help-42

[prisma mango]

Because not all the terms have a factor of x^2 that you can divide by

For you to divide by x^2 on the right side, that 5 would have to be 5x^2. If you did it that way on the left, you have 1 + 5/x^2

[calm coral]

@remote mural Has your question been resolved?

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[sage socket]

Can someone help me understand composition matrices

If you look at the photo I don’t understand how to find a second transformed vector

[calm coral]

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[tough marsh]

Help how do I do this

Yo?

[strange lichen]

Can’t help on this lol

[tough marsh]

Oh ok

[strange lichen]

Can’t be sending exam questions in here bro

[prisma mango]

That says final examination?

[tough marsh]

Oh ok mb

[strange lichen]

Yea nw

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[tough marsh]

Sorry srry

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[remote mural]

Help

[velvet osprey]

missing dx?

[remote mural]

yes

missing dx

i fogor 💀

but what should i do next?

what is the integral of 1

i know cos (2x) is sen (2x)

@velvet osprey

[sturdy zenith]

? The integral of a constant is the variable of integration

so like

$\int(C), dx=x$

[potent lotus]

KaiML

[sturdy zenith]

@remote mural

[remote mural]

so that would be

like that?

@sturdy zenith

[sturdy zenith]

Not sure what you mean by sen. do you mean sin?

putting aside 1/2 for now, seperate the integrals:

$\int 1 , dx - \int \cos{2x}, dx $

welp texit gave up on me

[remote mural]

yes

its sin

[sturdy zenith]

ok, then for cos 2x you need to do a u-sub

let u=2x, then du/dx=2

then :

$\frac{1}{2}\int \cos{u} , du$

du

lol

[potent lotus]

KaiML

[sturdy zenith]

yeh

and then integral of cos is sin, then plug u=2x back in

then it should be pretty straightforward

[velvet osprey]

Cx, not just x

[remote mural]

btw

in this case

[sturdy zenith]

welp yea forgot

[sturdy zenith]

at least in this case it is 1 so its x

my bad

[sturdy zenith]

?

anyway I wont be contactable for a while

[remote mural]

oh well

thanks

btw

.close

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[whole jasper]

Consider the line L given by x + y = 1. Find a parameterization of this line, and then find a
parameterization for the image of this line under T. Add the image of the line to your plot.

how do i parse this question? What does it want? (LInear algebra)

[whole jasper]

for reference, i drew a plot of a graph that was transformed. Idk how to draw this line?

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@whole jasper Has your question been resolved?

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[limber robin]

Hello! Please help me understand this

[limber robin]

I do not understand the table

[bright lagoon]

did you get this in a computer programming course?

[limber robin]

I wanna understand the table

[bright lagoon]

(I know this already)
Is this for Python?

[limber robin]

https://docs.blender.org/manual/en/latest/advanced/limits.html

[bright lagoon]

it hardly matters

oh! Blender 🙂

[limber robin]

Yes

Have you ever used Blender?

[lament girder]

I does

[bright lagoon]

so...it's telling you that if you have decimals that exceed that many decimal places power of 10, you will lose accuracy with decimal parts smaller than what is in the second column

[limber robin]

Be easy bro! Tell me the first row. What does this mean ?

10: 1/1,048,576th

[bright lagoon]

actually I understand why you are confused

and I am a little bit too

but I know this and this is what they are getting at

there is a fixed size (number of decimal places, at the end of it all) that any "number" can occupy with the "floating point" datatype

whether that be a number filling out decimals or a number that is in the millions or billions

[limber robin]

does that mean 1,048,576 x 10 is the answer for 10? and 1,048,576 is for 1?

[bright lagoon]

here's what I think it means

on the left column you have ascending powers of 10

starting from 1

climbing to 6

[limber robin]

yes

[bright lagoon]

I think it's saying that if you have a number with 2 decimal places (10^1), then the stored value will be accurate to a fractional part as small as 1/1,048,576

but if you get as large as the millions, the accuracy of the decimal part drops after you exceed1/16

[limber robin]

What does the Hint part says

[bright lagoon]

all of these numbers are powers of 2

on the right side

you lose bits the more decimal places you occupy

single-precision floating point is a 16-bit object (edited)

[limber robin]

" values within -5,000/+5,000 are typically reliable " which row is the reliable in the table?

[bright lagoon]

ah.

check out the figure in this wikipedia article

https://en.wikipedia.org/wiki/Single-precision_floating-point_format

some of those bits are RESERVED for exponentiation

[limber robin]

single float is 16 bits

[bright lagoon]

the largeness of the number

[limber robin]

double float is 32 bits

[bright lagoon]

yes, you have that right

[limber robin]

I wanna know what does the hint part says

[bright lagoon]

I wasn't paying attention

[bright lagoon]

I guess it's saying that outside that range you start noticing the loss of accuracy

Blender fields will only display up to 4 decimal digits

[limber robin]

like 0.1234?

[bright lagoon]

yes

[limber robin]

That mean this is the safe range?

