help-49

[calm halo]

And yes u are done, nothing more required

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[last slate]

Why do they set the x and y equation to y

could you just as well choose to set-up with x

[hard shard]

you said it yourself, they could just as well do it with the other variable

its whatever they chose to do

[last slate]

Ok just wanted to make sure

ty

[broken jolt]

there's no compelling reason to choose one over the other either

🤷‍♂️

[last slate]

Thanks guys

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[spring meadow]

Can i have some help with this question pls

[spring meadow]

I am stuck on part b

[midnight plank]

@spring meadow Has your question been resolved?

[spring meadow]

<@&286206848099549185>

[distant heron]

First it travels $d$ to go to the floor then $2\left( \frac{41}{50} \right) d$ when it first bounces and goes back to the floor then $2\left( \frac{41}{50} \right)^{2} d$ etc

[grand pond]

black_couscous

[distant heron]

When you write it down you get a geometric series

[spring meadow]

ik its a geometric but idk what to do with that after

[distant heron]

You've got $d + 2d\sum_{n \ge 1}\left( \frac{41}{50}\right)^{n}$

[grand pond]

black_couscous

[spring meadow]

sorry i dont mean to ask alot of questions but how have u gotten that

actually dont worry

i understand

[distant heron]

Don't worry you're here to understand. I just continued this

[spring meadow]

okay can u continue pls

@distant heron

[distant heron]

Then this is equal to $d + \frac{82d}{50}\cdot \frac{50}{9}$

[grand pond]

black_couscous

[spring meadow]

how do you find this out?

[distant heron]

I used the geometric series

[spring meadow]

i havnt done geometric series using sigma notation

would u mind explaining how to work it out quick please

[distant heron]

$1+ r + r^2 +... = \frac{1}{1-r}$ if $-1<r<1$

[grand pond]

black_couscous

[spring meadow]

oh yh

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[short chasm]

Yo a brother need help

[midnight plank]

Please don't occupy multiple help channels.

[tropic nexus]

in what way do you wish for this help

[terse spear]

Don't open multiple channels

.close

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[terse spear]

You have one #help-5

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[plucky canopy]

https://media.discordapp.net/attachments/936864151425413143/1032810976321478677/unknown.png

[plucky canopy]

ok so

i assume the first thing i do is 2(sqrt x + 11 - 2) on the top

i know usually for a square root you would multiply it by its opposite self but i havent seen something in the practice questions of a 2 variable sqrt before so im not sure how to go abouts doing this

[patent jolt]

https://tutorial.math.lamar.edu/Classes/CalcI/LimitsProperties.aspx

[plucky canopy]

2x + 22 - 16 / x + 3 ( sqrt 2x + 22 - 4)

[ebon nest]

not sure what u mean

is the issue is ur unsure about how to rationalize the numerator ?

[plucky canopy]

nvm nvm i solved it sorry

.close

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[dawn crescent]

How much should you deposit in a bank paying 2% compounded quarterly to accumulate an amount of P80,000 in 5 years and 9 months?

[ebon nest]

isnt there like a finance thing on ur calculator to this?

been a while but there's like PV PMT FV thing?

[vapid echo]

Yeah there is.

[noble eagle]

$A = P(1 + \frac{r}{n})^{nt}$

[grand pond]

Bestower

[vapid echo]

@dawn crescent

In this case it would be easier to have that $A = P(1+r)^t$ given that t = quarterly periods

[grand pond]

clip

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[topaz willow]

Is number 11 √30/ℼ??

[midnight plank]

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[topaz willow]

<@&286206848099549185>

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@topaz willow Has your question been resolved?

[velvet oar]

@topaz willow sqrt(30/pi), yeah

[topaz willow]

Ok

Thx

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[slender walrus]

they are not

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[midnight plank]

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@last slate Has your question been resolved?

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[flat halo]

computing residue of $f(z) = \Gamma [z] \Gamma [l-z], l \in \mathbb{N}$

[flat halo]

so i got residue at $a \in \mathbb{Z}, a \le 0$ as

[hard umbra]

wtf square brackets

[flat halo]

$\frac{\Gamma(l-a)}{\prod_{k=a}^{-1}k}$

[lyric charm]

\prod not \Prod

[flat halo]

can i simplify this? or its decent

[lyric charm]

i'd rewrite the product in denom as (-1)^a * |a|! ...

