help-0

[foggy pecan]

look here:

$\int_{}^{}\frac{dx}{x\sqrt{x^{2}-9}}\overset{\ast }{=}\\x=\frac{1}{t}\Leftrightarrow dx=-\frac{dt}{t^{2}}\\\overset{\ast }{=}\int_{}^{}\frac{t\text{ }dt}{-t^{2}\sqrt{\frac{1}{t^{2}}-9}}=-\int_{}^{}\frac{dt}{\sqrt{1-9t^{2}}}=-\int_{}^{}\frac{dt}{\sqrt{1-\left( 3t \right)^{2}}}\overset{\ast \ast }{=}\\u=3t\Leftrightarrow du=3dt\\\overset{\ast \ast}{= }-\frac{1}{3}\int_{}^{}\frac{du}{\sqrt{1-u^{2}}}=-\frac{1}{3}\arcsin\left( \frac{3}{x} \right)+C$

[ocean seal]

Joanna Angel

[lilac wave]

That's very useful and I appreciate your efforts.
But I'd like to ask, is there no way to end up with arcsec instead of arcsin ?

[tacit arch]

Why do you want an arcsec

[lilac wave]

It should be possible, and I'm interested in knowing how

[tacit arch]

Why do you say should

[lilac wave]

The answer is below the question, I just want to know how it works
Of course there might be a typo, and that's what I want to confirm, is it possible the get the answer below the question ? If so, how ?

[tacit arch]

You can check it by differentiating arcsec

,w diff (1/3) arcsec(x/3)

[ocean seal]

Results provided by WolframAlpha

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Requested by rie.mann

[tacit arch]

Looks correct yea

[lilac wave]

Give me a minute, let me wrap my head around it

[tacit arch]

If you haven't seen the proof before,
||https://proofwiki.org/wiki/Derivative_of_Arcsecant_Function||

[lilac wave]

I just don't know if it's actually the same problem, even it it is, I still don't know the method

[tacit arch]

Either do the proof backwards or memorize the derivative of arcsec

[lilac wave]

Aside from that, I still don't know how to integrate the original question, the question isn't to find the connection between them, the question is to integrate the problem and end up with an answer similar to the answer given

[tacit arch]

Did you read the first part

[lilac wave]

Oh

I'm not sure I fully understand it but I'll see what I can do

I'll close the cannel for now

.close

[lone heart]

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[cerulean junco]

so, rank of a matrix is the number of lineraly independent rows nR(A) or lineraly independent colums nC(A) in it ?

and nR(A) would be equal to nC(A)!?

[mortal trellis]

formally, the maximum number of linearly independent rows is the row rank and the maximum number of linearly independent columns is the column rank

it turns out they are both the same and thats why we just call them rank

[cerulean junco]

ohhhkay .....

[cerulean junco]

...and full rank ?
full rank when it has no LINEARLY DEPENDENT ROWS/COLUMNS ?

[mortal trellis]

full rank is when the rank equals either the number of rows of columns

cause then it cant have more

but note that if for example the matrix has 2 rows and 5 columns, then full rank means rank 2

[cerulean junco]

does thatean that no LD rows or colums?

[high oak]

I need help

[mortal trellis]

no, in my example there will be linearly dependent columns

there are 5 columns but rank is 2

[cerulean junco]

oh ...so in case of rectangular matrix it would have to be the least

[mortal trellis]

so only 2 of them can be independent

yes

[high oak]

Needs help with square root?

[cerulean junco]

..... it would have to match the least dim of the matrix, to be considered as full rank ? @mortal trellis

[high oak]

↘️:-D/|¦™®¥

=?

[cerulean junco]

Go and post your doubt/problem in another seperate #help channel.

[high oak]

No I'm asking if anyone needs help

[cerulean junco]

no I don't need help with sq root .
i need help with matrix .... i assume you know what that is !

[high oak]

What GRD u?

[cerulean junco]

come to general chat .... for these type pf conversation .
this is not the appropriate place .

[high oak]

Ohk?

[cerulean junco]

. @mortal trellis ?

[high oak]

I need help with the Sq root

Tbh

[mortal trellis]

I already answered yes

[cerulean junco]

oh mb ....i missed it .

[lone heart]

@cerulean junco Has your question been resolved?

[lone heart]

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[stuck fractal]

Q: Differential Vector Calculus

[ocean seal]

a) Able to show that flow is irrotational: General evaluation:
\[
\vec{\nabla}\times\vec{q}=(0-0){\textbf{\^\i}}-(-3z^2+3z^2){\textbf{\^\j}}+(4x-4x){\textbf{\^k}}
\]
b) Struggling with a conceptual understanding of what is wanted by the question, aware that $\text{grad}\;\phi$ might be helpful in this instance

[stuck fractal]

<@&286206848099549185> (sorry )

[foggy pecan]

look:

$\left{ \begin{array}{cl}
\frac{\partial\phi}{\partial\text{}x}=4xy-z^{3} \
\frac{\partial\phi}{\partial\text{}y}=2x^{2}\\frac{\partial\phi}{\partial\text{}z}=-3xz^{2}
\end{array} \right.\\\phi\left( x,y,z \right)=2x^{2}y-xz^{3}+\varphi\left( y,z \right)\Leftrightarrow \\\frac{\partial\phi}{\partial\text{}y}=2x^{2}+\frac{\partial\varphi}{\partial\text{}y}\Leftrightarrow \frac{\partial\varphi}{\partial\text{}y}=0\Leftrightarrow \varphi\left( y,z \right)=\psi\left( z \right)=C\\\phi\left( x,y,z \right)=2x^{2}y-xz^{3}+C$

[ocean seal]

Joanna Angel

[foggy pecan]

I did some calculations in my mind, I think you are familiar enough with this topic to be able to understand

[stuck fractal]

AHhhhhhhhhhhhhhhhhhhh

Okay, works like a puzzle piece

[foggy pecan]

smiles

[stuck fractal]

That's perfect thank you <3

[foggy pecan]

yvw )

[stuck fractal]

.close

[lone heart]

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[valid hazel]

.reopen

[minor bone]

does AA^-1 = I implies A^T has an inverse?

