Rationalize numerator".. how is it being rationalized? It's just 1 in the numerator

[mellow sapphire]

I wrote it out if u want to see it

[mystic minnow]

get a common denominator

actually subtract the fractions

[long cape]

please 🙂

[long cape]

Oh I see, so we are getting a common denominator

[mystic minnow]

^^^ there's a page of work you're skipping and then trying to understand what's happening

[long cape]

this is the same thing as "rationalizing"?

[long cape]

no it's OK, the steps are explained on Symbolab too. I am more concerned about the label "rationalize numerator", if that is considered a typo or not?

[mystic minnow]

no

[long cape]

seems to be this is just find a common denominator, which would be different I think?

we aren't removing radical signs

[mystic minnow]

yes you are

[long cape]

oh, for the numerator

[mystic minnow]

yes. So you're rationalizing the numerator

[long cape]

So basically this

I missed this explanation

multiplying by the conjugate

to rationalize the numerator

[mellow sapphire]

Yeah so you can do the limit

[long cape]

i suppose we could have rationalized the denominator instead.. either way, same answer

[long cape]

oh wait, maybe not? for limit would I need to rationalize the numerator?

to see if I can cancel anything

[mystic minnow]

there's a t multiplied onto the denominator, when you take the limit that could be a problem.

[long cape]

OK

.close

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[thorny onyx]

not sure what i will be dividing/multiplying at the end to get my limit

[pallid pine]

Divide both numerator and denominator

By u⁴

Then take limits

[thorny onyx]

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[neon badger]

Hi

[neon badger]

For all positive natural values of x find all integers N that in their decimal representation have x number of 2 in their digits and 1 time 5 in their digits and are a perfect square

It was in a math comp and I got 25 and 225 but I’m not sure if I’m correct

[smoky plover]

Yea but how do you prove for all x that these are the only sol

Might want to post this in #competition-math

[neon badger]

Ye probably

But imma leave it here too

<@&286206848099549185>

So my reasoning was that for it to be a perfect square it can’t end in 2 so the 5 must be at the end

After that

When x>2 when you try to factorise N you get a number with x-2 8s and a 9 at the end

Which is a prime number most of the time so it can’t be factorised so N isn’t a perfect square

[snow wagon]

but it can also end in any other number than 5 and the 5 being in any other digit right?

[neon badger]

No cause the number only has 2s and a 5

Which I asked and was verified by the overseer

Cause I had the same thought

[neon badger]

But this number can be factorised if x+2 I think is a multiple of 4

As it can be divided by seven in that case

But if I’m not mistaken the number after that division always comes out prime s

Is there a flaw in my line of thought?

…

Any luck

[smoky plover]

Oh, then consider rewriting it as 2(10^(x+2)-1)/9+5=k^2

10^(x+2)=(9k^2-44)/2

k=2n

10^(x+2)=18n^2-22

[neon badger]

What’s n

And k

I got confused

Ah I see k

[smoky plover]

Sorry it was wrong

$\frac{2(10^{x+2}-1)}{9}+3=k^2$

This is correct

Wait

[elfin berry]

kappa

[smoky plover]

Now we have to solve this Diophantine

[neon badger]

Ok

And how would one do that

[smoky plover]

Lemme try to

[neon badger]

Thanks

[marsh citrus]

@neon badger Has your question been resolved?

[marsh citrus]

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[still temple]

Is my solution correct

?

[stark trail]

!15mins

[marsh citrus]

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.

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

[still temple]

Anyone?

[snow wagon]

what is cis

do you mean cos?

[still temple]

Cos+isin

@snow wagon

[marsh citrus]

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[still temple]

.close

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[gusty flower]

hello

[gusty flower]

i need some help finding the derivative of 4xsinxcosx

[twilit arrow]

what have you tried

[gusty flower]

so i tried applying the product formula

and i got 4x * (cos(x)-sin(x)) + 4*(sin(x)cos(x))

would that be correct?

[twilit arrow]

i don't think so

what's the derivative of sin(x) * cos(x)?

[gusty flower]

cos(x)-sin(x)?

[twilit arrow]

how did you get that?

[gusty flower]

well bcz i thought d/dx of sin(x) is cos(x) and cos(x) is -sin(x)

[twilit arrow]

but you have cos times sin

so what rule must you apply?

[gusty flower]

product

[twilit arrow]

yes

[gusty flower]

oh shnap

so

d/dx (sin(x)cos(x)) = -sinx^2 - cos(x)^2?

[twilit arrow]

it's close

[gusty flower]

one sec

i did that in my head

would that be correct?

[twilit arrow]

no

oh

you changed it

um no

what's the derivative of sin

[gusty flower]

bruh

its cos(x)

[twilit arrow]

yes

so you should have cos^2(x) - sin^2(x)

[gusty flower]

ohhh

my bad i did that too fast

ok so now i would apply the product rule again?

since theres a 4x?

