help-42

[umbral wraith]

oh

oh yeah :/

sorry so 360 - 72.645?

[dull wagon]

if it was i wouldn't have bothered with trying to get you to draw that arc

yes

[umbral wraith]

NNooooo

i accidently didnt round it

i suck at maths 😭

can you help me with one last quesiton, please? I think I get it, just incase

wait @dull wagon I dont think ik how to do this ?

Is like this ?

[dull wagon]

poor diagram

ideally you'd want to draw this on a compass / coordinate axis

[umbral wraith]

o h

[dull wagon]

there's no sense of where anything is in what you drew

all the info should be clearly indicated in some way on your diagram

[umbral wraith]

is that more clear or no

[dull wagon]

no

[umbral wraith]

welp

[dull wagon]

if i don't see any of
coodinate plane / compass
any indication of north
starting / end positions of the boat
distance marked
its gonna be an immediate no

coodinate plane / compass
that's the first thing you should start with

[umbral wraith]

oh

im 1% battery now too bruh

I'd rather get a detention, its 1140pm here

I guess I'll just stop

[dull wagon]

go to sleep

[umbral wraith]

welp yeah ima get detention cos its due in 20mins

ok well see you, thanks

.close

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[nimble field]

How do you calculate this

[calm coral]

@nimble field Has your question been resolved?

[nimble field]

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[velvet mesa]

Hello
I have a question
Why we dont let "a" be negative in the exponential function (y = a to the power x)

[main marlin]

well you can actually but only for integer values

[velvet mesa]

Yes i know that

[main marlin]

but if we have $(-1)^{\frac12}$ you get $i$ which is a complex number

[potent lotus]

PajamaMamaLlama

[velvet mesa]

So the reason is that its just hard to show it on graphs?

[main marlin]

is because you can only graph integer values

[velvet mesa]

So they dont use it that much(at least in high school)

[main marlin]

like you can only graph $(-1)^x$ where $x\in\mathbb{Z}$

[potent lotus]

PajamaMamaLlama

[main marlin]

yep that's usually why in hs they teach it that way but complex numbers allow for it to be defined :)

,w (-1)^x

[velvet mesa]

Uh thank you very much

[potent lotus]

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Requested by XxMrFancyu2xX#6050

[main marlin]

notice how it oscillates, only where the imaginary part is 0

(at integer solutions)

is where it can be graphed on a normal xy plane

[velvet mesa]

Yes i got that thank you so much

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[grave blaze]

Given random variable $X$ where $P(X=k) = \frac{k}{10}$ for $k\in {1,2,3,4,5}$. What is $E|X-EX|$ and $Var X$

[potent lotus]

szahu420

[grave blaze]

I calculated that $E X = 5.5$ and that the $E|X-EX|$ is equal to that, I am not sure if that's correct

[potent lotus]

szahu420

[grave blaze]

also my variance is negative so I am not sure what I'm doing wrong

I got $EX|X - EX| = 3.6$

[potent lotus]

szahu420

[olive fox]

Are you sure the range of k is correct? it doesn't look like probability distribution

For k=1..4 it would be

[grave blaze]

oh yeah, I misread it is k = 1,..., 4

my bad

[humble fog]

ur $E[X]$ is correct, can u provide ur work on ur attempt of $E[X-E[X]]$?

[potent lotus]

Smol Fish

[grave blaze]

is that correct?

that's for $k \in {1,2,3,4,5}$ which is not really alongside what the task is asking

[potent lotus]

szahu420

[grave blaze]

Just let me know if the following is true $$E|X - EX| = \sum_{k \in {1,2,3,4, 5}}P(X = k)|k - 5.5|$$

[potent lotus]

szahu420

[humble fog]

i also did the same

[grave blaze]

all right thanks

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[grave blaze]

Given $EX = 5$ and $\sigma(X) =2$ what can we tell about $P(X \geq 10) \leq ?$

[potent lotus]

szahu420

[grave blaze]

We obviously have to use the Chebyshev inequality here, so $P(|X - 5| \geq 5) \leq 4/5$

But i don't know how to handle that absolute value

[potent lotus]

szahu420

[grave blaze]

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[raw saffron]

@buoyant thistle hello bob

[remote mural]

are you trying to graph 2-lnx/x^2

do you have f'(x)

[raw saffron]

do you guys know how to solve derivatives, limits, graph functions?

[remote mural]

can you write it again here

i cant see it well

no i just want f'(x)

what is the domain of f(x)

so x>0

then x^3 is also always positive

so the only thing you have to consider in f'(x) is 2lnx-5

can you graph 2lnx - 5?

its alot simpler than using a table

so are you able to do it

ok

is there a restriction in your question

then just try it out once

try graphing 2lnx-5

so thats f'(x)

check where its negative and positive

but x^3 is irrelevant

because its always positive

so you only need to consider 2lnx-5

no

you only need the critical points

to graph something

if you can, you should remove irrelevant parts in f'(x) so that you have something thats easily drawable

because you only need plus/minus sign of f'(x)

and thats easily checkable if you have an approximate graph of f'(x)

its simpler this way

why

you already know where the minimum is

you just do (e^5/2, -1/2*e^(5/2))

you dont have to approximate

well e^something is always positive so

multiplying -1/2 will give you a negative value

thats it

thats enough though

lemme show you

thats enough graphing

[remote mural]

this to be exact

why would you want to approximate the y value

that information is enough

no question will ask you to approximate -1/2e^5/2

youre not a calculator

by the way i suggest doing f(e^5/2) again

i think its a bit off

yes

e^5/2 is right

just try f(e^5/2) again

[calm coral]

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[hexed oriole]

It’s my last one

[calm coral]

@hexed oriole Has your question been resolved?

