help-10

[chrome crypt]

Ahhh okay. So just clarifying it assumes that the limit exists?

[tardy epoch]

yes

[chrome crypt]

So when you say making assumptions in cases where I don't know what the limit is and trying to find the limit, the only thing I can really do is clarify if the individual limits are real numbers and if thats true then the propoerty of the limit will hold and if it does not abide the definition then it won't hold.

wait so can I even use the properties of limits if I don't know what the values are of the limits?

[tardy epoch]

i don't know when that would be the case

particularly not in this problem

[chrome crypt]

THIS MAKES SO MUCH SENSE.

thank you. You have no idea what you just saved me from.

[obtuse pebble]

@chrome crypt Has your question been resolved?

[obtuse pebble]

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[timid silo]

what am i doing wrong? how do i keep ending up with 44/13 when it should be 38/13

[wooden cipher]

-(2x-2)

[timid silo]

?

[timid silo]

whats wrong with it?

[wooden cipher]

distribute the -

[timid silo]

what negative are you looking at?

?

[wooden cipher]

yes, that one

oh also when you multiplied by 2, you didnt distribute the 2 properly

[timid silo]

how so?

whats wrong with it?

oih it hsould be +4

ok nvm

awesome thanks

.close

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[pale scaffold]

i need help with question 9

not sure how to begin

[tardy epoch]

Calculate a1, a2, a3 and then find a pattern

[pale scaffold]

ok

a1 is 9

a2 is 16

a3 is 23

the difference is 7

formula is a_n = a_n-1 + 7

a1 = 9

thank you

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[fickle cargo]

Prove the logical equivalence

[runic void]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[fickle cargo]

1

I don't know where to begin

<@&286206848099549185> anyone help me out here

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[fickle cargo]

1. I don't know where to begin

[opaque current]

is that pvq or p(or)q

left hand side

first three symbols

[obtuse pebble]

@fickle cargo Has your question been resolved?

[fickle cargo]

First 2 are or and the next 2 are and

<@&286206848099549185> anyone help me out here

[fickle cargo]

are you there

[opaque current]

im here

im not very good with logic tho

[fickle cargo]

ok

what should i do now will anyone else come

[opaque current]

idk why nobody's coming

"equivalent to p or q or v"?

am i reading that right?

[opaque current]

right hand side

[fickle cargo]

p or q or r

[fickle cargo]

the right side

[opaque current]

oh thats an r

ok

[mellow panther]

Is this a "not"?

[fickle cargo]

yeah negation

[mellow panther]

Alright

So you don't know where to begin?

[fickle cargo]

yeah

[mellow panther]

So what properties of logic do you know

For example one simple one is that $\neg \neg P \equiv P$

Do you know any others?

[warm shale]

FuriousChocolate

[fickle cargo]

yeah

we can use demorgons law

[mellow panther]

Yep!

So where does that take us

[fickle cargo]

wait doing it

[mellow panther]

And just to be clear, the original problem is $p \lor q \lor ( \neg p \land \neg q \land r) \equiv p \lor q \lor r$ correct?

[warm shale]

FuriousChocolate

[fickle cargo]

oh no not that one

This one

yeah sorry i thought we were doing this one i have both of them doubts

[fickle cargo]

we can do this one first

[mellow panther]

$\sim \left[ \sim { ( p \lor q ) \land r } \lor \sim q \right] \land q$

[fickle cargo]

yeah after this what can we do

again de morgans law

[mellow panther]

Is this the problem?

[fickle cargo]

yes

[warm shale]

FuriousChocolate

[mellow panther]

Alright

[mellow panther]

So we'll start with this one

What rules can you apply to simplify it

$p \lor q \lor ( \neg p \land \neg q \land r) \equiv p \lor q \lor r$

[warm shale]

FuriousChocolate

[mellow panther]

@fickle cargo you still here?

[fickle cargo]

yeah

[fickle cargo]

i have no idea how to do this one

[mellow panther]

Ok

So theres some stuff in parentheses right

Can you think of any way to get rid of that?

[opaque current]

could we use associative and commutative laws to move stuff around or is that not a thing?

[mellow panther]

You can, but please try not to give answers away.

[fickle cargo]

yeah we can use that but i don't think there is any use doing that

[opaque current]

nah im just learning here

[fickle cargo]

could you tell me the first step to begin with

[mellow panther]

Alright

So first of all

We can't apply demorgans laws since theres three things inside the parentheses right

[fickle cargo]

ok

[mellow panther]

But do you remember that the or operator is assocative

Aka $P \lor Q \lor R \equiv (P \lor Q) \lor R$

[warm shale]

FuriousChocolate

[fickle cargo]

yeah then

How to move them

[mellow panther]

To move them?

[fickle cargo]

I still don't have any idea

[mellow panther]

Ok so look at this here

On the right you have an or statement between two things

Even though one of them is itself a statement

Its still between two things

So you've turned three things into two

Aka $\underline{P} \lor \underline{Q} \lor \underline{R} \equiv \underline{(P \lor Q)} \lor \underline{R}$

[warm shale]

FuriousChocolate

[mellow panther]

So if demorgans laws won't work on a triple and statement, can we somehow add parentheses to make a double and statement that we can work on?

