help-13

[stark shell]

I don’t think this question will require physics knowledge anyways tbh

So the equation you need to use is p=I^2 x R

[dull oxide]

You can check #old-network for physics servers

[stark shell]

And for 2 a I get p = 4.0x10^7

Oh oka

Ty

[soft owl]

the power loss is RI²

[soft owl]

yeah

but you need to express it by percentage

[stark shell]

Right

But

The answer in the textbook is 20%

[soft owl]

power loss *100/ total power

[stark shell]

Oh

Sorry

Ty

[soft owl]

anytime

[stark shell]

What would the total power be ?

The 200 MW?

[soft owl]

200 MW

[stark shell]

Ok

[soft owl]

200 * 10 ^6 W

[stark shell]

Right ok

[soft owl]

power loss = 4 * 10 ^7 W

$\frac{4 * 10^7 *100 }{2 * 10^8} = \frac{ 4 * 10^9 }{2 * 10^8} = 20$ %

[stark shell]

Thanks

I got it

[wraith dagger]

Mehdi_Moulati

[soft owl]

np

[stark shell]

Sheesh u from morroco

Lucky

[soft owl]

yep

[stark shell]

W

Another question

Do u understand the right hand rules

For electromagnetism

[soft owl]

yeah , i was

[stark shell]

Wdym

Oh like u don’t know it anymore ?

[soft owl]

It's been a long time since I used it

just ask the question

[soft owl]

I goes with the z-axis
and B rotate around z-axis
@stark shell Right ?

[stark shell]

X axis?

Ion know about those terms tbh

But here

If I’m gonna be honest I don’t understand any of these

I’m like really confused

[soft owl]

depend

question 1 ?

[stark shell]

Yes please

[stark shell]

Thank u so much

But there’s like 4 right hand rules

How do u know what to use

[soft owl]

you know the the direction of the current flow

and you need to find the direction of the magnetic

[soft owl]

so you will use this

[stark shell]

Oh right that makes sense Cus it’s the only thing they give u and it asks for magnetic field

[soft owl]

did you understand how to use it ?

[stark shell]

I do

But

What I don’t understand

Is the circles

[soft owl]

where ?

[stark shell]

The last 2

In 1)

[soft owl]

the circle with a point inside means to your direction

[stark shell]

Like up wards?

Coming out of the page

[soft owl]

The current flow is coming in your direction

[stark shell]

Right

Ok

And the x?

[soft owl]

the opposite

[stark shell]

Going into the page

Makes sense

[soft owl]

yes

[stark shell]

Tysm

[soft owl]

anytime bro

[stark shell]

Sorry if I’m bothering u

[soft owl]

no need to , bro

with my pleasure

[stark shell]

Direction of current is from north to south right

[soft owl]

where ?

which figure @stark shell ?

[stark shell]

3

the first one

sorry

and you would use the right hand rule where the palm is the direction of the force

imma brb if i take long

[soft owl]

question 3 1 figure ?

[soft owl]

direction of the magnetic filed is from north pole to south pole

Direction of current wire is out the page

so the force goes up

[cedar kiln]

@stark shell Has your question been resolved?

[cedar kiln]

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[stark shell]

kjd

[stark shell]

it would always be out of the page right?

or no

because if its the front of your palm

and your palm is facing up

the force will be up

[soft owl]

just in this case

[stark shell]

when wouldnt it be

nvm

i got it

[cedar kiln]

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[stark shell]

.close

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[frail trout]

I'm in desperate need of help my linear algebra final is in 3 hours and I don't understand kernel properly it seems like. I'm trying to practice a homework problem and I don't understand how functions with a constant input can be the formula for a linear map. Don't they atleast have to reference the variable x? I know the canononical basis, {(1, 0, 0), (0, 1, 0), (0, 0, 1)} and {1, x, x^2, x^3} but I need help with the rest. I can't find anything online or in my text book on these functions

[frail trout]

Also not asked in this problem, but can't the basis of the kernel just be the canonical basis of the input?

<@&286206848099549185>

[vagrant pewter]

Think about it like this, if there was a reference to the indeterminate x in the output of l then you wouldn’t have a real number you’d have a polynomial again but this is a map from the space of polynomials to the space of 3 dimensional real vectors

So I’m some sense your “variables” in the input of the function l are different polynomials

Like x+x^2 or x^3+10x+1

Proving that it is linear amounts to taking two different polynomials p(x) and q(x) and then showing that l(ap+bq)=al(p)+bl(q)

[frail trout]

hmmm but I don't understand how this notation communicates it, I would understand like l(ax^2+bx+c) = (2a, b-c, c)

but I just don't understand how the actual map is communicated

[vagrant pewter]

The easiest way to do so might be to assume $p(x)=a_1x^3+a_2x^2+a_3x+a_4$

[wraith dagger]

llspacebarll

[vagrant pewter]

And the values for these coefficients doesn’t matter but you would agree that this is what any given polynomial of the third degree looks like right?

[frail trout]

yea

[vagrant pewter]

Do the same for q and then you can plug these thing into your function l

Finding the kernel of l amount to finding which polynomials map to the zero vector under l

So you effectively have to find all p(x) such that l(p(x))=(0,0,0)

Btw P^3 is third degree or second degree polynomials?

