help-27

[indigo anchor]

uh huh

if this statement is true then shouldn't p + q = 0, and not p - q????

[last parrot]

Ok

Have you understood ewuation 1?

[indigo anchor]

yeah thats the original right

[last parrot]

The one next to "therefore"

[indigo anchor]

the (2) one? yes i understand it

[last parrot]

The denoted with number 2 right

[indigo anchor]

we are simply multiplying (1) with p^(1/3) to get (2)

[last parrot]

correct

[indigo anchor]

hmm

[last parrot]

But how to get 3 here, you multiply equation 1 by b and subtract equation 2 multiply by c

[indigo anchor]

yes im following now

i was stuck exactly after getting the equation (3)

[last parrot]

But already know that cube root of p is irrational, right?

[indigo anchor]

yes

[last parrot]

And b^2-ac with ab-c^2p are rationals

Since a b c p rational

[indigo anchor]

oh

true

[last parrot]

So if a rational multiply with an irrational and add another rational = 0, then the coefficient before the irrational must be 0

Then, b^2 =ac

[indigo anchor]

{(Q') x (Q) } + (Q) will always be Q' right?

[indigo anchor]

ohh i get it noww

OMG HOW DID I MISS THAT

SILLY ME

[last parrot]

Lmao

[indigo anchor]

omg im so embarrassed

[last parrot]

And ab = c^2p too

I guess annie went thru quite fast

[indigo anchor]

i think so

she did

thank you soo much for helping me here

i appreciate it

[last parrot]

No worries man

Type .close if finished

[indigo anchor]

yeah that i'll do

thanks again

.close

[devout snow]

Channel closed

Closed by @indigo anchor

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[last parrot]

All good

[devout snow]

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[indigo anchor]

sorry i closed the channel too fast

[indigo anchor]

can you continue from there

[last parrot]

What happened

[indigo anchor]

b^2 - ac = ab - (c^2)p = 0 right

how do we come to a = b = c = 0 from there

[last parrot]

No i didnt say this

[indigo anchor]

you didn't but isn't that true

[last parrot]

It is

[indigo anchor]

from equation (3), b^2 - ac MUST BE ZERO, which implies that ab - (c^2)p = 0

[last parrot]

Read this again

[indigo anchor]

so how do we come to a=b=c=0

yes

[last parrot]

We have c^4p^2 =(c^2p)^2 =(ab)^2

=a^2b^2

Correct]

?

[indigo anchor]

i understand c^4p^2 = (ab)^2, but how is (ab)^2 = (a^3)^2?

[last parrot]

You subs in b^2=ac

[last parrot]

Ignore this for now

[indigo anchor]

okay

[last parrot]

Please continue

You have a^2b^2, now you subs in b^2=ac, then?

[indigo anchor]

we get (a^3)c!

[last parrot]

Now, suppose c not equal to 0

[indigo anchor]

hmm

[last parrot]

In c^3p^2 = a^3, you divide both side by c^3

Then what do you have?

You almost there

[indigo anchor]

p^2 = a^3/c^3 = rational?

so pe is rational and irrational at the same time

[last parrot]

Now, take cube root both sjde

[indigo anchor]

which is not true

so c is 0?

[indigo anchor]

yeah we can do that too

[last parrot]

What happen if you take cube root both side here?

[indigo anchor]

we get p^(2/3) = a/c

[last parrot]

But, p^2/3 is irrational

[indigo anchor]

yeah

[last parrot]

And a/c is rational

So c must be 0

[indigo anchor]

yeah

[last parrot]

Now, you subs c back into c^4p^2 = a^3c

Then find that a=c=0

[indigo anchor]

p?

[last parrot]

Typo ;))

[indigo anchor]

np

[last parrot]

Then c^4p^2 = a^2b^2, subs c and a in

Find out b=0 also

Thats it

[indigo anchor]

we should put c=0 in b^2 = ac to get b = 0

[last parrot]

Either way but still correct

[indigo anchor]

alright

but when we initially put c = 0 in (c^4)(p^2) = (a^3)c, dont we get "0 = (a^3) * 0" which is true for all values of 'a'? then how come a = 0 is definite?

[coarse flume]

Substituting b=c=0 into the original equation we have a=0

[last parrot]

Add in c^3p^2 = a^3

Not all equations work btw

[indigo anchor]

oh that works!

[indigo anchor]

what does this mean

[last parrot]

Like this one

You cant find a from here

[indigo anchor]

ohh i get it

[last parrot]

But actually it still works

[indigo anchor]

yeah it does

[last parrot]

Disregard my quotes

You divide c both side

[indigo anchor]

i think we cant

cause c = 0

and they say it is forbidden to divide by 0

[last parrot]

But we havent known a c yet, dont we

[indigo anchor]

oh so we need to do this bofre everything?

[last parrot]

Correct

[indigo anchor]

oh

cool

[last parrot]

c is a variable tho

[indigo anchor]

hmm

thank you for explaing this question to me in great detail!

now i dont have more questions anymore

thanks again!

.close

[devout snow]

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[last parrot]

All good dawg

[devout snow]

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[fast flame]

This is the daily ACT Question from Gohar's guide. So my questions are: what is an ampitude and a period? and how do you even do sin (opposite/hypotenuse) for that equation?

[devout snow]

@fast flame Has your question been resolved?

[pine minnow]

its actually related to simple harmonic motion rather than trigonometry

[ocean haven]

no ?

[pine minnow]

mb i got a bit mixed up between general shm and sinusoidal wave it seems

[summer summit]

the general form for this is $y = A\sin (Bx + C)$

[woven radish]

Krish

[summer summit]

Amplitude is defined as $|A|$

[woven radish]

Krish

[summer summit]

period is defined as $\frac{2\pi}{|B|}$

[woven radish]

Krish

[summer summit]

@fast flame ^

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[fast flame]

thx bro

[fast flame]

thx as well

[summer summit]

ofc! just remember to type .close because you opened a new channel

[fast flame]

.close

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[devout snow]

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[barren imp]

I need help with completing the square

[faint gorge]

is there something specific you need help with that?

[barren imp]

just understanding it from the ground up

[faint gorge]

i have a blueprint

[barren imp]

I do not get

[faint gorge]

which step? maybe post your progress or something

[acoustic onyx]

okay you know about quadratic equations right?

