help-0

[slow grail]

I ended up with the row echelon form
12-32
01-70
0002

[alpine sable]

yh

.

[slow grail]

Not really

[alpine sable]

what is a pivot column of a matrix

[slow grail]

Okay

So it's where the leading 1's are

[slow grail]

The column where the leading 1 is

Sorry i have some connection issues

[alpine sable]

so which here

[alpine sable]

no worries

[slow grail]

So here its column 1 and 2

[alpine sable]

and the free would be?

[slow grail]

Column 3

[alpine sable]

so whats the answer

[slow grail]

So what im confused is

Does X_1 mean the column

[alpine sable]

pivots in alphabetical order
?

free variables in alphabetical order
?

[alpine sable]

yh

[slow grail]

So pivot are x_1 and x_2?

And x_3 is free variables

[alpine sable]

good

[slow grail]

A question

[alpine sable]

go on

[slow grail]

When we say free variable

Does it mean the whole column?

[slow grail]

Like here -3 and -7 and 0 are free variables?

[alpine sable]

normally i would say z is a free variable in this case

but since they went with x_3 that would be fine too since usually you'd have equations like
x_1 + 2x_y - 3x_3 = 2 and so one

[alpine sable]

uh those are constants

[slow grail]

Ooh okay

So why call it a free variable?

[alpine sable]

the best way to show you this is make you solve a parametric matrix equation

[slow grail]

Go on

[alpine sable]

would you like to try one

[slow grail]

Sure

[alpine sable]

okay

[alpine sable]

give me the solution space for this

[slow grail]

So augmented matrix first yeah

[alpine sable]

how would you tackle this

[alpine sable]

good

1 & 2 & -3 & 2\\
0 & 1 & -7 & 0\\
0 & 0 & 0 & 2
\end{pmatrix}$$```

[slow grail]

Okay so should i put the w first?

[alpine sable]

copy paste code and give the augmented matrix

[alpine sable]

alphabetical but any works ig

[slow grail]

2 & 1 & -3 & 0\\
1 & 3 & 2 & 0\\
4 & 2 & 0 & 12
\end{pmatrix}$$```

Tbh i have no idea how these code thing's work

[alpine sable]

$$\begin{pmatrix}
2 & 1 & -3 & 0\
1 & 3 & 2 & 0\
4 & 2 & 0 & 12
\end{pmatrix}$$

[ocean seal]

$$\begin{pmatrix}
2 & 1 & -3 & 0\\
1 & 3 & 2 & 0\\
4 & 2 & 0 & 12
\end{pmatrix}$$

[alpine sable]

just replace the numbers, if latex is a prob you can write on paper

[slow grail]

This is what i got

[alpine sable]

okay very good

what next

[slow grail]

Ok now i get it to echelon form

[alpine sable]

this might take a while

[slow grail]

Its not a must for reduced row right?

Just get it to echelon

[alpine sable]

yh i gave thumbs up

[alpine sable]

anyhow

as long as you're able to identify the free variables

and the pivots

also your definition of a pivot is a little bit limited, 'leading 1s' all you need are leading entries

[slow grail]

Ooh yeah

Mayn

This will take some time

[alpine sable]

i'll just reduce for you since you how to, to save up on time

[slow grail]

Please do

[alpine sable]

you do know how to right?

and comfortably so

[slow grail]

Yes i do

Main aim is to do row operations

[alpine sable]

okay

$$\begin{pmatrix}
2 & 0 & 1 & -1 & 0 \
0 & 3 & 3/2 & 1/2 & 0 \
0 & 0 & -3 & 5/3 & 12 \
0 & 0 & 0 & 0 &0
\end{pmatrix}$$

[slow grail]

Pivot points are 2,3 and -3

Free variables are y and z

Correct?

[alpine sable]

and what are the variables corresponding to those

[alpine sable]

❌

give the pivot columns in terms of the variables w,x,y,z same for free variables

[slow grail]

Sorry

Pivot column's are w x and y

Z is the free variable

[alpine sable]

good

now lets get solving

[slow grail]

Okay

[alpine sable]

we're going to let the free variable be some letter say s, s \in R and find the other variables w,x,y,z in terms of s

$$\begin{bmatrix}
w \
x \
y \
z \
\end{bmatrix} =$$

[alpine sable]

but lets start small

[slow grail]

Okay

[alpine sable]

Let z = s, s \in R

w= ?
x =? and so on

[alpine sable]

can you find w for me from this

[slow grail]

w= z-y

[alpine sable]

hmm?

[slow grail]

Sorry 1/2(z-y)

[alpine sable]

and what is z again

[slow grail]

S

[alpine sable]

so w=?

[slow grail]

w=1/2(s-y)

[alpine sable]

okay now find the rest

[slow grail]

I am in class right now 😬

Can i do it in like 5

[alpine sable]

the channel will close, do .close ping m later

[slow grail]

.close

[lone heart]

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[alpine sable]

i started by finding gradient of the perpendicular line with is 1/3
then found the point of intersection which is just 0=-3x+12
x=4

therefor intersection point is 4,0

then find equation of perpendicular line whihc is y=1/3x-4/3

A is just y intercept so set x to 0 and i get -4/3 therefor anser is [0,-4/3]

not sure where ive gone wrong

Ping plz

[fickle sandal]

but you put +4/3 in your answer?

[alpine sable]

Oh yea 💀💀

Ty

.close

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[raven girder]

How can I get something like this in to a form like this?

Can I use the binomial theorem for it?

