help-27

[devout snow]

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[devout snow]

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[devout snow]

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[finite shuttle]

i'm feeling reallyy dumb about his question

[finite shuttle]

i have to calculate the area between x²-6x+10 and y =5

i calculating 5 = x²-6x+10

and i'm gettin 2 and 3 as roots

but geogbera is givving me 1 and 5

and idk why

nvm

i was changing the other

.close

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[devout snow]

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[whole lantern]

Hi everyone, I'm taking a proofs course right now and just wanted to make sure I was calculating my infinite intersections/unions correctly, I haven't done this in a long time. Given: A_i = [1-1/i , 3 - 1/i) . I calculated the infinite union from i = 1 onwards to be [0,3) and the infinite intersection from i = 1 onwards to be [1 ,2). Wanted to just make sure I did the question correctly

[whole lantern]

$\bigcap^{\infty}_{i=1}[1-\frac{1}{i},3 - \frac{1}{i}) = [1,2)$

[woven radish]

aco

[whole lantern]

$\bigcup^{\infty}_{i=1}[1-\frac{1}{i},3 - \frac{1}{i}) = [0,3)$

[woven radish]

aco

[icy onyx]

looks alright

[whole lantern]

👍 Then for the section where I need to prove my result and I am required to show that both sets are subsets of eachother. I know that $\bigcup^{\infty}{i=1}[1-\frac{1}{i},3 - \frac{1}{i}) = A_i \cup \lim{n \to \infty}A_n = [0,3)$ and $\bigcap^{\infty}{i=1}[1-\frac{1}{i},3 - \frac{1}{i}) = A_i \cap \lim{n \to \infty}A_n = [1,2)$.

[woven radish]

aco

[whole lantern]

Let me just type up this last bit I am new to latex

In the case of the union, for $x \in [0,3) \rightarrow x \in \bigcup^{\infty}{i = 1}{A_n} $ can I use this chain of steps? $x \in [0,3) \equiv x \in [1-\frac{1}{1},3-\frac{1}{1})\cup \lim{n \to \infty}[1-\frac{1}{n}, 3 - \frac{1}{n}) \equiv x \in A_i \cup \lim_{n \to \infty}A_n \equiv x \in \bigcup^{\infty}_{i = 1}{A_n}$

[woven radish]

aco

[whole lantern]

I'm just trying to wrap my head around it since this is the first time i've worked on proving equalities between infinite intersections/unions

[foggy violet]

OMG! i hope i could have helped you :((

[whole lantern]

🙂 @foggy violet if you dont mind i'm still open to help

[foggy violet]

i havn't studied any of this yet so unfortunately i won't be able to help :((

[whole lantern]

no worries, do you know somewhere where I could ask this?

<@&286206848099549185>

[devout snow]

@whole lantern Has your question been resolved?

[devout snow]

@whole lantern Has your question been resolved?

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[frank kernel]

Hi! I‘m trying to solve this integral for alpha>0 over the unit circle. I managed to integrate for 0<alpha<1. But now I’m stuck on alpha >1, the solution says it goes to infinity, although I don’t quite understand why

[devout snow]

@frank kernel Has your question been resolved?

[frank kernel]

<@&286206848099549185>

[young spire]

the unit circle in $\mathbb{R}$ ? are you sure you aren't integrating in $\mathbb{R}^{2}$ instead ?

[woven radish]

BillyElKid

[devout snow]

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[frank kernel]

.reopen

[devout snow]

✅

[frank kernel]

This is part of the excersize, we‘re supposed to integrate in R, R^2 and R^3

The part I’m working on right now is in R

[young spire]

okay , then if $\alpha > 1$ what happens with this expression

[woven radish]

BillyElKid

[young spire]

for example if $\alpha =2$

[woven radish]

BillyElKid

[young spire]

is it really 0 ?

[frank kernel]

Ohhhhhhh

If alpha=2 I have -1/x in the red circle, so for x=0 it goes to infinity

Right?

[young spire]

yep

[frank kernel]

Thank you!!

[young spire]

kkat dont use this channel

dont worry

[frank kernel]

How can I close this channel?

[young spire]

type .close

[frank kernel]

Alright!

.close

[devout snow]

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[red bobcat]

$(1+\frac{1}{n})^{-n}$ find limit of this at infinity

[woven radish]

Erzis エルジス

[red bobcat]

how can I can I get rid of that minus?

it's obviously $e^x$ but I can't seem to get the $(1+\frac{a}{n})^n=e^a$ form

[woven radish]

Erzis エルジス

[red bobcat]

just imagine there is a limit at n -> infinity there

[devout snow]

@red bobcat Has your question been resolved?

[young spire]

$$((1+\frac{1}{n})^{n})^{-1}$$

[woven radish]

BillyElKid

[young spire]

what do you know about limits and continuity ?

you know that if $f(x)$ is continuous, then $$\lim_{x \to x_{1}} f(x) = f(x_{1})$$

[woven radish]

BillyElKid

[young spire]

?

[red bobcat]

thank you very much

I get it now

it was that simple

[young spire]

😉 no problem

[devout snow]

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[devout snow]

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[tacit rose]

.close

[devout snow]

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[devout snow]

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[formal bolt]

The sum of the digits of a two-digit number is 13. When we divide that two-digit number by the multiplication of our digits, we get 2 and the remainder 5. what is that number

[formal bolt]

Can you help me set up the equation

[young spire]

how many unknowns have a two-digit number?

[formal bolt]

I think 2

[young spire]

correct

[formal bolt]

Cause it s a sum

[young spire]

is just because it have two digit

so now think, this representation is useful to you = $xy$ ?

