help-39

[pulsar flax]

so what questions are you being asked

there is a lot you can say about vectors

[minor pebble]

should I send the sheets she gave in for practice?

so it is basically this

and I am almost at math level 0 cause I was off school for 3 years 🫡

I wanna learn about topics that would help me solve stuff like this, atleast for now

[pulsar flax]

generally khan academy is recommended for catching up to these kind of things

I think also organic chemistry tutor is another one that is usually recommended

see what works for you

[minor pebble]

I did try that but he seems to go more into different topics that I am not looking for now atleast, this does sound bad but I'm just on a tight clock and need to learn this for a test

[minor pebble]

I'll look into it man, thanks

[pulsar flax]

yeah both of them have a lot of topics

but they surely have plenty of videos about vectors and basic linear algebra in general

[minor pebble]

math genuinely seems alien after not touching it for years ngl

do you know any servers where people perhaps help on call or something like that?

[pulsar flax]

but after doing this for a while this can become another language you can talk

so it will eventually click

[pulsar flax]

sadly no

[minor pebble]

oh well, I do appreciate the response though man, have a good day

[pulsar flax]

but you probably can find tutors in your university/college

also, if you have any specific questions, you are welcome to ask here

either in a help channel or in #linear-algebra for example if your question is about linear algebra

[minor pebble]

got ya and I would ask my teachers if there wasn't a whole language barrier lmao

appreciate the help though

[minor pebble]

also am I supposed to close the channel now or is that to be done by the people who helped

wouldnt wanna keep it needlessly occupied

[pulsar flax]

yeah to close your channel you can do .close

[minor pebble]

.close

[pearl pond]

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[pearl pond]

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[worn finch]

the teacher didnt post the answer

so i think its c) but or d)

😭

[nocturne plover]

could you reason why you think it is c

[worn finch]

because the HA at y = 0 the top part is going to 0

but never reaches 0

but for d)

it never reaches 0 and HA looks like its at y = 1

[nocturne plover]

i agree

why not b?

[worn finch]

cause it doesnt decrease 1 infinity

it increases

[nocturne plover]

agreed

[worn finch]

answer is c then right

[nocturne plover]

yes

[worn finch]

ok thank u

.close

[pearl pond]

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[blazing iron]

Hi! How can i make an injective function, f; N x N -> N. I thought of a few functions but none of those could be applicable to all elements in N x N

[prime bramble]

perhaps it's easier to find an injection from N x N into Q?

you can find an injection into Q and then compose that with the standard injection from Q into N

[blazing iron]

hnm since theyre both infinitely countable sets i could show theyre equal right?

[prime bramble]

they aren't equal sets at all

one of them consists of 2-tuples of naturals, the other of rationals

[blazing iron]

wait arent the cardinality of N, Q and I equal?

[prime bramble]

I don't know what I is, but yes, the cardinalities of N and Q are equal

[blazing iron]

oh there was a misunderstanding sorry, I meant N and Q here

[prime bramble]

N and Q are still not "equal" though

equal cardinalities sure, but not the same set

[blazing iron]

oh since the elements arent gonna be equal right?

[prime bramble]

two sets are equal if their elements are equal yeah

[prime bramble]

on second thought, this seems to be a misleading and not a good hint

the bijection from Q to N usually goes through N x N first, not the other way around

that's my bad

[blazing iron]

wait i didnt get this

if the proof from Q to N goes that way then couldnt i use the same proof for N x N to N?

[prime bramble]

the typical way to write down a injection from Q to N is to find a bijection from Q to N x N and then compose that with an injection from N x N to N

it's circular, is what I mean

[prime bramble]

here's a better idea: think about mapping (x, y) in N x N to a product of powers of primes :p

for example : (x, y) -> ||2^x * 3^y||

[blazing iron]

wait thats so clever

[prime bramble]

I admit, I did not come up with that myself

but in hindsight, it's clearly the right thing to do

I suppose one way to come up with this map is to remember the fundamental theorem of arithmetic is a thing

[blazing iron]

yeah that does create an injection, wow I really need to do more of these problems so i can figure these kinds of function

[blazing iron]

so proving cardinality requires knowledge from everywhere?

by far its been the most challenging topic

[prime bramble]

not necessarily, but it's good to know some general things

there's probably a way to write down an injection that doesn't require mapping to a product of powers of primes

[blazing iron]

i looked it up and found something like cantor pairing function

but wasnt sure if i should use it since it was given by chatgpt and i was going for double injection proof

[prime bramble]

having just looked it up, I have no idea how I'd have come up with that

clearly, Cantor knew something I do not

[blazing iron]

yeah i looked at that formula and thought it wouldnt work for even numbers, then i looked at it again and it works for everything

[pearl pond]

@blazing iron Has your question been resolved?

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[fading nexus]

Idk what I’m doing wrong

[latent quail]

,rcw

[undone wedge]

It is possible to do this without using polynomial long division; you are told that g(x) = (x+2)^2(Ax+B)(Cx+D); so you can expand this out and set coefficients of powers of x equal to each other; this will yield a system of equations which you can use to then solve for A,B,C,D (I suggest this method since I do not remember how to do polynomial long division).

[fading nexus]

I don’t really know the other method

The way u do it is just doing long devesition twice idk where I went wrong

[ivory basin]

You can just do long division once

Just divide directly by x² + 4x + 4

As for where you went wrong gimme a moment

@fading nexus you've fucked up the first long division

132 - 20 is 112, you've written 122

[fading nexus]

Ohhh kk but now I think my teacher is wrong for the box because she did not make the -2x^2 negative

And she made the 56 plus?

[ivory swallow]

Just use synthetic division

[ivory basin]

They just took out a -1

[fading nexus]

Do u have to do that?

[ivory basin]

I mean it makes it cleaner

But no you don't

[fading nexus]

Do I need to find the a value?

