help-39

[copper umbra]

you may fix it to have r = 10000 or r = 10000000

[midnight haven]

if tht isnt possible, then okay x^2 cant be contained in a circle

[copper umbra]

yes y = x^2 can't be contained in a circle

because x goes to infty

and whatever circle of fixed radius you draw, y = x^2 will eventually move outside of the circle

[midnight haven]

well of course

if i can move at any speed forever

and you're stuck infront of me at some fixed non-changing distance

i will always end up being ahead of you at the end of time

[copper umbra]

so recap
definition of bounded: you draw a circle with fixed radius, and the whole set is inside the circle

[midnight haven]

fixed radius ok

the book just says "find any radius R"

maybe it seemed too trivial that R cant be infinite

[copper umbra]

well understand this as "find a fixed radius R" or "there exists a fixed radius R"

that would be better

[midnight haven]

yh it makes sense now

[copper umbra]

because definition of bounded would be $\exists R > 0$, such that $\forall x \in S, |x| < R$

[midnight haven]

what does double | mean?

[copper umbra]

it means the "size"/"length" of the object

for example you may say $||(1,4)|| = \sqrt{1^2+4^2} = \sqrt{17}$, distance of this point from (0,0)

but really depends on your course, maybe your notes use another notation

[midnight haven]

length is normaly with |a|

at least in my experience

[jolly parrot]

Dee3Cay

[copper umbra]

there you go :]

[midnight haven]

then we have if

[-1,4] in R^1 is bounded

surely i can make circle around this when this isnt growing

[copper umbra]

yes

but circle in R^1 are open intervals

so something like (-a,a)

so you can say a = 5
then $[-1,4] \subset (-5,5)$

[midnight haven]

yh but (-1,5)

[jolly parrot]

Dee3Cay

[copper umbra]

so bounded

[midnight haven]

yh nice

[copper umbra]

looks like this

[midnight haven]

yh

x+y<0

can i chose my x to be infinitely large here @copper umbra

[copper umbra]

well

what would your y be?

[midnight haven]

oh

can i say x<-y

[copper umbra]

-y+y = 0

it's not < 0

[midnight haven]

-y just has to be bigger than x

so i can make a circle bigger than this

right

other one is x^2 + 1/4*y^2 >= 1

i can also here make a circle tht contains this

[copper umbra]

do you think you can?

[midnight haven]

i do

[copper umbra]

ok what's the radius of the circle?

[midnight haven]

which one

[copper umbra]

for this one

[midnight haven]

R>1

[copper umbra]

you need to set R = ?

can't be R>1

[midnight haven]

ok 2

[copper umbra]

ok R = 2, I pick (x,y) = (3,0)

[midnight haven]

we'll take 4million

but doesnt it have to be 1

[copper umbra]

ok I pick (x,y) = (4million+1,0)

[midnight haven]

the radius of the eclipse

[copper umbra]

it's >=1

[midnight haven]

lmao

[copper umbra]

not =1

[midnight haven]

ok

i retract my statement

your honor

[copper umbra]

:]

so it should be unbounded

[midnight haven]

yes it cannot be contained within a circle

[copper umbra]

you can play this game with yourself and then you can decide whether it's bounded or unbounded

make a circle, and then pick a point that's outside of it

and see which side wins

if you can't pick a point outside of it then it's bounded

if you can always pick a point outside then it's unbounded

[midnight haven]

also

when you get asked a question like:
is this open/closed?
(x,y) R^2 | |x| and |y| =< 1

then it just means the two variables cant be negative

[pearl pond]

@midnight haven Has your question been resolved?

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[keen knoll]

Can someone help me with this integration by parts?

[pearl pond]

Please don't occupy multiple help channels.

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[pearl pond]

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[midnight haven]

someone help pls? my friend and I got confused <@&286206848099549185>

[midnight haven]

!15min

[pearl pond]

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.

[midnight haven]

what was the original question

[vocal sundial]

What is the question?

[midnight haven]

the questions are this for a) b) and c) and d)

[vocal sundial]

can you type the question?

[midnight haven]

Hey sorry

<@&286206848099549185> my friend and I were doing other tasks of hws

[vocal sundial]

ok

[midnight haven]

a) Show that 5a+2b=18
b) Write down another equation in terms of a and b
c) Solve the two equations simultaneously to find a and b
d) Hence find the area of the rectangle

@midnight haven @vocal sundial

[vocal sundial]

<@&286206848099549185>

[midnight haven]

Was there any figure given

?

That's all it says

for 2) The diagram shows a rectangle. All slides are measured in centimeters

[midnight haven]

In pencil I tried to work them out

but my friend and I not sure if they are correct

I tried this taking value ( a = 3, b = 4)

for a) ?

