help-13

[livid hound]

in this context, yes

[queen marsh]

That thing what you marked red is 30° ?

[livid hound]

yes

then determine the radius

[queen marsh]

how would I find the radius once I found the degree of S?

[livid hound]

using the definition of diameter and radius

finding the radius isn't relevant to the angle here

[cedar kiln]

@queen marsh Has your question been resolved?

[queen marsh]

Area of Triangle AOB = (1/2) * base * height = (1/2) * 6 *6 = 18 sq in. ???

[livid hound]

no

where are you getting 6 from

[queen marsh]

let me redo it really quick

s= 60/260 degrees x pi x 7^2 right?

i meant 360*

@livid hound is the area of the triangle 16?

[livid hound]

s= 60/260 degrees x pi x 7^2 right?
no

where's 7^2 coming from

yes, triangle area is 16

[cedar kiln]

@queen marsh Has your question been resolved?

[queen marsh]

So basically after finding the corner of that triangle what is connected to center of circle (30°) and knowing the diameter, get the radius which is simply 8 and these are two sides of the triangle between 30° corner, so now we can use the formula 1/2ab•sinα - 1/2•8•8•1/2=16= area of triangle AOB.
A bit harder with sector.
π=3.14, R=8, n=30°, so here goes the formula πR²/360°•n - 3.14•8²/360°•30°=~16,75= area of sector AOB.
Thats all

I believe that is correct right?

@livid hound

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[marble hull]

Could I get help with this

[cedar kiln]

Please don't occupy multiple help channels.

[marble hull]

Close

/close

(╯°□°)╯︵ ┻━┻

[median burrow]

It’s .close

[marble hull]

Oh

You trying to help me tho😅

.close

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[hot mulch]

Hi, ik this is pretty basic but im confused in how i have to make it ty

[cedar kiln]

@hot mulch Has your question been resolved?

[hot mulch]

No

[cedar kiln]

@hot mulch Has your question been resolved?

[arctic scaffold]

i suppose you need to find the area, right?

because they're rectangles, you can find it by multiplying both sides

in the first one: (x+3)(5x)

it'd be x(5x)+3(5x)

answer: || 5x²+15x ||

the same logic goes for the others

tho the second one in particular, you calculate the area of both rectangles and sum them at the end

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[calm drift]

Hey

[calm drift]

can someone explain this?

I have the answer but im not quite sure how to get there

and we are using angle sum identities to find the exact value of each

the answer for it is 2+√3

just dont know how to get there

[sly abyss]

Start by adding two unit circle degrees

That equal 75

[calm drift]

Well

i already have it like

set up

[sly abyss]

Do you know the formula

[calm drift]

My friend told me to do

cause he said its easier

[sly abyss]

Ye your friend loves to waste fucking time

[calm drift]

💀

[sly abyss]

And you should just be using the sum of angles formula for tangent

And not try to be proving the formula

[dire geode]

write 75 as a sum of two known special angles

,tex .sum diff trig

[wraith dagger]

riemann

[dire geode]

bottom one

[sly abyss]

Numerator is +, denom is -

If you were doing sum of diffirencrs the numerator would be -, and the denom would be +

[calm drift]

okay so

u could do like

tan(75) = tan(45)+tan(30)/1-tan(45)tan(30)

?

[sly abyss]

Ye

That’s all it is

[calm drift]

Okay

one sec

tan(75) = (1 + √3/3) / (1 - 1 * √3/3)

which would then be

2+√3

oh

that does make sense

okay ty

[cedar kiln]

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[nimble adder]

if 1/i=(-1)^-1/2, and you set that value to n, square both sides, simplify, then take the square root on both sides, you get 1/i=i
where did i go wrong?

[void sand]

i believe you cannot simply cancel out the sqrt and square

in the sqrt(n^2) part

[nimble adder]

why not?

it works for every other number

[void sand]

because then you can derive a contradiction

for example

2 = sqrt(4) = sqrt((-1)^2 * 4) = sqrt((-1)^2) * sqrt(4) = -sqrt(4) = -2

this is only possible because i took sqrt((-1)^2) and turned it into -1

by cancelling the sqrt and square

[nimble adder]

ohhhhhhhh okay

👍

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[timber turtle]

can someone please help me with this?

[cedar kiln]

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[craggy stratus]

How does ln(x) appear from this derivative?!

[dire geode]

Chain rule

,w diff a^x

[wraith dagger]

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Requested by rie.mann

[craggy stratus]

But if we have e^x

We just get e^x

[dire geode]

You don't

[craggy stratus]

So What are all the parts we get here then?

[dire geode]

x^x is not the same as e^x

[craggy stratus]

Yeah But how does it become ln..

That’s the part that confuses me :-;

[vagrant elbow]

they probably wrote it as $e^{x\ln x}$

[wraith dagger]

Norbert Baudin

[craggy stratus]

Like What are all the chain step rules to easily see this solution?

Is that the same even? o_O

[vagrant elbow]

yeah

[craggy stratus]

doesnt e ln and the E cancel so we just have x^x there?

Because x*lnx = ln(x^x) No?

[vagrant elbow]

yes that's the point

but this form makes it easier to differentiate

[craggy stratus]

oh so you’re saying from the ”first” derivative e^x^x we get e^x^x

• x^x ?

[vagrant elbow]

times the derivative of x^x

[crimson sedge]

Chain ruleee

[craggy stratus]

Yeah and the derivative of x^x in order to make it easier they rewrite it as e^x ln x

[vagrant elbow]

yes

[craggy stratus]

and then you get (lnx)+1?

