help-28

[cloud moon]

so for d you need to just say is it dependent or independent?

@torn jolt

[torn jolt]

oh

ig i ddint read the directions mb

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[cold crater]

Sup

[hot herald]

do you plan on asking a math question

[full forum]

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[gilded prairie]

So Im confused on how to start on this... appreciate your help 🙏

[lyric narwhal]

alright do u know basics of trig graphs

y=sin(x)

and so on

[gilded prairie]

yes its reading them directly has me confused

[lyric narwhal]

Its a sin graph

d(t)=a sin (wt)

in a normal sin graph "a" is the amplitute of the graph

in this image it isnt very clear so i wouldnt judge it by eye

however the period isnt hard to identify

do you know what the period is

[gilded prairie]

1/2

[lyric narwhal]

yes

the equation for the period is 2pi/n = period

however we have w so just substitute w for n

if we know the period is 1/2 we can find w

[gilded prairie]

4pi right? i crossed multiply 1/2 = 2pi/w

[lyric narwhal]

yes

so u have ur value of w now

because the applitute isnt clear and i wouldnt trust just reading it in this example

mb

didnt think about that well

[gilded prairie]

using the x intercepts for example?

[lyric narwhal]

ye

i dont it will work because you will always have a 0 y value

and then it means a = 0

which istn true

unless ur given any other infomation about a specific point i think you will have to just judge with your eye

[gilded prairie]

This is all the information my two guesses are 7 or 8

so either number willl be for A with sin (4pi* t)

[lyric narwhal]

yes that what i would think

stupid question

just counted the notches with a rule it seems to be 6.5

tbh im not sure

[gilded prairie]

me either but apparent its not either 7 or 8

[lyric narwhal]

try 6.5

[gilded prairie]

its not it either and i have no more chances left

I understand the concept better and thank you for helping me out

[lyric narwhal]

ye no worries

sorry i couldnt help more

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[kind jay]

$$\epsilon^{ijkl} =
\begin{cases}
+1 & \text{if} (i, j, k, l) \text{ is an even permutation of } (0, 1, 2, 3)\
-1 & \text{if} (i, j, k, l) \text{ is an odd permutation of } (0, 1, 2, 3)\
0 & otherwise
\end{cases}$$

[kind jay]

$\epsilon^{1230}$ is -1 right?

[glossy valve]

Fucktalogist

Fucktalogist

[kind jay]

https://tenor.com/view/schitts-creek-thats-correct-yes-nod-nodding-gif-10083239

[vital ravine]

ye

[kind jay]

thanks! t!@kind jay

[dull seal]

Asking to himself and answering to himself.

[kind jay]

np!

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[vital ravine]

i said yes!

[kind jay]

he said earlier!!! u copied him

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[runic spruce]

Stuck on a

[torn jolt]

https://www.google.com/search?q=how+to+find+the+midpoint+of+a+straight+line+in+a+chord&rlz=1C1AWFC_enGB941GB941&oq=how+to+find+the+midpoint+of+a+straight+line+in+a+chord&aqs=chrome..69i57j33i160l3j33i21.4153j0j7&sourceid=chrome&ie=UTF-8#kpvalbx=_CnApZOO3N4qC9u8P0uG84AI_31

watch dis video and if you still don get it, jus text again ^^

the easy one is c

to make an equation it would be another tangent that is perpendicular to ab

when you want to make an equation of a line that is perpendicular it would be the negative reciprocal

if AB is x-2y+3=0

you would firstly make it into y=mx+c form

2y=x+3

then it would be the negative reciprocal

[runic spruce]

wait for a though we dont have radius or the points

just the centre hm

[torn jolt]

which part we on about :^

[runic spruce]

finding midpoint of AB

[torn jolt]

okoko lemme checkk

[runic spruce]

i think i get this

its like 6 marks part a so something is up there

[torn jolt]

yeh yeh

give me like 10 mins

[runic spruce]

thanks!

[torn jolt]

imma work it out then explain ^^

[cold minnow]

um if i mayyyy

you can find the radius of the circle

youve been given the coordinates of A yup?

you can identify the y coordinate using the equation itself btw

[torn jolt]

yes yes

[cold minnow]

so calculate the distance between C and A

[torn jolt]

substitution??

[cold minnow]

that would be the radius

[runic spruce]

where is that

[cold minnow]

Sinceee A is a point that is intersecting the circle itself

in part A itself?

In part A it says

x coordinate of A

is = 3

[runic spruce]

oh i see what u mean

[cold minnow]

mhm

[runic spruce]

because its after a idk if we use that

ugh maybe its just poorly written

[cold minnow]

?

um wait for korljrn ig

please do tell me if you find another method tho!

[runic spruce]

ms hasn't used that

tryin to hard not to peep it and just figure it out😭

[torn jolt]

still working on it, i do not like these questions so 🙄

[runic spruce]

Same loool but my spec does so I have to🤣

[torn jolt]

i think i am more closer to killing myself than actually solving this

[runic spruce]

let me ping

<@&286206848099549185>

[torn jolt]

does it tell where the equation cuts off??