[fickle cargo]

Can you tell me the complete answer

[mellow panther]

Alright

So

You see this

$p \lor q \lor ( \neg p \land \neg q \land r) \equiv p \lor q \lor r$

[warm shale]

FuriousChocolate

[mellow panther]

We can regroup it like this

$p \lor \left(q \lor ((\neg p \land \neg q) \land r) \right) \equiv p \lor q \lor r$

[warm shale]

FuriousChocolate

[mellow panther]

And now distribute the q over (-p ^ -q) and r

Can you try that?

[fickle cargo]

I am going to class now can we continue this later

[mellow panther]

Alright

Good luck!

[opaque current]

how do contradictions and tautologies work?

[mellow panther]

What do you mean by that

Tautologies evaluate to simply being true statements

For example $P \lor \neg P$ is a tautology

[warm shale]

FuriousChocolate

[opaque current]

AAAAh

[mellow panther]

And contradictions, likewise, are simply false

[mellow panther]

What?

[opaque current]

i just got it

[mellow panther]

$P \land \neg P$ is a contradiction

[warm shale]

FuriousChocolate

[mellow panther]

Since P and not P is never true

[opaque current]

exactly

[mellow panther]

So for example $(P \land \neg P) \lor P$ is just $P$ since $false \lor P \equiv P$

[warm shale]

FuriousChocolate

[mellow panther]

Always the best feeling, isnt it

[opaque current]

truly is

[opaque current]

we can also use true in a similar way, right?

[mellow panther]

Yep

$true \land P \equiv P$

[opaque current]

and then u only get true

[warm shale]

FuriousChocolate

[opaque current]

oh

[mellow panther]

$true \lor P \equiv true$

[warm shale]

FuriousChocolate

[mellow panther]

Theres lots of things you can do

[opaque current]

yeah

ok i must have made a mistake somewhere bc i got "q or r equivalent to p or q or r"

[mellow panther]

From where

[opaque current]

the exercise we were doing just now

[mellow panther]

Yeah

If you think about the original statement

Oops

$p \lor q \lor ( \neg p \land \neg q \land r) \equiv p \lor q \lor r$

[warm shale]

FuriousChocolate

[mellow panther]

This one

You can see it must be true just by reasoning

clearly both are true if p is true

[opaque current]

it is

[mellow panther]

Clearly both are true if q is true

And if neither p nor q, but still r, then still both are true

[opaque current]

ok imma solve in a very detailed maner

oh i see the issue

yay

i solved it

[opaque current]

using this

and others

[mellow panther]

Nice!

[opaque current]

its weird, i knew like barely any logic before starting to solve this thing

[mellow panther]

Theres always more too

If you progress further in logic you'll get to things like $\forall$ and $\exists$ and how they interact with the other symbols

[warm shale]

FuriousChocolate

[obtuse pebble]

@fickle cargo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

what am i doing wrong?

the answer should be y/y-2 but somehow i keep getting

y/-4

[dire plinth]

u cant just take the squared away

[timid silo]

if youre dividing you can subtract no?

[dire plinth]

ye but u have the -4 and the y

[timid silo]

like for example x^5/x^3=x^2 correct?

[timid silo]

how does that matter?

or i mean, what should i do with it instead then?

[urban patrol]

wjud you just get rid of the 2 in front of y

[timid silo]

simplify

it shouldve been

y/-2

mb

[urban patrol]

hmm let me simplify this (removes random coefficients)

[dire plinth]

yea but u have to keep in mind rhat when u divide both the numerator and the denominator

the 2y and the -4 are involved too

[timid silo]

lol

[dire plinth]

factorise numerator and denominator

[wooden cipher]

Here, think about this:
(3^4)/(3^2) = 81/9 = 9 = 3^2 as expected

(3^2+3)/(3^2-4) = 12/5 which is not the same as (3+3)/(3-4) = 6/(-1)

[timid silo]

oh i see

[wooden cipher]

When you add and subtract things in the numerator and denominator, it affects what you can and cant cancel

[dire plinth]

^

instead of just directly removing the square

factor out the numerator and denominator

[timid silo]

so wait guys what would be the first step you take if you have

(y^2+2y)/(y^2-4)

[dire plinth]

so w the numerator

u have a y in both terms right

so u can factor it out

[urban patrol]

you want to get factors that multiply each other

thats when cancellation can happen

just based on the properties of addition and multiplication

[timid silo]

ok wait now im confused

[timid silo]

which y? are you talking about 2y or y^2?

[dire plinth]

both

wait yk how to factor right

[timid silo]

yeah

spent half a semester on that bull shit

[dire plinth]

ye so u can factor out the y from y^2 + 2y

[timid silo]

yes y(y + 2)

[mellow panther]

you know how y * (a + b) = ya + yb? So you can do that in reverse. If two things are being multiplied by y, say you have ya + yb, you can "take the y out" and rewrite it as y(a + b)

[dire plinth]

good

now factor the denominator

dyk difference of two squares?