[woven radish]

BillyElKid

[formal bolt]

I know I could set it like this x+y=13

But I need the whole equation

[young spire]

nice that is correct for the first 1

[formal bolt]

But don t know the second part

[young spire]

what can you infer from this

$$97=90+7=910^{1}+710^{0}$$

[woven radish]

BillyElKid

[formal bolt]

I don t know can you explain?

I guess you get the same result

[young spire]

yes but notice how i wrote it.

[formal bolt]

I need you just to help me set it up I know how to do it from there

[young spire]

so if you have a 2 digit number this is another representation of it $$10x+y$$ or not ?

[woven radish]

BillyElKid

[formal bolt]

It is

[young spire]

relax man, this discord is not for doing your work, i'm trying to hint you what you have to do

so what is this then

$$\frac{10x+y}{x\cdot y}$$

[woven radish]

BillyElKid

[formal bolt]

Ooo

I thought like this

x+y=13 2 digit number with / and then multiplication of numbers you get 2 and the remainder is 5

But tell me how did you get the 13

Don't i have to insert all this numbers in equation

I still Don t get it

[young spire]

you already told me that x+y=13 is your first equation

this will be a system

so you have to precise the second one

[formal bolt]

It says multiplication of numbers is 2

So x*y=2

[young spire]

nope it says this "divide that two-digit number by the multiplication of our digits"

[formal bolt]

The sum of the digits of a two-digit number is 13. When we divide that two-digit number by the product of our digits, we get 2 and the remainder 5. what is that number

[young spire]

so what was our two digit number? $10x+y$ mmm and the multiplication of our digits? $xy$ and then ?

[woven radish]

BillyElKid

[formal bolt]

Does it change if I put product

Cause I m bad at English so I use translation

[young spire]

aaaa ok my bad

the expression that the question is referring to is this $$\frac{10x+y}{x \cdot y}$$

[woven radish]

BillyElKid

[young spire]

the division between our two digit number ($10x+y$) and the product of his digits ($xy$)

[woven radish]

BillyElKid

[formal bolt]

I can show you a example but it s not the same as this one

The sum of the digits of a two-digit number is 3. When it is multiplied by a number composed of the same digits in reverse order, the number 252 is obtained. What is this number

Does this help

Adn then I did using quadratic equation

[young spire]

yeah but in that case the nature of the problem is different.

i'm trying to tell you to what algebraic expression, the sentence "divide that two-digit number by the multiplication of our digits" is equivalent

[formal bolt]

12 or 21?

1* 2

?

[young spire]

are you using a translator?

[formal bolt]

Yes

[young spire]

italian?

[young spire]

croatian

mmmm , chill then

i will give you your expression

[formal bolt]

This could be the second equation but what do I do with the remainder 5?

[young spire]

solve this

$$x+y=13 \quad (1)$$

and this

$$10x+y=2xy+5$$

[woven radish]

BillyElKid

[young spire]

yes that is almost correct, the remainder means this

[formal bolt]

Ok

[young spire]

if the remainder between $a $ and $b$ is $c$ then exists $n$ such that

$$a=nb+c$$

and $0 \leq c \leq b$

[formal bolt]

I m going to send you a photo when I m done

[young spire]

sure feel free to ping me, obviously i may not respond inmediately

[formal bolt]

Ok

[woven radish]

BillyElKid

[devout snow]

@formal bolt Has your question been resolved?

[formal bolt]

I think I've done it correctly

@young spire

[young spire]

and the value for $x$ ?

[woven radish]

BillyElKid

[young spire]

are you sure ? x must be a integer

y its fine tho

$y=5$ can you recover $x$ from the 1 equation , remember that is supposed to solve both

[woven radish]

BillyElKid

[formal bolt]

So I put 5 as y

x+5=13?

[young spire]

then?

[formal bolt]

Put it in the other equation

x=8

[young spire]

not really you can solve for x from that

exactly

so what is your number?

[formal bolt]

5?

[young spire]

the two digit number

[formal bolt]

Well I got 25/2

[young spire]

😆 i mean remember that we write our two digit number like $10x+y$

[woven radish]

BillyElKid

[formal bolt]

Oh

It s 85

[young spire]

$$8+5=13$$ and $$\frac{85}{8 \cdot 5}=\frac{80}{40}+\frac{5}{40}= 2 + \frac{5}{40}$$

[woven radish]

BillyElKid

[young spire]

so your number works, that is all.

[formal bolt]

So my result is 8

Right?

[young spire]

The sum of the digits of a two-digit number is 13. When we divide that two-digit number by the multiplication of our digits, we get 2 and the remainder 5. what is that number

what is that number ? 85 the two digit number

[formal bolt]

Ok

Thanks

[devout snow]

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[young spire]

no problem

[devout snow]

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[latent frost]

How would I go about proving this? I have reached the point of x ∈ P(A) or x ∈ P(B) and x ∈ P(C) but I'm not sure where to go from here

[devout snow]

@latent frost Has your question been resolved?

[spare crypt]

you sort of have to go 2 layers deep

you're using an arbitary x, but you also have to look at arbitrary elements of x

[latent frost]

What I did was said that x ∈ LHS which means that x ∈ of P(A) or P(B) and x ∈ P(C) but I think I get that from x being a subset of those sets

so I think im already looking at the elements of x if I'm understanding you correctly

[spare crypt]

well right now x looks like {a1,a2,a3} or something, it's a set

since a power set is a set of sets

when you look at elements of x you can start to do the equivalence like "y in x so y is in A or B and C, so x is in the power set of A or B and C"

[latent frost]

so I look at individual elements and say that as they are in A or B and C they must also be in the power set of A or B and C?

[spare crypt]

more that x is in the power set of A or B and C

since in a power set just means all its elements are in the specified set

[latent frost]

so if the elements of X are in A or B and C then X must be in the power set of A or B and C?