[runic spruce]

wdym

[torn jolt]

sometimes in questions it gives you where the equation has been cut off in the circle

i don think its given here so its a little harder

[runic spruce]

yeah its not given

[torn jolt]

okay still working out tho

i've tried but i don think ican get to the answer :((

hope someone else can help outt 👍

<@&286206848099549185>

[runic spruce]

I think I got it now

[torn jolt]

could you tell me what you did, i am eager :^^

[runic spruce]

find the perp bisector gradient using the eqn they gave

that passes through the centre so u can get hte equation of that line

then see where the two lines intersect

then u get coords

[torn jolt]

i completely forgot about the perp bisector 😭

i do apologise :))

glad you found the answer though!!!

[runic spruce]

thanks anyway

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[shut thorn]

python code for finding if all given points are collienear

[shut thorn]

where am i wrong

nvm got it

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[compact comet]

How to do this without a calculator?

[compact comet]

I can give the answer if it would help

[hollow wharf]

Think of how the second derivative helps identify points of inflection

[compact comet]

0, changes sign from left to right

[hollow wharf]

also is it just me or is this an ap exam

[compact comet]

Or right to left

[compact comet]

This is a q on a prac exam

[hollow wharf]

ah ic

i remember because i took the calc ab exam last year LOL

[compact comet]

Lol

[hollow wharf]

But yeah, you’re right

[compact comet]

Ok so

With that knowledge

And not being able to graph it

[hollow wharf]

This is part B mcq

this part allows graphing calcs

[compact comet]

How would i do it without a calculator in like 2 mins

Ik but our teacher gives us calculator section w/o calculator for tests

[hollow wharf]

check f first

[compact comet]

f”

or f

[hollow wharf]

what are the candidate POIs? Then, are they actually POIs?

[compact comet]

0,-1,2

[hollow wharf]

you sure?

[compact comet]

Nah

For g and h

They’re not

And I gotta see if thy change sign for f

At 2 and 0

POI

not 1

For f

[hollow wharf]

I should mention that the roots’ signs are actually flipped lol

but yes f does have exactly 2 points of inflection

looking at the choices, we don’t have to check g at all since we already know f works

so we just have to verify h

[compact comet]

How

Don’t I hav to set that shit to 0

So the group of factors would have to be 1

[hollow wharf]

yes

sadly I don’t know a clean approach to this other than graphing calc

[compact comet]

No multiplicity tricks or smithing

[hollow wharf]

Expanding the polynomial is just asking for trouble

[compact comet]

Also can you explain this one

Why is average rate of change= average value for f

Is it cuz like

6x3 gets us 18

6 is the slope

But the average rate of change is also the average value

Cuz if u get the beginning and end values and subtract and divide by the x
Change in x

Ur just distributing the values of f

Meaning average value

@hollow wharf

[hollow wharf]

Average value = $\frac{1}{b-a}\int_a^bf(x)dx$ if I recall correctly

[glossy valve]

CST

[hollow wharf]

rate of change is just the derivative

So what’s the derivative of g

(Recall FTC)

[compact comet]

F(x)

f(x)*

[hollow wharf]

yeah

so basically “average rate of change of g” is the same as “average value of f”

[compact comet]

So

Would avg rate of change of f

= avg value of f?

[hollow wharf]

no

[compact comet]

Why

[hollow wharf]

let’s say we have f(x)=x

consider from x=3 to x=5

What’s the average value of f(x)?

[compact comet]

5-3/2

1

[hollow wharf]

?

[compact comet]

4?

4 is the avg value?

[hollow wharf]

Yes

[compact comet]

Ok

[hollow wharf]

How about the average rate of change of f?

[compact comet]

1

[hollow wharf]

Ye

Clearly 4 =/= 1

[compact comet]

Yes

[hollow wharf]

So average rate of change of f is not the average value of f

[compact comet]

Right ok that makes sense

Ah i c

It’s because the avg rate of chang of g

Is just the avg amount the area under th curve increases

Over a time t

What I’m hella confused again

If by FTC

The integral is f(5)-f(1)

And the avg rate of change of g

Wait no

Nvm

Would the avg rate of change of g

Be g(5)-g(2)/3

Meaning f(2)+(f(5)/3

Which is literally just average value of f

@hollow wharf

[hollow wharf]

Use this to compute average value

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[amber phoenix]

Bruce, a store owner, would like to determine if a new advertising initiative has increased the proportion of sales he makes to women is more than 75%. To test this, he gathers information on 150 random sales and finds that 120 of those sales were made to women.

The following is the setup for this hypothesis test:

H0:p=0.75
Ha:p>0.75
The p-value for this hypothesis test is 0.079.

At the 1% significance level, choose the correct conclusion?

Select the correct answer below:

There is sufficient evidence to conclude that the proportion of sales he makes to women is more than 75%.

There is NOT sufficient evidence to conclude that the proportion of sales he makes to women is more than 75%.

There is sufficient evidence to conclude that the proportion of sales he makes to women is more than 50%.

There is NOT sufficient evidence to conclude that the proportion of sales he makes to women is more than 50%.