Thanks! You're awesome!

[midnight haven]

that's for a) only or?

No all parts

You can figure out all parts now ?

ohh cuz the teacher told us to write examples equations for a) and other equations for b) etc

Ok

if you have seen for how I wrote the examples

Got it

Wait

Thanks bro 👍

there's this here oof

if you forgot

[midnight haven]

?

[midnight haven]

Did you tried it or was it given in question ?

it was given in question

This exact figure ?

yeah

and for example for equation a) this is what me and my friend tried to do

but for b) we got confused then

3a= 5b+17

like even for these

Ok , you should have given the figure early !!

Lol I did but probably it took time to load or idk lol

It's this one

As you can see for 2) The diagram shows a rectangle. All slides are measured in centimeters

thank you bro fr 👍

No problem !!

so then I just need to work for c) and you don't know for d) ?

[midnight haven]

👍

No wait a little bit

Writing down those part

You're the best thank you! 👍

My friend told me to tell you thank you fr 👍

thanks 👍

@midnight haven Is this 5a or 3a or 8a?

3a=5b+17

[midnight haven]

Tyy 👍

Is it clear ?

I think so I'll check 👍

[midnight haven]

so this is for c)

and this for d) ?

Is this 5a+2c-D = 18? @midnight haven

No

Thats 5 a +2(-1)=18

Thanks bro 👍

[midnight haven]

so this is for c)

[midnight haven]

and that's for d)

right?

Ys area is 240

👍

Wanna be frnds ?

[midnight haven]

Yeah I added you before lol

Lol

I cut the 5 outs or?

[midnight haven]

I'm so bad to sometimes guess the numbers etc lol

15 a gets cancelled

and under them is (-) (+) ( division) ?

No sign change

Thanks! and under it it's 31b = - 31?

[midnight haven]

Sorry could you try to resend the image? It's a bit dark and glitchy

That's better thank you fr 👍

[midnight haven]

I added you

[pearl pond]

@midnight haven Has your question been resolved?

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[midnight haven]

.close

[pearl pond]

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[random hazel]

<@&286206848099549185>

[pearl pond]

@random hazel Has your question been resolved?

[pearl pond]

@random hazel Has your question been resolved?

[random hazel]

<@&286206848099549185>

[pearl pond]

@random hazel Has your question been resolved?

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[manic sparrow]

if ur doing a limit as x→∞ do u have to write $\lim_{x\to \infty^{-}}$

[pearl pond]

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[tulip cradle]

To test if a square matrix is orthogonal do I find the inverse and transpose

[timid spindle]

no you can just check AA^T = I

and A^TA

[hollow cloak]

s(infi) = a/(1-r) when r<1

[tulip cradle]

What is A the matrix itself?

[hollow cloak]

yes

and A^T = transpose of A

[tulip cradle]

I thought it would be the column

And also isn’t AA^T the same as A^TA?

I thought u check if transpose = inverse

[hollow cloak]

transpose isnt inverse

[timid spindle]

$A^T = A^{-1}$ is the same as $A^TA = I = AA^T$

[jolly parrot]

ΣΑCu

[tulip cradle]

I’m just going off my notes

Oh

Wait so

[timid spindle]

so you dont strictly need to calculate the inverse

[tulip cradle]

Those methods are the same method ?

[timid spindle]

they are equivalent things

[tulip cradle]

How do I know which to do hmmm

[hollow cloak]

inverse is adj(a)/|A| but in ortho the prop AA^T=I is for easy calc

[dusk dagger]

In general, those are not the same.

[hollow cloak]

unless they are ortho

[timid spindle]

generally finding the tranpose and doing two matrix mults is easier than finding an inverse

[dusk dagger]

$A^T = A^{-1}$ is a special case.

[jolly parrot]

chencking

[tulip cradle]

What is I here

[slim pelican]

the identity matrix of suitable order

[dusk dagger]

Inverses are unique so one matrix multiplication is enough.

[timid spindle]

right everything is square

[tulip cradle]

Still don’t get this 💀

[dusk dagger]

Multiplying a matrix by its transpose is not the same as multiplying the transpose by the matrix.

Matrix multiplication is not commutative.

[tulip cradle]

Ohh

Thanks

[dusk dagger]

If you know that AA^T = I (i.e. A is orthogonal) then it is. But that is a special case.

[tulip cradle]

So it’s not used as a test?

It would be used if they tell u it’s orthogonal and want u to prove?

[dusk dagger]

That's the definition of an orthogonal matrix.

It's just that in general A A^T = A^T A is not true for all matrices.

[tulip cradle]

Oh ok

Wait so I can do A times A^T and if that equals I it is orthogonal? And that’s a valid method for a test ?