[frail trout]

third degree

[vagrant pewter]

Because if it’s third degree then the dimension of it is one higher than the dimension of R3 and so you will have that the kernel of l is one dimensional

Or I think it might be more but idk for sure unless I computed it

But you probably could use the Rank-Nullity theorem to figure out dimension

[frail trout]

But what is q

[vagrant pewter]

Use different dummy variables

b_1

B_2

The name doesn’t matter

[frail trout]

no but I still like

don't get

what I derive from the actual formula given

like say

l(p(x)) = p(-1)

how do i interpret that as

[vagrant pewter]

Are you redefining l?

Because l is a function into R3

[frail trout]

l(ax+b) = a

[frail trout]

no

i'm writing a small question

just

because

I don't know where the actual coefficients of the map are in the problem

[vagrant pewter]

l(p(x))=(p(-1),p(0),p(2))

It doesn’t ever equal p(-1)

[frail trout]

before I've only ever seen stuff like l(a,b,c) = b-c+(2a-c)x

or whatever map of variables from one place to another basically

[vagrant pewter]

Yeah okay you’ve only seen maps from R^3 into P^3? Not the other way around

[frail trout]

yeah

[vagrant pewter]

l is a map that takes a polynomial of degree three

And it sends it to a real vector

Defined by the polynomial evaluated at -1, 0 and 2

Those are the x y and z coordinates of the vector that l outputs

It’s not gonna return another polynomial because it isn’t mapping you to a polynomial

[frail trout]

ohhhhh

[vagrant pewter]

It maps to a real vector

[frail trout]

thank you that helps!

a lot

[vagrant pewter]

I’m glad I can help, sometimes this stuff takes a second

[frail trout]

but i'm still confused I just don't know where yet

[vagrant pewter]

It’s very abstract

The idea for solving the question is do it one at a time start by showing that for any two polynomials p and q and real numbers a and b, $l(ap+bq)=al(p)+bl(q)$

[wraith dagger]

llspacebarll

[vagrant pewter]

So that is where now to actually solve this you want to describe the polynomials p and q in the most general manner possible

So define them as $p(x)=a_1x^3+a_2x^2+a_3x+a_4$

[wraith dagger]

llspacebarll

[vagrant pewter]

And here the coefficients are unknown but it doesn’t matter what they are

You can do a similar thing with q(x)

And the idea is now you have something concrete that you can plug into the map l

Then it comes down to doing some manipulations with real numbers since it’s easy enough to work with those, does that help at all?

[frail trout]

yea it helps a lot

but I think I'm still confused on where the actual p(-1) and stuff come into play,

i understand manipulating p and q but I don't understand where you put the numbers given to evalutate

[frail trout]

but I do really appreciate your help!

[vagrant pewter]

Lemme write it out and get your started

I think it’ll be more clear

Gimme a minute

[frail trout]

Okay, thanks!

[vagrant pewter]

Sorry for the shadow

But the idea is this

You plug in the polynomials to the mapping then you rearrange them to make it clearly a new polynomial

Since the sun of two polynomials is a polynomial

Then you apply the function l

Once you’ve done that rearrange and use the fact that we know that vectors in R^3 add like such

And try to show that the mapping l is linear

I wrote out at the top WTS that means what to show

So you wanna prove that fact

I didn’t finish the proof but it pretty much there

You just gotta do a few more manipulations

And if you compute l(p(x)) and l(q(x)) on their own it should become very clear

[frail trout]

Ohhh, so, if a_3 happened to be p(3) would you instead use 27 as the coefficient for the last equation?

[vagrant pewter]

I just plugged the 2 into x

For the third entry in the vector

[frail trout]

ohhh

[vagrant pewter]

It’s just evaluating the polynomial at a real number

[frail trout]

oaky

I think I actually understand now

this really helps

[vagrant pewter]

Now before I go I’ll help you start on the kernel part too

This is essentially what it amounts to, you want to find all polynomials p(x) such that l(p(x))=(0,0,0)

So I did a slight calculation using again a general form for p(x)

I just used coefficients a b c and d to emphasize that it doesn’t matter what you call them

[frail trout]

and then

[vagrant pewter]

And you basically just need to solve that system of 3 equations in 4 variables

[frail trout]

i can just solve this with normal row operations

[vagrant pewter]

Use a matrix or whatever method you like

[frail trout]

like any matrix right?

[vagrant pewter]

And you should know that you’ll have one free variable

So it mean that the kernel of l is a one dimensional space

It’s spanned and that why it says to give a basis vector for the kernel

Because technically there are infinitely many polynomials that mapped to zero by the map l

[frail trout]

and so

the basis

is just the kernel with the variable factored out right?

essentially because then its scalar

[vagrant pewter]

Well once you solve that system you’ll have one of the variables a b c or d be a free variable

So then once you go pack to your definition for p(x) you’ll plug in whatever you found for a b c and d

And then p(x) with those coefficients you solved for is your basis vector

And in fact you can just say that the basis has the free variable set to one since it’s a basis so the scaling factor doesn’t matter

[frail trout]

thank you a lot for all your help!

I think

.close

[cedar kiln]

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[frail trout]

.reopen

[cedar kiln]

✅

[frail trout]

oh wait also this is a dumb question

but

if I find the matrix associated with respect to a R^2 -> R^2 map is that supposed to be R^2 or M^2x2?