[calm coral]

@hexed oriole Has your question been resolved?

[maiden shale]

bro assume 2nd eqn to be true

mark few lattice points on the graph

make them satisfy the assumed eqn

and obtain the variables ...

[hexed oriole]

Yea idk how to do that

[maiden shale]

identify a lattice point for me

[maiden shale]

identify a lattice point for me

[hexed oriole]

I just want it solved man plz

It’s my last one

[maiden shale]

i aint gonna give u the answer

im telling u the method

step by step

[calm coral]

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[ebon quartz]

Having a hard time with this word problem, I'm fine doing regular problems- but word problems always stump me. I don't know how to set them up. I know Work = Rate x Time.

[ebon quartz]

I guessed on the other one, and I need to not do that.

[left bay]

If you can translate the problem into a set of equations, then you should be fine.

In this case, I would let r denote the repair rate of the repairman (with units hour per shoe), and I'd let a denote the repair rate of the assistant.

First of all, you are given that 1.5r = a, because the assistant takes 1.5 times as long as the repairman.

For the second information, you need to combine their shoe making power (so to speak). How many shoes per hour can they make combined?

The repairman can make 1/r shoes per hour, and the assistant can make 1/a shoes per hour, so in total
1/r + 1/a shoes per hour for the pair of them.

They need to make 10 shoes, which will take them 10/(1/r + 1/a) hours. This has to be equal to 6 hours, so your second equation is:

10/(1/r + 1/a) = 6

[ebon quartz]

What does the a stand for?

[left bay]

a is the repair rate of the assistant (in hours per shoe).

[ebon quartz]

Gotcha, thank you.

[left bay]

np

[ebon quartz]

Let me see if I wrote this correctly.

I still don't really understand.

[left bay]

What part?

[ebon quartz]

How to Solve it. Normally the equation is written differently,

I'm stuck, for the most part.

I'm doing my best to solve it, but I'm pretty sure I've been on this for an hour now. I apologize.

[left bay]

Ok, is there any particular part of what I wrote that doesn't make sense?

[ebon quartz]

Normally I'm used to it being written like a fraction, instead of /, I know that / means division which is a fraction, but the format is not lining up for me,

[left bay]

No worries

The repairman can make $\frac{1}{r}$ shoes per hour, and the assistant can make $\frac{1}{a}$ shoes per hour, so in total
$\frac{1}{r} + \frac{1}{a}$ shoes per hour for the pair of them.

They need to make 10 shoes, which will take them $\frac{10}{\frac{1}{r} + \frac{1}{a}}$ hours. This has to be equal to $6$ hours, so your second equation is:

$$\frac{10}{\frac{1}{r} + \frac{1}{a}} = 6$$

[potent lotus]

OneTrackPony

OneTrackPony

[ebon quartz]

Thank you.

so in other words, the repairman would make 1/r shoes an hour while the assistant makes 1/1.5r an hour.

I still have no idea how to solve this. It shouldn't be hard at all.

[left bay]

Yeah, that's right.

So it's like, I can make some number of shoes per hour, say 2 shoes per hour. You can make 3 shoes per hour. In total, we can make 2 + 3 = 5 shoes per hour.

For the same reason, since the repairman can make 1/r shoes per hour, and the assistant can make 1/1.5r shoes per hour, we just add them up to see how many shoes they combined can make per hour.

That's why I have above that they can combined make $\frac{1}{r} + \frac{1}{a}$ shoes per hour

[potent lotus]

OneTrackPony

[ebon quartz]

I understand, but normally you have numbers to combine. all we have is r and a.

r+a is just that, r+a

[left bay]

Yeah, for this particular problem, you have to juggle the symbols, which makes it harder.

They represent numbers, you just don't know what they are.

If it helps, you might assume some numbers while solving it, and them adjust them later.

[ebon quartz]

Thanks for the assitance.

[left bay]

np

Also, you have been given some options for what number the answer might be. You can try out those numbers in the problem to see if they fit.

For instance, one option is that it takes the repairman 1 hour to fix a pair of shoes on his own. You can then calculate how long it would take the assistant.

[calm coral]

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[lament inlet]

Hello, I need help with these two problems:

[lament inlet]

For Q8, I'm not sure which answer is incorrect and I'm not entirely sure how to do Q9

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[hallow summit]

What you refer when you mention (dy) ?

i mean

do you need to find the

derivate ?

ok

understood

[dull wagon]

no

you messed up in your attempt to apply chain rule

why are you multiplying the fraction by e^(2x^2) * 4x

why those specific expressions

what derivative are you using

[hallow summit]

the last one should be 2x

i mean

[dull wagon]

what did you take the derivative of to get e^(2x^2) * 4x

[hallow summit]

1/1+(e^x^2) times e^x^2 times 2x

dx

[dull wagon]

forget about the fraction component specificaly

[hallow summit]

consider e^x^2 as u

[dull wagon]

i'm asking about the chain rule component on the end

[hallow summit]

just subsitute

[dull wagon]

derviative of "what" = e^(2x^2) * 4x

[hallow summit]

and countinuesly derivate

[dull wagon]

now why are you taking the derivative of e^(2x^2)

that's the mistake you made

[hallow summit]

is equal to e^x^2 times 2x

[dull wagon]

but why are you taking the derivative of that

[hallow summit]

the derivate of e to the power of u is equal to e^u itself times the derivative of u

[dull wagon]

you're supposed to just use the derivative of the inner function, (not its square)

yes

[hallow summit]

you're done

it asks for dy

that is dy

what is the simplified

[dull wagon]

what's the intended answer?