[ivory basin]

,rccw

[ivory basin]

Are you writing in pencil btw

Cuz if so, please erase rather than overwrite

What are we trying to do now

[fading nexus]

I thought I did erase sorry I did long devisiton twice because the equation is a fourth degree. Once I finished that I use the box method to get the value into factored form

[ivory basin]

Oh hm

Yeah that looks correct

[fading nexus]

This is the answer key the only difference is the a value and one factor

[fading nexus]

Kk ty

[pearl pond]

@fading nexus Has your question been resolved?

[pearl pond]

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[elfin ridge]

question:

• there are 40 cards
• • 3 red, 37 blue
• there are 2 players
• • player one draws a card and keeps it
• • then player two draws a card and keeps it
• • then player one draws their second card, third total, and keeps it

what are the chances player one draws a red card as the first card?

• 3 in 40
• this one easy

what are the chances player two then draws a red card?

• something like 3 in 39 if P1 drew blue (which we dont like cause then P1 cant draw two reds), or 2 in 39 if P1 drew red
• idk how to combine

what are the chances player one then draws a red card as their second draw, the third card total?

• something like 3 in 38 if no reds drawn so far (which we dont like, since they "need" to have already drawn a red for our success)
• or 2 in 38 if P2 didnt draw a red (good) OR if P1 didnt draw a red (bad)
• or 1 in 38 if P1 and P2 both drew reds (good)

idk how to layer these
the "goal" is to figure out the chance for player ONE to draw two of the 3 reds, but this depends on if player two drew a red.. and if player one drew a red first

[elfin ridge]

norm id ask chat.. but chat is giving the same 1 in 260 chance for every variation of the question because it does not comprehend the question so it cannot help me learn how to find the solution.. but yall much smarter than chat and i need much smarter than chat for this

(possibly the answer is 1/260.. but it has given that answer for every type of question ive asked.. so that would be an extreme coincidence if it was correct for all i asked)

[pearl pond]

@elfin ridge Has your question been resolved?

[elfin ridge]

ive thought some more
at step one chances for success are 3/40
at step two we have a 2/39 to limit step three to 1/38 and a 37/39 to have step three be 2/38
so step three has ( 2/39ths of a 1/38th plus a 37/39th of a 2/38th ) times 3/40

i think?
if so then it is 1/260
( ((2/39)x(1/38)) + ((37/39)x(2/38)) ) x (3/40)
and boi is chat bad at explaining things

[pearl pond]

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[buoyant fulcrum]

67th

[buoyant fulcrum]

<@&286206848099549185>

[runic shoal]

@buoyant fulcrum how would you solve acos x + bsin x = c

[buoyant fulcrum]

Idk

[buoyant fulcrum]

I used to do some hit n trial in numerics
But it's variable

[runic shoal]

alright

so let's do an example

3cos x + 4sin x = 2

there are 2 main ways to solve this form equation but I'll only show one solution which is related to this problem

[buoyant fulcrum]

Square it?

[runic shoal]

the way is to divide the whole equation by sqrt(a^2 + b^2)

[buoyant fulcrum]

Ewe

[runic shoal]

in this case you get 3/5 cos x + 4/5 sin x = 2/5

[buoyant fulcrum]

Yea

Now it's doable

[runic shoal]

now next we solve for
sin t = 3/5
cos t = 4/5

the reason why we do that is that if you replace this into the equation i just wrote you get
sin t . cos x + cos t . sin x = 2/5
or sin (t+x) = 2/5 which is very solvable

[buoyant fulcrum]

Yeah

[runic shoal]

i can claim that there will always exists a t that satisfies both equations because it satisfies sin^2 t + cos^2 t = 1

[runic shoal]

now if we apply this method to this problem...

we get
$$\sin{t} = \frac{a}{\sqrt{a^2+b^2}}$$
$$\cos{t} = \frac{b}{\sqrt{a^2+b^2}}$$
$$\sin{(t+x)} = \frac{c}{\sqrt{a^2+b^2}}$$

[jolly parrot]

(De)Carbonized

[runic shoal]

now we have 2 solutions for sin(t+x) = that thing which is alpha and beta

since those 2 solutions are distinct then we shall have
$$t + \alpha = \arcsin{\frac{c}{\sqrt{a^2+b^2}} + 2k_{1} \pi$$
$$t + \beta = \pi - \arcsin{\frac{c}{\sqrt{a^2+b^2}} + 2k_{2} \pi$$

[jolly parrot]

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

[runic shoal]

@buoyant fulcrum you get this part?

[buoyant fulcrum]

No

[runic shoal]

where do you not understand?

[buoyant fulcrum]

Idk inverse trigonometry

[buoyant fulcrum]

Others I just got all

[runic shoal]

actually now i think the problem's wording is a bit vague

because this is a trig equation so it has infinitely many solutions

<@&268886789983436800>

[buoyant fulcrum]

Yeah

[runic shoal]

and if i pick 2 solutions that are 2pi difference apart then none of the options are correct

[buoyant fulcrum]

Shall I share the solution from material

[runic shoal]

so the distinct "solutions" the problem is talking about are actually a collection of series of solutions with 2pi period

[runic shoal]

nah

we're almost there

[runic shoal]

like this is what the "2 distinct solutions" the problem is talking about

alright continue

actually im just gonna jump straight into cos(alpha + beta)

$\cos{(\alpha+\beta)} = \cos{(\pi + 2(k_{1}+k_{2})\pi -2t)} = -\cos{(2t)}$

[jolly parrot]

(De)Carbonized

[runic shoal]

<@&268886789983436800>

@buoyant fulcrum you get this part?

[buoyant fulcrum]

Yes

[runic shoal]

we also have our sin t and cos t here

now use double angle identity for cosine and you're done

[buoyant fulcrum]

Ohk lemme now note things

[runic shoal]

there might be cleaner ways to solve this problem that i havent figured out yet

[runic shoal]

that is a disgusting solution

[buoyant fulcrum]

.close

[pearl pond]

Channel closed

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[buoyant fulcrum]

Thank You so much ur efforts were just so wow this time
@runic shoal

[runic shoal]

no problem

[runic shoal]

i dont get some of the steps here tbh

my solution doesnt even involve that much calculations

[pearl pond]

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[remote delta]

how to do

[remote delta]

huh

is anyone able to help me

[latent quail]

He’s trying to help you, have you tried anything beforehand?

