help-17

[thin vale]

no

positive definite matrix has x^T A x > 0

for all x

but you can test this by seeing if all eigenvalues > 0

which C has

[heavy plover]

sure then

[thin vale]

B has >= 0

so only semidefinite

and earlier we decided B was singular because of eigenvalue of 0

and C had all nonzero eigenvalues so it was nonsingular

that is where we are at @fluid obsidian

So now we want to find out what we know about B/C being either symmetric/orthogonal/diagonal

[heavy plover]

symmetric orthogonal and diagonal are left

[twin meteor]

fäf

[heavy plover]

i mean you could solve for B and C

[thin vale]

I am looking into symmetric

[heavy plover]

then its obvious

[thin vale]

poppylikecats

could you agree/disagree with this for me

if

$B=S \Lambda S^{-1}$ then $B^T=(S\Lambda S^{-1})^{T}=(S^T)^{T} \Lambda^{T} S^{T} = S \Lambda S^{-1}$

[heavy plover]

that does make sense

[twin meteor]

AustinU

[thin vale]

So $B^T=B$

[twin meteor]

AustinU

[thin vale]

B is symmetric

[heavy plover]

yes

[thin vale]

okay

when I do the same for C

[heavy plover]

i think it works for C too

[thin vale]

no

[fluid obsidian]

Yes

[thin vale]

but

[fluid obsidian]

The yes is not for you no

I was saying yes?

"When you do same for C" , "yes?"

[thin vale]

$C=S(\Lambda+I)^{-1} S^{-1}$ and if we transpose this we get$\newline$ $C^T=S(\Lambda+I)S^{-1} $

[twin meteor]

AustinU

[heavy plover]

seems to work

[thin vale]

the inverse sign

[heavy plover]

oh wait

[fluid obsidian]

Wait

Well why does the inverse sign go?

[heavy plover]

it would still be inverse

ur taking transpose

[fluid obsidian]

It's a diagonal matrix

[thin vale]

it would be

inverse

transpose

so just inverse

oh

it stays

[fluid obsidian]

Yes

[thin vale]

so C is symmetric aswell?

[fluid obsidian]

Yes

[thin vale]

Left is orthogonal and diagonal

[heavy plover]

for orthogonal you would have B(B^T) = I i think

[fluid obsidian]

C is not singular , C is symmetric, C is positive definite

[thin vale]

Do we know anything about the eigenvalues of orthogonal matrices?

[heavy plover]

since B^T = B then B^2 = I

if true then orthogonal

[fluid obsidian]

[thin vale]

alright well then $B(B^T)=S\Lambda S^{-1} (S\Lambda S^{-1})^{T}$

[twin meteor]

AustinU

[thin vale]

which is $S\Lambda S^{-1}S\Lambda S^{-1}$

[twin meteor]

AustinU

[thin vale]

which is $S\Lambda^{2} S^{-1}$

so B(B^T) only equals I when B=I

[heavy plover]

missing a squared symbol

[fluid obsidian]

How ?

[heavy plover]

on the lambda

i dont think you could diagonalise I like that

[twin meteor]

AustinU

[heavy plover]

unless lambda^2 = I

[thin vale]

I didn't diagonalize it like that, that is just what B(B^T) is

[heavy plover]

no i mean i dont think it can equal I

[thin vale]

Agreed

so B is not orthogonal

[fluid obsidian]

B isn't orthogonal

[heavy plover]

do the same for C

[fluid obsidian]

C isn't aswell

[heavy plover]

is that it

[thin vale]

no

still left with diagonal

[heavy plover]

ah

[thin vale]

diagonal matrices are orthogonal right? if normalized

so

not diagonal

for both

[fluid obsidian]

Well if it has to be orthogonal then the diagonal matrix in between has to be with ±1 diagonal entriws

[heavy plover]

yes i think diagonal matrices are orthogonal

[thin vale]

omg

guys

we got it right

https://tenor.com/view/catshakira-cat-dance-catdance-catdancing-gif-22503034

[heavy plover]

since you can make each row e1 e2 e3 etc when doing row reduction

[thin vale]

https://tenor.com/view/wedding-crasher-hro-gif-25636997

[heavy plover]

W

[thin vale]

https://media.discordapp.net/attachments/234802192014508033/1048635284272722031/pfEuieRu-1.gif

[fluid obsidian]

Can you show the response

[thin vale]

I've never been happier to get a linear algebra problem correct

[heavy plover]

https://tenor.com/view/jerma-live-reaction-live-jerma-reaction-jerma985-gif-25667361

[thin vale]

yes, it is rather underwhelming though

thank you both so much for your help btw!

[heavy plover]

np

[thin vale]

this is the answer

no working

:|

but that is what we got!

so all is great in the world

and it's a beautiful day

[fluid obsidian]

Wait it's positive semidefinite?