[barren imp]

good

leave that

[acoustic onyx]

ok

[jagged ledge]

its basically forcing a square term out of the terms you are given

so before anything , all square s are in the form of a^2 + 2ab +b^2. you can dee this for yourself when you expand (a+b)^2

lets look at x^2 + 2x

if you wanna complete the square then the a^2 term will equal the x^2 term, so a will equal x

then since we dont have a b^2 term, which we can see because we dont hsve anything without an x in the equation given, we must use the 2x which is the 2ab term

we know a equals x, so we can divide both by 2x, giving us that b equals one

so (x+1)^2 but if you expand that youll realise youve got an extra term, b^2 or 1

[jagged ledge]

so you must take that away from (x+1)^2 to give tou the original equation

so x^2 + 2x in completed swuare form is (x+1)^2-1

[devout snow]

@barren imp Has your question been resolved?

[barren imp]

I gtg , but i will write these in my notes

ty

[devout snow]

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[devout snow]

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[floral hollow]

this is high school mission,,,
When f(1 )equal fprime(1),
and in all x>=0 real number
f(x) >=fprime(x)

how can I get minor fprime (2)?

[floral hollow]

Yes

Minimum number that x=2 I think

More try needed?

Minimum of f(2)

Yes one point on the graph is f(2) I think

[last parrot]

!15m

[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.

[last parrot]

you see that $f(x) \geq f'(x)$

[woven radish]

Minhh

[last parrot]

but, for what type of function that its deriative would be smaller?

quick hint, this is some kind of exponential stuff

Next hint, Euler

[floral hollow]

Oh gosh

[last parrot]

what function have euler?

$e^{-x}$

[woven radish]

Minhh

[last parrot]

this would be obvious now, right?

since derivative is $-e^{-x}$

[woven radish]

Minhh

[last parrot]

and smaller than the f(x)

[floral hollow]

I feel dizzy

[last parrot]

well, we suppose $g(x) = e^{-x} f(x)$

[woven radish]

Minhh

[last parrot]

you find the derivatives

[floral hollow]

Okay

I learned derivative in high school, $g(x)$

[woven radish]

최선우

[last parrot]

nice

[hollow vine]

i'm just gonna jump in with no context but the thing that popped into my mind is that you can set up and equality case, and solve the differential equation

Minhh is right tho

[floral hollow]

My brain exposed cause I just thought about

[frosty crescent]

motivation: when we want to prove that the solutions to f'(x) = f(x) are f(x) = Ce^x, this is the approach usually taken lol

[hollow vine]

this a simple differential equation, like the most basic form

[floral hollow]

But there is more higher concept to understand math

[frosty crescent]

i wouldn’t say this is a diff eq, more of a diff inequality

[hollow vine]

$f(x)=f'(x)$

[woven radish]

IdelUser404

[floral hollow]

deferential equation is

[hollow vine]

a equation involving a function(s) and it's derivative(s)

such as the one above

[floral hollow]

1min

[hollow vine]

Example: Exponential growth differential equation

[
\frac{dy}{dx} = ky
]

General solution

[
y(x) = Ce^{kx}
]

Where:
y(x) is the unknown function
k is a constant
C is an arbitrary constant determined by initial conditions

[woven radish]

IdelUser404

[floral hollow]

Unknown function… I see

[hollow vine]

just an example it could look every different

[floral hollow]

In my question, there is three different valuables that $f(x)$

[woven radish]

최선우

[floral hollow]

Can I change may nicname back?

I try more my self

Next time I want to ask formal, good question

[hollow vine]

to solve such a inequality, you want to make the general form as a formula, so like in this case, $e^x$ is itself's derivative, so it fits this perfectly, with $f(x)=f'(x)$, the result will add the constant of integration C

[woven radish]

IdelUser404

[hollow vine]

i saw in your work you've been using polynomials, which won't work

[devout snow]

@floral hollow Has your question been resolved?

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[lavish vale]

Is it possible to work out Pr(B) from Pr(AnB)?

[lavish vale]

Please don't give me tips as it's against university policy

Just answering the initial question would be good

J9jef9joejf

I'd need Pr(A|B)

Which is what I'm trying to work out

Wait

Or I could do:

Pr(B) = Pr(AnB)/Pr(A)

[placid rover]

have you considered giving the answer is a big tip

[lavish vale]

I don't want you to give the answer

[last parrot]

Tips is hints

[lavish vale]

I want someone to say if it's possible or not to work out Pr(B) from the information I have

[half ermine]

P(B|A) = P(AnB)/P(A)

[placid rover]

they said no tips

[half ermine]

[lavish vale]

That won't give me Pr(B) tho

[half ermine]

Hmmm

What are all the given infos

Only p(a) and p(anb)?

[placid rover]

with what u have in your paper : yes

[lavish vale]

I've got 20 mins left to do it before it gets automatically handed in

left it to the last minute a bit

the procrastinator that I am

[hardy trail]

Uh oh, is this a test?

[lavish vale]

Weekly assignment

so kinda a test

is that against this server's policy as well?

[hardy trail]

Yep. #rules

[lavish vale]

shusduhfs

Bayes Theorm is killing me

[half ermine]

I think P(S) is an independent event???

That would give you p(s)

P(anb) = p(a)p(b)

[lavish vale]

Oh wait

lemme think

how would that work?

[half ermine]

It's like a property of baye's theorem

[placid rover]

Please stop helping the user

[half ermine]

😢😢😢

Even idk

[last parrot]

Close channel

[lavish vale]

anyway thanks guys

[placid rover]

^

[lavish vale]

.close

[devout snow]

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[half ermine]

Just tryna figure out 🥀

[placid rover]

that breaks server rules

[lavish vale]

we'll discuss it after it finishes

I'm gonna get like 80%

[half ermine]

Okay bro when u get answer dm it to me

It's an interesting question

[faint zinc]

@lavish vale a timed assignment? Not a test is it?

[lavish vale]

That's why we aren't discussing it

my bad

I forgot

[faint zinc]

welp, unfortunately, that's something I need to ban you for, sorry.

[lavish vale]

: (

[devout snow]

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[glass kraken]

Guys I need help with this question, this is my working so far but I don’t know what to do after this

[last parrot]

is the question 542?