[remote delta]

what do u mean

i read it i don’t get it

[tropic saddle]

what have you tried

[blissful cloak]

I think it's a laguage barrier issue?

[remote delta]

huh

[latent quail]

What’s your mother tongue?

[remote delta]

excuse me?

[tropic saddle]

what language(s) do you speak

[remote delta]

welsh

and english

[weary ledge]

What have you tried?

im not talking about the language, rather the question

[remote delta]

tried what

[weary ledge]

This question

what is your progress so far

[remote delta]

i read it but it just makes no sense to me

i think i have to divide 5 with the amount they make but in not sure

[weary ledge]

did you find the number of months in between 1st October 2011 and 31st Jan 2015

what is your reasoning for dividing by 5?

[remote delta]

since they cost 5p they would divide it with the amount they made to find out how many have been bought?

[weary ledge]

but thats not what is being asked here right?

[remote delta]

it’s not? oh

[weary ledge]

they've only asked an estimate of the amount given per month

[remote delta]

oh wait i think i get it

is it how much they estimate make a month

[weary ledge]

yes

[remote delta]

ahhhhh

[weary ledge]

that is what the question asks

the price of the bag is irrelevant to the question unless they ask an estimate of how many bags were sold

[remote delta]

is that 50 months

[weary ledge]

nope

[remote delta]

but 12 months in a year there is 4 years 12x4=48 + 2 since it’s october in 2011

[weary ledge]

theres not 4 years

[remote delta]

but it’s between 2011 and 2015

[weary ledge]

its 3 years plus additionl 3 months after october 2014

[remote delta]

oh..

39 then

[weary ledge]

yes

[remote delta]

ohhh

[weary ledge]

no wait

you have to incude one more month

[remote delta]

oh so 40

[weary ledge]

the entire month of January

[weary ledge]

do you understand why

we addded one more month

[remote delta]

since 2015 goes to january

[weary ledge]

good

now youve been given the range of the total amount given to charuty over 40 months

can you now make the range for the same but per month

[remote delta]

but i’m not to sure what to do since it says £16.8 million and £21.9 million

[weary ledge]

that is the lower and upper limit

youve been given that the value lies somewhere in between these 2

[remote delta]

so i found out what’s in the middle of

them

[weary ledge]

and?

[remote delta]

5.1 million

??

wait no

that’s not right

[weary ledge]

What is this?

did you jus subtract those values 😭

[remote delta]

i did…

[weary ledge]

for what reason

[remote delta]

LMAO

[weary ledge]

???

[remote delta]

idk

i don’t know how to find the middle

[weary ledge]

always do something with logic

find the average of 16.8 and 21.9

[remote delta]

18.5?

[weary ledge]

....

[remote delta]

i’m too stupid for this

.close

[pearl pond]

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[pearl pond]

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[twin pasture]

Would some one help

[pearl pond]

Please don't occupy multiple help channels.

Please stick to your channel.

[pulsar flax]

.close

[pearl pond]

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[frank violet]

test

[pearl pond]

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[robust lynx]

are we testing for meta

[shrewd mist]

.reopen

yes

[pearl pond]

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[topaz violet]

Working through questions for a proofs class right now, but the wording is confusing me, is the answer not just infinite different functions that can relate the two sets with a unique rule?

[unkempt yacht]

let $x_1, \dots, x_n$ be the elements of $X$; and $y_1, \dots, y_m$ be the elements of $Y$. i can have $f_1$ that maps $x_i$ to $y_1$, or $f_2$ mapping $x_i$ to $y_1$, but $x_1$ maps to $y_2$, etc. etc.

[compact ridge]

the rule is that each element in X must map to exactly one element in Y

[unkempt yacht]

in essence this is a combinatorics problem

[compact ridge]

(two different elements in X can map to the same value of Y, so you don't need to worry about overcounting)

[smoky musk]

Hint: there is a reason that the notation of the set of all functions between two sets are denoted like that. If you know which notation I'm referring to.

[topaz violet]

I dont think I know what the set of all functions looks like, but im pretty sure thats what the question wants me to find

[compact ridge]

I guess if you're familiar with functions being represented as arrows

[compact ridge]

we have an arrow that starts from $x_1$ (it exists because of this)

[jolly parrot]

south

[compact ridge]

to how many places in Y can this arrow go?

[topaz violet]

To just one

[unkempt yacht]

to one at a time

[compact ridge]

okay you should read this again

there's a lot of places to which this arrow can go

[topaz violet]

Wait does it mean functions with unique results?

Or mappings rather

[compact ridge]

wdym by unique results?

[topaz violet]

Say you have two different f’s

But they result in the same mapping

[unkempt yacht]

but f is a mapping

[smoky musk]

Then those two f is the same function

[topaz violet]

Im very confused now

[compact ridge]

yeah, if all the arrows are the same, then we count that as the same mapping

[topaz violet]

I thought the rule and the mapping were two separate things

[smoky musk]

In the set theory, they are the same thing.

Two functions f and g are identical if for all input x, f(x) = g(x)

[unkempt yacht]

think of functions as a set of pairs, whose first element is an element of X and second element is an element of Y

[topaz violet]

Alright that clears some things up

So then its just a matter of how many different ways can the n elements of X be paired with one of the m elements of Y

[smoky musk]

Yup

[topaz violet]

And then also proving that

I cant believe im this rusty at a basic stats problem pretty much but its something like n^m combinations right?

[unkempt yacht]

right track, but think again

[compact ridge]

you're so close

[topaz violet]

Do I just have it backwards

[unkempt yacht]

so what do you have now?