[bright lagoon]

they are

I am not sure based on what though

you can do an experiment

Matt Parker has done some of these

you try to store a number whose fractional value you know via float division and print what was stored

[limber robin]

oh ok! I will check it out. Thanks mate for the help. I really appreciate it. You have a good day Sir!

.close

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[atomic haven]

this is from the old midterm, what would be the appropriate answer for this?

[spark pasture]

Hello

[remote mural]

lol use #help-0

be quick before someone claims it

[jaunty nexus]

Maybe #❓how-to-get-help would be a better start

[atomic haven]

bruh

[jaunty nexus]

Can you tell the limit of right ride and left side of inéquation ?

[atomic haven]

no I can not

i think I got this one wrong sadly

[jaunty nexus]

Let's take for example the left side, -1-t²

[atomic haven]

'tis -1

and the other side is 0

so like its broken idk how to do it

[jaunty nexus]

So you can tell ^^

[jaunty nexus]

I think that's the point :p of the inequality being kinda "broken"

[atomic haven]

what do u think the answer is

[jaunty nexus]

What's important is what you think 👀 (cause I know the answer)

[pallid halo]

all you know is that your function lies between these two curves:

[atomic haven]

its an old midterm I already took

prof is just molasses at posting answers

thus I am here

[jaunty nexus]

😅

[atomic haven]

plz i just wanna know

[jaunty nexus]

Okay so

Why did you said it's broken ?

[pallid halo]

it's easy to come up with functions that lie between those curves which have different limits as t->0, or no limit at all

[atomic haven]

so its I?

i said

cus the squeeze theorm is squezing nothing

[pallid halo]

why would you say I?

[atomic haven]

left and right do not exist

[jaunty nexus]

Yep. Cause the inequality does not gives enough information

[pallid halo]

that would be J then

[jaunty nexus]

Exactly.

Yeah J mb

[atomic haven]

so one person says I

NOOOOOOOOOO

i answered i

rip

[pallid halo]

the limit may or may not exist

depends on the specific function f

[atomic haven]

i dont get that

but 2)

[pallid halo]

for example, f(x) = 0 satisfies the conditions

[atomic haven]

f*ck multiple choice math midterms wtf 0 marks for that

[pallid halo]

but so does a function where you just assign random values that lie between those two curves

[atomic haven]

i dont get that

:(

oh

[pallid halo]

take for example (1/2)(sin(1/x) - 1) (assign any value you like at x=0), that function satisfies the inequalities but does not have a left or right limit as x->0

[atomic haven]

would a squeeze theorm ever prove a limit does not exist? I suppose not

yeah i see u can have like a peicewise function that satisfies it too i assume

[pallid halo]

it's not really a squeeze theorem here, all you really have is that the function fails to be squeezed as x->0

[atomic haven]

oh

[pallid halo]

it has enough room to do all sorts of wild things

[atomic haven]

hmm

whats the point of giving the interval

i feel like this comes into play somehow

man im actually malding rn tho why do multiple choice midterms exist for math wtf

[pallid halo]

all that really matters is that the domain includes an interval around t=0 (but not necessarily t=0 itself)

[pallid halo]

extreme laziness of instructor/grader, imo

[atomic haven]

it does not even come with the benefit of having it marked fast

[atomic haven]

ah gotcha so this is just an indication of bait

cus my prof is 100% the type of guy to

a) be like hey, find a point on this graph blah blah blah

but the point isent even on the graph

or b) probs be like find the limit as x approaches 0 but the interval does not include 0

he's literally setting up everyone to fail

[pallid halo]

it is a bit odd that he went to the effort of coming up with 8 numerical answers

in addition to "not enough info" and "does not exist"

[atomic haven]

ima rant here some more thats irrelvent

but thats not even the worst part

physics for me is marked with multiple choice

and its the type of stuff where
a) the multiple choice answers are SMALL errors like missed a negative somewhere, dident add this ect
b) the other answers rely on the previous

i feel like everyone is trying to make everyone fail

[pallid halo]

classic assholery

i recall an E&M exam i once had that was multiple choice with your same a) and b) properties, and it was only 10 questions long

so you could really wreck your score by just getting a sign wrong somewhere

like that answer and the next 3 would be wrong, and bingo, you failed

[atomic haven]

lmaoo why do they make it so randomly hard for like no sake

in addition why do they make the tests like always down the the last minute

its literally like "yeah it took me like 45 minutes to complete the exam so ill be nice and add another 5m for the students" - every prof

anyways

do u wanna help me with some linear algebra :D ?

[pallid halo]

i was just reading/watching a linear algebra video about AlphaZero apparently discovering some new matrix multiplication algorithms that improve slightly on the fastest known ones in existence, for certain matrix sizes

[atomic haven]

excellent

[pallid halo]

discovered using reinforcement learning

[atomic haven]

well I am a first year noob who barely knows what subspace is so maybe u can help

[pallid halo]

ha sure, fire away

[atomic haven]

Okay so this is homework, but people are allowed to help me as long as like they guide me through it/do a very similar problem as an example or something but just dont drop the answer on it

[pallid halo]

hmm, do you know about null spaces?