[glass kraken]

Yes

[last parrot]

you can find x1 y1 as trig

[glass kraken]

I did, like the last step I wrote

[last parrot]

then you subs P in L

subs x1 y1 in L

[glass kraken]

Ohh I see, right , then I can convert the whole trig expression to be in terms of either x1 or y1

[last parrot]

knew that : = x+y=4

[glass kraken]

I’ll try that

[last parrot]

ye ye

[glass kraken]

Ok now I’m getting a weird mess in terms of tan and cos

I’m sure we only need to find slope of M

But idk how to here

[last parrot]

have you found in term of sin cos

after you subs in point A in L

[glass kraken]

But A does not lie on L right ?

[last parrot]

hmm i mean there is misunderstanding here

Let the line L cut at AM, note this point be P

then you should have $x_1 = 1 + d \cos \theta = 1 + \sqrt{\frac{2}{3}} \cos \theta$

[woven radish]

Minhh

[last parrot]

and $y_1 = 2 + d \sin \theta = 2 + \sqrt{\frac{2}{3}} \sin \theta$

[woven radish]

Minhh

[last parrot]

now you subs these values in line L

[glass kraken]

Hmm I see

I’m getting

Cos + sin = root 3 by root 2

[winged tapir]

Do you know how to solve this type of equation?

[glass kraken]

Hmm divide n multiply with root 2??

[winged tapir]

✅

[last parrot]

correct right there

or you make sin + cos = sin (theta + something)

[glass kraken]

Got it finallyyyyy

Thanks a ton u guys

.close

[devout snow]

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[last parrot]

No problem

[devout snow]

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[glass kraken]

Guys why is the answer here D and not A??

[faint gorge]

Show yout work

[glass kraken]

If p and q coincide (as the circles are intersecting) wouldnt the distance between the two points become zero

[ocean haven]

are you sure they intersect?

[glass kraken]

Oh wait mbbb

[placid rover]

first thing you should do is probably complete the squares on the equations?

[glass kraken]

Nah Idts that was required

I messed up with calculations ar the start n thought radius was 4 instrad of 2

[placid rover]

i don't see how you're drawing such a diagram without doing so

[faint gorge]

But it's a good approach to not draw them wrong

[glass kraken]

Hmm theres just a direct formula we can use to find centres and radii so I used that

Thanks yall

.close

[devout snow]

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[placid rover]

the thing is, that formula literally is gonna come from completing the square, and i don't really see the point of memorising such a thing

whatever suits you

completing the square on the equation will directly give you the centres and radii

[glass kraken]

Yes I agree, it’s just in competitive exams it’s wiser to use certain results weve already proved

Because we literally have 1 hour or less for 25 math questions depending on how fast or slow u solve other subjects

[devout snow]

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[placid rover]

you should try both ways. a student familiar enough with completing the square should be able to do so at about the same speed as utilizing a formula there. It would take under 10 seconds.

Completing the square is a far more useful general skill that pops up in many places.

[placid rover]

.close

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[glass kraken]

I totally agree with that, but memorising formulas gives a lot of ppl an extra edge (in objectives)

[glass kraken]

.close

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[teal tapir]

can someone please help me understand this paragraph i have no idea what they're talking about i just saw a formula suddenly that came out of nowhere

[gloomy aurora]

defined by

they define rotational there.

[teal tapir]

yeah but what does it mean

like what does rot x F mean

what's the rot vector

[gloomy aurora]

they just defined it

[gloomy aurora]

oh i feel the notation, should be more aptly $\overrightarrow{\mathrm{rot} F}$

[woven radish]

Annie Maqionde

[teal tapir]

so logically the vector rot has as coordinates the partial derivatives

huh

but it's partial derivatives with for no function

i mean when you calculate the determinant u'll get it eventually but what does a vector having derivative of nothing as coordinates mean

[trail eagle]

Strictly speaking, $$\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$$ is an operator on functions. You give it a function and it spits out a vector field. This gives the usual notion of the gradient as $$\text{grad} f(x,y,z) = \nabla f(x,y,z) =\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right).$$

From this we can define other quantities, like $\text{rot} \vec{F}(x,y,z) = \nabla \times \vec{F}(x,y,z)$, which if you remember the cross product is exactly the determinant you're calculating in that picture.

[faint gorge]

You can think rot as a vector function that takes up a vector F in its argument

[woven radish]

Azyrashacorki

[teal tapir]

should i know what a vector field is?

actually i just started this chapter about partial derivatives then suddenly this paragraph came out of nowhere after differentials

[trail eagle]

A vector field is exactly what is given to you as $\vec{F}$

[woven radish]

Azyrashacorki

[trail eagle]

It gives you a vector for every input (x,y,z)

[teal tapir]

i haven't took anything about vector functions or vector fields

[teal tapir]

yea i saw it it's like a 3d space full with vectors at each (x,y,z)

[trail eagle]

They call it just a vector in this paragraph which it is for every input but such functions that associate to every point a vector is called a vector field.

[gloomy aurora]

to sum up, vector fields output vectors

[teal tapir]

i'm confused now why i'm taking this without taking vector fields before

i barely can understand the rest now

[trail eagle]

Well they call it a vector and you can think of it as such regardless of whether you give it the name of "vector field" because it's just a vector that depends on some inputs.

[teal tapir]

oh alright

can you explain a bit more about the gradient operator like what does it actually do when multiplying it with a function

cause there's this paragraph right after

[gloomy aurora]

$\mathrm{grad}f = \nabla f = \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)$

[woven radish]

Annie Maqionde

[teal tapir]

yeah i know this i meant like geometrically what does it mean

[gloomy aurora]

soo like its a vector that points in the direction of the maximum (steepest) "ascent"(rate of increase) of the function

[teal tapir]

i just don't like to memorize things without understanding them

[raw cedar]

It’s important to know what are you memorizing.

Do you get it better?

[teal tapir]

ohhh okay so if for example it we take the exponential function it's gonna point upwards?

the gradf

vector

[teal tapir]

yup

[raw cedar]

No, because it’s for xy no for z directions.

[teal tapir]

something to do with this thing?

[raw cedar]

Yeah.

[raw cedar]

Do you have more questions?

[teal tapir]

no but for some reason i'm having trouble understanding its concept maybe because i just started multivariable calculus

[raw cedar]

Is there a specific point that makes it harder for you to understand?

