help-27

[zealous bronze]

19th q

.

This is the diagram I drew

[silk panther]

<@&268886789983436800>

[velvet geode]

<@&268886789983436800>

[velvet geode]

wherever I go
I see him

[zealous bronze]

😔

If I give the steps am at

[velvet geode]

must be the wind

[zealous bronze]

From there there can help ?

[velvet geode]

wdym

[lusty mantle]

hey so like dont post it again?

[velvet geode]

mb gng i thought im giving y'all proof 🥀

what if he deleted it nd i got in trouble

[zealous bronze]

1 min

[velvet geode]

alrr, time starts now 🕰️

[zealous bronze]

I jst need to prove hae is equal to gcf right

[velvet geode]

whts hae nd gcf

[zealous bronze]

Before that I need to prove angle hae = angle gcf

[zealous bronze]

A side

[velvet geode]

ah

(still didnt get it)

[mossy flicker]

It's actually pretty easy

Should I walk you through?

[zealous bronze]

Yeah pls

[mossy flicker]

What you have drawn is brilliant

This

[zealous bronze]

Ehehe thanks

[celest lantern]

I believe you have to prove using vectors here.

[zealous bronze]

From here i needed to make angle hae and gcf as equal

[mossy flicker]

Name AO as some x

Then EB will also be x

True ?

@zealous bronze

[zealous bronze]

How

We jst know its parallel

Nt equal r8ght ?

[mossy flicker]

Parallelogram property

[zealous bronze]

Opp sides are equal ?

[mossy flicker]

Parallel and equal yeah bro

[zealous bronze]

Oh fine

Then next step

[mossy flicker]

Then name OB as some t

So AE also t

[zealous bronze]

Ok

[mossy flicker]

Now do same for all these

[zealous bronze]

K did

Next?

[mossy flicker]

Then see length of EF

Which is EB + BF

[zealous bronze]

2x

[mossy flicker]

What no wait

Not 2x

Not necessarily

[zealous bronze]

How

[zealous bronze]

I mean x +t ? Is right t ?

[mossy flicker]

You'll be getting something like this

[mossy flicker]

Analyse the picture once

[zealous bronze]

Bf is z?

[zealous bronze]

Ya did

[mossy flicker]

As I've assumed OC as some z

So BF in parallelogram

[zealous bronze]

Opp sides are equal here

[mossy flicker]

Yeah

[zealous bronze]

But still how to prove its parallel?

[mossy flicker]

Bf is parallel to oc

Oc is parallel to DG

So BF parallel to DG

[zealous bronze]

Yeah

[mossy flicker]

Same for all

And see EH

K+t

See FG

K+t

[zealous bronze]

Yeahhh

[mossy flicker]

Equal and parallel

[zealous bronze]

Yaaa

But still ef is nt parallel to hg right ?

[mossy flicker]

It is

[zealous bronze]

We still don't even know that ef is a line or not

Before that we need to prove angle b is 180 degreee

[mossy flicker]

I would say that by symmetry it would be perfect

[zealous bronze]

How ? It will be like we r assuming it

Nt proving

I have a way to prove but a stuck at middle

Can u help me in that ?

[mossy flicker]

Yeah br fs

Where stuck

[zealous bronze]

Here u tried

To prove angle is equal to 180

[mossy flicker]

Lemme try too give me 2 mins

[zealous bronze]

I jst wanted to prove angle hae is equal to gcf

If those angles are equal
Then we will consider triangle hae and gcf by sas we can prove that it is congruent .

Then we can say he =gf by cpct

[mossy flicker]

Hmm

Wait you've written BAO

= 180 - AOB

[zealous bronze]

Ai says both angle are equal .
If u say last line and last 3rd line u can say Its equal ?

[zealous bronze]

Hao

As adjacent angles sum up to 180 degree

[mossy flicker]

2md line

[zealous bronze]

Yes It is correct only ?

When u add 2nd and 3rd line u will get angle a right ?

[mossy flicker]

AOB is triangle

[zealous bronze]

Yes . How does this connect here ?

[mossy flicker]

Sum of two angles is not equal to 180

In triangle

Sum of all three is 180

So BAO is not equal to 180 -AOB no cap on god

[zealous bronze]

It is nt bao

It is hao

[mossy flicker]

Hao in both lines ?

[zealous bronze]

It is hao
I think u read I wrong

[zealous bronze]

Yaaa

[zealous bronze]

May be photo wasn't clear sorry

[zealous bronze]

Eao

In 2nd here

[mossy flicker]

Bruh

Ok wait I'll check rest

Han then

Ohk I get where you're confused

That AOD +AOB not necessarily 180

?

Ok done

@zealous bronze

[zealous bronze]

Yes it isn't

I didn't say taht it is 180

[zealous bronze]

I never mentioned that

[mossy flicker]

Have to use vectors

You know vectors?