[dusk dagger]

That's the definition. A square matrix is orthogonal if its transpose is its inverse.

[tulip cradle]

Ok

I think I get it

Thx

.close

[pearl pond]

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[manic sparrow]

,w limit as t→∞ of ∫ from -t to t of xdx

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jashxdlol

[manic sparrow]

,w ∫ from -∞ to ∞ of xdx

how???

why they different

[glacial sequoia]

You don't have to approach infinity and -infinity the same way

[gusty garden]

find the natural number n if....

[pearl pond]

@manic sparrow Has your question been resolved?

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[real moon]

I'm upposed to create a graph that follows the form: f(x) =ax+b/cx+d. The answer for this graph is 2x-3/x-3. But why is -3 on the numerator -3 when x-int is 1.5?
https://cdn.discordapp.com/attachments/997197680386265110/1166473288508256286/IMG_1916.jpg?ex=654a9de8&is=653828e8&hm=24e404706810319326e9a546db0e964fb009ea868135c7284971a06b1dd4486f&

[real moon]

x ints = b/a. 1.5 = 3/2

[pearl pond]

@real moon Has your question been resolved?

[real moon]

.close

[pearl pond]

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[wild fable]

why the perimeter not 2x + 2y + (pi•r)/2

[timid spindle]

missing the bottom of the quarter circle

[wild fable]

What bottom

[timid spindle]

the side that isnt curved

also why 2x

oh x = radius

then it seems fine to me

[wild fable]

I’m not understanding isn’t it a rectangle attached to a quarter circle? And if a circle is 2pi•r then a quarter is (2pi•r)/4

[timid spindle]

yes

[wild fable]

So why isn’t it correct den

[timid spindle]

who is saying its not correct

[karmic hound]

r=x no ?

[wild fable]

Didn’t you say it was wrong cuz it wasn’t curved or something

[timid spindle]

.

[wild fable]

Oh

[timid spindle]

i had pointed out you hadnt added the extra r from the bottom of the quarter circle, but since r=x and one x from the rectangle isnt included its fine

you possibly made two mistakes that cancelled out

[wild fable]

lmoa

Alright

So do I set 2x + 2y + (pi•r)/2 = 36 or something? But the perimeter in the question doesn’t have any y variable so what do I do for that?

It only has an x

[rigid mist]

you are provided with the area of this bed

you can get a relation from that

[rigid mist]

and then put y in terms of x in this

you should be getting the right answer then

i think i sound correct

[wild fable]

I don’t think I understand

[rigid mist]

what is the area of this whole garden bed given to you?

[wild fable]

36

[rigid mist]

now get a expression in terms of x and y for the area of this bed

...

[wild fable]

2x + 2y + (pi•r)/2 = A where A is area of the garden bed?

[rigid mist]

dude this the perimeter

[wild fable]

Bruh

2x + 2y + (pi•r)/2 = P

done

[rigid mist]

i asked for the area of this water beam

[wild fable]

Oh

lol

x•y + (pi•r^2)/4

[rigid mist]

and this is equal to what?

[wild fable]

Um

36

[rigid mist]

btw put r=x

[wild fable]

What’s r

[rigid mist]

radius

=x

,, \frac{\pi x^2}{4}+xy=36

[jolly parrot]

Yajatjamal

[wild fable]

Why does r = x

[karmic hound]

see the figure

[wild fable]

Is the dotted line of the quarter circle r?

[karmic hound]

yes

[wild fable]

that’s r?

[rigid mist]

yea

[wild fable]

Okay

[wild fable]

So now what do we do with this

[rigid mist]

what do you think

[wild fable]

Solve for x

Or y

Not sure

[rigid mist]

solve for y

[rigid mist]

and then the y you get, put in this

[wild fable]

Why do we do y btw?

[karmic hound]

because you want x at the end

[wild fable]

Oh facts facts

So xy = 36 - (pi•x^2)/4

y = (36 - (pi•x^2)/4)/x

[karmic hound]

ye

[rigid mist]

also dude what abou the physics answer

[wild fable]

It’s gone my boy disintegrated

Lost in time

So subbing back y

2x + 2(36 - (pi•x^2)/4)/x) + (pi•r)/2?