[trail eagle]

Well the exponential function has only one variable, so there's only two directions to pick from : left or right. But indeed, the gradient vector in this case would point to the right, which is the direction in which e^x increases.
It's more complicated for functions of multiple variables because if you're standing at a point on the graph of the function, there are lots of directions for you to move towards

The gradient tells you which direction to move towards so your function increases the quickest. You may want to choose another direction and wonder how much the function is changing in that direction, and this is what the paragraph on directional derivatives refers to.

[raw cedar]

Yeah, the granadient make it easier and faster.

[teal tapir]

i think i got it

it'll probably become way easier after i solve more problems

i'm still doing the course

[raw cedar]

Yeah you can do examples and if you have questions ask here.

With practice you will get it fast.

[teal tapir]

but there's one thing here idk if i got it right
we say that a vector F has a potential if the gradient of a function is equal to the vector F?

[teal tapir]

yeah hopefully there are way more paragraphs to do

[raw cedar]

Exactly.

[teal tapir]

alright, thanks @raw cedar @trail eagle @gloomy aurora for the help

.close

[devout snow]

Channel closed

Closed by @teal tapir

Use .reopen if this was a mistake.

[raw cedar]

You are welcome. Have a nice day!

[devout snow]

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[indigo anchor]

Hey guys

Is my proof valid?

[faint gorge]

nice handwriting

[raw cedar]

You have a clearly schematic write.

[rare kernel]

thought u would hv said nice pfp

[raw cedar]

You state that if $a \neq c$ and $a \neq d$, then $a, b, c, d$ must be rational squares. This is an assumption, you need to proof that.

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

oh. i have not heard that a lot.

[indigo anchor]

oh

how do i prove it?

[raw cedar]

You can isolate a square root.

[indigo anchor]

quadratic?

i dont understand

[raw cedar]

Then square both to finally regroup the rational terms. And analyze.

[raw cedar]

Square.

[indigo anchor]

hmm

[river lance]

hey crackers whats a good website to learn math and what comes after algebra 1

[indigo anchor]

are the (i) and (ii) parts correct in my proof? i mean are they valid? cause i think they are not formal enough.

[shy osprey]

#discussion

[indigo anchor]

you should discuss that in some other channel

[raw cedar]

Use another channel for that.

🙂

[river lance]

oh ok

[indigo anchor]

what do you think

[raw cedar]

Are not enough.

[indigo anchor]

hmm

[raw cedar]

I recommend you start with: $\sqrt{a} + \sqrt{b} = \sqrt{c} + \sqrt{d}$

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

for the first two parts?

[raw cedar]

Yeah.

[indigo anchor]

how do i continue from there

[raw cedar]

You can eliminate the square roots.

Plowing both sides squared.

[indigo anchor]

hmm

ok ill try that

[little jasper]

Help

Can someone help me

Pleased

[raw cedar]

You can ask in other channel for help like #help-25.

[little jasper]

I did

[raw cedar]

Oh, I see, let me see your problem.

[indigo anchor]

is this valid? (For first two parts)

[raw cedar]

Is so good but you go from $a+b=c+d$ and $ab=cd$ directly to ${a=c \wedge b=d} \vee {a=d \wedge b=c}$. Although it is true, you must explain why.

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

oh i forgot to mention that

[raw cedar]

Np, it's normal.

[indigo anchor]

hmm thanks

to me it looks like that's an obvious thing

but surely there is a way to present it formally

how do i do that

?

[raw cedar]

You can use the difference of squares.

[indigo anchor]

difference of squares?

interesting

[raw cedar]

Yeah, try it.

[indigo anchor]

so i tried it

i couldn't do it

cant find where to use the difference of squares

[raw cedar]

We can divide in 2 the original equation and have this: $$\sqrt{a} - \sqrt{c} = \sqrt{d} - \sqrt{b}$$

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

how?

oh

[raw cedar]

Here we apply the square difference by multiplying and dividing each side by its conjugate.

Like here: $$\frac{(\sqrt{a} - \sqrt{c})(\sqrt{a} + \sqrt{c})}{\sqrt{a} + \sqrt{c}} = \frac{(\sqrt{d} - \sqrt{b})(\sqrt{d} + \sqrt{b})}{\sqrt{d} + \sqrt{b}}$$

[indigo anchor]

yeah yeah yeah i got that

ok ill try again

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[raw cedar]

Try it.

[indigo anchor]

ok

[raw cedar]

You could?

[indigo anchor]

yeah almost on the last step

[raw cedar]

Ok, do you have more questions?

[indigo anchor]

i may have them as the question is not yet finished

[raw cedar]

Ok, do you know how to continue?

[indigo anchor]

i think i do but i may get stuck somewhere

[raw cedar]

Np, continue and if you get stuck send here where and I can help you.

🙂

[indigo anchor]

okayy

I proved the (ii) part

(i) will be done similarly right?

[raw cedar]

Good!

[indigo anchor]

Thank You!

[raw cedar]

Yeah, try it.

[indigo anchor]

ohh wait wait wait

so instead of making roota - rootc = rootd - rootb from the original equation, we can make roota - rootd = rootc - rootb

and the prove the (i) part

right???

[raw cedar]

Right, you get it.

[indigo anchor]

yesss

thank you so much

[raw cedar]

You-re welcome.

[indigo anchor]

but i have yet not solved the (iii) part

[raw cedar]

Yes, you need to solve that to finish.

[indigo anchor]

yes

ok i'll try the third part now

[raw cedar]

Try it.

[raw cedar]

You could?

[indigo anchor]

still trying

[raw cedar]

Ok, try, take your time.

[indigo anchor]

am i on the right track?

[raw cedar]

Yeah, but you can factorise the expression, right?

[indigo anchor]

which one?

[raw cedar]

$$\sqrt{ad} + \sqrt{ac} - \sqrt{bd} - \sqrt{bc} \in \mathbb{Q}$$

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

yeah i can

k ill do it

[raw cedar]

👍🏻

[indigo anchor]

im getting $$(\sqrt{d} + \sqrt{c})(\sqrt{a} - \sqrt{b})$$

[woven radish]

Anshuman

[indigo anchor]

(omg i used this feature for the first time!!!)

[raw cedar]

Good.

Then you can apply square diference.

[indigo anchor]

how

[raw cedar]

Remeber $a^2 - b^2 = (a + b)(a − b)$

[indigo anchor]

yeah

right

[raw cedar]

Use that.