[lone heart]

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[unkempt notch]

hello i have a test in an hour and i dont know how to do this

[unkempt notch]

HELP!

😢

[flat roost]

i guess find df(x)/dx first?

do you know how to find df(x)/dx

[unkempt notch]

where do you get the d from

[flat roost]

well it says average rate of change

differentiating the function first is all i can think of lol

[alpine sable]

no

the average rate of change is not the derivative

[unkempt notch]

here is the lesson

[alpine sable]

[
\underbrace{\f{\m f{x+h} - \m fx}h}_{\t{average rate of change}}
]

[alpine sable]

you can use that

[unkempt notch]

i don’t understand the lesson

[flat roost]

what is h

[alpine sable]

okay so like

do u remember the slope formula for a line

the one you learnt back in middle school

[unkempt notch]

y=mx+b

[alpine sable]

yeah

[unkempt notch]

im still in highschool

[alpine sable]

how do u find m

[unkempt notch]

i dont know

[alpine sable]

so like uh

[
m = \f{\Delta y}{\Delta x} = \f{y_2 - y_1}{x_2-x_1}
]

[alpine sable]

does this ring a bell now

[unkempt notch]

oh yeah

yay

[alpine sable]

yea

rise over a run

do u understand how that works

[unkempt notch]

yes

up and then left or right

or down first

[alpine sable]

ok

so do u know what a secant line is

[unkempt notch]

yes but i forgot

so no

im gonna fail this test

im scared

[alpine sable]

gimme one second ill make something

[unkempt notch]

thank you so much for helping

[alpine sable]

\begin{tikzpicture}
    \draw[<->] (-5, 0) --  (5,0) node [right] {$x$}; 
    \draw[<->] (0, -5) -- (0,5) node [above] {$y$}; 
    \fill (1, 0) circle [radius=2pt] node [above right] {$(1,0)$};
    \fill (0, -3) circle [radius=2pt] node [above left] {$(0,-3)$};
    \draw [<->] (-0.5, -4.5) -- (2, 3) node[midway, below right=9pt] {rise};
    \draw [dashed, thick] (0, -3) -- (1, -3) node[midway, below] {run};
    \draw [dashed, thick] (1, -3) -- (1, 0);
    \draw [thick, blue] (0,-3) -- (1,0) node[midway, above left = 15 pt]{slope};
\end{tikzpicture}

[unkempt notch]

WOW

[alpine sable]

so like this is an illustration of the thing

[unkempt notch]

wait a sec

how did you get that

[alpine sable]

its an example

like not related to what u r doing

im just trying to explain

[unkempt notch]

ohhh

okay

yes

[alpine sable]

but basically whats happening here is the run = x_2 - x_1 and rise = y_2 - y_1

and then dividing them you get the "steepness" of the line aka the slope (y_2 - y_1)/(x_2-x_1)

now this relates to the "secant" line similarly

[unkempt notch]

so basically you need to find how steep it is with y2-y1/ x2-x1

?

[alpine sable]

yeah

thats the idea of a slope

but like

\usetikzlibrary{decorations.pathreplacing}
\begin{tikzpicture}[scale=1.75,cap=round]
\tikzset{axes/.style={}}
\begin{scope}[style=axes]
\draw[->] (-.5,0) -- (4.5,0) node[below] {$x$};
\draw[->] (0,-.5)-- (0,3) node[left] {$y$};
\foreach \x/\xtext in {1.5/x_{1}, 3/x_{2}}
 \draw[xshift=\x cm] (0pt,2pt) -- (0pt,-2pt) 
node[below,fill=bg,font=\normalsize]
  {$\xtext$};    
\foreach \y/\ytext in {1/y_{1}=f(x_{1}), 2.125/y_{1}=f(x_{2})}
  \draw[yshift=\y cm] (2pt,0pt) -- (-2pt,0pt) 
  node[left,fill=bg,font=\normalsize]
  {$\ytext$};
 \draw[domain=.5:3.25,smooth,variable=\x,red,<->,thick] plot ({\x},{.5*(\x-1.5)*(\x-1.5)+1});
\filldraw[bg] (1.5,1) circle (1pt) node[above] {\scriptsize $P$};
\filldraw[bg] (3,2.125) circle (1pt) node[left] {\scriptsize $Q$};
\draw[thick,blue!50,shorten >=-.5cm,shorten <=-.5cm] (1.5,1)--(3,2.125) 
node[midway,left] {\scriptsize Secant Line};
 \draw[blue!50,thick,dashed] (1.5,1)--(3,1)--(3,2.125);
 \draw[blue!50] (3,1.1)--(2.9,1.1)--(2.9,1);
 \draw[decoration={brace,mirror,raise=5pt},decorate,blue!50]
    (1.5,-.250) -- node[below=6pt] {$x_{2}-x_{1}$} (3,-.250);
 \draw[decoration={brace,mirror, raise=5pt},decorate,blue!50]
    (3,1) -- node[right=6pt] {$f(x_{2})-f(x_{1})$} (3,2.215);
\filldraw[bg] (1.5,1) circle (1pt) node[above] {\scriptsize $P$};
\filldraw[bg] (3,2.125) circle (1pt) node[left] {\scriptsize $Q$};
\end{scope}
\end{tikzpicture}

[alpine sable]

(credits to that one tex stackexchange dude for this one)

but this is a secant line

a secant line is like

a line that crosses two points for a function

then you can try to find the "steepness" of that line by finding its slope

we can apply the same idea of finding rise and run

so like

"rise" would be f(x_2) - f_(x_1)