[vagrant pewter]

Well it’s a 2x2 matrix

It’s not so sure what you mean though

R^2 and M^2x2 are vector spaces

So a 2x2 matrix is in M^2x2

[frail trout]

"write the matrix associated to l
with respect to the standard basis of IR^2,"

so I would normally write that as a 2x2 matrix

[vagrant pewter]

Well in this case it’s a 4x3 matrix I’m pretty sure

[frail trout]

but I was confused by the with respect to the standard basis wording

[vagrant pewter]

Or sorry 3x4

It outputs a 3 dimensional vector and takes as input a 4 dimensional vector

[frail trout]

this one yeah, but I'm looking at a different easier problem right now

[vagrant pewter]

It always will depend on the dimension of the spaces you are mapping between

If it’s between R2 and R2 then yes the matrix is 2x2

A matrix is just a fancy compact way of writing a linear map

[frail trout]

okay thank you, I just wanted to check!

[vagrant pewter]

At least that’s how I like to think about it

[frail trout]

yea for me matrixes are easy to think about

because I learnt to code a long time ago

but I get confused

the less explicit things are

[vagrant pewter]

Yeah fair enough

[frail trout]

thanks for your help!

.close

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[viral mural]

Incredibly easy question, I'm aware but I just want a second opinion so that I know definitely that the answer key is wrong

[viral mural]

could somebody do this and say what they get?

AB is 4/3 times AP, right?

I.e., AP = (4/3)AB?

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[long swan]

AP/AB = 3/4
AP = 3/4 AP

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[meager glade]

What does ; mean

(this is a question asking to find the tangent of a curve)

[south juniper]

for x=-2

tangent to the curve at x=-2

[meager glade]

Ohh okay

Thanks boss

.close

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[crimson sedge]

on the basis of which we concluded here that the rank of the matrix is 2

[vast beacon]

hi

what is question

[crimson sedge]

why rank of this matix is 2?

based on what

the professor told me that it is clearly visible here that the rank is 2

[vast beacon]

What ? what is the Lesson

[crimson sedge]

same here how is rank 2

lesson is rank matrix

and this is the easiest assignment

[vast beacon]

let me think

answer

https://lens.google.com/search?p=AXAp4wjMsxwoBRaYgXeVBpBigw6iPVOZmGz5uAE3jf-4IRsXuHJXRuXGxbAc1LAqmeVZqcrrkFY4BFBtNijEFW_R8n3uTt5MLD3hAqa9UnqEjcEDSub5CJmQ32UxrQWDQn2M-hTkfJ1liAl7N5EJ0yMEvdUolpChPmy9EjBfpfGjPFSKP_ljizKH7nqxA2uo6GxHzV5bkgGXwKw25UGPfbpENg_y-nn4XpNed5Jt1WVT6euUauo0ElQ3Cf9SqlTEI6KdGlhwGmfvGd8TEGdwUs1PF2x4TzXSKQ5QahKGmPJijSu4PYULS3KwQaDa1qKU8L0T1ceEwpdbspU0CP86sz4%3D&ep=gisbubb&hl=en-IR&re=df&st=1673519091138

[vast beacon]

https://lens.google.com/search?p=AXAp4wjMsxwoBRaYgXeVBpBigw6iPVOZmGz5uAE3jf-4IRsXuHJXRuXGxbAc1LAqmeVZqcrrkFY4BFBtNijEFW_R8n3uTt5MLD3hAqa9UnqEjcEDSub5CJmQ32UxrQWDQn2M-hTkfJ1liAl7N5EJ0yMEvdUolpChPmy9EjBfpfGjPFSKP_ljizKH7nqxA2uo6GxHzV5bkgGXwKw25UGPfbpENg_y-nn4XpNed5Jt1WVT6euUauo0ElQ3Cf9SqlTEI6KdGlhwGmfvGd8TEGdwUs1PF2x4TzXSKQ5QahKGmPJijSu4PYULS3KwQaDa1qKU8L0T1ceEwpdbspU0CP86sz4%3D&ep=gisbubb&hl=en-IR&re=df&st=1673519091138

[crimson sedge]

if it is not maybe because the first column is linearly independent of the other two, so the rank is 2 because of that?

[crimson sedge]

I do not have access to this service.

[vast beacon]

ahhhhh

search that in google image

https://lens.google.com/

[vast beacon]

what country you are>

[crimson sedge]

Serb

[vast beacon]

ok its not filter search that picture on it

write .close

[crimson sedge]

what i do here?

why i need to write close?

I did not get an answer to the questions

maybe someone knows

[vast beacon]

https://www.google.com/imghp?hl=en&tab=ri&ogbl

search in this

I do not understand the question

[cerulean sail]

Basically you have this after a row operation

[or row operations]

If you were to do all the work to get it into REF, you'd e.g. divide the second row by -20, and then after that do the first row - 6* the second row [after the step before which makes that -20 into a 1]

From that point, you'll see that the matrix would be in the form $\pmqty{1 & 0 & ? \ 0 & 1 & ?}$

[wraith dagger]

chartbit

[cerulean sail]

But basically all the work up there kinda shows that the rows we have are linearly independent anyway, so you wouldn't have needed to do all that work

If any of that helps?

[crimson sedge]

it looks like I'll stay on this problem for a while longer

.close

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[steel hearth]

Derivative changes sign also around point x = 0, but that’s an excluded value for the function.

I spotted it by accident and without that would’ve drawn the wrong function. What is the general check that no values like that were left out of the solution? How to check if derivative really changes sign around 0?

My idea was to take limit first approaching from the left then from the right and to see if they approach the same value + are of different signs. Is that correct way to do it?