[indigo anchor]

where

[raw cedar]

Think about that $$\sqrt{c} + \sqrt{d} = \sqrt{a} + \sqrt{b}$$

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

but its d c and a b? arent they different? if it wre d c and d c the i could do that

[indigo anchor]

oh

so we get $a - b$ is a rational number right

[woven radish]

Anshuman

[raw cedar]

Right!

Do you have more questions?

[indigo anchor]

so from that logic $c - d$ is also rational

[woven radish]

Anshuman

[indigo anchor]

but what do we do from these informations now

when its given in the question that $a, b, c, d$ are rationals, isn't it already understood that $a - b$ and $c - d$ are rationals???

[woven radish]

Anshuman

[indigo anchor]

so what is the profit of doing these big calculations

[raw cedar]

Proving that $ab$ is rational is trivial. The importance of the expression $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = R$.

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[raw cedar]

This implies that $\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}$ is a rational number, right?

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

yes it does

[raw cedar]

If we have :
$\sqrt{a}-\sqrt{b} = k(\sqrt{a}+\sqrt{b}) \implies \sqrt{a}(1-k) = \sqrt{b}(1+k)$

[woven radish]

Ga³¹Br³⁵I⁵³9000✞

[indigo anchor]

do we need to do the componendo dividendo thing now?

[indigo anchor]

i dont understand this

[raw cedar]

Yes, use it, it-s the faster way.

[indigo anchor]

then we directly get $\sqrt{\frac{a}{b}}$ is rational

[woven radish]

Anshuman

[indigo anchor]

which means $\frac{\sqrt{a}}{\sqrt{b}}$ is rational

[woven radish]

Anshuman

[indigo anchor]

since $\sqrt{\frac{a}{b}}$ is rational, and we know that a and b are rational:
$\implies \frac{a}{b}$ is a square of a rational

why did i get an error

[supple knot]

\implies it outside of the $$

[indigo anchor]

ohh

my bad

[woven radish]

Anshuman

[indigo anchor]

this, in turn, implies that, $a$ and $b$ are both are squares of rationals

[woven radish]

Anshuman

[indigo anchor]

which further implies that $c$ and $d$ also are squares of rationals

[woven radish]

Anshuman

[indigo anchor]

hence all the other fractions concerned in the original question are rationals

am i right?

[raw cedar]

Yeah.

You are right.

[indigo anchor]

okay

ill write one final proof for you guys to check

then ill be done

[raw cedar]

[devout snow]

@indigo anchor Has your question been resolved?

[tender flare]

2+2=4

[indigo anchor]

thanks for telling

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[indigo anchor]

.reopen

[devout snow]

✅ Original question: #help-27 message

[indigo anchor]

looks like i cant paste the message its too long

wait

[indigo anchor]

ALRIGHT IM DONE

THIS FEELS LIKE MY LIFE'S WORK

@raw cedar hey can you check the final proof please

anyone else?

[raw cedar]

Correct!

[indigo anchor]

Thank you soo much for explaining this problem to me step by step!

[raw cedar]

You are welcome.

[raw cedar]

Do you have more questions?

[lunar harbor]

You've shown that the product of these two numbers is rational, but that doesn't necessarily mean that each of them is rational - take a=2, b=18, c=8, d=8

[raw cedar]

Oh, yeah. I didn't see that.

Ty.

[indigo anchor]

now that you mention it

i need to take that into account too

[lunar harbor]

Here, you implicitly use that a-c and b-d are nonzero. You should address this edge case separately.

[indigo anchor]

hmm

[lunar harbor]

alright imma go eat so I'll leave y'all to it

[raw cedar]

Check that. The other looks correctly for me.

😉

[indigo anchor]

yessir

so for the first flaw in my proof, where i forogt to take into account the edge cases, if infact, a-c and b-d are nonzero, then (i) is already satisfied

even if one of them is zero, the other one automatically satisfies because of the (1) equation

[ocean haven]

enjoy

[devout snow]

@indigo anchor Has your question been resolved?

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[indigo anchor]

thank you @raw cedar and others!

[devout snow]

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[raw cedar]

You are welcome. Have a nice day!

[raw cedar]

.close

[devout snow]

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[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

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[devout snow]

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[quaint apex]

im struggling how to draw this

[devout snow]

@quaint apex Has your question been resolved?

[devout snow]

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[vestal dirge]

Let $f: \mathbb{C} \to \mathbb{C}$ be a complex-valued function defined by:
[ f(z) = \left(\frac{3}{2}\right)^z - \left(\frac{1}{2}\right)^z - z ]

The objective is to prove the existence of complex roots in the set $\mathbb{C} \setminus \mathbb{R}$.

[woven radish]

tobi

[vestal dirge]

How do I proof this?

[devout snow]

@vestal dirge Has your question been resolved?

[vestal dirge]

.close

[devout snow]

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[devout snow]

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[pure cedar]

if two tangents meet outside circle, their length is equal, is the converse also true?

[ripe grove]

[topaz axle]

if they have length at all

then they meet

[gloomy aurora]

If two tangents lenghts are equal, they need not necessarily meet

[raw cedar]

Correct, it’s not necessary.

[pure cedar]

here , i know that down one is tangent, but idk about up one

[topaz axle]

they are lines

they don;t have length

except if they meet, there's a distance

if there's no second point, they just don;t have length

[last parrot]

https://tenor.com/view/riddler-paul-dano-new-rockstars-thumbnail-does-he-know-he-doesnt-know-gif-27552850

[topaz axle]

i don;t know geometry no

[pure cedar]

this is the question basically

i know that its equal

what i dont know is that if it is a tangent or not

[clever flume]

oh my god

i got the andwer calculation

[gloomy aurora]

since the original image has literally no points labelled, i'd labelled them(its much easier for helpers to help then). then BD=DE

[topaz axle]

it's not tangent

[clever flume]

[pure cedar]

i got it, how did the guy take it to be a tangent

[topaz axle]

AD is not, tangents are gentle and sweet

[pure cedar]

i know that BD=DE , cuz they are radii of same circle
what i dont know is that if DE is a tangent or not

[gloomy aurora]

DE appears to be a tangent: there's a single point marked.

[pure cedar]

i js thought they joined the point D to E without tangency

[gloomy aurora]

ok i just realized there's some congruence for some pair of triangles that shows DE is a tangent, if im not wrong.

