help-38

[fleet torrent]

I think it's \partial i just dont wanna type that 6 times

[balmy hornet]

it is

copy and paste maybe

[fleet torrent]

anyway, your $u$ in this case is $x+y$

[solid kiln]

Oliver

[balmy hornet]

oooh

[fleet torrent]

$\frac{\partial u}{\partial x}$ - do you know how to solve this? and of course the asnwer for $y$ is the same

[solid kiln]

Oliver

[balmy hornet]

it doesn't like none of the above

[fleet torrent]

It isn't none of the above

[fleet torrent]

You can just plug in to here, if you understand what each component in the product on RHS means

[balmy hornet]

it wants f'(x+y)

idk

[fleet torrent]

Yes

[balmy hornet]

v

[pallid lake]

what does that mean anyways

[balmy hornet]

idk

[pallid lake]

like df/d(x+y)?

[fleet torrent]

Right. so you have that formula which is the multiariable chain rule. Do you know $\frac{\partial u}{\partial x}$

[solid kiln]

Oliver

[fleet torrent]

If I told you $u=x+y$

[pallid lake]

not really clear

[solid kiln]

Oliver

[pallid lake]

terrible notation for this kind of stuff

[fleet torrent]

I disagree, this question is completely clear

[pallid lake]

that's ur opinion

no issues

[fleet torrent]

This is a standard multivariable calculus question

[pallid lake]

I never had this kind of notation

I had the D notation

but alright now it's solved so

[balmy hornet]

i don't understand it though

[fleet torrent]

I cannot help you if you don't answer these questions

If you don't understand that is fine, if you do then I can proceed to the rest

[pallid lake]

well we pretty much take the derivative of f(x+y) with respect to (x+y), and then use the chain rule

[balmy hornet]

but I'm confused because idk what f(u) = ?

but

[fleet torrent]

You don't need to

The point of this is that this result works for any arbitrary function

[balmy hornet]

I guess
d/dx f(u) is f'(u) or du/dx?

[fleet torrent]

I feel like there is a misunderstanding

Where's this coming from

$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}$

[solid kiln]

Oliver

[fleet torrent]

when $z=f(u)$

[solid kiln]

Oliver

[fleet torrent]

So plug in $u=x+y$ and see if you can solve that, and if not let me know what you cannot do

[solid kiln]

Oliver

[balmy hornet]

$f'(u) = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}$

?

[solid kiln]

smeagol

[balmy hornet]

or

[fleet torrent]

No

This formula is for the derivative of z, not f

$\frac{\partial f}{\partial u} = f'(u)$

[solid kiln]

Oliver

[supple copper]

why are we using partials on a single variable function

[balmy hornet]

that's what's confusing me

it asks for partials

but its

x + y

not x, y

[nimble niche]

What is "f" and "u" that yall are talking about

[balmy hornet]

I'm confused for b

so I asked

[viral merlin]

there is a function f , which took the input x+y , its not a representation of a multivariable function
there is a function f(x,y) such that it is f(x+y)
something like this

[nimble niche]

Yeah you are correct with dz/dx, it'll be f'(x+y) (1+0) which is just f'(x+y)

Same with dz/dy

[nimble niche]

Yeah

[balmy hornet]

ooh ok

I just had to do chain rule

[nimble niche]

Your notation should have $\partial$ instead of $d$ though

[solid kiln]

Executor (ask on server b4 DM)

[nimble niche]

But otherwise is fine

[balmy hornet]

gotcha

thank you

[nimble niche]

yw

[balmy hornet]

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[limpid belfry]

I dont understand the bottom part.. why consider A,C,B,D?

[limpid belfry]

also isnt |CA|/|CB| = |DA|/|DB| the harmonic thingy for ABCD ? or am i dumb and theyre equal

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@limpid belfry Has your question been resolved?

[limpid belfry]

<@&286206848099549185>

[fervent drift]

isnt that what its saing?

saying*

[limpid belfry]

it says ACBD which is a different ordering so i dont get it

[fervent drift]

true

honestly i have no idea what its saying

[limpid belfry]

wuh

💔

helpp

[fervent drift]

oh right

i think

because they need to go across

i.e. cross ratio

[limpid belfry]

wdym

💔

thats not what a cross ratio is

[fervent drift]

they are taking the magnitude

so AC is smae as CA

then if you just cross multiply u get the particular quad they want

they just used ACBD to look prettier on the ratios

[limpid belfry]

ehh

wdymm

they said A,C,B,D is used often in olyms for some reason

pleh

i still dont get it ;-;

[fervent drift]

its just to make a pretty ratio

i think

visually

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[sharp stream]

um

im rlly confused

can someone explain what i gotta do

[quick tusk]

and AB = BC

[sharp stream]

right

thxxx

OH Wait

im so dumb lol

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[scenic cloak]

What exactly does this look like? Can someone walk me through an example?

[scenic cloak]

This is their example which I still can't visualize

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[quaint whale]

Hiii! Uh does anyone here do know how to do microeconomics? I have a question

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[quaint whale]

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[half marsh]

.what is your question about it?

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[modest rain]

how do i solve this ?

[modest rain]

i tried to do it like
i know for the whole thing to be 0 the separate terms should be 0
so i tried to do ksinxcosx - 1 = 0

so ksinxcosx = 1
i dont know what to do after

[fierce lake]

Why only ksinxcosx - 1 = 0?

What made you think that (sinx + cosx)^2 is always 0 for all x?