[amber phoenix]

can someone help please

<@&286206848099549185>

[ember shadow]

!15mins

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@amber phoenix Has your question been resolved?

[amber phoenix]

no

<@&286206848099549185>

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@amber phoenix Has your question been resolved?

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@amber phoenix Has your question been resolved?

[wispy sage]

if we are considering the 1% significance level, we only care if the p-value is bigger or smaller than 0.01. Since the p-value is greater than 0.01, we do not have enough evidence to reject the null hypothesis. In other words, there is not sufficient evidence to conclude the proportion of sales made to women is more than 75%

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[dull thicket]

Im trying to determine the function for the sum of this power series

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Please don't occupy multiple help channels.

[light sonnet]

You already have #help-17 , don't open multiple channels

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[compact comet]

Why is this D not C

[compact comet]

Isn’t it maclaurin of e^2x

Derivative so 2e^2x

Wait nvm I get it

If u take derivative of maclaurin straight up u get D

But why can’t u just take it off the pre maclaurin

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[compact comet]

Oh

.reopen

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✅

[compact comet]

@regal lake

[regal lake]

Thank u :))

[regal lake]

Write the f(x) by expansion.
Derivate it and reduce the terms to standard form

[compact comet]

Ok that makes sense

My question is

Why couldn’t u derívate pre

Then expand

[keen spruce]

p sure u can do both whatever works better for you

[regal lake]

Yeahh, we get the same result, verified.

[compact comet]

How would u do derivation on e^2x then expand

I did it and got wrong answer

[regal lake]

Okkk

[regal lake]

It is not the expansion of e^2x but e^(x^2)-1

[compact comet]

Oh helll nah ur right

Ur right

[compact comet]

Why -1

[regal lake]

Write expansion of e^(x^2) and compare it with expansion of f(x)

[compact comet]

Of f(x)?

Wdym

[regal lake]

The question. Look at it.

U have a f(x)

[compact comet]

Right that makes sense

Since e^x expansion has to have +1 to account for x=0

[regal lake]

Yeahh exactly.

[compact comet]

Ok so

I’m still confused on how we can go from differentiation of e^x^2

To the 2nx

Cuz we get 2xe^x^2

Which is 2(x^2n+1)/n! No?

[regal lake]

2x.e^(x^2)
= 2x.[1 + f(x)]
= 2x + 2x.f(x)
= 2x + 2x³ + x⁵ + x⁷/3 + ....
= f'(x)
(Option D)

[compact comet]

Oh shi

Huh

Hmm

CN you talk derivative of x^2n?

God damn wtf is wrong with me

It’s so much easier to just derivative f 💀💀💀

Than try and mold e^x^2 into that

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[tacit eagle]

how did they get 1/8 on the bottom right?

[normal tree]

chain rule

[tacit eagle]

how do you do a chain rule of a exponent fraction

i forgot

need refresher

https://tenor.com/view/crack-cold-one-crack-a-cold-one-gif-8817237

[normal tree]

it doesn't really matter that it's an exponential or anything

the chain rule is the same

[teal aurora]

just determine f and g and plug it in the thing

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@tacit eagle Has your question been resolved?

[tacit eagle]

bro

i dont know what to do

i have everything except the 1/8 at the end

walk me through step by step how to get that

please math wizards

[teal aurora]

f(t)=30000 * 2^t

g(t)=t/8

(f(g(t)))'=f'(g(t)) * g'(t)

[tacit eagle]

id prefer you didnt use derogatory words. people will get offended

💀

[teal aurora]

💀

do you know how to find f'(g(t))?

[tacit eagle]

i mean ik where they are

i dont go looking for them

[teal aurora]

ok stop

[tacit eagle]

ok

i think i do

let me write it down

is g' = -t/64

[teal aurora]

no

[tacit eagle]

and f' = 30000 ln(2) * 2^t

[teal aurora]

ok write down the arguments aswell

of the function

like where you're evaluating it

[tacit eagle]

which arguments

[teal aurora]

like instead of g' type g'(?)

so i know youre not messing things up

[tacit eagle]

aint nobody got time for that

jk one sec

and f'(t) = 30000 ln(2) * 2^t
g'(t) = -t/64

[teal aurora]

ok f'(t) is correct but not g'(t)

for g'(t) remember 1/8 is just a constant

its like the derivative of g(t)=at

which would be what?

[tacit eagle]

yooooooooooooooo

dope

i cant believe

i forgot this stuff

so ancient

ye

1/8

[teal aurora]

exactly

you have the derivative of f evaluated at t

but in the chain rule you are looking for the derivative of f evaluated at g(t)

so that would be?

[tacit eagle]

30000 ln(2) *2 ^t/8

then u multiply by g' outside

and u add the 1/8

30000 ln(2) *2 ^t/8 times 1/8

[teal aurora]

yes

[tacit eagle]

is e correct?

this is a similar problem

just different numbers

i found p'(t): fg' g then found what t = (2004-1994) then plugged that into t into p'(t) and got 571

3998

i found it

[teal aurora]

,w log(2)/7*15000e^((log(2)*10)/7)

[glossy valve]

Results provided by WolframAlpha

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

Requested by Køter#5504

[tacit eagle]

awesome

thanks koter

ye we need all those decimals

to be exact

😄

how would you find a's formula using p_o e ^-tk ?