[atomic haven]

Ive never heard the term null space

[pallid halo]

ok

[atomic haven]

All what I know about sub spaces is

a) it is closed under addition
b) closed under scalar multiplication
c) contains 0 vector

[pallid halo]

first of all, you can simplify the equation

$\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}1 \ 0 \ 1\end{bmatrix} = \begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}2 \ -1 \ 1\end{bmatrix}$ is the same as $\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}-1 \ 1 \ 0\end{bmatrix} = 0$

god i hate typsetting matrices

then with the simpler form, it's easier to verify the three conditions

you can see right away if you plug in x=y=z=0 that the equation is true for the zero vector, so c) is satisfied

you can check the other two directly

[atomic haven]

hmmm not sure what you did here

[pallid halo]

basically using the following property of dot products: $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$

[potent lotus]

Bungo

[pallid halo]

you don't have to make that simplification if you don't want to, it just makes the rest of the work a bit easier

[atomic haven]

hmm

I see the property you used but did you add them correctly?

[potent lotus]

Bungo

[pallid halo]

(sorry the RHS of the second equation should be the scalar zero, not the zero vector, i fixed it)

[atomic haven]

ah okay so you moved the matricies to all one side

i kinda forgot u can do that

/did not know for some reason I thought U could not

[pallid halo]

showing more steps: $\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}1 \ 0 \ 1\end{bmatrix} = \begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}2 \ -1 \ 1\end{bmatrix}$
is the same as
$\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}1 \ 0 \ 1\end{bmatrix} - \begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}2 \ -1 \ 1\end{bmatrix} = 0$
is the same as
$\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \left(\begin{bmatrix}1 \ 0 \ 1\end{bmatrix} - \begin{bmatrix}2 \ -1 \ 1\end{bmatrix}\right) = 0$
is the same as
$\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}-1 \ 1 \ 0\end{bmatrix} = 0$

[potent lotus]

Bungo

[pallid halo]

btw, once you learn about null spaces you can rewrite that last dot product as $\begin{bmatrix}-1 & 1 & 0\end{bmatrix} \begin{bmatrix}x \ y \ z\end{bmatrix} = 0$

[potent lotus]

Bungo

[atomic haven]

oh boy

[pallid halo]

and therefore the set of vectors satisfying the equation is just the null space of the matrix $\begin{bmatrix}-1 & 1 & 0\end{bmatrix}$

[atomic haven]

not today i already suck at linear algebra

[potent lotus]

Bungo

[atomic haven]

gotcha okay so

I can move terms

[pallid halo]

btw, geometrically this is just a plane that passes through the origin and which has (-1, 1, 0) as a normal vector

and any plane passing through the origin is a subspace

[atomic haven]

man wtf how do you know that I wish to harness your skills thats like 90% of this class so far

[pallid halo]

haha, i've studied more linear algebra than you, that's all

[atomic haven]

u just figured that out in 5 minutes that would take me 5 days

[pallid halo]

it will eventually be obvious to you haha

[atomic haven]

btw I dont see

how there would ever be a case where like any vector in general could not be transformed to contain the 0 vector

that might not make sense but what im trying to say is

how would you test that c) is ever false

[pallid halo]

suppose that instead of your equation we instead had $$\begin{bmatrix}x \ y \ z\end{bmatrix} \cdot \begin{bmatrix}-1 \ 1 \ 0\end{bmatrix} = 1$$ (so a 1 on the RHS instead of a 0)

[potent lotus]

Bungo

[pallid halo]

then if you plug in x=y=z=0 you get 0 on the LHS but 1 on the RHS, so the zero vector does not satisfy the equation

hence the set of solutions to this equation is not a subspace

(it is a slightly more general thing called an affine space, which is a plane that does not necessarily pass through the origin)

[atomic haven]

interesting, now I am intrigued on why a subspace has to contain 0

[pallid halo]

basically because a subspace is itself a vector space by definition

and all vector spaces contain 0

as to why that's a useful definition, it's basically because null spaces and images of are very useful when investigating linear maps, and null spaces and images are always subspaces

[atomic haven]

gotcha

alrighty so, we have proven c)

now for b) closed under scalar multiplication

my understanding of that is

not the best

I think its where

[pallid halo]

for that, you want to show that if (x,y,z) is a solution to the equation, then so is c(x,y,z) where c is an arbitrary scalar

[atomic haven]

if you have some scalar C and you scale the subspace in question its still contained in the subspace? - but it does not make alot of sense cus like when wouldent it/would it

[pallid halo]

if you go back to my last example, with the 1 on the RHS, notice that (x,y,z) = (1,2,0) is a solution, but c(1,2,0) is not in general e.g. take c=0

copying it here for reference

[atomic haven]

ok sorry this is a low level question but is an unfortunate hole in my knowledge, how would I know that (1,2,0) is a solution

[pallid halo]

what's the dot product of (1,2,0) and (-1,1,0)

[atomic haven]

oh i see it is 1

[pallid halo]

right

[atomic haven]

how would you find solutions?