I thought beacuse you're probably in 9th I'll use geometry only but

Not possibul I think not easily

[zealous bronze]

Am in 12th

But with geometry we can prove ittt

[mossy flicker]

Damn

Okay well I'm also in 12th

[zealous bronze]

How can we prove hae and gcf by angle equal

[mossy flicker]

Then it's probably vectors but let's try geometry just for a teeny tiny more time

[mossy flicker]

Just 2 mins please

[zealous bronze]

Sure

[zealous bronze]

From India?

[zealous bronze]

If u see last 3 lines then it might be bit easy to understand

[mossy flicker]

Yeah

[zealous bronze]

Preparing for which competitive exam ?

[mossy flicker]

Jee

But listen

Use mid point theroem

@zealous bronze

Now everything is sorted guys

[zealous bronze]

Mid point ?

[mossy flicker]

Mid point theroem class 10th

If you're from India

[zealous bronze]

Didn't understand

[zealous bronze]

Yes

[mossy flicker]

Look look

[mossy flicker]

See this

Once

[zealous bronze]

I waned this

[mossy flicker]

No need of extra mental pressure

I mean straight up prive

That hae is equal to gcf

Just not by angles

@zealous bronze ?

[devout snow]

@zealous bronze Has your question been resolved?

[zealous bronze]

2 min

[mossy flicker]

Ok

[zealous bronze]

For now forget abt wt ever u said

If this is the case . Then are both angle equal am asking .
Ai says yes but when it is explaining logic am nt able to understand

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[zealous bronze]

.reopen

[devout snow]

✅ Original question: #help-27 message

[zealous bronze]

If angle hae is 360 - ( aod + aob)
And gcf is aod + aob
Then is hae and gcf equal ?

<@&286206848099549185>

[mossy flicker]

Ok I will forget and try understanding your reasoning

Ai says yes

[zealous bronze]

Yes pls 😌

[zealous bronze]

Ya

Shall share its response ?

[mossy flicker]

Yeah fs you can share personally the ss

No AI here

[zealous bronze]

Shall I t3xt u from another discord id then?

[hasty cargo]

???

Wtf

No ai allowed here

[zealous bronze]

Ok

[mossy flicker]

Yes

[mossy flicker]

.

[mossy flicker]

You can from this one or any u like

[zealous bronze]

Once

Understand this perspective

[zealous bronze]

I sent u friend request

[devout snow]

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.

@zealous bronze Has your question been resolved?

[hasty cargo]

Err what

Is moving to dms and abandoning the help channel allowed

[hollow bolt]

another reason behind this

[hasty cargo]

And to prevent any misinformation to be spread

Or even worse shit that im not going to talk about

[hollow bolt]

lol

[devout snow]

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[devout wave]

can some one help me ques 11.3 . When i try to derivative i got x^-1/2 then 0 ^ -1/2 is not exist

[jagged harbor]

show your work?

it looks like you're trying to center it at 0, not 4 as it says to

[devout wave]

am i right , do i need to turn back to x ?

-(2n+1)/2

@jagged harbor

.close

[devout snow]

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[devout snow]

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[trail cairn]

Can I use this

[trail cairn]

@vestal surge

[vestal surge]

okay

find y in terms of x

using the perimeter given

[trail cairn]

Hmm

Done

[trail cairn]

Put in area now?

[vestal surge]

substitute and find area in terms of X

[trail cairn]

Now?

[vestal surge]

ye

good enough

but instead of finding common denominator

id say

split into 3 terms

[trail cairn]

Ok

[vestal surge]

$\text{Area} = 5x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2$

[woven radish]

Velt

[trail cairn]

Got it

[vestal surge]

got this?

[trail cairn]

Yea

[vestal surge]

take x common

[trail cairn]

Writtten

Ok

Got

No need to LaTeX everytime man.

Ik the soln

What now?

[vestal surge]

then you find common denominator for 3/2x^2 and sqrt(3)/4 x^2

[trail cairn]

Ok

[vestal surge]

$\text{Area} = x \cdot \left[ 5 - \left( \frac{6 - \sqrt{3}}{4} \right) x \right]$

[woven radish]

Velt

[vestal surge]

do we have this

[trail cairn]

Yup

[trail cairn]

Damn it's clean.

Now?

[vestal surge]

so now how can we manipulate it for AM-GM condition

any ideas

?

[trail cairn]

Idk tht

Leme google

[vestal surge]

..

do you know what am-gm inequality is?

[trail cairn]

Now I do

[vestal surge]

....

[trail cairn]

Gemini-ed

[vestal surge]

........