[alpine sable]

you can use rational root theorem

find a root

divide so you get a quadratic

and then the rest should be easy I'd say

[raven girder]

Okay ty

.close

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[glad anchor]

Can anyone explain a method for me to solve 6x^2 - 11x - 10 ??? Thank you 🙏

[alpine sable]

you mean factor I presume?

[glad anchor]

Uhh yeah

[alpine sable]

okay so like

when you solve for quadratics

[glad anchor]

U know like make it into 2 brackets

( )( )

Like this?

[alpine sable]

For a quadratic $\m fx = ax^2 +bx +c$, can you find me two numbers, $u,v$ that satisfy the two equations below [
u\cdot v = a \cdot c \
u+v = b
]

[alpine sable]

yes

okay so in other words

[alpine sable]

what two numbers multiply to -60, and add up to -11

[glad anchor]

That's the problem here

I'm so confused

[alpine sable]

so?

like

start by the "multiply to 60"

just tell me two numbers that multiply to 60

let's trial and error this

[glad anchor]

1, 60
2, 30
3, 20
5, 12
6, 10
10, 6

[alpine sable]

okay now there is also a negative

so like

[glad anchor]

Ohh

Yeah right

[alpine sable]

it's okay

let's see what adds up to -11

like

for example for 1,60

either numbers can be negative

[glad anchor]

-1, 60
-2, 30
-3, 20
-5, 12
-6, 10
-10, 6

[alpine sable]

but 60-1 = 59 and 1-60 = -59

yeah

it could be the opposite too

so we have to consider everything here

but let's start in the middle

what's -6 +10

[alpine sable]

also you're missing one

[glad anchor]

4 and 12?

-4 12?

[alpine sable]

,calc 4*12

[ocean seal]

Result:

48

[alpine sable]

not really 60

[glad anchor]

Ohh

Sorry

I'm crazy

[alpine sable]

but you are close

[glad anchor]

4 and 15

[alpine sable]

yes

[glad anchor]

😂

[alpine sable]

okay so

-4+15 = ?

[glad anchor]

Wait

Yeah

11

[alpine sable]

yeah but we want -11

so

[glad anchor]

4 + -15

[alpine sable]

yes

okay great

[glad anchor]

Ohhhhhh

[alpine sable]

now I want you to pay close attention

[glad anchor]

Ok

[alpine sable]

of what we are going to do here

[
u\cdot v = a \cdot c \
u+v = b
]

[alpine sable]

those are the two equations we used right

u and v are the numbers we just found

[glad anchor]

Ok

[alpine sable]

so

for the second equation

[glad anchor]

U = 4
V = -15

[alpine sable]

can you multiply both sides by x

u+v = b

what happens if you multiply both sides by x

[glad anchor]

-11x = bx?

[alpine sable]

let's keep everything in like letters

I'm just trying to give you the intuition to it

[glad anchor]

Ok

[alpine sable]

so you can understand later on

so, in letters, what would it be?

[glad anchor]

Oh ok

Ux+ vx = bx

?

[alpine sable]

yes!

great

now

[
\m fx= ax^2+ \c b{bx} + c
]
and we have [
ux + vx = \c b{bx}
]

[alpine sable]

what do you think is something we can change in the above?

[glad anchor]

I don't know

[alpine sable]

basically

[glad anchor]

What is f(X)

?

[alpine sable]

what if we put ux+vx in the first equation

[glad anchor]

Is that a function?

[alpine sable]

yes

we can have ax^2 + ux + vx + c

[glad anchor]

Ohh yes

[alpine sable]

does that make sense ?

[glad anchor]

So we've broken down the bx

[alpine sable]

yess

exactly

so like

with your quadratic

[glad anchor]

Into UX and vx

[alpine sable]

6x^2-11x-10

we can do the same!

[glad anchor]

ux*

[alpine sable]

so bx is -11x

we found u = 4 and v = -15

so what do you think we have

[glad anchor]

6x^2 +4x -15x -10

So

[alpine sable]

exactly!

[glad anchor]

(6x+4)(6x-15)

And do we find their highest common factor?

So 6 and 4's highest common factor?

[alpine sable]

I feel like there is a better way for u to do this

do you wanna see?

[glad anchor]

Ok

[alpine sable]

how do you factor 6x^2 +4x

factor fully

[glad anchor]

2x(3x+2)

?

[alpine sable]

yes!

okay great

how do you factor -15x-10

factor fully

[glad anchor]

5(-3x-2)

?

[alpine sable]

yeah but

can you pull out the - too

[glad anchor]

Ok

-5(3x+2)

Is it like this?

[alpine sable]

yes

[glad anchor]

Ok

[alpine sable]

now

2x(3x+2) -5(3x+2) is what you have

[glad anchor]

Yep

[alpine sable]

do you see anything you can factor

[glad anchor]

No

[alpine sable]

okay

so let's say

we make z= 3x+2

we are just hiding the 3x+2 behind z

so we have 2xz -5z

do you still not see anything you can factor

[glad anchor]

10xz^2 ?

[alpine sable]

noo

okay even simpler

how do you factor z-5z?

[glad anchor]

Wait

Wdym by factor

Do I just do the same thing I did before?

Like

Z(-5z)?

Or

[alpine sable]

like greatest common factor

[alpine sable]

I'm asking u for the greatest common factor in this

[glad anchor]

Isn't 2 and 5's greatest Common factor 1?