[pallid halo]

but if i multiply (1,2,0) by a scalar (other than 1) then that also scales the dot product by the same factor

[pallid halo]

i just found one by picking x=1 and working out what y and z would work

there are infinitely many solutions

[atomic haven]

mmm, what if the numbers were big and ugly

[pallid halo]

basically pick any number you like for 2 of the coordinates and solve for the third

example:

we want to find a solution to (x,y,z) . (-1,1,0) = 1, let's try letting y=1 and z=1

then the dot product becomes -x + 1

and that needs to be equal to 1

so -x + 1 = 1

solve for x, you get x=0

so (x,y,z) = (0,1,1) is a solution

[atomic haven]

Ooh okay I see

Im just repasting it here

gotcha so now I know what b) and c) means

so a subspace must contain the 0 vector (because a subspace is essentially a vector space)
and a subspace must be closed under scalar multiplication

[pallid halo]

yes

[atomic haven]

wait well err where do I go from here because I mean, I dont think it equaling 0 is closed under scalar multiplication

[pallid halo]

no those are two separate conditions

you have to show that both are true

although technically you can deduce that it contains the zero vector from the fact that is closed under scalar multiplication

because you can choose the scalar to be zero

[atomic haven]

right

[pallid halo]

however that assumes that it contains at least one vector

[atomic haven]

I think a subspace has to contain atleast one vector

atleast for the level im at

[pallid halo]

right

but you're trying to show that this is a subspace

you don't already know that it is

so to do that

you show:

c) it contains the zero vector, i.e. it contains at least one vector

b) it is closed under scalar multiplication

a) it is closed under vector addition

[atomic haven]

indeed indeed

okay

so for b ima just try randomly expanding it out and seeing what I get

I get x + z = 2x -y +z

so

-x = -y , x = y

z = 0

[pallid halo]

wait

you can't conclude this part: ```
-x = -y , x = y
z = 0

this part is true: x + z = 2x -y +z

but all this tells you (rearrange and simplify) is that x = y

so any vector (x,y,z) with x=y is a solution

and those are the only solutions

[atomic haven]

hmm

[pallid halo]

so to prove that the set of solutions is closed under scalar multiplication, basically it boils down to showing that if x=y then cx = cy for any scalar c

[atomic haven]

Im not sure where to go from here, I can substitute x = y into the matrix but idk what good that does me

hmmm gotcha one second let me ponder

do I just be like

c(x) = c(y)

[pallid halo]

basically yea

you have x=y

multiply both sides by c, you get cx = cy

[atomic haven]

oh wonderful

[pallid halo]

hence if (x,y,z) is a solution, so is c(x,y,z)

[atomic haven]

I don't really understand that

all I see what we proved is that x = y and thus c(x) = c(y)

and thus its closed for (x,y,z)

[pallid halo]

you established that (x,y,z) is a solution if and only if x=y

i.e. a vector is a solution if its first two coordinates are equal

so suppose (x,y,z) is a solution, which means x=y

we want to show that c(x,y,z) is a solution

now c(x,y,z) is the same as (cx,cy,cz)

it will be a solution if its first two coordinates are equal, i.e. if cx = cy

but this is true

because we know that x=y

and we can just multiply that by c to see that cx = cy

[atomic haven]

ah gotcha!

this brings me back to the days where i was learning algebra for the first time in idk 3rd grade and I was asking my dad for help on how to solve x-3=0 and was super lost and embarrassed

this feels like that

you have been super helpful so far and i fr appreciate it

[pallid halo]

it's a different way of thinking than you have most likely done so far

a bit more abstract

takes some getting used to

but you will get the hang of it after a while

[atomic haven]

wholesome moment

but success I now understand c) and b)

now it is time to prove closed under addition

[pallid halo]

you can proceed similarly to how you did the scalar multiplication

suppose you have two vectors (x,y,z) and (u,v,w) that are solutions

therefore x=y and u=v

the goal is to show that (x,y,z) + (u,v,w) is a solution

how would you do that?

[atomic haven]

hmm

so

basically I need to show

(x,y,z) + (u,v,w) = 1

i think?

hmm

[pallid halo]

no it's simpler than that

remember that you determined the condition for a solution: the first two coordinates are equal

so you want to find out:

are the first two coordinates of (x,y,z) + (u,v,w) equal?

[atomic haven]

arent they?

i was thinking that

[pallid halo]

you tell me

[atomic haven]

ima say yes

[pallid halo]

they are, but you have to show why

[atomic haven]

because

wait nvm my explanation I was thinking does not show it

1 sec lemme think

if

x = y is a solution for the first

and

u = v is a solution for the second (x = y)

they are then equal

i think

[pallid halo]

go back to the question you need to answer: are the first two coordinates of (x,y,z) + (u,v,w) equal?

what is the first coordinate of (x,y,z) + (u,v,w)?