[unkempt notch]

:(

[alpine sable]

and "run" would be x_2 - x_1

if you dont understand its okay i can try explaining more

[unkempt notch]

i dont think we learned that

[alpine sable]

no you definitely did haha

[alpine sable]

its written in ur page

its this pretty much

[unkempt notch]

we only learned the formula part

[alpine sable]

yeah

[unkempt notch]

LOL my flop drawing

[alpine sable]

[unkempt notch]

wait i think i understand the lesson a little

you just plug in the numbers

and then put the numbers in the formula into the equation

[alpine sable]

like

[unkempt notch]

no you put the equation in the formula that has the numbers

the f(x)

wait let me try solving the question

[alpine sable]

dont learn to do it this way

its really not good for you

[unkempt notch]

i only have 20 minutes before school starts

[alpine sable]

plugging and chugging into a formula will make u like very confused in the future probs

oh lol

[unkempt notch]

i should have not left this at the last second

but i was also doing bio

ughh

[alpine sable]

[unkempt notch]

ok lemme try brb

[alpine sable]

we can continue this after ur school is over u dont have to cram everything right this second lol

unless you have an exam

then uh

you are kinda fucked

[lone heart]

@unkempt notch Has your question been resolved?

[unkempt notch]

i have a test fr in 30 minutes

i’m getting ready for school 😞

[alpine sable]

good luck

[unkempt notch]

my exam is in 2 weeks

i seriously need help with math

[alpine sable]

whats the difference between a test and an exam

[unkempt notch]

test is just for the unit

and exam is for all the units

[alpine sable]

oh i see ok

[unkempt notch]

the whole course is the exam

i have a 67% in math rn…

😢

[alpine sable]

might be able to salvage that

[unkempt notch]

its extremely bad

[alpine sable]

anyways like this convervastion kinda digressed. Do you still need help with ur thingy?

[unkempt notch]

yes

one sec is solving it

[alpine sable]

ok

[unkempt notch]

i had to go to the bathroom

😭

[alpine sable]

lmao no worries

[unkempt notch]

yay

[alpine sable]

yeah

thats right

[unkempt notch]

YAY

okay next part

[alpine sable]

yeah

the instantaneous chagn

its basically like

[unkempt notch]

this

[alpine sable]

yuh that wokrs

[unkempt notch]

i don’t have time to finish it but i think i got it

😍

yay

[alpine sable]

wOooo

good job

[unkempt notch]

OKAY CYA I HAVE TO GO NOW

[alpine sable]

byee

type .close to close this

[unkempt notch]

THANK YOU FOR YOUR HELP

okay :)

[alpine sable]

no worriesss

[unkempt notch]

.close

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[alpine sable]

good luck with school

[unkempt notch]

thank you

[lone heart]

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[fallow wadi]

id start by making 1/2C x (3A+B^-1)^-1

can this be more simplified or not?

[summer dirge]

1/2C?

[fallow wadi]

yes

[summer dirge]

You can’t divide by a matrix though??

[fallow wadi]

C/2

[summer dirge]

Oh

[summer dirge]

I don’t think this is the correct approach

Actually, nvm

That’s fine

[fallow wadi]

can i do smth more with it?

@summer dirge

Or anyone?

https://cdn.discordapp.com/attachments/490557019623915520/1196435021834235934/IMG_0324.jpg?ex=65b79df1&is=65a528f1&hm=8304edccbf32fef68a692521a3fb69e37669f66ca899e717ac86a92f28651623&

this is the original problem

@vale wigeon maybe you know

i need someone to check, whether the simplification is CORRECT

[lone heart]

@fallow wadi Has your question been resolved?

[lone heart]

@fallow wadi Has your question been resolved?

[lone heart]

@fallow wadi Has your question been resolved?

[stuck fractal]

,wolf c/2(3A+B^-1)^-1=((3A+B^-1)2c^-1)^-1

[ocean seal]

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Requested by antoan.io

[stuck fractal]

Yes

It is correct

There are nicer forms though haha

i.e.
[
\frac{bc}{6ab+2}
]

[stuck fractal]

or [
\frac{bc}{2(3ab+1)}
]

[stuck fractal]

Oh wait you're doing matricies I see why you want it that way

[fallow wadi]

@stuck fractalso

it cant be simplified more

[fallow wadi]

.close

[lone heart]

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[thick cipher]

hi guys

[thick cipher]

intermediate value theorem, it is specifically asking us if thre is a 0 in an interval

since this is sin, what do i do next?

[lone heart]

@thick cipher Has your question been resolved?

[lone heart]

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[lyric seal]

can someone explain to me the beginning of this proof, I cant understand how B(m,n) := φ_m(n)

[golden canyon]

They defined B(m, n) to be phi_m applied to n

[lyric seal]

yeah wait i think i missed the point of the phi_m, thank you

.close

[lone heart]

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[keen orbit]

I want to say that the loop invariant is Arr[i..j]

but that won't be true before the loop runs

or maybe it should be Arr[i+1..j-1]

[lone heart]

@keen orbit Has your question been resolved?

[lone heart]

@keen orbit Has your question been resolved?

[stable warren]

um i think u should go to the Computer Science group?

[lone heart]

@keen orbit Has your question been resolved?

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[tropic mango]

Hello

[tropic mango]

Is this allowed how I did it?

[vale wigeon]

allowed? yes. helpful? questionable.

errr

should be f'(x) = x cos(x) and f(x) = cos(x) - x sin(x) tho

[tropic mango]

Oh right right

[tropic mango]

So I can do that?

[vale wigeon]

you can but i doubt it'll help you.

[tropic mango]

Why wouldn't it?