[pure cedar]

cant we use the fact that equal tangents intersect

since they are equal and intersecting , they must be tangents

[gloomy aurora]

its the equivalent of what I said :D.

[gloomy aurora]

that requires certain justification, though.

[pure cedar]

thx , imma get it now

.close

[devout snow]

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[devout snow]

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[devout snow]

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[rare kernel]

br

[devout snow]

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[plush bolt]

The coefficient of x^70 in x ^ 2 * (1 + x) ^ 98 + x ^ 3 * (1 + x) ^ 97 + x ^ 4 * (1 + x) ^ 96 +...+x^ 54 (1+x)^ 46 is 99Cp-46Cq. Then a possible value of p + q is:

[plush bolt]

?help

[dull parrot]

the expression of the terms looks like $x^k(1+x)^{100-k}$ from $k = 2$ to $k = 54$

[woven radish]

MxRgD

[dull parrot]

i'm pretty sure the sum is just a geometric progression

[plush bolt]

nope, got this question from binomial theorem chapter

[dull parrot]

hmm, the common ratio $\frac{x}{1+x}$ works here no?

[woven radish]

MxRgD

[dull parrot]

although I guess if it's for binomial theorem, it'll involve that

[plush bolt]

please can u solve it until u get final answer?

[dull parrot]

I was thinking you can reduce the sum into a formula since its just a geometric progression

then find the coefficent of x^70 using that formula and binomial theorem

[plush bolt]

i didnt get it

[dull parrot]

Just try reduce the sum

using this, and see what you can find

as a hint

[plush bolt]

how to find number of terms bro?

[ocean haven]

use this to reduce the expression

then find the coefficient of x^70

[plush bolt]

bro i got the answer

thanks for the help

.close

[devout snow]

Channel closed

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[ocean haven]

do open new channels

[devout snow]

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[robust bobcat]

for number 10 b. I have a couple of questions regarding my attempt. I have proven continuity at $h = 0$ and I know that continuity at a point doesn't imply differentiability. So $\lim_{h\to0} Q(h) = Q(0)$ but if we have continuity at that point can I not also write it as $lim_{h+x\to h} Q(h+x) = Q(h)$. Now if that is the we can rewrite the limit as $\lim_{x\to 0} Q(h+x) = Q(h)$. We have $\lim_{x\to0} Q(h+x) = $\lim_{x\to0} Q(h)$ subtracting the left side limit and then dividing both sides by $\lim_{x\to0} 1/x$ we then have $\lim_{x\to0} \frac{Q(h+x)-Q(h)}{x} = 0*\lim{x\to0} \frac{1}{x} $. Regarding this I have two questions so the limit of 1/x as x goes to 0 does not exist but then what happens when you multiply it by 0? Also I know that the limit as x goes to 0 of Q(h) is Q(h) since it is not a function of x. But isn't h just a dummy variable could I not replace it with x and then my previous statement might not hold.

[woven radish]

BigBen
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[inland carbon]

you cannot split the limit when the limit doesn't exist

$\lim_{x\to0} \frac{Q(h+x)-Q(h)}{x} = 0*\lim_{x\to0} \frac{1}{x}$ this step

[woven radish]

bloubbloub

[sand dove]

when the product of limits (like 0 * 1/0) is not defined, you're not allowed to make the split

because it can give way to any limit you want (like x/x, x^2/x, x/x^2, etc...)

[robust bobcat]

by split you mean that I can't multiply it on both sides

[fallen tartan]

Limits do NOT work like this:

\frac{\lim f(x)}{\lim g(x)} = \lim \frac{f(x)}{g(x)}

This only works if both limits exist and are finite

[sand dove]

by split we mean transforming lim(a * b) into lim(a) * lim(b)

You can't do so unless the product lim(a) * lim(b) makes sense

[sand dove]

here you have 0 * "basically infinity"

makes no sense

[robust bobcat]

what about my second question?

[sand dove]

h is a constant compared to x

so

you can't have "h = x"

You're looking at continuity and differentiability at a fixed point "h"

so you're not gonna make h vary

[fallen tartan]

• x → the variable going to 0
• h → a fixed point

So:

\lim_{x \to 0} Q(h)

means:

“take a constant value Q(h)” → result is just Q(h)

But if you replace h with x, you change the meaning completely:

\lim_{x \to 0} Q(x)

Now it’s a real limit, not a constant

[robust bobcat]

I see it now. Thanks

[fallen tartan]

Continuity → numerator → 0
• You cannot divide by \lim 1/x
• 0 \cdot \text{undefined} is undefined
• Continuity does NOT imply derivative = 0
• h is fixed, not a dummy like x

[robust bobcat]

.solved

[devout snow]

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[dense summit]

I need ideas in 3

[modern lance]

What does it say, can you translate the problem to english

[dense summit]

We need to prove that
(3-√5)ⁿ+(3+√5)ⁿ divisible by 2ⁿ

[rain summit]

hm

[dense summit]

I have Idea but I'm not sure

[rain summit]

$a^n + b^n$ is divisible by $a + b$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

am i seeing this correctly or not tho

[dense summit]

I don't think it's true
Take a=2,b=3,n=2
a+b=5
a²+b²=4+9=13

[rain summit]

oh yea

mb

$u_n$ is always guaranteed an integer or smth?

[woven radish]

1 divided by 0 equals Infinity

[modern lance]

Try induction

But first find a recurrence relation between u_n and its previous term

[rain summit]

it's stated in the question did they

[modern lance]

Yes it always is

[inland carbon]

did you use the hint

[dense summit]

What about the binomial formula

$$
\begin{aligned}
(3+\sqrt{5})^n &= \sum_{k=0}^{n} \binom{n}{k} , 3^{,n-k} (\sqrt{5})^k \
(3-\sqrt{5})^n &= \sum_{k=0}^{n} \binom{n}{k} , 3^{,n-k} (-\sqrt{5})^k
\end{aligned}
$$

[woven radish]

William James Moriarty

[dense summit]

If we sum both of them

[modern lance]

That works yes

[dense summit]

That's will be 2 × the sum 0 to n
If we assume n is even

$$
\begin{aligned}
(3+\sqrt{5})^n + (3-\sqrt{5})^n
&= \sum_{k=0}^{n} \binom{n}{k} 3^{n-k}(\sqrt{5})^k

• \sum_{k=0}^{n} \binom{n}{k} 3^{n-k}(-\sqrt{5})^k \
  &= \sum_{\substack{k=0 \ k\ \text{even}}}^{n} 2 \binom{n}{k} 3^{n-k} 5^{k/2}
  \end{aligned}
  $$

[woven radish]

William James Moriarty

[dense summit]

But that's work for 2 only not 2ⁿ

[modern lance]

Well you probably have to do some nasty algebraic steps

[desert tree]

Won't it work for 2^n with some modular arithmetic?