[atomic haven]

wait hold on

x + u = y + v

so

[pallid halo]

right

you need to show whether that equation is true

given that x=y and u=v

[atomic haven]

y + v = y + v

:o

[pallid halo]

well yeah

but how does that help

[atomic haven]

ah right

ermm may I request another hint

:)

[pallid halo]

well if you know that x=y and u=v

does x+u = y+v ?

try adding the two equations x=y and u=v together

[atomic haven]

I get to the same point x + u = y +v

if i add them

[pallid halo]

right

so it's true

[atomic haven]

ohh I see I see

okay sweet :D

now I should be able to tackle this problem on my own tomorrow, thanks alot Bungo!!

I will try this on my own tomorrow (go to bed now) and see where I get, do you mind If I send you a FR and maybe ask you some questions later on? All good if not, just you were insanely helpful today :)

[pallid halo]

sure, i mean i can't guarantee that i'll be available to answer, so your best bet is a help channel if you need quick help

[atomic haven]

'tis 1 am for me now, what about for you?

[pallid halo]

3am here

heading to bed shortly haha

[atomic haven]

goodnight fellow night owl, and again thanks for the help :)

[pallid halo]

cheers, good luck!

[atomic haven]

.close

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[worldly plover]

I have the inequality sqrt(1 + 4x^2) > 2|x|. I noticed that whenever you change the |x| s.t. x < 0, you flip the inequality so that sqrt(1 + 4x^2) < -2x. But, this is clearly incorrect because the left hand side is positive and the right hand side is negative. So, when are you allowed to flip inequalities when absolute values?

[runic laurel]

wdym?

[worldly plover]

@runic laurel https://www.shmoop.com/alg2-equations-inequalities/av-inequalities.html

When you have an absolute value equation, you flip the inequality whenever x < 0.

Or is that only when you’re solving for x?

[worldly plover]

Right, and I’m consider the case when x < 0.

So, I change |x| to -x

Yeah, I agree.

But, what if x <= 0?

[ancient thistle]

omg im actually scuffing this

[ancient thistle]

never type latex on a phone

[worldly plover]

Lol you’re good. So, why is it that I see sometimes that the inequality gets flipped for absolute value equations?

[ancient thistle]

the inequality is never flipped

you only flip the inequality when you have a order reversing transformation

[worldly plover]

Oh, it’s only flipped if you divide the negative, but adding the negative in doesn’t flip it right?

[ancient thistle]

you don't add a negative

[worldly plover]

Like changing x to -x won’t flip it

I know

[ancient thistle]

there is no changing

|x| is -x when x<0

[worldly plover]

|x| when x < 0 is -x

Yes

[ancient thistle]

there is no changing

you don't go like

2 > 1 so 2 < -(-1)

that makes no sense

[worldly plover]

Okay, for some reason I thought you flipped inequalities when going from |x| to -x

But that clearly makes no sense

You only flip it when you divide by negative

Or multiply by negative

Okay, thanks

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[small prawn]

what is the probability that the face with 6 points will appear for the first time on the fourth throw of a perfect die?

[small prawn]

For a 6 to appear for the first time on the fourth throw, the die must display a number other than 6 on the first three throws followed by a 6 on the fourth throw. Try calculating that probability

[remote mural]

p = 5/6 • 5/6 • 5/6 • 1/6

[small prawn]

5/6 is the probability of getting another number,not 6, right?

[remote mural]

Yes, as there are 5 other numbers and 6 numbers total.

[small prawn]

thanks

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[wise stag]

why is this wrong

[wise stag]

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[kind prawn]

I don't know why you would be asked to do something like this without linear algebra

[dim shadow]

Isolate y

[hasty wraith]

Is it possible to find the value ?

Or like express x y z in terms of a and b

a b x y and z

This kind of linear algebra has a name?

Since i havent learn it

Do you have any examples?

Sorry, i dont know that language

Determine the number of solutions of the following system depending on the values ​​of the parameters a and b:

I used google translate

[whole obsidian]

I assume you know gaussian elimination and stuff?

[whole obsidian]

so basically, our goal is to just calculate the rank of the augmented matrix, i.e by putting one of the row as a pivot row, and see whether or not it's true as an equation

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

@quartz juniper Has your question been resolved?

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[light wharf]

,rotate

[light wharf]

Proof questions are such a pain

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[remote mural]

is this correct

i just reversed (-2,3)

and did (3,-2)

but idk what the (3) inside the equation does to it

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[pine topaz]

uhm

density=mass/volume no?

but they work with area

[bronze rune]

valid

[pine topaz]

i think i would do this via integration

[tribal ember]

$$\dd{m} = \delta(r) \dd{A}$$
$$M = \int_0^5 \delta(r) \dd{A}(r)$$
it remains to find the area element if you integrate using small circles

[potent lotus]

nvx

[pine topaz]

also, polar coordinates are your friend

@bronze rune do you need further help?

for example with the integral

[bronze rune]

nah this should work

appreciate it

[pine topaz]

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[terse badge]

how do i write h=a(t-0)(t-10) in vertex and standard form

[terse badge]

its in factored form how do i change it

<@&286206848099549185>

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[upbeat bear]

i dont understand this please help

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[fossil pebble]

How do I do 54

[dense wasp]

have you graphed it

[fossil pebble]

Yes

[dense wasp]

what does it look like

[fossil pebble]

Wavy

[dense wasp]

graph cosine

[fossil pebble]

of 3pi/2 + x?