[spare crypt]

that's right

straight from the definition

[latent frost]

ok that helps thank you!

.close

[devout snow]

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[formal bolt]

If the edge of the cube is reduced by 2, the volume of the cube is reduced by 98. By how much is the area of ​​the cube reduced

[formal bolt]

I need help with setting up the equation

And i don t know which formula to use

[timber pebble]

what is the formula for volume of a cube?

and surface area of a cube?

[formal bolt]

I think it s this

[timber pebble]

you should know it more intuitively

whats the formula for area of a square

[formal bolt]

Well 6a ^2

[timber pebble]

6a^2 is the surface area of a cube

whats the volume of a cube with side length a?

[formal bolt]

If I put it in formula

This is what I get

[timber pebble]

i think youre trying too fast to use a calculator

whats the volume of a cube with side length a?

[formal bolt]

I don't know then

[timber pebble]

,w volume of a cube

[woven radish]

Results provided by WolframAlpha

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

Requested by jan Niku#2119

[timber pebble]

area of a square is a^2

volume of a cube is a^3

so if the volume of a cube with side length a is a^3, whats the volume of a cube with side length a-2?

[formal bolt]

-1?

[timber pebble]

its not a number

itd be in terms of a

notice how if the side length is a, then the volume is a^3

[devout snow]

@formal bolt Has your question been resolved?

[devout snow]

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[bronze schooner]

ohh got it

[devout snow]

@bronze schooner Has your question been resolved?

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[proven saddle]

Theres this math question that has been stumping me for an hour now

a box is in the shape of a rectangular prism. Its volume is x^3 - 2x^2 + x.
determine the algebraic expressions for the dimensions

[hybrid elk]

factor out the common x, so you get $x\left(x^2-2x+1\right)$

[woven radish]

light

[hybrid elk]

now, just factor $\left(x^2-2x+1\right)$

[woven radish]

light

[proven saddle]

yeah i did that part

[hybrid elk]

right, so just factor out the part above

[proven saddle]

ok

.close

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[tribal valve]

Needs to be a 2x2 answer even though 7A would be a 3x1

[main gull]

What do the arrows do?

Change the size of the matrix?

[tribal valve]

Nope, they just tell you what's a row and what's a column.

[main gull]

But why are there arrows that goes up and down?

Can you click it?

[tribal valve]

Eee, that's embarrasing, because I tried dragging it earlier and it just came off as an image and I immediately assumed it was just there for show.

Clicking it works.

Thank you for your time.

.close

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[devout snow]

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[strong crystal]

I already answered this. A bit confused. What would the answer had been and why?

[mortal haven]

$(a^b)(a^c) = a^{b+c}$

[strong crystal]

I think its the parenthesis?

[woven radish]

kiid

[strong crystal]

.close

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[mighty ferry]

Discrete Math: I will post a screenshot of the problem here, involves relations, equivalence relations and equivalence classes.

I just dont understand where to start with this. I do not get how to count equivalence classes on a relation that isnt defined.

https://cdn.discordapp.com/attachments/565401702384009216/907469664223895553/unknown.png
https://cdn.discordapp.com/attachments/565401702384009216/907469772726362112/unknown.png

[long fog]

Can you send a bit clear photo, it's hard to see subscipt

[mighty ferry]

yup

is that better

@long fog

[mighty ferry]

<@&286206848099549185> i can ping yall now right? its been over 15 i think

im in compsci but i hate this class

oh wait i need to be in uni level topics

.close

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[granite tulip]

Hi

For the variables x, y and z, it is known that z = x2y and 2x+y = 50.
Find the maximum value of z if:
x∈[0,25]

[granite tulip]

It ried few, but the answer should be 12500/27

[devout snow]

@granite tulip Has your question been resolved?

[devout snow]

@granite tulip Has your question been resolved?

[raw rock]

ok so
if 2x + y = 50, then y = 50 - 2x
then you just put it into first formula:
z = x^2 * (50 - 2x) = 50x^2 - 2x^3
solving questions with finding maximums of something is most easily done by calculating differential of that something

so we do just that

more exactly, we need to find places where diff is zero

[woven radish]

it's Sam

[long fog]

Sorry I didn't see you answered

[raw rock]

happens

I'll finish my explanation still

[long fog]

Yes

[raw rock]

because these are the places where functions stops whatever it was doing (getting bigger or smaller) and does the opposite thing

therefore these are the mins/maxes of a function

that's pretty easy when diff is a quadratic

100x - 6x^2 = 0
x(100 - 6x) = 0

x1 = 0, x2 = 100/6 = 16 wholes and 2/3

diff before 0 is negative, therefore it's a min point

diff before 16 2/3 is positive, therefore 16 2/3 is a max point, therefore we just need to calculate z putting 16 2/3 as x

[granite tulip]

Thank you for the explanation, let me try to solve it by using your explanation

[raw rock]

if you need help with stuff (like, you don't understand why diff of 50x^2 is 100x), just say

[granite tulip]

I got the correct answer!

Thank you @raw rock @long fog

.close

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[blissful fable]

Is there any way I can skip or represent in a shorter way the “lim” part when doing limits?

[polar bolt]

If you are just rewriting the expression, for example rationalizing or whatever then you can do all that without having lim at every equality

Then once rewritten to the form you like you can add it

But otherwise no

[blissful fable]

So preferably I'd want to write it every time right?