[mossy summit]

srry

[fierce lake]

Np

[mossy summit]

misread it lol

[modest rain]

for the whole thing to be 0 i thought maybe its like 0 + 0 = 0 so the separate terms (cosx...)^2 = 0 qnd ksinxcosx-1 = 0

[mossy summit]

1 + 2sinxcosx +ksinxcox -1 =0

k = -2 ig

[fierce lake]

But (sinx + cosx)^2 is not 0 for all x though

[mossy summit]

yeh

its 1+ 2sinxcosx

[modest rain]

can u explain 😅

[modest rain]

how did u solve after that to get -2

[mossy summit]

expandit

sin^2x + cos^2x +2sinxcosx

thats 1 +2sinxcosx

1 geets canclled in that eq

and only

2sinxcox + ksinxcosx = 0 is left

taking it to that side

-2sinxcosx = ksinxcosx

assuming only principle values ig

[modest rain]

ohh ok i got it

but i didn't get the 0 part

[mossy summit]

cancl sinx and cosx

[modest rain]

this

[mossy summit]

like

[fierce lake]

(sinx + cosx)^2 = 0
Sinx + Cosx = 0
Sinx = -Cosx
Which is satisfied only when x is of the form (2n+3)pi/4

[mossy summit]

ur saying 1 + 2sinxcosx = 0

that makes sin2x = -1

which is only true for some values

not for all x

[modest rain]

ohh okay got it thanks!

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[mossy summit]

hape it helps byee

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[humble hill]

Could someone remind me of the rules of integration? I haven’t done it in a long time and I kinda forgot how to integrate from Product rule & quotient rule

[round imp]

https://byjus.com/maths/integration-rules/

hope this helps

covers IBP and u sub

You could also learn bprp DI method

[humble hill]

Tysm

[humble hill]

Oh… I don’t think I’ve heard of that method yet

[round imp]

It is useful

You can go on youtube

And search "integration DI method", i think it will pop up

[humble hill]

Thanks

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[round imp]

np

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[misty cosmos]

hi did i draw the triangle correct as in the secondary triangle

[misty cosmos]

i used 5.5sin30 to find height which was 2.75 so h<b<c, so I think there is ambigious case but Im not sure if this is drawn correct

[trim lichen]

i dont think so

[brave pike]

Anyone can help

[trim lichen]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[trim lichen]

you have your own channel, wait there

[brave pike]

Ok

[trim lichen]

@misty cosmos this is what your diagram would have looked like

copying the numbers from yours

[misty cosmos]

ohhh

wait does that mean the triangle is the same except for the fact that there iis an isosceles iinside it?

@trim lichen

[trim lichen]

that's

a weird way of putting it i think

the point is that the data you've got cannot distinguish between triangles ABC and ABC'

hence the ambiguity in the ambiguous case

[misty cosmos]

wait in what way cannot it not be dinstinguishable

[trim lichen]

can you say that again but without the accidental double negative

[misty cosmos]

oh mb wait do u mean iin terms of the angle like sin(66.4) and sin(133.6)?

like that they are both positive

@trim lichen

or sin(theta) and sin(180-theta)?

[trim lichen]

you're kind of overthinking it

[misty cosmos]

😭

alr then we just subtract 180 bc its flat so 180-66.4 to form our second triangle

[misty cosmos]

hey can u check if drew it right this time

[trim lichen]

MS Paint has line and text tools jsyk

but yes this looks ok assuming the sine law was applied correctly

[misty cosmos]

yessss

thanks

[misty cosmos]

o lol

!close

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it's .close

[misty cosmos]

.close

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[spiral ocean]

combine

the fractions

it looks like things will cancel

[ornate token]

Hold up

Lemme try

[short shore]

how do i simplify the bottom part?

[spiral ocean]

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

[solid kiln]

k

[short shore]

oh my gosh

wait

let me

[ornate token]

Just

Rationalise

Divide a multiple by 1 + root(1-x)

For first

[spiral ocean]

just do butterfly

[ornate token]

Bro just rationalise it will be done

Don’t do this

[spiral ocean]

wdym

ohh

the negative sign

[short shore]

root(1-x^2)?

[spiral ocean]

i missed it

ye rationalise first

[short shore]

hmm

[spiral ocean]

,w simplift 1/(1-sqrt{1-x}) + 1/(1+\sqrt{1+x})

[solid kiln]

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Requested by nuclearinteraction

[spiral ocean]

for this

factor a^(something) out

things will cancel

[short shore]

please bear with me, are there any steps in questions with irrational numbers?

i dont really understand the results of these

[spiral ocean]

not yet

[ \frac{1}{1-\sqrt{1-x}} = \frac{1}{1-\sqrt{1-x}}\cdot \frac{1+\sqrt{1-x}}{1+\sqrt{1-x}} = \frac{1+\sqrt{1-x}}{1-1+x} = \frac{1+\sqrt{1-x}}{x}]

[solid kiln]

k

[spiral ocean]

rationalising the first term

[short shore]

there should be a +2 at the top right? and how do you do something like (1-root(1-x))(1+root(1+x))

oh i see

you rationalize them individually

okay okay

that makes so much more sense

and since they have a common denominator you just add them tgt

[spiral ocean]

yup

in the end u'll have

this

now to the a^ mess

[short shore]

let me try solving

this thingy

[spiral ocean]

is there anything u can factor out so that the top looks somewhat famaliar to the bottom

[short shore]

hmm

let me see

[short shore]

how did we remove the 2 up top?

man i am bad with irrational numbers like these

[spiral ocean]

what w

2

[short shore]

there will be two positive ones after we rationalize right?