[topaz violet]

I was writing down a small set of pairs to try and remember what is is and with n=3 and m=2 I got 8 different combinations or m^n

Unless I missed a combination

[unkempt yacht]

m^n is correct

[topaz violet]

Alright

The proof for that would just be needing to prove that formula I think

[unkempt yacht]

that is, the number of possible mappings $X \to Y$ is $|Y|^{|X|}$, where $|X|$ denotes the cardinality of $X$

[unkempt yacht]

this is also the notation xwtek was referring about; that is, the set of all mappings $X \to Y$ is called $Y^X$

[unkempt yacht]

you can reasonably show it works by simply counting

x1 can pair with y1 to ym, so m ways. same with x2 to xn

[topaz violet]

I dont think my professor would accept that

[unkempt yacht]

why wouldn't they?

[topaz violet]

I think he’d want me to fully prove it works for every pair of sets

Hes rather strict about it

[unkempt yacht]

this is already rigorous enough, you don't need to go mega abstract just for the sake of it

in fact you can do combinatorial proofs by pulling the ball-and-urn card and it would still be an honest-to-goodness proof as long as the reasoning is well-structured

[topaz violet]

Sorry im just trying to think if theres a way he’d criticize it

Hes seriously strict about it to an annoying degree

Which I get

I feel like Id need to do a form of induction or something

Because n and m arent infinite

[unkempt yacht]

very very overkill lol

[unkempt yacht]

by "strict" i mean that your prof requires your proof to have "rigor"

that is, you don't pull symbols out of thin airs, your reasoning is airtight, and your overall proof is well structured

[topaz violet]

Im just 90% sure he’d say something along the lines of how do you know it works for sets larger than n and m and that some really large number that I havent calculated doesnt do something different

[unkempt yacht]

well if you take some time to think about what i said here, whether m > n or whatever does not matter

[unkempt yacht]

Your general steps should be as follows:
\begin{itemize}
  \item Define $x_1, x_2, \dots, x_n$ as the elements of $X$ and $y_1, y_2, \dots, y_m$ as the elements of $Y$
  \item For each $x_i$, you have $m$ ways to map it to some $y_j$. Thus, how many total mappings do you have?
  \item Using the above line of reasoning, you can see that it is obvious that this gives no shit to whether $m > n$ or $m < n$ or $m = n$
\end{itemize}

[topaz violet]

Alright

Sorry for being a bit difficult about this btw its just that my teacher is rather difficult about these things

And I also am trying to present this in front of the class so while being criticized by him doesnt bother me much it does still get a bit annoying

Its a whole thing

But thanks for the help

I think this works and I understand it so

Do I gotta do something to close the request

[unkempt yacht]

.close

[pearl pond]

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[karmic leaf]

Hello, I am confused here about if I apply the chain rule to x^1/2. so I know this is first a product so I wrote it in the product form, then I did the chain rule to x^2 (am I right to do that) and do I apply it to all exponents? In general as well

[dusty flame]

holy

${\dv{x} (g(x^2) \cdot \sqrt{x}) = \sqrt{x }\dv{x} g(x^2) + g(x^2) \dv{x} \sqrt{x}}$

uhh

where did g go

[karmic leaf]

I just used different letter

[dusty flame]

huh...

[karmic leaf]

What are you taking about? Isnt this a product rule I just put it in the that form

[jolly parrot]

k

[karmic leaf]

Yes that is what I did

[dusty flame]

u did this right?

im kinda confused with the notation ;-;

[karmic leaf]

This is the d of the first times second plus first d of the second

Then I did the chain rule

I am just asking when I apply the chain rule that’s my question

[dusty flame]

here

[karmic leaf]

What about the square root of c

*x

Which I write as x^1/2 so is that not a chain rule

[dusty flame]

sqrt(x) can be differentiated with js power rule

[karmic leaf]

Why is x^2 chain and that only power that what I was confused about

[dusty flame]

because g(x^2) involves two functions g(x) and x^2

whereas x^(1/2) is js x^(1/2)

[karmic leaf]

Oh ok can you set this problem up for me to start

So which do I do first, can I write the product rule first then use chain rule in that

[dusty flame]

so u are here

and u will try to simplify each of these two terms individually

${g(x^2) \dv{x} \sqrt{x}}$ isnt a problem

[jolly parrot]

k

[dusty flame]

its js power rule

however

${\sqrt{x} \dv{x} g(x^2)}$ requires chain rule

[jolly parrot]

k

[dusty flame]

why?

cuz g(x^2) is a composite function of g(x) and x^2

[dusty flame]

product rule then chain rule to evulate one of the two terms generated by the product rule

[karmic leaf]

Ohhh ok

Tysm I know how to do it now

Have a good day/night

.close

[pearl pond]

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[pearl pond]

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[harsh ruin]

Is this correct?

[harsh ruin]

Pls ignore the background

[spiral pivot]

Can you give the original question please? It's difficult to decipher your handwriting.

[harsh ruin]

Ok

c

[pearl pond]

@harsh ruin Has your question been resolved?

[warm tiger]

Why are you solving like this

[pearl pond]

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[harsh ruin]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[harsh ruin]

Because I don't know the nth derivative of tan-1 x

[warm current]

do you know the first derivative?

[pearl pond]

@harsh ruin Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[robust lynx]

<@&268886789983436800> scams

[pearl pond]

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[glacial forum]

I'm having trouble figuring out how to start this problem here. Should the sections be vertical or horizontal? I also have a Simpson's rule numerical integration spreadsheet

[spiral pivot]

You may choose whichever is more convenient for you

In fact, it might be useful to do both, and compare your answers to see if they agree.

[sharp vigil]

you could also consider rotating the image to make diagonal sections

[glacial forum]

.close

[pearl pond]

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[glacial forum]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[glacial forum]

.close

[pearl pond]

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[pearl pond]

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[glacial forum]

When I'm approximating this area using the Trapezoid method, how do I need to adjust my values for the scalar 100ft at the bottom? I need to use a spreadsheet to show my work too which I'm confused about too-

[glacial forum]

And how do I figure out the function itself?

[pearl pond]

@glacial forum Has your question been resolved?

[glacial forum]

<@&286206848099549185> ;-;

[calm bone]

is this going towards your grade?

[glacial forum]

eventually, I just wanna know how to incorporate the scalar

this is from an assignment that was just assigned earlier today

.close

[pearl pond]

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[glacial forum]

smh

[pearl pond]

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[unkempt yacht]

please check my work for (d)

[pearl pond]

@unkempt yacht Has your question been resolved?