[twilit arrow]

you apply it twice

you do product rule on 4x and cos(x)sin(x)

and then product rule on cos(x)sin(x)

[gusty flower]

wait hold up

so product rule on the original and then product rule again?

is that because g(x) has a product within itself?

[twilit arrow]

yes

[gusty flower]

so

rn i have 4x(cos^2x-sin^2x) + 4(cosxsinx)

and i apply the product rule to the second half again correct?

[twilit arrow]

no, because you're not taking the derivative

[gusty flower]

so what should should i do now?

[twilit arrow]

nothing, unless you want to simplify

in which case recall your double angle formula

[gusty flower]

ah okay

thanks

.close

[marsh citrus]

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[still temple]

can someone help me figure out how where im supposed to find the standard dev?

[main idol]

It's a binomial distribution

[still temple]

okay thank you

[still temple]

im confused tho how would i implement that

[main idol]

You don't implement it

You follow the formulas for the binomial distribution

And plug in your parameters

Start by reading the binomial distribution

[still temple]

can you show me the first few steps im so in over my head with this class

@main idol

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[main idol]

Look these up

[still temple]

okay

[marsh citrus]

@still temple Has your question been resolved?

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[primal ore]

Help asap.

[primal ore]

What are the bounds

for x: 0 and 4
for y: 0 and x^3

?

I got this but it's wrong. What did I do wrong?

[main idol]

Why asap

[primal ore]

Due soon

Help man

Riemann sum

Help me Bernhard Riemann

[mellow sapphire]

Switch the two integrals

You can’t do the one you posted

[primal ore]

Switch them?

[mellow sapphire]

Yeah

Dy first

Then do

Dx

[mellow sapphire]

Yeah try that

[primal ore]

Yeah it worked

but why?

[mellow sapphire]

You can’t do integrals with a different variable on the outside

Cuz you can’t solve it

[primal ore]

Wdym

[mellow sapphire]

If you reverse order of integration

You switch the two

And you can do a integral with respect to y if one bound has x in it

It will leave x in the answer

[primal ore]

Thank you mate

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[still temple]

I am struggling with a tough combinatorics problem:

Find in how many ways a sequence of length n (containing only letters a, b, c, d, e, f, g, h, i) can be arranged, given the following rules:

• the sequence must start with either a, b, c, d or g.
• the sequence must end with either a, b, c, f or i.
• after a, b, c, f and i, there must be either an a, b, c, d or g (or the end of the sequence).
• after d and e, there must be either an e or f.
• after g and h, there must be either a h or i.

I have no idea how to even get started with this problem. I know the basics of combinatorics, such as factorials, permutations etc., but I still cannot solve this problem.

[sleek lake]

interesting

[fathom ridge]

My dating standards are lower

[sleek lake]

hwat

you mean higher

[fathom ridge]

@still temple when you say n letters, do you mean n letters, or the letters in those brackets

[fathom ridge]

No they're lower 😔

[still temple]

The sequence's length is n.

[fathom ridge]

But they can only contain the letters in those brackets?

[still temple]

Yes. The sequence only contains the letters in the brackets.

[sleek lake]

so you go (a|b|c)* then you maybe go (deeeeeef | ghhhhhhhi) and then you maybe start over

that's not very complex but how to count that

[still temple]

If by "start over" you mean going back to either a, b, c, d or g, yes.

[fathom ridge]

This is a mad question

I think you need to get set theory involved with the venn diagrams

[sleek lake]

let's look at n=4

dfgi
gidf
def_
_def
ghi_
_ghi
ghhi
deef
_df_
_gi_
__gi
__df
gi__
df__

amazing

[still temple]

What do the underscores mean ?

[sleek lake]

abc

any mix

[still temple]

Oh, okay.

[sleek lake]

also 4 underscores

and also gigi and dfdf

so you sorta fit any number of these df things, you can leave empty spots, and they can be any length but at least 2

that doesn't give me ideas how to count it

but it's still like, within reach sorta

[still temple]

I was thinking of creating a list of auxiliary sequences (let's name them f), representing how many choices we can make at a given point.

f(n) can only have five values
If f(n) = 3, it means that at position n, our sequence contains either a, b or c.
If f(n) = 1, it means that at position n, our sequence contains d.
If f(n) = 1', it means that at position n, our sequence contains g.
If f(n) = 2, it means that at position n, our sequence contains either e or f.
If f(n) = 2', it means that at position n, our sequence contains either h or i.

Now, given a length n, we can see how many of those auxiliary sequences we can build - then, for each sequence, we multiply all of its values, and then we add the results obtained from each sequence.

The reason I marked some numbers as prime is so that the sequences (1, 2, 2, 3) and (1', 2', 2', 3) count as different sequences, but when multiplying all terms, they count as the same numbers.