[glad anchor]

Wait is the answer (3x+2)(2x-5)

Cz I found the common factor of (6x+4)(6x-15)

HCF of 6 and 4 is 2
HCF of 6 and 15 is 3

Sorry, I think I'm rlly bad at this 💀

🥲

😞

😢

@alpine sable

It's fine nvm

[alpine sable]

sorry something happened irl

did you figure out something

[glad anchor]

Hi

Oh, sorry for pinging 💀

[alpine sable]

no worries lmao

[glad anchor]

So this

[alpine sable]

yes but what about the variables

[alpine sable]

yes it is!

[glad anchor]

Like this?

[alpine sable]

yeah that works

also

the thing I was going for is

[glad anchor]

Ohh ok

[alpine sable]

here

you can factor out the 3x+2

to get

(3x+2)(2x-5)

make z be 3x+2

2xz-5z

you can factor z out as it's the gcf

so

z(2x-5)

[glad anchor]

Ohh so the common factor is z

[alpine sable]

yes

[glad anchor]

Ohhhhhh

I thought numbers

😂

[alpine sable]

lmfao yeah nw

[glad anchor]

Oh ok, I get it now

Ig? 😂

Thank you for helping me 😊🙏

[alpine sable]

hope the process like makes sense to u now tho

aye aye

[glad anchor]

👍 yep

[alpine sable]

have a nice day

[glad anchor]

You too!!!! 😊

Thank you!

❤️

[alpine sable]

no problemoo

[glad anchor]

😁

Do I just put .close? 😂

.close

[lone heart]

Channel closed

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[alpine sable]

yes

[glad anchor]

Ok

[lone heart]

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[raven girder]

When I write down the matrix to find the Eigenvector for an Eigenvalue, but all the rows are linearly dependent, what does that mean?

[ruby current]

well they should be linearly dependent

otherwise it's not an eigenvalue

[alpine sable]

yeah if your A-Iλ matrix does not have dependent rows you did something wrong

[raven girder]

uhhhhh

okay maybe I said something wrong

I mean no matter what operation I do, I will get only all zero rows

[alpine sable]

can we see the problem

[raven girder]

you know what I mean?

ya wait

Sorry

if I add 2* first row to second row then I get all zeroes

[alpine sable]

yeah its fine now find the eigenspace

[ruby current]

,w eigenvalues {{-2, -2, 1}, {2, 3, -2}, {0, 0, -1}}

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by tusharmurali

[raven girder]

so the two rows express the same thing?

which means I just take the first row

and then try to figure it out or something

like x_1 = 2 * x_2 - x_3

[ruby current]

you solve the homogeneous system (A - lambda*I) x = 0

[raven girder]

but how

[ruby current]

row reduce and solve

[alpine sable]

you did that already to get the eigenvalues,

now you're going to sub those into the A-Iλ
then reduce each of those matrices as Tushar said

[raven girder]

but guys

how do I reduce the matrix

when the first row is a multiple of the second row

do you know what I mean

[ruby current]

then you get two rows of zeros

and two free variables

[raven girder]

oh hmm

so does that mean we have three free variables?

because all rows are 0

[ruby current]

how are you getting all rows zero

what are your operations

[raven girder]

oh just two rows right

wait

[ruby current]

row reduction preserves solutions

so you should intuitively see that it's impossible to get all zeros (that would mean everything is a solution)

because you have at least one nontrivial equation

[raven girder]

I'll say I just read about this free variable thing because I totally forgot about it. Am I doing it correctly?

now I just wonder what the Eigenvector would look like

(-2a+b, 1, 1) with a,b in R?

or (-2a+b, a, b)?

🤔

[ruby current]

can you find a basis for all such vectors?

this is correct btw

the eigenspace is all vectors of the form (-2a+b, a, b), where a,b in R

but you can express this as the span of two vectors

[ocean seal]

Tushar

[ruby current]

I'm simply separating out the terms containing a and b

now try factoring out the constants a and b

then you should be able to express it as a span of two vectors

[raven girder]

a(-2,1,0) + b(1,01)

[ruby current]

yep

so the eigenspace is $$\left{a\begin{bmatrix}-2 \ 1 \ 0\end{bmatrix} + b\begin{bmatrix}1 \ 0 \ 1 \end{bmatrix} : a,b \in \bR\right}$$

[ocean seal]

Tushar

[ruby current]

which you should recognize as a span

[raven girder]

Can I say the Eigenvector is v_1 = (-2a+b, a, b)?

Or is that incorrect

[ruby current]

well that's the form of every eigenvector

if you're only asked for one eigenvector, you can choose anything of that form

but above you have found a basis for the eigenspace

[ocean seal]

Tushar

[ruby current]

remember that the eigenspace is the space of all eigenvectors corresponding to an eigenvalue

so you have two eigenvectors right there

what exactly does your problem ask for?

[raven girder]

it asks for all eigenvalues and eigenvectors

[ruby current]

do you have a picture

I suspect they probably want these

[raven girder]

it's german

[ruby current]

ok sure

probably just the ones that span the eigenspace

wolfram gives these too

there is no such thing as "all eigenvectors" because any scalar multiple of an eigenvector is still an eigenvector

,w eigenvectors {{-2, -2, 1}, {2, 3, -2}, {0, 0, -1}}

[ocean seal]

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Requested by tusharmurali

[ruby current]

wolfram gives the same ones you found

see v2 and v3

those correspond to lambda = -1

you repeat this process for lambda = 2

[raven girder]

🤔

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@raven girder Has your question been resolved?

[raven girder]

.close

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[modern spoke]

how to solve this question?