[dense wasp]

no

just x

[fossil pebble]

it’s the exact opposite

[dense wasp]

yes

exactly

to reflect things across the x axis you can multiply by -1

[fossil pebble]

so then

• cos(x) = sin(3pi/2 + x)

?

[calm coral]

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@fossil pebble Has your question been resolved?

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[remote mural]

This one has got me rattled, I don't exactly know where to begin here, can I get a hand

[pure kayak]

tan(t-pi) is tan(t) translated along the x axis by +pi, as tan repeats every pi, it should just be the same i believe

[remote mural]

Yeah I think that's correct, I tried tinkering around and I think that's right

Thanks.

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[final mulch]

.

nvm

.close

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[clever ruin]

not sure how to solve this

[clever ruin]

what i did so far

11.280 and 2.506 are wrong

[thin tendon]

,w plot (2pi)^(-1/2)e^(-x^2/9)

[potent lotus]

Results provided by WolframAlpha

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

Requested by Benjamin#3570

[thin tendon]

right?

[clever ruin]

where did -1/2 come from

[thin tendon]

1/sqrt(a) = a^-1/2

[clever ruin]

ohh ok

then how do we proceed after seeing the graph

[thin tendon]

and we rotate this around the y axis

[clever ruin]

but is my formula wrong

[thin tendon]

i'm not sure what you are trying to do, there should be a double integral i think

because you integrate a volume

not an area

[clever ruin]

i followed along this vid

https://college.cengage.com/geyser/cengage_9780538738101/html/watchit_player/?asset=larcalc9_07_03_013&prod=larcalc9

it works for him but not for me

[thin tendon]

oh my bad then i didnt understand what they wanted

or maybe i did but their way of answering is weird

[clever ruin]

i dont understand how he went to solve it at the end

[thin tendon]

they just use a formula for the revolution

around the y axis

[clever ruin]

i tried putting in the last answer i had in a calculator and got it wrong

[thin tendon]

and dont make you set up the actual double integral

[clever ruin]

makes sense

how did the the square of x disappear

the 3 also disappeared in the vid

-x^2/3

and he got v = sqrt 2pi (1 - 1/sqrt of e

[thin tendon]

so assuming the formula, this is correct

now lets integrate this

since whats inside the exponential is a x²

we dont need to do much

either you can eyeball it if you feel confident

or you can use u-sub

y = -x²/9

or y= x²/9

[clever ruin]

ok ok

and then what do we do after with the u sub @thin tendon

[thin tendon]

what do you get?

after the u sub it should be straightforward

[clever ruin]

do we subsitute it

i dont know how to u sub

[thin tendon]

but are you supposed to know?

[clever ruin]

i dont think so

[thin tendon]

okay then we'll do it the eyeball way

you have something of the form e^x²

in your integral

[clever ruin]

yes

[thin tendon]

since the derivative of e^x is itself

the antiderivative will also involve your exponential

it will likely be e^-x²/9

times something

[clever ruin]

ok ok

[thin tendon]

let me write it for 2 min it'll be clearer

[clever ruin]

1.317 is this the right answer

[clever ruin]

ok ok

the 9/2 is outside now

[thin tendon]

do you get what i did?

can you explain it back?

[clever ruin]

u used u substitute

u = -x^2/9

[thin tendon]

kind of

but not really

this is a more direct way

[clever ruin]

hmm

[thin tendon]

like i couldve done all this without talking about a function u

this way

[clever ruin]

u used chain rule to get -9/2 outside

[thin tendon]

the key point is what's at the beginning

i notice that the x

is almost like the derivative of what's inside the exponential function

so i make appear the (-2/9)

by multiplying by (-9/2) *(-2/9)

which is equal to 1

[clever ruin]

okay

[thin tendon]

and then i exactly have

[clever ruin]

so its 1/e now

[thin tendon]

the derivative of whats inside the exponential

and thx to chain rule

i know

[clever ruin]

do we sqrt the e

[thin tendon]

why would we

[thin tendon]

?

[clever ruin]

where did the e go then

[clever ruin]

( 1 - 1/sqrt e)

[thin tendon]

explain

dont just throw formulas at me

why is this there

[clever ruin]

i dont know why we are using chain rule when i think they are asking to use shell method

[thin tendon]

shell method is

to set up the integral

then we have to calculate it

[clever ruin]

ok

[thin tendon]

do you get this?

[clever ruin]

no

[thin tendon]

first equality is ok?

[clever ruin]

yea

[thin tendon]

then we get the -9/2 outside of the integral

and whats left inside is exatcly the derivative of e^(-x²/9)

so we can integrate easily

[clever ruin]

okay

[thin tendon]

oooh nevermind

i forgot to write an x

this is it

[clever ruin]

now it makes more sense, how do we go about simplifying it

[thin tendon]

do you get those two steps fully?