[spiral ocean]

yup

[short shore]

sorry there must be something wrong with my calculations

how did we remove those

[spiral ocean]

[ \frac{1}{1-\sqrt{1-x}} = \frac{1}{1+\sqrt{1+x}}\cdot \frac{1-\sqrt{1+x}}{1-\sqrt{1+x}} = \frac{1-\sqrt{1+x}}{1-1-x} = \frac{1-\sqrt{1+x}}{-x} = \frac{\sqrt{1+x} - 1}{x}]

[solid kiln]

k

[spiral ocean]

[ \frac{\sqrt{1+x} - 1}{x}]

[solid kiln]

k

[spiral ocean]

this should be ur end result

for rationalisation of the second one

[short shore]

thank you

i will go eat for now, please hold this channel

[trim jolt]

@short shore Has your question been resolved?

[short shore]

im back

i may have found a way to solve this.

[short shore]

omg

[short shore]

so this is the question right

[short shore]

is the answer 4?

[spiral ocean]

i havent found it yet

gimme a sec

[short shore]

oh okay okay

my bad

[spiral ocean]

,w {[sqrt((1-\sqrt(3)(2))) + sqrt((1+\sqrt(3)(2)))]/[sqrt(3)/2]}^2

[solid kiln]

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Requested by nuclearinteraction

[short shore]

i need to learn how to do that

[spiral ocean]

could u show

ur derivation

[lusty delta]

yes

[short shore]

i can send pics

omg

ok wait

sorry i dont really know how to do digital stuff

i have it on paper

[forest needle]

I think it should be 4/3

[short shore]

where did i go wrong

man

oops sorry it sent twice

[forest needle]

In the 1st line you didn't distribute the negative in the rationalisation of the 2nd term

[short shore]

first line..

oh dear

can u see any other mistakes

ill go make coffee

[lusty delta]

this is a bit more calculations than necessary

the final answer does come out to 4 though

[forest needle]

Ok no mb you're correct

It's 4 mb mb

[lusty delta]

$\frac{\sqrt{1 - x} + \sqrt{1 + x}}{x} = \sqrt{\left(\frac{\sqrt{1 - x} + \sqrt{1 + x}}{x}\right)^2} = \sqrt{\frac{2 + 2\sqrt{1 - x^2}}{x^2}}$

[solid kiln]

Mqnic_

[lusty delta]

for positive x

[short shore]

thank God.

[short shore]

this question took longer than any of the other ones, i was wondering if there would be a shorcut

[lusty delta]

remember this method btw

[short shore]

for sure, thanks

[short shore]

aa sorry i forgot

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[hardy basalt]

I've been training to solve this for the past few mins but I couldnt get it

[undone gorge]

There is a formula you can use, do you know that?

[hardy basalt]

no

[quick viper]

hey

i am new here i need some help with maths

[undone gorge]

Open your own channel please

[quick viper]

i am new to dicord just need some help from you guys

[undone gorge]

So, what kind of method have you been taught? Could you explain it to me?

[hardy basalt]

i wasent in class

but what i did rn is

[undone gorge]

See #❓how-to-get-help

[hardy basalt]

for example I toke y=-2 and the other y in point m which is 4 so -2-4 then I got -6 then I added -6+-2 and i got -8 which is right but

when i tried the same thing on another point i didnt get the same answer

thats why i dont get it

[quick viper]

can you guys help me

[hardy basalt]

OPEN YOUR OWN CHANNEL

[quick viper]

how can i open my own channel i need help not want to help someone

[hardy basalt]

Go to help 8 or help 15 there is no one there

[quick viper]

why

[hardy basalt]

wdym why

.close

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[quick viper]

everyone is helping each outher na

okk tell me how to open my own channel

[undone gorge]

Well this seems like just a fluke to me, I am sorry

[hardy basalt]

its alr

[undone gorge]

You can imagine it like this:
For example K(-1,0). Now, the reflection is about x-axis as y=-2, so there is gonna be no change in x coordinate

[forest niche]

#❓how-to-get-help

[undone gorge]

Now, imagine a mirror at y=-2, the distance of point from the mirror should be equal to the reflected point.

[hardy basalt]

step one is to find the distance between y and -10 right?

then what

[undone gorge]

So, the distance(shortest, i.e the straing perpendicular line) of K(-1,0) from y=-2 is 2. Now you have to find the point with the same perpendicular distance but the direction is below y=-2, I.e y- coordinate<-2

So, the x coordinate remains same. As in K'(-1,y).
Now, you can see y should be -4 for it to fit the criteria

Or you can do this too:
The distance between them is 2. Double that, cuz mirror. So, 4 is the distance between the point and it's reflection. Now, we know x coordinate is gonna be same, So K(-1,0)--->K'(-1,0-4) . Minus, cuz we have to go 4 down

[undone gorge]

I like this method more personally

You can try this with N(2,1).
Distance from y=-2: 3
So, distance between point and reflection: 6
N(2,1)---->N'(2,1-6)---->N'(2,-5)

@hardy basalt

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[rotund owl]

How to prove x e^x >= e^x -1 without derivative

[wraith hinge]

X>=-1

Wait

[rotund owl]

Close to 0

Not in R

[stoic garden]

$$0 \geq e^x - 1 - xe^x $$
$$1 \geq e^x \qty(1 - x)$$

[rotund owl]

Yea same thing

[wraith hinge]

X>=-1e^-x

[rotund owl]

Not same rhing

Only first one is correct

[stoic garden]

Hold on

Oh i cant math

[rotund owl]

But how you prove it

[solid kiln]

@stoic garden

[rotund owl]

Yea

[stoic garden]

Hmm

[wraith hinge]

Or you can divide e^x on both sides?