[hard umbra]

match the gamma function in the numerator with a factorial

[flat halo]

$\frac{\Gamma(l-a)}{(-1)^a |a|!}$

[flat halo]

hm?

so what do i do :c

[hard umbra]

you might have a bad representation

[flat halo]

do u think i just calculated it wrong n i shld show how i got it

[hard umbra]

no as in

your calculation might just be unhelpful

[flat halo]

lol

[hard umbra]

you want to like

shift the gamma functions out of the region you're trying to calculate the residue

so that it's holomorphic where you're taking the residue

and then in the process of shifting you should get a bunch of factors in the denominator

[flat halo]

yea thats what i did

so like

[hard umbra]

oh ok

[flat halo]

$(z-a) \Gamma(z-a+1) \Gamma(l-z) / [z(z+1)(z+1)...(z-a)]$

[hard umbra]

rewrite the numerator as

a factorial?

and cancel maybe

but then

it doesn't necessarily make it better

you can probably just leave it

[flat halo]

oh yea

i can rewrite it as a factorial

ok so i guess the final form is still ugly either way?

[hard umbra]

well you can also rewrite the denom as a gamma function

it's gonna look like that

[flat halo]

idk in another q the final bit is like this

[hard umbra]

cuz you get part of a factorial popping up

[flat halo]

is this any better

[hard umbra]

oh boy

that looks kinda terrible

[flat halo]

also is prod vs factorial any better

or either is fine

in terms of simplified anw

ans

[hard umbra]

factorial is smaller in the denom of a fraction

[flat halo]

wdym

[hard umbra]

that's probably one benefit lol

like

if you write the product symbol

it's massive in the denom of a fraction

so either you do a ^-1 exponent

[flat halo]

but isit considered simplified or nay

[hard umbra]

or you write it as a factorial

well

depends

if you can write it as factorial it's probably better

[flat halo]

this mayb

[flat halo]

hm oki

thankss

[hard umbra]

that's a factorial

[flat halo]

:c

lol

cuz it ends at -1 right

so if it starts at 1

[hard umbra]

yeah

[flat halo]

or ends at -1

then i can rewrite it

[hard umbra]

maybe id write (-a)!

instead of |a|!

[flat halo]

wait

thats a thing?

[flat halo]

this

[hard umbra]

well i mean

a is a negative integer isn't it

so -a is a positive integer

[flat halo]

OH WAIT

lol

ok

thanks

Breakthrough

[hard umbra]

(l-a)Pa

[flat halo]

Level up Divine Pulse Layer 8

amazing

hm?

[hard umbra]

like

nPk

lol

oh wait there should be a -1

same same

[flat halo]

y r u talkin abt permutation rn

im so confused

ok well

my main dish has been served

so

.close

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[keen turret]

Help

[last slate]

send your question

[keen turret]

Ok

This one?

[tropic nexus]

take square root

and don't forget ±

[keen turret]

Ok

[keen turret]

Hmmmm

[keen turret]

That's it?

[tropic nexus]

I believe so

[keen turret]

Ok..

Is this correct?

...

[keen turret]

After 15m I should ping helpers?

[last slate]

idk but that's what it says

[keen turret]

Ok.

[jade stag]

Hello

Whatcha need help w?

[keen turret]

Yo

[keen turret]

This one

[jade stag]

Ur answers are sooooooooo close

Just has to be negative

[keen turret]

So close?

[jade stag]

-4/3

[keen turret]

Ok

[jade stag]

Anything else?

[keen turret]

Hmmm wait.

[jade stag]

Waiting…

…

[keen turret]

Here

[jade stag]

Need a walk through?

[keen turret]

Yup

I think I'm stuck at 1 & 2

[jade stag]

Aight

[keen turret]

It's not easy for me yet.

[jade stag]

So I think “what factors of -27 add up to +6”

[keen turret]

Which number?

[jade stag]

“There’s -1 27 to get 26, -3 9 to get +6, oh wait that’s the answer”

[keen turret]

...

[jade stag]

And then I rewrite it as (x-3)(x+9)

[keen turret]

Ok

[jade stag]

So (x-3)(x+9) = 0

Oh u got it? Ok

U can factor out an x

[keen turret]

A little

[keen turret]

Ok

[jade stag]

So x(6x-8)=0

U got this one too?

[keen turret]

Yes I think

[jade stag]

(x-3)(x+9) = 0
Expression is zero if either factor is zero, so x-3 =0 and x+9=0

And solve for x

Now u all good?

Anymore problems that’s bothering u? Lemme at’m

[keen turret]

will this channel get removed?

[keen turret]

ok

[jade stag]

If it times out yes

If u do .close the channel is also removed

It times out after 15 minutes of no activity I think

But I gtg. Later bro

[keen turret]

ok cya later

Like this?

[midnight plank]

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@keen turret Has your question been resolved?

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[earnest prawn]

This is a question from an exam I had a few days ago

[earnest prawn]

I was not able to solve it, what kind of series test should have I done?

Integral or?

I need to find out if the series converges or diverges

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[last slate]

Want a refresher on how c would work again

[last slate]

i suppose it is when they are identical, but how would i relate that to here

[strong lava]

You do want both lines to overlap (lines in the x1 x2 plane ofc). Therefore you want the same slope and a shared point

[last slate]

hm so

ah i guess this means

h is 2

okay thats that

k is 9?