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[potent galleon]

i forgor how to do doubling period

i thought it's just initial size * 2

[ivory cairn]

pretty sure it's how long it takes for the population to double

[potent galleon]

hmmmmm

this is the

thing i have so far

[ivory cairn]

so it would be to find t when P = 2 * initial size

[potent galleon]

like this?

[ivory cairn]

no, take this and set it equal to 2 * starting amount

solve it for t

[potent galleon]

ohhh

i see

\

sad

[ivory cairn]

2 * starting amount
Not just 2

[potent galleon]

THX

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[errant portal]

Draw a 2 card hand from a standard 52 card deck.
1.) What is the probability they are both red cards? (26/52) x (25/51) = 325/1326 or 25/102
2.) What is the probability they are both Kings? (4/52) x (3/51) = 3/663
3.) What is the probability they are both red or they are both Kings? 25/102 + 3/663 = 211/884
4.) What is the probability there is at least one black card? 1 - 25/102 = 77/102
5.) What is the probability that at least one is a King? 1 - 188/221 = 33/221
6.) What is the probability that at least one is a King or at least one is a black card? 33/221 + 77/102 - 3/663 = 1799/4422 or 257/663
7.) What is the probability that at least one is a King and at least one is a black card? 3/663
8.) What is the probability that at least one is a King if at least one is a black card? 3/663) / (77/102) = 34/221.

[errant portal]

Can someone check this?

[edgy geyser]

Why don't u use combination and permutation?

[errant portal]

I need to solve using probability

[edgy geyser]

Hmm

[wintry lion]

(3) isn't quite right:

Pr(A or B) = Pr(A) + Pr(B) - Pr(A and B)

[errant portal]

25/102 + 1/221 = 55/221 This is what I did

Is it right now

[wintry lion]

You need to subtract the case where they are both red and both kings

[errant portal]

Wdym

[wintry lion]

So, what you have computed is

Pr(both red) + Pr(both kings)
using Q1 and Q2. But there is a small overcounting here: you have double counted the case where the two cards are the King of Hearts and the King of Diamonds, so you need to subtract that probability.

[errant portal]

What do i subtract it by

[wintry lion]

Well, what is the probability that you draw both of the red kings?

[errant portal]

3/663

[wintry lion]

No, that is the probability that you draw 2 kings, not the probability that you draw 2 red kings.

It is the same kind of computation as you did for Q2, except now instead of 4 target cards there are only 2.

[errant portal]

Is it 1/221

[wintry lion]

Not quite. Write it as a product, like you did for Q2

[errant portal]

325/1326 + 3/1326 - 1/1326 ?

Is this what u wanted

[wintry lion]

That is the correct term for the subtraction, but I think you made a typo there, as the second term should be 6/1326

(i.e., the answer to Q2)

[errant portal]

Are the rest of them right

325/1326 + 6/1326 - 1/1326 like this right for 3?

Then u solve

[wintry lion]

1, 2, 4, and 5 are right. 6 to 8 would take me a bit longer to check

[wintry lion]

Yeah

[errant portal]

Alr

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[torn jolt]

how do i solve this problem
find natural number n such that 3^n + 4 is a perfect square

[torn jolt]

oh i know how to solve it

[storm goblet]

What's was the solution

[torn jolt]

we have 3^n + 4 = x^2 ( x is a natural number )

then we have 3^n = (x-2)(x+2)

since n is natural number

we can call x-2 = 3^a and x+2=3^b (a<b and a+b=n)

then we have 3^b - 3^a = (x+2 )- (x-2) = 4

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[merry holly]

Hello guys, today we learnt about basic concepts of chemistry in which prof said that even tho there are 118 elements and some elements might get discovered in the future, there are infinitely many compounds which can be found by reacting these compounds together

[shut thorn]

Not really

It is just an exaggeration calling it infinite

Of course it can't be infinite

[merry holly]

I was wondering that if there are limited number of elements, then there should be limited number of compounds formed by them too

[shut thorn]

It's like same thing as saying a bag of salt has infinite salt

It is just hard to count and a very large number

It is not infinity

[merry holly]

But if we had to calculate all those compounds how would we calculate it using mathematics?

[merry holly]

I agree

[shut thorn]

You cant actually

You need a lot of chemistry knowledge

[cloud moon]

sazada in chemistry later you will learn for there to be more elements you need to add protons and neutrons on an atom and heavier they are the unstable they get, we already tried to make these elements cause they don't exist in earth and they are so unstable they decay into smaller elements

[shut thorn]

To know which elements react with what and some elements combine differently with the same element too

[merry holly]

Can we calculate them if we consider all of the compounds formed stable?

