help-13

[fair mortar]

10/root 2 = b

side is

20 / root 2

perimeter is 80 /root 2

1/root 2 = a/10

a is 10/root 2

ap is 800/2

1/2 ap is 200

which is correct

i dont know what is wrong here

im getting it right

[smoky terrace]

so this formula is right?

[fair mortar]

not that

the method i used is right

[smoky terrace]

which one then

[fair mortar]

i messed up somewhere i dont know where

[smoky terrace]

oh right

[fair mortar]

maybe try to come up with one yourself

[smoky terrace]

i'll do that

[fair mortar]

you can refer the above messages for the moethod

method

sorry for not being helpful

[smoky terrace]

thanks for the help

[fair mortar]

yw

[smoky terrace]

.close

[cedar kiln]

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[neat gyro]

Hello?

[neat gyro]

Can anyone help me pls?

https://curriculum.gov.mt/en/Examination-Papers/Secondary-Papers/2016/Documents/Mathematics/Maths - Form 3/Maths - Form 3 - Track 3/maths_f3_t3_nc_2016.pdf

Can anyone help w this?

[inland ocean]

Is it a test?

[neat gyro]

https://curriculum.gov.mt/en/Examination-Papers/Secondary-Papers/2016/Documents/Mathematics/Maths - Form 3/Maths - Form 3 - Track 3/maths_f3_t3_main_2016.pdf

And this

[neat gyro]

No it's just a sheet my parents gave me

Because I didn't go to school this week and this is what we learned and I got no clue what any of that stuff is

[modest lichen]

what do you need help on

[neat gyro]

All of it

[modest lichen]

all?

[neat gyro]

Yea

[modest lichen]

have you tried

[neat gyro]

Yea

Some of it is confusing

[modest lichen]

im not going to do a whole 30min paper for you

[neat gyro]

Fair

[modest lichen]

do what you can

and ask for what you cant do

if you dont understand the concepts i can explain them

[inland ocean]

We can help with the confusing part

Atleast

[neat gyro]

Okay

There's a question it says show the inequality of 500 < w < 800 ?

[inland ocean]

Yeah

It is actually

[neat gyro]

So what is that..

[inland ocean]

$550\leq w < 800$

[wraith dagger]

ColdTee

[neat gyro]

What is it?

[inland ocean]

And you have to show it on the number line

[neat gyro]

Detail.

Yep

[inland ocean]

It means the value of w lies between 550 and 800

[550,800)

Closed interval 550 open interval 800

[neat gyro]

So I write this? (550,800?)

[inland ocean]

No

It is [550,800)

And to show it on the number line

If im not wrong it was represented by a black dot meaning that point is included and a hollow dot meaning that point is not included

Black dot => closed interval
Hollow dot => open interval

And draw a line between those two points

To represent the value

Of w

[neat gyro]

Okay

[cedar kiln]

@neat gyro Has your question been resolved?

[cedar kiln]

@neat gyro Has your question been resolved?

[inland ocean]

Okay so

What is confusing you?

[cedar kiln]

@neat gyro Has your question been resolved?

[neat gyro]

I figured it out

[cedar kiln]

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[old mural]

Hello! Im not the best at math and I would like some help to learn how to solve it! Its the simeltemions equations im not sure if i wrote it correctly but yeah.
If someone can just help me get through one of them and teach me how too that would be lovely!
So one of the equations are:
y = 2x + 1
y = -x + 2

[dull oxide]

There's a concept in mathematics about equality

If two things are equal to the same thing, then those two things are equal to eachother

In your case, the two things are 2x+1 and -x+2, and the same thing is y

That is, 2x+1 and -x+2 are both equal to y, so it will follow that 2x+1 and -x+2 are equal to each other

[old mural]

Im not sure i completely understand? Hm

[dull oxide]

That's alright, let's work through it.

Where do you start to get confused?

[old mural]

Well im just overall confused in how to solve it

[dull oxide]

Sure. We're trying to find an x and y value that will satisfy both equations. Does that make sense?

[old mural]

that i understand

[dull oxide]

Cool.

And would you know how to solve something smaller, like 2x+3=-1?

[old mural]

yeah i should be able too not too sure tho ive done something similar

[dull oxide]

Something similar?

[old mural]

yeah things like x + 5 = 6x stuff like that

im not sure if it is the same thing

sorry i aint the smartest but im trying to become it 🥲 Ive got to start somewhere

[dull oxide]

No worries

[dull oxide]

Okay we'll work on this because the strategy will actually be required to solve your problem

To solve x+5=6x, we need to find a value for x such that x+5 and 6x come out to the same number

[old mural]

this is x = 1 right?

[dull oxide]

Yes

Did you just try random numbers or did you actually solve?

[old mural]

i solved it

wanna see?

[dull oxide]

Good. Then you have everything you need to solve your problem

[dull oxide]

I trust you. You're fine

[old mural]

alright then! :D

[old mural]

soo what next?

[dull oxide]

What you need now is to understand this idea

It's basically the idea that equality is transferable

If A is the same as B, and B is the same as C, it must be that A is the same as C

[old mural]

I understand that

[dull oxide]

Cool. So we will use that in your problem

[old mural]

Alrighty

[dull oxide]

I described it here

Does that bit make sense?

[old mural]

uhh yeah they are the same or equal correct?

[dull oxide]

exactly

[old mural]

or well they arent the same but they equal to the same outcome right?