[cedar kiln]

@steel hearth Has your question been resolved?

[steel hearth]

<@&286206848099549185> would appreciate help here

[cedar kiln]

@steel hearth Has your question been resolved?

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[dark gale]

Can someone help me, I got stuck

[dark gale]

I’m doing integration by parts

When it comes to the last part the integration of (1/3)x^3 times cos x dx

Do I have to use the substitution method?

To solve it?

[lethal schooner]

differentiate x^2 and integrate sinx

parts is good here

[dark gale]

So let my U= x^2 and my dV= sin x?

[lethal schooner]

indeed

[dark gale]

It can’t be solve like I’m doing it? Just wondering

[lethal schooner]

Yes, because you will never receive a constant from sin(x) as it will be binded to our integral

hence it can never be solved as such

use the late rule

L = Logs

A = Algebraic expressions

T = Trig

E = Exponents

In that order

[dark gale]

The late rule is for deciding your U or dV value?

[lethal schooner]

U value

[dark gale]

Thank you a lot @lethal schooner then. I’ll finish it on my own

[lethal schooner]

no problem mon ami

[dark gale]

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[dark gale]

.reopen

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✅

[dark gale]

Damn it I got almost into the same situation

Don’t I have to use substitution method to solve for that -cosx xdx?

Nvm I figured it out

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[crimson sedge]

show your working

[long swan]

sqrt(x^2) = ?

Answer that first

[crimson sedge]

^

[long swan]

Wrong

sqrt(x^2) = |x|

Wdym

You just take the intersection of the two inequalities

x <= 5, or x>=6

I'm not seeing your confusion

Actually check your arithmetic again

-x-2 >= 11 - (-x+3)
-x >= 13 + x - 3
-x >= 10 + x
-2x >= 10
x <= -5

-5 is less than -2, so it's in the interval in which you made your assumption

But you are right

If you get a solution that violates your assumption, then it's not a solution

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[wicked helm]

solution pls

[south cobalt]

Hi, this is grade 9 math can someone help? step by step

[daring ivy]

For each of the sequences, you have to make an equation relating all the elements in the sequence

[wicked helm]

then what?

where can i get the b+c

[daring ivy]

For the arithmetic, you use the common difference to do that
So if x is the common difference,
2+x=b and b+x=c
Then after making x subject of the formula and comparing the two equations you'd get 2b=c+2 which will be the equation for the arithmetic sequence and your first equation

[daring ivy]

The working is quite long
You'd need to the step above for all three sequences first then work it like a sort of simultaneous equation by substituting

Should I continue?

[wicked helm]

yes pls

[daring ivy]

Ok
For the geometric sequence, y can be the common ratio
so,
by=c and cy=d
y=c/b and y=d/c
equating them and then cross multiplying, you'd get
c^2=bd which is your second equation

[wicked helm]

umm, i think i have solved it

it's 19 right?

[daring ivy]

Yep

Did you use a different way?

[wicked helm]

same way

algebraic manipulation

tnx

.close

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[daring ivy]

Ok
no prob

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[deep pendant]

Thank you

it's ok ill open a new channel

sorry olivia you can ask your question

[cedar kiln]

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[deep pendant]

.close

[mossy mango]

@cloud tinsel

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[south turret]

help on ii

[south turret]

feel so close but can't get it

how to remove ln 2

lnx/ln2?

probably should remove all ln at the same time lol

y=x-2?

a

did I do it

I think I did

woohooo

.close

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[crimson sedge]

Is it impossible to have a vertex in 2D with more than three regions assuming that all the regions are adjacent to each other and without one of the regions wrapping around at least one of the others?

[hidden osprey]

draw a line connecting two opposite points of the square

[patent cape]

already answered

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

to give more background on my question as I don't just confuse the person reading; suppose you have a map consisting of 9 countries of the following:

Suppose you want to colour every country in a way such that you use the least amount of colours, with the constraint that no bordering country can have the same colour so that the distinction between them does not disappear

[crimson sedge]

Intuitively, it is quite easy to see what the answer is going to be after a bit of thinking with it, but I am trying to build a logical, formal argument which goes back to my question here

[patent cape]

i remember watching a video on this topic, there was a dude who proved it but ill try and find it rq

[patent cape]

yep look up the 4 color theorem

https://arxiv.org/ftp/arxiv/papers/0905/0905.3713.pdf#:~:text=A computer-assisted proof of,a thousand hours [1].

aw

i kinda dont want to send a file here

but the link looks shady too

its a computer assisted proof but its still something

[crimson sedge]

Ah I see yeah thats what i needed thank you

have a nice one

.close

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[sage merlin]

how would i calculate the area ?

[daring ivy]

Since 4/9 is covered by gravel, can you say what fraction is covered by lawn

[sage merlin]

i need the area of the garden covered by lawn

[daring ivy]

Yh
but first you need to find the fraction of the garden covered by lawn
that fraction would be used to multiply the total area of the garden to get the answer

Or
you could find the area covered by gravel by multiplying 4/9 by the total area of the garden
then subtract that from the total area of the garden

Either way works

[cedar kiln]

@sage merlin Has your question been resolved?

[cedar kiln]

@sage merlin Has your question been resolved?

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[red grove]

how come 2(x-7)

doesnt make 2x - 14

how come x remains unchanged when multiplied by 2

[modern compass]

it can. but you're also dividing by 2

what happens when you multiply by 2 and divide by 2?