I've seen someone write an L

IDK if it's his/her convention or it's widely accepted

[polar bolt]

Like I said if you are applying limit laws between equalties then you need the limit at every equality

Idk what you mean with L

[blissful fable]

Like an abreviation I guess

[blissful fable]

Basically this then, ok, well thanks… Don't worry about the L stuff

[polar bolt]

For example you can do this $2x-x=x \xrightarrow[x\to \infty]{}\infty$

[woven radish]

ScapeProf

[devout snow]

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[dim quartz]

Hi

Uhm yeah, im rn starting Calc 1 and my teacher sent us a joke question (I guess) today on our homework, I thought it would be cool if I managed to like get at least close to whatever the answer is.

Also, I did a little research and this seems to be something of Calc 2 (?) but I'm not sure. Anyway, I was hoping you guys could help me cuz I'm pretty much new to all of this.

[devout snow]

@dim quartz Has your question been resolved?

[pseudo basin]

i think it might be possible to do this with some kind of vieta's theorem fuckery

@dim quartz are you familiar with that by any chance?

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[hollow sonnet]

okay i have asked this before but i just realised i am still stuck on this

[hollow sonnet]

https://cdn.discordapp.com/attachments/903481101744484362/907539835756093490/IMG_3581.jpg

how would i find the area of this?

[restive river]

do you know the formula for a trapezoidal prism

[devout snow]

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[hollow sonnet]

.reopen

[devout snow]

✅

[hollow sonnet]

nope

[restive river]

what about a trapezium

[hollow sonnet]

yea

[restive river]

you can work out the trapezium base and then multiply it by the width

[devout snow]

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[potent ravine]

could somebody explain why that is the case?

[pseudo basin]

$P(X \geq i) = \sum_{j=i}^{\infty} P(X=j)$

[woven radish]

Ann

[pseudo basin]

oh but the sum in that formula of yours should start with i=1 i think

yeah bc you need to count P(X=1) once, P(X=2) twice, P(X=3) three times, etc.

@trail glade please do not post in occupied channels.

[trail glade]

Yeah I’m sorry, i just realised that-

[potent ravine]

@pseudo basin hmm ok thanks but what throws me off is that it is expected value, we're not multiplying with the actual value of the r.v. ?

[pseudo basin]

wdym

it can be proven that $\sum_{i=1}^{\infty} P(X \geq i) = \sum_{j=1}^{\infty} j P(X=j)$

[woven radish]

Ann

[pseudo basin]

and the right-hand side is definitionally E(X)

[potent ravine]

yeah rhs I get but how is the left equal to that?

[pseudo basin]

do you agree with this?

[potent ravine]

yeah I guess, adding all the probabilities greater is the same as all the greater

[pseudo basin]

$\sum_{i=1}^{\infty} P(X \geq i) = \sum_{i=1}^{\infty} \sum_{j=i}^{\infty} P(X=j) \ = \sum_{1 \leq i \leq j} P(X = j) \ = \sum_{j=1}^{\infty} \sum_{i=1}^j P(X = j) \ = \sum_{j=1}^{\infty} \underbrace{P(X=j) + P(X=j) + \dots + P(X=j)}{j \text{ copies}} = \sum{j=1}^{\infty} jP(X=j)$

[woven radish]

Ann

[potent ravine]

yeah I'm lost at the first line

[pseudo basin]

okay maybe i can explain this differently

i'll write p_i for P(X=i) to reduce clutter

P(X >= 1) = p_1 + p_2 + p_3 + p_4 + ...
P(X >= 2) =       p_2 + p_3 + p_4 + ...
P(X >= 3) =             p_3 + p_4 + ...
P(X >= 4) =                   p_4 + ...

and so on

does that make it any clearer?

[devout snow]

@potent ravine Has your question been resolved?

[potent ravine]

Yeah ok so you count p2 twice and p3 three times and so on

[devout snow]

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[potent ravine]

@pseudo basin ok I get it now from thinking about it a bit harder. Thanks 🙂

[pseudo basin]

alright

.close

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[merry crest]

Since D and A are both complex numbers, D - A would be a complex number too? to express it in polar coordinates, id need magnitude and direction. R is already the magnitude. I dont know how to compute the direction. Could someone help? I need to solve for D-A and its time derivative. Any help would be appreciated! I'm still in 11th grade, some of these concepts are still new to me. The angle between R and positive x axis is (theta + phi)

[merry crest]

<@&286206848099549185>

[magic haven]

U don't need <@&286206848099549185> to solve this

U need einstein

[merry crest]

ha it said after 15 minutes i could tag them

LMAO

it isnt all that hard, im close to the answer but it just seems awfully inuitive

<@&286206848099549185> anybody?

[devout snow]

@merry crest Has your question been resolved?

[long fog]

If I had been exposed to such maths 5 years ago I probably would've given up on maths

[quiet radish]

it's india bro

we have fucked up eduaction system

we have to solve these questions in less than 3 minutes

[devout snow]

@merry crest Has your question been resolved?

[merry crest]

did any of yall get anywhere?

[quiet radish]

umm i know it's not relevant but are u preparing for the olympiads?

[merry crest]

nope

[merry crest]

<@&286206848099549185> ???

[long fog]

But you don't know where I'm from

And if you know then I'm among those "we"

[merry crest]

someone, please, help. im going insane over this one question

[quiet radish]

try doubtnut or something

[quiet radish]

lol.maybe bc your username is sam.

[signal sand]

Where did u find this question

[long fog]

I had my own name before but yk we have long names so meh

[quiet radish]

why sam tho?

[merry crest]

HA does it matter

[long fog]

First three letters of me name

[long fog]

Seems like this brings out nostalgia for you?