[ivory basin]

Parentheses on the first line pls

The entire thing is negated

Same with the second line before the ∃y

Other than that looks good

The english could have been phrased a bit better ig

[unkempt yacht]

what other way can it be interpreted as tho

[ivory basin]

Fair actually

I just like to have more parentheses rather than less

Cuz you never know when your prof is gonna be in a bad mood

[unkempt yacht]

my prof for this course is usually really positive haha, i'm sure it will be acceptable

[ivory basin]

Fair enough

You could phrase the English slightly better as follows:

There is a movie actor who has never been in a movie with KB and has never been in a movie with anyone who's been in a movie with KB

This is a slight manipulation of the last step

∃x ¬M(x,KB) ∧ ∀y M(x,y) → ¬M(y,KB)

[unkempt yacht]

oh you turned the lor into implication, good idea

thanks for the help

.close

[pearl pond]

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[ivory basin]

. This channel took a minute to go to available lmao

[weary star]

Trying to help a friend of mine, but I dont remember much about stats since it was nearly 6 years ago. How do I help them?

[weary star]

Assuming X and Y are independent? and S = X+ Y. So S is a normal RV of N(1 + u, 3)?

So P(0 <= S < 3) > 0.44?

Then I recall we do some standardization to N(0, 1) correct?

then I forget

this is their work and honestly I cant interpret it because its been so long

this is the answer, they are confused how they find this

<@&286206848099549185>

[pearl pond]

@weary star Has your question been resolved?

[pearl pond]

@weary star Has your question been resolved?

[pearl pond]

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[pearl pond]

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[wraith wraith]

is it even possible to use systems to solve using vietas

[wraith wraith]

cuz even if u plug the numbers in

I noticed it just gives me the identical equation in a different form

HALP PEOPLE

[vital estuary]

you dont even have a specific problem

[wraith wraith]

i dont need one right

im just curious

[lean oak]

such are help channels that it's better for you to have an example problem to illustrate where you're stuck at anyway.

[wraith wraith]

ok so

why do u need to cancel out the i

why can't you just leave it

and how do u simplify i

it

[lean oak]

what has this to do with the original question about vieta?

[wraith wraith]

and also

[wraith wraith]

I'll bring it up later

for now these are some other questions i have

[lean oak]

I seem to recall another helper having explained this image to you already.

[wraith wraith]

it didn't make sense whatsoever

i'm a slow learner ok?

[lean oak]

then what do you remember about their explanation?

[wraith wraith]

they said there is a rule that makes it so you can't multiply sqrt's in imaginary land

but idk why

i'm not just going to take a blatant axiom bc it isn't one I need an explanation

[lean oak]

please do not spam ping.

[wraith wraith]

^^^^

[wraith wraith]

did I?

[lean oak]

yes. you just did again.

[wraith wraith]

where

[lean oak]

turn off pinging on replies.

[wraith wraith]

I didn't tag

so may I get help

[lean oak]

so what about this explanation did you not get?

you marked the problem as solved right after this helper offered their explanation.

it should then imply you got what he meant.

[wraith wraith]

because I had to go

i don't understand how

like

it doesn't explain this scenario of imaginary at all

it is just talking about +- stuff

and I understood that

basically u do +- when YOU take the sqrt

but if the sqrt comes WITH THE PROBLEM then u just take the nonnegative one

isn't that correct ^^^

[lean oak]

so far so good.

[wraith wraith]

yes

but this doesn't relate to imaginary

so its not that i don't get the explanation it is that it isn't relevant

Anyone?

I require help

@lean oak I don't wanna tag but ur the only person helping

@vital estuary

[crystal dew]

!vol

[pearl pond]

Helpers are just people volunteering their time to help you. Be polite and patient.

[quick star]

😭

[quick star]

to get it into standard form

where you can distinguish more clearly between the Real and imaginary part and well that allows you to carry out "operations" directly

just like how you would naturally simplify 4/8 into 1/2

it's just a stylistic thing to an extent (?)

[pearl pond]

@wraith wraith Has your question been resolved?

[quick yacht]

$\sqrt{-1}\sqrt{-1}=\sqrt{(-1)^2}$ is false

[jolly parrot]

Cycadellic

[quick yacht]

In general we do not have that $(x^a)^b=(x^b)^a$ for complex numbers

[jolly parrot]

Cycadellic

[quick yacht]

The reason stems from roots of unity

[quick yacht]

You dont leave the i in the denominator because we require a+bi to call it a complex number

[pearl pond]

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[pearl pond]

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[cloud shuttle]

why the vertex form of a quadratic function contain x-h, why doesn't it have x+h?

[random ermine]

what value of x minimizes (x-h)^2 ?

[cloud shuttle]

Positive values,

[random ermine]

ok let's say h=1

so f(x) = (x-1)^2

what is f(0)

[cloud shuttle]

1?

[random ermine]

yes

f(1)?

[cloud shuttle]

0

[random ermine]

f(2)?

[cloud shuttle]

1

[random ermine]

yep

[cloud shuttle]

Oh

[random ermine]

can f ever be negative?

[cloud shuttle]

No

[random ermine]

so what's the minimum of f

[cloud shuttle]

0

[random ermine]

if it's never negative

yes

and what value of x gave us (x-1)^2=0?

[cloud shuttle]

+h?

[random ermine]

?

[cloud shuttle]

1

[random ermine]

yes

so h minimzes (x-h)^2

so the minimum is at x=h

[toxic lichen]

play around with it on desmos

[random ermine]

i.e. that's where the vertex is

[toxic lichen]

see that y=(x-h)^2 has vertex at (h, 0) always

[cloud shuttle]

Ok, I will

Chat gpt told me that we are actually measuring distance from the current place and the desired place

Ok, I will first visit desmos.

.close

[pearl pond]

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[frosty grail]

hello! how is this true?