[hallow summit]

expression?

[dull wagon]

me neither

[hallow summit]

listen, often at my college it happens when the results in the books and not quite accurate or even corret to the actual answer

[dull wagon]

that seems to be the solution for 33,

[hallow summit]

the answer that we get

it cannot be more simplier

[dull wagon]

similar shapes, not the same

[hallow summit]

idk

it's not like we did much

[calm coral]

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[novel sonnet]

I'm not sure how to start this one

[leaden iron]

What kind level is this?

[novel sonnet]

like class?

vector calc

[calm coral]

@novel sonnet Has your question been resolved?

[calm coral]

@novel sonnet Has your question been resolved?

[mint ingot]

can you start by finding the equation of the new line?

Hint: you can work out the direction it goes, and you know a point on the line

@novel sonnet

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[idle marten]

I'm wondering if the vocabulary "over 100" means, more than

[pure kayak]

i would say so

[idle marten]

basically is the inequality 100<12d + 34
or 100=<12d+34

[muted lotus]

Over means over

[fleet verge]

The first one

[muted lotus]

So more than

[idle marten]

ok ty

.close

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[ember coral]

hi

[ember coral]

Is anyone able to explain why if you multiply say .8 by itself it becomes .64 it doesnt seem right

[hot creek]

uhm

[half jolt]

think of it as 8/10

square that, you get 64/100

[hot creek]

you are adding "8" 8 times

[ember coral]

thanks for the help thart makes sense now

.close

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[dusk quiver]

Hi, I was wondering if I can help solving this integral using the integration by parts method

[exotic falcon]

What do you need help with? It’s fairly straightforward

[dusk quiver]

so i need to learn how to evaluate the integral, but my answer was off

[exotic falcon]

Show your work then

[dusk quiver]

the method i need to learn is the integration by parts

okay one sec

[exotic falcon]

There are integral calculators online that will show you step by step for these btw, and lots of YouTube videos

[dusk quiver]

i used a youtube video to follow along but i still ended up getting it wrong

[compact drum]

this screams at me "introduce a new variable"

[exotic falcon]

It’s very simple IBP

[compact drum]

yeah, but new variable is fancy

[exotic falcon]

Your algebra is wrong meelohh

3rd to last line

What happened to the x

You had 3x * ….

And just deleted it

[dusk quiver]

oh yeah i think i added it back to the 2nd to last line right

[exotic falcon]

No you didn’t

Because you deleted the 1/6 too

Redo your algebra, starting at the 4th to last line (because that line is correct)

[dusk quiver]

ok ok ill redo it rlq

this is the answer but idk how to get the /2 and /12 but im assuming its my algebra

[exotic falcon]

It is, literally just double check your work

You randomly dropped like so much stuff out of no where

So just don’t do that and you’ll get it right

[dusk quiver]

yep..

[compact drum]

alright

[dusk quiver]

gotta stop doing that

[compact drum]

since you got the result

here is another method:

a bit longer in this case, but really nice

[dusk quiver]

😁

[compact drum]

the trick here is to swap integration and differentiation, which we cant always do

[dusk quiver]

i dont think im there yet

i just strated calc 2

sooo 🙏 but ill keep it

[exotic falcon]

It’s a bad method meelohh ignore it

[compact drum]

whaaat

[exotic falcon]

Requires excessive justification that you don’t know about

[compact drum]

thre are some integrals which you can't really solve otherwise

[exotic falcon]

Martin, OP is clearly, and like they just said, just in basic calc2

[compact drum]

😦

[exotic falcon]

When they’re supposed to be learning basic IBP

Let them learn IBP

Don’t suggest some other more complex method when they’re doing the basics

It’s fun and cool and all, but it really isn’t helping them learn for their class

[dusk quiver]

thanks @exotic falcon for the second opinion 🙏

.close

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[split pawn]

A pension fund seeks to invest 1 million dollars. To satisfy its current and future obligations, the fund must earn $54 000 per year in interest. The fund has two sorts of investments; stocks paying 6% per year and bonds paying 4% per year. How much should the pension fund invest in each type of investment in order to earn the desired income

[split pawn]

(Linear system word problem)

[hexed grove]

well, first we need an equation describing the how much the fund must earn as a function of time, then we need some functions for the 6% interest and for the 4% interest

then, we should be able to do linear combinations on the principals to solve the problem

[calm coral]

@split pawn Has your question been resolved?

[split pawn]

Can u like solve it or na

[calm coral]

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[glad sinew]

why is lower bound y^2 and upper bound y+2

[glad sinew]

ohhh

ok

.close

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[ornate wadi]

im supposed to ingerate by substitution but the first x confuses me what should i do with it?

[ornate wadi]

<@&268886789983436800>

[obsidian ingot]

could sm1 check help-33 please

[tired fiber]

can you find the form g'(x) f(g(x)) here?

[ornate wadi]

is it
2x * sin(x^2)

[tired fiber]

Yup

[ornate wadi]

if yes, what happened to the x before?

[tired fiber]

x before?