[strong lava]

Yes

Another way to see it is indeed to make the equations the same, i.e. show they're equivalent

[last slate]

oh interesting, it also has NO solutions for any number that is h = 2 and not equal to k = 9

[strong lava]

Because parallel lines

[last slate]

ah i see

[strong lava]

All concepts that are further developed in linear algebra btw

[last slate]

same slope, different intercepts would indicate they are in fact parallel, same for both would be just both being overlapping thus having infinitely many solutions

fair enough

[last slate]

which is the course i am taking right now 😄

very fun, just building up my basis again

[strong lava]

If you want I can somewhat expand on this

Though preferably knowing kernels already would be a +

[last slate]

ah yes, i would like to prove it mathematically

[strong lava]

Linear algebra is this field of studying complicated properties of simple things

Linear applications are indeed very simple

You only need to know how they act on a basis to know how they act on any vector

Did you see kernels yet ? I'd assume you haven't

Ok actually, due to being in an affine vector space, almost all of linear algebra doesn't directly apply lol

But the concepts still carry over

[last slate]

oh funnily enough

apparently it is one of the few fields of maths where you could say that is 'fully understood' from what i read

kind of shocked me to a degree, but it is interesting

[last slate]

Nope, i can show you our syllabus

[strong lava]

It's much easier to study in finite dimension, but the existence of a basis for any infinite dimensional vector space makes it manageable

[last slate]

this can be counted as an introductory course, so yes

[strong lava]

Sounds like intro to linalg

Yes

[last slate]

just exited the lecture an hour ago, just talking about null spaces for now

so thats as far as my knowledge goes

[strong lava]

Another name for kernel

Actually

[last slate]

oh really?

we just got introduced to it today haha

[last slate]

I see

[strong lava]

To try and apply linalg to this problem, I think it's fair to ask "could there be finitely many (but > 1) solutions" ?

[last slate]

that sounds like a contradicting statement for the layman i won't lie

how can something that could be anything be represented finitely?

[strong lava]

Why can't there be exactly 3 solutions to this equation?

Is what I meant

[last slate]

You mean for our original question problem?

[strong lava]

Yes

[last slate]

hmm

3 solutions would suggest that they would intersect at three different points

that's not really..possible for simple lines, i am guessing?

[strong lava]

Yes

[last slate]

interesting analogy, i like it

[strong lava]

It's not an analogy

[last slate]

yeah it is the stuff itself, more like it

[strong lava]

These equations define lines in the x1 x2 plane

[strong lava]

To be able to apply linalg to this, I'm just going to shift the coordinate space around, say shifting the origin 3 units to right. That way we have x + hy = 0

Where x and y are our knew coordinates

[last slate]

Also, I find it interesting how you can construct problems for things above 3 dimensions even if you can't really...represent those dimensions, right?

makes sense

[strong lava]

You learn to either take your examples in 2 or 3D, or you just do away with such visualizations. With enough habit you just don't need them to understand the nature of stuff

Which is fortunate, because having an exam that is equivalent to the study of a 9D cone would be rough if we could only rely on visualizing it that way

[last slate]

yeah exactly

what would 9 dimensions even be..

it is so astonishing how you can represent things in maths that the human brain definitely cannot feel or see

[strong lava]

Then the 2nd equation becomes 3x + 6y = k'. Since it's an equation, we divide it by 3 to get x + 2y = c (and we need to find c)

[strong lava]

It's 3^2. That's how the 9 came up. A linear application over R^3 is represented by a point in 9D space

In the exam I'm thinking of, we were considering some of these applications, and it ends up being a cone in the 9D space in which you can represent them

[strong lava]

We want to find x and y that verify both equations. Clearly c = 0 is the only solution (working backwards you'd get k=9 back)

But then, every point on one line is on the other, because they're described by the same equation

So there has to be infinitely many solutions

In linear algebra, that generalizes to the intersection of null spaces

If instead I had f(x) = g(x) = 0, where f, g are linear in some n dimensional vector space

Which you can just think of as R^n basically without loss of generality

[strong lava]

Then the vectors x that are solutions both to f(x) = 0 and g(x) = 0 are therefore in Ker f and in Ker g (by definition of the null space

That means x is solution iff x is in Ker f intersected with Ker g.
But an intersection (even infinite) of vector spaces is a vector space

Therefore the set of solutions is a vector space: a point, a line, a plane, etc...