[main timber]

does anything lead you to believe it isn't?

also, what is x constrained to this doesn't really matter for your question actually, could matter to steps you haven't showed

[frosty grail]

i am trying to simplify it, but im stuck and dont know what to do

i tried doing
sinx/(sin^2x + cos^2x) - cosx/(sin^2x + cos^2x)

but i dont know how that would lead to the answer

im struggling with the fractions

i dont know what im allowed to do

and what im not allowed to do

im not sure how that simplification is possible

[split void]

its a result of the pythagoras theorem

the denominator is 1

[frosty grail]

OH

i totally forgot about that omg

THANK U OMG

😭

[crystal dew]

nice handwriting, by the way

[frosty grail]

thank you so much, my parents complain its too small lol

.close

[pearl pond]

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[pearl pond]

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[zenith gyro]

hello

[split void]

this is a help channel. if you want to talk in general, go to discussion or discussion-2

[loud sluice]

bro said one word 😭

[pearl pond]

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[steady blade]

hello guys sorry interrupting i still struggling understand exponential and arithmetic can anyone explaining to me like i was 5 years old

[hazy quail]

2² = 2 x 2 = 4

Tada

[autumn fossil]

do u have any specific examples of what u struggle with?

[hazy quail]

你能否举例说明你在哪些方面遇到困难？

[split void]

乘方运算是重复乘法。请解释一下你哪里感到困惑了。

[split void]

its just repeated multiplication

like $2^2 = 2*2 = 4$

[jolly parrot]

Annie Maqionde

[split void]

and $2^3 = 2 \times 2 \times 2 = 8$

[jolly parrot]

Annie Maqionde

[gaunt scroll]

generally

any number with an exponential n, say 2^3 or 3^2, just means multiply by itself n times

[steady blade]

woah

[split void]

so $3^2 = 3 \times 3 = 9$

[jolly parrot]

Annie Maqionde

[split void]

and $3^3 = 3 \times 3 \times 3 = 27$

[jolly parrot]

Annie Maqionde

[split void]

so on so forth

[gaunt scroll]

需不需要解释一下分数次方和负数次方的意思

比如说-1次方和1/2次方

[split void]

1/2 是平方根

1/2 is just the square root

like $9^0.5 = 3$

[jolly parrot]

Annie Maqionde

[split void]

sorry

[steady blade]

so exponential like multipling something but the answer more big than ordinary multiplying?

[split void]

$9^{0.5} = 3$

[jolly parrot]

Annie Maqionde

[split void]

and $16^{0.5} = 4$

[jolly parrot]

Annie Maqionde

[hazy quail]

4^½ = sqrt(4)

[gaunt scroll]

uh, almost there

exponential just means multiply by itself, n times, so for example 3 to the power of 4 means

3 times 3, but do that 4 times

[split void]

yes

if its $a^b$, you multiply $a$ by itself, $b$ times

[jolly parrot]

Annie Maqionde

[gaunt scroll]

so you would get 3x3x3x3

[split void]

so take $4^2$. we multiply 4 by itself 2 times, so we get: $4 \times 4 = 16$

[jolly parrot]

Annie Maqionde

[steady blade]

woah like 3³= 27 because 3 . 3 . = 9 and 9 . 3 = 27?

[hazy quail]

Yes

[gaunt scroll]

lovely you got it

[split void]

yes.

[steady blade]

and if the power of multiply is minus become a 1/a ? like 2^-3 = 1/2^3

[split void]

yes

correct

so $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$

[latent quail]

seems like OP cannot read Mandarin

[jolly parrot]

Annie Maqionde

[steady blade]

seems beutiful

[split void]

I was confused as to what language OP will understand since the username is in mandarin

[latent quail]

makes sense

[steady blade]

i just love china and trying to learn it but still dont have timebecause . i need prepared to university

[hazy quail]

For real?

[steady blade]

(just loving the educational)

and yeah back to the problem . i still dont understand about exponential growth function

[split void]

$P(t) = P(0) \times e^{kt}$

[jolly parrot]

Annie Maqionde

[split void]

thats the general formula

[steady blade]

and what is P, t= P (0) x e^kt?

[gaunt scroll]

也不一定是e,任何其他常数应该都行，只不过exp 的底是去e

[split void]

是的，但我们通常与e合作。

为了简单起见

[gaunt scroll]

may be a bit too much, let me break it down

generally, exponential functions mean you have it in the form of c^x, where c can be any number, lets take 2 for example, you would have a function of 2^x

then as x increases, you would get the following

2^1=2
2^2=4
2^3=8
2^4=16 and so on

in this case she took an alphabet e, which actually corresponds to a constant known as euler's number, approx 2.718

the k simply means a constant you multiply to the t, and t is your variable, so your t would increase

[steady blade]

ok i am understand that's make sense like if a^2 means a . a = b but for euler number seems to far from my theory but that is worth of explanation

[gaunt scroll]

yes, you can ignore that and use more fundamental integers instead, e often shows up more in other topics (such as calculus, differential equations etc which are far out of this scope)

[steady blade]

yeah that's right that's to far because is grade 11 or 12 student and im still grade 10

[gaunt scroll]

understandable, hopefully your queries have been answered to

[steady blade]

and i still struggle to understand about exponential growth function @gaunt scroll

[gaunt scroll]

Feel free to keep this open, or send a direct message for further clarifications to me, and if other users also are willing to

[wary gale]

Bruh

tell me ur Chinese?

[pearl pond]

@steady blade Has your question been resolved?

[steady blade]

can semeone explain me about exponential growth function like i was a 5 years old?

... im still dont understand about Formula: (y=ab^{x})  (a): The initial amount  (b): The growth factor. For growth, (b) must be greater than 1. If (b) is between 0 and 1, it represents decay. Formula: (y=a(1+r)^{t}) (a): The initial amount or value (r): The growth rate (expressed as a decimal) (t): The time period (e.g., years, months)   <@&286206848099549185>

[jolly parrot]

伊赞×派扬

[split void]

yes

what can't you understand here?

[split void]

idts he is

[toxic mountain]

Hi

[steady blade]

how that formula works? and can you explain that?