[shut thorn]

So we can't calculate it with just maths

[merry holly]

Just in theory i mean

[cloud moon]

theoretically you can pump more protons neutrons and electrons for new Elements but these elements don't look like your average oxygen and carbon elements they are so unstable they decay

[merry holly]

But lets just talk about the 118 elements for now

[cloud moon]

elements just differ from atomic number and on the table they are listed by that number starting top left

[merry holly]

You are right

[cloud moon]

but studying them more in depth their order also hides more properties

that's why periodic table doesn't look like a normal rectangular table

but i know where this question comes from i remember too when i vaguely knew about elements i had the same question

like you expect new elements to have special properties like oxygen or hydrogen and we just need to make them

[merry holly]

Mhm for that exact reason i want to consider all 118 elements perfectly stable as well as their compounds being stable, i know this might be far from reality but i am just curious if this was the case what would be the number of compounds formed by paring these elements and pairing the compounds formed by elements together, thats all

[cloud moon]

but that's not the case new elements behave very similar to each other in that they are so unstable that's the main reason you don't find the 119th element ln nature

[cloud moon]

i mean you can come up with hypothetical rules and for these rules come up with an equation but it's pointless

[merry holly]

How would i come up with an equation?

[cloud moon]

so lets say you have 3 elements

[merry holly]

And yes agreed it might be pointless, but i am just curious, i hope you understand

[cloud moon]

A B and C

[merry holly]

Okay

[cloud moon]

if we set the rule to be a molecule is just 2 basic elements non repeating

we would have AB BC and AC

[merry holly]

Yes

[cloud moon]

this is combinations and permutations google them and you will be able to come up with equations to find possible combinations for n elements

[merry holly]

Hmm

[cloud moon]

things change if order matters like AC CA would be the same when it comes to molecules

H2O exists but O2H Doesn't

but ye don't waste your time xd

[merry holly]

But then if we would pair up these compound to react together, it will form 3 more new compounds right? And at the end the cycle will repeat forming more and more compounds.

[cloud moon]

yes but in chemistry people have studied the reactions and it's not as simple as 3 compounds react this many times together

some elements and molecules can have multiple reactions depending on conditions and some others have 0 reactions with whatever conditions

[merry holly]

Interesting

So that means the cycle does end somewhere

But to find that somewhere one must have chemistry knowledge as well

[cloud moon]

if you get there you will see how things get much more complicated

[merry holly]

Oh lol, i still wonder why people havent thought about finding that end of cycle like just for curiosity sake, as science is basically human being's curiosity which ended up benifiting the entire human race.

it was not like 5000 ago we weren't living, we were but now its just better quality

But anyways i appriciate your time, have a good day ahead

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[silk ice]

hel

[silk ice]

I know if we find two nonparallel vectors in the plane, their cross product will be a normal vector to the plane.

and from given line

first vector

a= <-1,2,-3>

how to find second vector ?

namely b

[vapid barn]

Let t = 0, what point do you get on the line

[silk ice]

4,-1,0

[vapid barn]

Now do

[silk ice]

but this is on the line

[vapid barn]

(3,5,-1)-(4,-1,0)

[silk ice]

1,-6,1

but why ?

[vapid barn]

Do you agree that (3,5,-1) is not on the line?

[vapid barn]

Also recheck this

[silk ice]

-1,6,-1

4,-1,0 on the line

but

3,5,-1

is not

[vapid barn]

Yep

[silk ice]

youre creating a new line

[vapid barn]

So the vector pointing from (4,-1,0) to (3,5,-1) should be within the plane, but not within the line

[silk ice]

which is in the plane and same time not parallel to the a

yep

thx

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[silk ice]

We have two normal vectors

n1=<1,2,3> and n2= <2,-1,1>

I know I should take cross product of these to attain a parallel line to the plane

but I really couldn't understand exactly

why ?

[dusty hill]

Is the question: Why does a cross product of two vectors help attain a parallel plane?

[silk ice]

ye

parallel vector

to the plane

you can't get a plane with cross product but vector

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[sharp rover]

On the written portion, the highlighted equation on the bottom is the correct answer, above it is what I got, the digital is the picture of the problem. I'm still not sure what I'm doing wrong?

[spring heart]

Divide the numerical coefficients by 2

[sharp rover]

Hmm could I have a broader explanation please?

[spring heart]

So the answer is 2x³ -26x -24

Divide all of the terms by 2

[sharp rover]

Oh and then distribute the -2?

[spring heart]

No

(2x³)/2= x³
(-26x)/2 = -13x
(-24)/2 = -12

so I think the answer should be
X³-13x-12

[ember shadow]

no because you must have f(-2) = 12

[sharp rover]

Yea I got this at first but I also had to distribute the x

[spring heart]

Yup

But it says there SIMPLIFY

if you do f(-2) in 2x³ -26x -24 answer will be 12

But it says SIMPLIFY your answer

Divide that -2 to all the terms

[sharp rover]

Hm

So that would give me the x³-13x-12?

Or wait

Actually nvm I'm just confused :(

[spring heart]

Doesn't it have an answer key?

[sharp rover]

Yea

The answer is the equation that's filled in

That's why it has a little red tag on the corner

So I was assuming the x³+26x+24 just needed to be multiplied by a +2, but that wasn't the a I was left with

I got a -2?