Oh wait sorry

[heavy plover]

https://cdn.discordapp.com/emojis/742646798589624351.gif?size=160&quality=lossless

[thin vale]

yes that is what we decided earlier

[fluid obsidian]

Saw for B

[thin vale]

that was like the second thing we did

alright alright alright

[heavy plover]

we got it all

[fluid obsidian]

I thought it was for C

[heavy plover]

no B had a 0 eigenvector

so we said semidefinite

[fluid obsidian]

Yes

I thought it was about C

[thin vale]

this deserves some kind of party

but, I have more problems to finish

so hopefully I can do them, and I won't see you guys anymore tonight

but I probably can't do them

[fluid obsidian]

You are done for the day Austin

[thin vale]

and I probably will see you later

XD

can't stop wont stop faf

[heavy plover]

ill be here probably

if its linear algebra

[fluid obsidian]

Maths have to love you back

[thin vale]

can I tag you? if I have another one soon poppy

[fluid obsidian]

It's not a one sided relationship

[thin vale]

it better

[fluid obsidian]

Yeah

[thin vale]

alright, thanks so much again guys. This was great to finally get finished

cya around!

[heavy plover]

sure

[thin vale]

.close

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[thin vale]

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[glad socket]

Yo I’m really confused with question 5. I have no clue what to do

[lone niche]

you have 2 points

that's enough info

[sour cedar]

do yk slope-intercept form?

i mean

point slope forn

form*

[glad socket]

nope

[lone niche]

slope intercept is cringe anyway

point slope clears

[formal rock]

Mood

[sour cedar]

invalid opinions

[lone niche]

$(x_1,y_1) \Longrightarrow y - y_1 = m(x - x_1)$

[twin meteor]

zfnQRZJT

[lone niche]

you have a point and you know the slope m

[glad socket]

ye

[lone niche]

does that help

[glad socket]

ye i think i got it

[lone niche]

alr

[glad socket]

u can close now

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[violet lodge]

How do u do this?

[vocal sleet]

@violet lodge Has your question been resolved?

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[fathom hedge]

how would you do this one

im completely stumped

[paper depot]

in a parallelogram, two of the vectors joining one point to another are the same

or rather

let me try to say this in a notation free manner

if you take the vector joining two of your points, and the vector joining the other two points, and if you were lucky enough to pick them in the right order to make them two sides of the parallelogram and facing the same direction,

then those vectors will be equal.

your problem has the complication that you don't know which order to pick those points in.

but there are only so many ways you could do that.

[fathom hedge]

yes

[paper depot]

brute force maybe?

theres uh. 12 ways to pick these?

[fathom hedge]

wait that actually did work

there are 6 possible parralel combinations

[paper depot]

yeah 6

my bad

CD=EF
CD=FE
CE=DF
CE=FD
CF=DE
CF=ED

these six

[fathom hedge]

cd and ef, de and cf, and ce df

I just did like cd = n * ef

solved for x and z

for each of the three

noticed two had the same solutions

so it must be those

[vocal sleet]

@fathom hedge Has your question been resolved?

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@fathom hedge Has your question been resolved?

[vocal sleet]

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[mossy sandal]

i figured the first one

using distance formula

i dont know how to find the equation

i know the format is y = mx + c

but idk how to find the y intercept (c) or slope

[timber gale]

to find the slope rise/run. if youve found c anyways

[mossy sandal]

how would you find c

from the question given

[timber gale]

yeah im stumped on that

[mossy sandal]

slope is x2 - x1/y2 -y1

ig i can do that

[timber gale]

hmm

i think its perpendicular no?

[mossy sandal]

it should be

right angled isosceles

triangle

let me draw a diagram

[timber gale]

yeah then you have the slope for that since the gradient for perpendicular is inversed

[mossy sandal]

what..

[timber gale]

I swear ive done this recently let me jog it up brb

[timber gale]

what in the diagram is that

[mossy sandal]

the diagram

[timber gale]

you didnt get point c tho lol

[mossy sandal]

i made

i didnt

i just

rough drew it

[timber gale]

if you can find the slope of ab then you can find the slope of bc

[mossy sandal]

bc?

but wont it have a different slope

[timber gale]

since its perpendicular hence the right angle

[mossy sandal]

oh yeah

what about the intercept

[timber gale]

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/parallel-and-perpendicular-lines-2/#:~:text=Perpendicular lines do not have,slope of the other line.

[timber gale]

lol once youve found the slope just start drawing it

[mossy sandal]

i need some working for marks

if it were up to me i wouldve used desmos for the entire thing

[timber gale]

let me work it out in a bit

[mossy sandal]

alr

[timber gale]

but you need to understand it conceptionally

desmos will draw the graph for you. when you need to draw the graph yourself which is why your stumped here

[mossy sandal]

i can still draw it by myself but

[timber gale]

time?

[mossy sandal]

i guess they dont take drawings

you have to prove it

prove that the y intercept is that

this is a practice question

[timber gale]

hmmm

thats kinda icky the one i drew

@mossy sandal the gradient for c is -2 i believe

[mossy sandal]

bc?