[split void]

I mean

how does the first one work?

$y = ab^x$

[jolly parrot]

Annie Maqionde

[split void]

so if $b>1$,

[jolly parrot]

Annie Maqionde

[split void]

then $y>a$

[jolly parrot]

Annie Maqionde

[split void]

so that's growth. growth means that the variable $y$ is increasing as $x$ increases

[jolly parrot]

Annie Maqionde

[split void]

if $b<1$

[jolly parrot]

Annie Maqionde

[split void]

you can find out by yourself that $y<a$

[jolly parrot]

Annie Maqionde

[split void]

and that's decaying

since $y$ is reducing

[jolly parrot]

Annie Maqionde

[steady blade]

thanks about your explanation

.closed

.close

[pearl pond]

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[pearl pond]

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[brittle quail]

I need help finding the solution to the following problem:
Two monic polynomials (that is, polynomials whose leading coefficient is 1) with integer coefficients p(x) and q(x), are such that gcd(p,q) = (x-1)(x-2), and lcm(p,q) = (x-1)^2 * (x-2)^3 * (x-3)(x+1), Moreover, the degree of p(x) is less than or equal to the degree of q(x), in how many different ways can p(x) be chosen?

[brittle quail]

tha answer is supposed to be 8 but i dont know what to do tbh

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[molten parcel]

what if i get +inf from one limit and -infinity from the other limit

[molten parcel]

it would still be an assymptote right

[west sapphire]

yes

[sharp vigil]

you could even get +infinity on one limit and a finite number on the other and it would still be a vertical asymptote

you only need one sided limit for it to be true (hence the "or")

[pearl pond]

@molten parcel Has your question been resolved?

[molten parcel]

give example of such a function

[sharp vigil]

e^(1/x)

[vital crescent]

there are many: you and write a piecewise function 1/x if x>0 and 0 otherwise. so literally taking one whose limit is finite at 0- and one whose limit if infinite at 0+ and glue them together

[molten parcel]

.close

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[unique ibex]

this is an exact ODE, which you can solve for with the initial condition and then get the answer

[shell current]

i got the f(x)

cant integrate it

perhaps its wrong

[shell current]

ion know dat, i solved with LDE

[unique ibex]

could you show your work?

and if next time you do this up front, people will help you faster

[shell current]

alr gimme a sec

[shell current]

basically i cant intergrate further

for the required

[unique ibex]

is the right side this?
[ \frac{x^4 + 2x}{\sqrt{1-x^2}]

[jolly parrot]

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

[shell current]

yep

no

x^5

plus x^2

o wait

i didnt

[unique ibex]

I mean on the first line

[shell current]

put

5 there

[unique ibex]

I wasn't sure, because it's a bit unclear

[shell current]

yea

das right then

[unique ibex]

so it's supposed to be x^5?

[shell current]

in the first line i;ve just rewritten the question

so it's x^4

[unique ibex]

oh right

[shell current]

✅ what u wrote for the 1st time

[unique ibex]

hmm it's a bit weird because you have sqrt(x^2-1) divided by sqrt(1-x^2) on the right side and those are both defined only when x = 1

[unique ibex]

anyway, you can't just cancel out sqrt(x^2-1) with sqrt(1-x^2) when you move from the first line here to the third, but I don't know how to fix this myself

I think you need to use the integrating factor sqrt(1-x^2) on both sides instead

[shell current]

i didnt

i used a method

wait

.close

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[wooden flax]

An exercise in my homework says to find the limit of the series \frac{n^{n}}{n!}.
I tried to use the ratio test, and I got that the ratio is equal to (1+\frac{1}{n})^n and I know that this limit as n approaches infinity is e.
In the lecture I was told that if the limit of the ratio is smaller than 1 then the series converges to 0, and of the limit of the ratio is larger than 1 then the series diverges to infinity. So apparently the series diverges to infinity (which is true) since e>1.
But on Wikipedia there is a different definition:
If the absolute value of the limit inferior of the ratio is larger than 1 then the series diverges. If the absolute value of the limit superior of the ratio is smaller than 1 then the series converges absolutely. And since the infimum of the ratio is 2, 2>1, the series diverges.

Am I missunderstanding something?

[tropic saddle]

if a sequence converges, then limsup = liminf = lim

[wooden flax]

If a sequence converges absolutely, does it necessarily mean that it converges to 0?

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[versed mica]

this is the more general statement and the one you saw in your class is the one more often used in standard calculus courses

[versed mica]

you mean the series?

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[molten parcel]

are 0 and 2 the inflection points?

[molten parcel]

(even tho the 2nd derivative is undef)

[sharp vigil]

that depends on your definition of "point of inflection"

can you post your textbook's definition?

[pearl pond]

@molten parcel Has your question been resolved?

[molten parcel]

it says either 2nd derivative is 0 or it fails to exist

so ig 0 and 2 are inflection pts

[sharp vigil]

that's not the definition

that's something which would have to be true based on the definition

[bitter vortex]

inflection points are where second derivative changes sign

[plush bramble]

https://www.mathsisfun.com/calculus/inflection-points.html

[bitter vortex]

that is quite literally what i said btw

second derivative changing sign is the same thing as changing concavity

[pearl pond]

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[minor tapir]

can someone help me

how do I isolate for y

[rough forge]

doesnt seem possible

you'd be better off to leave it implicit or solve for x

[sturdy bane]

@minor tapir bro why are u on an alt 😭

[minor tapir]

huh...

im so confused\

[minor tapir]

WTF HOWD YK ITS ME

LMFAOOO

yea alt acc so i dont get distacted by other stuff

[sturdy bane]

maybe cuz ur name is catsandkitties2

[minor tapir]

LMAOOOO

[sturdy bane]

he’s saying ur answer is already good basically

[minor tapir]

im so surprised you remember me

[minor tapir]

thanks

[vagrant trout]

they mean you can't isolate for y

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[thorny stream]

i need to show that when c = 2, the above formula is the same as this

i rewrote the first formula as the following:

but im not really sure where to proceed

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[thorny stream]

<@&286206848099549185>

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@thorny stream Has your question been resolved?