[ember shadow]

at the top on the right you have a "-12 = ..." when it should be a +12

[sharp rover]

OH WAIT I THINK I GOT IT

oh yea I do

[spring heart]

Yes you got this

[sharp rover]

Sorry I genuinely mess up on this a lot

Okay I just have to plug in the -2 into the equation? Rather than having the -2 to the front and distributing

At least that's what the bots saying

[ember shadow]

yes, plug in -2, set it equal to 12, then solve for the coefficient

[sharp rover]

Ooooh okay okay

Alright I understand it now

Thank you sooo much!

I'll be closing the channel so u can help someone else ^^ Have a good one

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[solid saffron]

Does anyone know how to solve this?

[hallow walrus]

$\frac{a}{b}:\frac{c}{d}::ad:bc$

[glossy valve]

kheerii

[solid saffron]

Yeah, I know

But how to solve?

[hallow walrus]

what do you mean "solve"?

you can't solve a ratio

[solid saffron]

Yes, I do

I dont remember how to solve

[wintry torrent]

do you mean simplify?

[solid saffron]

Sure!!

[solid saffron]

How to?

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[solid saffron]

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[void arch]

I don’t quite understand why they’re checking its well defined? And what that even means in the context of this function?

[torn jolt]

checking 'well defined' usually means to check that any two representations of the same preimage element maps to the same thing

here, the worry is that $\alpha(g^n)\neq h^n$ for $n$ bigger than the order of $g$

[glossy valve]

Toby

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[bronze trench]

3𝑥2 − 14𝑥 + 5, is there a way I can find how to break down -14x the fastest

[gritty rose]

what does "to break down -14x " mean

show the instructions to the problem

[bronze trench]

Uhhh idk how to explain lemme get example

𝑥2 − 4𝑥 − 12, you can break this expression down to x^2-6x+2x+12 so it can be factored into x(x-6)+2(x-6)

[gritty rose]

oh

you look at multiples of 3 and 5

3, 6, 9.

then 3x^2 - 9x - 5x + 5 can be grouped

or do they need to be the same?

pain i forgot negative

try -15x + 1x

[bronze trench]

Wait but 5*3 = 15

If multiply the coefficients the answer is not positive 15

Bro ngl this prob not factorable

I got that tingling feeeling

[gritty rose]

did you try this

[bronze trench]

Bro the worksheet says (3𝑥 − 1)(𝑥 + 5) is correct this worksheet mad doo doo cheeks

[bronze trench]

Well that wouldn’t add up to positive 15

The worksheet meant -5 at the end

So ur right

[gritty rose]

,w expand (3x-1)(x+5)

[glossy valve]

Results provided by WolframAlpha

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

Requested by riemann#2529

[gritty rose]

pain

[bronze trench]

I need some website with good worksheets u know any?

8-10th grade stuff

[light sonnet]

Google

[bronze trench]

Ok

Anyways thx rieman for help

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[low sundial]

very broad question and sorry if this isnt really what this chat is for - what does ergodicity mean, informally/simply?

[low sundial]

i know its the idea that a point in a moving system will eventually visit all parts of the space it is moving

is there like a simple 2d/3d example of this?

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[sly sail]

Any idea where I screwed up? I don't seem to be able to factorise (k+1)²(k+1) into (1 + ..... k)² to give me the expression I'm looking for (which is (1 + .... k+1)²)

[hot herald]

what's the original question

[sly sail]

Ii

[drifting dune]

(1 + ..... k)² is not the same as (1 + .... k+1)²)

[sly sail]

Am I doing induction wrongly?

[sly sail]

Yes, it isn't. I didn't state that in my working tho did I?

[drifting dune]

no?

oh

mb

[sly sail]

It's fine, side effects of my bad handwriting

[drifting dune]

there is a visual proof or something

[sly sail]

?

[hot herald]

1 sec

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@sly sail Has your question been resolved?

[hot herald]

considering
$$(1+\dots + k + (k+1))^2 = (1 + \dots + k)^2 + \red{2(1 + \dots + k)}\blue{(k+1)} + (k+1)^2$$
gives you a clearer idea of how to approach this \
partially expanding the $(k+1)^2(k+1)$ to $k(k+1)^2 + (k+1)^2$ \
leads to
$$LHS = (1 + \dots + k)^2 + (k+1)^2 + \red{k(k+1)}\blue{(k+1)}$$

[glossy valve]

ℝamonov

[hot herald]

then consider sum of the first k integers for the red part

[sly sail]

Kk gimme a moment I'm helping someone else out

[sly sail]

Oh.... wait..

I just need to show that $k(k+1)^2 = 2(k+1)(1+....k)$

[glossy valve]

Kiameimon | Welt Rene

[sly sail]

Or more specifically, $2(1+...+k) = k(k+1)$....

[glossy valve]

Kiameimon | Welt Rene

[sly sail]

Question is.... how? I can't seem to do it

Oh WAIT.....

Wait a second....

I have a fishy feel like I'm supposed to use the fact that sum of integers is n(n+1)/2...