[timid silo]

yes its some more factoring stuff

[dire plinth]

yep so 4 is a square number

so u can factor it into the form (a+b)(a-b)

[timid silo]

so it should be (y+4)(y-4)

[mellow panther]

No

4 is 2^2

Not 4^2

[timid silo]

oh shit

i meant

2

[dire plinth]

ye

[timid silo]

thanks lol

[dire plinth]

anyway now w ur fraction

u get

[mellow panther]

So write out the whole fraction now

What do you get

[timid silo]

boom

[dire plinth]

yes

now once u get that u can cancel the (y + 2) from top and bottom

[mellow panther]

So in the same way

If you have a/b * y/y you get ay/by

But y/y is just 1

So a/b * 1 = ay/by

[dire plinth]

in other words u can imagine as if you split the fraction

and make it become (y+2)/(y+2) * y/y-2

[dire plinth]

yes

[mellow panther]

Yep

[timid silo]

boom?

[dire plinth]

good

[timid silo]

awesome

so always gotta factor first huh

[dire plinth]

if

[polar fossil]

you can do this specifically because everything is being multiplied/divided

[dire plinth]

its being added or subtracted

[mellow panther]

Right

So for example if you have (10+1)/20

You cant simplify to (5 + 1)/10

Since the 1 is there

[timid silo]

oh lol

that makes a lot more sense

[mellow panther]

If you had (5 + 1)/10 * 2/2, you'd get (10 + 2)/20, which is not what you started with

[timid silo]

i see

alright thank guys!!

much appreciated

.close

[obtuse pebble]

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[timid silo]

Hi. What's a good course on Intuitionistic Type Theory? I prefer a good textbook for self-study. Prefereably not the HoTT book. I didn't understand anything from that book. I already know a bit of untyped lambda calculus.

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

found one

book is called Type Theory and Formal Proof: An Introduction

way better than TAPL and HoTT imho

!close

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[light basin]

does this count as a gaussian integral, if it is then is it solvable using polar coordinates?

[tacit scarab]

the gaussian integral is $\int_{-\infty}^\infty e^{-x^2}dx$

[warm shale]

WhereWolf

[tacit scarab]

only when it is integrated over the whole real line

[light basin]

so it isnt solvable when it isnt integrated over the whole line?

[tacit scarab]

yes

[light basin]

ok thx

.close

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[obtuse pebble]

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[indigo folio]

Eee

[indigo folio]

are there any easier ways to solve this 😭

[zenith spade]

Well, it seems (-1,-6) is the vertex

You also have both roots

[obtuse pebble]

@indigo folio Has your question been resolved?

[indigo folio]

I solved that one my bad

?

how do i do this

[dire plinth]

question 13?

[indigo folio]

yes

[dire plinth]

okay so lets take the general form

ax^2 + bx + c = 0

do you know how to complete the square?

like w any given quadratic

[dire plinth]

for the third line

[indigo folio]

?

[dire plinth]

oh wait no i see what you did

[indigo folio]

sorry it’s a bit chunked up

yesss

i got it

[dire plinth]

bring the fraction over to the right side

ye

[indigo folio]

sheesh

That was smooth

I appreciate your help lol

goodbye

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[willow acorn]

Not even sure where to start, any suggestions?

[hidden compass]

Start by making those two lines perpendicular, as it's saying. What does the relationship btw the slopes have to be if you want the lines to be perpendicular?

[willow acorn]

I’m stuck from here

I think my p value is wrong

[willow acorn]

What do I do with the x’s?

[hidden compass]

Wdym?

[hidden compass]

Answer to this first, then we go on

[willow acorn]

I did in the top right

How m is related to p

[hidden compass]

Ahn I didn't notice that sorry

[willow acorn]

All good

It’s messy

[hidden compass]

I just saw the first photo

Let me se what you did, so that I can help you in a better way

[willow acorn]

Alright thanks

[hidden compass]

Mmh I don't see where you found the x-coordinates of L and M

[willow acorn]

How do I do that?

[hidden compass]

You wrote the correct equations, line = 0 but you didn't solve those for x

[hidden compass]

It's the same as solving 2x - 1 = 0

A first degree equation

[willow acorn]

I got x = 6p/p^2+1

Like this or is this wrong?

[hidden compass]

That's not an x coordinate, neither of L nor of M

I don't understand actual why you did that

[hidden compass]

Yes, that's wrong

[willow acorn]

Wait I’m stupid one sec

[hidden compass]

Yes, that's it 👍

[hidden compass]

Yep

[willow acorn]

Got it

Thank you my friend

[hidden compass]

You're welcome 🤗

[willow acorn]

Thanks for not just giving me the answer too I appreciate it

Cya around

.close

[obtuse pebble]

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[slow spade]

I'm trying to compute the limit of $f_{n}(x)=\sqrt{nx}\arctan\frac{1}{nx}$ as $n\to\infty$ and for $x>0$. I'd be grateful for any hints. I'm stuck on the fact that we have an indeterminate form $[\infty\cdot 0]$ as $n\to\infty$.