Much better than working with cos(x)^2

[vale wigeon]

feels like it will just complicate the integral afterwaard.

you'll get x sin^2(x) somewhere in there.

[tropic mango]

Oh

Hm

My solution has to be

So what would be an appropriate way to do it?

[vale wigeon]

write cos^2(x) as (1 + cos(2x))/2

then you have x/2 + x cos(2x)/2

x/2 integrates easily and x cos(2x)/2 integrates with a much cleaner ibp

[tropic mango]

So that means I'd have to work with two integrals?

[vale wigeon]

sure

but one of them's trivial

so i don't see what the issue is

[tropic mango]

I am just bad at math

That's the issue 😂

Stuff like this I can do

I also have to do it with the partial integration method

I will be trying and see where I go

[lone heart]

@tropic mango Has your question been resolved?

[tropic mango]

I am still figuring this out

[tropic mango]

So far got like this

This is truly pain

I don't know how to continue

[tropic mango]

<@&286206848099549185> I kind of am stuck here :c

[lethal solstice]

i think it'd be easier to use f(x) = x instead

[tropic mango]

Oh

So

g'(x) = cos(2x) ---> g(x) = sin(2x)/2
f(x) = x ---> f'(x) = 1
Like this?

[lethal solstice]

yeah

[tropic mango]

Hm well alright

I try it with that

Should I leave the other integral alone or also convert it?

[lethal solstice]

the 1/2 int xdx?

[tropic mango]

Yep

[lethal solstice]

that one is fine because you can solve that immediately

[tropic mango]

1/2 * 1/2 * x^2 = 1/4 * x^2 = x^2/4

This?

[lethal solstice]

nice, yeah

[tropic mango]

Thanks. There are moments I can be very good, but other times dumb

[lethal solstice]

that's alright it happens to everyone

[tropic mango]

I guess so

[tropic mango]

@lethal solstice thank you

[lethal solstice]

np

[tropic mango]

.close

[lone heart]

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• Be polite and have a nice day!

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[paper pollen]

Hello I just have a quick question
the answer key for this question says that the answer is B
but C is the only one thats making sense
is there something that I'm missing?

[lone heart]

@paper pollen Has your question been resolved?

[paper pollen]

<@&286206848099549185>

[stiff yew]

It has to be exactly 2 wins you know that it starts with win,win,loss. So you have to calculate how many different arrangements there are, when taking 2 losses out of a total of 5 remaining games. Which is the binomial coefficient 5 over 2, so the answer is 10.

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[hasty arch]

https://cdn.discordapp.com/attachments/741761765691621387/1196586995741360128/image.png?ex=65b82b7b&is=65a5b67b&hm=d23b432e76fc2664840d4cc61c8a3b90eaf0809ab5537136a0dde72d2dff3ba0&

[hasty arch]

I don't know how to find the value of m in this

I understand the form y-y1 = m(x-x1)

I used ln(2x)/4x to find 0.173 but can't find m

[brave moth]

wait what xD

isn't m just 0.0767?

based on your calculations

@hasty arch

[ivory fern]

You have to use the derivative

[hasty arch]

how lol

alrighty well I'm not sure where to go from there

don't we have to deal with ln(2x) first with chain rule before using quotient?

I also thought about putting it into the form ln(2x) (4x)^-1 but idk if thats possible

[hasty arch]

btw this picture is from my teacher, she didn't show any work lol

[brave moth]

ah so you want to actually find that m = 0.0767

missunderstood

[ivory fern]

Do you mean that you don't know how to find the derivative, or what to do with the derivative to find m?

[hasty arch]

not sure how to find the derivative

I know to find m we substitute x=1 into the derivative to find it

[ivory fern]

Yes, you can do this

Or the quotient

[hasty arch]

so then how would I do it either way

f(x) = ln(x), f'(x) = 1/x

so then 1/x (2x) x 2 cuz d/dx (2x) = 2

so ln(2x) just becomes 4

[ivory fern]

$\dfrac{d}{dx}\dfrac{\log 2x}{4x}=\dfrac{\tfrac{d}{dx}\log(2x)\cdot 4x-\log(2x)\tfrac{d}{dx}(4x)}{(4x)^2}$

[ocean seal]

d

[hasty arch]

is this in the form f'g - fg' / g^2 ?

[ivory fern]

Why 1/x (2x) times 2 instead of 1/x times 2?

[ivory fern]

Yes

[hasty arch]

it's ln(2x), I thought to use chain rule

oh

wait I see it now lmao

[hasty arch]

or at least this way

ln(2x) = f , 4x = g

[ivory fern]

Exactly

[ivory fern]

Do you know how to continue from here?

[hasty arch]

I'm lost on how we get 1/x from ln(2x)

I know the derivative of ln(x) is 1/x, where does the 2 in (2x) go?

[ivory fern]

log(2x)' = 1/(2x) times (2x)' = 1/(2x) times 2 = 1/x
Also, log(2x) = log x + log 2, so log(2x)' = (log x)' + (log 2)' = 1/x

[hasty arch]

oh so chain rule is used here

[ivory fern]

Chain rule is used always

[hasty arch]

right right ok

[ivory fern]

Except for when there is only one function

[hasty arch]

OH 2/2x = 1/x

got it

so then 1/x * 4x - ln(2x)(4) / 4x^2

using quotient rule ^

f' = 1/x , g = 4x | f = ln(2x), g' = 4

[ivory fern]

Yes
But please use parentheses

[hasty arch]

lmao my bad, gotta learn to use texit again for better clarification too

thank you!

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[vocal tapir]

Why does ln(1+1/n)=(1/n)-(1/2n^2)+...