Oho...

LETZGOOO IT ACTUALLY IS

Thanks for assisting @hot herald

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[fresh compass]

Hey! I don’t really know how to do these. Could anyone help ?

I kinda get the idea I just can’t do it

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[rocky vale]

@fresh compass The general strategy is to try to evaluate the integral in terms of p. It will be a limit because of the infinite boundaries

and then figure out which p allow that limit to exist

[fresh compass]

Tysm! I just solved it

[rocky vale]

awesome! np 👍

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[maiden narwhal]

What is your solution?

Looks like that's correct.

It's the same.

[kind jay]

yes

[maiden narwhal]

Multiply your solution with $\sqrt{x}$ to get the same

[glossy valve]

octopyska

[maiden narwhal]

and make it a fraction.

Multiply the -3sqrt(X).

With sqrt(x)/sqrt(x)

also do the same with -4.

If only I was better at using Latex kek wait lemme try in Overleaf.

Ok I'll just try to tell you through words.

First simplify the the left part by doing dividing it by 2. You get (3-x)/sqrt(x).

Then the right part becomes -2sqrt(x) -4

multiply the right part by sqrt(x)/sqrt(x).

And you get combine it into one fraction.

Hopefully that makes sense. @slate ivy

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[craggy sandal]

Compute the Dirichlet density of the set of primes p that end with 22221 in the usual decimal notation

[onyx glen]

do you have a definition of Dirichlet density on hand

[craggy sandal]

Yes, it can be found here https://en.wikipedia.org/wiki/Dirichlet_density

[onyx glen]

okay, sorry, i had to go do stuff irl

... i could ask a friend of mine who i know knows stuff about this

[craggy sandal]

Sure

[onyx glen]

the wiki page says:

If a subset of primes A has a natural density, given by the limit of

(number of elements of A less than N)/(number of primes less than N)

then it also has a Dirichlet density, and the two densities are the same.

although.... hm

[craggy sandal]

<@&286206848099549185>

[onyx glen]

it feels like the natural density of A would be something like 1/phi(100,000) but i am unsure.

or at least, from a purely intuitive and non-rigorous standpoint, the primes are spread approximately uniformly across all residue classes coprime with 100,000

of which there are phi(100,000) by defn

so... what, 40,000?

[craggy sandal]

Note, this is useful too https://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions

So here, I presume; all primes are of the form 22221 + 100000k for k = 0,1,...

So I guess we have from Dirichlet's theorem that the dirichlet density is 1/phi(100000)

[onyx glen]

ok, yeah, we do.

[craggy sandal]

So that's 1/40000

So I guess the answer is 0.000025?

Okay nice

Thank you

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[torn jolt]

Please explain: http://mathonline.wikidot.com/norm-minimization

[clear lily]

try chatgpt

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[fast peak]

be more specific about what you want explained

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✅

[torn jolt]

This line:

Recall from the Orthogonal Projection Operators page that if V is an inner product space and U is a subspace of V such that V=U⊕U⊥ then any v∈V can be written as the sum v=u+w where u∈U and w∈U⊥ and the orthogonal projection operator of V onto U is PU∈(V) defined as PU(v)=u for all v∈V.

[fast peak]

what about it. that stuff new to you?

there is a link you can click on which goes into more detail

[torn jolt]

and the orthogonal projection operator of V onto U is PU∈(V) defined as PU(v)=u for all v∈V.

[fast peak]

how about instead of just citing stuff you actually say what you dont understand about it

this is just a definition

there is not much more to explain

its exactly as it is stated there

[torn jolt]

I see. Is there any article or book which I can refer to understand more about norm minimization?

[fast peak]

any book on linear algebra should cover that

[torn jolt]

I see. Many Thanks @fast peak 🙂

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[vital ravine]

why is this invalid?
https://cdn.discordapp.com/attachments/903486480985489478/1092861023729483797/image.png

[torn jolt]

what confuses you

[vital ravine]

why this is wrong

and that the integral should not converge

[torn jolt]

$f(x) = f(-x) \implies \int_{-a}^a f(x) \dd{x} = 2\int_0^a f(x) \dd{x}$ not 0

[glossy valve]

null

[torn jolt]

look at the graph

[vital ravine]

1/x^3 is an odd function

its not f(x) = f(-x)

its f(-x) = -f(x)

[torn jolt]

yeah, my bad, i thought it was 1/x^2

regardless, it diverges

,w graph 1/x^3

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[torn jolt]

it clearly diverges

[vital ravine]

how?

i mean ya

but its still an odd function

so both sides should cancel each other out

[topaz valley]

there is no cancelling out

the integral doesn't exist to cancel anything out

you can't subtract doesn't exist from doesn't exist to get 0

[vital ravine]

the limit as x goes to infinity of (x-x) is 0

so why is it that wrong here

[topaz valley]

that's not how an integral is defined

[vital ravine]

is there a proof that states that this shouldnt converge?