[dull oxide]

Yeah that's a good way to think about it too

[old mural]

alright i get it

[dull oxide]

In general, yes, they are different things, but you want to find the specific value for x that makes the same, or makes it so they both get to the same outcome, as you said

So do you think you know how to solve it now?

[old mural]

hmm yes probably yeah ill try and solve it

if i cant figure it out can i come back here?

[dull oxide]

yes

[old mural]

Thank you very much! :D

[cedar kiln]

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[cedar kiln]

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[velvet timber]

What is wrong 😭 answer (1,-1)

[velvet timber]

"by using Cramer's rule find s.s for each:"

[obsidian coral]

You went from a1 a2 to c1 b1

In the matrix

It needs to be consistent

So meaning you changed it from a2 to b1

Instead of keeping it a2

If that makes sense

[cedar kiln]

@velvet timber Has your question been resolved?

[velvet timber]

im confused

@obsidian coral

.close

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[jaunty mural]

Help, this server is unavailable

[jaunty mural]

@ Helpers

[red pumice]

hm

[crimson sedge]

lol

[red pumice]

whats wrong with the server

[crimson sedge]

mods live in australia

its april fools there already

¯_(ツ)_/¯

[red pumice]

lmfao

[jaunty mural]

help

@ helpers

<@&1022875670466023534>

[crimson sedge]

shuwu-chan haii

[crystal raptor]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[wraith dagger]

Toby

[jaunty mural]

[crystal raptor]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[crimson sedge]

.coose

[jaunty mural]

what the hell is this

[crimson sedge]

differential geometry

[jaunty mural]

whys this so pixelated. @pseudo hinge help

[pseudo hinge]

Zoom in

Is it still pixelated?

[jaunty mural]

yes.

[crystal raptor]

zoom out

[jaunty mural]

is that a blumming image

screenshot

[pseudo hinge]

Also you can only have so much resolution on an image with this much detail

And I'm sure TeXit doesn't want too much resolution since that would take more time to send

[jaunty mural]

bad bot

pls ban

dont u see this

omg no its discord

[crystal raptor]

!help

[cedar kiln]

Please read #❓how-to-get-help

[jaunty mural]

<@&1091183017051050074>

[pseudo hinge]

Oh fuck it's pingable.

[crimson sedge]

lol

[cedar kiln]

@jaunty mural Has your question been resolved?

[crystal raptor]

hello?

,w hello

[dull oxide]

!help

[cedar kiln]

Please read #❓how-to-get-help

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iCaird#6258

[dull oxide]

Stick to your own channel

[crystal raptor]

ban this mod pls

[dull oxide]

Do not challenge my power. I don't know how to do the command, and nothing is showing up when I type, /, but I'm very confident I have it.

[crimson sedge]

mod abuse

[dull oxide]

It's purely theoretical

[crystal raptor]

do it

you wouldnt

[cedar kiln]

@jaunty mural Has your question been resolved?

[jaunty mural]

omg

shut up

[jaunty mural]

<@&1091183017051050074> this user is spamming here

[jaunty mural]

this annoying person

[cedar kiln]

@jaunty mural Has your question been resolved?

[solemn torrent]

Since when have any of you been mods

[slender parcel]

lol

[cedar kiln]

@jaunty mural Has your question been resolved?

[gilded elm]

@jaunty mural

i nid halp

[obsidian coral]

\ban DerpZ#9141

[cedar kiln]

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[frosty dirge]

what do i do?

young modulus

[digital cliff]

do you know the formula for youngs modulus

[frosty dirge]

hi

sorry, is it alright if i @ u tomorrow

my brain is too tired rn

i’m going to sleep

[digital cliff]

you can if you want, i cant guarantee ill be available though

[frosty dirge]

thank u!

i’m sorry again, thought i would be able to do math rn but i need sleep

[digital cliff]

no worries

[frosty dirge]

.close

[cedar kiln]

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[wary latch]

Hi there. I have a question on some conceptual stuff from vector calculus. I'm confused on how many dimensions are for a function like f(x,y,z) = x^2 + y^2 - z^2
Is this 3 dimensional?

If it is 3 dimensional, what is f(x)=x^2?

[rapid minnow]

It would be in 4 dimensions because the function is a function in 3 dimensions (3 variables) that outputs in the 4th dimension. Kinda difficult to conceptualize 4 dimensions intuitively tho

[wary latch]

Ok, thank goodness. I was getting so confused because chatgpt kept telling me it was 3 dimensional. lol

[rapid minnow]

Bruh chatgpt for maths 💀

That thing plays chess like a cocky 5 year old

[wary latch]

😂

so the gradient of a function like f(x,y)=x^2+y^2 is (2x) i + (2y) j

Is there a way to visualize this?

[lethal jackal]

it's a function from R^3 -> R, so the domain is 3-dimensional

the graph is 4-dimensional, because it's a subset of R x R^3

[wary latch]

Oh I see. Thanks!

[cosmic steppe]

Oh a contour plot, yes

And you'll find that the gradient is always perpendicular to the level set

[wary latch]

Oh ok. I'll check it out. Thanks 🙂

[rapid minnow]

I think the gradient would look something like a V or an X shape of some sort since the graph is a circle which has decreasing slope on the right, the tangent line at the very top has slope 0, and then starts to increase is slope.