Therefore as soon as there's a nonzero solution, because the equation is linear, all of its multiples are and there's infinitely many solutions

[last slate]

Sorry, currently unable to read, but i will make sure to read this in my free time if you don't mind

[strong lava]

For example, a plane is described by an equation, like x + 2y - 3z = 0. This is actually a consequence of the fact hyperplanes (in finite dimension, these are n-1 dimensional subspaces of an n dimensional space) are null spaces for a linear form (that is also unique up to a multiplicative constant) (a linear form maps a vector to a scalar). In this example, the linear function is f(x,y,z) = x + 2y - 3z

That's hinted at by the fact that all kernels are subspaces

Now of course, all of this is total overkill for your exercise, where it is sufficient to divide the 2nd equation by 3 and notice x can only be a solution to both if the constants are equal (exercise: why is that?)

[midnight plank]

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[midnight plank]

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[thick parcel]

Pls halp ;;;

[muted steeple]

What’s the whole question?

[lyric charm]

i think that is the whole question

it's implied that what follows is a blank to be filled in

anyway drawing a position-time graph might help

[muted steeple]

There’s something written in the next line

[meager basin]

Wait

Hey guys, has anyone seen that problem on Irodov's physics problems... in fact the first problem, its kinda the same....

[tribal carbon]

irodov

[hard umbra]

[hard umbra]

i see

[tribal carbon]

that a jee popular book

[fringe wind]

Holy jesus jee PPL here?

[midnight plank]

@thick parcel Has your question been resolved?

[midnight plank]

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[obsidian sequoia]

Write the following as an inequality.
x is less than or equal to 5 and greater than
Use x only once in your inequality,.
8

[obsidian sequoia]

Anser

[bold mango]

try using these to solve the question

[midnight plank]

@obsidian sequoia Has your question been resolved?

[obsidian sequoia]

Give answer pls🀄️🀄️🀄️

[meager basin]

Do not insist the helpers on simply giving out the answer.

[midnight plank]

@obsidian sequoia Has your question been resolved?

[midnight plank]

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[shut bronze]

shouldnt this be -2/3 since modulus of -i gives you i and when you square it you get i^2 which is -1?

[obtuse basin]

|-i| = |i| = 1

Modulus of i isn't the same as i^2

|i^2| isn't the same as i^2

Nonetheless it's 1

[sleek cloud]

I think because |z|=sqrt[a^2+b^2] for z=a+bi

[obtuse basin]

Yes

[midnight plank]

@shut bronze Has your question been resolved?

[shut bronze]

oh, yh got it, modulus of a complex number is sqrt(real_part^2 + imaginary_part^2)

thanks

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[west fiber]

well if the number you're multiplying by doesn't have the base as a factor you can't simplify it anymore

or if the base isn't simplified fully

like a square number

for exmaple

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[opaque drum]

HEELP

[opaque drum]

PLSSSSSS

if i have a ramp and i have a car, how can i make the ramp so the car has a high speed but low acceleration
also how do i make it so there is low speed but high acceleration

[keen herald]

What feature of a ramp would cause a car to accelerate really fast?

[midnight plank]

@opaque drum Has your question been resolved?

[opaque drum]

Idk

I was thinking of maybe sharp turns

Cuz even if speed stayed constant it accelerates?

@keen herald

[keen herald]

I don't think you'd need to go as far as to have the ramp change instantaneously

I imagine a very steep ramp would have higher acceleration than a very shallow ramp

[opaque drum]

yes but

what would have higher acceleration but low speed

[midnight plank]

@opaque drum Has your question been resolved?

[next rover]

this feels like it's both high speed and low acceleration

or maybe it would be considered low speed, because it loses speed with time too

if we keep it as simple as we can, the speed is proportional to height, the acceleration is the slope, the shape is just a line

[midnight plank]

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[normal citrus]

where does the p come from?

[solemn fable]

They multiplied and divided with p!

[normal citrus]

hum

(p+1)...(p+n) = (p+n)!

is it true?

[tired shadow]

No

But p!(p + 1)...(p + n) = (p + n)!

[normal citrus]

so (p+1)...(p+n) = (p+n)!/p!

honestly i don't get why

[vale canyon]

(p+1)(p+2)...(p+n) multiplies all integers from p+1 to p+n

but (n+p)! multiplies all integers from 1 to n+p

so to remove all number from (n+p)! to get to (p+1)...(p+n), you have to divide by all 1*2*3*...*p, which is p!

[normal citrus]

ok ok it's clearer, thx

.close

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[last slate]

How can i simplify this into

This

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[broken wren]

I havent come across a problem like this before

[broken wren]

I know sin(A+B) = Sin(a)+Cos(b)

but we have 2 cos so we cant really use that

<@&286206848099549185>

[muted widget]

i found a formula for this it might help

cos(x)cos(y) = 1/2(cos(x-y)+cos(x+y))

i think you carry the 6 for the ride

[midnight plank]

@broken wren Has your question been resolved?