[vagrant elbow]

times x^x

as the derivative of x^x

[craggy stratus]

o_o

Man i hate this so much

[vagrant elbow]

$\dv{x}(x^x) = \dv{x}(e^{x\ln x}) = e^{x\ln x}\dv{x}(x\ln x) = x^x(\ln x +1)$

[wraith dagger]

Norbert Baudin

[craggy stratus]

ok i think i see

Nevermind i’m still not sure What happens later on q.q

e^x^x
When derived we get e^x^x * x^x
Then we look at the next part of e^x^x which is x^x and derive it
we rewrite it as e^xlnx and get e^xlnx * xlnx, and then we have x * ln(x), the derivative of x is 1 so we get ln(x) and the derivative of ln(x) is 1/x so we get x/x = 1
Summed up we have e^x^x * x^x * ln(x)*1?

.close

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[inner carbon]

Can anyone Please help me prove the AM GM inequalities?

[inner carbon]

<@&286206848099549185>

[cedar kiln]

@inner carbon Has your question been resolved?

[cedar kiln]

@inner carbon Has your question been resolved?

[crimson sedge]

for two variables: $(a-b)^2$ => 0 by trivial inequality. Add $4ab$ to both sides to get $a^2 +2ab + b^2$ => $4ab$. Square root both sides and divide by 2.

[wraith dagger]

Nandan

[inner carbon]

tell for the five terms please

[cedar kiln]

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[inner carbon]

.reopen

[cedar kiln]

✅

[inner carbon]

<@&286206848099549185>

[wraith dagger]

Snöwdinger

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@inner carbon Has your question been resolved?

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[fervent karma]

Which line did I mess up

[zenith sail]

Are you sure you wanted to add 121/9 ?

[fervent karma]

oh- we're not supposed to add that?

[zenith sail]

Not quite

You want to add (b/2)^2

you added b^2

[fervent karma]

ohhh, okay, lemme try it again 😅

[zenith sail]

Notice later on when you changed it to (x+11/6)^2

but $(x+\frac{11}{6})^2 = x^2 + \frac{11}{3}x + \frac{121}{36}$, not $x^2 + \frac{11}{3}x + \frac{121}{9}$

[wraith dagger]

tatpoj

[zenith sail]

Sorry for rearranging so much. I'm bat at latex lol

[fervent karma]

It's okay haha ion even know how that bot works

So I got: x^2 + 11x/3 + 121/9 = 4/3 + 11/6 and then i go on from here, would this be right?

[zenith sail]

No, not quite

[fervent karma]

oh wait why is it over 36?

[zenith sail]

Tell me if you agree with this: $(x+\frac{11}{6})^2 = x^2 + \frac{11}{3}x + \frac{121}{36}$

[wraith dagger]

tatpoj

[fervent karma]

Ahhh we do 11/3 x 1/2 and then ^2 it?

[zenith sail]

right

[zenith sail]

so, I take it you do agree with this? lol

[fervent karma]

yes haha

ill re try it again

[zenith sail]

alright 👍

[fervent karma]

im somehow getting x = 8/6 as the final answer

[zenith sail]

can you show your work?

There should be two solutions

[zenith sail]

,rotate

[fervent karma]

ooops why is this sidewards

[zenith sail]

no worries lol

[fervent karma]

i prolly misplaced a number

[zenith sail]

Two quick things

In this line, you added 121/36 to the left side, which is correct

but that means you should have added 121/36 to the right side as well

Whatever you do on one side, you must do to both

so instead of adding 11/6, that should also be 121/36

[fervent karma]

Ohhh i see, i thought im supposed to put the root on the right

[zenith sail]

Nope, there aren't any special rules for this or anything. Just like always, if you add something to one side, you must add the same thing on the other

[fervent karma]

Oooo okay

[zenith sail]

The only thing that is a little different about quadratics, is that when you take the square root, you need to add the +- sign on one side. Like in this step

That's how you get both solutions

[fervent karma]

So for that line, would it be sqrt (x + 11/6)^2 = 4/3 + sqrt of 121/36 or do we add 4/3 and 121/36 first?

[zenith sail]

add 4/3 and 121/36 first

then take the square root of both sides

[fervent karma]

Ooo okie

I got x = 1/3, -4

[zenith sail]

me too 👍

nice job

[fervent karma]

Lez goo

Can you perhaps make something like this about the flow of the equation

bcs im still a little lost with the steps, it's okay if you cant

[zenith sail]

I think it's honestly probably easier to understand in words

The thing you add to both sides should be half the middle term, squared

[fervent karma]

Oh okay, that'll work too

[zenith sail]

so, (b/2)^2

where b is the number on the second term

Like, here, you had 11/3, so you added ((11/3)/2)^2, which was 121/36

Add that to both sides, then take the square root

And don't forget the +/- sign when you take the square root

That's pretty much it I think

[fervent karma]

Got it, thanks, this makes things easier 🫡

[zenith sail]

Awesome, happy to help 🫡

[fervent karma]

.close

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[smoky cosmos]

set theory- If X is subset of A or X is subset of B then X is subset of A union B, but the reverse is not true, however if X is subset of A and X is subset of B then X is subset of A intersection B and the reverse is also true in this case. why is there unsymmetry here, hmmm.

[tropic oxide]

do you want a counterexample?

[smoky cosmos]

first i wanna know if what i said is correct or not

there is this unsymmetry ?

[tropic oxide]

yes it is

here is an example where X is a subset of A union B but not of A or B individually:

A = {1,2,3,4,5}
B = {3,4,5,6,7}
X = {1,3,6}

[smoky cosmos]

IF we take my orginal statement and replace or with and we get this "anti symmetric" statement so there is a symmetry here ----- If X is subset of A AND X is subset of B then X is a subset of A union B. and reverse is also true, but if X is subset of A or X is a subset of B then X is not neccesarily subset of A intersection B, but the reverse is true.