[thorny stream]

<@&286206848099549185>

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[void grail]

How to find limit of this please?

[pulsar flax]

✨ Factoring ✨

[void grail]

(x+1)(x+2)
/
(x+1)(x+3)

[fallen sundial]

yep that should do it

[void grail]

Thanks

I think it's the ε-δ I don't yet understand how to with first order language

[steep saddle]

the what

[void grail]

∀ ε ∃ δ

[steep saddle]

oh the epsilon definition

[void grail]

∀ε>0 ∃δ>0: 0<|x+1|<δ ⇒ |(x+2)/(x+3) - 1/2|<ε.

[pulsar flax]

Do you need to use epsilon-delta for this?

[void grail]

I think so

[viscid shale]

hopital 🏥

[smoky musk]

I don't think you need to use epsilon-delta here.

[void grail]

$\frac{(x+1)(x+2)}{(x+1)(x+3)} \to \frac{0}{0}$

[jolly parrot]

kytsu

[void grail]

at x→-1, cancel (x+1) first → (x+2)/(x+3).

L’Hôpital: f' = 1, g' = 1 → limit = 1/2 🐙📏

[void grail]

Why not?

[smoky musk]

There is an easier way

[viscid shale]

Multiple in fact

[cloud zephyr]

at which point are you struggling with for epsilon-delta ?

[void grail]

I don't yet easily see what to look for

[cloud zephyr]

well, it's possible to use something like delta = min{a constant, something in terms of epsilon}

[void grail]

the distance around -1 is the delta? |-1-δ| > 0 ⟹ |f(x)-ε|>0 ?

maybe |f(x)-L|>0

$0<|x - (-1)|<\delta \implies \Big|\frac{(x+2)}{(x+3)} - \frac12\Big|<\varepsilon$

[jolly parrot]

kytsu

[void grail]

     |<───2δ──>|
-1-δ ○────○────○ -1+δ
         -1

[pearl pond]

@void grail Has your question been resolved?

[void grail]

   L+ε ○
       |
       |
       ○ L
       |
       |
   L-ε ○

[cloud zephyr]

sorry, my boss called me just now

[cloud zephyr]

and yea, I'm back
I'll write something on the notepad and see if those make sense to you

this is my thinking process

[pearl pond]

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[pearl pond]

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[magic ruin]

I know the short method..
i wanna know the basic method

[toxic lichen]

,rccw

[toxic lichen]

call the previous number of wickets x
then the player used to have a total of 11.125*x runs in his career

now he has x+7 wickets and 11.125x + 38.5 runs, and the new average is 11.125-1.125 = 10.0

(11.125x + 38.5)/(x+7)=10

can't get much more basic than this i think

[magic ruin]

cool

got it

Teacher solved it using alligation..

thank you ann

.close

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[narrow crystal]

x and y are two positve integers which differ by 2020^3 , how many values are possible for the HCF of x and y

[narrow crystal]

the answer here is to just factorize 2020^3 into 2^6 x 5^3 x 101^3

then 7 x 4 x 4

112

only thing i dont understand is why does it work?

[split void]

$gcd(x,y) = gcd(x, x-y)$

[jolly parrot]

Annie Maqionde

[split void]

Assuming $x>y$

[jolly parrot]

Annie Maqionde

[narrow crystal]

so if y was greater js switch it around right

[split void]

yes

[narrow crystal]

anyways why does this work?

[toxic lichen]

no such assumption is needed

unless you're the kind of person who staunchly refuses to consider gcd(5, -10) to be a real thing

[split void]

I mean just for simplification.

And negative number GCDs could be a source of confusion.

[split void]

uh there was a proof

so every factor of x and y is a factor of (x-y)

[narrow crystal]

okok lemme go search a video about it

kk i understand now

ty

.close

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[fading ledge]

What's this question related about?

[pearl pond]

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[fervent galleon]

I have the following passage in a book:

[fervent galleon]

I noticed the y is used both as a vector and as a non-vector.

Can someone that has the ability verify that they are used correctly there?

I ask because I have found errors in this book, and I'm not entirely sure whether this passage is 100% correct or not.

I'm specially curious about the use of the vector in the differential dy.

[final ember]

$d\vec{y}$ is basically saying they're integrating over a path of y

[jolly parrot]

Katrro

[final ember]

but the next integral he's substituted in the values of y for the bounds

that's why it changes into a non vector form

[fervent galleon]

ok. Thanks for verifying it.

.close

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[jolly kernel]

I need to find y_p

[jolly kernel]

(particular solution)

[serene sentinel]

is that 4y or xy

[jolly kernel]

it's hyperbolic sin -- sinh (x)

The answer key said

So I'm not sure what I did wrong

[toxic lichen]

derivative of cosh is +sinh

not -sinh

[jolly kernel]

ooh

so they never go negative?

[eternal tulip]

the derviative of cosh(x) is sinh(x) and derviative of sinh(x) is cosh(x)

[jolly kernel]

gotcha thank you

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[eternal tulip]

you can verify it using the definitions of sinh and cosh respectively

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[pastel granite]

quick question cuz i forgot, for a function to be true, can there be 2 x values to one y value, or 2 y values 2 one x value?

[plush bramble]

what

[pulsar flax]

a function must be single-valued

[plush bramble]

oh you mean a well-defined function

[pastel granite]

a question like this

[pulsar flax]

meaning every x has a unique y value, so a y value must exist and be unique

[pastel granite]

perfect, thanks

was the deleted message wrong?

[tame cave]

2 x values are possible for 1 y value but 2 y values are not possible for 1 x

[tame cave]

yes

[pastel granite]

okok thank you

[tame cave]

for example y = x^2

x=2 and x=-2 both give value 4

[pastel granite]

ahh

yes

okay thank you very much

.close

[tame cave]

np!

[pearl pond]

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[jolly kernel]

Finding particular solution

Then I get 0 = 1 so something is wrong 😅

[sharp vigil]

have you checked what the homogeneous solutions are?