[ornate wadi]

this guy

[tired fiber]

it's still there, i dont understand

[ornate wadi]

but we used the x^2 from sin(x^2), took the derivative and put it on the left side

isn't that 2x different from the x in my picture?

[tired fiber]

it should be

the same

You took u = x^2 right?

[ornate wadi]

yes

[tired fiber]

du = 2x dx, correct?

[ornate wadi]

makes sense

yes

[tired fiber]

we can multiply outside by 1/2 and inside by 2 to get this form or you can subtitute dx with du/2x

outside and inside the integral

[ornate wadi]

oh i didn't know you can do that

okok i think i have too many gaps in my mind

i'm speedrunning this topic right now so i should learn more on my own

thank you

[tired fiber]

What specfically?

[ornate wadi]

like multipliying the outside and inside by 0.5 and 2 respectfully

[tired fiber]

I see, it's just we are multiplying by 2/2 and constants are free to do whatever in intregral

[ornate wadi]

i think i now relealize what's happening

[ornate wadi]

we are trying to mimic this

[tired fiber]

Yup

[ornate wadi]

aaaaaa

[tired fiber]

or more specically the du = 2x dx in our subsitution

You can go about it as you please ig

[ornate wadi]

yeaaa

[tired fiber]

im off, if you are done with this question you should .close, better starting a new channel for any more questions

[ornate wadi]

oh ok

didn't know how to close it

.close

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[wintry glacier]

is there an easy way to do this

[wintry glacier]

or do i need to take 9 dot products

D is an orthogonal base

[pallid halo]

what are the d_i from the previous exercise

[wintry glacier]

B is an orthogonal base

sorry i should say D is orthonormal

at least thats kinda nice

i can always write the matrix as a vector and the same for |d_i > < d_j| but that would require finding 9 matrices and then taking 9 dot products with a bunch of roots and stuff so i feel like there should be an easier way

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[brazen elbow]

$\int \frac{2x-1}{x^2} e^{2x}dx$

[potent lotus]

FungusDesu

[brazen elbow]

putting these on wolframalpha got me some Ei function, is there a way without using that?

[fleet verge]

This one?

I think not, let me see

No this is by parts and it’s normal

Maybe u copied wrong in wolfram

[brazen elbow]

nope i forgot to clarify the result didnt have Ei, but the intermediate steps did

they just cancel out

[fleet verge]

I didn’t have that, can u show what u see?

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[lost ice]

is there a way to do this

that isn't 4 pages long

like can i notice something

without being a genius

otherwise id just differentiate, plug and chug and find recursive soln

[calm coral]

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[remote mural]

i think you are just meant to plug in the solutions yeah

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[flat pumice]

Option A?

[civic dirge]

ye

[calm coral]

@flat pumice Has your question been resolved?

[flat pumice]

How to solve it?

[civic dirge]

What does T do to the basis vectors?

[flat pumice]

Transformation

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[civic dirge]

I mean that's technically correct lol

I mean "where do the vectors go"

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[misty blade]

how do u solve this? I would just throw it into my calculator but i know that’s not the proper way to do it

[misty blade]

it’s just the negative exponent that’s throwing me off

another formula was p = a/ (1+i)^n

my teacher mentioned using exponential rules but i don’t really remember

[nocturne zephyr]

a^-n = 1/a^n

you are looking for the negative exponent rule

since the whole term (1+ ...) has that -48 as an exponent

you can turn it into a fraction basically to have a positive expoenent

[misty blade]

what would that look like

[nocturne zephyr]

look like where

[misty blade]

if i turned it into a fraction to have a positive exponent

in the equation

[misty blade]

Oh ok i figured

[nocturne zephyr]

basically this form

[misty blade]

jeepers ok hold on

[nocturne zephyr]

your number is going to be just shy of 1500

practically dividing by 1

[misty blade]

oh okay i just realized i was doing it fine 😭😭

I just thought the answer was wrong

thnx for clarification

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[shadow sinew]

hello, this function definition is slightly confusing to me. I know that it passes through the point (a, an) but I don't have a single clue where to begin with graphing it.

[pallid halo]

shouldn't that be An indefinite article

[calm coral]

@shadow sinew Has your question been resolved?

[shadow sinew]

wdym

[spark stratus]

dont think so because itd be a bit circular to use an in its own definition

[calm coral]

@shadow sinew Has your question been resolved?

[spark stratus]

try to find some context for this letter before the letter n, since the current definition is too open to pin down specifics
since the letter is missing from my computer, I've never seen it in my entire life, but it seems to be the key to comprehending the definition of the Indefinite Rticle
you could then use this definition to find functions through (, n) though lines simply work from the get-go

[shadow sinew]

got it

thank you

[spark stratus]

np

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[main tulip]

guysh i forgor how to do this

[glad plover]

cosine rule

[main tulip]

oh shi i completely forgot abt that

ty

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[remote mural]

Hello, I am learning about relations and function
I have a question, is the equivalence class donated as [a] = { x ∈ A | (x,a) ∈ R} or
[a] = {x ∈ A: (a, x) ∈ R}
i have find [a] = {x ∈ A: (a, x) ∈ R} formula on some websites and i am confused

[remote mural]

.close

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[jade mantle]

I got this gap in my knowledge where i'm not sure how to derive functions without using software for it.
I got these tables explaining what to do when deriving different functions.
Or maybe it lies more in not understanding what makes a composite function?
I just don't get how

[jade mantle]

Does not become

[unique jackal]

you need to apply chain rule

You do have a composite function

[jade mantle]

Yeah but i don't see why it's a composite function

I got a gap in my knowledge somewhere

[unique jackal]

e^(something)

e is the outer function, the something is the inner function

for example, e^(sinx) is a composite function

E(S(x))

[jade mantle]

So..
something to the power of something
is a composite function..?