[red grove]

ok so step by step

id have

2x -14

then x -7

oh

hahaha

ok i see thank u

i hate when they skip steps, confuses us struggling students

thank u for your time

.close

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[glass hinge]

Need to check if I’m doing it rig by

Right

[tropic oxide]

why are you integrating with respect to n

[glass hinge]

That is just how I write my x

[tropic oxide]

well you shouldn't write your x this way if you want ppl not to misread it

[glass hinge]

okay gotchu

[tropic oxide]

anyway $\paren{2x - \frac{x^3}{3}} \cdot \frac1x \neq -\frac{2x^3}{3}$

[wraith dagger]

Ann

[tropic oxide]

the cancellation you did there is incorrect

[glass hinge]

Oh so I cannot cancel there

So I leave the back as like that

And integrate

[cedar kiln]

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[viral breach]

log3(x)=log5(2x) 3 & 5 are bases

[viral breach]

How would I solve this?

I believe I have to use the change of base formula but not sure how to apply it

[mighty drift]

ln(x)/ln(3) = ln(2x)/ln(5)
use ln(2x) = ln x + ln 2 and it's a linear equation in ln x, which you can find and then find x

[viral breach]

How would I solve it algreabixally

[mighty drift]

You know how to solve linear equations

[viral breach]

Kinda

Might’ve forgotten

[mighty drift]

This is of the form ax + b = cx + d

Just with different names

[mighty drift]

You're not supposed to say that about middle school problems when studying logarithms. This is a level where you just invent the method because you're supposed to be good enough to do that

[viral breach]

I solved but just got 0

[mighty drift]

x = 0 shouldn't be possible because ln x isn't defined then

[viral breach]

Yup

I tried

Combing the ln s on the right side

To ln 5x

Then i have ln 2x = ln 5x

Which doesn’t make sense

[mighty drift]

I don't know what you did there but it's wrong. Also that would give no solution, not x = 0

[viral breach]

Sorry I got ln(2x) = ln(x) + ln(2)

Combine the right

Then I just get ln(2x) = ln(2x)

[mighty drift]

It's technically correct

[mighty drift]

Friendly reminder that I gave you a method

[viral breach]

Yes so now I have ln(2x) = ln(x) + ln2

But I don’t know where to go from here

The teacher got 4.4 roughly

Yes I used that to get ln2x = ln x + ln 2

[mighty drift]

ln(x) (1/ln 3 - 1/ln 5) = ln 2 / ln 5

[viral breach]

I understand this

But where would I go from here

Is what I’m not sure

[carmine bronze]

Isolate ln(x) by algebraic means then use the identity e^ln(x) = x.

[viral breach]

.close

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[marsh bluff]

I'm not sure how to integrate $\frac{1}{(1+x^{2})^{2}}$. I checked using an online calculator, and it looks like I need to use some trignometric substitution, which leads to $\frac{1}{sec^{4}(u)cos^{2}(u)}$. I can't see what substitution has been used though. It looks like $1+x^{2}=sec^{2}(u)cos(u)$, but I can't see how you're supposed to come up with this when doing the question

[wraith dagger]

William VII

[crystal raptor]

x = tan(u)

[marsh bluff]

What's the intuition behind choosing that substitution?

[crystal raptor]

Because 1 + tan(u)^2 = sec(u)^2

You kill the sum in the denominator

And things cancel nicely

Anything like something + (something else)^2 screams trig or hyp trig subs

[marsh bluff]

Alright, that makes sense

I'll go from here and see if I can work out the rest of the question

[cedar kiln]

@marsh bluff Has your question been resolved?

[marsh bluff]

Managed to get the answer, thanks for the help

.close

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[soft cloak]

Hello, I'm doing some homework relating to Markov chains and I'm asked to prove one inequality that I just can't figure out. The Markov chain in question has finitely many states $E$ with transition matrix $P$.

It is assumed that there exists $\alpha > 0$ and a probability vector $q$ such that $\forall i,j\in E, P_{ij} \ge \alpha q(j)$. Given any two probability measures $\mu$ and $\nu$, I must show that $|\mu P - \nu P| \le (1-\alpha)|\mu - \nu|$.

[wraith dagger]

Syst3ms

Syst3ms

[soft cloak]

I've tinkered and tried a lot of things, but the given inequality always seems the wrong way, on top of inequalities and absolute values not mixing together. I'd appreciate all the help I can get.

[cedar kiln]

@soft cloak Has your question been resolved?

[cedar kiln]

@soft cloak Has your question been resolved?

[sage roost]

.help

[cedar kiln]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[sage roost]

.unsolved

.reopen

bruh

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@soft cloak Has your question been resolved?

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[red grove]

i did it in a different order and got a wrong answer

(-7) +3

-4

[gaunt hamlet]

PEMDAS

[red grove]

1/2(-4)

[gaunt hamlet]

Multiplication before addition

[red grove]

ahhh so i fuked up

okok simple fix

thanks

[vagrant elbow]

There is only one order

Technically 2

[red grove]

okok

wait what

[vagrant elbow]

PEMDAS and BODMAS

You're probably following PEMDAS if you're in the US or whatever

[gaunt hamlet]

Also BEDMAS

[red grove]

bodmas lol

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[tardy hatch]

would love some help with this 🙂

[lethal jackal]

shift + windows + S lets you take screenshots

[tardy hatch]

thank you, but i don’t use discord on laptop unfortunatekt

[cedar kiln]

@tardy hatch Has your question been resolved?