[stoic garden]

Why do you not want to use the derivative?

[rotund owl]

This is part of proving e^x derivative

[wraith hinge]

And then X>= -1e^-x

[rotund owl]

So using derivative is circular reasoning

Same reason we cant use lhospjtal for e^x -1/x when x goes to 0

[stark bison]

Are you using the limit definition of e then?

[rotund owl]

Wdym

Im trying to use squeeze theorem

[wraith hinge]

@rotund owl

[stark bison]

We can't show anything about e if we don't define it

[stoic garden]

I wonder if you can say something like:

• e^x (1 - x) = 1 at, and only at, x = 0.
• e^x (1 - x) is continuous for all x E R.
• On either side of x = 0 (by testing two random x-values), we know that e^x (1 - x) < 0.

[stark bison]

One of the definitions is that e^x is its own derivative, but clearly we aren't using that one

[wraith hinge]

Can anyone listen to me

[rotund owl]

Wdym

[rotund owl]

I dont understand what after that

[stoic garden]

We sometimes define $e$ to be the number such that $\dv{x} e^x = e^x$.

[solid kiln]

@stoic garden

[rotund owl]

@wraith hinge you forgot a 1

[rotund owl]

Thats definition of e

[wraith hinge]

(E^x)X>=(e^x) (1-e^-x) so then divide (e^x) it becomes x>=1-e^-x

[stark bison]

But you said we don't know that e^x is its own derivative yet

Why do you need this inequality then?

[rotund owl]

Where did you find that?

[wraith hinge]

What do you mean

[rotund owl]

To prove (e^x)' = e^x

[rotund owl]

It's xe^x >= e^x -1

[stark bison]

Where e is what?

[gusty plank]

If you're using that definition then you don't need to prove that...

[rotund owl]

Im not though

[wraith hinge]

The minus one in the exponent?

[gusty plank]

Why did you say you were then

[rotund owl]

wdym were e is what

The constant e

Named after Euler

[stark bison]

Which definition of e are you using

[rotund owl]

Where did i say that

[gusty plank]

Right here

[rotund owl]

Does it say im using definition of e?

Still cant see where i say that

[gusty plank]

You need to define something to use it, and you said "Thats definition of e"

So what else could that imply if not that you're using that definition

[rotund owl]

What he sent i just said is a definition of e

[wraith hinge]

Yes exactly mine is correct way

Dude

The -1 is now -1e^-x

[rotund owl]

Can you send it to chat gpt and ask it to send in latex

[stark bison]

taebek is right, they didn't claim to use that definition, but telling which definition is being used would be nice

[rotund owl]

@bright quarry why are you spending your time reading messages and reacting with skull

[bright quarry]

because you’re goofy

[fierce lake]

So you wanna prove by first principle

[rotund owl]

Oh we got a smart guy here

[bright quarry]

word

[rotund owl]

So smart he spends his time reacting to messages of goofy people

[bright quarry]

exactly

[rotund owl]

By limit defintion of derivative

[wraith hinge]

You're right! Let's correct that. The original inequality you meant is likely:

$(e^x)x \geq (e^x)(1 - e^{-x})$

Now it's correct — the right-hand side includes the term.

Dividing both sides by (which is always positive), we get:

$x \geq 1 - e^{-x}$

Let me know if you'd like this plotted or solved further!

Tf

[limpid dawn]

mlk forgot the $

[rotund owl]

No ask it like send in latex for discord bot

And it sends the format

[rotund owl]

Ontos

[gusty plank]

dont use gpt ...

!nogpt

[trim jolt]

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[solid kiln]

Daniel

[rotund owl]

Its to get it in latex form 🙏

[wraith hinge]

I gave my own response he said I should translate it to latex using chatgpt

[gusty plank]

I see

[wraith hinge]

Why did I get warned?

[gusty plank]

You didn't

[wraith hinge]

See

[rotund owl]

And you prove that how

[wraith hinge]

Prove about what

[rotund owl]

The last thing you got to

It isnt a known inequality or anything

[wraith hinge]

Divide both sides by e^x

[rotund owl]

Yea after that

[wraith hinge]

You want to find an exact integer?

What X value is?

[rotund owl]

No

I want to prove that the inequality stands close to 0

Like there is δ such that for xE(-δ,δ) the inequality is true

[wraith hinge]

You need to do limits here?

[rotund owl]

Yea i need to prove the inequality fir squeeze theorem

[fierce lake]

Yeah so f(x + h) - f(x) / h as h tends to 0 right

[rotund owl]

Yea so e^x * lim h->0 (e^h -1)/h

The inner limit i can find with squeeze theorem if i can prove h<= e^h -1 <= he^h

Left side is easy second side idk how

[limpid dawn]

Maybe Bernoulli inequality?

[rotund owl]

Idk what that is

[limpid dawn]

https://math.stackexchange.com/questions/1070524/calculate-lim-x-to-0-ex-1-x-without-using-lhôpitals-rule

[rotund owl]

Oh intresting yea

With x/1-x

Does proof of that go through e^x derivative at all

Or that limit

Bernillis one i meant

[limpid dawn]

Bernoullis' inequality is proven via induction, if that's what you are asking

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@rotund owl Has your question been resolved?