[vocal tapir]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[neat mango]

This is a specific example of Taylor expansion. Here, the function f(1+1/n) = ln(1+1/n) is expanded about the point x = 1 (where f(1) = ln(1) = 0)

[vocal tapir]

ohh so expanded at 1

[neat mango]

Exactly

[vocal tapir]

alright ty

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[vocal tapir]

wait

[vocal tapir]

it doesn't add up

I tried it manually

doesn't give 1/n-1/2n^2

even with the rule of e^x>x^p>ln(x) in means of growth of functions

where you write off the slower growing function

[maiden glen]

I think that's actually the expansion at infinity

[vocal tapir]

yes

that's right

but how like

I've never expanded a function besides at point 0

how would I expand a function at infinity?

[maiden glen]

substitute x = 1/n and expand x at 0

[vocal tapir]

that's clever

[maiden glen]

that's why all the ns are in the denominator

when you swap the xs back

[vocal tapir]

it all makes sense. That wasn't explained in our lecture

but

what does it have to do with infinity?

[maiden glen]

well, you can also think of it as taking a limit of expansions as the point of expansion goes to infinity

you get the same thing as substituting

[vocal tapir]

right

if x->0

x=1/y -> y inf

[maiden glen]

yeah

[vocal tapir]

alright tysm @maiden glen

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[magic dragon]

trying to find the point of intersection between L1 and L2

[magic dragon]

do i multiply both sides by (1/7, 1/3, 1/1)?

to isolate s

or can i not do that

or is there a better way of doing this entirely

[maiden glen]

expanding your final row yields a system of 3 equations in 2 variables

solve using whatever simultaneous equations method you like

if the system is consistent, then those'll be the values of s and t for the intersection

if it's inconsistent then your lines are parallel or skew

[magic dragon]

does expanding mean distributing the s and t within the vector?

[maiden glen]

well

I mean compare the coordinates

so your first equation is 7s = 3 + 3t

etc.

[magic dragon]

ah yeah ok

so i'd have 3 equations, 1 for every axis

[maiden glen]

but only 2 variables

so pick a pair and solve them simultaneously

[magic dragon]

this is good right

and getting t is not neccessary, i would just need to plug in s for one of the equations to get the point of intersection right?

[maiden glen]

well

yes, if you already know they intersect

but if you don't know this, you need to solve the other pairs of equations (2 and 3) and (1 and 3)

and check that the values of s or t you get are consistent

[magic dragon]

thanks

could you also just check if they are parallel

[maiden glen]

lines can be non-parallel and still not intersect

[magic dragon]

oh yeah that's true

[maiden glen]

(they are called skew in that case)

[magic dragon]

oki thank you

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[strange smelt]

https://youtu.be/JiAbdbPmRzM

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@strange smelt Has your question been resolved?

[tacit arch]

.close

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[old sluice]

Yo

[old sluice]

I was wondering why would my answer be wrong if I divided the equation circled in red by -1

[chilly crag]

what are you trying to do here?

like, what is the goal

[old sluice]

Because I scanned my answer and it said it was wrong but once I divided the second equation by -1 it worked

[acoustic marten]

he

hello can i get some help

[old sluice]

Elemination method

[old sluice]

Go somewhere else

[chilly crag]

you mean, you want to get y=...?

[acoustic marten]

damn

[chilly crag]

@acoustic marten #❓how-to-get-help

[old sluice]

Yes

[chilly crag]

alright

we have 3x+2y=-1

[old sluice]

I don’t know how to explain it

No no

I already got the answer I just don’t understand why i can’t multiply -1 by either one

[chilly crag]

you can

you get
-3x-2y=1

but that doesnt help you in your pursuit of y=...

doesnt change the end result

[old sluice]

Yea that’s why it said it was wrong

But it was right when I did it by the 4x

I’m just trying to understand why and how i identify this next time

[chilly crag]

ok

for that, let's look at our equation

3x+2y=-1

step one is looking where our variables are

in this case, the y is only once in the equation

otherwise we would have to use the distributive rule

step 2:
get rid of everything else on the side of the equation with the y

in this case, there is a 3x there as well

last, we get rid of the 2 by dividing

multiplying or dividing by -1 is used to change the sign from + to - and from - to +

so we would want to do that, when we have stuff like 3x-2y=-1

because in the end we want the y to be +y and not -y

[old sluice]

Ohhhhh so everytime I get the y as a negative I know my answer is wrong?

[chilly crag]

well

it is not wrong

[old sluice]

Yea I was about to say nvm bc on this I got a negative for the y and it’s correct

[chilly crag]

if you solve something and get
-y=3x+5 for example
then that equation can be true, it just isnt fully solved

[old sluice]

I just want to know why can’t I plug in -1 into the equation with the arrow pointing at it??

[chilly crag]

what do you mean with "plug in -1"?

[old sluice]

Do you not see the -1 I’m distributing into the second equation?

That’s what I’m talking about

[chilly crag]

ah that is what you mean

yeah thats totally fine

[old sluice]

And I’m asking why can’t I’m plug it into the first one

[chilly crag]

you can

if we do that, we get:
-3x-2y=1
4x+2y=-6
adding them, we get
x=-5
which is what you got before as well

[maiden glen]

(for reference, to "plug in" some number usually means to substitute it into a variable, hence the confusion)

[old sluice]

Alr thank you I did that at first but idk why I got it wrong thanks

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[bright marsh]

Hi this is my math problem. I got stuck trying to find the moles of the excess Oxygen.

[bright marsh]

<@&286206848099549185>

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@bright marsh Has your question been resolved?

[bright marsh]

<@&286206848099549185>

[fervent kestrel]

I can't see that question, is there a way for you to rewrite it or send a photo that does not have a transparent background

[bright marsh]

• An unknown gas consists of 80.0% carbon and 20.0% hydrogen. A 4.60 g sample of this gas occupies a 2.50 L volume at 25.00°C and 152 kPa.
a) If this sample was combusted with 20 g of oxygen gas, what is the volume of the unreacted excess reagent after combustion when the temp has climbed to 515oC ~ same pressure?
• b) As the reaction cools below 1000C, the water vapour produced by the combustion reaction will start to react with the carbon dioxide to create carbonic acid. What is the pH of this acid?