[tropic oxide]

what is your point

or what is your doubt

[smoky cosmos]

am i right ?

[tropic oxide]

you are going in circles and you have lost me

[reef whale]

If you wanted a truly “anti” statement, you would replace “subset” with “superset”

[smoky cosmos]

all i am asking is am i right

[reef whale]

They replaced “union” and “or” with “intersection” and “and” and wondering why it’s not equivalent

[tropic oxide]

the real question is why should that lead to an equivalence at all

[smoky cosmos]

so if we take all these 8 possibilities making a check table of it they are anti symmetric, they form the truth table of

is this a deep mathematical symmetry or just lingustic biased garbage

ok i am going insane time to switch subjects

sorry

THANKS FOR HELP

[cedar kiln]

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[lofty sparrow]

Hello, another explanation type thing. Idk how to do it in writing so uhh

How does 2(5(sqrt2)) equal 10(sqrt2)

Does multiplication NOT touch radicals, like are radicals only affected through exponents only?

[cedar kiln]

@lofty sparrow Has your question been resolved?

[lofty sparrow]

<@&286206848099549185>

[glad fog]

$2(5\sqrt{2})$ is the same as writing $2 \times 5 \times \sqrt{2}$

[lofty sparrow]

im following

yeah

[wraith dagger]

hhhapz

[glad fog]

you can take things in and out of a radical if you properly modify it based on how a radical works

like, for example:

$2\times5\sqrt{2} = 2 \times \sqrt{2 \times 5 \times 5}$

[wraith dagger]

hhhapz

[lofty sparrow]

hm

uhh

maybe i need to study more on radicals

[glad fog]

i wouldn't personally get too hung up about this

[lofty sparrow]

oka

[lofty sparrow]

so 2 doesnt get distributed to the sqrt2

[glad fog]

you could

what is the opposite of the sqrt ?

[lofty sparrow]

squaring

[glad fog]

yeah

so when you put something into the sqrt

you square it

$x = \sqrt{x^2}$, and $x \times \sqrt{2} = \sqrt {2x^2}$

[lofty sparrow]

i follow

so its best to touch radicals only with radicals/exponents

[wraith dagger]

hhhapz

[lofty sparrow]

ohhh like u put something in u square it

[glad fog]

thats all i did in the example with the 2*5*5

i put in 5^2

but i just wrote it as 5*5

[lofty sparrow]

i see

okay thank you very much

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[livid hound]

geoemetric serires formula

[crimson sedge]

[
\sum_{i=1}^n a_1 r^{i-1} = a_1\p{\f{1-r^n}{1-r}}
]

[mystic wren]

ty

.close

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[gentle abyss]

how can i have a?

[gentle abyss]

<@&286206848099549185>

[shadow hedge]

what's the question

[gentle abyss]

.

[shadow hedge]

what does how can I have a mean

[gentle abyss]

how can i get a*

[mighty shuttle]

you can't. You should have been given a, or perhaps the question is to find the value of a for which the limit is -1/7?

[gentle abyss]

yeah find a so the sol of limit is -1/7

[shadow hedge]

use l'hopitals

[gentle abyss]

it would be two lhopitals?

[shadow hedge]

maybe, try it once and see what you get

[gentle abyss]

i get this, but numerator would be 1-1

[gray sandal]

use hospital rule

[shadow hedge]

there's a sign error

(sin error)

[gentle abyss]

+2sin(2x) right?

[shadow hedge]

yeah

[gentle abyss]

i still having the problema that 1-1(1+x) is 0

.close

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[oak jasper]

One message removed from a suspended account.

[upbeat monolith]

#❓how-to-get-help

[split pike]

!occupied

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[worldly flint]

.close

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[halcyon granite]

Hey guys. I'm trying to evaluate

\begin{align*}
I_0 = \int \frac{x^2}{(1+x^2)^2} , dx = , ?
\end{align*}

This is definitely doable with integration of rational functions, but I'd like to try w/o it.

We can notice that
\begin{align*}
\int \frac{x^2}{(1+x^2)^2} , dx = \underbrace{\int \frac{1+x^2}{(1+x^2)^2} , dx}{\arctan x} - \underbrace{\int \frac{1}{(1+x^2)^2} , dx}{I_1}
\end{align*}

But what could be done about $I_1$?

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

maybe a clever substitution would work

[mighty shuttle]

maybe x= tan(u)?

yeah, that should work

oops, sorry for the ping

[halcyon granite]

hmm, we do them a bit different here in europe, so just so that we are on the same page. you mean something like

$$
\int \frac{1}{(1+x^2)} , d(\arctan x)
$$
?

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

and then mb int by parts...

[mighty shuttle]

no x=tan(u)

so $dx=sec^2(u)du$

[wraith dagger]

Why am. I here

[halcyon granite]

so, this way then?

$$
\int \frac{1}{(1+x^2)^2} \cdot \cos^2 x, d(\tan x)
$$

[mighty shuttle]

where did the cos come from

[halcyon granite]

I mean, $d(\tan x) = \frac{1}{\cos^2 x} , dx$

[wraith dagger]

Sweet Tea 🧋

Sweet Tea 🧋

[mighty shuttle]

true, yes

uh, I'm not sure I follow, you are performing a u-sub, right?

[halcyon granite]

idk 🙈

we call it here 'putting stuff under the differential'

prob u-sub ig

I thought that was what you suggested

[mighty shuttle]

I did, which would lead you to $\frac{sec^2(u)du)}{(1+tan^2(u)^2)}$

[halcyon granite]

$\frac{sec^2(u)du}{(1+tan^2(u)^2)}$

[wraith dagger]

Sweet Tea 🧋

Why am. I here

[mighty shuttle]

yeah, the integral of this

I think this is what you were trying to do, right?