[rotund owl]

Ight then that works

Thanks

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[fierce lake]

You wanna find this without L hospital?

Nvm it’s done

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[cobalt wraith]

Helo

[cobalt wraith]

Who is right?

I am supposed to idenify if graphs are misleading

[rotund owl]

Its a mistake to use Lhospital

[cobalt wraith]

Helo

[cobalt wraith]

Can you help?

[rotund owl]

What subject is this

[cobalt wraith]

Ill pay by telling discussions thag you are very good guy

Its math

[rotund owl]

I mean the last is true

The x-axis has inconsistent intervals

[cobalt wraith]

thanks

[cobalt wraith]

Is all graphs supposed to have consistant intervals?

[rotund owl]

No

But I'm guessing thats what is considered misleading

Idk what class this is

[cobalt wraith]

Its math grade 8

[rotund owl]

Don't know any theory about misleading graphs

Maybe its economics

[cobalt wraith]

Do tou know what it means hy thousands?

[rotund owl]

Wdym

It means that the values it says are the number time 1000

So 15.20$ is 15.2k

[cobalt wraith]

Its not cents?

[dusty sleet]

If it says thousands, it's thousands. Why would they have to be cents if there's "in thousands" written?

[rotund owl]

In the graphical representation it says 15$ and 20 cent

But on the top says in thousands

Meaning its that times 1000

So 15200$

[cobalt wraith]

O thank s

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[glacial crescent]

what am I doing wrong??💔💔

[glacial crescent]

the answer is 3 but idk im not getting there

[glacial crescent]

bro im actually tweaking what is wrong wit me 😭

thank you

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[wraith hinge]

Help

[wraith hinge]

Anyone know?

I got like 7 mins left

Timed hw

Anyone?

Pls

Or is it (3/5,inf)

[nimble niche]

Show your work

[topaz fog]

Up means derivative positive concave up means second derivative positive right

[wraith hinge]

I figured it out it’s (3/5,inf)

Tysm

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[topaz fog]

(3/5,inf) or (-inf,3/5)?

[wraith hinge]

(3/5 , inf)

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[dry vortex]

Okay so planar straight line real number polygon, any plane, projected by lines to eyes intersecting view plane, does x and y on plane correspond to z off plane? Does this apply for any planes? Why or why not? Does this apply for more dimensional where only what is on view plane projected views? X E.

[dry vortex]

Should I just mathematically test myself?

Please ask if I am unclear. X E.

Alright, ping me if you want me to know. X E.

[proper kernel]

What do you mean by “x and y on plane correspond to z off plane”?

If you only have a point on the view plane, draw a line through that point to see all the possible positions the point originally could’ve been before being projected onto the view plane

[dry vortex]

I mean like triangle ABC unprojected corresponds to DEF projected, if mid point of AB and BC and mid point of that then mid point of DE and EF and mid point of that defines z, right?

[dry vortex]

That does not work for me, in many dimensional, just want z value for an x and y without going through every dimension. X E.

Sorry if I was just annoying. X E.

[proper kernel]

You want a version of z that goes in and out from the eyes and through the viewplane?

[dry vortex]

Something like that, calculate z from x and y. X E.

Based on values of z for vertices. If I am wrong, don't go deriving for me please. X E.

[proper kernel]

It’s impossible to calculate z only knowing its position on the screen
In what direction are x and y facing?

[dry vortex]

x and y dimensions. X E.

++. X E.

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[dry vortex]

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[cunning scaffold]

How do I get from top line to bottom line

[burnt mulch]

That’s just wrong unless there’s some other relationship with s

,w graph (1/s +2)/(1/s), s/(s+2)

[solid kiln]

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[cunning scaffold]

Hm

[austere cedar]

Neat to see they're tangent, though

[cunning scaffold]

The graph does make sense

👍

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[jagged wharf]

Super simple statement probably, but

A set U is open iff it is a neighborhood of itself.

The definitions I'm given is that a set is open if it is a member of the topology of some topological space X. Moreover, a neighborhood of a set U is a set A subset X such that U is contained in the interior of A. I have also been given the definition that the interior of a set A is the union of all open sets contained in A.

Now if a set is a neighborhood of itself, it must equal its interior, which is open. On the other hand, if U is open, then how do I proceed given the definitions I'm given?

[trim jolt]

@jagged wharf Has your question been resolved?

[cosmic trout]

Then since U contains U and U is open, U is in the interior of U, so U is a neghbourhood of itself

[jagged wharf]

Ah yes, now that you edited, it makes sense 🙂 Thank you!

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[nocturne mural]

Is it possible to construct a trapezium ABCD with A and D are right angles such that AB = 25, BC = 20 and AC = 15?

[nocturne mural]

Because I got this question that I have to construct this diagram

But can't seem to find a possible way to construct it

[trim lichen]

are you talking like compass and straightedge

@nocturne mural

[nocturne mural]

no

like

i can't seem to construct a trapezium with those lengths

but with a bit of help

i can now

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[slate barn]

Find the power series expansion of the function
[ f(x) = \frac{1}{x + 1} ]
centered at ( x_0 = 3 ).\\
My solution:

[f(x)=\frac{1}{x+1}=\frac{1}{1-(-x)}=\frac{1}{1-(-(x-3+3))}=\frac{1}{1-(-(x-3)-3)}]
so
[f(x)=\frac{1}{1-(-(x-3)-3)}=\sum^{+\infty}_{k=0}(-(x-3)-3)^k]

[solid kiln]

Slowaq

[slate barn]

Is my solution correct?