You could put it in geogebra and mess around with the visuals to get an idea

[cedar kiln]

@wary latch Has your question been resolved?

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[robust crater]

,rotate

[robust crater]

This is what i have so far for this problem

I cant figure out how to go from xn = 2^3 (xo)+(1+2+4)

to getting the closed form

[violet flume]

tell your teacher what latex is

please

"not the value, but the size"

[robust crater]

Lol its the book sorry

[cosmic steppe]

Your teacher has a skill issue

[robust crater]

Lmfao my Actual teacher gives us extra credit for submitting homework if we do it in LaTex

[violet flume]

you should do it in latex

[robust crater]

im just tryin to do the PROBLEM

[violet flume]

ah

thats a whole nother story

[violet flume]

whats your guess for the closed form? is this it?

[robust crater]

no I havent got the closed form yet I need to get from having x0 to just n

[violet flume]

well x_0 = 1

[robust crater]

but I dont know how I go from x0 to like 2n-1

[violet flume]

so just substitute

[robust crater]

well theres alot more to do than that

[violet flume]

oh i misread your

is it 1+2+4+...+n?

[robust crater]

yes

[violet flume]

but n might not be a power of 2?

have you checked this form

[robust crater]

it probably is

I mean I dont know how to write that formula correctly but I can see its increasing by a power of two so does the front of the formula\

[violet flume]

im noticing something here

do these numbers look familiar to you?

1, 3, 7, ...

almost like 2, 4, 8, ...

@robust crater say it were $x_n = 2x_{n-1}$

[wraith dagger]

jan Niku

[violet flume]

can you tell what closed form youd get out?

idk im just trying to nudge you into the answer

[robust crater]

I actually wrote the formula for this earlier as 2^n+1-1 n=0

for a seperate problem

for this sequence

[violet flume]

huh?

[violet flume]

do you recognize this sequence?

[robust crater]

yes

[violet flume]

so whats the closed form

[robust crater]

2n-1

[violet flume]

2^n - 1

but yea

[robust crater]

I have to bridge the gap from 2^3(x0)+4+2+1

to 2n-1

[violet flume]

what class is this for

[robust crater]

Discrete

[violet flume]

so my guess is

they care more about your proof?

[robust crater]

yes

[violet flume]

it doesnt matter where you got the answer

the proof is what matters

you still have to do strong induction

[robust crater]

He wants to find the the closed form - plug it into the recursive definition and simplify and check it works for the initial condition

and to skip the induction

because were covering induction in this chapter not the one this assignment was assigned in

I want to show how to get closed form because im probably going to have to reduce to closed form on the test but then from there I need to simplify the closed form and check initial

[violet flume]

could you not just use the normal geometric form here?

to be honest its been years since ive done one of these the 'proper' way so im a little rusty but

wouldnt you just recognize its geometric

so isolate the n-1 term

[robust crater]

Im not sure what that is im sorry this book is literally the worst thing ive ever seen

[violet flume]

lol

[robust crater]

there's 1 single example for this in the entire book

and that also includes the only explanation of this

[violet flume]

oh man then were gonna have to learn together

heres what i remember is like

you do a few terms

[robust crater]

I literally have the answer on chegg rn

[violet flume]

you recognize the form of the sequence

[robust crater]

and all the steps

[violet flume]

its usually geometric or arithmetic

[robust crater]

I just have no idea what im looking at and im trying to learn it not just rip it off

I think its arithmetic

[violet flume]

chegg is pointless

its not

[robust crater]

chegg is fantastic

[violet flume]

arithmetic sequences have terms with a common difference

[robust crater]

ah

[violet flume]

geometric sequences have terms with a common ratio

[robust crater]

which is x2

[violet flume]

right

now how you handle that +1

i havent done this in this way in a long time

lemme see

[robust crater]

so right now i have 2^3(x0)+(1+2+4...n) and I know its going to be a power of 2 closed form in some way

[violet flume]

well i got to like

$x_n = 2x_{n-1}+1$ right

[wraith dagger]

jan Niku

[violet flume]

let $k$ be some integer up to n

[wraith dagger]

jan Niku

[violet flume]

well first

$x_n = 2(2x_{n-2} + 1)+1$

[wraith dagger]

jan Niku

[violet flume]

is a good example that

$x_n = 2(2^{k-1}x_{n-k} + k - 1) + 1$

i believe

[wraith dagger]

jan Niku

[robust crater]

I think youre going to a more complex place with this

[violet flume]

this almost works out

if you let k=n

[robust crater]

Here's cheggs info

that i'm trying to decipher like runes

[violet flume]

you get $x_n = 2^n - 1 + 2n$

[wraith dagger]

jan Niku

[violet flume]

[robust crater]

this seems to be about where I was mentally

[violet flume]

wait

[robust crater]

both sides are a factor of 2 one is just being sequentially added

[violet flume]

why is he using the sum formula

fling save us

[robust crater]

when N>=1 does that mean x0=1 or that you start at x1?

or does that mean the pattern grows

or that n will always be greater than 1

[carmine pawn]

Was is this?

Looks like Z transform?

[violet flume]

this is what i was asking about

$x_n = 2x_{n-1}+1$

no no

[wraith dagger]

jan Niku

[violet flume]

its discrete math

[carmine pawn]

Ooooh

[violet flume]

do you remember

[carmine pawn]

No sorry

[violet flume]

[robust crater]

hahahahaha

I just dont get what im trying to find

[carmine pawn]

Heya i liked it tho

[violet flume]

i guess like

if you just say that

[carmine pawn]

When youre helping someone feel free to tag me!