[warm shale]

sunside

[royal basin]

arctan(t) ~ t as t -> 0

and your 1/(nx) certainly does approach 0

[slow spade]

that helps, thank you

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[vernal folio]

I'm learning about infinite series right now

Are these rules I'm supposed to know???

Or properties??

[sage geode]

Yeah, it's better than proving a limit whenever you see one

[vernal folio]

So these are kind o flike

of like

Sequence properties?

Like how there are exponential properties??

[sage geode]

I'd rather call them limit properties or smth

[tacit scarab]

true

[vernal folio]

o

okeyokey

[obtuse pebble]

@vernal folio Has your question been resolved?

[sage geode]

Do you have any other questions? @vernal folio

[vernal folio]

oops i NEVER Posted it

caps

its okay

.close

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[compact pebble]

I was wondering if there's a simpler way of writing the condition for a binary relation R between sets X and Y to be a function from X to Y
I came up with this but it's pretty long
\begin{align*}
R \text{ is a function from X to Y}\newline \iff \forall x [x \in X \iff \exists y (y \in Y \land (x,y) \in R \land \lnot (\exists z(z \ne y \land (x,z) \in R)))]
\end{align*}

[compact pebble]

ummm

did texit cut it

[sage geode]

Yeah if it doesn't fit then texit cuts it

[compact pebble]

is there a way to add a newline

without separating it into multiple $$ $$

[earnest ruin]

u too ok?

[wild swallow]

do \begin{align*} you math here \end{align*}

instead of $$ math $$

[warm shale]

Frisk17

[earnest ruin]

ok

[compact pebble]

looks like that worked, thanks

[sage geode]

Hm, can't you just say the relation R is a function from X to Y whenever $((x, y1) \in R \land (x, y2) \in R) \Rightarrow y1 = y2$?

[warm shale]

A Lonely Bean

[compact pebble]

isn't the condition that all elements of X have to have a correponding element in Y also a necessary condition

since a binary relation doesn't have to contain (x,y) for all x

unlike a function

actually wait I can probably replace this condition with that horrific not exists z thing I put

[wild swallow]

theres a way to phrase it that uses set equality

Let $\Delta_Y$ denote the diagonal relation on $Y$.

Then $R \subsete X \by Y$ if and only if
[
\comp R = \comp {\Delta_Y} \compose R.
]

[sage geode]

What's a diagonal relation?

[compact pebble]

\begin{align*}
R \text{ is a function from X to Y} \iff \forall x [\exists y ((x,y) \in R \land \forall z ((x,z) \in R \implies z = y))]
\end{align*}

[warm shale]

Frisk17

[compact pebble]

this is correct right?

[wild swallow]

[
\Delta_Y = \set {(y, y) \mid y \in Y}
]

[sage geode]

Oh, that's what I was thinking

[wild swallow]

i think this looks right

[compact pebble]

what is the circle operation?

[wild swallow]

composition

[compact pebble]

also why is this a diagonal relation
what's diagonal about it

[wild swallow]

draw it in the cartesian plane

[compact pebble]

ah

makes sense

thanks for the answers

.close

[obtuse pebble]

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[timid silo]

There is no solution to this but how am I supposed to find that out?

Am i just supposed to try and solve for X and if I cant theres no solution?

[high lily]

pretty much

[timid silo]

oh sick lol

alright thanks

.close

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[glad fable]

please someone help me

[high lily]

jpeg pls

[obtuse pebble]

@glad fable Has your question been resolved?

[glad fable]

nope

[polar fossil]

[obtuse pebble]

@glad fable Has your question been resolved?

[glad fable]

no

[obtuse pebble]

@glad fable Has your question been resolved?

[obtuse pebble]

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[willow acorn]

What’s the process

How do you reflect something in the line y = -1

[runic void]

.close

[obtuse pebble]

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[willow acorn]

Mb @runic void

[obtuse pebble]

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[round wing]

I just came back from a test and this question has been stuck on my mind

[round wing]

We were given 3 vertices (3,5) (5,3) and (3,0) a triangle

and we were asked to rotate it 90 degrees counter clockwise using a standard matrix

I remember the book had something like (cos(90)x - sin(90)y , sin(90)x + cos(90)y)

but the new vertices I got, the shape was bigger

I thought about it for 10 minutes before running out of itme and just submitting it

[polar fossil]

well it shouldn't be bigger

[round wing]

yes

[polar fossil]

try doing it now

[round wing]

Im trying to find the section of the book where I saw it, cant find it so far

Ok found it

this is exactly what I did

instead with 90 deg

[polar fossil]

okay

and what did you get?

[round wing]

let me redo it, dont remember the values

(-5,3) (-3,5) (0,3)

[polar fossil]

yep (also, you should be able to do that without a calculator)

[round wing]

its reassurance

dont wanna mess up on a test

[polar fossil]

ok

anyway you said the area got bigger?