[topaz valley]

well yes

,w integral from 0 to 2 of x^-3

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[topaz valley]

,w integral from -2 to 0 of x^-3

[glossy valve]

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[topaz valley]

[topaz valley]

it's not 0

it's undefined

infinity - infinity is undefined

[vital ravine]

im saying as it approaches 0

the area is still just 0

[topaz valley]

no

that's not how the integral is defined

both integrals from -2 to 0 and 0 to 2 must exist

[vital ravine]

wait i have a question

[topaz valley]

you are not allowed to take the limit of the end points approaching 0 together

[vital ravine]

one second

is it that this is only allowed when both integrals converge?

[topaz valley]

that equality holds iff all 3 integrals exist

[vital ravine]

ohhh

[topaz valley]

and taken as definition for the LHS if the RHS is well defined but the LHS is improper

[vital ravine]

so i assumed that this second one exists

[topaz valley]

no

the equality isn't applicable here because you don't even get 0 on the RHS

[vital ravine]

ye thats what i said

[topaz valley]

you get infinity - infinity which isn't well defined

[vital ravine]

this is what i meant

i assumed that the second one was 0 cuz i did this in my mind

[vital ravine]

but the two on the right are infinityish

[topaz valley]

the problem is there is no zero at all

[vital ravine]

thats what i said

[topaz valley]

the RHS is rubbish

[vital ravine]

there is no 0 cuz these two on the right dont exist

[topaz valley]

yes

[vital ravine]

so i am correct

thank you snow

[topaz valley]

ok good

[vital ravine]

i got a cool gif for this

one sec

https://tenor.com/view/cristiano-ronaldo-award-winning-walking-gif-24664103

[topaz valley]

[vital ravine]

ronaldo is me

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[lament elm]

can someone help

[atomic blade]

Let u = ln(x)

[lament elm]

what i did was substitute lnx as u

[lament elm]

yeah it became 2^u du then

[atomic blade]

Did you change the bounds

[lament elm]

when do i need to change the bounds?

[atomic blade]

Whenever you do a u-sub

Because u = ln(x), the new bounds will be 0 and ln(2)

[lament elm]

hmm i understand

[atomic blade]

Also recall that the derivative of a^x is a^x ln(a)

[lament elm]

so would the answer be

(2^ln2 - 1)/ ln2

[atomic blade]

Why would there be a -1

The antiderivative is 2^u / ln(2)

Oh I see

Yeah

[lament elm]

because i substrated the 2 bounds

[atomic blade]

Yeah I see

[lament elm]

the answer is correct like that right?

if so i have 1 more integralquestion

[atomic blade]

Yeah

I gtg

[lament elm]

alright thanks anyways!

[limber flicker]

partial fraction decomp should be an obvious way

[lament elm]

ohh

so if i make the fraction x^2 ( x + 1)

then solve for u = x + 1 ?

[limber flicker]

nono

you can write x^3+x as a pair of factors

then take the partial fraction of that

[lament elm]

hmm could u help wıth that

[limber flicker]

idk if you mean x(x^2 + 1)?

[lament elm]

maybe give a hint

[lament elm]

yeah nvm it wont work

i miswrote it

[limber flicker]

👍

do you know what I mean by partial fraction decomposition ?

[lament elm]

yeah

i dont know how to do it in this question though

[limber flicker]

how so?

[lament elm]

honestly no idea

idk how i decompose the fraction into parts

[limber flicker]

what factorpartial fraction would you write for the factor x ?

A / x ?

should be in your notes if you've covered them

[lament elm]

im trying 1 second

[limber flicker]

and note what type of factor (x^2 + 1) is

[lament elm]

a/x + (bx+c)/(x^2+1) ?

[limber flicker]

yes

very nice

[lament elm]

that makes it

1/X - x/(x^2+1)

?

c= 0 , a=1 , b=-1

[limber flicker]

yes

only nice integrals left over

[lament elm]

is lnx - 1/2 ln (x^2 + 1) + C the final answer?

[limber flicker]

hmm

|x| |x^2+1|

not (x)

(x^2+1)

[lament elm]

ohh i see

but what about the 2nd one

isnt x squared always positive

is it really needed

[limber flicker]

no

[lament elm]

alright

tysm for the help

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[hexed hearth]

how do we graph this function

[fast peak]

remember what log means

what is log_a(a)

[hexed hearth]

1

but what would be the domain

just 1?

[fast peak]

well for which a is log_a defined

and for which a is log(a) defined

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[grizzled basin]

For Lagrange Error Bounds, what is the point of the interval from the center to whatever x value (a) maximizes the (n+1)th derivative of f?

[grizzled basin]

i dont see its purpose because the "a" x-value will always be the endpoint of the interval

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[grizzled basin]

💀

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@grizzled basin Has your question been resolved?

[grizzled basin]

💀

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@grizzled basin Has your question been resolved?

[grizzled basin]

💀

[storm saffron]

https://youtube.com/shorts/9zRyLxylu9Y?feature=share

[grizzled basin]

cool

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[dim geyser]

Question. Can you derivate discrete function?

[harsh carbon]

But I think for discrete you work with differences or divided differences

So actually no derivative

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[patent grotto]

can someone explain to me what im doing at #6