What does this equal doe

[karmic hound]

P

[wild fable]

Okay

So now we should solve for P and we magically get the answer then

[karmic hound]

now you just need to simplify

[wild fable]

Yes

[rigid mist]

im worried about the pi tho

or maybe you have to put 22/7 in it

[karmic hound]

no nothing to worry about

[wild fable]

I did it and everything simplified fine

Got the answer

[karmic hound]

nice

[wild fable]

Thanks

.close

[pearl pond]

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[heady oak]

how would i graph 1/x^2-5

[heady oak]

this one

wouldnt it just be 5

x is not equal to 5

x is not equal to -5

yea

0

wait we messed up

the x-ints are like 2.3 or sum like that

and -2.3

OH MB

I ALR SOLVED THIS

I needed help with another one

1/x^2+2

that was the one

mb

yea but for this

there's no x-intercepts

ok wait

lemme send the problem

howd u get 0.5 as y-int

i see

om so we have 1/2

then what

@wind salmon what do we do aftrer

[pearl pond]

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[misty thicket]

why does 3/i^-78 = -3. whats the process of getting to the answer.

[cursive wraith]

i^(4k+m) = i^m

[midnight haven]

Noticing that i^4=1, I guess

[feral leaf]

And 1/i = -i

[midnight haven]

$i^{4k+m}=i^{4k}i^m=(i^4)^ki^m=1^ki^m=i^m$

[jaunty mantle]

i to the 78 is -1 because 78 remainder 4 gives 2 which is -1

[jolly parrot]

jan Lapiwin

[jaunty mantle]

Three over negative 1 is negative three

[misty thicket]

so 78-76 = 2 and i^2 is -1?

.close

[pearl pond]

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[celest lintel]

hi can you teach me how to do the equation 4x-4+=-4x+4?

[regal herald]

it was written like that, with the random +?

anyway

what do you think you should do

[pearl pond]

@celest lintel Has your question been resolved?

[pearl pond]

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[weary tide]

Why is the horizontal shift wrong?

[tiny egret]

because its 4x

gotta factor that 4 out

[weary tide]

And

Sorry

How do you factor a 4 out of a fraction

I dont think I have ever done that

Or if I did I some how forgot

.close

[pearl pond]

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[oak lynx]

I'm 14 and just started learning analysis, a random adult started bullying me (literally lmao) and calling my first work wrong, instead of focusing on this I'd like to use this moment to improve my proof

this was a proof intended to prove that the cardinality of an interval between 0 and 1 within reals and the cardinality of all reals is equal, is this work incorrect or correct, could you also explain why, and perhaps if it is correct, tell me a better or simpler way I could've proved this fact rather than making a function from scratch

this was my work (I hope the handwriting isn't abysmal lol):

[mild island]

Hole shit how

I can't help, but ig good jobb

[oak lynx]

okay lol

[mild island]

What are you trying to prove

[oak lynx]

I wrote it on the post

someone bullied me over this saying that it's wrong pseudo math lmao so I've moved on to these professionals to help me

[brisk scarab]

finding an explicit bijection between a bounded (open) interval and the reals

[warm current]

I didn't read the whole thing, but tangent function as your bijection is the correct approach

[oak lynx]

I just specified any interval and that would have been enough proof

[oak lynx]

that's good to hear

[warm current]

Here's a real challenge for you if you want to spend some time on it: prove (a, b] and [a, b] also have same cardinality as R

[oak lynx]

sounds good

I was worried the whole thing was awfully wrong and that there was some more general way to prove it rather than a function lol

is there any simpler way? is this how you would show it on paper yourself

[warm current]

Edit : prove [0, 1) has the same cardinality as (0,1). That's the easiest way to go about it

[brisk scarab]

I'm pretty sure using tan is one of the simplest ways, given it's just rescaling a function we already know is bijective and is bijective bounded domain

technically there may be shenanigans with the definition of tan

[oak lynx]

lol yeah

[warm current]

Tangent function is the simplest. The are other ways, but I do not know them.

[brisk scarab]

but if you're actually doing proper real analysis that shouldn't be too much of an issue

[oak lynx]

okay thanks guys

[oak lynx]

I'll work on this then lol

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[zenith basin]

can someone help me with 10a?

[zenith basin]

I only got one of the answers

x = 344

but idk how to get the other answer

Working out

[pearl pond]

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[zenith basin]

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[dim rover]

hey fellas

[dim rover]

im not sure where to start here

[midnight haven]

what do tangents of circles have in common with the radius

of a specific point

[dim rover]

The tangent makes a right angle at the point of tangency with the radius of a circle.