[trail cairn]

AM gm

AM gm only

[night forum]

I think it might have been easier if he did some substitution earlier

Like

Instead of dealing with 3x and 2y

[vestal surge]

lemme latex what i thought, it seems sigma

[night forum]

He can just call it a and b

And replace 3x and 2y wherever

[trail cairn]

🙏 🙏 Tell me the way by AM GM pls

Some1

[vestal surge]

$\text{Area} = \frac{4}{6 - \sqrt{3}} \cdot \left[ \left( \frac{6 - \sqrt{3}}{4} \right) x \right] \cdot \left[ 5 - \left( \frac{6 - \sqrt{3}}{4} \right) x \right]$

[woven radish]

Velt

[vestal surge]

perfectly balanced

[trail cairn]

RHS of Am gm

?

[vestal surge]

beautiful am-gm

[trail cairn]

Its RHS?

[vestal surge]

no

its the area

i multiplied it by 1

[trail cairn]

Bruh how we use AM GM

Here

[vestal surge]

and wrote 1 as $= \frac{4}{6 - \sqrt{3}} \cdot \left( \frac{6 - \sqrt{3}}{4} \right)$

[woven radish]

Velt

[vestal surge]

ok

[trail cairn]

Fast pls

What now

[trail cairn]

Aft this

[vestal surge]

because we say A = $\left( \frac{6 - \sqrt{3}}{4} \right) x$ and B = $5 - \left( \frac{6 - \sqrt{3}}{4} \right) x$

[woven radish]

Velt

[vestal surge]

chill latex takes time

but makes it beautiful

[trail cairn]

Oh np

[trail cairn]

👍

[vestal surge]

now adding them cancels out X completely so you get the maximum product which happens only when $A = B$

[woven radish]

Velt

[vestal surge]

so you simply do $\left( \frac{6 - \sqrt{3}}{4} \right) x$ = $5 - \left( \frac{6 - \sqrt{3}}{4} \right) x$

[trail cairn]

Hmm

[woven radish]

Velt

[trail cairn]

Why A=Band not A>B?

[night forum]

Am>gm except when a=b

[trail cairn]

Oooh

[vestal surge]

test out values

its a much better way to learn

[trail cairn]

Yeh

[vestal surge]

try 3 results

[trail cairn]

I got it

[vestal surge]

A = 5, B = 5
A = 6, B = 4
A = 9, B = 1

[night forum]

You could also try deriving this inequality urself

[vestal surge]

as the gap increases the product decreases

so thats why max product is when A = B

[trail cairn]

Alr

[trail cairn]

Doing

[vestal surge]

keep the answer in terms of the irrational

no need to insert value and try to simplify it further

its just more dirty division if you do

[vestal surge]

yeah

[vestal surge]

thats your ans

[trail cairn]

Learned smthing new tday

I'll study in detail am gm later

K gn

.close

[devout snow]

Channel closed

Closed by @trail cairn

Use .reopen if this was a mistake.

[vestal surge]

sure

[opaque forum]

bro grinding human eye and the colorful world in the background

[trail cairn]

Lol

Its physics

[trail cairn]

Y not friend?

[vestal surge]

sure y not

[trail cairn]

Iv sent

Many days bfore so....

[vestal surge]

i sent

[trail cairn]

👍 🙂

[opaque forum]

I know I'm in CBSE too

[devout snow]

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[mystic scarab]

Is the answer 0?

[wheat pawn]

is there any reason why you cannot simply calculate the roots?

[mystic scarab]

Actually yeah I think it is doable with such little numbers

Like there's just a power of 3 and a second degree equation

[desert tree]

Well you can simplify: alpha^3 = 3alpha-alpha^2 and then alpha^2 = 3 - alpha, and the same process for beta

So that everything has power 1 and becomes much easier to do calculations with

[mystic scarab]

Yes it turns out you get a nice expression that can be simplified using Vieta in the end

[devout snow]

@wary bison Has your question been resolved?

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[severe plaza]

Heeeeelp heeeellp absolute values

[severe plaza]

Heeelp me please

[neon folio]

fih

[uncut crow]

[severe plaza]

This is 100% incorrect

[neon folio]

,rccw

[severe plaza]

https://tenor.com/view/spongebob-spongebob-meme-spongebob-squarepants-spongebob-explode-spongebob-explosion-gif-22938834

[neon folio]

which one

[severe plaza]

8m

9

10

Oops

AND EVEN MORE ON THE BACK

[neon folio]

3 * absx = 15

this one?

[severe plaza]

Yes

[neon folio]

ok

try dividing both sides by 3

[severe plaza]

Oh

WOW THE ABSOLITE VALUE OF X=5

[neon folio]

yes

[severe plaza]

What if its within the lines?