[alpine sable]

what have you tried

[silk dome]

#help-33

[modern spoke]

.close

.close

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[lime bronze]

Is the directrix of the hyperbola really "a^2/c" units from the center, where "a" is the distance from the center to one of its vertices and "c" is the distance from the center to one of the foci? or is it the midpoint from the center to a vertex?

[lime bronze]

if so can someone link me a credible source i can use that states it?

[lone heart]

@lime bronze Has your question been resolved?

[lime bronze]

<@&286206848099549185> anyone free?

[lone heart]

@lime bronze Has your question been resolved?

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@lime bronze Has your question been resolved?

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[somber mica]

Need help proving this with delta epsilon proofs

[open marten]

Please mention me later when you find solution id cv you don't mind

[somber mica]

ok

<@&286206848099549185>

[limpid turret]

Start by setting up the epsilon delta requirements

[somber mica]

Yea I got the formal defn down and everything

like absolute value of 18-x^2 - 2 < epsilon

[limpid turret]

18-x^2?

[somber mica]

18/x^2 - 2 mb

[limpid turret]

$\abs{\frac{18}{x^2}-2}<\varepsilon$

[ocean seal]

SWR

[somber mica]

anyways I factored it out and got 2 times (x+3) times -1 (x-3)/x^2

I figured out what x + 3 and x^2 are bounded by

then I got it less than -14delta/16 and set that equal to epsilon

[limpid turret]

$2\abs{\frac{-(x+3)(x-3)}{x^2}}<\varepsilon$

[ocean seal]

SWR

[somber mica]

oh

i Just left the 2 inside

[limpid turret]

doesn't matter

[limpid turret]

What are thet bounded by?

[somber mica]

anyaways I figured x +3 is bounded by 7

and x^2 is bounded by

4

I used 1

then I got less than -14delta/16?

[limpid turret]

How do you figure this?

[somber mica]

Well you set absolute value of x-3 less than 1 than use inequalities to show x + 3

example from my prof

I was wondering if it was ok to have delta as negative in this case

[limpid turret]

Oh I see

You're assuming $\delta\le1$, thus $\abs{x-3}\le1$, or $-1\le x-3 \le1$, leading to $2\le x\le4$

[ocean seal]

SWR

[limpid turret]

no

[somber mica]

yea then I messed up

[limpid turret]

How did you get x² bounded by 4? That doesn't seem right.

[somber mica]

absolute value of x-3 is less than 1 right

so you want to show x^2 by itself in the middle

wait hang on

[limpid turret]

Oh did you mean x² is lower bounded by 4?

[somber mica]

yea than -1 < x - 3 < 1 add 3 to everything 2 < x < 4

then square both sides and take the max

so 16

[limpid turret]

$4\le x^2\le16$

[ocean seal]

SWR

[somber mica]

yea exactly

just less than not less than or equal to

[limpid turret]

right

[somber mica]

so we can substitue the values bounded by here?

that would give less than -14/delta/16 which we know is wrong

[limpid turret]

Yes, but be sure to use the correct bounds

[limpid turret]

!show

[lone heart]

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

[somber mica]

alright ill show you one sec

sorry forgot to rotate

[limpid turret]

,rccw

[somber mica]

idk where I went wrong

[limpid turret]

Why did your absolute values disappear?

[somber mica]

I fully simplified so I dropped the bars?

or should I still have it?

ok

[limpid turret]

x=3.1 will show you why

[somber mica]

is there anywhere I went wrong in the proof?

[limpid turret]

Everything is correct up to here

But that bit there looks incomplete.

[somber mica]

yea cause that is where I stopped

[limpid turret]

So far, you have it right then.

Your delta selection, I do not agree with though

[somber mica]

I never had a case where delta was negative so that was why

[limpid turret]

And you shouldn't, no.

[somber mica]

This is my friends method what do you think?

[somber mica]

if it was a quadratic I could have used the triangle inequality

but I can't

oh wait I think I can

apple absolute value to x then bound x, will it work?

[limpid turret]

Close. Final line should be $\delta=\min(\frac{\varepsilon}{2}, 1)$

[somber mica]

wait can u explain?

[ocean seal]

SWR

[limpid turret]

Also this line should be $\abs{x-3}<\frac{x^2}{2\abs{x+3}}\varepsilon$

[ocean seal]

SWR

[somber mica]

so I guess this method works better

where did the 3 come from?

[limpid turret]

Your friends proof was assuming $\delta\le 1$, so that must be part of their final result.

[ocean seal]

SWR

[limpid turret]

The limt: $x\rightarrow 3$

[ocean seal]

SWR

[somber mica]

so this would be the correct method?

[limpid turret]

Eseentially, yeah

[somber mica]

by the triangle inequality, can u apply absolute value to x^2 and x on the bottom?

that's how he used 3

[limpid turret]

wdym?

[somber mica]

idk if it would work but apply absolute values to those 2 so you can sub in a number that absolute value of x is bounded by

his method works by isolating x-3 I can figure what x^2 and x+3 are bounded by? Delta would not be negative

[limpid turret]

Actually, I think there's an error here

[somber mica]

exactly

[limpid turret]

Choose $\varepsilon=2$

[somber mica]

They are very strict with format of proofs so I cannot just do that

[ocean seal]

SWR

[limpid turret]

Then by your friend's logic, $\delta=1$

[ocean seal]

SWR

[limpid turret]

Which means $\abs{x-3}<1$. So we can choose, say, $x=2.01$

[ocean seal]

SWR

[limpid turret]

,w abs(18/(2.01^2)-2)<2

[ocean seal]

Results provided by WolframAlpha

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Requested by spelwerdzrite

[somber mica]

yea true

[limpid turret]

Limit is not satisfied.