[broken wren]

so we would get like 3(cos(27s-2s)+cos(27s+2s))

[muted widget]

yeah i think so

[broken wren]

so just submit that?

or 3cos(27s-2s) + 3cos(27s+2s))

[broken wren]

well that was wrong

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[torn marsh]

in b why is f' 0.4? shouldn't it be negative

[blazing sparrow]

realisticaly yes

or in my mmind at least

but there are weird cases where increasing the price of a product makes you sell more

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[midnight plank]

@solar dragon Has your question been resolved?

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[void widget]

Hello I'm having a problem with an image

[crude dune]

Yes

P r2

R = 14

[short nova]

Find the area of the (square?) - find the area of the circle = answer

[crude dune]

Multiply by pi

Area of sqaure is 28*28

Minus circle

[void widget]

I know both already

Why do you minus?

[short nova]

so what’s the issue

Okay let me ask

[crude dune]

To get the blue area

[short nova]

Never mind @crude dune can take over

[void widget]

Ok thanks

[crude dune]

You got it?

[void widget]

Yh I just wanted to why minus

Cause most classmates was telling me to add

But thanks

[crude dune]

Np

[void widget]

I so confused

[void widget]

Any suggestions what to say?

[crude dune]

Oh

Yeah

Is the sqaure or circle shaded reigon

First

Im assuming sqaure is shaded

Cause if circle is shaded you just do pi r sqaured and ignore sqaure

[void widget]

It is

[fallow scarab]

<@&268886789983436800>

[short nova]

bruh

[crude dune]

Just say the shaded region is definietely the sqaure because it wouldnt make sense if it wasnt

[void widget]

Thanks lad

[crude dune]

And if you were looking for whole area it is just the square ignore the circle

[void widget]

Cause I asked like 4 helpers the same question and they're doing minus so I don't know who is wrong...

.close

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[steel night]

so for b)

it would just be t distributed with n-1 degrees of freedom right

i feel like thats too easy though

or am i paranoid

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[real ice]

How would I go about solving this?

[steel night]

by hand?

[real ice]

i need a step by step on how to answer

[steel night]

oh yeah sure

[real ice]

Kinda confused on how to go about this

[steel night]

$\frac{1}{1.67 * 10^{-24}} = \frac{1}{1.67} \frac{1}{10^-24} = \frac{1}{1.67} \frac{10^0}{10^{-24}} = \frac{1}{1.67} 10^{0--24}$

what the hell

[real ice]

are you using LaTex?

[steel night]

trying to haha

[grand pond]

Gijs

[steel night]

$= \frac{1}{1.67} * 10^{24}$

[grand pond]

Gijs

[real ice]

so 10^24/1.67?

[steel night]

yes

[real ice]

how would that look without scientific notation

[steel night]

,calc 10^24/1.67

[grand pond]

Result:

5.9880239520958e+23

[real ice]

huh

thank you

i get it

[steel night]

this isnt eactly scientific notation though

[real ice]

i think my math teacher will accept it

[steel night]

oh for sure

[real ice]

.close

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[pearl fern]

what seems to be the problem here?

[pearl fern]

My answer aren't in the choices

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[dark fulcrum]

An isosceles triangle has hypotenuse of 15, determine the length of other 2 equal sides

[dark fulcrum]

I want to know is hypotenuse only exists in right triangles?

Or in isosceles triangles it’s just the different leangth side

I don’t want exact answer I want some hint

[midnight plank]

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[midnight plank]

@dark fulcrum Has your question been resolved?

[past quiver]

yeah, hypotenuse only exists in a right triangle

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[last slate]

Finally

[last slate]

.close

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[fresh crow]

If I have a/b = 2^m where m is an integer

[fresh crow]

Then do I know for sure that a and b are themselves integer powers of two?

[past quiver]

they can be or they may not be

but there are many ways that we can assemble them to be

[fresh crow]

Ahh, darn. I thought I was on the right track

I'm trying to show that if a/b = 2^m, then b/a = 2^n is also true (m and n are integers)

[past quiver]

that's super easy

no need to calculate a or b

[fresh crow]

Would you care to share so that the slower students in the class can follow along? XD

[past quiver]

b/a = (a/b)^-1

[fresh crow]

Hmmmmm

This?

[past quiver]

yeah

[midnight plank]

@fresh crow Has your question been resolved?

[fresh crow]

Rockin!

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[toxic scroll]

for the forward direction, i'm not sure what linear transformation to choose

i wanted to go with A(h)=0

but i dont think i can pull out absolute values like that

[fallow scarab]

A is a linear transformation. that means A(x) = mx + b for some constants m and b

[toxic scroll]

Right

I understand that

but how would you pick a smart A(h) such that the limit is 0?

any hints?

[fallow scarab]

write f as its taylor series

[toxic scroll]

okay

so

$f(x)=\sum^{\infty}_{n=0}\frac{f^{(n)}(a)}{n!}(x-a)^n$

[grand pond]

kawaii_pear

[fallow scarab]

only up to quadratic terms are needed

[toxic scroll]

Oh

so $f(x)=(x-a)+f(a)(x-a)+\frac{f'(a)}{2}(x-a)^2+\frac{f''(a)}{6}(x-a)^3$?