[neon folio]

what lines

[severe plaza]

Do i just flip the subtraction symbol

[neon folio]

well

you split it up into cases

either the thing iside the abs value is 7

or

its -7

correct?

[severe plaza]

Oh yeah

Correct

[neon folio]

right

so 2x-1 = 7

and

2x-1 = -7

now you solve each one

and get your 2 solutions

[severe plaza]

Wait i thought an absolute value cant be equal to a negative value

https://tenor.com/view/spongebob-spongebob-meme-spongebob-squarepants-spongebob-explode-spongebob-explosion-gif-22938834

[neon folio]

the inside of the abs value can be negative

abs(-7) = 7

[severe plaza]

Oh right

Im solving x

[neon folio]

right

[severe plaza]

Well

2x-1=-7 is incorrect

X=-3

So x=4 only

[neon folio]

wdym

why isnt x=-3 valid

[severe plaza]

because when I input it into the equation

[severe plaza]

2(-3) -1 = -7

[neon folio]

yes

[severe plaza]

-9 - 1 = -7

-10 = -7

https://tenor.com/view/spongebob-spongebob-meme-spongebob-squarepants-spongebob-explode-spongebob-explosion-gif-22938834

[neon folio]

2*-3 = -6

[severe plaza]

Oh

[neon folio]

-6 -1 = -7

[severe plaza]

OOPS

Ok now both are

Good

[neon folio]

[severe plaza]

x=4 and x=-3

Now

Let me attempt this first

And ill update you

In 5 mins

[neon folio]

ok

[severe plaza]

3. x=9 and x=3 although I got |-3| =3, thats still correct I think

Um

[neon folio]

es

yes

its correct

[severe plaza]

Well

x=-3 is incorrect

In 4.

But x=-1 is correct

[severe plaza]

|-3 + 4 | = 5 + 2(-3)
| 1 | = 5 - 6
| 1 | = -1

[neon folio]

yes its incorrect

[severe plaza]

Erm

I got a radical

Im insure if thats allowed

Sorry if I sound like an idiot

I missed the whole unit

https://tenor.com/view/patrick-drooling-patrick-star-spongebob-movie-drooling-slime-gif-525340425450564005

[neon folio]

you should get integer solutions

show me your work

[devout snow]

@severe plaza Has your question been resolved?

[severe plaza]

Oops

Um

Wait

Imm @ you in an hour or 2

[neon folio]

imma be asleep by then

im sure someone else can help though

[burnt lark]

u can @ me ill be around

[lunar harbor]

.pin

[devout snow]

@severe plaza Has your question been resolved?

[devout snow]

@severe plaza Has your question been resolved?

[burnt lark]

are you still working on them? im here if u need

[cursive crater]

Woah 1 hour 17 minutes answering

[devout snow]

@severe plaza Has your question been resolved?

[severe plaza]

my bad I probably wont get to it today

Ill open a new channel tomorrow

[devout snow]

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[serene seal]

Hey! I'm trying to solve this integral using u substitution, and I'm a little confused how we got from u to the integral on the bottom. Does anyone have any ideas? Thanks!

[safe knoll]

rearrange the du term so u have dx = . . .

substitute that and u would get smthing like
$$\int \frac{1}{x^2} \dd u$$

[woven radish]

JustToPro

[safe knoll]

can find the x^2 term using ur u-sub and then substitute again

[serene seal]

Interesting, why does this work? It's a little unintuitive 😅

[safe knoll]

why wouldnt it

u sub is just used to convert a integral from x variable to u variable , in hopes of simplification

tho personally , i wouldnt use that substitution

[serene seal]

Is there something better you would recommend?

[safe knoll]

ye

x = 2sec theta

[serene seal]

how would that work?

[safe knoll]

dx would become 2 sec theta tan theta

subbing back in
$$ \int \frac{\dd \theta}{2 \sec \theta \cdot \sqrt {2^2 \sec ^2 \theta - 4}} \cdot 2 \sec \theta \cdot \tan \theta$$

[woven radish]

JustToPro

[safe knoll]

and then just basic simplification from here

whenever we have integral with sqrt {a^2 - x^2} or sqrt {x^2 - a^2} , i like to use trig sub

[serene seal]

ooh ok then

I should prolly look into that then

[safe knoll]

for the 1st , its x = asin theta and the 2nd one is x = asectheta

[safe knoll]

oh u havent learnt trig subs?

[serene seal]

we have, I'm just behind

[safe knoll]

oh rip

would suggest to watch a video on trig sub , its kinda nice

[serene seal]

ty for the help!

[safe knoll]

np

[serene seal]

ill look into that fs

thank you!