[clever ruin]

yes

[thin tendon]

okay then now you sub in the values

of 1 and 0

[clever ruin]

sub in with what other values

or do we just remove them

[thin tendon]

do you not know whatr the brackets with 0 and 1 written on the right corners mean?

this

[clever ruin]

exactly

[thin tendon]

you dont know?

[clever ruin]

yes

[thin tendon]

yeah now it makes me question if you know what an integral is

how would you calculate this?

[clever ruin]

with a calculator

[thin tendon]

arent you in a calculus course?

[clever ruin]

calculus 2

im a cs major

[thin tendon]

you dont know how to integrate polynomials?

[clever ruin]

yes

[thin tendon]

what are you doing currently in your class

like whats the lesson?

[clever ruin]

integrals

[thin tendon]

its a bit wide

[clever ruin]

geometric sum

[thin tendon]

are you at the very beginning of it?

[clever ruin]

middle

we using hooks law

finding indefinite integrals

[thin tendon]

isnt hooks law something about springs and physics?

[clever ruin]

yes

[thin tendon]

ok w/e

i cant teach you a whole integrals course sorry

[clever ruin]

it's alright Benjamin you tried your best with me

I stopped learning math in 9th grade

[thin tendon]

yeah its hard to come back

but if you try to pass by the fundamentals it'll be impossible later on

[clever ruin]

this will be the last class for calc

[thin tendon]

definition of an integral

integrating classic functions

[clever ruin]

im doing khan academy starting from algebra

and then will move on to trig and calc 1 and 2

i learned absolute value yesterday

[thin tendon]

oh you are self taught then

then definitely take your time

[clever ruin]

no im in a college class

but self learning rn to catch up

[thin tendon]

i suggest you find someone irl

too help you

a tutoring programm or a maths club or something

[clever ruin]

definitely, it's hard to get these concepts by myself

[clever ruin]

my tutor just solves the questions for me

[thin tendon]

its not that its hard

its that you are supposed to have prior knowledge and abilities

and since everything moves fast

you dont have time to come back to basics

[clever ruin]

true

[thin tendon]

and must learn methods to solve things

without understanding anything

try essence of calculus by 3b1B

on youtube

its very useful to get an insight on whats happening

[clever ruin]

will watch him

[thin tendon]

https://www.youtube.com/playlist?list=PL0-GT3co4r2wlh6UHTUeQsrf3mlS2lk6x

the beginning is mostly about derivatives

[clever ruin]

i think i watched a video by him

[thin tendon]

but since an integral is the reverse operation

it'll be useful

[clever ruin]

where there is a society of dominant ppl and submissive

red and blue

and how it works

it was fun

[clever ruin]

bet

[thin tendon]

i'm not sure but many people use the code he developped to animate maths

[clever ruin]

oh alright

.close

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[dawn narwhal]

I have a question

[dawn narwhal]

x = 6y - 1 / y + 3

is our equation

and for what value y takes we cannot find x

ok i already solved the equation answer is 6

my question is

apparently when u get the ratio of the coefficients of y

basically 6/1 here

u get the answer

why

wheres the proof. why does it work

does it always work

is it cuz the slope or whatever

[cosmic totem]

What is 6. This is diaphantine equation

[dawn narwhal]

what do you mean

i already told you

if x is 6 it would mean we cannot find y

thats what the question says

<@&286206848099549185>

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[remote mural]

How do you set this up

Do you need to use polar coordinates

This is what I got for the set-up but I’m not really sure why, I’m just converting to polar coordinates and plugging in what looks right without a solid understanding of why you use triple integral instead of double, and why the bounds go from 0

[calm coral]

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[tight thorn]

You use triple integral because it's a 3 dimensional solid, with sides that are funky

bounds go from zero on theta because 0 -> 2pi will cover the whole circle, on r because stops at 0, and on z because the xy plane is the bottom bound

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[keen sundial]

How does one do this

[calm coral]

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[keen sundial]

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[lethal radish]

still having trouble with this problem, even when i use the f inverse prime of a formula i still can't tell which one is wrong

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Please don't occupy multiple help channels.

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[remote mural]

.close

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[remote mural]

.open

.reopen

[mellow crater]

what's your question

[remote mural]

how does arg(z-4) - pi/2 mean that x = 4

[mellow crater]

let z = x+yi
arg(x + yi - 4) = pi/2
arg(x - 4 + yi) = pi/2
tan(y/(x-4)) = pi/2
y/(x-4) = arctan(pi/2)
since arctan(pi/2) tends to infinity we need to have x = 4 and y > 0

[remote mural]

that makes sense

because 0/0 is also infinity?

[mellow crater]

notice that if x = 4 then we have arg(yi)

now you can see it's true for y > 0

because

[mellow crater]

in fact we have something positive divided by 0^+

it tends to inf

also

[remote mural]

okay makes sense

but

isnt it arg(y/(x-4)) = tan(pi/2)?

[mellow crater]

nope, arg(z) is tan(y/x)

we don't do tan( . ) on both sides

[calm coral]

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[remote mural]

.

so

[remote mural]

im confused on the last part

the tans

i get why its tan(y/x)

im just struggling to get my head around it

if arg(z) is tan(y/x) then why is tan(arg(z)) = the angle

is the angle arg(z)?