The advantage of this is that these auxiliary sequences have easier rules, and I assume that calculating the number of such sequences isn't as hard. However, I still can't calculate the number of these sequences either.

[still temple]

No, this is wrong. Sorry. Using underscores instead of a, b and c is still the easiest method.

[sleek lake]

i think i have an idea

[still temple]

Ok, I have another idea: I'll forget about g, h and i for now, and I'll focus only on sequences with underscores and d, e and f.

[sleek lake]

to any sequence you can append an underscore
a sequence ending with f/i can be extended, f becomes ef
and you can append a whole df to extend by 2

that should give a recurrence

[still temple]

Hmmm, I'm not sure I understand.

[sleek lake]

it's like your idea i think

different sequences based on what's last

in the end we will ad them up

you pretty much understand

cuz i haven't made it work yet

[still temple]

Correction: f(n) can only have three values:

• 3 means that it has a, b or c.
• 1 means that it has d, e, or f (it counts as only one choice, because whether it is d, e or f depends on the position on the chain).
• 1' is the same as 1 except it is for g, h and i.

[sleek lake]

Nu(n) = 3 (Nu(n−1) + Nf(n−1) + Ni(n−1))
Nf(n) = Nf(n−1) + Nu(n−2) + Nf(n−2) + Ni(n−2)
Ni(n) = Ni(n−1) + Nu(n−2) + Nf(n−2) + Ni(n−2)

that's it

[still temple]

It's wrong to count the e and f as two choices, because which is which is already determined by the auxiliary sequence. This is why I got rid of the f(n) = 2 and f(n) = 2'.

[sleek lake]

so Nu means ends in underscore

um

not that this solves it, but i think it helps

[still temple]

Hmmmm.

[still temple]

What does that n argument mean ?

[sleek lake]

the length

[still temple]

Oh, Nu(n) means the number of ways a sequence which ends in underscore can be arranged ?

[sleek lake]

yes

[still temple]

Okay. Thanks.

I'll think about the recursive relations.

[sleek lake]

no idea how to even proceed

[marsh citrus]

@still temple Has your question been resolved?

[sleek lake]

we just need someone to solve that

or tell us that it's not simple enough

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[unreal socket]

I need help with breaking down a statistics problem. The first question is asking what is the probability of the sampled households directly affected are greater than 8

[unreal socket]

Per my notes, I'm supposed to find the (x(bar) - mean)/(SD*(sqrt(N))

I know N is supposed to be Sample size (25) but I dont know the mean and SD, in which I think I may be using the wrong formula for this one

[marsh citrus]

@unreal socket Has your question been resolved?

[unreal socket]

.close

[marsh citrus]

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[wispy garnet]

Could use a hand with question. I think I'm supposed to parameterize, but I haven't really learnt how. I think I learn best through examples so solving this would help a lot! Here's my attempt without parameterization: (I believe 0<x<sqrt(72) and -6<y<6 are the appropriate limits?)

[devout mauve]

general advice, dont upload random files

[still temple]

(discord doesnt like heic images, use png instead)

[wispy garnet]

my bad

[marsh citrus]

@wispy garnet Has your question been resolved?

[wispy garnet]

any <@&286206848099549185> out there who aren't as dead in the water as i am? :p

[marsh citrus]

@wispy garnet Has your question been resolved?

[marsh citrus]

@wispy garnet Has your question been resolved?

[red meadow]

Same case for me been waiting for hours

[marsh citrus]

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[quaint arrow]

i dont understand where the 1 + (n+1) - 3 comes from

[echo bough]

this is a recursion proof: in blue is the n gon so it has n(n-3)/2 diagonals

in pink and black are the added diagonals

when you consider the extra vertex

[marsh citrus]

@quaint arrow Has your question been resolved?

[quaint arrow]

i still dont really understand

[echo bough]

they consider a n+1 convex gon

pick a vertex

and say that the remaining n vertices

form a n convex gon

by induction

this has n(n-3) diagonals

then they count how many diagonals arre added when you add the vertex back

[quaint arrow]

okay wait, can we just use a pentagon for analogy purposes please?

[echo bough]

yeah ok

[echo bough]

but is your problem with induction proofs

or this one in particular?

[quaint arrow]

so if i pick the tip, there are 4 remaining vertices

[echo bough]

yep

[quaint arrow]

im just hoping an analogy with a pentagon will let me understand where the 1 + (n+1) - 3 comes from

[echo bough]

the pink ones are the diagonals that start at the picked vertex

[quaint arrow]

so now im i forgetting about that top vertex and im a really only looking at a square?