[quiet radish]

could u pls explain what we have to find.i will try to ans

[signal sand]

I want to find more like this
its super interesting
and no i havent solved it

[signal sand]

Nope

this is what im going through as well

[merry crest]

oh, lol try MIT

[signal sand]

being a 12th grade student in india isnt fun

[merry crest]

find D-A in terms of theta, phi, and R

[long fog]

You both are on the same way then

[merry crest]

im in 11th grade and it's a nightmare

[long fog]

Both indian in hs

[signal sand]

This seems to be solvable with some physics differential equation memery but im not sure

[signal sand]

But i didnt see such a question in my 11th

[merry crest]

yo that's only the first subsection

[signal sand]

[merry crest]

the question is crazy after that

[signal sand]

wtf

[merry crest]

if i solve this, i can do the rest

[signal sand]

try using vectors

[signal sand]

:confused_psyduck:

i think i will go do something normal
i will solve this stuff when i know lagrangian mechanics

[merry crest]

that's the whole question

[signal sand]

thats one fuckton of scary in a single question

[merry crest]

this doesnt involve langrangian mechanics

[long fog]

I guess some people can solve it here but looking at that big question they rather not

[signal sand]

i know it doesnt

but it can

[quiet radish]

fuck man why would u want to suffer? as if the existing pain isn't enough

[merry crest]

intuitively, i guess Re^(i*(phi+theta)) should be the right answer

[signal sand]

I find it fun

[signal sand]

[long fog]

Great thoughts

[signal sand]

wait how is the alligator moving in this question?

is it chasing the duck?

[merry crest]

yeah

it's chasing the duck

[signal sand]

Try seeing if pursuit curves work

they might

or just be intuitive and see how the vectors are changing

D will only change direction

but A will also change magnitude

[inland widget]

is this solved?

[quiet radish]

no

[devout snow]

@merry crest Has your question been resolved?

[devout snow]

@merry crest Has your question been resolved?

[devout snow]

@merry crest Has your question been resolved?

[devout snow]

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[safe echo]

hey can someone help

[merry crest]

i was right about to claim this

[safe echo]

eh

[merry crest]

ggs ha you beat me to it

[safe echo]

haha

<@&286206848099549185>

[merry crest]

well for part b) just compute for its derivative and take the positive integer

[safe echo]

yeah

i need helpw iht b

B

idk how to do B

i derive then do quadratic formula and i still get it wrong

[merry crest]

once you find b) you can sub it in the original eq and youll get a value for volume

[safe echo]

how do i find b

any help?

[merry crest]

did you get 12x^2 - 440x +2400?

[safe echo]

ya

then i tried to find x

[merry crest]

ha you could always just plot the curve or use the quadratic formula, did you do that already?

[safe echo]

yeah

i did quadratic

but got it wrong

[merry crest]

answer's 30 for the maximum x value

[safe echo]

30

alright

so

[merry crest]

sub 30 in eq 1

4(30^3) - 220(30^2) + 2400*30

[safe echo]

woah

i got

-18000

tht dosent seem right

[merry crest]

check again LMAO

[safe echo]

hmm

i still got

18000

:/

[safe echo]

i think we're subbing into the wrong equation

idk

@merry crest help pls

[merry crest]

yeah, hi

[safe echo]

idk wht to do

[merry crest]

x is not 30

[safe echo]

then how do i find x?

[merry crest]

it's 20/3

cause

if x = 30 then b = 30 - 2x. b cannot equal to 30 - 60 --> -30

you cant have a negative value for breadth

also you can deduce whether it's the maxima or minima by finding its second derivative

if the derivative is less than 0, it's a maxima

in this case f''(x) is 24x - 240

sub for 20/3

[safe echo]

so

its max

bcs its -80

?

[merry crest]

160-440

the second derivative is -280

f(x) is maximum when x = 20/3

sub for 20/3 in the original eq to find the maximum volume

6.6667

do you follow me?

[safe echo]

yeah but im still confused on where 20/3 came from

liek

which equation

[merry crest]

oh

solve the quadratic

youll get 30 and 20/3

[safe echo]

oh

[merry crest]

two values for x

[safe echo]

20/3 is negative

the negative one

okay

[merry crest]

nope

30 is the negative one

20/3 gives you a positive answer

4(6.66^3) - 220(6..66^2) + 2400*6.66

your answer's 7524 approximately

[safe echo]

alright

ok

i see now

thanksss

.close

[devout snow]

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[merry crest]

so ive solved for D-A now

it just happens to be Re^(i*(phi+theta))

since phi+theta is the argument of D-A and not just A alone

but now

ive to solve for its time derivative

any help would be appreciated

[devout snow]

@merry crest Has your question been resolved?

[merry crest]

<@&286206848099549185>

[versed summit]

hey! im struggling with some graph questions is someone able to help? ill send a photo of the worksheet

im struggling with task 2, 5 and 6

[inland widget]

Now I wasn't able to help you yesterday but it seems like you have been here really long
I am kinda assuming(because I am low on time here) you know how to calculate dR/dt and d(phi)/dt the only bit that could be missing I think would be d(theta)/dt now for that you will need to use the vector method (i.e. take components of velocity v of D in perpendicular and parallel with respect DA ) and then you can calculate the angular velocity of the length DA to get your d(theta)/dt
I hope this makes sense also if this does not help you enough and you don't get help here I would recommend going down to #calculus

[merry crest]

oh alright! ill try #calculus out, thank you for your time. I appreciate it :)

.close

[devout snow]

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[versed summit]

.reopen

[lusty stratus]

in this question how do we know d is 32?

and how do we know that its a - cos?

@spare crypt could u help me i gota test tom

<@&286206848099549185>

oh wait

d is = to 32 because its the middle

right?

.close

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[devout snow]

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[shut pollen]

I have this very simple question

what I got so far is...

5z - 3 = 3z + 21
5z = 3z + 24
z = (3z + 24) / 5

and honestly I think I am doing it wrong.