[midnight haven]

yes

find the equation of the radius first

and then you will know the slope of the tangent

[dim rover]

doesnt it give me that

find the equation of the radius

[midnight haven]

radius not circle

[dim rover]

do u mean like r = c/2pi

[midnight haven]

circle is xx+yy=25 but what about the radius passing through (4,3)

no

the equation of the specfiic radius

that passes through the center (0,0) and (4,3)

you know how to do that

just equation of a line

actually just find the slope

[dim rover]

i mean yeah that would be a line with a slope of 4/3

[midnight haven]

3/4 rather

[dim rover]

yeah

[midnight haven]

so whats the slope of tangent

[dim rover]

3

?

wait

fucj

[midnight haven]

what is the slope of a perpendicular line

its negative recriprocal

so -4/3

now you have the slope and point

just use point slope formula now

[dim rover]

x1 - x2/ y1 - y2?

or the other way around

[midnight haven]

other way works too

if this for geometry or what

[dim rover]

nah calculus im not sure why this is here

[midnight haven]

ok bud

i think thats cap

[dim rover]

?

why would i lie

[midnight haven]

you fr in calc?

bro u need to review your algebra rly bad

[midnight haven]

i thought u were being sarcastic

oh dang

ok yea

you already have slope and the point

just find the line now

[dim rover]

ive not encountered a single problem where i needed to use these rules my man

thats why im here

[midnight haven]

u def know point slope form

u guranteed found equation of tangent a lot

for derivatives

[dim rover]

oh yeah

for sure

many times

[midnight haven]

same thing

btw

there is a calculus way

to solve the problem

should of told me

differentiate both sides

and plug in x and y

do you kno implict differentiation?

[dim rover]

ehhh

[midnight haven]

if not just solve for y

and plug in the x and y

dy/dx is the slope

ok well u should prob know now

if u need more help @meeeeeee

[dim rover]

well y = sqrt(25 - x^2)

[midnight haven]

yea

so dy/dx is

(-x)/(sqrt(25-xx))

chain rule

btw its plus or minus

but its theplus one since y is positive

u good now?

[dim rover]

dude what the fuck r u saying am i taking the derivative of the left side of the equation

like where did you get negative x

[midnight haven]

you can do that

chain rule wdym

you know chain rule right?

[dim rover]

you skipped like 2 steps

yes

[midnight haven]

i just gave the answer

but u can do it urself

[dim rover]

dis shit is ridiculous, i feel r*tarted

[midnight haven]

ok well peace out now

ask helpers

if u still need help

[pearl pond]

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[pearl pond]

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[modern storm]

What are the steps to find a different order of integration given a function's limits? (3-D)

[modern storm]

Example: how to change from [ \int_{0}^{1} \int_{y}^{1} \int_{0}^{y} f(x,y,z), dz , dx, dy ] to , dy , dz , dx

[jolly parrot]

Scriptkitty

[modern storm]

**Just made up a problem lol

[uneven smelt]

changing order of integration, you can think of it as you are integrating over a volume and the function is the density function at any given point. so to change the order of integration, one way would be to draw out how the 3d object would look like and accordingly parametrise it,

sorry, I might be wrong as I forgot about how multivariable integration worked with variables in them

[modern storm]

I tried doing that and I got (for my example problem) [ \int_{0}^{1} \int_{0}^{x} \int_{z}^{1} f(x,y,z) ,dy ,dz, ,dx ] but if I put that into desmos, it's \textit{not} same volume as the original equation

sorry I'm new to latex lol idk how to italicize

[jolly parrot]

Scriptkitty

[uneven smelt]

sorry, could you show your working?

I got the bounds, z<=y<=x

the other integrals, I got the same

I got it from,
0<z<y (take < to be <=, lazy)
y<x<1
0<y<1

so from there you can see that z<y and also, y<x thus, we get,
z<y<x

@modern storm

[modern storm]

That's what I thought at first, but when I graphed the bounds:

It appears that in the y-direction it's from z to 1

actually, let me redo the graph

So, for the red lines (the y-direction) looks like the surfaces are going from y=z to y=1

and in the z-direction it looks like the curve bound is z=0 to x

and in the x direction the point bound is 0 to 1

That was my logic

Yes or no? @uneven smelt

[pearl pond]

@modern storm Has your question been resolved?

[uneven smelt]

@modern storm your diagram looks good.

[uneven smelt]

try looking at the xy plane and then look at the yz plane, you should get, z<y<x

I took awhile looking at your diagram (⁄ ⁄•⁄ω⁄•⁄ ⁄)

[pearl pond]

@modern storm Has your question been resolved?

[modern storm]

Okay, Thank you!

I'm just going to take a closer look at this tomorrow

Thank you

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[winter thistle]

The answer is 21.9 the textbook says but I don’t understand how

I got the sf area of the cylinder first I got 3pi and minus the overlapping part so 2.75pi

[arctic siren]

Are they overlapping? textbook diagrams like this are never too reliable, but it looks like the cylinder ends at the edge of the triangular prism

[winter thistle]

The triangular prism I got 20 and without the overlapping part I got 20-.25pi=19.2

[winter thistle]

I think they are

It looks like the cylinder is in the middle

Right under it

[arctic siren]

Idk, to me it looks like the triangle is just balanced on top. See if you get the right answer with no overlap, just in case

[winter thistle]

With no overlap I still got it wrong

[arctic siren]

What did you get?