[somber mica]

the general rule is to get e in terms of delta

can't seem to do it here though

delta in terms of e I meant

[limpid turret]

Everything up to here is fine but change the last (x+3) to |x+3|

After this, you need to use the bounds you found for x² and |x+3|

[somber mica]

yea I was thinking

just to go over it

16epsilon/14

are u allowed to reduce?

[limpid turret]

This is not correct

[somber mica]

really?

[limpid turret]

yes

[somber mica]

I thought x^2 is bounded by 16

[limpid turret]

yes

bounded above, to be specific

[somber mica]

x + 3 is bounded by 7

yea

[limpid turret]

bounded above, yes

[somber mica]

then 16 / 2 * 7 epislon right?

[limpid turret]

Consider this: If $a<b$, then is $\frac{1}{a}<\frac{1}{b}$?

[ocean seal]

SWR

[somber mica]

swap the direction of the inequality but yea

[limpid turret]

That's the key error you are making in your delta calculation then

[somber mica]

but I am not taking the reciprocal of anything?

[limpid turret]

$\abs{x+3}$ is in the denominator.

[ocean seal]

SWR

[limpid turret]

You are saying "$|x+3|<7$, therefore $\frac{x^2}{2\abs{x+3}}<\frac{x^2}{2(7)}$", but this statement is actually false.

[somber mica]

hmm

[ocean seal]

SWR

[somber mica]

but its less than 7, cant we use that then?

[limpid turret]

refer here

[somber mica]

if we swap the direction the proof would not work

[limpid turret]

No it would not

You need a lower bound for |x+3|

[somber mica]

so 5 then?

how about x^2 then same thing?

[limpid turret]

x² is on the numerator, so it's fine as is

[somber mica]

so be 16/10? epsilon

[limpid turret]

works for me

[somber mica]

it is possible to have 3 values for delta?

delta = min (5, 16, 10/16 epsilon)?

[limpid turret]

why the 5 and 16? Where are they coming from?

[somber mica]

oops just 1 mb

can u explain in this example, why they are not taking the lower bound?

Problem solved here

How did they get 2e/7 though

I figured it out bound x^2 to 4

why do we take the lower bound

[lone heart]

@somber mica Has your question been resolved?

[limpid turret]

I made a few errors here and there, so I did a full rewrite

[lone heart]

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[eager iron]

derivative of this..

[eager iron]

if i'm supposed to take the derivative of this

well i think the issue is that i viewed 1/5 as a constant

but in the context of derivatives - a constant is a number with a + or - to it, right?

and then my next question is what would the "correct" way to do it be?

because what i did was

and then at the last step just add the 1*5 again

and even though the answer is correct i don't feel like that's how it's supposed to be done..

because if we only have the x^2/x^2-4x+5

[quasi vector]

why not?

looks good to me

[eager iron]

then its f'*g-g'*f / g^2

but what happens to the 5 in that case? is it not accounted for at all!?

[quasi vector]

you just divide it by 5

the final answer

i didn't check your algebra btw, but the first step is correc

t

you just didnt write the 1/5

[eager iron]

so i can just put the 1/5 to the side?

[quasi vector]

$\frac{d}{dx}[c\times f(x)]=c\frac{d}{dx}(f(x))$

[eager iron]

and then in the end do the end result multiplied by 1/5?

[ocean seal]

kheerii

[eager iron]

but normally constants get cancelled when doing derivative, right?

but constants mean + and -

[quasi vector]

not when they are being multiplied to some other function

[eager iron]

ok but if we had

then it would be 1/2 * d/dx (x^3/x^2-4x+5) and just remove the 5? or?

should that also go outside the derivative and wait until its done?

urfgh

[quasi vector]

yes

you're right

[quasi vector]

this

[eager iron]

ok so essentially if it's something multiplied to the function it stays until afterwards, but all + and - constants get removed?

because that would make sense then

[quasi vector]

yeah, any constant being added can be removed

[eager iron]

ok awesome, thank you a ton mate

have a nice day/evening <3

.close

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[untold wolf]

10. Suppose in a certain region, an earthquake occurs every 600 days (on the average). After an earthquake occurs, what is the probability that it will take more than 1000 days for the next earthquake to occur? (hint use exponential distribution).

My professor never went over exponential distributions in class, so my understanding is that 600 is lambda and 1000 is x, but everytime I do the formula or use my calcuator I get 0.

[lone heart]

@untold wolf Has your question been resolved?

[main saddle]

i tried 360-215=145 but nothing worked

oh

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[merry pelican]

I encountered something really weird doing precalc review... I'm in calc but I never took a precalc class, so maybe there's something I'm missing here, but even mathway gives the answer I gave. What am I missing? Second picture is what they consider correct.

[merry pelican]

full question

[main saddle]

.reopen

[vale wigeon]

wrong command usage. .reopen is for when you have a follow-up question in your channel.

this one is now occupied by somebody else, so if you have a question of your own, please open your own channel.

[vale wigeon]

this looks like a bug on their part

[main saddle]

it says to reopen if it was a mistake

[vale wigeon]

you were late to the punch sorry

it timed out

[main saddle]

since my question havent been answered yet

[merry pelican]

Awesome, thank you. I figured as much but thought maybe I missed something.

.close

[bleak geyser]

Make a new channel :)

[lone heart]

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[main saddle]

k

[lone heart]

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[alpine sable]

I need help with #4

I solved for E already (E=55)

Is C = S - E?