.close

[devout snow]

Channel closed

Closed by @serene seal

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[safe knoll]

tho is this from the book or smthing?

dont know why they didnt use trig sub

[serene seal]

yea it's partway through the solution of a bigger integral in the book

its in the u sub section

[safe knoll]

oh , hmm

i wouldve just waited till after trig sub to do smthing like that integral

but whatever :D

[serene seal]

🤷

[young spade]

Orrr, you can fall into the dark side and compute the partial fraction with imaginary roots.

[serene seal]

How does that work?

[young spade]

Do you understand both concepts separately?

[serene seal]

I've dealt with imaginary roots before, if that's what you mean

[young spade]

And what about partial fraction decomposition?

[serene seal]

conceptually yes, it's been a while though

were working on it in class rn actually

[young spade]

Well, you usually use it when you have real roots, but there isnt anything actually stopping you from using it w/ complex numbers

[serene seal]

don't the complex numbers get in the way though? I'm a little confused

[safe knoll]

crazy recommendation

tho if the guy understands , ig its fine

[young spade]

Quite literally just use the partial fraction, compute the antiderivative of 1/(x+a), which is ln(x+a)

And fun enough, once you evaluate what those roots actually are, you get back the correct answer too. -> entirely contained in the reals

[serene seal]

huh

interesting

ill have to look into that

is there a practical reason why someone would do this though, other than fun?

[young spade]

If you dont remember how to do trig subs?

or which one, for that

[serene seal]

makes sense

thanks for the idea!

[young spade]

np, good luck

[devout snow]

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[proper otter]

Can someone help me with finishing this up

[last parrot]

The beak can be applied the modulus function

[proper otter]

huh

[last parrot]

y=|x|

[summer summit]

i'm unfortunately busy right now but that face looks extremely funny lol

[proper otter]

this is what we can use

[last parrot]

Use number 7 for lower beak

[proper otter]

isnt absolute for pointy shi

[last parrot]

It is

Try draw it out

[proper otter]

beks smooth

bet

[devout snow]

@proper otter Has your question been resolved?

[devout snow]

@proper otter Has your question been resolved?

[proper otter]

.close

[devout snow]

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[proper otter]

.reopen

[devout snow]

✅ Original question: #help-27 message

[proper otter]

what do i do here

what function

[last parrot]

y=|x|

and a linear function

[proper otter]

y=mx+b

ax^{2}+bx+c

im using this equation for the green

but how can i make it face red

[devout snow]

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[last parrot]

I have a quick question, if I have matrix A, says I want to find the basis of row(A) and col(A), then why is taking the basis row(A) we take from A's RREF form, but col(A) is taking from it's original form?

[vital edge]

You can also take the RCEF for col(A)

[last parrot]

Wait I can use rows from RREF as basis for A because it is equivalent right

But columns changed and not equivalent

[vital edge]

basis for A doesnt make sense. Do you mean basis for row(A)

[vital edge]

Yes, but you could do RCEF for that

[last parrot]

Yea I meant basis row (A)

[vital edge]

Which is basically doing RREF on the transpose and then transposing it back

(or just using elementary column operations)

[last parrot]

I have never tried RCEF before, but i think its the same for RREF but transpose it

[vital edge]

Yes, transpose, do RREF and transpose back

[last parrot]

Alright thanks, it solved my doubt

.close

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[silk panther]

hello, help me check answer

[silk panther]

thanks in advance

[devout snow]

@silk panther Has your question been resolved?

[silk panther]

<@&286206848099549185>

[runic prawn]

this is not a valid proof

every circle is uncountable, yes

and P is countable

hint: consider the lengths of elements of P

[silk panther]

the lengths?

[runic prawn]

yes, like length of a vector

[silk panther]

err

lemme see

[runic prawn]

a circle of radius r is the set of points of length r

[silk panther]

yes

but P is not a circle

lets say P = {(a,b) : a,b in R}

so since P is countable, lengths of a and b is some number c

is it like that

[runic prawn]

Let Q = {||v||: v in P}

Q is a countable subset of [0, infty)

[0, infty) is uncountable

so we can find some r in (0, infty)\Q

[silk panther]

ye

[runic prawn]

the circle or radius r is what we desire

for every point v in C, ||v|| = r but no point in P has length r so v is not in P

[silk panther]

yes

but ngl I kept asking questions here and even if I understand the solution, I actually wanna know how you guys thought abt the steps

like thought process

so like for this question, my first thought is to think P is all those points forming like a circle, and since it's countable, there is always a "bigger circle" covering P

[runic prawn]

imagine the points on R^2

[silk panther]

thats how I think

[runic prawn]

ok but what if P is unbounded

[silk panther]

ohh then I can also imagine a circle between the distance of two points in P

man idk my thought process is kinda lame rn

[wicked rover]

what if P has no isolated points, eg Q^2

[silk panther]

well either ways it's still countable, can't cover the whole R^2, so there must exist one point not in P