[winter thistle]

I got 20+3pi which is still way above 21.9

[arctic siren]

What are your full surface areas for the cylinder and triangle independently?

[winter thistle]

It’s surface area not volume

I got

3pi for the cylinder

And 20 for the triangular prism

Wait No not 20m

I got

14

Oh I got 21.9 now

I understand what I did wrong

[arctic siren]

👍

[winter thistle]

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[midnight haven]

.reopen

how does this work

[midnight haven]

Hello

I need help with ellipses

as in can anyone tell me how to construct an ellipse using it's focal property

that's it

Thank you

[earnest stratus]

You can cut a cone in a slanting manner. Like an acute angle with the axis.

If that's what you meant.

[pearl pond]

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[severe siren]

13 and 17

[severe siren]

nvm bruh

.close

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[toxic lichen]

do not troll.

[midnight haven]

<@&268886789983436800>

[unborn abyss]

.close

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[pulsar musk]

how to solve x

its trig

[midnight haven]

yeah and what trig identity can you use

[pulsar musk]

dont i need 2 sides to use anything from trig?

sorry its our first lesson on it

[midnight haven]

that involves x

[pulsar musk]

sin and cos?

[midnight haven]

what do those mean

sin = ?/?
cos = ?/?

[pulsar musk]

u cant use sin

right

[midnight haven]

.

[pulsar musk]

cos = 17/x

we dont have the opposite side tho

[midnight haven]

what does cos mean

[pulsar musk]

so isnt cos the only thing we can use

its adjacesnt over hypotnuese

[midnight haven]

okay lets look at the angle 48

[midnight haven]

cos 48 = ?/?

[pulsar musk]

17/x"

?

no

x/17

[midnight haven]

is 17 adjacent to 48

so cos48 = x/17

solve for x

[pulsar musk]

yes

just use calculator_

?

[midnight haven]

$$\cos{48} = \frac{x}{17}$$

[pulsar musk]

oh

times both sides by 17?

[midnight haven]

yeah and what are you going to plug into the calculator

[pulsar musk]

isnt it just 48 x 17?

or like 17cos48

or something

[midnight haven]

good

any other questions

[pulsar musk]

nope thats all

thank you so much

[midnight haven]

yw

[midnight haven]

.close to free the channel

[pulsar musk]

.close

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[pearl pond]

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[languid quest]

Are (f + g)(x) and f(x) + g(x) the same thing?

[pearl pond]

@languid quest Has your question been resolved?

[earnest stratus]

Yes.

@languid quest

[languid quest]

Ty

Also how do I justify that x^2 + 1 has a domain from -infinity to infinity?

[pearl pond]

@languid quest Has your question been resolved?

[pearl pond]

@languid quest Has your question been resolved?

[brazen meteor]

Not sure, what the meaning of domain is:
Possible x values: Trivial
Possible y values: x^2 cant be smaller than 0 for real numbers so the domain would be [1;infinity] (for x in R)
In case the exercise sais to solve with complex numbers, ||you have to remember that i² = -1 and thus you can get negative values from x^2||

[languid quest]

Ok ty

[brazen meteor]

Did it say to solve for complex numbers ?

[pearl pond]

@languid quest Has your question been resolved?

[languid quest]

No it's just the domain

[pearl pond]

@languid quest Has your question been resolved?

[languid quest]

How do I justify that x exists for all real numbers of x in f(x) = x^2

?

[pearl pond]

@languid quest Has your question been resolved?

[hasty lake]

its graph is continous

I do not know if that is a justification

[languid quest]

Ty it's good enough

[pearl pond]

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[still fiber]

I don’t understand how the chain rule in b works

Like I just don’t understand how to find each part

Where is the z

I guess dz/dx and dz/dy I found in a

So is dx/dv and dy/dv just the derivatives of the two new given functions?

[vale meadow]

let me explain

for now just assume z = x^3*y..making it easier for you to understand

it's given that x = t^2 and y = t + 1

now tell me can you find dx/dt and dy/dt?

[still fiber]

Yea

I found that in a

[vale meadow]

oh my bad.. I am not well versed with PDs..you can ask anyone else in that case

[still fiber]

It’s ok I figured it out

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[lethal eagle]

could someone explain the second question?

[lethal eagle]

the simplified form is :-

[pearl pond]

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@lethal eagle Has your question been resolved?

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[shy ginkgo]

in how many possible ways can you toss a set of 2 coins?