[halcyon granite]

. give me a min plz to make sense of it, I didn't have much sleep today xD

[mighty shuttle]

now notice $sec^2(u)= (1+tan^2(u))$

[wraith dagger]

Why am. I here

[mighty shuttle]

basically what you're doing is finding dx in terms of a new variable du

[halcyon granite]

but wait, $$x=\tan u \implies u = \arctan x \implies du = \frac{1}{x^2+1} , dx$$

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

or $(x^2+1)du=dx$, now express x only in terms of u

[wraith dagger]

Why am. I here

[halcyon granite]

but we don't have $u$ there, only $d, u$

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

no, I'm multiplying both sides by $x^2+1$

[wraith dagger]

Why am. I here

[mighty shuttle]

and then expressing x as a function of u

[halcyon granite]

umm

where did u do that?

[mighty shuttle]

so (x^2+1)du=dx, right?

[halcyon granite]

ok, maybe let's start from the beginning xD I stopped following

[mighty shuttle]

sure, from where?

[halcyon granite]

ok, integrating $$I_1 = \int \frac{1}{(1+x^2)^2} , dx$$

what do you want to put inside $d (\ldots)$?

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

let x= tan(u)

so $d(tan(u))$

[wraith dagger]

Why am. I here

[halcyon granite]

hmm, so we basically have

$$
\int \frac{1}{(1+u^2)^2} , d(\tan u) = \int \frac{1}{(1+u^2)^2} \cdot \frac{1}{\cos^2 u} , du
$$

and to make it the function we are interested in we gotta multiply by $\cos^2 u$, hence giving us

$$
I_1 = \int \frac{1}{(1+u^2)} \cdot \cos^2 u , d(\tan u)
$$

is that what we are doing?

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

the first part is right

not 1+u^2 though

1+tan^2(x)

remember x=tan(u)

not u

[halcyon granite]

I'm just a bit confsed, since we never did that before, but that's probably ok.

So we do

$$
\int \frac{1}{(1+\tan^2 u)^2} , d(\tan u)
$$

but how does that relate to
$$
\int \frac{1}{(1+x^2)^2} , dx
$$

are they equal then?

[mighty shuttle]

x is tan(u)

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

what's d(tan(u))?

wrt u

[halcyon granite]

$\frac{1}{\cos ^2 u} , dx$

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

or not

oh gosh

[mighty shuttle]

yes, and what is that in terms of tan(u)

[halcyon granite]

tan of what?

tan of x or u?

[mighty shuttle]

$1+tan^2(u)= sec^2(u)$

[wraith dagger]

Why am. I here

[mighty shuttle]

use this

[halcyon granite]

so it becomes $\int \cos^4 u , d(\tan u)$?

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

uh, yeah, I think so, though why the tan(u) is still inside the d() is confusing me

[halcyon granite]

then $$\int \cos^2 u , dx$$

[mighty shuttle]

I would simply write x= tan(u) so differentiating both sides wrt u we have dx/du=sec^2(u)

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

yup

[halcyon granite]

but x was $\tan u$...

[wraith dagger]

Sweet Tea 🧋

[mighty shuttle]

wait du, right

not dx

I don't really understand your method, sorry

[halcyon granite]

yea, me too (mine and yours xDDD)

[mighty shuttle]

the best way I can explain what I'm doing is to ask you to think you're finding the area under a curve of scale X with small subdivisions dx. now we let x=tan(u) . do you follow until here ?

[halcyon granite]

I'd rather come back later and see if someone can explain it in terms of what I'm familiar with, if you don't mind 👉 👈 🥺

I just have a test in a few days – don't wanna confuse myself even more by learning a technique we didn't learn and probably never will (since in eu I think we do u-sub a bit differently)

[halcyon granite]

well, but indefinite integral isn't definite, is it?

[mighty shuttle]

sure, I understand.ATB for your exam!

[cedar kiln]

@halcyon granite Has your question been resolved?

[cedar kiln]

@halcyon granite Has your question been resolved?

[raw gulch]

read and analyse this:

$\int_{}^{}\frac{dx}{\left( 1+x^{2} \right)^{2}}=\int_{}^{}\frac{\left( 1+x^{2} \right)-x^{2}}{\left( 1+x^{2} \right)^{2}}\text{ }dx=\\=\arctan\text{}x-\frac{1}{2}\int_{}^{}x\cdot \frac{2x\text{ }dx}{\left( 1+x^{2} \right)^{2}}=\\=\arctan\text{}x-\frac{1}{2}\left[ x\cdot \left( -\frac{1}{1+x^{2}} \right)-\int_{}^{}\frac{-\text{ }dx}{1+x^{2}} \right]=\\=\frac{1}{2}\arctan\text{}x+\frac{x}{2\left( 1+x^{2} \right)}+C$

[wraith dagger]

Joanna Angel

[cedar kiln]

@halcyon granite Has your question been resolved?