[winter patrol]

wheres x coming from

[shut pollen]

Oh sorry typo, it was suppoused to be z

there i fixed it

[winter patrol]

have you solved linear equations before?

[shut pollen]

I have, but very simple ones

[winter patrol]

would you be able to solve
k - 5 = 7

[shut pollen]

k - 5 = 7
= k = 7 + 5
= k = 12

[winter patrol]

inappropriate use of = signs at the start of the second and third lines

[shut pollen]

oh

[winter patrol]

but yes k=12

what about
3m = 9
?

[shut pollen]

3m = 9
= m = 9 / 3
m = 3

[shut pollen]

Ohh

It does

[winter patrol]

inappropriate use of = sign at the start of the second line

[shut pollen]

hmm, I see.

but how

[winter patrol]

apply those two ideas to what you had in the second line of your original question

[shut pollen]

but like

what did i do wrong with the = sign, like i used it wrong. But like how?

[winter patrol]

$a = b\
, , , =c$ \
implies that $a=c$

[woven radish]

ℝamonov

[shut pollen]

oh

that does make sense 🤔

thanks imma close this.

.close

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[languid badge]

Can I say that anything that can be in parentheses can also be represented as x? Is that called the associative property?

[devout snow]

@languid badge Has your question been resolved?

[devout snow]

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[devout snow]

@hallow dirge Has your question been resolved?

[devout snow]

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[restive river]

so i have a function f defined as

[restive river]

$f: \mathbb{R}^{+}_{0} \rightarrow (0, 1], \ f(x) = \frac{1}{\sqrt{1+x^2}}$

[woven radish]

mhD

[restive river]

I am supposed to find f-1 which I did and i defined it as

$f^{-1}: (0, 1] \rightarrow \mathbb{R}^{+}_{0} , \ f(x) = \sqrt{\frac{1}{x^2}-1}$

[woven radish]

mhD

[restive river]

now im asked to find the identity function by calculating the compositon of f \circ f-1

its easy to plug everything in but im stuck at this

$f \circ f^{-1} = f(f^{-1}(x)) = \frac{1}{\sqrt{1+\sqrt{\frac{1}{x^2}-1}}}$

[woven radish]

mhD

[restive river]

how am I suppose to find f \circ f-1 = x?

[visual meadow]

you forgot to square it when you input f inverse into f

[restive river]

yyyyyeeee

[visual meadow]

$f \circ f^{-1} = f(f^{-1}(x)) = \frac{1}{\sqrt{1+(\sqrt{\frac{1}{x^2}-1})^2}}$

[woven radish]

hyrflood

[restive river]

im blind i literally found it out the sec you sent your message

[visual meadow]

xD

[restive river]

that square is a game changer

thanks anyway

[visual meadow]

[devout snow]

@restive river Has your question been resolved?

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[grizzled iron]

Hi. I am studying a class on probability and would like some clarification on random variables. I have random variables $X_1$ and $X_2$. My understanding is that random variable are functions so given an initial probability space $\Omega$, $X_1 : \omega \in \Omega \rightarrow X_1(\omega) \in \Omega^{1}$ and $X_2 : \omega \in \Omega \rightarrow X_2(\omega) \in \Omega^{2}$.

So when asked for a set max$(X_1, X_2) < 2$, I would assume that this would be a set of ordered pairs of numbers from $\Omega^{1}$ and $\Omega^{2}$ (which $X_1$ and $X_2$ map to), where the max number in the pair is no greater than $1$. In that case, the way this is written is a little bit confusing: why is it written as max$(X_1, X_2)$ if $X_1$ and $X_2$ as random variables are actually functions? Would max$(\Omega^{1}, \Omega^{2})$ not make more sense?

Thanks

[woven radish]

danield_159

[grizzled iron]

I suppose they don't have to necessarily be functions but they are at least mappings, I guess.

[devout snow]

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[white drum]

where did these two terms come from?

[devout snow]

@white drum Has your question been resolved?

[devout snow]

@white drum Has your question been resolved?

[shy wing]

i don’t even know what’s going on in that problem lol

[white drum]

Its a part of the solution that i dont understand

Like theyre differentiating both sides, fair enough. But then those two terms pop up and i dont know how they got them

[shy wing]

no i know how to differentiate an integral

i just don’t understand their process

[white drum]

ohh ight, well, me neither lol

[shy wing]

i’m honestly confused if the x in the integral is the same as the x in the derivative

i would assume that the x in the integral is just some unknown constant

this seems intentionally confusing

[white drum]

yeah i was thinking those two terms i marked came from the x bound, i just dont see how they got those terms from it

[devout snow]

@white drum Has your question been resolved?

[wooden verge]

7

[devout snow]

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[dusty moss]

How do I solve this?

[torn bramble]

Rs. is a currency?

[dusty moss]

Yes, Indian currency. Forgot to tell about it, my bad

Consider it as $300 for your reference if you want

[reef bough]

Do you understand that the equation for the fuel cost is
fuelCost = (multiplier * speed^2) + 300

[devout snow]

@dusty moss Has your question been resolved?

[reef bough]

ok I might have to rethink this a little

[devout snow]

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[junior cloak]

so these are two questions on my homework and i just want to make sure i came up with adequate examples
https://i.imgur.com/jvUqDx4.png
https://i.imgur.com/4X3Faww.png

[junior cloak]

4. $a_n=b_n=(-1)^n\implies (a_nb_n)=(1)$ which is convergent

[woven radish]

nix

[junior cloak]

8. $f(x)=g(x)=(-1)^{\floor{x}}\implies f(x)g(x)=1^{\floor{x}}=1$ which is continuous $\forall x$

[woven radish]

nix

[prisma rivet]

The a_n and b_n in your solution problem 4 aren't divergent sequences

[junior cloak]

youre saying (-1)^n is not divergent?