[wicked rover]

my point is in that case its very hard to imagine a circle between points

[silk panther]

ok

brb

back

alr so here is my thought: Define a set for every point in R^2 that forms a circle

say that set is uncountable

then conclude there exists such point

[wicked rover]

i dont even understand what ur claim is

btw i typed smth while u were gone

[silk panther]

dang

[wicked rover]

@silk panther this is not my solution but i could imagine how we could discover it

we want the circle to avoid P. equivalently we want the radius to avoid lengths of its elements, so we remove all the lengths from (0,infty). the result is nonempty bc (0,infty) is uncountable and P is countable. nonempty result means a suitable radius exists

[silk panther]

actually what is ||v||

||v| |

[wicked rover]

length of v

[silk panther]

ok

wait why do we need length of v

[wicked rover]

we want the circle to avoid P. equivalently we want the radius to avoid lengths of elements of P

[silk panther]

im confused by lengths of elements of P, is it like the longest distance between two points

[wicked rover]

length means distance from origin

[silk panther]

ok

I see so we want to get a length that is longer than the ones in Q

[wicked rover]

nowhere does the solution say longer

[silk panther]

what do we want exactly? Do we want to find a length that's not in Q?

[wicked rover]

yes thats why we remove Q from (0,infty)

[silk panther]

ok i got the idea

[wicked rover]

awesome

hope its all clear, gotta head out now, have a good one

[silk panther]

thx cya

.solved

[devout snow]

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[swift badge]

I’m kind of confused with this question

[swift badge]

Coz I’d assume it’d be like this

But it’s wrong

[sacred sedge]

well it is bigger than the lower bound

[swift badge]

ohhh

u are right

[sacred sedge]

and smaller than the upper bound

[swift badge]

1 would be greater than x

i mean

yeah

ohh okl i see now

thanks

😄

[sacred sedge]

[swift badge]

wait actually how would u do this

[sacred sedge]

the most outer integral says

0 ≤ x ≤ 1

[swift badge]

wait wat

[sacred sedge]

?

[swift badge]

i dont rlly understand

what am i supposed to do

[sacred sedge]

what are having trouble with?

[swift badge]

idk where to start

[sacred sedge]

is it still this question?

[swift badge]

yah

[sacred sedge]

look at the dz dy dx

so the inner most is z = ...

middle is y = ...

and x = ... for the outermost

[swift badge]

o

so now we're going through x first

wat do we do with this

[sacred sedge]

so x is the outermost integral

[swift badge]

yes

wait wat isnt x the inner one now

in this new one

[sacred sedge]

the order of integration in the question is different to what you gave

[swift badge]

they changed it from this one

part i and ii

[swift badge]

wait ik its swithced here

[sacred sedge]

oh mb

[swift badge]

yah

[sacred sedge]

I did not see that

[swift badge]

hahahaha i shuda been more clean

clear

[sacred sedge]

lets do i) since I think it is still not correct

[swift badge]

yah ik its switched

[sacred sedge]

alr

[swift badge]

i just couldnt be bothered taking another ss lol

wait ill do it now so its not confusing

[sacred sedge]

that's fine

so you have z = y, y = x^2 , x =1

[sacred sedge]

does these make sense

[swift badge]

yeah

[sacred sedge]

now when we change from dz dy dx to dx dz dy

we write out the bounds in terms of equations

like I did

z = y

[swift badge]

oh yea

[sacred sedge]

y = x^2

[swift badge]

ok

[sacred sedge]

x = 1

so from there

our inner most is x

and y =x^2 is our equation from before, so can you tell me x in terms of y

[swift badge]

y = 1

[sacred sedge]

not quite

we have a function y = x^2 from before

try to manipulate this

[swift badge]

y = sqrt(x)

[sacred sedge]

yup

wait

the other way

x = sqrt(y)

[swift badge]

oh

yah

[sacred sedge]

so we have our bound for inner most integral to be

0 ≤ x ≤ sqrt(y)

can you do that for the middle and outermost?

[swift badge]

can i have y in it both times

[sacred sedge]

oh ye

then it should be from 1 to sqrt(y)

[swift badge]

wait why

[sacred sedge]

since sqrt(y)≤x

from before

and x≥1

[swift badge]

whatever ill just lose 2 marks

on the test

...

[sacred sedge]

ok look

from before we had

0≤x≤1

0≤y≤x^2

[swift badge]

yeah

[sacred sedge]

to put it together

we have x ≤1 and sqrt(y)≤x

or it is sqrt(y) ≤ x ≤ 1

maybe this is easier to see the bounds

@swift badge

we can see for the other ones as well

[swift badge]

yag

[sacred sedge]

from before we had z ≤ y , z ≥ 0

so 0 ≤ z ≤ y

this is the same as that before

does that make sense

@swift badge

[devout snow]

@swift badge Has your question been resolved?