Using $\sum_{k=0}^{n} \binom{n}{k} = 2^{n-1}$, when k is even

[woven radish]

Sarin

[modern lance]

Uh what

Is that even true for all n, also how would that help

[desert tree]

Yes for the first question

[ebon coyote]

The hint suggests you want to do a stepped induction fyi

[modern lance]

I'm not the op

[ebon coyote]

What I mean is the recurrence relation you're suggesting isn't sufficient

i.e. you want to find a recurrence relation between the nth, (n+1)th and (n+2)th term

[modern lance]

Huh

[ebon coyote]

(I can't really verify if this hint is correct, but it does say that here)

"Hint: [We can] Show that: [...]"

[desert tree]

3 and 5 are both 1 mod 2 so the sum becomes the sum of binomial coefficents mod 2. I don't know too much about modular arithmetic though and could be wrong

[modern lance]

Oh I meant terms

[modern lance]

That only useful for mod 2, but we need to show it's 0 mod 2^n

There's probably a lot of thing going on in n choose k

That lead to it divisible by 2^n

[desert tree]

Ok, thanks for clarifying

[dense summit]

This is not right
I verify that
The right is u_(n+2)=6u_(n+1)-4u_n
I'll try with that by induction

[modern lance]

Wait ykw, that works yeah

Mb

You already handle n choose k with that sum

[modern lance]

I forgot this

[desert tree]

Also if you were wondering, you can prove this using the expansions of (1+x)^n and (1-x)^n

[dense summit]

I try to get a relation but it don't work

[dense summit]

Taylor series?

[ebon coyote]

omg fuxk off

[modern lance]

<@&268886789983436800>

[desert tree]

No just binomial expansion

[dense summit]

I do that but It don't works

[desert tree]

What do you mean?

Are you talking about your question in general or the sum I provided?

[dense summit]

The sum

[dense summit]

This sum

[desert tree]

Oh, I was talking about a different sum. Your question mentioned induction so you should probably do that

[dense summit]

I get the solution by different way

I use the golden ration

$$
3 + \sqrt{5} = 2(\phi + 1), \quad 3 - \sqrt{5} = 2(2 - \phi)
$$

$$
(3+\sqrt{5})^n + (3-\sqrt{5})^n = 2^n \big( (\phi+1)^n + (2-\phi)^n \big)
$$

[woven radish]

William James Moriarty

[dense summit]

We can use the binomial expansion here

$$
(\phi+1)^n = \sum_{k=0}^{n} \binom{n}{k} \phi^k 1^{,n-k} = \sum_{k=0}^{n} \binom{n}{k} \phi^k
$$

$$
(2-\phi)^n = \sum_{k=0}^{n} \binom{n}{k} 2^{,n-k} (-\phi)^k = \sum_{k=0}^{n} \binom{n}{k} 2^{,n-k} (-1)^k \phi^k
$$

$$
(\phi+1)^n + (2-\phi)^n = \sum_{k=0}^{n} \binom{n}{k} \phi^k \left( 1 + (-1)^k 2^{,n-k} \right)
$$

[woven radish]

William James Moriarty

[modern lance]

Yeah well that just single handly solve the question

[desert tree]

Don't you have to prove that this is an integer?

[dense summit]

Yeah That's don't work

$$
\phi = \frac{1+\sqrt{5}}{2}, \quad \bar{\phi} = \frac{1-\sqrt{5}}{2}
$$

$$
3 + \sqrt{5} = 2(\phi + 1), \quad 3 - \sqrt{5} = 2(\bar{\phi} + 1)
$$

$$
(3+\sqrt{5})^n + (3-\sqrt{5})^n = 2^n \big( (\phi+1)^n + (\bar{\phi}+1)^n \big)
$$

$$
(\phi+1)^n + (\bar{\phi}+1)^n = \sum_{k=0}^{n} \binom{n}{k} (\phi^k + \bar{\phi}^k)
$$

[woven radish]

William James Moriarty

[modern lance]

Huh wait

Just stop for a moment

[dense summit]

This is the binet formula

Where φⁿ+φ^-n is an integer

[modern lance]

Well good enough ig

I haven't verified the last step tho

[dense summit]

$$
(\phi+1)^n = \sum_{k=0}^{n} \binom{n}{k} \phi^k \cdot 1^{,n-k} = \sum_{k=0}^{n} \binom{n}{k} \phi^k
$$

$$
(\bar{\phi}+1)^n = \sum_{k=0}^{n} \binom{n}{k} \bar{\phi}^k \cdot 1^{,n-k} = \sum_{k=0}^{n} \binom{n}{k} \bar{\phi}^k
$$

$$
(\phi+1)^n + (\bar{\phi}+1)^n = \sum_{k=0}^{n} \binom{n}{k} (\phi^k + \bar{\phi}^k)
$$

[woven radish]

William James Moriarty

[dense summit]

That's it

[modern lance]

fascinating, yeah you solved it

[dense summit]

Yeah
I think I need to know the prove of binet formula

It's elegant

Thanks guys

.solved

[devout snow]

Channel closed

Closed by @dense summit

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[modern lance]

Well you're supposed to use induction lmao that's why you haven't taught binet formula

Btw we ain't that helpful here tbh

[desert tree]

Yeah you did by yourself, good job

[dense summit]

Yes I need use the induction but it's boring
I'll try to solve it also by induction

[devout snow]

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[indigo anchor]

If $a, b, c, d \in \mathbb{Q}$
And: $$\sqrt{a} + \sqrt{b} = \sqrt{c} + \sqrt{d}$$
And: $$(a \neq c \land b \neq d) \land (a \neq d \land b \neq c)$$
Then prove that: $$\sqrt{\frac{a}{b}}, \sqrt{\frac{a}{c}}, \sqrt{\frac{a}{d}}, \sqrt{\frac{b}{c}}, \sqrt{\frac{b}{d}}, \sqrt{\frac{c}{d}} \in \mathbb{Q}$$

[stone stump]

what about it

just find examples?