[violet flume]

$x_{n-1}$ right

[wraith dagger]

jan Niku

[violet flume]

say that $x_{n-1} = rx_{n-2}$

[wraith dagger]

jan Niku

[violet flume]

but thats a lie

[carmine pawn]

My brain auto erases a lot of stuff once a course is over lol

[violet flume]

well they teach you that stupid way in these classes

that you never use again

[robust crater]

my brain auto erases everything directly after the class ends

and my teacher uses his power of autism to infodump like a madman without actually SAYING anything useful (He's a great person though)

behold some of his notes

[violet flume]

yea ive had teachers like this before lol

[robust crater]

I just need to know how to replicate getting the closed form for the test

I can do the rest

[violet flume]

you know i get your result now

[robust crater]

my 2^3(x0) nonsense?

[violet flume]

its not nonsense

$x_n = 2^nx_0 + \sum_{i=0}^{n-1}2^i$

[wraith dagger]

jan Niku

[robust crater]

NOT THE SIGMA

Thats my next chapter

that I have to learn tonight aswell

[violet flume]

ah

you know

[robust crater]

i got sick and got behind a whole month and i have until my teacher wakes up to get caught up

[violet flume]

this might be why this person is using that formula

for the sum

now that i think about it

x_0 is 1 right

so that x_n is just

2^n plus 2^n-1

thats partial sum of a geometric sequence

https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3Dsq4ayQC4R_8&psig=AOvVaw2nfSer2AVFmKlG9E7iPc7W&ust=1680413769254000&source=images&cd=vfe&ved=0CA8QjRxqFwoTCPDelqH7h_4CFQAAAAAdAAAAABAE

[robust crater]

yes

I dont fully get how we derived 2n-1

[violet flume]

so $x_n = \sum_0^n2^i = \frac{x_0(1-2^n}{1-2}$?

[wraith dagger]

jan Niku

[violet flume]

yea, it works

[robust crater]

jesus

[violet flume]

interestiiingg

okay so lets step backwards

$x_n = 2^n + \dots + 2^0$

[wraith dagger]

jan Niku

[violet flume]

does this make sense?

[robust crater]

uhy

[violet flume]

dont give up now lol ive just parsed the chegg answer

uhy?

[robust crater]

Uh*

[violet flume]

start from like

[robust crater]

yes the next answer is the last answer plus the current power of 2

right

[violet flume]

well you can get here algebraically

you dont have to know anything

start from $x_n = 2x_{n-1}+2^0$

[wraith dagger]

jan Niku

[violet flume]

substitute

$x_n = 2(2x_{n-2}+2^0)+2^0$

[wraith dagger]

jan Niku

$x_n = 2(2x_{n-2}+2^0)+2^0$

[violet flume]

expand

$x_n = 2^2x_{n-2}+2^1+2^0$

[wraith dagger]

jan Niku

[robust crater]

should be +2^1

[violet flume]

nah

not yet

[robust crater]

oh

[violet flume]

not while its in the parens

do you see the pattern emerging?

try to write it out for x_n-3 if not

[robust crater]

yeah each next power of two makes the inside variables grow by a power aswell

[violet flume]

you should get $x_n = 2^3x_{n-3}+2^2 + 2^1 + 2^0$

[wraith dagger]

jan Niku

you should get $x_n = 2^3x_{n-3}+2^2 + 2^1 + 2^0$

[robust crater]

which is what I have now

[violet flume]

take this process to its conclusion

[robust crater]

2^3(x0)+(4+2+1

well it can grow outwards indefinitely

[violet flume]

you get $x_n = 2^nx_0 + 2^{n-1} + \dots + 2^0$

[wraith dagger]

jan Niku

you get $x_n = 2^nx_0 + 2^{n-1} + \dots + 2^0$

[violet flume]

eventually we can subtract enough to get x_n-somethign to be x_n-n right

we just have to repeat this process n-1 times

[violet flume]

and well get this.

[robust crater]

I guess this is just too large of a leap in the logic for my brain to follow

[violet flume]

you have to trust the pattern i guess

you can do like

x_n-4 right

[robust crater]

this doesnt account for +1 anymore

[violet flume]

and x_n-5

[violet flume]

it does

[robust crater]

where

[violet flume]

2^0 is 1

im not sure if thats what you mean

[robust crater]

it just makes no sense period

[robust crater]

reading this is like staring into a foglight

[violet flume]

i dont believe you

since you reproduced these results basically

before we even started talking

im just trying to make you believe it

[robust crater]

2^n(xn-1) gets what power the first two is

[violet flume]

this is the jump, but its not that much of a jump

do some more steps

[violet flume]

you have to feel it as counting down

[robust crater]

what is this menagerie right here

[violet flume]

what if i wrote it this way

[robust crater]

we need parenthesis

2nx0)+(2n-1

[violet flume]

remember this

$x_n = 2^3x_{n-3}+2^2 + 2^1 + 2^0$

[wraith dagger]

jan Niku

[violet flume]

how about

[robust crater]

i just got it

[violet flume]