[grand pond]

kawaii_pear

[toxic scroll]

i may have messed up something its been a while

[fallow scarab]

yea that's wrong. look it up in a book or wiki

[fallow scarab]

this is correct

[toxic scroll]

wait

i think i found another way of doing this

A(h)=f'(a)*h

since we can manipulate the definition of f'(a)

[fallow scarab]

that's equivalent if you taylor expand f about a+h

[toxic scroll]

Ohh i see

okay

what about for the other direction then?

nvm

i got it

thank you

.close

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[mint carbon]

Hi, I'm doing ParabolaVertices-VertexFormula

[mint carbon]

I would like to know if it's correct so far and the steps to solve it

My problem is when substituting the X as a fraction and solving it

[midnight plank]

@mint carbon Has your question been resolved?

[mint carbon]

ok well this was helpful

.close

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[steady sandal]

hi

[steady sandal]

or can you only use the squeeze theorem to prove a sequence converges

[midnight plank]

@steady sandal Has your question been resolved?

[midnight plank]

@steady sandal Has your question been resolved?

[fiery topaz]

Hi

[midnight plank]

@steady sandal Has your question been resolved?

[clever reef]

it oscilates, it doesnt diverge

[steady sandal]

diverge was the correct answer though

if a sequence oscillates doesn’t that mean it diverges

i mean it’s oscillating between -1 and 1 i think as n approaches infinity

so it definitely diverges

[clever reef]

no, -1^n doesnt diverge. mb, the numerator oscilates between negative and positive infinity

so ig u say thats diverging

[midnight plank]

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[hollow geode]

if we want to get the sin expression

do we multiply by x and devise by 2

or like the 2nd expression (as the teacher did )

multiplied by 2 and devided by x

im confused

[open hatch]

1st line does not equal the second line

There's a sign error

[hollow geode]

inside parenthesis

?

the minus sign is outside

[open hatch]

Yes

[hollow geode]

minus the fraction not only the exponential

yes we put it outside

to have 2i in the denominator

we have to multiply by x/2 right ?

[open hatch]

Ok first of all do you mean to say e^(ix) + e^(-ix) in the second line?

[hollow geode]

what

[fallow scarab]

-a + b = -(a - b)

if you're gonna factor out a - sign

[open hatch]

Is this a fourier transform btw

[hollow geode]

no its not

should be - (x/2)

am i right

[open hatch]

there seems to be a sign error again

here

[open hatch]

you don't need this

[open hatch]

here

[hollow geode]

would the sign be there if it was x/2

[open hatch]

what. if you just remove the minus sign then it's correct

IDK what "it" is

[hollow geode]

We wanted to get the sin expression

thats why we need to multiply by x/2

but its 2/x

[open hatch]

do you do physics?

[hollow geode]

yes

[open hatch]

i can tell

smh

[hollow geode]

im offended

anyways

this is what i mean

[open hatch]

second line is zero

[hollow geode]

omg ignore the sign

[open hatch]

hmm

[open hatch]

correct

[hollow geode]

i don't think so

when we want to get the sin expression

we multiply by x/2

and then 2/x

thats why at the end we got

[open hatch]

consider using equals signs

[hollow geode]

alright

i give up on this

i'll come back for another integral

stuck on some basic maths

thanks

.close

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[vital nacelle]

Let ABC be a right-angled triangle with \angle A=90. If AB=500 cm and \angle C=30, BC=?

[cedar bone]

do you know

thats either law of sines or law of cosines

i think cosines

[vital nacelle]

I didn't learn it

[cedar bone]

just google it

its an equation

[slender walrus]

have you drawn a diagram

thats either law of sines or law of cosines
overkill

[vital nacelle]

I have.

[slender walrus]

can you show your diagram

[winter steppe]

$sin(30)=\frac{1}{2}=\frac{500}{BC}$

Its just right angle trig

[grand pond]

Tech support

[midnight plank]

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[carmine narwhal]

Hi can someone help me? My answer is indeterminate 🥺 should i consider the fact that my function has asymptotes? I dont know how to approach this problem 🥹🥹

[halcyon trellis]

U sub is the way to go

[solemn fable]

This is rhe same question isnt it

[solemn fable]

They asked to integrate it to pi, function isnt even defined at pi

[halcyon trellis]

Ah

[velvet compass]

mhm it isn't even improper integral

[velvet compass]

but

[halcyon trellis]

Technically you can still get a complex answer but I don’t think that’s what you’re looking for here

[velvet compass]

it's possible to solve it over complex

[carmine narwhal]

Is splitting integral not possible? Like from 0 to 1 then from 1 to pi?