[prisma rivet]

yes

[junior cloak]

it is though. it does not converge to any one value

[prisma rivet]

that's not what divergent means

[junior cloak]

...

[prisma rivet]

divergent means that it goes on to infinity, right?

[junior cloak]

not necessarily

[prisma rivet]

sorry, I just looked it up..

my bad

[junior cloak]

so it would be correct then?

[prisma rivet]

yes, I think both of your examples are correct

[junior cloak]

okay cool

.close

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[lean seal]

why can the textbook arbitrarily say: if x>p/2

[lean seal]

is it just because in the limit, x is going to be |x-p|<delta and delta is small

so that assumption is allowed ?

[prisma rivet]

@lean seal since p is positive and x -> p, then its safe to assume x > p/2

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[restive river]

Hello, I do not know how to move further.

What I did so far was:
Make a perimeter equation.
3000 = 2L + W
Make W = 3000 - 2L
Use that in A = L*W for:
A = 3000L - 2L^2
Took the derivative of A:
A' = 3000 - 4L
L = 3000/4

Am I doing this right?

[devout snow]

@restive river Has your question been resolved?

[leaden raven]

yes, but if you want to be absolutely sure this method is correct, you could check if this is maxima and not minima

[restive river]

Okay, so I want to get W = the same way as I got L = ?

[leaden raven]

no, i mean to check if function on left side is bigger than zero and on right side is smaller than zero

[restive river]

oh yeah i just mean

i want to proceed the same way I got L in order to get W?

to finish it?

[devout snow]

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[rustic igloo]

I need help in my proof. In the case that f prime of c is less than zero.

We move to consider the limit of f(x)-f(c)/x-c as x approaches c which is equal to f prime of c

We then precisely define that limit as for all epsilon>0 there exists a delta>0 such that 0< l x-c l < delta -> 0<l f(x)-f(c)/x-c -L l < epsilon

This is the part i am not sure of. Can i set my epsilon to equal to -L/2 because L is a negative number and as such the epsilon we choose would end up being positive and greater than zero?

[polar bolt]

Yes L is negative

The part leading up to this could be written better (and idk why you are calling anything prime)

[devout snow]

@rustic igloo Has your question been resolved?

[rustic igloo]

Oh sorry, i meant the derivative of f(x)

Can you look at my complete proof and verify that i didn't butcher the epislon delta proof?

[polar bolt]

Sure

[rustic igloo]

@polar bolt

[polar bolt]

Looks fine

[rustic igloo]

Thanks!

.close

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[shrewd nacelle]

Help

[shrewd nacelle]

I’m doing a probability worksheet in class which involved stuff such as standard deviation

And need answers badly

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@shrewd nacelle Has your question been resolved?

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@shrewd nacelle Has your question been resolved?

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[full gulch]

[\sum\limits_{i = 1}^{n} 1/i-1/(i+1)]
How can I prove that this converges to 1-1/(n+1)?

[full gulch]

https://cdn.discordapp.com/attachments/840667252793802752/909579489862897704/485902339543269382.png

I found an explaination that says to assume 1-1/(n+1) is true, then say a_(k+1) = a_k + 1/(k+1)-1/(k+2)
But by doing it this way, if I didnt know it converges to 1-1/(n+1) it wouldnt work

[devout snow]

@full gulch Has your question been resolved?

[spare crypt]

what can you do with this sum?

[devout snow]

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[weak rapids]

<@&286206848099549185> \

[graceful cosmos]

Thoughts on the question?

[weak rapids]

i imagine its based on constraints

with respect to an objective function

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[devout snow]

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[merry shore]

@copper mango

[merry shore]

how did the foil method that we had became $x^2 - 4rx + 3r^2 = x^2 - bx + 48$?

[woven radish]

Emmanuel??

[copper mango]

cause you were told the 2 roots of x^2-b+48..

[merry shore]

so like where did the substitution happen

[copper mango]

what?

what substitution?

[merry shore]

how did x^2 - 3xr - rx - 3r^2 became somthing like this

[copper mango]

You were given the 2 roots

$(x-r)(x-3r)=x^2-bx+48$

[woven radish]

Mosh

[merry shore]

is that the formula?

[copper mango]

No

it's called quadratics having many forms

I equated the standard form with the factored form.

[merry shore]

ohh

the exand and equate

but when i solve it

there's no real solution

x^2 - b(4)x + 48 = 0

[copper mango]

As I stated already, b=16

,w graph y=x^2-16x+48

[woven radish]

Results provided by WolframAlpha

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

Requested by Omegabet_#9771

[merry shore]

i get it now it's 4 and 12

[copper mango]

the roots are 4 and 12, yes

however we already concluded that r=4, so this shouldn't be new info

[merry shore]

yes because sqrt of 16

tyyy mosh im so sorry for the inconvenience hehe tysm

.close

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[restive river]

hey would someone be able to help me with a power series question? just learning it so its a super simple one but i would love to have it explained

[restive river]

so i have this function and i need to represent it as a power series

and starting at 0, it should be (-1)^n * x^n * (n+1), right?

but the question has it starting at 1 and i cant seem to get the input correct

<@&286206848099549185>

[devout snow]

@restive river Has your question been resolved?