This is the question

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@bright marsh Has your question been resolved?

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[alpine sable]

How do i find the local minimum/maximum on this graph.

Is this correct:

Local Maximum: (30, 100.385)
Local Minimum: (0, 0) & (80.765, 0)

Link to graph: https://www.desmos.com/calculator/q3fworvvuc

[stark crater]

There are more local mins and maxes

You listed global mins and maxes only

[alpine sable]

So what are the local max and mins and how do i find them

like where purple and black intersect is that an example?

[stark crater]

Yeah that is a local min.

Local means in a small interval

So any point that looks higher,lower than its neighboring points are local max/min

[alpine sable]

alright

and the instantaneous rate of change at a local min/max should be 0?

.close

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[leaden robin]

If X is an integer, then 3x+4 is even. Prove the following conditional statement using the contrapositive method

[vapid shuttle]

?

[echo socket]

You can't prove what's false though

[vapid shuttle]

x=3

[tiny nexus]

wait is this a question, seems like it doesnt make sense since its not a true statement?

if x=5, 3x+4=19
if x=7, 3x+4=25,
so if x is odd the statement does not hold?

[leaden robin]

its a proof question

sorry for not explaining

[summer dirge]

but it's not true??

let x = 1

then 3(1) + 4 = 7

which is not even

[leaden robin]

you cant do that with proofs

[summer dirge]

huh?

[leaden robin]

have you done proofs ?

its a series

[summer dirge]

yes

what do you mean

can you explain?

[leaden robin]

the contra positive method is when you have a statmenet where A=>B where now not A => not B

in this example

[summer dirge]

no i know what contrapositive is

[leaden robin]

the question would be changed to if 3x+4 is odd then x is not an integer

[summer dirge]

it doesn't change the fact that the original statement is not true though

[leaden robin]

yea its not

but you have to prove it

not by subsituting values

[summer dirge]

this isnt true either

[summer dirge]

we cant prove what isnt true??

i substituted a value as a counterexample, not to prove it

[summer dirge]

again, let 3x + 4 = 7, then x = 1 which is an integer

thus falsifying the claim

[leaden robin]

ill give you another example

im not saying your wrong

but i need a proof

if x^2-6x+5 is even then x is odd

ive already solved this so ill show you what type of answer i need

prove the following stament by contrapositive

if x is even then x^2-6x+5 is odd

x = 2k where x is a element of an integer Z

x^2-6x+5 = 2k^2 - 6 (2k) + 5

2k^2-12k+5

2(2k^2-6k) + 5

[summer dirge]

4k^2, but yeah

[leaden robin]

let 2k^2-6k = m

2m , m is a element of an integer

2m+5

[summer dirge]

either way though, i dont see why this is relevant??

[leaden robin]

as the contrapositive is true then the og statment aswell

[leaden robin]

the format of the answer

[summer dirge]

yes, but your claim is true in this case

it is not in the original question

[leaden robin]

the question is irrelevant but i wanna show the format

[summer dirge]

so there is nothing to prove

[summer dirge]

no, i recognize the format lol

i have done these proofs before too

[leaden robin]

yeah but how do i answer this ?

If X is an integer, then 3x+4 is even. Prove the following conditional statement using the contrapositive method

[summer dirge]

you literally cant

its just not true

[leaden robin]

well

[summer dirge]

unless perhaps you meant that if x is an even integer, then 3x+4 is even?

[leaden robin]

no

thats the og question

[summer dirge]

well the og question is wrong

[leaden robin]

ill just ask my teach thx for the help

.close

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[ebon meadow]

im learning about integrals using Riemann sum. how would i evaluate this? the thing on the right is a hint my TA gave

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[ebon meadow]

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[alpine sable]

Can someone help me on where to begin

[alpine sable]

(will pay nitro)

[quasi vector]

which question do u need help on?

[alpine sable]

11

both

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[brittle ember]

For question 11 I think you can use conservation of momentum

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[drifting warren]

I solved this one the way they said to and got (mq - pn)(ad - bc) which is correct although I dont see why you cannot solve the normal way with (am+bp)(cn+dq) - (cm+dp)(an+bq)

[native cloud]

Does it say (am+bp)(cn+dq) - (cm+dp)(an+bq) is incorrect?

[drifting warren]

the two answers are not equal

Doesn't say its incorrect but I think it is

[native cloud]

How did they even get (mq - pn)(ad - bc)

[drifting warren]

The theorem

I can send work if it helps

[native cloud]

Those gamma symbols aren't eigenvalues, right?

And are we solving for det(∆)?

[drifting warren]

No they are not eigenvalues and yes

The work if it helps

<@&286206848099549185>

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[drifting warren]

<@&286206848099549185>

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@drifting warren Has your question been resolved?

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@drifting warren Has your question been resolved?

[drifting warren]

.close

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[exotic canopy]

trying to understand what transitive (sub)groups are. in this definition, what does the N_n mean?

[exotic canopy]

$\bN_n$

[ocean seal]

artemetra

[exotic canopy]

is it the same as $\bN^n$?

[ocean seal]

artemetra

[exotic canopy]

ping when replying

[vale wigeon]

@exotic canopy no it would seem that it means 1:n

{1,2,...,n}

[exotic canopy]

oh i see

that makes more sense

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[karmic ravine]

1
Write verbal statements for the meaning of:
a fxjx>5, x2Rg
b fxjx63, x2Zg
d fxj26x64, x2Zg
e ftj1<t<5g

[exotic canopy]

is this ocr

maybe send a picture?...