[halcyon granite]

hm, sorry for a possibly dumb question, but what happened in between lines 2 and 3? @raw gulch

looks line integration by parts, but what did u put under the differential then?

other than that everything is clear to me

wait a second, I think I get it – the answer to my question is probably $, d(-\frac{1}{1+x^2})$

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

Ok, I tried to fill-in the gaps (for myself), here's what I've got:

\begin{align*}
\int \frac{x^2}{(1+x^2)^2} , dx = \int x \cdot \underbrace{\frac{x}{(1+x^2)^2}}_{\text{aiming to put it under the dif.}} , dx = , ?
\end{align*}

\begin{align*}
\int \frac{x}{(1+x^2)^2} , dx = \frac{1}{2} \int \frac{1}{(1+x^2)^2} , d(1+x^2)=-\frac{1}{2} \cdot \frac{1}{1+x^2}
\end{align*}

Hence now we can compute the initial integral by parts:
\begin{align*}
-\frac{1}{2} \int x \cdot \frac{x}{(1+x^2)^2} , dx &= -\frac{1}{2}\int x , d\left(\frac{1}{1+x^2}\right) \
&= -\frac{1}{2} \cdot \frac{x}{1+x^2}+\frac{1}{2} \int \frac{1}{1+x^2} , dx \
&= -\frac{1}{2} \cdot \frac{x}{1+x^2} + \frac{1}{2} \arctan(x)
\end{align*}

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

And then indeed the rest matches your answer:

\begin{align*}
I_0 = \arctan x - \frac{1}{2} \arctan x + \frac{1}{2} \cdot \frac{x}{1+x^2} = \frac{1}{2} \arctan{x} + \frac{x}{2(1+x^2)} + C
\end{align*}

[wraith dagger]

Sweet Tea 🧋

[halcyon granite]

Thank you sooooo much @raw gulch

.close

[cedar kiln]

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[rancid steeple]

Hello! Would anyone be able to walk me through how to do a few problems?

[hoary condor]

?

what problem

[rancid steeple]

Question B if possible!

[warped coyote]

!status

[cedar kiln]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[rancid steeple]

7

[warped coyote]

So what do you want?

Don’t dm me, post it here

[rancid steeple]

I was hoping youd be able to run me through some problems im studying. Im new to the subject and have no base knowledge lol

[hoary condor]

; - ? idk how too

[warped coyote]

Read this first https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/triple-integrals-topic/a/triple-integrals

[cedar kiln]

@rancid steeple Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[velvet hemlock]

Any idea on how to prove this

[velvet hemlock]

any hints?

[idle tusk]

what's the formula for $\mathbb{P}(X \cup Y)$?

[wraith dagger]

artemetra

[velvet hemlock]

P(x) + P(y)

Ik this xd

[humble karma]

It's not true though.

[velvet hemlock]

?

it is true lol

commutative

propert

y

[humble karma]

It's not. Say I roll a dice, the probability that I get a 1 OR and odd number isn't the probabilities of both these events added

[humble karma]

That's not what commutative means.

[velvet hemlock]

then what comm mreans

means

[idle tusk]

it is true for independent events

but the events aren't necessarily independent

[velvet hemlock]

ik that?

lmfao

so k go its P(x) + P(y) - what is similar in both

good?

[humble karma]

In general, you have to use the inclusion-exclusion principle

[velvet hemlock]

.-.

[humble karma]

Yes that

[velvet hemlock]

ik..

I said

already

[humble karma]

I was writing down something and you wrote it, if you knew it then why not use it in the first place?

[velvet hemlock]

k

next time

how this is even going to help me

proving lol

[idle tusk]

yeah and how do you write "what is similar in both"

[velvet hemlock]

idk how to do it here

[idle tusk]

$\cup$ or $\cap$

[wraith dagger]

artemetra

[velvet hemlock]

..

bro srsly

ik

cap

[idle tusk]

yes awesome

and now

[idle tusk]

$X=(A \cap B^c)$\
$Y=(A^c \cap B)$

[wraith dagger]

artemetra

[idle tusk]

so use the formula

[velvet hemlock]

still haha

i tried tho

[idle tusk]

where did you get stuck?

[velvet hemlock]

bro u can try urself

u will get nothing

u will just loop

[idle tusk]

...

are you here to get help

[velvet hemlock]

what?

then why im asking

?

[idle tusk]

or complain that you can't get to something and ask me to do it for you to reach the same point

[velvet hemlock]

ask you?

u know that i got the solution?

i can use drive?

you know that?

[idle tusk]

drive?

[velvet hemlock]

but im not gonna check

ye?

[idle tusk]

what's drive

[velvet hemlock]

bro

google drive?

u dont know

[idle tusk]

ah

[velvet hemlock]

l0l

[idle tusk]

yes i do know

[velvet hemlock]

good

hint?

l0l

im not gonna look on drive

useless

[idle tusk]

well i can't give you a hint when i don't know what hint you need

[velvet hemlock]

..

ok then how i can solve?

the thing u said the formula

is not really working

since u always loop

back to the start

or where u start

[idle tusk]

bruv

[velvet hemlock]

which ios uselss

[idle tusk]

it literally isn't

[velvet hemlock]

is*

it is?

[idle tusk]

can you show where you get stuck

put some effort, pls

[velvet hemlock]

wtf ur talking about haha

will send you sec

[idle tusk]

yep

,rccw

[idle tusk]

now

as per the $P(A \cap B^c)$

[wraith dagger]

artemetra

[velvet hemlock]

bro this is like to do P(A(A cap B))..

\

not helping

[idle tusk]

oh wait i get it now

the venn diagram for this is very helpful

[velvet hemlock]

ik..

im using it.. tho

[idle tusk]

this is equivalent to $P(A \cup B) - P(A \cap B)$

[wraith dagger]

artemetra

[velvet hemlock]

ye or just like this lol

[idle tusk]

won't help

[velvet hemlock]

its the same

[idle tusk]

do you see how that's true?