[mellow crater]

we don't take tan of arg(z)

arg(z) is already tan

in this case we take arctan on both sides and in this case arctan(tan(y/(x-4)) = y/(x-4) because it's pi/2

$$\arg (z) = \tan \Big(\frac{y}{x}\Big)=\varphi \ \text{(angle)}$$

[potent lotus]

Modus

[remote mural]

so the original equation is saying arg(z) = angle

and we can say arg(z) = tan(y/x)

to find y and x?

[mellow crater]

yes

[remote mural]

okay ty

.close

[calm coral]

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[remote mural]

this is the reason i was confused

why is it tan of the angle here?

instead of tan(y/x)

[mellow crater]

because they've skipped the step

[remote mural]

but if tan(y/x) = pi/3

then y/x = arctan(pi/3)

not tan(pi/3) ?

[mellow crater]

aa sorry, mb I've told you it's tan(y/x), it should be tan^(-1)(y/x)

[remote mural]

okay yes that makes more sense

$$\arg (z) = \tan^-1 \Big(\frac{y}{x}\Big)=\varphi \ \text{(angle)}$$

[potent lotus]

kirkychicken

[mellow crater]

yep

[remote mural]

okay yes thank you

[calm coral]

@remote mural Has your question been resolved?

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[remote mural]

I had a math test today and there was a question i could not really wrap my head around nor understand how to even progress on and i would really like to solve it and understand it. You needed to decide the variable a so that the equation was correct.

[dull wagon]

can you express the left side as a single power of x

[remote mural]

alright

isn't the left side then

@dull wagon ?

[dull wagon]

yes

can you convert the right side to exponent form

[remote mural]

x^(1/3)

there we go

thanks for helping me come somewhere

in advance

[dull wagon]

and you should have an idea of how to continue

[remote mural]

hmm

no i still need more help, don't know how to continue from this one

fyi i'm 16 and i've just began highschool in sweden, things are a little sluggish from me and math atm

[dull wagon]

it should be quite Intuitive

[remote mural]

imma see if geogebra can solve this one for me

[dull wagon]

don't overthink this

or use calculator assistance

[remote mural]

im thinking Solve(2a - 8 = 1/3)

[dull wagon]

not quite

[remote mural]

yeah no

[dull wagon]

the power on the left is 8-2a,

not 2a-8

[remote mural]

right

[dull wagon]

you should have the equation
8-2a = 1/3

[remote mural]

right this should explain why

geogebra ain't helpinh

but i think i got an idea

8-2a = 1/3 |:2

4-a = 1/6

-4

-a = 1/6 - 4

-a = 1/6 - 24/6

-a = -23/6 |: -1

a = 23/6

nice

Seem right?

Someone?

Thanks alot @dull wagon

.close

[calm coral]

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[calm coral]

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[celest epoch]

im very confused with this and cant find any help with google

ive tried answers R, 3R, R/3, and R^3

[calm coral]

@celest epoch Has your question been resolved?

[celest epoch]

<@&286206848099549185>

[calm coral]

@celest epoch Has your question been resolved?

[celest epoch]

.close

[calm coral]

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[paper prairie]

Hey

[paper prairie]

So I have these 6 graphs, where the x axis is gpu temperature, the y axis is cpu temperature, and each graph is a different core affinity

All of them are described by a 2nd order polynomial function

And they appear to be, somewhat, linearly connected

How do I get all 6 of these functions, as one function, that has both the GPU temperature, and core affinity, as one function, to determine CPU temperature

[calm coral]

@paper prairie Has your question been resolved?

[calm coral]

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[frank valve]

“How many distinct ways can five letters be posted in four distinct mailboxes?”

[frank valve]

Apparently the answer is 3^5 but idk how

[calm coral]

@frank valve Has your question been resolved?

[frank valve]

<@&286206848099549185>

.close

[calm coral]

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[calm coral]

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[autumn tusk]

You have an expression a + b = c, where
a = c - b
b = c - a
c = a + b

If you eliminate the expression (a + b = c where a = c - b, b = c - a and c = a + b you get -2a - 2b + 2c = 0) and the given question (a + b = c) you get answer no. 1 . Then continue simplifying the expression with (a = c - b, b = c - a and c = a + b) . Every time you simplify the expression eliminate with the answer no. 1, 2, 3 till infinity.

[narrow geode]

man thats a lot of a b c
xD

[autumn tusk]

lol

[dull wagon]

wdym by

you get answer no. 1 .

[autumn tusk]

nah

u

should keep eliminating till infinity

[dull wagon]

not what i asked

answer to what

what are you saying is 1

also a+b=c is an equation not an expression

[autumn tusk]

answer no.1 is

when u

[mellow crater]

it has a little sense cause a = c - b, b = c - a, c = a + b implies a + b = c

[autumn tusk]

plug in the values of

a , b, c

in the question

that is answer no 1

[dull wagon]

answer to WHAT

banana is 1?