"a fxjx>5, x2Rg b fxjx63, x2Zg" is illegible

[karmic ravine]

sorry the question is not pasting right one sec

exercie c

im not quiet under how to right them in verbal statements

[exotic canopy]

okay that's much better lol

[karmic ravine]

haha sorry

[exotic canopy]

write in plain english what that set is

[karmic ravine]

im not sure how to translate it

[exotic canopy]

it's "the set of all values that can be such that y ___"

fill in the blank

0 < y < 6

[thorny patio]

E.g. for the first one:

Set of all Numbers x, such that x is greater than 5 and x is a member of Real Numbers

[karmic ravine]

for 1 A
is it "The set of all values that can be such that X"

im not sure how to right it beyond that point

ohh

[thorny patio]

The stuff on the left is objects in the set, the vertical bar is the phrase "such that" the stuff to the right of the bar are the conditions imposed

[karmic ravine]

is this the answer or is their more

i think i understnd if it is

[thorny patio]

That would suffice for an answer in my eyes

Unless your book or teacher has a very specific way of wanting these statements

[thorny patio]

Should be "element of" not member of

And if you want to be a little more precise, element of the set of all real numbers

But other than that there is no more to describe

[karmic ravine]

for B
Set of all numbers x, such that x is less or equal to 3

thats all ive got

[thorny patio]

So that one says also x is in Z

Which means

x is an element of the set of all Integers

[karmic ravine]

this is like a different language damn

[thorny patio]

Yea kinda haha

You will quickly get used.to.it tho

[karmic ravine]

Set of all numbers x, such that x is less or equal to 3

[karmic ravine]

is that the "x E z" symbol?

[thorny patio]

And such that x is an element of the set of all integers

Yea that "e" symbol

Means that the previous object is part of the named set

So

x "E" Z

Says x is an element of Z

Z is a famous well known set

The set of all integers

(Comes from the German word for integer, Zallen)

[karmic ravine]

Set of all numbers x, such that x is less or equal to 3 so that one says also x is in Z and such that x is an element of the set of all intergers

this is tricky

[thorny patio]

Yea so it would be

Set of all numbers x s.t. x is less than or equal to 3 and s.t. x is an element of the set of all integers

That covers everything

[limpid turret]

$x\in\bZ$

[ocean seal]

SWR

[thorny patio]

Thank you lol

[karmic ravine]

oh thanks

[thorny patio]

I gotta learn the text commands in here some time

[karmic ravine]

alr i think im slowly getting there

im gonna try 3

c

[thorny patio]

Hey btw

Dont dox yourself

In your screenshot

[karmic ravine]

oh mb

its not my account tho so dw

[thorny patio]

I mean i dont think its against the rules xD

[karmic ravine]

haha

the set of values that can be such that y is greater than 0 and less than 6

i gave it a go but im still not quite sure

im gonna put it here

[thorny patio]

So the first thing you want to do is describe the thing on the left

[karmic ravine]

yes

[thorny patio]

Well it's values or numbers called y

[karmic ravine]

right

[thorny patio]

So name those first

The set of all numbers y

The set of all numbers t

Etc etc

Then once you describe the thing on the left

You decribe the conditions or restrictions they abide by

Is it all possible objects y?

[karmic ravine]

OHHH

[thorny patio]

No its just those such that the following is true

[karmic ravine]

The set of all numbers y such that o is greater than y and less than 6

[thorny patio]

Yea!

[karmic ravine]

thank goood

thank you so much

[thorny patio]

Well you should say

...and y is less than 6 for that one

But that is nitpicking

You got the basics i think

[karmic ravine]

are you able to change my answer around so that its perfect just so i can see where i went wrong

[thorny patio]

The set of all numbers y such that 0 is greater than y, and y is less than 6

[karmic ravine]

ohh right

[thorny patio]

Compound inequalities are like Compound sentences

But the set notation part you did perfect

[karmic ravine]

is d the same kind of thing but a bit longer?

thanks alot

[thorny patio]

Yea it's exactly the same

Just two conditions

So you have to name them both

[karmic ravine]

the set of all numbers x such that x is greater or equal to 2, and x is lesser or equal to 4, and s.t x is an element of the set intergers.

im not sure about this one lol

[thorny patio]

Yea thats it!

[karmic ravine]

YESS

[thorny patio]

Just say set of all.integers

Cause there are tons of integer sets

Theres set of all.positive integers, set of negative integers, set of even integers, set of odd integers, set of integers that are divisible by x number lol

[karmic ravine]

ohh right

[karmic ravine]

are u able to implement that into my answer again please

[thorny patio]

the set of all numbers x such that x is greater or equal to 2, and x is lesser or equal to 4, and s.t x is an element of the set of all intergers.

[karmic ravine]

i appreciate your patience lol

[thorny patio]

No worries

[karmic ravine]

alr thanks alot

[thorny patio]

I wish someone explained this stuff to me i had to figure it out on the job

[karmic ravine]

yeah wow mustve been ridiculously hard by yourself i was stuck on this topic for ages and watched so many yt videos

alr e

[thorny patio]

I had a really good professor for my first analysis course

So that helped a lot

Very gentle approach to teaching the theory stuff

[karmic ravine]

the set of all numbers t such that t is greater than 1 and t is lesser than 5

well thats good

[thorny patio]

Yea! You got it

[karmic ravine]

alr awesome

last one finally

this one has an "or" in it so is it the same kinda thing but extended?