$x_n = 2^nx_{n-n} + 2^{n-1} + \dots + 2^0$

[wraith dagger]

jan Niku

[robust crater]

but you needed a -1 in yours

its just a very very ugly version

and not the closed form

[violet flume]

no no

[robust crater]

its the same as mine but my version is ugly

[violet flume]

but you need to believe this

[robust crater]

I understand it

[violet flume]

do you believe $x_n = 2^nx_0 + 2^{n-1} + \dots + 2^0$

[robust crater]

its just not the closed form

[wraith dagger]

jan Niku

[violet flume]

okay

lets plug in x_0 = 1

$x_n = 2^n + 2^{n-1} + \dots + 2^0$

[wraith dagger]

jan Niku

[violet flume]

this is called a partial sum of a geometric series

it has a known closed form

[robust crater]

should be 2-1

[violet flume]

this is how you make the jump

no, it shouldnt lol

[robust crater]

like

x1 = 2-1

2^x1-1 = x0 = 1 so 2(^1)-1

so it proves it

[robust crater]

and 2n-1+1 is the closed form

[violet flume]

stop

it doesnt prove anything

it should be +

please we have to finish this haha

i have to go to bed

well really i wanna take a sleeping pill

[robust crater]

Okay go to bed

I got it

[violet flume]

nah i want you to get it

[robust crater]

I just cant convey my getting it to you because I dont get it but I understand what we're reaching at

[violet flume]

i thikn you might be misinterpreting

you need the geometric series formula

[robust crater]

I dont know what that is

[violet flume]

if you arent using that you have made a mistake

[robust crater]

it wasnt taught to us

bro flew through this at mach 12 and it will only show up 1 time

[violet flume]

its usually known from calculus

or precalc

[robust crater]

calc wasnt required to take this

it was college algebra then discrete

[violet flume]

it just says

if you take some number

say 5 right

and you take

okay 5^4 + 5^3 + ...

all the way down to 5^0

what do you get

[robust crater]

factorial?

[violet flume]

its a form for that

no, its called a geometric series

[robust crater]

well there was no 5^5

[violet flume]

series because its a sum

[robust crater]

nvm nvm

yes okay

[violet flume]

and geometric because each term of the sum differs by that ratio

[robust crater]

I was confused by us interpreting the formula as one that grows out

[violet flume]

so we write $r^n + \dots + r^0 = \frac{1-r^n}{1-r}$

[wraith dagger]

jan Niku

[violet flume]

this is the known form

(really theres an initial term piece here but its not important for this problem)

well we have

$x_n = 2^n + \dots + 2^0$

[wraith dagger]

jan Niku

[violet flume]

and $2^n + \dots + 2^0 = \frac{1-2^n}{1-2}$

[wraith dagger]

jan Niku

[violet flume]

then $x_n = (-1)(1-2^n)$

[wraith dagger]

jan Niku

[robust crater]

how tf does the fraction become -1(1-2n)

[violet flume]

look 1-2 on the bottom

so dividing by -1

...

[robust crater]

geometric series arent covered in this chapter

until AFTER

this problem

this problem isnt meant to fuck with all this

[violet flume]

well then i have no idea man

[robust crater]

ones later will be designed for it

Im tellin ya i think I got what we're getting at

[violet flume]

okay

[robust crater]

I appreciate you getting me here

[violet flume]

i believe you

good luck

[robust crater]

now enjoy your night ty

.close

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[dense locust]

Help

[dense locust]

Question b, isnt supposed to be 30cos22.4?

Why is it written 30sin22.4

Nvm

Book error

.close

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[charred bay]

May i know how to solve this matriks use gauss elimination ,i got stuct in here

[crimson sedge]

https://atozmath.com/MatrixEv.aspx?q=ludecomp

try this

step by step solution of gauss

[cedar kiln]

@charred bay Has your question been resolved?

[cedar kiln]

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[quartz compass]

how do I find the mixed 2nd order partial derivative of an implicit function?

[quartz compass]

I managed to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ but I don't know how to proceed from there

[wraith dagger]

mesmeriSe

[quartz compass]

I used the implicit function theorem (?) thing,

$\frac{\partial z}{\partial x} = \frac{\frac{-\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$

[wraith dagger]

mesmeriSe

[crimson sedge]

take the function and take $\frac{\partial}{\partial x}\frac{\partial}{\partial y} z$

[quartz compass]

right i thought so but it doesnt work

[wraith dagger]

Jester

[quartz compass]

oh

so i got $\frac{\partial z}{\partial x} = -1+xe^{-(2x+2z)}$ which I confirmed is correct

[wraith dagger]

mesmeriSe

[quartz compass]

but if i were to take partial y of that, wouldn't it be 0?

[crimson sedge]

now if you take partial y of that you get 0 which is the answer to the second

yeah

[quartz compass]

but it's not..

it's supposed to be $3\frac{x}{y}e^{(-4x-4z)}$

[wraith dagger]

mesmeriSe

[crimson sedge]

oh well because your first differntation is not correct

you are taking partial z / partial x

[quartz compass]

i think it's correct, and the solution says this too

oh

[quartz compass]

isnt that what i want to be doing?

i take partial z/ partial x. Then I take partial y of the result

[crimson sedge]

oh i see where you went wrong

[quartz compass]

??