[velvet hemlock]

bro this is the same

as the pic

i send you

u just rewrite in another way

[idle tusk]

yes

and that's helpful

[velvet hemlock]

but wait no lol its wrong

must be like this first:

(AcupB)cup(AcapB)

then u use P on that

[idle tusk]

yes

but those are independent

[velvet hemlock]

ye u will loop l0l

[idle tusk]

bruv

[velvet hemlock]

im srs

try iot

it

i already did that

i swear

[idle tusk]

you just use my formula here

and you get to the result

that's it

[velvet hemlock]

lets see

what is :
(AcapB)cup(AcupB)

this will never works tho

[idle tusk]

i did it

let me rewrite it in latex

[velvet hemlock]

dont show me ur solution

lol

useless tho

i told u already

i got the solution

[idle tusk]

ok

[velvet hemlock]

can u answer this pls

[velvet hemlock]

..

[idle tusk]

$(A\cap B)\cup (A\cup B)$

[wraith dagger]

artemetra

[velvet hemlock]

ye?

no

in the middle

its cap

[idle tusk]

$=A\cap B$

[velvet hemlock]

no

[wraith dagger]

artemetra

[velvet hemlock]

its not A

no

wrong

use venn

there is no way

[idle tusk]

u sure?

$(A\cap B)\cap (A\cup B) = A \cap B$

[velvet hemlock]

yes

im 100% sure

101% even

[wraith dagger]

artemetra

[idle tusk]

yep

[velvet hemlock]

no wrong

lol

[idle tusk]

what is it, then

[velvet hemlock]

oh

nvm

[idle tusk]

lmao

[velvet hemlock]

lol

[velvet hemlock]

can you explain how u got to this lol

[idle tusk]

venn

[velvet hemlock]

how haha

this is not equal to this

[velvet hemlock]

...

[idle tusk]

this is what your original question is asking for

[velvet hemlock]

but b^c

[idle tusk]

right?

[velvet hemlock]

B^c

Ye

i agree

[idle tusk]

great

so

that's the same as

[velvet hemlock]

how this is equal to

B^c cap A

[idle tusk]

minus

[velvet hemlock]

?

what is this

this is A b

[idle tusk]

that

[velvet hemlock]

What is B^c ..

[idle tusk]

what are you talking about

[velvet hemlock]

its what not in B

[idle tusk]

this is a different part of the question

forget that

[velvet hemlock]

??

what bro

im lost now

[idle tusk]

bro

[idle tusk]

this area

[idle tusk]

is this (union)

[idle tusk]

minus that (intersection)

so $P(\text{whatever was in your question}) = P(A \cup B) - P(A \cap B)$

[wraith dagger]

artemetra

[idle tusk]

do you agree?

[velvet hemlock]

i send pic

its loading

there is no way its true

[idle tusk]

can't read

[velvet hemlock]

send another one loading sec

[idle tusk]

that's false

and i don't see where i said that

[velvet hemlock]

then how ....

thats what u said

go back

and read

[idle tusk]

i hope someone else will be able to help you cuz i gtg now

[velvet hemlock]

lol u can go

[idle tusk]

but forget everything i said

[velvet hemlock]

all g

[idle tusk]

except that

[velvet hemlock]

ye cause its false

what u have done is false tho

also using venn is not formally eanough

enough

[humble karma]

That doesn't hold. What does hold is B' n A = A \ (A n B), so in particular P( B' n A) = P(A) - P(A n B).

[velvet hemlock]

what?

ik bro..

omg

lemme tag it sec i said that

[velvet hemlock]

.....

[velvet hemlock]

what u said in the end is not formally

enough

how and why

btw dont send a solution tho

because i have one

just saying

<@&286206848099549185>

[humble karma]

It's because P(A n B) + P(A n B') = P(A)., in particular from the fact that A = (A n B') u (A n B) which is a union of disjoint sets

[velvet hemlock]

prove what ur saying

[humble karma]

Those are basic facts. Try it yourself, some x can't be in both AnB' and AnB

Because x can't be in both B and B'

[velvet hemlock]

ummm

make sense

now

but wait idk

still lol..

how u go P(A n B) + P(A n B') = P(A).

lol

[humble karma]

Take the probability of the last set equation I gave you

[velvet hemlock]

ik that A n B cup A n B ' = a

A

[humble karma]

Since it's a union of disjoint sets you can take just the sum of the probabilities

[warped coyote]

If A and B or A and not B, ie A if B or not B is just A

That’s all @humble karma is saying

[velvet hemlock]

so :

A = (AnB')u(A'nB)

no still wrong lol

[warped coyote]

A=(AnB)u(AnB’)

Draw the venn diagram

[velvet hemlock]

venn is not formally enough tho

cant use it

wants 100% formal

[warped coyote]

Not in your proof, but so you can understand what you’re trying to do

[velvet hemlock]

ik i can prove this by proving if x in A then etc..

[humble karma]

Prove it yourself then

[velvet hemlock]

.close

[cedar kiln]

Channel closed

Closed by @velvet hemlock

Use .reopen if this was a mistake.

[humble karma]

.reopen

[cedar kiln]

✅

[humble karma]

For record, the reason why this took so long is that we can't read your mind and know what you're taking as truth or not. Usually, you don't build the whole theory from the ground up when proving something like that.

.close

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[velvet hemlock]

thank you @humble karma

• @idle tusk

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

can osmebody help me with tyhis pls:

pls somebody

i need help

[dire thorn]

ur supposed to wait 15 min before pinging helpers

[crimson sedge]

oh mb

[dire thorn]

What part do you need help with

[crimson sedge]

all of them

[dire thorn]

well the answers to a is in b and the answer to b is in c

[crimson sedge]

wdyyyym

wdym*

[dire thorn]

4/5 of the answers are given

[crimson sedge]

elaborate pls

[dire thorn]

empty

filled

2nd slot empty

but its filled in part d

to find the final one write out the formula for mean and fill in everything you can

[crimson sedge]

ok

so its...