[echo bough]

this is how induction works

[quaint arrow]

oh okay

[echo bough]

the square has 4(4-3)/2 diagonals

but now how many are added if you consider the 5th vertex

[quaint arrow]

3 more

so by adding 2 extra sides, we get 3 more diagonals

right?

or 1 extra vertex = 3 more diagonals

[echo bough]

its true for the pentagon

but probazbly it depends on n

and thats why we do it on the general case

to not do each case

[quaint arrow]

right, but for the pentagon, is it 1(1-3)/2, which is -1, + 4?

oh wait it's not linear

that won't work

[echo bough]

wdym?

[echo bough]

this doesnt make sense iml not in your head

[quaint arrow]

i was saying if we plug in 1 (since 1 extra vertex) into the formula, we get 1(1-3)/2

But im guessing that this calculation is meaningless

[echo bough]

why would you do this calculation?

just try it in the general case

if you add one vertex to an n gon

how many diagonals are added?

theres one for the two adjacent vertices

that were adjacent to each other before

but arent anymore

[quaint arrow]

is it n-2?

[echo bough]

why?

[quaint arrow]

since it can touch all the vertices besides the 2 it's adjacent with

[echo bough]

yeah right

[echo bough]

but dont forget the last one

between the two neighbours

[quaint arrow]

oh right, that one is tricky

so n-1

[echo bough]

those are the pink ones

[echo bough]

this is the black one

[quaint arrow]

ohhhh. Thank you so much. I understand it now

[echo bough]

then its just calculating

[quaint arrow]

Right. I really appreciate you walking me through the problem. Thanks a million!

.close

[marsh citrus]

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[static quarry]

where are you stuck?

tangent = what over what?

ok, which side is opposite

and which side is adjacent

no..

34 is the hypotenuse

the missing side

what theorem tells you how the sides of a right triangle are related?

yep

try using that

how did you get that

it can't be longer than 34

the hypotenuse is the longest of the three sides

no..

suppose you call the unknown side Y

then you have Y^2 + 30^2 = 34^2 right?

solve that for Y

where are you getting 2056 in the first place

can you show me what you did

you don't add them

Y^2 + 30^2 = 34^2

how do you get Y^2 isolated on one side?

yea

that's Y^2

so how do you get Y from that

yep

well which one?

ah right

16 for Y

then the tangent is opp/adj

you sure you're checking the answer for the right problem?

wait one moment

which one is opp and which one is adj?

you want opp/adj

opp is 30

adj is 16

[marsh citrus]

@hard portal Has your question been resolved?

[marsh citrus]

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[rotund yacht]

i forgot to write my work

but i tried it again with 315 and i got something else

154/2

77cos(315)=54.44

77-54.44=22.56

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[vast shale]

What do

https://tenor.com/view/dog-doggo-cute-math-formulas-gif-17580986

[empty hill]

You can construct a triangle to find out $h = 80 - 77 \cos \theta.$

[elfin berry]

Shao

[marsh citrus]

@rotund yacht Has your question been resolved?

[rotund yacht]

where did u get 80 from

[rotund yacht]

ahh

[empty hill]

Diameter is 154, so the radius is 77

[rotund yacht]

yes and

the distance is 3

thank you

makes sense

[empty hill]

Yuppp

[rotund yacht]

.close

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[oblique widget]

hey

[oblique widget]

oh oops

[oak void]

lol

[oak void]

is the range not -1/4 <y<1/4?

question B btw

<@&286206848099549185>

https://tenor.com/view/bean-mr-middle-finger-gif-25557078

https://tenor.com/view/lonely-gif-18978208

[marsh citrus]

@oak void Has your question been resolved?

[ebon orchid]

@oak void

[oak void]

wb the range of y?

[ebon orchid]

(0 to +infinity)

[oak void]

\

[oak void]

^

[ebon orchid]

o to +infinity

[oak void]

1/4 is not zero

[ebon orchid]

sec function cannot be less than 1 ..

[ebon orchid]

what?

[oak void]

it says the range is y> 1/4

thats the answer

[ebon orchid]

gemme a sec

yea you right

my bad

it was mine mis-calc

it is 1/4 to + inf

[oak void]

.close

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[weary temple]

Could use a small bit of help. Given is a countable partition {A_i}_{i in N} in X. Also given is the sigma-algebra F which just consists of every possible countable union of A_i. How can I show that if B is in F that the complement is too? Any hints would be appreciated. I did try looking into applying DeMorgens law but that just gives me an intersection of complements and I am not sure that's helpful for anything

[rugged cobalt]

can you give me the definition of a sigma-algebra

[weary temple]

3 properties. the sigma algebra contains the whole set. If an element B is in the sigma algebra then so is the complement of B and finally taking an union of a countable subset of the sigma algebra is in the sigma algebra

[rugged cobalt]

so what is there to really show?

you do just use demorgans law

and then use the definition.

[weary temple]

Right except the fact that the sigma algebra contains countable unions?

Taking demorgans law gives me intersections

[devout mauve]

when you say every possible countable union, hopefully you also include finite unions, yes?