[round wing]

yes, I plotted the vertices

rough sketch

Nevermind

I just replootted it

it looks fine

I think I just messed up somewhere, mentally in the test

it was 2+ hrs nonstop so

I wrote down the right vertices but wrong diagram

[obtuse pebble]

@round wing Has your question been resolved?

[round wing]

.close

[obtuse pebble]

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[obtuse pebble]

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Channel closed

Closed due to the original message being deleted

[dapper venture]

How would I start to solve this?

[polar fossil]

square both sides

[dapper venture]

so it would be 2x + 3 + x + 1 (with square roots) = 7x + 4

no I wrote that wrong

You have to do (sqrt (2x + 3) + sqrt (x+1))(sqrt (2x + 3) + sqrt (x+1)) = 7x + 4

but I don't know how to do sqrt 2x + 3 multiplied by sqrt x + 1

$ sqrt(2x+3) * sqrt(x+1) $

[polar fossil]

$\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ if both $a \geq 0$ and $b \geq 0$

[warm shale]

Hayley

[dapper venture]

oh thanks I completely forgot about that

.close

.close

/close

why isn't it closing

.reopen

.close

[polar fossil]

weird

bot might be broken, I'm not sure it was ever opened?

[tranquil sonnet]

You could also multiply by sqrt(2x-3)-sqrt(x+1)

[obtuse pebble]

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[neon ridge]

I need someone to checkover my steps/answer please and thank you.

[arctic drum]

It's good

[neon ridge]

Are you a helper?

[nocturne minnow]

Anyone can help

[neon ridge]

Ahh okay, ty!

[neon ridge]

could you also check it over?

[wet moss]

bro does NOT trust you lol

[nocturne minnow]

As the person above said, it's good, trust them

[arctic drum]

I'm in an engeeniring program bro

Just trust me

[neon ridge]

my fault bro 😭

[arctic drum]

Chemical engineering I do more than basics math

[wet moss]

engineer, dont trust him

[arctic drum]

Lol

[neon ridge]

i’m so confused do i trust him 😭

[arctic drum]

I'm studying engeeniring trust me

[nocturne minnow]

$e = 3 = \pi$

[warm shale]

dldh06

[wet moss]

true

[neon ridge]

.closed

[arctic drum]

Yes

[wet moss]

1st lecture

[nocturne minnow]

.close

[arctic drum]

And π^2=g

[neon ridge]

.close

[obtuse pebble]

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[craggy crystal]

Hi, I'm doing some factoring summer work before my honors alg 2 class and I have completely forgotten how to factor, the problems that i'm struggling with are 18m^2-156m+288 and 7x^2-12x-27. I would really appreciate some help!

[nocturne minnow]

Look up some resources to refresh your memory

[craggy crystal]

I have, and I kinda understand how to do basic factoring but this is more complicated it seems

i got this from google but I don't really understand how any of it works

[nocturne minnow]

It's applying factoring by grouping

https://www.youtube.com/watch?v=4v-EQIxqpMQ&ab_channel=TheOrganicChemistryTutor

[craggy crystal]

thanks!

ok his whole method involved multiplying the number before the x^2 by the third number but on that calculator it completely ignores that and I also don't understand where the -8m-18m came from. I get that those 2 numbers add to -26 but they don't multiply to 48

[nocturne minnow]

The calculator doesn't ignore it

It doesn't show it

If you look the number before x^2 is 3, and the third number is 48, if you multiply those together, you get 8 * 18

[craggy crystal]

ohh

[fringe rivet]

Did u get the anwser or do you need more help

If so I got you

[craggy crystal]

a little bit

for 7x^2-12x-27 I'm stuck at factoring the two pairs

the two pairs being (7x^2-12x-27)+(-21x-27)

I got x(7x+9) perfectly fine but what I don't get is how (-21x-27) common factored is -3(7x+9)

Like why is -3 used as the factor instead of positive 3?

[nocturne minnow]

So you can have the sign of the x term is positive

(-21x-27) common factored is -3(7x+9)

Notice how inside the parentheses, it has a positive x term

You can use 3, and get 3(-7x - 9) but negatives aren't the easiest to deal with

[craggy crystal]

so it really doesn't matter at the end of the day? using -3 just makes it easier?

[nocturne minnow]

Yeah

[craggy crystal]

When you're placing the common factors into the equation does it matter which order?

on the problem I'm doing (4-^2+5p+1) the common factors are 4p and 1p so does it matter if it's 4p^2+4p+p+1 or 4p^2+p+4p+1

[nocturne minnow]

No because addition is commutative

Where a + b = b + a

[craggy crystal]

damn its like I'm forgetting basic math

thanks

so at the end of this problem I have 4p(1p+1)+(1p+1) after factoring, how do I finish this?

[nocturne minnow]

Did you watch that video?