8

ok now how can i do these

1. The sum of 3 consecutive numbers is 63. What is the largest number?

2. Sulaymaan scored 95 in physics, 90 in math, and a score in chemistry that is half his score in geography. The mean of the 4 scores was 80. What is his score in Geography?

3. Alice scored a total of 250 points in mathematics, physics and English. She scored 10 more marks in mathematics than English, and her physics score was twice her English score. What were her 3 scores?

4. There are bicycles (2 wheels) and cars (4 wheels) in a parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and bicycles are there?

5. The difference of two numbers is 20 and their sum is 70. Find the mean of the two numbers. (There is a way to do this question without finding the two numbers)

6. James has rent due soon. His rent is $10 more than half of his earnings. The mean of his rent and his earnings is $5000. How much is his rent and earnings?

[dire thorn]

it is not 8 that would give a mean of 7.4 but you need a mean of 8

[crimson sedge]

sorry

i meant

11

mb

[dire thorn]

ye

[crimson sedge]

can you help me with the questions above too

[dire thorn]

Yes lets start with 1

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[crimson sedge]

1. The sum of 3 consecutive numbers is 63. What is the largest number?

2. Sulaymaan scored 95 in physics, 90 in math, and a score in chemistry that is half his score in geography. The mean of the 4 scores was 80. What is his score in Geography?

3. Alice scored a total of 250 points in mathematics, physics and English. She scored 10 more marks in mathematics than English, and her physics score was twice her English score. What were her 3 scores?

4. There are bicycles (2 wheels) and cars (4 wheels) in a parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and bicycles are there?

5. The difference of two numbers is 20 and their sum is 70. Find the mean of the two numbers. (There is a way to do this question without finding the two numbers)

6. James has rent due soon. His rent is $10 more than half of his earnings. The mean of his rent and his earnings is $5000. How much is his rent and earnings?

[dire thorn]

????

[crimson sedge]

those ones

[dire thorn]

.close

[cedar kiln]

Channel closed

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[dire thorn]

I am confused

[crimson sedge]

i need help with those ones too\

[dire thorn]

Ok you need the sum of 3 consecutive numbers to equal 63

[crimson sedge]

k

[dire thorn]

Lets call these numbers a,b, and c

[crimson sedge]

ok

[dire thorn]

So a+b+c=63

How can we write b in terms of a (remember they are consecutive)

[crimson sedge]

a=b*2?

[dire thorn]

consecutive means each number is 1 greater than the previous

[crimson sedge]

oh yeah fogot they have to be consecutive

um

[north acorn]

Let's call the first number x. Then the next two consecutive numbers are x+1 and x+2. The sum of the three consecutive numbers is x + (x+1) + (x+2) = 63. Combining like terms, we get 3x + 3 = 63. Subtracting 3 from both sides gives 3x = 60. Dividing both sides by 3 gives x = 20. So, the three consecutive numbers are ||20, 21, and 22.||

but try solving to see if you can get

[cedar kiln]

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[lyric galleon]

Yo

[dire thorn]

<@&268886789983436800>

sub 1 second

[slate lintel]

i had just opened the thread when you pinged lol

[lyric galleon]

That’s horrible

Um

So my question

I think I’m traumatized let me recollect myself

[slate lintel]

sorry

[lyric galleon]

It’s not your fault but I have a question

The red is the answer and this is factoring

Can the two in parentheses be interchangeable? Or is this the fixed answer

[slate lintel]

you can swap them around yeah :)

[warped coyote]

Yes they can be switched around, and you can prove it by multiplying it each way!

[slate lintel]

since multiplication is commutative

$\gsq\cdot\psq = \psq\cdot\gsq$

[wraith dagger]

hayley

[lyric galleon]

I see. Thank you everyone!

[warped coyote]

Is this base latex?

[slate lintel]

nope, $\textcolor{green}{\blacksquare}$ for you

[wraith dagger]

hayley

[warped coyote]

Ouch okay thanks

[cedar kiln]

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[brave wedge]

wondering what im doing wrong here or if i could get a hint

[brave wedge]

if k=-1/x^2 dont we get x-x=0 which is where it would divide into two regions?

[cedar kiln]

@brave wedge Has your question been resolved?

[brave wedge]

.close

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[cedar kiln]

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[static fern]

idk where to start, perhaps a certain identity would be useful here??

[cedar kiln]

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[eternal seal]

isnt the probability like P(A|(1-C))P(B|(1-C))

[cedar kiln]

@eternal seal Has your question been resolved?

[cedar kiln]

@eternal seal Has your question been resolved?

[eternal seal]

https://tenor.com/view/mr-smith-mr-and-mrs-smith-angry-frustrated-brad-pitt-gif-11041739

.close

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[tidal plover]

Is this maths problem?

[sacred heath]

wtf

[coral jewel]

seems so

[sacred heath]

what grade are u bro

[tidal plover]

wut?

[coral jewel]

though after a bit of research, im wondering why 2 is correct instead of 1

[tidal plover]

university sister

[sacred heath]

ummmmmm

[coral jewel]

https://en.wikipedia.org/wiki/Malthusian_growth_model

[tidal plover]

which books has this kind of topics??

I meant I have not read such topics yet?

[coral jewel]

you could look into this page and see the references tab

it should cite some resources regarding Malthus equation

[tidal plover]

.close

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[astral sky]

Hi

[astral sky]

I have a question

I'm in 9th

and have no idea how to solve this question

[tropic oxide]

what is this, a picture for ants?

[astral sky]

sorry

[tropic oxide]

The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 km/h?