[tardy hatch]

any chance you could help ?

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[gaunt mesa]

trying to do some studying. i already know the answer to this and i figured out one length its just no matter what i do i cant find the second one. not looking for an answer im looking for a way to solve it

[brisk dirge]

Use cosine law

[gaunt mesa]

alright ill try that

[brisk dirge]

You'll get a quadratic in a to solve

[gaunt mesa]

idk why i didnt think of that lol i was using sine law

[livid hound]

sine law is fine too

but is longer

[gaunt mesa]

astonishing

wow im dumb thats way easier

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[past wigeon]

how do i do this

[cedar kiln]

@past wigeon Has your question been resolved?

[past wigeon]

<@&286206848099549185>

[arctic grail]

use pythogorean theorem to get AB

then use pythog to get MC and MD

and multiply sum of everything with 12

@past wigeon

[past wigeon]

when i use pythag for MD i get a decimal

i do 8^2 + 12^2 = c^2

[arctic grail]

ok

why is that an issue

@past wigeon

[past wigeon]

i got an answer

okay i think im good

thanks

.close

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[cedar kiln]

@crimson sedge Has your question been resolved?

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@crimson sedge Has your question been resolved?

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[dark coyote]

Can someone explain to me how that equals 256

[frosty ocean]

Do you understand this?

[dark coyote]

Yea

Thank you

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[shadow verge]

https://gyazo.com/07364ec879e8ff2b8864c241668d7906

[shadow verge]

i am not able to solve this using l hopital

does the limit doesnt exist

cause

(-1)^n^2

[manic acorn]

divide numerator and denominator by n^2

[shadow verge]

its just infinity/ infinity tho

[manic acorn]

even after division?

[shadow verge]

no wait

the numerator is weird

its like (-1)^n^2

[manic acorn]

yea

[shadow verge]

idk if it will be 1 or -1

so the limit doesnt exist right?

[manic acorn]

would it matter

[shadow verge]

ouu

so like -1/infinity or 1/infinity

if i use l hopital i cant differentiate the numerator

[manic acorn]

you cant use l hopital after the division

[shadow verge]

oh yeah cause it doesnt give 0/0 or infinity/infinity

[brisk dirge]

degree of denominator is higher so answer will be zero

[shadow verge]

ohhh

okay i get it

thank you

.close

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[charred bay]

May i know how to calculate eigenvector , i'm not sure my answer is correct

[dry swallow]

You wrote Eigenvector is lambda = -1 ?

[charred bay]

Yes

[charred bay]

Lambda = -1 or lambda = 3 right? I find this from eigenvalue @dry swallow

[dry swallow]

But do you understand that a vector cant be of dimension 1 if your matrix is 2x2?

[charred bay]

Hmm wdym?

[dry swallow]

Your matrix is 2x2 right?

[charred bay]

Yep

May i ask , is it true for eigenvalue?

[dry swallow]

So any vector that can be multiplied with that matrix is 2x1, right?

[dry swallow]

Whats true?

[charred bay]

Correct for lambda

Eigenvalue

[dry swallow]

Eigenvalue is lambda = -1 you mean?

[charred bay]

Yes

And lambda = 3

[dry swallow]

Correct

Not an eigenvector though

[charred bay]

Yes cuz idk how to calculate it

What the first step to calculate? I watchinf video on yt and it doesn't help me

[dry swallow]

Just a moment, I became a little rusty
I'm p sure what you did down there is correct, I need a moment to check though

[charred bay]

Oke thanks i will wait :)) i'm also not sure for calculate eigenvector

[dry swallow]

Essentially you're looking for a vector that does
Av = a * v
So
a * v - A * v = 0
So
(I * a - A)v = 0

What you did is good
You may be convinced if you plug in that vector into the matrix and see if the outcome is that vector multiplied by the eigenvalue

[charred bay]

Wait , is this for eigenvector right?

[dry swallow]

Yeah

[dry swallow]

Essentially, what you need to do is that the (already calculated) (I * a - A) matrix
And multiply it by a "general" vector: (x, y)
All of this equals zero
Solve for x and y

Thats what you did, no?

[charred bay]

Wait how to solve x and y

[dry swallow]

Its a single variable for each equation, you'll get an answer

Just plug it in

Ah I see the issue I think.

[charred bay]

I'm still don't understand ^_^

[dry swallow]

Its essentially what you did, but your conclusion is wrong

You said 4x = 0, and 8x = 0
So the vector is (1,2), care to explain that?

[charred bay]

That cuz 4/4 = 1
8/4 = 2

🤔

[dry swallow]

Why are you dividing by 4 though?

[charred bay]

Too make it simplier

[dry swallow]

But if 4x = 0
Then dividing by 4 implies:
x = 0

No?

Ditto for 8x
8x = 0 ->
2x = 0 - >
x = 0

[charred bay]

Yes

4 . 0 = 0

[dry swallow]

Cool
So what is x?

As in, x = ?

[charred bay]

0?

[dry swallow]

Very good

[charred bay]

Y= 0?

[dry swallow]

Is that what the equation says?

[charred bay]

Umm idk

[dry swallow]

4x = 0
8x = 0
so
y = 0?

Probably not, right?
We didnt really reach a conclusion regarding y

[charred bay]

Wait🤔

Idk

[dry swallow]

Well let me put it this way

Does the equation say anything about y?