[woven radish]

Anshuman

[indigo anchor]

hey guys

i would like to get me proof reviewed

i mean checked

[stone stump]

I dont like "by symmetry". sqrt(ac) might behave differently compared to sqrt(ab) due to signs

but eh, probably works out

[indigo anchor]

so are you letting it slip or is it actually valid?

[stone stump]

assuming I knew the proof I might let it slip

you should maybe instead write "analogously"

[indigo anchor]

oh

so the wording huh

okay ill keep that in mind

other than that the proof checks out right

no errors?

[stone stump]

up until that point it looks good

[indigo anchor]

hmm

[stone stump]

I can understand why you didnt do the other ones aswell

I imagine its essentially identical

[indigo anchor]

yes

plus doing it 6 times is not ideal

[stone stump]

at that point its a question for what audience you are writing the proof

[indigo anchor]

makes sense

actually it was a solved example in a textbook which i could not understand properly as in the textbook it was like just 3-4 lines long. so i was trying to prove it my way

so i did and wanted to know if it checks out

[stone stump]

did they do it this way or some other way

[indigo anchor]

i still dont understand them tbh

ignore the "case 1" "case 2". they were easy subparts. the question asked in this help channel was the last subpart

this is the og question

u there?

[stone stump]

ok so they really just skipped all the work you did

from what I see

[indigo anchor]

oh

i see

i can see some correlation between their and my work but they just basically went "obviously, obviously, obviously, blahblah"

anyways thanks

.close

.close

[devout snow]

Channel closed

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[devout snow]

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[silver breach]

can anyone translate this? i cant read it xD

[silver breach]

the black font words

[indigo anchor]

maybe a pharmacist could help

[ocean haven]

what language is that

[hollow merlin]

The first picture seems to be about the seven bridges of Konigsburg

[indigo anchor]

guys

[silver breach]

ohhh it says bridgeee

[indigo anchor]

i know

[silver breach]

i thought it was badges

[hollow merlin]

It's a famous problem in graph theory

[ocean haven]

it is possible to traverse each bridge only once and get back to start

[woven vale]

it doesn't help that one of the words is german, does it

summary is: königsberg is four islands connected by seven bridges (purple). is it possible to traverse each bridge once & only once, and make it back to the start?

it's modelled as a graph in the second slide. an easy necessary condition is, in order to traverse a landmass (vertex), you must enter by one bridge (edge) and leave by another, different edge. (and at the start, you leave by one bridge, then enter by another)
=> every vertex must have even degree
=> assume connected

TONCAS: The Obvious Necessary Condition is also Sufficient

[ocean haven]

was sorrounded by pregel river and four land masses seven bridges connectingthe land masses

city of konisberg above that maybe

[silver breach]

omg thats a w for was

[silver breach]

THANK YOU THAT MAKE SENSE

[woven vale]

no problem! unfortunately, graph theory professors are Just Like That when it comes to bad handwriting

i remember in my graph theory course, i was confused why there was a lonely $\int$ symbol - turns out it was $\gamma$

[woven radish]

حسیب ♥

[ocean haven]

https://tenor.com/view/leo-dicaprio-djngo-gif-18718603

[silver breach]

yeah xD i thought i saw a integral too but it was lambda.. i miss a day and i didnt take any notes to read later

.close

[devout snow]

Channel closed

Closed by @silver breach

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

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[cursive sage]

im soryy that this is in french but i think i need a lil help in this

[cursive sage]

how can i prove the existence of only 2 solutions

[devout snow]

@cursive sage Has your question been resolved?

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

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[wet dirge]

hey im having trouble with this type of promblems

[wet dirge]

i understand slope and how it's rise over run but i dont know how to compare them

[versed juniper]

whatevers bigger grows faster

[wet dirge]

but if it like 2 over 3 compared to 3 over 5 how would i do it

[versed juniper]

they're just fractions

find a common denominator

rewrite both in that common denominator

compare numerators

[wet dirge]

would i but able to multiply it to have the same denominator and choose the one with the bigger top number?

i would choose the one with the bigger numerator?

nevermind i got it

thanks for the help

close.

.close

[devout snow]

Channel closed

Closed by @wet dirge

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[devout snow]

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[opaque forum]

After integrating, I'm getting a positive answer, although the answer is negative.

[opaque forum]

Where did I do wrong?

my workflow :
xsquare = u
2x dx = du
dx = du/2
substitute

1/2 integral(e^-u du)

= 1/2 e^-x2

[last parrot]

is this the gpt guy yesterday

[opaque forum]

That was gemini

[last parrot]

so you was using AI 💀

[opaque forum]

Yeah so where did I get the sign wrong

[opaque forum]

Not to solve it or smthg I was spectating

Am I not allowed to? (just experimentally)

please tell me abt the sign issue here

[last parrot]

wait let me write down

[winter patrol]

not allowed to use it for others,
not allowed to represent it as your own work
are allowed to use it for yourself although we advise against that

[errant harbor]

the integral of e^-u is -e^-u

[opaque forum]

Oh.

Okay I'll remeber that

Tysm

[opaque forum]

Didnt try to show it as my own either, it was clearly ai

but still got it

.close

[devout snow]

Channel closed

Closed by @opaque forum

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[last parrot]

should make e^x^2 be e^-x^2 first

then make u = -x^2

[opaque forum]

Oh okay

Ty

[devout snow]

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[primal badge]

Can someone help me with this

[primal badge]

I am struggling with the angles

Between them

Can someone help me real quick I have an exam to catch 😭🥀

[chilly owl]

damn twin your cooked

[primal badge]

💔 this is the hardest at most which will come

[chilly owl]

why does your handwriting look so good

how bru

[primal badge]

💔 yk the torture

Smh

Of assignments trained me

[chilly owl]

damn

[primal badge]

Can u help me. Bro

My exam is in

30 minutes

[chilly owl]

dude

[hot terrace]

u hv the gradient right

[chilly owl]

i skipper half of 7th grade and im skipping 8th grade

im going directly to 9nth

im cooked too

[primal badge]

Yes

[primal badge]

Oh it’s the 10 th grade