[crimson sedge]

when you take the partial of y you forgot partial z / partial x

because z is not constant, only x is

[quartz compass]

ohhh and to take partial z/ partial y we have to do it implicitly so

$\frac{\partial^2 z}{\partial y\partial x}=\frac{\partial}{\partial y}(-1+xe^{-(2x+2z)})=\frac{\partial}{\partial z}(-2z)xe^{-(2x+2z)}=-2\frac{\partial z}{\partial y}xe^{-(2x+2z)$

[wraith dagger]

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

[crimson sedge]

yeah

[cedar kiln]

@quartz compass Has your question been resolved?

[cedar kiln]

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[raven quiver]

We construct an acute triangle ABC with the three heights intersecting at H. Prove that HA+HB+HC<AB+AC

I cant use any trig or pythagoras, just triagular theorems (except the above)

[cedar kiln]

@raven quiver Has your question been resolved?

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[quartz salmon]

Currently stuck on a problem in number theory. The problem is as follows:
\\
"A theorem in number theory states that if $a$ and $b$ are positive integers and $p$ is a prime number that divides the product $ab$, then $p$ must divide $a$ or $p$ must divide $b$. From this theorem, it follows that:
\\
(a) If $p$ divides $ab$ and $p$ divides $a$, then $p$ does not divide $b$. \
(b) If $p$ divides $a^2b$, then $p$ must divide $a$ or $p$ must divide $b$.\
(c) If $p$ divides $a$ and $p$ divides $b$, then $p$ divides $ab$.\
(d) None of the conclusions (a) through (c) follow from the theorem."
\\
I am having trouble determining which of the options is true. I initially chose option (c) but I was told that the correct answer is actually option (b). I am unsure why this is the case.

[wraith dagger]

dgh

[radiant topaz]

b and c both seem right but you have to choose option that follows from theorem

[quartz salmon]

So we can see that a and b are both coprime right? Because they only have the number 1 as a common divisor?

[radiant topaz]

how?

[quartz salmon]

nvm

but (c) is just https://en.wikipedia.org/wiki/Euclid's_lemma

Can someone explain how (b) is true?

[mighty drift]

A theorem can also be applied multiple times in a row. Feel free to split cases

[cedar kiln]

@quartz salmon Has your question been resolved?

[quartz salmon]

Could you elaborate on that

[mighty drift]

Either p divides a^2 or p divides b
Apply the theorem again on p | a^2

[astral bay]

c is not euclid's lemma, euclid's lemma is the theorem we've been given

c is completely obvious, in fact you barely need any of the hypotheses: if a divides b, then a divides bc, for any a,b,c

[quartz salmon]

So c is incorrect because it’s just restating what we’ve been given?

[astral bay]

it's not restating what we've been given

it's "incorrect" in the sense that the statement is true, but it doesn't really follow from the theorem we've been given, it's true for completely unrelated reasons

whereas euclid's lemma, the theorem we were given, is actually useful in proving b

(a is false so it definitely doesn't follow from the true theorem we were given)

...i mean, how would you prove c?

[quartz salmon]

We would break it down

Instead of a^2 b

We write aab

Then it follows that p | a^2b

From p | a

[astral bay]

...what is this a proof of?

[quartz salmon]

Of euclid’s lemma

[astral bay]

why are you proving euclid's lemma?

...what is euclid's lemma? what's the actual statement?

[quartz salmon]

We are confirming it’s true by choosing an option that is a consequence of its validity

[astral bay]

...i'm not sure what that means but i don't think it's what the question was asking for

[quartz salmon]

Okay well I’m not sure what you’re asking me for

I got the answer already anyway

.close

[cedar kiln]

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[astral bay]

...i'm not convinced you do have the answer but ok then

[cedar kiln]

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[silent hazel]

I'm working on a question and was wondering if there is a way to express this without using the summation operator. if possible, I want to do it on my own but just wondering if there is a reasonable way to simplify it. Thanks

[silent hazel]

$\sum_{k=1}^{m} \frac{k-1}{n} (-\frac{1}{n})^{m-k}$

[wraith dagger]

PerpetualOwl

[solemn torrent]

like find an expression that equals to that or?

[silent hazel]

yeah

m and n are arbitrary constants

[solemn torrent]

do you have any exemplars

[silent hazel]

of m and n? like they are positive integers

[solemn torrent]

i know

i meant

like nvm

it's gonna be long and tedious

[silent hazel]

yeah don't spend too much time on it just wanted to know if there is something i'm missing

[solemn torrent]

well you can algebraically manipulate it

smth smth

geometric progression

[silent hazel]

yeah but the k in the numerator is the issue

mixed arithmetic and geometric

[solemn torrent]

let's see what you've done

[silent hazel]

$(-\frac{1}{n})^m\sum_{k=1}^{m} \frac{k-1}{n} (-n)^{k}$

[wraith dagger]

PerpetualOwl

[silent hazel]

this follows

but then the issue of n^k being mixed with k persists

$(-\frac{1}{n})^m\left(\sum_{k=1}^{m} \frac{k}{n} (-n)^{k} - \sum_{k=1}^{m} (-n)^{k}\right)$

[wraith dagger]

PerpetualOwl

[solemn torrent]

okay let's just start from the beginning

what

[silent hazel]

uhhh

[solemn torrent]

there

So what you wanna notice is

You got a nice a geometric series

So obv you know a geometric progession can be generally expressed as

[silent hazel]

yeah

[solemn torrent]

ye and what you basically are finding is an expression for sum a_i

So if you just compare what you have in the summand

to a_mr^(n-m)