[craggy crystal]

I did, though the calculator is throwing me off

would it just be (4p+1)(1p+1)?

nvm it is

[nocturne minnow]

Yes

[obtuse pebble]

@craggy crystal Has your question been resolved?

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[obtuse pebble]

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[opaque galleon]

Set up an iterated double integral or a sum/difference of iterated double integrals that gives the volume of:

the solid in the first octant bounded by the parabolic cylinder z = 9 − x^2 and the planes z = 0, y = 3, and x + y = 3.

How do I start? I tried sketching but still no idea how to do it

[lethal sand]

can you show what you have sketched?

[opaque galleon]

i am shy to show my hand writing

😦

[lethal sand]

I can't really guide you if I don't know where you are at

[opaque galleon]

okay wait

i will do it in paint

[lethal sand]

sure

[royal shard]

i would start by setting up the integration for x and y

[lethal sand]

what is the orange plane supposed to show?

I suppose you wanted to draw y=3 there?

that would be the plane perpendicular to the y-axis and intersects it at y=3

[opaque galleon]

nah that is z=0

then blue is y=3

[lethal sand]

ah i see

[royal shard]

there we have something like this

[lethal sand]

next time you should try to label things so it doesn't look ambiguous

anyway, since the problem given us a solid bounded by surfaces, but it asks us to use double integrals to evaluate the volume, what you want to do now is to use your sketching to rephrase the question such that it asks to evaluate the volume under a surface on a region

can you do this for me?

[royal shard]

here you can rotate the thing around in 3d if that helps
https://www.mathcha.io/editor/zx6pEhElSdyhOq80w1S4JQrVqHeOjvLqCzKYxvd

[opaque galleon]

where did you get x =3

[opaque galleon]

👍

[royal shard]

you got the condition x+y=3
therefore y=3-x

[opaque galleon]

no i mean x= 3

[royal shard]

uhm

you mean in the graph?

why there is a 3?

[opaque galleon]

so I tried to graph it

this is what i got

yours has a line

[royal shard]

ahhh ok i see

[lethal sand]

yes, this is what the contours of y=3 and x+y=3 are. The picutre martin sent is the region under the the surface z=9-x^2 that you want to integrate over

which is why I asked you to use your sketch to rephrase the question

[opaque galleon]

okay

ill do that rn

[royal shard]

this is all assuming when z=0
in that case we can use z=9-x^2
with z=0 we get 0=9-x^2
9=x^2, x=3

[opaque galleon]

set up double integral of the solid whose shadow is bounded by y=-x+3 and y=3 on the xy plane and bounded above by z= 9-x^2

[opaque galleon]

yeah yeah yeah i see it now too when i rephrased the question

set up double integral of the solid whose shadow is bounded by y=-x+3 and y=3 x= 3 on the xy plane and bounded above by z= 9-x^2

[lethal sand]

right, so I suppose you can pick it up from here?

[opaque galleon]

yeah

but still the question is weird af tho

like if i had a 3d graphing software

[lethal sand]

nothing is weird about it. The question gave you a solid that is bounded by surfaces. Whereas in a double integral, you need at least a surface over some kind of region

[opaque galleon]

it woulda been easier

so when i got a question like that

maybe i should try to get what it's shadow look like first

on the xy plane right?

[lethal sand]

and you can do this with just the xy cartesian planes

yes, just draw a bunch of contours to see what you are dealing with

[opaque galleon]

so I don't gotta draw it in 3d

yep

thanks

.close

[obtuse pebble]

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[graceful wind]

a

[graceful wind]

can someone solve this and show work pls

i dont know where to start

i know a^3-b^3 = (a-b)(a^2+ab+b^2)

[gilded needle]

what seems promising

[graceful wind]

but -8 is not raised to the power of 3 what would i do

[polar fossil]

it is too

[graceful wind]

bruh

i was thinking square for a sec

[gilded needle]

-8 is something raised to the power of 3

[graceful wind]

yeah i jsut realized

yeah i just like thought of that

thanks

lmao

.close

[obtuse pebble]

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[opaque galleon]

Let S be the surface defined parametrically by

x=vcosw,
y=wcosw,
z=v+w where (v,w) ∈R.
Find the equation of the tangent plane S at the point (v,w)=(0,π).

[opaque galleon]

I am having trouble finding an expression for F(x,y,z)

I already know that when (v,w) = (0, pi), (x,y,z) = (0, -pi, pi)

x=vcosw
y=wcosw
z=v+w

x+y = z cosw

I am stuck here

<@&286206848099549185>

I am stuck at finding an expression for F(x,y,z)

😦

🙏 🙏 🙏

[lethal sand]

well firstly, F(x,y,z) is not like what you think. It doesn't represent the surface parameterised above. F(x,y,z) would rather show you a hypersurface, something that doesn't quite exist in our dimension.
Anyway, why do you think you need to find an implicit function f(x,y) that represents our surface?

We don't need that, we can find the tangent with what we are given

[opaque galleon]

so I can get the gradient

[lethal sand]

are you familiar with vector-valued function?