[unique jackal]

Yeah, because exponentials are functions too

maybe the easier way to think about it is to start with e^x

let’s write that as E(x)

The x in E(x) is a placeholder for an input

i can replace it with another variable, an entire function, a smiley emoji, doesn’t matter

In this case, if we let P(x) = x^3 -12x, then we replace the x in E(x) with P(x)

So we have E(P(x)), which is composite

does that make any sense or am I being incoherent

[jade mantle]

Tough to answer. Because it does make sense. However i still don't understand it enough to apply it. Yet i'm unsure how to formulate a question which answer will make me understand.

Could you explain when something is not a composite function, then? Because currently it seems that most functions I would work with will be composites.

[unique jackal]

There are three main ways to combine functions

Addition, multiplication, and composition

[jade mantle]

so anytime you lift anything to the power of anything
it becomes a composite?

[unique jackal]

Composition is probably the hardest to recognize, because the other ones are fairly obvious I think

[unique jackal]

Yes, because exponential functions are functions too

You can raise a base m to the power of a function

m^(function)

m^x is itself a function, so we’re left with M(F(x))

unfortunately, I’m not sure I can explain any better, not because there isn’t a better way to answer, but because my pedagogical skills are still developing lol

my apologies

[jade mantle]

No worries, you're doing miles better than my teacher anyway! xd
🙂

Ok, just for my own insight
Dw, i don't expect you to solve it, but how would I go about deriving this function then? (See bottom image)
I'll go about my own thought process here
Because it seems to me then i'd need to use the chain rule 3 times here(?)
5 becomes 0
5x^3 is a function
e^3x is a function
2^3 becomes 0
cos(x) is a function

Admittedly i'm not sure where to go from here, as there's more than 2 products, but here goes:
I can imagine f'(x)=
(5x^3)' * e^3x * cos(x) +
5x^3 * (e^3x)' * cos(x) +
5x^3 * e^3x * (cos(x))'

[spark stratus]

you got it right that (abcd)' = a'bcd + ab'cd + abc'd + abcd'

however thats not used for f(x) here because its just two separate pairs of products

each one has f'g + fg' done on it to split it

[jade mantle]

If you don't mind, could you please walk me through how to solve it, just as I did above, and perhaps add why.

[spark stratus]

is that $f(x)=5+5x^3\cdot e^{3x}+2^3\cdot\cos(x)$

[potent lotus]

mtt07734

[spark stratus]

(making sure youre not multiplying all three together)

[jade mantle]

The function i sent an image of was purely for examplary purpose. If it makes a major difference if they're added or multiplied, would you care to explain that as well?
But yes, the function is correct

[spark stratus]

then that can come later

first, lets consider what the rules look like:

(fg)' = f'g + fg'

(f/g)' = (f'g - fg') / g^2

(f(g))' = f'(g) g'

all of these act on something done on the outside

for example in (5x^3 * e^3x)', the first thing done would be product rule since * is the outermost operation

it wouldnt be the 3 * x buried deep in there

think of outer/inner as like ((5 * (x ^ 3)) * (e ^ (3 * x)))'

so that product rule does: ((5*(x^3))** * (e^(3*x)))'** = (5*(x^3))' * (e^(3*x)) + ( 5*(x^3)) * (e^(3*x))'

[calm coral]

@jade mantle Has your question been resolved?

[jade mantle]

Give me a few more min, bot
Also, tysm ING and mtt. I'll most likely send another question here after I've tried to work with this a bit more

[spark stratus]

np

[desert cargo]

Do you need help?

@jade mantle

????

Just answer?

@jade mantle

[jade mantle]

Sorry for hogging up the channel!
I think I just need to sit with it for a bit as I don't yet understand but I'm not sure what question to even ask.

[desert cargo]

Wait I am about to get answer for derivative question

@jade mantle

Answer is

15x^2e^3x+15x^3e^3x-8sinx

[jade mantle]

Thing is, I didn't really need the result. I just need to know how to get there without a CAS tool

[jade mantle]

How would i progress through solving this function?

.close

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[brisk briar]

x^3-6x^2+12

[crimson cliff]

$f(x)=x^{3}-6x^{2}+12$

[potent lotus]

Dubs

[crimson cliff]

and?

[brisk briar]

(Q1, how did you do that lol)

Q2 how do you find the minimal value (and the corresponding argument) of this function?

[crimson cliff]

$$ math goes here $$```

[brisk briar]

gotcha, thanks

[crimson cliff]

or single dollars

double dollars is to align to middle

[crimson cliff]

have you learned about derivatives ?

[brisk briar]

in school not yet

have done some on my own

[crimson cliff]

using derivatives; you can do second derivative test

[brisk briar]

i got 3(x-4)x as the first derivitive

[crimson cliff]

There is a picture, i can give for some intuition

to really know whats going on, i suggest you to read more on this

[brisk briar]

3(x-4)x=0
6x-12>0

So x=4?

[crimson cliff]

there is a more important question

[brisk briar]

hm?

[crimson cliff]

in what interval are you trying to find the maximum?