Can you express it like that

and there's like only one way to do that

[quartz salmon]

are we typing random things in latex

[solemn torrent]

no

okay let (k-1)/n be equiv to a_m and -1/n be equiv to r and m-k be equiv to n-m

let's express a_m as a non fractional coefficient

and like you said n is some arbitrary constant

we could apply summation distributive law

take the constant out

[solemn torrent]

blackboard (double struck) N sub (subscript aka suffix) 1 is the set of natural numbers from 1 onwards, equivalent to the set of positive integers, some authors define natural numbers including or excluding 0

are you following me or you confused

[silent hazel]

sure makes sense, the k and n^k is still an issue

[solemn torrent]

ye right but

let's just forget about that for now

[silent hazel]

ok

[solemn torrent]

$\sum^m_{k\in\mathbb N_1}\frac{k-1}n\left(-n^{-1}\right)^{m-k}=\frac1n\sum^m_{k\in\mathbb N_1}(k-1)(-n^{-1})^{m-k}$

omfg

[solemn torrent]

cool

now let's focus on the RHS

particularly the summand

$\sum^m_{k\in\mathbb N_1}(k-1)(-n^{-1})^{m-k}$

[solemn torrent]

what can you express this as?

[silent hazel]

uhhhh

you can multiply out the k - 1 into two summations and the one with the -1 has a solution

[solemn torrent]

what

let's still work with a single sum

and keep on the geometric series track

\begin{equation*}
\sum^n_{k=1}ar^{k-1}=
    \begin{cases}
        a\left(\frac{1-r^n}{1-r}\right), & r\neq1\\
        an, & r=1
    \end{cases}
\end{equation*}

obviously if you know your summation rules, idex shifting and all that

that's what i want you to apply

summation rules and geometric series type shi

[silent hazel]

yup

[solemn torrent]

$\sum^m_{k\in\mathbb N_1}(k-1)(-n^{-1})^{m-k}=\sum^{m-1}_{j=0}j\left(-n^{-1}\right)^j$

[solemn torrent]

ye?

confusing asl or?

[silent hazel]

makes sense

[solemn torrent]

honestly there's many ways to do this but let's continue down this path

To evaluate the sum we differentiate both sides of the equation

$\dv{x}\sum^{m-1}_{j=0}x^j=\dv{x}\frac{1-x^m}{1-x}$

[solemn torrent]

and ye more geometric series rules which you can prove for good practice

$\sum^{m-1}_{j=0}jx^{j-1}=\frac{mx^{m-1}(1-x)+x^m}{(1-x)^2}$

[silent hazel]

thats actually a very interesting way of approaching it

[solemn torrent]

Ikr i hope harvard gives me a full time scholarship

let x=-n^{-1}

and ye

3 am

goodnight cuh

[silent hazel]

cya xd

[solemn torrent]

i'm kidding cu

finish yo shit

[silent hazel]

thanks

.close

[cedar kiln]

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[solemn torrent]

@toxic epoch

balls

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[coral valley]

why the value of tan x and sec x is different?

[south tundra]

Why would you expect them to be the same?

[coral valley]

I thought that you can also find the angle using sec?

hyp/adj

[south tundra]

tan is opp/adj and sec is hyp/adj

[coral valley]

yes, hyp/adj = 25/20

sec x = 25/20

[south tundra]

And opp/adj = 15/20, tanx = 15/20

Is there still a confusion?

[coral valley]

nvm, forgot to convert degree to radians in my calculator, sorry

.close

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[silver arrow]

Hey.

[cedar kiln]

Please don't occupy multiple help channels.

[silver arrow]

Somone can help mee

Interval
start
10
11
12
13
14
14,5
15
16

Interval
done
10
11
12
13
14
14,5
15
16

hyppighed

18
29
45
40
16
11
25
16

frekvens

0,09%
0,145%
0,225
0,2%
0,08%
0,055%
0,125%
0,08%

sum frekvens

0,09%
0,235%
0,46%
0,66%
0,74%
0,795%
0,92%
1

[latent bloom]

What is this

[silver arrow]

How can I make a sum curve diagram

[latent bloom]

Could you translate this please?

[silver arrow]

its like this

ohh yea

Interval
10-11
11-12
12-13
13-14
14-14.5
14.5-15
15-16
16-17

Frequency
18
29
45
40
16
11
25
16

frequency

0.09%
0.145%
0.225
0.2%
0.08%
0.055%
0.125%
0.08%

sum frequency

0.09%
0.235%
0.46%
0.66%
0.74%
0.795%
0.92%
1

[latent bloom]

Hmm

Frequency
Relative percentage
Cumulative percentage

[silver arrow]

??

Where she can the graph is made

[latent bloom]

Okay first

Please make a table

[silver arrow]

how

I have

what know

[latent bloom]

You have the separated columns of the table

[silver arrow]

yes

Yes, but what do I do now?

Yes, but what do I do now?

@latent bloom

<@&286206848099549185>

this good?

Hallo

[cedar kiln]

@silver arrow Has your question been resolved?

[silver arrow]

@latent bloom

<@&286206848099549185>

[silver arrow]

...

[cedar kiln]

@silver arrow Has your question been resolved?

[silver arrow]

Close this ticket.