Or there isn’t a standalone power part?

[prime crystal]

your answer isnt centered at x=3

-(x-3)-3 = 0 => x = 0

try f(x) = 1/(x-3)+4

[slate barn]

😕 yes I figured but I hoped that I could somehow justify my answer

[prime crystal]

nah

[slate barn]

But thanks anyway

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[royal patio]

If A--> B then notA-->notB or notB-->notA ??

[halcyon oar]

huh

type in latex form

[trim lichen]

the latter.

[royal patio]

Its pretty simple

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[torpid lantern]

the green bit why do we devide by 60 we r going from secs to minutes should we time by 60?

[ionic pendant]

if you have 60 seconds then you divide that by 60 to get 1 minute

minutes are longer, so as a rule of thumb, converting to minutes should give you a smaller number

[torpid lantern]

i see ty

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[lunar stirrup]

i got the answer which is 3 by using this formula

sorry tan inverse x + tan inverse y not -

now my doubt is why arent we taking the other cases for this formula where xy>1?

or does this formula include all solutions which may or may not satisfy and we just have to put it back in the original equation to check?

[empty orchid]

I feel like this has something to do with the fact that arctan's range is only defined on (-pi/2, pi/2)

[lunar stirrup]

oh ok but arctan can also be negative right

[lunar stirrup]

also can u answer this?

[empty orchid]

See, this is why I'm not 100% sure

My gut tells me its the same thing as like you would if you square both sides of x=3, you're "adding some solutions", but I could be completely bullshitting

[lunar stirrup]

ohh ok yeah that kind of makes sense...so we get extra solutions

should i wait for someone to confirm or should i close the channel?

[empty orchid]

Confirm

[lunar stirrup]

👍

[empty orchid]

I have a warning at the end of the message 💀

[lunar stirrup]

oh lmao

im asking cause every problem i see where u have to use tan inverse x + tan inverse y formula i start getting scared whether ill miss solutions

[main sigil]

the formula only works if xy < 1 if that's the question

so you cant use it for xy > 1, because it just doesnt work anymore

[lunar stirrup]

nono like im asking why arent we considering all these 3 cases why only the first case

[main sigil]

ohh

you could consider them

you'd find out that there are no sols

[main sigil]

look at (2) e.g.

when xy > 1, arctan(x) + arctan(y) = pi + arctan(sth)

pi + arctan(sth) is gonna haver range (pi/2, 3pi/2)

[main sigil]

which is completely outside of the range of a single arctan here

arctan(x) + arctan(y) is gonna be > pi/2 for x, y > 0; xy > 1

and so it wont equal arctan of any number

[lunar stirrup]

ohh ok right and even for the 3rd case its the same

[main sigil]

yes, same thing

arctan(x) + arctan(y) is gonna be < -pi/2 for x, y < 0; xy > 1

[lunar stirrup]

so it is only for this particular question that we arent considering all the cases and what i said here is wrong?

[main sigil]

again, that can never equal arctan of anything

[main sigil]

I feel like the other cases are kind of considered implicitly

the authors probably assumed that it's clear that the other 2 cases wont yield any solutions

so they didnt bother talking about them

but i didnt see the author solution so i cant say

[lunar stirrup]

oh so we can just ignore the other cases and check if our solutions work in the original equation

[main sigil]

in this case, they dont yield any new solutions, so yeah

if you wanted to be absolutely sure, you'd have to consider all cases

[lunar stirrup]

damn bro i cant afford to do that in an exam

unless its necessary

[main sigil]

It's not like you have to do the algebra

[main sigil]

you just have to realize this

[main sigil]

and this

if arctan(x) + arctan(y) > pi/2, then surely it cant equal arctan of any number

[lunar stirrup]

oh okay

i think i got it now

thank you

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[lunar stirrup]

so for this to happen,
x^2+4x+a^2 - a >=0 right

so discriminant has to be less than 0

so a^2-a-4>=0

now the question ive been asked is :
"sum of all real values of a for which f(x) is onto is"

im not able to understand how they got D=0?

[gentle crag]

f : R --> [0,p/2]
is that π/2?

[lunar stirrup]

yes

[gentle crag]

but arccot never takes 0 as a value, how can it be onto?

[lunar stirrup]

its (0,pi/2]

question has a typo

[gentle crag]

So their answer is (D)?

[lunar stirrup]

yes its D

sorry

a^2-a-4>=0 but this is the inequality we are getting right? cause disciminant <=0?

[gentle crag]

Thing is, we don't just want its values to be ≥ 0, we want it to be surjective onto [0,∞)

[lunar stirrup]

i didnt get that

surjective onto?

and what is "it"

[gentle crag]

in order to get arccot to reach all values in (0,π/2], what's inside has to reach all values in [0, ∞)

[lunar stirrup]

yes

[lunar stirrup]

thats the first condition i applied

[gentle crag]

so you want that polynomial to be surjective onto [0,∞), not just positive

[lunar stirrup]

OH

[gentle crag]

if Δ < 0, the polynomial is strictly positive

[lunar stirrup]

i kind of see what ur getting at

so for that to be onto the polynomial must be onto as well

[gentle crag]

onto [0,∞) specifically

[lunar stirrup]

yes

[gentle crag]

“onto” is kinda meaningless in the void

[lunar stirrup]

but how do we ensure that?