[charred summit]

I think yes

[alpine sable]

Here are the formulas

[charred summit]

So use the first one

Oh

[alpine sable]

Yeah

I’m trying to find C

Cost

[charred summit]

Oh why don't you use the second law

[alpine sable]

M$?

[charred summit]

Nevermind I meant the 5th one

[alpine sable]

Oh

<@&286206848099549185>

[charred summit]

?

Use the 5th fromula

To get M$

[alpine sable]

For C?

[charred summit]

You can't get it directly here

[alpine sable]

So I need to solve for M$ first?

[charred summit]

You need to work with the most right given value

[charred summit]

Yeah

[alpine sable]

Ok

[charred summit]

Then using M$
Use the second fromula
Which will give you the P

Then use the first fromula to get C

[alpine sable]

But how can I find M$?

?

<@&286206848099549185>

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@alpine sable Has your question been resolved?

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[kind patio]

im perfectly fine with derivatives but

for some reason I cant wrap my head about how they got this

I know theres some chain rule Im supposed to do but I do not get it

[rigid sail]

Exactly, you do a change of variables. For example, let u=sqrt(x-2)

[kind patio]

Im at the point of the u substitution part

and I already derived the u to get 1/2(sqrtx-2)

but Im not sure how I get rid of the fraction

[rigid sail]

Use that if u = sqrt(x-2), then x= u^2+2

=> dx = 2u du

[wheat isle]

or just use the fact that u^-1/2

[kind patio]

ohh i get ti

it thank you

.close

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[onyx plaza]

Can someone help me with this

[onyx plaza]

How did we get -23/60

[hot bluff]

adding the first 5 terms of the sequence

[ivory pivot]

are you evil?

[onyx plaza]

Yes

[ivory pivot]

nice

[onyx plaza]

Italian

[ivory pivot]

a little bit

[onyx plaza]

It doesn't make sense to me

@hot bluff i don't get it

[hot bluff]

show the beginning of the example

[onyx plaza]

That is the beginning

[hot bluff]

that is (c)

there's no (a) or original example problem?

[onyx plaza]

C

Yes

[tall crystal]

60 is the common denominator for 2,3,4,5 and 6

[hot bluff]

still not showing the entire problem.

[onyx plaza]

That is it

Let me show you

[hot bluff]

right, so there's a formula for the nth term. then they want the sum of the first 5 terms

[onyx plaza]

What about 23

[hot bluff]

find x for every n 1-5 and add?

[tall crystal]

Your goal is for the denominator of all the 5 fractions to be the same.

the common denominator for 2,3,4,5 and 6 is 60

you multiply the entire fractions by a certain number to achieve this

the first fraction,for an example, was multiplied by 30 in order to get the denominator to be 60

[onyx plaza]

I get that part

[hot bluff]

[onyx plaza]

What about the numberator

[tall crystal]

when you multiply a fraction you multiply both numbers

basically -1/2 x 30 is equal to -30/60

[onyx plaza]

Dude

Wtf

Lol

[tall crystal]

both numbers get multiplied not just the denominator

[onyx plaza]

Thank you so much for the help!!!!

Close

.close

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[flat halo]

if angle 2 and angle 4 are a linear pair

does that mean 3 and 4 is also a linear pair

like are consecutive interior angles linear pairs

[modern jolt]

are line a and line b parallel? and line c and line d parallel?

i take it linear pair angle is angles that add up to 180

so

since angle 2 = angle 1

and angle 1 = angle 3

we can get that angle 2 = angle 3

and because we know that angle 2 and angle 4 is linear pair

so angle 3 and angle 4 add up to 180 as well

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@flat halo Has your question been resolved?

[lone heart]

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[radiant bay]

Anyone know what this sign is or means?

[ocean whale]

Proportional

[fickle heath]

https://en.wiktionary.org/wiki/∝

[modern jolt]

isnt that alpha?

[gray isle]

not enough context to know the intended symbol

looks like prop to,
and it may also be how some people write alpha

[keen mason]

id say its alpha too

could also be propto

$\propto$

[ocean seal]

Mortta

[keen mason]

$\alpha$

[ocean seal]

Mortta

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[tropic spoke]

.open

[onyx finch]

hey

[onyx finch]

can anyone please help me with a linear algebra question?

i just need someone to check my work

<@&286206848099549185>

[placid zinc]

.close

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[gusty umbra]

how do i do this

[lone heart]

Please don't occupy multiple help channels.

[gusty umbra]

i do know the final answer has 2024! and (1+2+...+2024)

but i dont know how to continue from tht

[lone heart]

@gusty umbra Has your question been resolved?

[lofty gorge]

rewrite the operation in a nicer form of $a*b=(a+1)(b+1)-1$, and then see what happens when you repeat it

[ocean seal]

Edward II

[gusty umbra]

ooh wait

2023 * 2024 = 2024 x 2025 - 1
2022 * (2024 x 2025 - 1) = 2023 x 2024 x 2025 - 1
2021 * (2023 x 2024 x 2025 - 1) = 2022 x 2023 x 2024 x 2025 - 1
...
1 * n = 2 x 3 x ... x 2025 - 1

is it 2025! - 1?

thank you!!

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[cyan steeple]

Plug in the values for S and x and solve the equation for k

[alpine sable]

nvm im stupid i didnt read the part where it said money was in hundreds

kept getting a ridiculously small number

[cyan steeple]

oh ok

[alpine sable]

ty tho

[cyan steeple]

np

[alpine sable]

.close

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@remote depot Has your question been resolved?