[brisk briar]

wasnt specified

[crimson cliff]

because if you really look for the entire real numbers

it's unbounded

it's a polynomial expression, i grows cubically

[brisk briar]

(only the minimum is asked for)

[brisk briar]

wouldnt the maximum be x=0 though?

[crimson cliff]

i think the minimum is also unbounded

[brisk briar]

They go on to positive and negative inf

[crimson cliff]

yes, check the graph

0 is not the minimum

[brisk briar]

maximum

[crimson cliff]

no.

the functions grows

isn't f(10) > f(0) ?

[brisk briar]

well yea

but i mean a local maxiumum/minimum

i wasnt specific- im sorry

[crimson cliff]

so there is no global maximum and global minimum

[brisk briar]

i see it, yea

[crimson cliff]

so for local max and min, you need to specify an interval

a function only attains max and min a closed intervals

say between [-1,1]

f(0) is the max

[brisk briar]

- A local minimum of a function is a point where the function value is smaller than at nearby points, but possibly greater than at a distant point.
- A global minimum is a point where the function value is smaller than at all other feasible points.

[brisk briar]

so there must be an interval stated in order to gen local min max?

[crimson cliff]

when we use the term local, we mean local in some closed interval

yes, what is local elsewise

the locality you specify is different from what i suggest

[brisk briar]

maybe it would be easier like this. This is the exact problem I'm going over:

i got up to 3-4

[crimson cliff]

what is [3-1] ?

[brisk briar]

1

[crimson cliff]

whats the notation?

[brisk briar]

its spaces to fill on the answer sheet

its a question from an exam

treat the [3-X] as blank spaces to enter your answer

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[brisk briar]

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[fringe reef]

as you can see d is negative

so eventually the terms of the series will become negative

your task is to find what's the last positive term before it starts becoming negative

4k+1 = 25

$25-8n > 0$

[potent lotus]

artemetra

[fringe reef]

solve for n and make sure it's an integer

then substitute the largest possible value of n in the general formula for n-th term

yes

your a_1 is 25

you have n and d

[calm coral]

@remote mural Has your question been resolved?

[untold drum]

how can i help?

what equation? where does n come from?

there is no n in the example.

you have to find k (what you did) and you have to finde the smallrst positive value.

you had k = 6, so the three conescutive 41, 33 and 25. as the differnce is -8, the next elements of the sequence are 17, 9, 1, -7 wo the smallest positive value is 1.

how often can you subtract 8 from 25 to be positive?

so whats the smallest multipy of 8 less then 25?

sorry, not the smallest multiply, the biggest multiply. and the biggest multiply of 8 less then 25 is 24, so 25-24 = 1 -> the smallest positive value in te sequence is 1.

its just what i said. 25-24=1

i dont know. ask him/her.

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[jade scaffold]

Does this mean I'm supposed to substitute the given P function into the differential equation? I'm very confused on what the question is asking here.

[calm coral]

@jade scaffold Has your question been resolved?

[leaden thunder]

Yea I agree I have no idea what the question is asking

You could interpret it as plugging in the given P(t) into the DE

Or you could also interpret it as solving the IVP to produce P(t)

Ask your professor to clarify

[calm coral]

@jade scaffold Has your question been resolved?

[jade scaffold]

its getting too late😭

shits due tn

i think it might be solving the ivp to get p(t) but i really don't know

<@&286206848099549185> does anyone else have a better idea on what this means

or happens to concurrently be enrolled in ma266 at purdue

[calm coral]

@jade scaffold Has your question been resolved?

[past zinc]

verify whether the given solution is actually a solution to the given initial problem, it is the most logical at this point

[calm coral]

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[wanton dove]

where does the -1 go ? shouldnt there be another -(5/6)

[wanton dove]

the assignment is whats P(Omega) of a Dice roll where we want to know how likely it is for "6" to land on an even number of throws

[pallid halo]

they renumbered the sum to start at k=0

[wanton dove]

yes but can you just do that? doesnt there need to be somewhere the 5/6 goes?

[pallid halo]

no, write out the first few terms of each sum

you'll see they are equivalent

basically did a change of variables:
j = k-1

and then renamed j back to k

[wanton dove]

ohh i guess its because 1-1 is the same as ^0 ?

[pallid halo]

yes

right

[wanton dove]

oh thank you! didnt think of it that way

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[wanton dove]

Hey (me again) so this time the question is "What are the odds for that a dice lands on 6 on an uneven amount of throws"

I figured out what the probability is for the throws at all and tried it but comparing with the Solutions showed i was wrong about 2 parts (yellow)

1: Why is it 2k-1 instead of k-1 like ive shown above?
2: How does it become 2k-1-1 and why

[pallid halo]

that's definitely not right

that first equality

[wanton dove]

is it not?

[pallid halo]

well even ignoring that they're summing over i but using k in the expression

just check the i=1 terms for both

[wanton dove]

oh that was my bad actually

[pallid halo]

for the first one it's (1/6)(5/6)
for the second its' (1/6)(1)

[wanton dove]

i usually use i and the solution used k so i mixed it up

[pallid halo]

wait, did you write this or someone else?

[wanton dove]

ill send the actual solution wait

[pallid halo]

ok

[wanton dove]

so i was right about the k-1 part i guess but how they turned that to (2l-1)-1 is a mystery to me

[pallid halo]

oh well that's a key difference

[wanton dove]

or why even

[pallid halo]

here they're only summing over odd values of k

[wanton dove]

i apologize for the confusion

[pallid halo]

nw

so they're using the fact that any odd number k can be written in the form 2l-1, where l is an integer

[wanton dove]

ok so say the index is = 1, then it would be (2*1-1)-1 which becomes 0 though right?