[gentle crag]

2nd degree polynomials are just parabolas

you want a parabola to reach all heights from 0 onward and those only

so it touches 0 only once

so Δ = 0

[lunar stirrup]

huh i didnt get that

oh

wait nvm i got it

okay got it thanks a lot bro

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[lunar stirrup]

.reopen

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✅

[gentle crag]

sad

what up

[wraith hinge]

where can i get help regarding jacobians and multivariable calc?

[lunar stirrup]

one second i need to process

[urban hull]

#multivariable-calculus

[lunar stirrup]

@gentle crag so the explanation i have is:
it must only have one x value so that the other x value can be equal to the corresponding y value so that the function is onto and all y values are covered from [0,infinity)

is this correct?

[gentle crag]

I'm sorry but what (?)

[lunar stirrup]

💀

[gentle crag]

write down some subjects

[lunar stirrup]

then i dont think i got it

one sec let me send a pic

[gentle crag]

look at this parabola

can you tell me the sign of the discriminant?

[lunar stirrup]

<0

[lunar stirrup]

im not able to understand why 2 places where its 0 wont work

[gentle crag]

okay
this is not allowed because this parabola also takes negative values, but the original function is specifically defined to take value in (0,π/2]
and for y < 0 you get arccot(y) > π/2

[lunar stirrup]

ok right

just think x>=0

[gentle crag]

for x ≥ 0 your polynomial still takes negative values

but still, the domain of the function is the whole ℝ, if you change the domain you're changing the exercise

[lunar stirrup]

okay

[lunar stirrup]

continue with this pls

[gentle crag]

this is a parabola with Δ < 0, and would correspond to some of your proposed values of α

is this parabola onto [0,∞)?

[lunar stirrup]

no

OHH

okay now i got it

for sure

thank u

[gentle crag]

No parabola with Δ < 0 is onto [0,∞)

if you want one onto, you must take Δ = 0

[lunar stirrup]

so that it touches x axis

[gentle crag]

yes

because you want y=0 to be a value

[lunar stirrup]

yeah

[gentle crag]

but it can't go lower then the x-axis, because negative values are not allowed

[lunar stirrup]

yeah 👍 thanks lmao u were really patient

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[vocal epoch]

can someone explain how my teachers getting p(x) in laymans terms

[vocal epoch]

i dont get these questions

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@vocal epoch Has your question been resolved?

[delicate bobcat]

What part didn't you understand?

x is the amount of iPhones needed and it depends on n (we get it directly from the question)

x = 200 + 20n, since at the beginning it says that a store has been selling 200 pieces

and it increses by 20 with each $10 rabate, hence 20n

p(x) is already explained, it's the price (i.e. how much money a store makes)

and p(x) = 350-10n, because each piece has been selling for $350

and taking n $10 rabates into account we have to subtract it (because rabate means a store earn $10 less for each piece)

After this, your teacher solved the 1st equation for n and then put it into p(x)

That's all

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[torn shale]

when we want the domain of quadratic inequality, we get their roots, put them on the number line, and say that hte domain is either the inbetween or whats on the two sides of teh roots. when did we do that and when did we do that

[torn shale]

actually let me rephrase

okay done

[delicate bobcat]

You mean, how do we know if we're supposed to take that part "between" and when the other part?

[torn shale]

yes

[nimble niche]

Start from + on the rightmost side

[torn shale]

i dont quite understand?

[nimble niche]

R1 and R2 are the two roots of the quadratic, which are distinct and real

So the quadratic is positive for any value of x greater than R2

and negative for any value of x between R1 and R2

While it is positive for any value of x lesser than R1 as well

At R1, R2 it is zero since they are roots

[torn shale]

yes

ooooh

that makes lense

that maeks sooo much sense

ty

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[nimble niche]

https://www.youtube.com/watch?v=xdGdFT2DHp0 Here's a tutorial incase you need it

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[gray kiln]

so basically, I solved the question but idk if it's the correct approach or not,
there are a total of 4200 aliens and a total of 2022 planets,
total number of path = 4200!
we have defined,
total number of safe path ~ the number of trailing zeros of the binary counterpart of total number of path
i.e. total number of safe path = total number of zero's in binary equivalent of 4200!

we know,
for any number N to have k trailing zeros, N/2^k must be an integer

using this,
we define, total number of safe path = 2022!/2^k, where k belongs to N and is the maximum value such that 2022!/2^k belongs to Z

we have,
number of safe path = 2022!/2^k
it can be simplified as the gif(2022/2) + gif(1011/2)..........
which is equal to 2014

i.e number of safe path = 2014

since the number of aliens = 4200

number of path repeated = least integerfunction(4200/2014) = 3

Hence it has been proved that there exist atleast 3 aliens that took the same path.

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[hallow fox]

Claim

[hallow fox]

I think i am at status 3 or 4

Is there someone who can help me??

[queen dew]

18?

[hallow fox]

Ye

[queen dew]

For the numerator: you can write the second term as sin((A-C)-C), now expand this with the formula for sin(x-y)

Then simplify the numerator

Then do the same for denominator

[hallow fox]

Can't i do something in my done solution

[queen dew]

There must be, but it will just become larger and larger

[hallow fox]

Ok then what to do

[queen dew]

This

[hallow fox]

I don't know anything about sin((a-b)-c)

[queen dew]

Consider (a-b) as an angle and c as the other angle

Use sin(x-y) = sinxcosy-cosxsiny, where x = (a-b) and y = c

[hallow fox]

Ik only sin(a-b)

Ok let me try

Do i do same with denominator

[queen dew]

Yes

[hallow fox]

Ok

Beta = b-c and c= y?