[opaque galleon]

yeah

in calc 2

I was pro at that

[lethal sand]

great, and that should mean you also know that for a vector function f(v,w) the partials f_v and f_w are also two other vector functions that is tangent to f(v,w)

[opaque galleon]

ok i see

do I have to do f_v x f_w

[lethal sand]

yes, and that would give you the normal to the surface

[opaque galleon]

nice

[obtuse pebble]

@opaque galleon Has your question been resolved?

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[daring burrow]

Im stuck with how to approach this problem in general. What would be the first for something such as this?

[pine sail]

What kind of a sequence do you think this is?

[daring burrow]

I think this would be considered geometric

[pine sail]

Do you know what the common ratio is?

[daring burrow]

It looks like the common ratio is -1/5

would that be considered constant throughout? if so it would be arithmetic instead?

[pine sail]

What do you mean?

[royal basin]

why "considered"

do you think being a geometric series is a matter of "consideration"?

[daring burrow]

I should choose my words a bit better

Arithmetic is when the terms of a sequence differ by a constant.

Geometric is if the ratio between successive terms is constant

So in this case the sequence given is definitely geometric

Sorry, I go through a lot of self doubt often

[pine sail]

Do you know the formula for the sum of a geometric progression?

[daring burrow]

Is it Sn=a(r^n -1)/r-1

[frank monolith]

how many terms are there?

[obtuse pebble]

@daring burrow Has your question been resolved?

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[errant lark]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[royal basin]

well if you take the question at its word then this area is infinite as it consists of an infinite number of chunks each with positive area

but this is most likely what was meant...

[astral aurora]

How to find the area then?

Any idea

[royal basin]

it's just an integral lol

[obtuse pebble]

@astral aurora Has your question been resolved?

[astral aurora]

Yes it is

But what will be limits

0 to pi/4 ?

[royal basin]

0 to 3pi/4

3pi/4 is where the graph crosses the x axis for the first time

[astral aurora]

I was taking the highest point which is at pi/4 no?

,w integration sinx +cosx limit 0 to 3pi/4

[warm shale]

Results provided by WolframAlpha

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[astral aurora]

,w integration sinx+cosx dy 0 to pi/4

[warm shale]

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

Requested by arjunn5589

[astral aurora]

Sinx+cosx =0
Tanx=-1
Yes at 90+45 it will be -1

,w integration sinx +cosx limit 0 to 3pi/4

[warm shale]

Results provided by WolframAlpha

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

Requested by arjunn5589

[obtuse pebble]

@astral aurora Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[hazy sky]

if possible, factor each of the denominators, then reduce to
same denominator “as small as possible”

[hazy sky]

hey i tried different ways of factoring the denominators but i couldnt find then i looked at the answer but i still cant find a way to factor it

[tacit scarab]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[hazy sky]

i didnt write anything cause every time i tried to factor it in my head i saw it wouldnt work

i tried factoring by x

[tacit scarab]

just try

[hazy sky]

to have x-1

everywhere

but i couldnt get it on the last one x²-1

[tacit scarab]

first simplify $\frac{x^2}{x^2+x}$

[warm shale]

WhereWolf

[hazy sky]

x

i didnt even realize that

[tacit scarab]

.

what do you ge

t

[hazy sky]

u get x ?

[tacit scarab]

you factor out x

then simplify

[hazy sky]

1

[tacit scarab]

???

[hazy sky]

how should i factor out x

[tacit scarab]

$x^2+x=x(x+1)$

bro

[warm shale]

WhereWolf

[hazy sky]

but i already did that yeah

[tacit scarab]

..

[hazy sky]

but i thought u wanted to delete x^2

[tacit scarab]

show me everything you did

you can't just type x and expect me to know what you are doing

[hazy sky]

yeah

[tacit scarab]

so how to simplify $\frac{x^2}{x(x+1)}$

[warm shale]

WhereWolf

[hazy sky]

1-(1/x+1) ?

[tacit scarab]

what

$1-(\frac{1}{x+1})$???

[warm shale]

WhereWolf

[south cargo]

[tacit scarab]

why

why do you turn that to this form

[hazy sky]

x/x+1

[tacit scarab]

yes

x/(x+1) is better

$F=\frac{x}{x+1}-\frac{3x+1}{x-1}+\frac{x-2}{x^2-1}$

[warm shale]

WhereWolf

[tacit scarab]

what can you still factor?

[hazy sky]

x^2-1 ?

i did that

but i dont think thats good

[tacit scarab]

wut

no

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

[warm shale]

WhereWolf

[tacit scarab]

also if you tried that please tell me beforehand

so I know where you are stuck

[hazy sky]

i think i stucked everywhere tbh

so i should make this become a²-b²?

[tacit scarab]

simplify x^2-1

or turn x+1 and x-1 to x^2-1

[hazy sky]

thats what u mean?