[devout snow]

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[devout snow]

Channel closed

Channel closed due to the original message being deleted.
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[sturdy fiber]

uhhh

okay

:|

[devout wave]

actually i try to seperate

i was struggling with n^2 * x^n

and i think i will intergrate twice

[sturdy fiber]

wait which part i opened the channel as soon as u deleted the message ;-;

[devout wave]

9.2

[sturdy fiber]

integrate?? show your working please

[devout wave]

wait

[sturdy fiber]

because we can just take sigma x^n, differentiate it, multiply both sides by x and differentiate again to get n squared and then multiply by x and we are done with the lhs

[warped fulcrum]

There is a closed form specifically for n²x^n

[devout wave]

what form i dont really know

[devout wave]

ok will try

[warped fulcrum]

Its x(1+x)/(1-x)^3

[devout snow]

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[devout wave]

i think we should intergrate the function if u derivative we will get n^ 4

[sturdy fiber]

its better for them to derive this answer rather than learn it and not understand

[sturdy fiber]

okay wait so

[warped fulcrum]

Fair enough, my bad

[sturdy fiber]

instead of this

you should take y= sigma x^n

[celest lantern]

Can you show the original question?

Ah wait, nevermind.

It was closed because of timeout.

[sturdy fiber]

yea

[sturdy fiber]

and then just follow this after taking y as i said

because already starting with n squared gives us n^4 which is irrelevant here

[devout wave]

yes i found the form

[sturdy fiber]

wrong, you multiplied by x in the RHS but divided by x in LHS

[devout wave]

yeah yeah my mistake

[sturdy fiber]

apart from that this is the correct way 👍

[devout wave]

im really sorry but is this the answer for Sum n^2 * x^n

@sturdy fiber

[sturdy fiber]

perfect! now we already know the series for sigma x^n which we can apply here!

[devout wave]

lovely work

thank u

.close

[devout snow]

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[devout wave]

.reopen

[devout snow]

✅ Original question: #help-27 message

[sturdy fiber]

yes?

[devout wave]

is this bottom up method . i mean in exam we do top down from n^2 *x^n to the answer

but now we do from answer to n^2 * x^n

[sturdy fiber]

oh

[devout wave]

note that i have sum n^2 * x ^n and try to find the answer. And what we are doing here is create something to = n^2 * x ^n, not n^2 * x ^n to = something

just a discussion

i get what u teach me

[sturdy fiber]

we will just reverse everything we have done so far, instead of differentiating, we integrate, and instead of multiplying both sides by x, we divide both sides by x

[pure stone]

<@&268886789983436800> advertising/possible scam

[devout wave]

this is my original idea =))

[sturdy fiber]

oh

welp

mb :p

[devout wave]

.close

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[ocean haven]

derivative is the slope of secant as the two points tend to move close to each other. so what is derivative at a point

[summer summit]

the derivative is the slope of the tangent line at that point

[ocean haven]

but it never speaks of the point

[celest lantern]

Wouldn't at that point it's already have the derivative formed from 2 points that is very small?

[uncut crow]

think of it like a limit. the secant lines “converge to a line” when the points get closer together

[ocean haven]

ah

[uncut crow]

that line they converge to is the tangent line

[ocean haven]

but would it be correct to say "at the point"

[uncut crow]

ambiguous question but i can probably say sure

[ocean haven]

mm ok

ty

.close

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[severe prairie]

Why is AM that?

[severe prairie]

Heres the og question

[lunar harbor]

any arbitrary point on the line can be represented by the parametric equations

You have $x=4-\lambda$, $y=2\lambda$, $z=1+3\lambda$, so you can let $M$ have those coordinates

[woven radish]

Civil Service Pigeon

[lunar harbor]

and ofc $\overrightarrow{AM}$ is just subtracting the position vector of $M$ from the position vector of $A$

[woven radish]

Civil Service Pigeon

[severe prairie]

Ahh ok

So any point on the line is just the equation of the line

And theyre tryna find lamda

[lunar harbor]

yes that's the whole point of the equation of a line

mhm they're trying to find $\lambda$ for that specific point $M$

[woven radish]

Civil Service Pigeon

[severe prairie]

When they find lamda, when you sub in the lamda that will get you point M

[lunar harbor]

correct

also it's lambda, not lamda

[severe prairie]

And how come you equate it to 0

[severe prairie]

Mb

Is it bc you use cross product

And cos90 is 0

[lunar harbor]

shortest distance is obtained by the perpendicular

orthogonal vectors have dot product zero

[lunar harbor]

because of this

[lunar harbor]

cross product seems a bit high tech here imo

[severe prairie]

Oh lol sorry

I always confuse the 2

[lunar harbor]

rip

[severe prairie]

Ahaha

Alr ty bro

.close

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[lunar harbor]

[severe prairie]

For MN, i think it would be OM-ON

But i don’t quite get why

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[severe prairie]

@lunar harbor come back goat💔

[supple knot]

!noping

[devout snow]

Please do not ping individual helpers unprompted.