[fervent vapor]

Say .Close

[plush warren]

.close

[cerulean rose]

https://tenor.com/view/family-guy-stewie-griffin-gone-crazy-insane-letter-l-gif-17135412

and big

my kevin

[cedar kiln]

@silver arrow Has your question been resolved?

[cedar kiln]

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[short blade]

draw R

the theta and r bound should become clear

then use the substitution r^2 = x^2 + y^2 into the integral

[cedar kiln]

@clear grail Has your question been resolved?

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[wraith dagger]

-Znar

[slim ice]

Hello

How do u want me to help

[buoyant latch]

What do you mean it doesn’t have an inverse

[forest nova]

for it to have a inverse, $2x^2 + 6x$ needs to be injective, but clearly $2(0)^2 + 6(0) = 2(-3)^2 + 6(-3)$, hence it doesn't have an inverse

[wraith dagger]

The Universe

[buoyant latch]

Well the inverse just isn’t a function

Injective means every output is mapped by at most 1 input

[forest nova]

yea

that's bijective

[buoyant latch]

You can have inputs with no output or outputs with no inputs

[forest nova]

injectivity means that \
$f(x) = f(y) \implies x = y$

[wraith dagger]

The Universe

[buoyant latch]

Because it’s a quadratic

Well, a non degenerate quadratic to be specific

All non degenerate quadratics

[forest nova]

non degenerate ones at least

$ax^2 + bx + c$ for $a\ne 0$

[wraith dagger]

The Universe

[buoyant latch]

A degenerate quadratic would be 0x² + 2x + 1

Yeah it’s a quadratic by name but not really

Well…

Hence non degenerate

Like if you put 3 points on top of another they can still form a triangle

But it’s just a degenerate triangle

[forest nova]

f(0) = f(-3) = 0

[buoyant latch]

Hold up

Wikipedia says an inverse only exists for bijective functions

[buoyant latch]

So this is not quite right

[forest nova]

I meant yea the inverse would be undefined for certain values

[buoyant latch]

Yes

But it’s good to give an example

[buoyant latch]

Like this

I know it’s only 1 mark so you probably don’t need much detail

f is not bijective and thus does not have an inverse,
f(0) = f(-3) but 0 ≠ 3 (a requirement for injectivity)

Look at f(x)

Bijective means it’s also injective means only 1 input can map to 1 each output

[buoyant latch]

Which is what this means

But in f we found x=0 and x=-3 both these inputs map to 0 but are not equal to each other

Yes

[cedar kiln]

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[buoyant latch]

It’s not the only reason but it is one of the reasons

[cedar kiln]

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[calm hedge]

ive gone absolutely braindead

[calm hedge]

so i go from

(x)^3-x(3)^2

how does that go into x(x-3)(x+3)

[livid hound]

first thing to look for when factorising is whether all terms have a common factor

and if so, factor that out first

[calm hedge]

yep

so

x(x^2-9)

ohh i see, then difference of two squares

[calm hedge]

oh hang on

then this is used if its ax^3-b

or something

correct me if im wrong

[livid hound]

wdym

what's used

[cedar kiln]

@calm hedge Has your question been resolved?

[calm hedge]

because

i see people taking out a third power

like similar to the example i did before

i gtg now, thanks for your help

[cedar kiln]

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[cedar kiln]

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[brittle pivot]

Question 15

[brittle pivot]

What is this graph called

I’m guessing the transformations are it’s up 30 and a horizontal dilation of 2

But I’m not sure

[cedar kiln]

@brittle pivot Has your question been resolved?

[brittle pivot]

.close

[cedar kiln]

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[sick mirage]

Hi did I get this right?

I'm second guessing myself because the original vector is [ v1 v2] and adding any arbitrary vector could be a subspace

[zenith elbow]

I think it is

Since the x value can be anything we don't need to worry about addition or multiplication taking it out of the subspace

And the y coordinate is never affected

It being non empty is trivial

Wait sorry I'm wrong adding two of them will make y = - 10

So you're right

[sick mirage]

Ahh ok - yeah that's what I thought since the 0 vector is not included in {x, -5}

is that a good reason too?

[cedar kiln]

@sick mirage Has your question been resolved?

[cedar kiln]

@sick mirage Has your question been resolved?

[mighty drift]

yes it is

[sick mirage]

it is a subspace?

[mighty drift]

no

[mighty drift]

it is a good reason

[sick mirage]

ok so I was correct

gotcha ty much!

did I get all correct?

[mighty drift]

4/5

[sick mirage]

3rd quad one is wrong?

[mighty drift]

5th one

[sick mirage]

that was the one I was uncertain of

ahh true

[mighty drift]

also doesn't contain 0

[sick mirage]

+3 makes it not - got it

right?

[mighty drift]

+3 makes it an affine space, which isn't a subspace (an affine space is of the form a + E with E a vector space and a a vector, and therefore is only a vector space if a = 0)

[sick mirage]

I think I understand - so fails the Identity test?

[mighty drift]

if you call it that

[sick mirage]

!close

.close

[cedar kiln]

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Channel closed

Closed due to the original message being deleted

[tropic oxide]

@gusty mason you fucked up and deleted your original message, now you need to abandon this channel and pick a new one.

[gusty mason]

Yeah I know I'm typing the latex of my question

[cedar kiln]

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[wary anchor]

I've got B and C the other way around, would it matter?

[subtle raven]

No I think you're okay

[wary anchor]

i got it the other way around and got it wrong