[queen dew]

yes

Numerator: $2\sin{(A-C)}\cos{C} - \sin{((A-C)-C)} = 2\sin{(A-C)}\cos{C} - (\sin{(A-C)}\cos{C} - \cos{(A-C)}\sin{C}))$

Like this for numerator

[hallow fox]

,rotate

[hallow fox]

Is it correct??

Can i seprate 2c in both other equations?

So it can become x-c and b-c

[queen dew]

No

Woth the second term

[hallow fox]

So what to do?

[queen dew]

For the sin(A-2C) use this method

[hallow fox]

Which?

[solid kiln]

@queen dew

[hallow fox]

My screen time will hit in few minutes

Can u just givve me hints

[queen dew]

Like this see

The second term in the numerator is sin(A-2C) which can be written as sin((A-C)-C)

[hallow fox]

Ok

So i had not to convert 2sin(a-c)cosc into 2sun((a-c)-c)

[queen dew]

No

[hallow fox]

Ok then bye 1 min left

[queen dew]

And similarly for the second term of denominator

[hallow fox]

Thx i will try it

For secound term i have to?

Only

[queen dew]

sin(B-2C) = sin((B-C)-C)

[hallow fox]

Ol

[queen dew]

The second term

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[fierce lake]

Also

!status

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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

[fresh pendant]

1

[fierce lake]

I think you gotta approximate thr growth of |sint|

Take mean of it

[fresh pendant]

It's a periodic function

[fierce lake]

Yeah

Basically, it performs inf many cycles for x tending to inf

So average it out to 2/pi

[fresh pendant]

Do I need to apply the L'h rule since both the numerator and denominator are infinitely large?

[main sigil]

isnt the integral constant?

oh wait

[fierce lake]

You can but how would you find derivative of |sinx|

[main sigil]

the symbol is x or pi?

[fierce lake]

X

[main sigil]

why'd you need that? You can differentiate the whole integral

[fierce lake]

Really? How would you differentiate |sint|

Oh wait

Mb

Mb

Im dumb

No but wait

Yeah no I'm stupid

@fresh pendant Ignore everything i said and use LH

[main sigil]

oh the limit of the quotient of derivatives is DNE

that means we cant use lhop

i guess we will have to integrate then

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[jagged phoenix]

Try substituting u=t/x?

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[blazing geode]

How to do 5129-3330

[full dock]

subtract

[blazing geode]

I have my own way of doing it and I’m not sure when it is going to trip me over

[full dock]

what is your way

[late flume]

Is it a mental math question

[blazing geode]

3330 + 670 + 1129=5,129

[blazing geode]

What’s your thoughts if you were asked to do mental math on this

[blazing geode]

So the result is 1799

[late flume]

I would do 5129-3000

Then 300

Then 30

[blazing geode]

I have been doing this like all the time

I’m not sure if it is a good habit

Especially when I’m doing test, this might kill me

[late flume]

It is quite fast

[blazing geode]

Yes, I need to weigh the pros and cons

The downside is when I get nervous, all of this stuff might just fk me up

[late flume]

Have you done something like mental abacus

[blazing geode]

No

[late flume]

I highly recommend doing it

The learning curve is steep

But I know people who are really fast at arithmetic

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[olive fractal]

i cant figure this out.

Tom gives Mike 2 5 dollar bills.
Mike wants to buy a vape for himself, which costs 15 dollars
He takes 10 out of the atm, and pays using one of the 5 dollars bills from Tom
The other 5 dollars he accidentaly lost, but he wants to pay Mike back
He takes another 10 dollars from the atm to pay Tom back
Mike only lost 20 dollars in total from his own money, how?
Where is the other 5 dollars? how does Tom get 10 dollars back and Mike only loses 5?

[olive fractal]

I have tried to write this down and visualise it but every time it doesnt make sense

[trim lichen]

wait

The other $5 he accidentally lost, but he wants to pay Mike back

sorry, who wants to pay back whom?

[olive fractal]

Mike wants to give Tom his money back

He lost one of the bills its gone forever

Dropped it somewhere

[trim lichen]

do you have a picture of the original statement just so we are sure there is no deliberate word twisting

[olive fractal]

Its a real life thing that ahppened to me lol

[nimble shuttle]

yeah okay that's right

no wait

that's confusing

[trim lichen]

mike's financial losses are:

• $15 paid on the vape
• $5 lost
• $10 paid back to tom
  while his gains are:
• $10 borrowed from tom

-30+10 = -20

[olive fractal]

how did he pay back 10 and only lost 5 from his own pocket

?

[trim lichen]

i do not understand your question

the payback and the loss-into-nothing are separate expenses

[olive fractal]

the money he owed was 10, no? how did he only pay back 5 from his own pocket

[nimble shuttle]

mike has 10 from tom
mike takes 10 from atm so now he has 20
mike spends 10 from his money --
5 from tom
now tom has only 5 and wants to pay tom back 10
mike loses the 5
so mike takes out a ten
gives to tom
Mike spent another 10 ----
20 in total

it is right

the thing is in the mike paid everyone he owes no one anything
What did he pay 15 for the vape and lost a 5