[shy ginkgo]

ik # of outcomes = n1 * n2

so 2 * 2

but i thought it was 16 bc u can have (H, H), (H, T), (T, H), (T, T)

4 * 4 = 16

can someone explain why the first version is right and the second part is wrong

oh wait

im dumb

[shy ginkgo]

this is 2 coins.... so 4 lol

ok ima close this now lmao

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[wintry anvil]

I have no clue what to do here.

[glacial sequoia]

Start by getting it in center radius form

[calm wing]

can you turn this into a (x-a)^2 + (y-a)^2 = r form?

[glacial sequoia]

Complete the squares

[wintry anvil]

How do I do that?

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[twin sphinx]

Is there any way to now for what n a real Dirichlet character modulo p such that p Is prime, will be -1 or 1

[twin sphinx]

Ik it’s related to quadratic residues but like how can I find some kinda sequence within quadratic residues

Cause im trying to find a way to rewrite the zeta sum or a dirichlet L function, as a sum of reciprocals of APs which are connected to the n for which the firichlet character is -1 or 1

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[vague pollen]

is this very wrong

[inland ivy]

the first integral is correct

but $\int \frac1{u^2-u^4} du \ne \ln|u^2-u^4|$

[jolly parrot]

kheerii

[inland ivy]

that rule only applies to linear functions in the denominator

the same problem is in the third integral

you will need to perform partial fractions

[vague pollen]

but for the first integral

i dont see it in the answer

[inland ivy]

oh yea

[inland ivy]

even the first one is wrong

mb

[vague pollen]

[inland ivy]

you made the same mistake in all of them

[vague pollen]

so partial fraction for everything after subbing u = e^x?

[inland ivy]

yep

you can do that in the first step, it might be easier

[vague pollen]

okay

[inland ivy]

i mean it will be less steps

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@vague pollen Has your question been resolved?

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[pearl pond]

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<@&286206848099549185>

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[merry steeple]

I know I can use the arc length formula to get the distance

but how do I find a and b

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[merry steeple]

.reopen

[pearl pond]

✅

[merry steeple]

<@&286206848099549185>

[pearl pond]

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[primal bluff]

did they make a mistake here? I get eigenvector v_1=[-1/4 1]

[fluid axle]

it's fine

there's infinitely many eigenvectors here, not just [1 -4]

and any of them will do except 0

[primal bluff]

any linearly independent ones

[fluid axle]

if you have eigenspaces with dimension more than 1 yes

[primal bluff]

yeah so I need to calculate the eigenspace

but shouldnt all eigenvectors be in the eigenspace

[fluid axle]

yes

[primal bluff]

like how did they get another eigenvector? maybe its quickerp

[fluid axle]

you just pick some representatives of the eigenspace

aka, a basis

[primal bluff]

how did they get an eigenvector that is not in this basis

[fluid axle]

[-1/4 1] = -1/4 [1 -4]

[1 -4] is in the eigenspace

[primal bluff]

and 1/4 falls away as it is just a scalar multiple?

[fluid axle]

the choice of basis isn't unique

again

[fluid axle]

well if v is an eigenvector, then kv is also an eigenvector for any k != 0

{[-1/4 1]}, {[-1 4]} are two different bases of the same eigenspace

[primal bluff]

makes sense, thanks a lot

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[pearl pond]

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[lusty dagger]

what is going on here

[lusty dagger]

how does u = sqrt(x) suddenly turn into 12u^2

[last summit]

what is the replacement for dx?

[lusty dagger]

2udu

[last summit]

ok

so

..

6 stays

sqrt(x) -> u

dx -> 2u

what happens?

12u^2

du

[lusty dagger]

oh bc the formula f(u(x))u'(x)?

[last summit]

no

because you are literally just substituting

you have 6sqrt(x)dx

replace sqrt(x) and dx with what they are in terms of u

and you will literally just get 12u^2

[lusty dagger]

thanks

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[lyric inlet]

can someone help me with

B please

[pearl pond]

What step are you on?
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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

[lyric inlet]

3 and 4

mainly

4/

[pearl pond]

@lyric inlet Has your question been resolved?

[pearl pond]

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[flint viper]

how do i start approaching this question

[flint viper]

i think i can assume the expressions inside the radicals are greater than 0?

[nimble osprey]

think geometrically

[flint viper]

hmm it looks like the formula for getting the distance for two points

[nimble osprey]

yes

do you know triangle inequality ?

[flint viper]

ohhhhh

sum of two sides is greater than the other side

so distance from (-4, 10) to (1,-2)

that would be 13

thank you so much

.clos

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[spice coyote]

I NEED HELP! Linear Transformation

[spice coyote]

IDK if I did this well?

[pearl pond]

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[spice coyote]

.reopen

[pearl pond]

✅

[nimble osprey]

i don't even read

[spice coyote]

will u help?