[pallid halo]

yea

[wanton dove]

wouldnt that be an even amount of throws again

[pallid halo]

same is true in the previous sum

when k = 1, the exponent k-1 is zero

[pallid halo]

well which variable represents the number of throws?

[wanton dove]

doesnt k do that?

so if its k=1 its my first throw?

[pallid halo]

and are these probabilities true?

if so then that explains the k-1

kth throw --> exponent = k-1

[wanton dove]

oh so k-1 is my actual "throws"

[pallid halo]

well k-1 is the number of unsuccessful throws

each has probability 5/6

then you get one last throw which is successful

which had probability 1/6

[wanton dove]

where does the (2k-1)-1 come from though thats my biggest confusion

what function does it bring

[pallid halo]

you mean the l here?

[wanton dove]

yeah they rewrote it weirdly enough

[pallid halo]

we're doing that because we don't know how to sum over just k = 1, 3, 5, 7

so we introduce a new variable l

which goes over all positive integers, not just odd integers

and then we need to adjust the exponent to compensate

the change of variables is:
l = (k+1)/2

which is always an integer because k is odd

so k+1 is even

[wanton dove]

wouldnt k+ make more sense then

[pallid halo]

and when you divide an even number by 2 you get an integer

[wanton dove]

is it harder to work with?

[pallid halo]

k+ what

[wanton dove]

k+1

or l+1 in that picture

[pallid halo]

try writing out the first three exponents in each sum

[wanton dove]

ah nvm im so confused i think im confusing you hahaha

ok will try

[pallid halo]

for the first sum you have k = 1, 3, 5, so the exponent is 0, 2, 4

for the second sum you have l = 1, 2, 3 so the exponent is 0, 2, 4

the key is that (1) the exponents are the same either way and (2) summing over l = 1,2,3, ... is easier because we have formulas for that

we don't have formulas for k = 1, 3, 5,...

that's the motivation here

[wanton dove]

oh so the geometric series im applying here wouldnt work without that gotcha

[pallid halo]

right

[wanton dove]

ah this is all so abstract but i thank you the second time

[pallid halo]

this is a fairly standard trick but it's confusing the first time you see it

[wanton dove]

its the last class before i am able to graduate so i hope itll not drag out longer

thank you!

[pallid halo]

ah well gl!

cheers

[wanton dove]

.close

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[flat pumice]

x^2 -8x+15-2x+7
x^2-10x+22=0

5+√3, 5-√3

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[viral mica]

https://stackoverflow.com/questions/77917279/how-to-calculate-the-expected-value-of-the-estimator-simplification-with-combin
Can someone help with this

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[pine coral]

help

[pine coral]

hi im in maths methods unit 1&2

i need help with a question

tag me pls

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[pine coral]

<@&286206848099549185>

[inner delta]

Try getting a common denominator

On the left side

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[pine coral]

.close

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[wintry echo]

I didn't understand why in the Bolzano-Weierstrass theorem a succession of increasing natural numbers is formed.

[wintry echo]

help pls

[half jolt]

I mean that's the construction

You try to find an increasing subsequence (or decreasing) if i remember correctly

Then that subsequence must converge

Wait, why natural numbers though?

Are you sure you don't mean the indexes

[half jolt]

Actually i think thats wrong ignore that

[wintry echo]

so when I divide the intervals I have to make sure that I always have a number n greater than all those in the previous interval?

[half jolt]

You create a convergent subsequence by finding which indexes to take

That might be the increasing thing you are talking about, the index sequence

[wintry echo]

the problem is that I have numbers n that alternate in size

[half jolt]

What's n?

Can you show me the proof you are looking at?

And i'll try to explain it

[wintry echo]

ok

but only this part??

[half jolt]

Sure

If i need more context ill tell u

[wintry echo]

I got to the point that I have two sequences Ak and Bk, where Ak<Bk

and Bk-Ak = (B-A)/2^k

[half jolt]

Alright

[wintry echo]

the interval [a_1,b_1] contains the terms of the sequence a_n; therefore the first integer n_1 exists, such that a_n_1 ∈ [A1,B1]. For the same reason there exists a first integer n_2, among all natural numbers larger than n_1, so a_n_2∈[A2,B2]. So as to create n_1<n_2....<n_k...<...
so a_n_k ∈[Ak,Bk] . Since Bk-Ak =(B-A)/2^k . That is, Ak<=a_n_k<=Bk = Ak +(B-A)/2^k

[half jolt]

If you construct a subsequence with elements in the nested intervals, you will get a convergent subsequence

You can even just take a_n to be the subsequence

Or b_n

Well actually

No nvm

You cant lmao why did i say that

[wintry echo]

[A...,B...] are the intervals I obtained, dividing the previous ones with C=(B_k/A_k)/2

[half jolt]

Yeah sure

You construct a sequence of increasing indexes, n_k

So in [a_1, b_1]

Take n_1

So that a_n_1 is in that interval

[wintry echo]

yes

[half jolt]

And then u get the next, smaller interval

Pick another element in it, but with bigger index than the last one

U can do that since there are infinitely many elements there

[wintry echo]

but why n has a bigger index

if i choose like the bigger n of [a_1,b_1], How do I get a bigger n, in the next interval?

so i mean if i choose this number b_1]

[half jolt]

Just choose it, u know it exists

There are infinitely many elements in that smaller interval

[wintry echo]

but are smaller then this n

[half jolt]

There must be an element in there with index bigger than n

[half jolt]

There are only n-1 indexes smaller than n

[wintry echo]

example pls

but therefore increasing in the sense that for every interval there is a number n in it?