[thorny patio]

The difference is that

[karmic ravine]

oh alr

[thorny patio]

We say or instead of and

In this case either condition is acceptable

The listing with commas in earlier problems suggest theyre all simulataneous conditions

The use of the word or is just like in english

Either n is less than 2 or it is greater than or equal to 6

[karmic ravine]

the set of all numbers n such that n is greater than 2 and the difference is that n is lesser or equal to 6

i think i mightve butchered that one

[thorny patio]

Yea it's like this:
The set of all numbers n such that n is less than 2 OR n is greater than or equal to 6

You dont need to change much about the structure of the answer just use the word "or" to link the conditions

[karmic ravine]

ohhh alr

how do i tell if its lesser or greater again'

[thorny patio]

the alligator eats the bigger number lmao

1 < 2

1 is less than 2

5 > 4

5 is greater than 4

[karmic ravine]

haha thats works well

[thorny patio]

Imagine the symbols are the jaws of an alligator thsts how i remembered it as a kid

[karmic ravine]

yeah lol

ima use that one

[thorny patio]

And probably still think of it subconsciously

[karmic ravine]

if i close this will i be able to scroll up and my wokring out still be there?

[charred jewel]

or just imagine the number line, smaller number is always at the left of the bigger number, hence the arrow points left

[thorny patio]

Yea it will be there idk if it stays forever but it's not erased right away

[karmic ravine]

oh yeah true

[charred jewel]

a ruler is a good way to show a number line

[karmic ravine]

alr i think thats all i will prob be back but if i dont see you thanks you so much youve helped be ALOT and i really appreciate it.

[thorny patio]

Yw!

Gl with your class

[karmic ravine]

ty

.close

[lone heart]

Channel closed

Closed by @karmic ravine

Use .reopen if this was a mistake.

Available help channel!

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Remember:
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[stiff badger]

What is the equation of the line through (5, 2) with x-intercept thrice the y-intercept?

[stiff badger]

im losing my mind again help

[north hemlock]

what have u tried

also is that kazuha in ur pfp?

he looks different for some reason

anyways, send ur work u got so far

if any at all

[stiff badger]

i actually dont hav

also yee its kazuhaa

idunno what equation form i should use

[north hemlock]

which ones do u know

notice you're given a slope and a point

[stiff badger]

the point slope form and the slope intercept form

[north hemlock]

well sorry, ur not directly given a slope

so let's do that real quick

pick an x-int thta's 3x the y-int

[stiff badger]

theres this intercept form that idk how to use as well

[stiff badger]

what

[north hemlock]

"x-intercept thrice the y-intercept"

this part

[stiff badger]

okay okay

whats the value of the x intercept tho

[north hemlock]

well imagine a line going thru that point and u have a few it could reasonably be

[stiff badger]

okay then

[stiff badger]

?

[north hemlock]

okay so say u have a y-intercept called a

then you want an x-intercept, 3x "more"

right?

so our equation so far is

y=mx+a

the x-intercept is of course -a/m

do you follow so far?

all i did was set y to 0

[stiff badger]

how'd it become -a/m

[ocean seal]

Juke | ping me if no response

[north hemlock]

does that make it clear?

x-intercepts are just when y is 0

so I set it to 0 and solve for x

[stiff badger]

okay

[north hemlock]

ye

and a is the y-int we have

so we want that x-int times 3 to be the y-int

[ocean seal]

Juke | ping me if no response

[north hemlock]

does this make sense?

[stiff badger]

thats the xinte thrice the yinte?

[north hemlock]

bc the x-int is thrice the y-int, ye

3 times the y int is the x int

thta's all i wrote here

then you solve this for m

thta's ur slope

then u use the point they give you (5, 2)

and then u can solve the missing parts

$\frac{-a}{m} = 3a \implies \frac{-a}{am} = 3 \implies \frac{-1}{m} = 3$

[ocean seal]

Juke | ping me if no response

[north hemlock]

do u see how I am solving for m here?

[stiff badger]

yea yea

[north hemlock]

yeah so what do u get for m

it's just the last step I didn't write

[stiff badger]

wait

[north hemlock]

waiting

[stiff badger]

is m -1/3??

[north hemlock]

yuh

so the info we have so far then is

y=(-1/3)x + a

now we need to find the y-int

we do this by plugging in our point that was given

(5, 2)
2= (-1/3)(5)+a

solve for a

[stiff badger]

so we used the y=mx+b to find the slope?

[north hemlock]

that plus this

[north hemlock]

i meant to reply to this one

the fact that -a/m = 3a

that was important

[stiff badger]

wait so is this like

the last thing to solve

[north hemlock]

ye

[stiff badger]

so I'll find a?

[north hemlock]

yea

then u have m and a

so u would write it y=mx+a

but this time u replace the m and a with what u get those to be

we did m already, which is (-1/3)

we're getting a rn

[stiff badger]

okay

[north hemlock]

lmk if u need help with that

[stiff badger]

a is 11/3?

[north hemlock]

yesh

so know u have ur final answer

y=(-1/3)x+11/3

[stiff badger]

okay wait

[north hemlock]

waiting

[stiff badger]

so its

3y+x-11=0?

[north hemlock]

one sec

yes

why would you write it like that though...? ( out of curiousity )

[stiff badger]

idunno how to write it

[north hemlock]

this is fine

[north hemlock]

this is too

but that's an interesting way to write it lol

to each their own

whichever u pref

or ig, whichever ur teacher prefers

[stiff badger]

hehe

is it done now

[north hemlock]

yuh

[stiff badger]

is it correct

[north hemlock]

yeah

[stiff badger]

so i just repeat the same equations?

[north hemlock]

11/3 times 3 is 11, so that's fine, and the line go thrus ur point, both conditions met, ggez

[stiff badger]

does this work on the other forms as well?

[north hemlock]

for similar problems? yea

[north hemlock]

it should.... but tbh it's easier to see it with y=mx+a since it's clear what your slope is (m) and what a is (y-int)

[stiff badger]

okeii

[north hemlock]

yuh

[stiff badger]

thank uuu

[north hemlock]

np

glhf