[nimble osprey]

if you find f(1) f(x) f(x^2) its done

f(2(x^2+x-2)-2x^2-1)=f(2x-5)...but now f(2x-5-2(x+1))=f(-7) ..so you get f(-7) and you can compute f(1)

so you can compute f(x)=f(x+1-1)

then you can compute f(x^2)=f(x^2+x-2-x+2)

[spice coyote]

but im not sure about 2 things.

1- I dont really know how to find them, A person told me I cant operate thos Transformations to get them as i did at the image.
2- when I tryed i get first T(7)= (something) IDK if i have to do 7T(1) or T(7/7)

[nimble osprey]

if for example f(7)=(2,3,0) then by R-linearity 7f(1)=f(7)=(2,3,0) divide both sides by 7 ... f(1)=(2/7,3/7,0)

i have to go

good luck

[spice coyote]

oh, so i think i find f(1) well?

[spice coyote]

wdym? F(x) = F(x+1)-f(1)?

[pearl pond]

@spice coyote Has your question been resolved?

[pearl pond]

@spice coyote Has your question been resolved?

[pearl pond]

@spice coyote Has your question been resolved?

[pearl pond]

@spice coyote Has your question been resolved?

[pearl pond]

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[pearl pond]

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[rough trail]

q9

[toxic lichen]

,rccw

[toxic lichen]

!status

[pearl pond]

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

[rough trail]

and am pretty sure we cant use quadratic equation

[toxic lichen]

do you mean you can't use the quadratic formula?

[rough trail]

yeah

[toxic lichen]

however just mentioning a quadratic equation, without solving it, is fine.

[rough trail]

ok

so

i know how to get x² + 1/x²

[toxic lichen]

hm actually hold on

[rough trail]

idek how to get x² - 1/x²

[brisk pike]

Easy

[toxic lichen]

i think there might not be enough info to get its value?

[rough trail]

how

[rough trail]

oh

but its my school book so it prolly isnt wrong

idek what to do next

[toxic lichen]

,rccw

[toxic lichen]

ok so like you can square this and get x^4 + 2 + 1/x^4

then subtract 4 from THAT and get x^4 - 2 + 1/x^4, which equals (x^2 - 1/x^2)^2

trouble is you don't know whether x^2 - 1/x^2 is positive or negative

[rough trail]

ohhhhh

ok

makes sense

tysm

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[tidal otter]

Given a set S we say that a set T is nice if T $\subseteq$ S and T $\neq \phi$. Suppose A,B,C are nice sets. Show that A $\cup \big($ B $\cap$ C$\big)$ is nice but A $\cap \big($ B $\cup$ C$\big)$ need not to be nice.

[jolly parrot]

Normed

[real scarab]

phi is the empty set?

[tidal otter]

Yes

[real scarab]

$\emptyset$

[jolly parrot]

Benjamin

[real scarab]

did you try something?

[tidal otter]

I don't understand how to even approach it

[real scarab]

you need two properties

included inside S and non emptu

just try to see how these properties change

when you do intersection and union

A and B are nice

is AUB nice?

[tidal otter]

Need not be

[real scarab]

why?

i want a proof

give a counter example if it doesnt need to be

if AUB is nice then prove it formally

[tidal otter]

Alright let me try

AUB is non empty since both A and B are non-empty.
AUB is a subset of S iff for all x in AUB, x in S
Let's say there exist some y in AUB s.t y doesn't belong to S
this implies y belongs to either A or B but doesn't belong to S
This is a contradiction as all elements of A and B belongs to S
Thus AUB is a subset of S
Hence AUB is a nice set

This works right?

[hearty grove]

That works for showing AUB is nice, yes.

[tidal otter]

Alright B \cap C need not be nice since these can be disjoint subsets of S

[hearty grove]

Yes, B intersect C may or not be nice

[real scarab]

finish the proof

what happens if they are disjoint

this will help you for the global AU(BnC)

[tidal otter]

You mean if all three are disjoint?

[real scarab]

no if B and C are

its not hard

just state it

[tidal otter]

If B and C are disjoint then this is true

i.e B \cap C is not nice if B and C are disjoint subsets

AU(B \cap C) is non empty since A is non empty

And say AU(B \cap C) is not a subset of S i.e there exist some x in AU(B \cap C) s.t x not in S this implies x in A or B \cap C but not in S
if x in A then this leads to a contradiction
and if x in B \cap C then x in B and C but not in S and this also leads to a contradiction, thus AU(B \cap C) is a subset of S. Hence AU(B \cap C) is nice

And A\cap(BUC) need not be nice since BUC is nice and A is nice but A and B\capC may be disjoint

Alright I think this proof works
Thanks

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[pearl pond]

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