[full gulch]

my first instinct is to cancel out the -1/(n+1) and 1/(n+1)

[spare crypt]

yup that solves it

[restive river]

looks like telescoping

[full gulch]

but the resultant doesnt give me something that looks like 1-1/(n+1)

i get -2/(n^2-2n)

[full gulch]

ah sorry I meant from just this

https://i.imgur.com/0EVlx0G.png

I get -2/(n^2-2n) from just this, rather than my ans

[spare crypt]

you have to cancel way more terms

[full gulch]

Yeah, but how do you cancel out the terms before it?

since there could be many

[spare crypt]

so like (1/1-1/2) + (1/2-1/3) +(1/3-1/4) + (1/4-1/5)

see what you can cancel

@restive river replace n with n-1 in what you wrote to shift it up to starting at 1

and add a minus sign because of the (-1)^n

[full gulch]

oh i think im seeing it better now
everything here would cancel and in this case it seems like n=5

[spare crypt]

yeah

so just the first and last are left over

which is what you want

[full gulch]

Thanks

[restive river]

im not sure what you mean, that "n" power for the first box is written in for us, the first box is the input for 1 being non-alternating and -1 being alternating, and then the second box is for the series itself

which im not sure why its left as just an "n" power since we differentiated

[spare crypt]

I mean for (-1)^n * x^n * (n+1)

to make it start at 1, it'd be x^(n-1) * (n)

with a - in front from the alternating

[restive river]

i tried that and it gave me an incorrect response :/

[spare crypt]

-x^(n-1) * n did?

[restive river]

wait, negative x ?

[spare crypt]

yeah the (-1)^n should technically be (-1)^(n-1)

[restive river]

x^(n-1) * n returned an incorrect response

[spare crypt]

which spits out a negative sign

[restive river]

ohh so then it'd cast onto the x ?

[spare crypt]

yeah, it's weird but it seems necessary for this format

[restive river]

yeah this format is less than ideal

that was correct, thank u v much dawg

.close

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[hidden rampart]

hi can I have halp pls

[hidden rampart]

<@&268886789983436800>

[wicked rover]

pls don't ping mods for hw help

[hidden rampart]

R u a mod

[wicked rover]

yes

[hidden rampart]

Can u halp pls 🥺

[languid ocean]

do you know the definition of an arithmetic sequence?

[hidden rampart]

Yeah

[hidden rampart]

I know the formula

it’s a^n=a1+(n-1)d
Idk how to write subscripts so ye

[languid ocean]

well... for the subscript, you can use text_{whatever you want as a subscript}
for example a_{n}

but anyways

we can find d here

right?

[hidden rampart]

oh cooool thx

[hidden rampart]

I can’t :(

[languid ocean]

so we have the first and 25th terms

[hidden rampart]

@languid ocean I can’t see how to find the distance through terms that are not in order

[hidden rampart]

Yeeeee

sowwy ^-^

[languid ocean]

a1 = 12

a25 = -60

plug those into the formula

[hidden rampart]

wait what

ok so

[languid ocean]

try it

(if it doesn't work out, please send your work so we can see whats wrong)

[hidden rampart]

$a_{n}$ = 12+(n+1)(-75)

oop the subscript thing didn’t work

anyways!

[languid ocean]

surround it with dollar signs
and texit will automatically render it for you

(there is another way, but i just find the dollar signs more convenient)

[woven radish]

squidi

[hidden rampart]

oooo nice

anyways that’s my try @languid ocean

is it correct

[languid ocean]

that's not what you're supposed to do...

you have n, right?

[hidden rampart]

yeeee

[languid ocean]

(do you know what n represents here?)

[hidden rampart]

yeah the position

right

[languid ocean]

so we should replace n with the position that we have

we have the 25th term

so n = 25

(i'm multitasking - ping me when you're back pls)

[hidden rampart]

ohhh OHHH but wat about the distance

@languid ocean

[languid ocean]

d isn't for distance...

[hidden rampart]

the distance is what I’m confused

[languid ocean]

it's the difference between 2 terms

[hidden rampart]

OH

[languid ocean]

*consecutive

terms

[hidden rampart]

yeah consecutive

[hidden rampart]

that’s what I’m asking for :( @languid ocean

[languid ocean]

so we have $a_n=-60$\
also n=25

[woven radish]

Adavocowana

[languid ocean]

plug that in the formula

[hidden rampart]

Cuz I tried to find the distance between terms that aren’t consecutive

[hidden rampart]

wait what

[languid ocean]

plug those numbers in the formula and you'll be able to find d

[hidden rampart]

I’m so confused

OH THE NUMBERS

WAIT LEMME TRY

-60 = 12 + (25-1)d

SO I JUST HAVE TO FIND D :))

[languid ocean]

yeah that's it

[hidden rampart]

thank you both sm! @next herald @languid ocean

[devout snow]

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[restive river]

Hello

[restive river]

A large rectangular conference room has a lengh of 12 cm and breadth 9cm its volume m3 find its heigjt

Need to submit on 10 mins

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[pseudo basin]

@restive river what's this "Need to submit on 10 mins" business? are you in a test?

also

conference room
length 12 cm

[restive river]

But oh well

Google classroom

[pseudo basin]

mkay whatever

those ten minutes are up

[restive river]

Ya, but i can submit as late

[pseudo basin]

can you post the problem again but preferably as a picture? i think you've fucked up in copying it over.

[restive river]

lol ok

[pseudo basin]

this person has left.

.close

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[devout snow]

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[cyan sentinel]

I have to solve by factoring

[cyan sentinel]

What I did was factor the original equation to become A(w)=2w(-w+24)

I then took that and made A(w)=0 and subbed in

I then did (-w+24)=0 and isolated w to get w=24

is that right?

[bright ridge]

w = 0 should also give A(w) = 0

[restive river]

You also have 2w in the factored equation, what value of w there could make A(w) = 0?

[bright ridge]

for b you can differentiate it

[cyan sentinel]

please elaborate on part b)

[bright ridge]

do you know that derivative is zero at maxima and minima