[charred bay]

I think
8y =0
y= 0

I think 8y ^_^

[dry swallow]

No, because thats not what we got right?

Wr only know that
8x = 0
4x = 0

[charred bay]

Why 8 can be x not y?

[dry swallow]

Can you make it be y somehow?

[charred bay]

Cuz 8 has same row as y

🤔🤔

[dry swallow]

Yeah, but thats not the calculation you did

You said
(I * lambda - A) * (x, y) = 0 no?

[charred bay]

Yep

[dry swallow]

Cool

And then you got
4x = 0
And
8x = 0

Did you?

[charred bay]

Yes

[dry swallow]

Cool
Solve for x, what do you get?

[charred bay]

0

[dry swallow]

Amazing

So we know that, in the general vector we plugged in (x,y) : x = 0

If I were to simplify it for now:
(x , y) = (0, y)

Because x = 0

Are you following?

[charred bay]

Yess

[dry swallow]

Very good

[charred bay]

Cuz y doesn't have constant so just y ? And y = 1?

[dry swallow]

So, you might be asking:
"But how do we find y"

Right

[dry swallow]

Not even y = 1

Just any y you want

It can be 1, it can be pi, it can be -1000

If you're looking for an eigenvector, then yeah
(0, 1) would work

[charred bay]

Okay , then may i ask ,How do we know the constant of x or the constant of x?

[dry swallow]

Constant of x?
You found it, its 0

[charred bay]

Yepp

How do we know is x or y

[dry swallow]

Because you got x

Not sure I understans

for any vector of the form (a,b)
a = 0
No matter what you call a

You called a "x" this time

You may call it anything else, but the first variable will always be 0

[charred bay]

Owwwhh okay

I want to find x and y if lamda = 3

[dry swallow]

Right, just plug it in then

As you did just there with lambda = -1

Ping me when you're done if you want to post the result

[charred bay]

I just found like this ^_^

Idk what should i do next

[charred bay]

x doesn't have value? @dry swallow

[dry swallow]

Yeah

But rather, x is anything

It doesnt have a specific* value would be better to say

[charred bay]

the y = -2

[dry swallow]

Uhh lets see

[charred bay]

So can i assume x = 1

[charred bay]

Cus 8/-4? ^_^

[dry swallow]

If you're looking for just one eigenvector, you may plug in x = 1

[charred bay]

Okay

[dry swallow]

Wait

But thats not what the equation says
Care to show me all of your calculations?

[charred bay]

y = 8/-4 ?

[charred bay]

I think

[dry swallow]

Can you show me the rest of the calculations for this?

Just multiply the matrix with the vector

[dry swallow]

No, thats not how you multiply matrices is it?

[charred bay]

I haven't plug the lamda

[dry swallow]

You did

[charred bay]

Same

[charred bay]

Same as this

[dry swallow]

How did you reach x = 1?

I see 8 and -4 in your matrix

[charred bay]

Cuz x doesn't have value / constant

[dry swallow]

Thats not true though

Do you know how to multiply a matrix?

[charred bay]

Det?

[dry swallow]

No

Multiply a matrix

As in, multiply matrices together

[charred bay]

Hmm

[dry swallow]

I'll spoil it for you if you dont mind
What you got is:
0 = 0
And
8x - 4y = 0

Not!
8y = 4y

Do you understand why?

[charred bay]

Not really

[dry swallow]

Thats... how you multiply it

Im afraid you're lacking some fundamental knowledge friend

[charred bay]

Wait how it can be 8x?

[dry swallow]

Thats just how you multiply the two

You need to look up how to multiply matrices together before you can solve this
At least the basics

[charred bay]

Wait multiply 1 matrix?

[dry swallow]

No, multiply one matrix with another

[charred bay]

Oh with this matrix

[dry swallow]

Yeah lol

(x,y) is also a matrix incase u forgot

[charred bay]

I'm confused about this linear algebra :((

[dry swallow]

Yeah its hard to grasp

[charred bay]

(x,y) is the first matrix?

[dry swallow]

Second matrix from the left

[charred bay]

Hmmm

So how to calculate this x and y

[dry swallow]

You multiply the matrices together

[charred bay]

:((

[dry swallow]

Im sorry love, you cant solve equations before you know basic operations

[charred bay]

Hmm

[cedar kiln]

@charred bay Has your question been resolved?

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[cedar kiln]

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[lethal schooner]

I have to justify whether this is surjective or not

I know what a function being surjective is

And due to this I have the function being surjective at the moment

I'd just like someone to tell me if that is correct, as I am a bit suspicious

on how to write this out without giving any specific examples

[crimson sedge]

What exactly are you thinking about to prove it's surjective?

[lethal schooner]

just using the general rule that the set of naturals are infinite

from 1 to infinity

since our codomain is equivalent to range

as the domain is also N

hence the number of digits are 1,2,3 and so on

[crimson sedge]

Hmmm

Let's say we don't consider 0 in N

One way to prove a function is surjective is to take an element of the Codomain N and show you can find an element n in the domain so that c(n)=N

[lethal schooner]

but that's a specific example

[crimson sedge]

Sorry i don't really know if it's "codomain" or "range" as we don't use those terms in my country

[lethal schooner]

it's okay

i get you what you meant is range

[crimson sedge]

Alright so

Basically let's say we take N a natural number

Can we find n natural so that c(n)=N?