[supple knot]

if you're gonna open a new help channel

[fierce viper]

MN = MO + ON
= -OM + ON
= ON - OM
so you're not correct in saying MN = OM - ON

[severe prairie]

Ah oops

But why

Ohhh

MO+ON

Makes sense

[fierce viper]

yes.

[devout snow]

@severe prairie Has your question been resolved?

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[knotty pulsar]

would it be 2/9 or 1/9?

[solid perch]

please show your calculations and/or why you think it's either.

[knotty pulsar]

so if there is 9 marbles and if there is a possibility of choosing a red and blue one with replacement as the problem says then its either 2/9 meaning the red and blue one or 1/9 meaning one of them was put back

[potent dirge]

I took the “replacement” thing to mean that the marbles are put back in right after you select them

[knotty pulsar]

yep

[potent dirge]

So, since each event doesn’t depend on the previous event, you just multiply the probabilities

[potent dirge]

So, what’s the probability of selecting a red marble?

[knotty pulsar]

1/9

[potent dirge]

No

How many red marbles are there?

[knotty pulsar]

wait my mistake it would be 1/3

[potent dirge]

Exactly!

And then, how many blue marbles are there

[knotty pulsar]

1/4

[potent dirge]

No.

How many marbles in total?

[knotty pulsar]

1/9

[potent dirge]

No

How many blue marbles?

[knotty pulsar]

4 blue

[potent dirge]

Out of how many total?

[knotty pulsar]

4 out of 9 total marbals

[potent dirge]

So:

4/9

[knotty pulsar]

4/9

[potent dirge]

And, like I said, since each event doesn’t depend on the previous event, we just multiply the probabilities

[knotty pulsar]

so the answer is 4/27?

[potent dirge]

Bingo!!

[knotty pulsar]

so now its asking the same question but instead of one blue and one red its two blue without replacement

[potent dirge]

Ok. So there are still nine in total, yes?

[knotty pulsar]

yes

[potent dirge]

So, when where are nine marbles, what’s the probability to draw one blue marble?

[knotty pulsar]

1/9

[potent dirge]

How many blue marbles again?

[knotty pulsar]

4

4/9

[potent dirge]

So now, after this we’ve removed one blue marble

I’ll be so impressed if you get this. I’ll be so happy.

What’s the probability of drawing another blue marble

[knotty pulsar]

3/9

[potent dirge]

Are there still 9 marbles in the bag?

[knotty pulsar]

yes

[potent dirge]

Are you sure?

[knotty pulsar]

3/8

[potent dirge]

You said without replacement this time

[potent dirge]

Bingo

So now, just multiply they’re probabilities

@knotty pulsar, if you’re done, close the channel

[knotty pulsar]

ok

.close

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[cerulean ruin]

Prove that if $rank(x) = \omega$, then x is infinite

[woven radish]

toast

[cerulean ruin]

I don't know if this is trivial

but im stuck lol

[celest lantern]

In which context?

[cerulean ruin]

uh

rank is elast ordinal such that x subset V_alpha

never mind

.solved

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[celest lantern]

Uh.. Okay?

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[solid trellis]

I was curious if there was any website or program where I could experiment with different transformations and like shape the grid myself. 3blue has an engine, but I want to like do it with my mouse

[devout snow]

@solid trellis Has your question been resolved?

[solid trellis]

No

[devout snow]

@solid trellis Has your question been resolved?

[gray coyote]

https://shad.io/MatVis/

@solid trellis

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[valid rock]

I need help with trying to find the domain and range of rational expressions, would anyone know how to help me here?

[neon folio]

well for the domain you just need to find where the functions arent defined

in the case of fractions

this happens when the denominator is 0

and in the case of radicals

[valid rock]

Uh-huh.

[neon folio]

this happens when the inside of the radical is <0

[valid rock]

Right.

[neon folio]

right so do that

[valid rock]

In case you were wondering, I am relearning precalc by studying math on my own.

For Calc II.

BUT I'm having trouble with stuff like the domain and range since it's been years since I last took it.

[neon folio]

ok

lets take the first one

ok?

[valid rock]

Any number BUT three I assume?

[neon folio]

yes

so domain is

R - {3}

[valid rock]

Uh-huh.

[neon folio]

and for the range

do you know how to find horizontal asymptotes?

[vital edge]

-# just wanted to comment: this is only true for square roots (and other even power roots). Cube roots can take negative values as input