[astral sky]

Yes

[tropic oxide]

right

so you're saying you don't even know where to start?

[astral sky]

I mean we find the side and then the diagonal

but im getting something like square root of 0.08

after converting hectares to km square

[tropic oxide]

show your work?

might be something like an arithmetic error

but 8 ha = 0.08 km^2 is correct.

[astral sky]

and we find the square root right?

[tropic oxide]

yes, that'll give us the sidelength of the field in kilometers.

sqrt(0.08) can be simplified a bunch.

consider 0.08 = 8/100 = 2/25, for instance.

[astral sky]

so thats root 2 / 5

[crimson sedge]

=0.4

[opaque root]

No, 2/5 is 0.4, sqrt(2)/5 is not

[crimson sedge]

4/5?

i am a Chinese.

[opaque root]

Where do you get 4 from?

[crimson sedge]

emm...

i maybe need you speak Chinese.

emm..

sorry,i maybe can't know meaning of a lot of sentences

emm...

[crimson sedge]

2*2

[cedar kiln]

@astral sky Has your question been resolved?

[tropic oxide]

that's the sidelength of the field yes

sorry for the delay on my part.

@crimson sedge for your knowledge: sqrt means square root (平方根)

[crimson sedge]

thank you.

[tropic oxide]

(i don't speak chinese, i only looked up the word in wiktionary)

[crimson sedge]

@tropic oxidei will add you friend.

[tropic oxide]

no thanks.

[crimson sedge]

ok...

[cedar kiln]

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[polar hearth]

how to continue from (ii)

[cedar kiln]

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[cosmic quartz]

There are a number of students on a school bus. At the first stop, 3 students get off and 7 others get on. Nina and her two sisters get off at the second stop. The remaining 12 students get off at the third stop. How many students were on the bus for the first stop?

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14

[livid hound]

!status

[cedar kiln]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[cosmic quartz]

2

[cedar kiln]

@cosmic quartz Has your question been resolved?

[cosmic quartz]

<@&286206848099549185>

[halcyon swan]

Yes

What is your doubt?

[livid hound]

if you have work, show it

[cosmic quartz]

I got 3+7+3-12 but idk if its correct

[halcyon swan]

3+-7 = 10

10 + 3 =13 not 12

[cosmic quartz]

no wait

[halcyon swan]

Which question are you talking about

[cosmic quartz]

.

i got

3 + 3 + 5

[pallid trout]

total 11 (-3+7-3-12) , (negative shows get off students)

[halcyon swan]

Initial: At the beginning, an unknown number of students (x) are on the bus.
First Stop: 3 students get off and 7 get on, so the number of students becomes x - 3 + 7 = x + 4.
Second Stop: Nina and her two sisters get off, reducing the number to x + 4 - 3 = x + 1.
Third Stop: All remaining students (x + 1) get off, resulting in an empty bus.
Now, we know the final number of students is 12 (from the third stop). Therefore, we can set up an equation:

x + 1 = 12

Solving for x, we get:

x = 11

Therefore, there were (A) 11 students on the bus for the first stop.

I hope I am right

[cosmic quartz]

yes i understand

i also got some other question

How many of the following four statements about natural numbers are true?

(1) Of three consecutive odd numbers, exactly two are prime

(2) Of three consecutive odd numbers, at least two are prime

(3) Of three consecutive odd numbers, at least one is prime

(4) Of three consecutive odd numbers, at least one is not prime.

[pallid trout]

3, 4,.

[cosmic quartz]

what

.close

[cedar kiln]

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[pearl bolt]

the question is to prove this is true for any 3 dimensional vectors a,b,c ... i have been staring at it for a while now, not sure what to do

[pearl bolt]

@velvet ice please post in another help channel so we both get an answer and noone gets skipped

[velvet ice]

ok sorry i don't know how it works

[pearl bolt]

turns out i have to prove something called the grassmann identity, im gonna google how to prove it

[pallid trout]

solve separately, vector and scaler multiplication formulae , it is prove of vector and scaler equality (cross product and dot product)

[pearl bolt]

with the matrices and all that?

@pallid trout

[pallid trout]

you can solve via matrixes, or generally ,

[cedar kiln]

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[cedar kiln]

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[drowsy hare]

can someone help me with this?

[cedar kiln]

@drowsy hare Has your question been resolved?

[crimson sedge]

Discord is having media issues. Please transcribe your issue in text so we can see your problem.

[drowsy hare]

I do not know hwo to use the latex bot

but ill try words

A double integral with an outer integral, bound from 4 to 1 and an inner integral, bound from sqrt(x) to 1, with an integrand of (x^2 + y^2)dxdy

[drowsy hare]

The questions asks me to sketch the region of integration and calculate the integral, and then to reverse its order of integration and calculate it again.

[cedar kiln]

@drowsy hare Has your question been resolved?

[drowsy hare]

<@&286206848099549185>

[sweet hound]

Can you send the question again the discord problem is solved now

[sweet hound]

I can't send what i did

Idk why

But basically you need to switch the dx and the dy for the first question

And the second one you just need to calculate this double integral as it is

[drowsy hare]

this would not give an actual answer tho would it?

cuz then it would have variables in its solution

[sweet hound]

Yeah just x but i think it is what they want you to show , maybe since they are equal you can solve for x

[drowsy hare]

hmmm

i could see that maybe

but i have the teachers solution here

[sweet hound]

So why you posted this question if you have already the answer lol

[drowsy hare]

cuz i dont understand the solution

[sweet hound]

I don't understand it either

Xd

[drowsy hare]

yeah, thats what i mean

[sweet hound]

You can just ask your teacher for more clarification of his answer