Knowing N is the number of digits of n

[lethal schooner]

yep

we can

[crimson sedge]

So tell me what n is ?

Or at least one solution for n

[lethal schooner]

1,2 and so on

[crimson sedge]

Hmm

Alright let's say we have N=6

What could be n?

[lethal schooner]

anything from 100000 to 999999

[crimson sedge]

Alright

Know let's try to find a general form for n using N

Let's try with multiple N

Solutions for N=1? N=2?

[lethal schooner]

1 to 9

11-99

[crimson sedge]

10

[lethal schooner]

yeh my bad

10

[crimson sedge]

Alright N=3 maybe?

[lethal schooner]

100 - 999

[crimson sedge]

So 1 is solution for N=1 , 10 for N=2, 100 for N=3 can you see a pattern?

[lethal schooner]

yeah

n*10

for all N

[crimson sedge]

No you need n using N

So n=N*10? It doesn't work for N=3

[lethal schooner]

of yeah

oh

um

i get your explanation

therefore this means it's not surjective?

[crimson sedge]

Hmm no i think it is

[inland ocean]

how about this $f(x) = y, f(x) = x^2 \ x^2 = y, x = \sqrt{y} \ f(x) = \sqrt{y}^2, f(x) = y \$ since there are no negatives numbers in N, this should prove it right.

[wraith dagger]

ColdTee

[lethal schooner]

i thought it was at the beginning

it must be surjective though

lol i hate proof

[crimson sedge]

We're using the number of digits of x² not x² i think @inland ocean

Alright let's take it slow alright it's not hard

[lethal schooner]

this might work if we sub in x as root y

[inland ocean]

yes but if it can be proved for x^2 it should work for the number of digits too ig

[crimson sedge]

Hmm i can't say.

[lethal schooner]

i mean it's definitely surjective actually

but you can't do this by exhaustion it'll take years lol

[crimson sedge]

My solution is that n=10^(N-1) is solution for any natural number N so that c(n)=N

[inland ocean]

yes it is, it is many one surjective.

[lethal schooner]

why does that work though? we only have powers of 10 here

which doesn't satisfy all natural N

oh wait

[crimson sedge]

Ho yeah i forgot about the square

[inland ocean]

but n = 10^(N - 1) if N = 2 its only justified for 10 but N = 2 when n ranges from 10 to 99

[crimson sedge]

N=3 so n=100 so it would work if it wasn't for the ²

We need one solution not all

[lethal schooner]

well one general solution

[inland ocean]

hm that might work then

[lethal schooner]

I was thinking to use cardinalities

but we don't have sets here

[inland ocean]

but i still think as long as we prove it for x^2 it still means we have proved it for the number of digits as well

since there will be no element in the co domain not related to the domain

[crimson sedge]

You mean if you prove x² is surjective it's the same for c(n)?

[inland ocean]

yes when its mapped from N to N

[crimson sedge]

What about n=Sqrt(10^(N-1))

$\begin{aligned}N\in \mathbb{N} ,\exists n\in N,c\left( n\right) =N,\ n=\sqrt{10^{N-1}}\end{aligned}$

[wraith dagger]

Zamarus

[crimson sedge]

$\begin{aligned}n^{2}=10^{N-1}\ digits\left( n^{2}\right) =N\end{aligned}$

[wraith dagger]

Zamarus

[inland ocean]

oh

yeah

but -1 isnt in our domain

so

[crimson sedge]

wdym

N is never -1

[inland ocean]

i meant n

[crimson sedge]

Yeah n is a square root i doubt i can be -1 either

We said we consider 0 not in N

Else we have a problem with 0 digits numbers

In that case it's not surjective i guess

[inland ocean]

if we consider 0 its definitely not surjective

[crimson sedge]

Maybe you're right and it's just a small trap but some people learn that 0 is not in N so idk

@lethal schooner What do you think?

[lethal schooner]

yeah in this question we look at N>=1

i still think it's surjective

but just needed a general form which nobody in my class can seem to get

[crimson sedge]

I gave it to you i believe

[lethal schooner]

yeah i think it's correct

i've tried it out and it seems to work for current cases

[crimson sedge]

To find it it's a lot of "feeling" it

[lethal schooner]

yeah

[crimson sedge]

So i can't really explain sorry

[lethal schooner]

for sure

its fine

thanks fgor your help

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[wind maple]

Can anyoneplzhelpme with q9

[tropic oxide]

what is the goal here?

[dull oxide]

I think it's to prove the angles equal

[tropic oxide]

let OP speak.

@wind maple

[wind maple]

So we have to find the unknown lenght

and prove angle

are equal

[tropic oxide]

... you have to prove the angles BAC and DBC are equal?

that's strange.

[dull oxide]

Indeed

[tropic oxide]

if you do not know that they are equal, then you cannot make any progress here period.

[wind maple]

It says angel are congruent

so

[tropic oxide]

show us the entire page.

[wind maple]

I am sure we need to find the m and n

[tropic oxide]

show us the entire page. there have to be written instructions on it somewhere.

[wind maple]

okay

IT SAYS

Find the sides marked withletters

in q1 to 11

[latent bloom]

[dull oxide]

I think it's given then that those angles are equal

[wind maple]

Can you tell me the vertex

that are corresponding

with other shape

[tropic oxide]

maybe i was not clear enough? when i said show, i expected a photo.

[wind maple]

ok