Im pretty sure the answer there can be found using complex process but i dont know how to

[velvet compass]

this is even correct result (according to WolframAlpha)

[carmine narwhal]

Coz they wont be giving that if it’s not answerable

[halcyon trellis]

Just plug in pi and 0

[carmine narwhal]

Yup but the answer is indeterminate

That’s my problem

They want definite answer 😔😔

[solemn fable]

The answer isnt valid at least in real number cuz the argument for sin inv is greater than 1

[halcyon trellis]

It is possible over complex

[velvet compass]

it's complex number then

[carmine narwhal]

How can i solve it? Like the process. Coz i really dont have idea

[halcyon trellis]

Use the complex definition of sin

[midnight plank]

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[last slate]

help

[midnight plank]

Please don't occupy multiple help channels.

[warm magnet]

Just post the question

[last slate]

how to find X

<@&286206848099549185>

[warm magnet]

Dude you're breaking rules

Chill

Is there a typo

Is it supposed to say "BD bisects ABE"?

[last slate]

idk man

that’s what i said

[warm magnet]

Well with the information given, it's not possible

[last slate]

i think it is

[warm magnet]

Not unless I can get clarification about a possible typo

Then no, it's not possible

[last slate]

just assume it is please

[warm magnet]

Well then it should say BD bisects ABE

[last slate]

he makes typos a lot

[warm magnet]

So which angles are congruent

Given the information

[last slate]

3 and 4?

[warm magnet]

What else

[last slate]

1 and 2

@warm magnet

[warm magnet]

You don't have to ping me every goddamn secon

But yes

But now let's look at the first question

[last slate]

sorry I’ve been stuck on this for 45 minutes

[warm magnet]

Its asking you find m4

You know that m3 = m4

[last slate]

yea

[warm magnet]

So set up that equation

[last slate]

ok

[warm magnet]

They give you equations for m3 and m4

[last slate]

9x+10?

[warm magnet]

What, no

Just set up an equation

m3 = m4

Legit substitute with the information they gave you

[last slate]

ok

[last slate]

4x + 10 + 5x = ??????

4x + 10 = 5x*

[warm magnet]

Yes

Solve for x

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[midnight plank]

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[heavy falcon]

Hi

[heavy falcon]

I have a question

How to prove that the vector 0 such that for every vector x, x+0=x , is a unique vector

??

[last slate]

assume there is another zero vector 0', then using this axiom and commutativity of addition, can you show that 0=0'?

[heavy falcon]

Give me a sec to do this

I am stuck

I think that I am stuck because I am stupid

So for now I assumed that there exists 0' different than 0 such that x+0'=x

[last slate]

start like this
0 = 0+0' = ...

[heavy falcon]

Ok so if I replace 0' by x+y where x and y are additive inverses

[last slate]

no no, no need to introduce any x or y

[heavy falcon]

Then 0=0+(x+y)=(0+x)+y=x+y=0'

[heavy falcon]

I think I have a mistake here

Since I can't say x+0=x

[last slate]

just look at 0+0', whys it 0'?

[heavy falcon]

Wdym by that

Bc its different than 0

[last slate]

i mean, why is 0+0'=0'?

[heavy falcon]

Bc 0 is the vector such that 0+x=x and in our case x=0'

[last slate]

so,
0=0+0'=0' right?

[heavy falcon]

Ok I have a question is 0 here already defined to be the vector such that x+0=0

Bc if that's the case then we can do this

[last slate]

yes, we assume
x+0=x and
x+0'=x

[heavy falcon]

But in this case 0'=0

[last slate]

just as a tiny detail, i would add
0=0+0'=0'+0=0'

[heavy falcon]

Oh ok

But I have a question

Can't we work it as follows

x=x+0'=x+0

By assumption

So x+0'=x+0 then 0=0'

[last slate]

yes that should work ..

but you gotta show why 0=0' follows from this

just substitute x=0 i guess

[heavy falcon]

Bc x=x

Then subtract x from both sides

[last slate]

subtracting x from both sides would work yes, but doing it axiom by axiom is a bit of work, depends how detailed you want it

[heavy falcon]

Or in other words add the additive inverse of x to both sides call it y

[heavy falcon]

Actually I am studying linear algebra and this is the 2nd section of the first chapter so I probably do it using the axiom

[last slate]

if you just put x=0 it will work out much easier

[heavy falcon]

But if I plug x=0 isnt that just working in a particular case

Oh wait nvm

To obtain 0=0+0' where 0' is the vector such that x+0'=x I plug x=0 to get the first equation

[last slate]

we have
x+0'=x+0 for all x, so in particular it works for x=0 too, and everything that follows from there is true too
the result 0=0' doesnt depend on any x

[heavy falcon]

Oh okk

Tysm for your time and sorry if I annoyed you

Have a very nice day

.close

[midnight plank]

Channel closed

Closed by @heavy falcon

Use .reopen if this was a mistake.