[placid zinc]

Where's the bottom line coming from? It looks like you put a value in for d

But yeah everything above looks good

This didn't parse for me. Are you saying your solution set is incorrect?

[ocean whale]

Ignore me, I typed in something wrong

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[bitter meadow]

In a factory that produces light bulbs, quality control has identified a defect rate of 5%. That is, 5% of the light bulbs produced are defective, while the remaining 95% are of acceptable quality. You receive a shipment of 30 light bulbs. The probability that exactly 2 light bulbs in the shipment are defective is equal to ....

[placid zinc]

Know the binomial distribution?

[bitter meadow]

Yes

[bitter meadow]

=BINOM.DIST(2,30,0.05, FALSE)

so that gives me

[placid zinc]

What's with the FALSE?

[bitter meadow]

False, because its equal to

I am doing it on excel

[placid zinc]

,w C(30,2)(0.05)^2(0.95)^28

[ocean seal]

Results provided by WolframAlpha

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Requested by kaynex

[placid zinc]

Yeah we seem to agree

[bitter meadow]

Thank you so much, what I am stuck at after undrstanding this

is what would the probability that between 5 and 15, inclusive, light bulbs in the shipment are defective is equal to ....

i tried finding for 4 and then 15 and substracting, but something seems wrong

[placid zinc]

Do you have a CDF function on Excel?

Is that what happens when you put "TRUE" lol

[bitter meadow]

sorry, whats CDF?

false would mean probabilty mass function

and then true, would mean cumulative dis function

[placid zinc]

The binomial PDF is the probability that exactly x successes happen.

The binomial CDF is the probability that x or less successes happen

Okay so yeah, using TRUE, and subtracting 15 - 4 should work

Let me look at the excel function a little closer maybe

[bitter meadow]

i am getting 0.0156

[placid zinc]

That seems quite small

Are you using TRUE?

[bitter meadow]

=BINOM.DIST(15,30,0.05,TRUE) - =BINOM.DIST(4,30,0.05,TRUE)

[placid zinc]

Huh. Well, maybe that's just the answer then. Do you have a reason to think it's wrong?

[bitter meadow]

Yes, because the previous questions asks. " The probability that at least 5 light bulbs in the shipment are defective is equal to"

And the answer also gives me 0.0156

=1 - BINOM.DIST(5 - 1, 30, 0.05, TRUE)

[placid zinc]

The probability that up to 15 bulbs are defective is likely near 100%

So these two matching isn't that weird

[bitter meadow]

so both answers are 0.0156?

[bitter meadow]

update, the other answer for at least 5 bulbs is 0.9844 i think

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@bitter meadow Has your question been resolved?

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@bitter meadow Has your question been resolved?

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[lone heart]

@next chasm Has your question been resolved?

[next chasm]

<@&286206848099549185>

[placid zinc]

Where ~ means "the same cardinality"?

[next chasm]

Yep

So I need to construct a bijection

But idk how

[sour dove]

I would say list out each element between the two sets

it's small enough you can prove by construction

[next chasm]

Yeah

But i need to find such a function

Thats the prob

[sour dove]

well you're trying to prove or disprove that statement.

I wouldn't worry about the function yet, just see if it's possible to have a bijection in the first place

[next chasm]

Its trivial that surjection exists

So we need to prove injection

[placid zinc]

Even before "proofing", let's think.

Is there a reasonable way to take a real number, and split it into two real numbers?

Is there a reasonable way to take two real numbers, and combine them into one?

[next chasm]

Um sure

ax + by

[placid zinc]

How can we use this to split a real number into two real numbers?

[next chasm]

Generate two coefficients such that they produce values from [0, 1]

So 0.5x + 0.5y = z

Can we say that it is a bijection

@placid zinc

@sour dove can you help?

[lone heart]

@next chasm Has your question been resolved?

[placid zinc]

So using that map,
(0.3, 0.7) goes to 0.5
(0.4, 0.6) goes to 0.5
The map isn't injective

@next chasm

But let's say we did this:
(0.3, 0.7) maps to 0.37
(0.4, 0.6) maps to 0.46

Can we make something like that work for more reals?

[next chasm]

Ahhhh ok

Wait

What abt when y = x = 1

Wait no

Sure

z = f(x, y) = x + 0.1y with restrictions

z = f(x, y) = x + 0.1y for x<=0.9 y<=1, or x <=0.99, y <= 0.1, or, x <= 0.999, y <= 0.01, …

Is this viable

@placid zinc

[placid zinc]

So you won't be able to write this as an equation

That's true for most bijection questions

[next chasm]

Yeah sure

But still

Im describing it as one

So at least there should be some limitstions placed

Or proved rigorously that it does not exist

[next chasm]

Yo?

@placid zinc

What do you propose

[lone heart]

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[lone heart]

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[austere compass]

hey i have this exercise:

Let (\mathcal{I} := [t_0,T[ \subseteq \mathbb{R}), and functions (\alpha, \beta \in \mathcal{C}(\mathcal{I},\mathbb{R})) be given.\
\
i) Suppose (\phi \in \mathcal{C}^1(\mathcal{I},\mathbb{R})) with
$$
\phi'(t) \leq \alpha(t) + \beta(t)\phi(t)
$$
for all (t \in \mathcal{I}). Show that then, for all (t \in \mathcal{I}), the following inequality holds:
$$
\phi(t) \leq \phi(t_0)e^{\int^t_{t_0} \beta(\tau)d\tau} + \int^t_{t_0} \alpha(s)e^{\int^t_s \beta(\tau)d\tau} ds
$$

so ive gotten to this point:

that we rewrite the inequation to $\varphi'(t) - \beta(t)\varphi(t) \leq \alpha(t)$, and then define a function $F$, so that $(F\cdot\phi)'(t) \leq F(t)\alpha(t)$. then i should integrate it between $t_0$ and $t$.

but my problem is: how can i prove that this inequality $(F\cdot\phi)'(t) \leq F(t)\alpha(t)$ holds? I've defined F as (F(t) := e^{-\int^t_{t_0} \beta(\tau) d\tau})

[ocean seal]

Levens

[lone heart]

@austere compass Has your question been resolved?