[tacit scarab]

no

you can't do that

x^2-1 = x^2-1^2

[hazy sky]

ah yeah

im so lost lmao

(x+1)(x-1)

[tacit scarab]

$F=\frac{x}{x+1}-\frac{3x+1}{x-1}+\frac{x-2}{(x+1)(x-1)}$

[warm shale]

WhereWolf

[tacit scarab]

now make all denominator the same

[hazy sky]

(x+1)(x-1)

[tacit scarab]

yes

make all denominators (x+1)(x-1)

[hazy sky]

i must multiply above aswell right,

[tacit scarab]

yes

keep going

[tacit scarab]

now that the denominators are the same you can comine them

[hazy sky]

-2x²-4x-3/(x-1)(x+1) ?

[tacit scarab]

no

the first term should be $\frac{x^2-x}{(x+1)(x-1)}$

[warm shale]

WhereWolf

[tacit scarab]

the last term should be $\frac{x-2}{(x+1)(x-1)}$

[warm shale]

WhereWolf

[hazy sky]

thats what i did here

then last i added them

[tacit scarab]

,w simplify (x^2)/(x^2+1)-(3x+1)/(x-1)+(x-2)/(x^2-1)

[warm shale]

Results provided by WolframAlpha

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Requested by _wherewolf_

[tacit scarab]

hmm weird

[hazy sky]

,w factor ((x^2)/(x^2+x))-((3x+1)/(x-1))+((x-2)/(x^2-1))

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by jaozes

[tacit scarab]

yeah we can't simlify anymore

[hazy sky]

how did i find -2x²-4x-3

[tacit scarab]

here

[hazy sky]

its bcs on the result sent its -(2x²+4x+3) ?

[tacit scarab]

?

[tacit scarab]

same

[hazy sky]

oh is it?

[tacit scarab]

factor out -1

[hazy sky]

ah ok thats what i was thinkingb

it looks so simple when ur done with it but i know i wouldnt be able to do another one like this

idk its hard to find the way u must do it every time

it always changes

what could make it easier

[tacit scarab]

first factor everything you could

then simplify the fractions

make denominator the same

and combine everything

finally factor and simplify the combined fraction

[hazy sky]

i always begin by trying to find a way i could make denominator smae

for example on this one i can start by doing x²-2² on the second one right

then make it (a-b)(a+b)

[tacit scarab]

yes

[hazy sky]

when im here what should i do?

[tacit scarab]

make denominator the same

not that (2-x) = -(x-2)

[hazy sky]

thats what i was thinking about

but there are then 3 times (x-2)

what do i do with the others left ?

what can i do with the (x+2) and (x+1)

[tacit scarab]

multiply them

[hazy sky]

so i should have (x+2)(x+1)(x-2) everywhere ?

[tacit scarab]

yes

[obtuse pebble]

@hazy sky Has your question been resolved?

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[obtuse pebble]

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[wary temple]

i think that a multiple of a abundant number is a abundant number, but I don't know how to prove it.

[wary temple]

pls solve for odd numbers

[sacred root]

You think or is it true?

[wary temple]

it is true

nope, 945 * 3 = 2853

[sacred root]

Ye I mistook multiple and divisor..

[wary temple]

2835 is abudant number

[wary temple]

no big deal

[sacred root]

Well I've some useful info, about divisor...but I'm not sure if it will help it...wait let me provide it

[wary temple]

okey

[sacred root]

Perhaps this maybe help, but I may not be able to help you this is beyond my level

#advanced-number-theory

Maybe ask here...

[wary temple]

thanks you, im try to solve it)

[wary temple]

I'll give it a try

thanks)

[sacred root]

👍

[wary temple]

.close

[obtuse pebble]

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Closed by @wary temple

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[timid silo]

anyone able to explain to me how this was found?

[timid silo]

is there some sort of analysis going on there that i dont see?

im going to the gym to take a break from this. if you know anything please ping me so i can see it! thanks!!!

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I don't know how to do this integral so id appreciate any help.

Ive tried the sinxcosx identity, but I dont know how to do that with x being only in cos

[obtuse pebble]

@timid silo Has your question been resolved?

[sage geode]

How is f(x) defined though?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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[timid silo]

i wanna ask a question

[jagged onyx]

post a pic of it

[timid silo]

Q62

[jagged onyx]

Have you tried solving it?

[timid silo]

yes

but

[jagged onyx]

send your work

[timid silo]

couldnt plot a graph

for sin2^x

[jagged onyx]

...

[timid silo]

i kinda couldnt think what to do after simplifying pi^x +pi^-x

[jagged onyx]

try making a quadratic out of (pi)^x

[timid silo]

okay wait lemem try

[jagged onyx]

lemem

ok

[timid silo]

shut

[jagged onyx]

ok

ok

try

[timid silo]

report?

[jagged onyx]

srsly

just try solving it

the question

[timid silo]

i am you r not letting me

[jagged onyx]

do it

ok

bye

[timid silo]

someone gonna help?

[jagged onyx]

try finding values of x for which sin^2 (2^x) -4 is greater than or equal to 0

good luck

[timid silo]

no values

coz

[jagged onyx]

there you go

[timid silo]

sin belongs

from 1 to