[half jolt]

You need to know the difference between an element in the sequence and an index

[wintry echo]

index i think is like n_1 this 1 is an index?

[half jolt]

No

Like

Sure

But

Its hard to type a lot on phone

Maybe someone else can help

[wintry echo]

wait

maybe i fix

[half jolt]

If this channel if still open when i get home ill try

[wintry echo]

when I choose a number in the interval, it will necessarily have a greater one, because n is unlimited and since my interval has infinite elements???

[half jolt]

You choose an index n_1 so that a_1 is in the interval

Then, take the smaller interval

[wintry echo]

last question

[half jolt]

Pick a next index n_2 so that a_n_2 is in that interval

And n_2 > n_1

U can do that since there are infinitely many elements in that interval, so just take one with bigger index

You know it exists

[wintry echo]

so it doesn't matter the size of the number, when he talks about an increasing n sequence, does he mean increasing in the sense of indices?

like not the number 1<2<3<4, but n_1<n_2....<n_k

[half jolt]

Yes

[wintry echo]

ah ok

i understand

thank you so much

[half jolt]

No problem

[wintry echo]

.close

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[calm coral]

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[heady tree]

Find the rate at which Rs.4000 will give Rs.630.50 as compound interest in 9 months , interest being compounded quarterly

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Please don't occupy multiple help channels.

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[heady tree]

Close.

[marble escarp]

.close

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[gray oxide]

do i simply swap the x'es with the c?

[pallid halo]

what c

,rotate

[gray oxide]

the 4

[pallid halo]

all operations here are continuous, so yea

[gray oxide]

the answer is so big

[pallid halo]

a mere 5 digits

[gray oxide]

oh alright

thx

.close

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[gray oxide]

help how do i solve this ;p

[naive thorn]

try to factor by $(x-1)$ the numerator

[potent lotus]

José

[leaden thunder]

Use synthetic or polynomial division

[pallid halo]

or of course you can l'hopital that bad boy if you're allowed to

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[dark relic]

Need help with this question, not sure how to go about this or what to do

[gray oxide]

took me an hour to learn, but it was worth it.Holy how cool that formula is

[pallid halo]

i guess if you don't see any easy shortcut, you can always brute-force it

like write B in terms of its elements, and then explicitly multiply out AB and BA

[gray oxide]

i applied the rule on this, is 1 correct?

[pallid halo]

can you open a new channel? this one timed out and is now owned by someone else

[gray oxide]

oh sorry

[pallid halo]

nw

[calm coral]

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[terse valley]

can anyone help me with part c

[pallid halo]

how are 8 + 64/x and x + 8 related?

[terse valley]

im not too sure

[pallid halo]

well you can start by multiplying 8 + 64/x by x to get rid of the denominator, what does that give you

[terse valley]

8x

is it

a x 8 b

is that the answer

nevermind

[lunar meteor]

8x + 64

and look there is something in common between 64 and 8 to simplify the equation

[terse valley]

8

[lunar meteor]

exactly

[terse valley]

so its 8ab

[lunar meteor]

XD

[terse valley]

the thing is ive got a picture of another classmate and the teacher said the correct answer is 3 + b-a

[lunar meteor]

i don't think thats true

btw is log2(8+X) related to your question ?

[terse valley]

yes

its b

b is 8+x

a is x

[lunar meteor]

so yeeh, your teacher is wrong

look focous you simplyfied it until you got 8x+64 right

so you can divide by 8

and you will get the answer

[terse valley]

oh thanks

i thought that was correct too

just seeing the teachers work bamboozled me

[lunar meteor]

sorry

[terse valley]

thanks

[lunar meteor]

your teacher is true

[terse valley]

is she

[lunar meteor]

i though the question was 8+64/b

now i see its 8+64/x

[lunar meteor]

so let me tell you how to simplify it
you see log2(8+64/x) is equal to log2 (8 . (1+8/x)) right?

now you can use log properties
you will get log2 (8) + log2 (1 +8/x)

log2 (8) equals to 3 as you know

so its 3 + log2 ( 1+8/x) which you can write it as log2 (x+ 8)/x you can use log properties again
you will get 3 + log2 (x + 8 ) - log2 (x)
you know that log2 (x + 8 ) = b
and log2 (x) = a
and thats your question solved

sorry i messed up at the start i thought was over b , tho i read its x

[calm coral]

@terse valley Has your question been resolved?

[terse valley]

thank you very much

it makes sense now

[calm coral]

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[chrome plume]

I am not sure where to begin with

I have restated sqr x to the power of 1/x

[glad plover]

could consider the logarithm

[chrome plume]

how would logarithm help?

[glad plover]

logarithm gets rid of the weird power

if you can find the limit of the logarithm, if the world is nice then this will be the logarithm of the limit

[chrome plume]

but how do get the b?

[chrome plume]

since i don't wknow what it is equal to

[glad plover]

$$\ln\left( \left( 1 + \sin(x) \right)^{\frac{1}{x}} \right)= \frac{1}{x} \ln(1 + \sin(x))$$

[potent lotus]

Kaisheng21