[vale crag]

hey

well you have that $\phi'(t) - \beta(t)\phi(t) \leq \alpha(t)$

[ocean seal]

aPlatypus

[vale crag]

now we do the integrating factor trick that I told you about the other time

$F(t)\phi'(t) - \beta(t)F(t)\phi(t) \leq F(t)\alpha(t)$

[ocean seal]

aPlatypus

[vale crag]

the sign of the inequality never changes at any point cause F is always positive

[austere compass]

oh hi! i was trying to find you 😄

[vale crag]

and now what makes the integrating factor method worthwhile is that $$F'(t) = -\beta(t)F(t)$$ you should check it if you're not aware of it already

[austere compass]

ohh yeah i just did that i thought it was wrong

[ocean seal]

aPlatypus

[vale crag]

(typo'd)

[austere compass]

and now we integrate

[vale crag]

(F ° phi)'(t) <= a(t)phi(t) ? yeah

[austere compass]

yup

Would u have an idea for the second part of the question?

ii) Now (\beta \geq 0) applies (pointwise). Conclude from part (i): (\varphi \in \mathcal{C}(\mathcal{I},\mathbb{R})) with
$$
\varphi(t) \leq \alpha(t) + \int^t_{t_0} \beta(s) \varphi(s) ds
$$
for all (t \in \mathcal {I}), the estimate applies to all (t \in \mathcal {I}).
$$
\varphi(t) \leq \alpha(t) + \int^t_{t_0} \alpha(s) \beta (s) e^{\int^t_s \beta(\tau)d\tau} ds.
$$

[ocean seal]

Levens

[lone heart]

@austere compass Has your question been resolved?

[vale crag]

yeah idk

[lone heart]

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[cosmic scarab]

Computer Programming Problem:-
You are given n magical fish with a lifespan of 0 to 6 days. When a fish dies, it reproduces two new fishes, one with a lifespan of 8 days and the other with a lifespan of 6 days. You are given a set of fishes with lifespans from 0 to 6 days for the first cycle. You need to tell how many fishes you will have by the end of the nth day.

Visual representation of a single fish with a lifespan of 6 days in the beginning:-

Days -> | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10| 11| 12| 13| 14| 15| 16| 17| 18|
--------+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
 L    8 |   |   |   |   |   |   |   | F |   |   |   |   |   |   | F |   |   | F |
 i    7 |   |   |   |   |   |   |   |   | F |   |   |   |   |   |   |   | F |   |
 f    6 | F |   |   |   |   |   |   | F |   | F |   |   |   |   | F |   |   |   |
 e    5 |   | F |   |   |   |   |   |   | F |   | F |   |   |   |   |   | F | FF|
 s    4 |   |   | F |   |   |   |   |   |   | F |   | F |   |   |   |   |   |   |
 p    3 |   |   |   | F |   |   |   |   |   |   | F |   | F |   |   |   |   | F |
 a    2 |   |   |   |   | F |   |   |   |   |   |   | F |   | F |   |   |   |   |
 n    1 |   |   |   |   |   | F |   |   |   |   |   |   | F |   | F |   |   |   |
 ↓    0 |   |   |   |   |   |   | F |   |   |   |   |   |   | F |   | F |   |   |
--------+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
          1                           1                                       1
                                      1                                       2
                                                                              1
Noticed Pascal's Triangle pattern which holds upto row 3,

Problem growth is exponential (n^d), and starting fishes adds alot of depth, so input case for starting fishes and no. of days is limited.
Problem Tags: Number Theory, Geometry (for visualization and pattern finding), etc ...

[cosmic scarab]

Need help in devising a mathematical formula to solve it in O(1) or O(n) if we can make a linear solution

[lone heart]

@cosmic scarab Has your question been resolved?

[cosmic scarab]

<@&286206848099549185>

[cosmic scarab]

channel occupied, delete this message

[lone heart]

@cosmic scarab Has your question been resolved?

[lone heart]

@cosmic scarab Has your question been resolved?

[cosmic scarab]

<@&286206848099549185>

[lone heart]

@cosmic scarab Has your question been resolved?

[fickle heath]

When a fish dies, it reproduces two new fishes
I interpret that as spawning the new ones on the same day the old one dies

Is the visual representation given or did you make it?

In any case, you're right to notice Pascal's triangle: it holds, it's just slanted in time

At the nodes are the number of fish that die (and so they each spawn one fish in each branch)

@cosmic scarab

[fickle heath]

Feel free to ping me but I probably won't respond immediately

[lone heart]

@cosmic scarab Has your question been resolved?

[cosmic scarab]

Yea its already past midnight here

[cosmic scarab]

I noticed the pattern, but it fails the Pascal’s triangle after row 0-3
Ik its reasoning as well, but didnt expected it to make a return in a fancy way

I made a linear solution which works but its slow in computing because as time passes the number of computations increases alot