[last slate]

np, thx

[heavy falcon]

np

.reopen

[midnight plank]

✅

[heavy falcon]

I have another question

If I want to show that y such that x+y=0 is unique then I do a similar procedure ?

[last slate]

yeah assume x+y=0 and x+y'=0 and show y=y'

[heavy falcon]

So x+y=x+y' then I plug x=0 to get y=y'

?

[last slate]

hmm i think that wouldnt work in this case because y depends on x, like you would get
x+y=0
=> 0+y=0

[heavy falcon]

Oh ok

[last slate]

but instead of plugging x=0 theres another thing you can do to get rid of the x

[heavy falcon]

Add y on both sides

Is that right

If I do this then I get x+y+y=x+y'+y then 0+y=0+y' which means that y=y'

[last slate]

yeah, like
x+y=x+y'
y+x=y'+x
(y+x)+y=(y'+x)+y
y+(x+y)=y'+(x+y)
...

thats lots of work again unfortunately

theres also the "smart" way
y=y+0=y+(x+y')=...
can you see how to continue?

[heavy falcon]

But right after the last step you wrote comes y+0=y'+0 and y+0=y which is the same for y'+0=y' then y=y'

[heavy falcon]

Yea =(y+x)+y'=y'

[last slate]

yeah, for these kinds of problems theres always the short path, but sometimes its hard to think of it

[heavy falcon]

It's nice to think of the short path

You helped me a lot tysm

Self studying is nice but needs effort

These are simple but I am stupid atm

[last slate]

that will come over time, once you have much more complicated concepts this will be easy

[heavy falcon]

Self studying isnt hard it's just harder than studying in the class

That's why it needs more work and effort than studying in class

[heavy falcon]

These concepts should already be too easy but idk why I was stuck at them

Maybe because I didn't even think about them before since I saw them as pretty much obvious ideas which don't need to be proved

[last slate]

but now you learned them

[heavy falcon]

Yea thanks to you

I am grateful for you

You gave me from your time

I gtg sleep gnnnn and have a very nice dayyyy/nightttt bc idk whats the time in your country rn XD

Cya

[last slate]

bye

[heavy falcon]

.close

[midnight plank]

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[midnight plank]

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[umbral cradle]

Number 9 down

[umbral cradle]

Am I able to add the x to the power down and remove the lnx2 with lnx4 to have 1/lnx2

[midnight plank]

@umbral cradle Has your question been resolved?

[umbral cradle]

<@&286206848099549185>

[umbral cradle]

.close

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[carmine ferry]

that is the only issue I am having with this problem why is it not -(x+s)^3+x- (x^3+x)/S

.close

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[uneven sandal]

What does it mean for something to be an orderd field

[uneven sandal]

that 2 objects are comparable?
with < or = or > ?

I guess

[midnight plank]

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[cosmic storm]

Is this right???

How is f’(x) equal to 0 i don’t get it

[last slate]

because cos is a function of t not of x

hence the derivative with respect to x is 0

[cosmic storm]

Ohhhhh

Ic

Ty

.close

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[gusty trellis]

why is the bigger power negligible ?

[gusty trellis]

if there is a more simple explanation i would appreciate that

[broken anchor]

What's the confusion exactly

When x gets small larger exponents go to 0 faster

I can't really explain it simpler than that without knowing what exactly you're stuck on

[flat halo]

$\lim_{x \to 0} x^a = 0$

a>0

[gusty trellis]

if its x^6 and x^5 if x=1 they both will be 1, so wont they reach 0 in the same speed ?

[flat halo]

um

it said x->0

not

x->1 or x->c

u cant just change that to x->1

[gusty trellis]

but in a graph representation, how will they go to zero faster if they are both x^m and x^n

[flat halo]

er

just look at

$\frac{x^n}{x^m}$

[flat halo]

thats the defn of it anw

[gusty trellis]

so basically they will hit 0 at the same time, but limit here is for the area before literal 0 right ?

[flat halo]

in which case u get some this with a>0

[broken anchor]

Yeah what matters is the behavior close to but not at 0

[flat halo]

duh, exactly = 0 when x=0

[flat halo]

but this

hit 0 doesnt matter

like it can literally never hit 0

thats not what we are concerned about here

[gusty trellis]

we are concerned about this ?

[flat halo]

yes

[broken anchor]

Well the goal is to help us evaluate limits

[flat halo]

so we just some deleted neighbourhood of the point we are evaluating

[gusty trellis]

so the larger the exponent, the faster it will go to 0 when x is decreasing to 0

which makes the bigger exponent negligible

[broken anchor]

Yeah basically

[flat halo]

um vanish faster

[gusty trellis]

but when x tends to infinity, the smallwr is negligible if the limit is 0 because it grows slower ?