[timber gale]

yes since its perpendicular

[mossy sandal]

does the sign change

[timber gale]

i have a question myself

[mossy sandal]

yup

[timber gale]

it just goes inverse litteraly

https://www.varsitytutors.com/act_math-help/how-to-find-the-slope-of-a-perpendicular-line#:~:text=Perpendicular lines have slopes that,reciprocal is therefore 1%2F2.

[timber gale]

if the gradient is -2 is the denominator gonna be positive or negative

[mossy sandal]

so you cant find the y intercept without any working?

[timber gale]

y intercept for what?

bc?

or ac

what if there is no y intercepts...

wait unless it intersects through the triangle i guess??

[timber gale]

wait what

they just said an iscosceles has a 90 degree angle

oh shoot it can

im so goofy looking at the wrong but similar diagrams

[mossy sandal]

whats c gradient

[timber gale]

bc is -2

mb i was just doing it ig

[mossy sandal]

whats c gradient though

[timber gale]

its the bc

[mossy sandal]

gradient of line bc?

ah

[timber gale]

i just didnt write b

..

[mossy sandal]

i see

also why did u make 2/4 -> -1/2

[timber gale]

you can further simplify gradients

is 4/2 not 2?

[mossy sandal]

it is

but it became negative

nvm

i read one of the equal to thing as a negative sign

[timber gale]

negative since perpendicular

[timber gale]

dont mind my weird working out

[mossy sandal]

ab is 2/4

so that means bc is

-1/2

right

[timber gale]

no

AB is the 2/4 or the 1/2

-2 is the BC which is perpendicular to it thus making the 90 degree angle

[mossy sandal]

wait

[timber gale]

-2 x 1/2 = -1 thus perpendicular

[mossy sandal]

if ab is 1/2

wont bc just be negative of that

because of the perpendicular law you mentioned

[timber gale]

yes thats why its -2

[mossy sandal]

-2 is reciprocated

you are supposed to reciprocate and switch the symbol as well?

[timber gale]

yes

[mossy sandal]

ohh

ok -2 is bc

how can i use that for c

actually wait

i can probably use the slope to find where c is right

but then how do u get intercept 😢

[timber gale]

which c

just the vertex c?

or you mean ac

[mossy sandal]

intercept

[timber gale]

which

y or x

[mossy sandal]

y

[timber gale]

the question is

[mossy sandal]

in y = mx + c

for the equation of bc

[timber gale]

does bc intersect?

through b and just go through

[mossy sandal]

if it stops at point a no

[timber gale]

if so then the y intercept using the slope is 6

[mossy sandal]

ok so no y intercept at all

for bc

so its just

y = -2x

let me check the answer ke

key

nvm

its -2x + 6

[timber gale]

lol i said 6

🙃

[mossy sandal]

how do u get

6

[timber gale]

the slope

[mossy sandal]

i kinda get it

but how do u get the point exactly

[timber gale]

which point

6 exactly?

[mossy sandal]

the y intercept

yes

[timber gale]

rise over run

[mossy sandal]

like the working needed for it

[mossy sandal]

thats slope though right

[timber gale]

-2/1 but im going up 2 and then 1 square or x over

[mossy sandal]

oh

you did it like that

[timber gale]

YES CUZ thats how the slope works lol

[mossy sandal]

theres probably a formula

[timber gale]

No

[mossy sandal]

to figure out at what point the y intercept is using the slope

[timber gale]

for what

[mossy sandal]

or not

[timber gale]

oiadh gaiohgih

prolly.

[mossy sandal]

imagine if the slope was a decimal lol

[timber gale]

it wont be

[mossy sandal]

thats why im kinda concerned

[timber gale]

itll be a fraction

Bro

decimals can be fractions

a slope is legit used as a fraction

[mossy sandal]

what if the point is far away

from the axes

[timber gale]

you mean larger numbers?

[mossy sandal]

yup

theres probably a formula at that point right

[timber gale]

yes you need to substitute and do algebra not a formula

[mossy sandal]

is there a source somewhere

with those steps

[timber gale]

.

https://www.wikihow.com/Find-the-Y-Intercept

[mossy sandal]

not for this question but just samples

[timber gale]

wikihow.

[mossy sandal]

oh bruh

its just

[timber gale]

or google

[mossy sandal]

make x 0

[timber gale]

yes

if you have the equation>?

[mossy sandal]

actually thats for the whole equation yeah

[timber gale]

oh yeah im so dumb lol but yes

[mossy sandal]

https://www.khanacademy.org/math/algebra-home/alg-linear-eq-func/alg-writing-slope-intercept-equations/v/finding-y-intercept-given-slope-and-point

[timber gale]

if you have the equation but just missing the y int

[mossy sandal]

maybe this has my answer

[timber gale]

just make x = 0

[timber gale]

what answer you stilll looking for

or you tryna understand it

[mossy sandal]

s

without just drawing a line and calculating it visually

[timber gale]

the whole gdamn line uses the slope

[timber gale]

if you want it fast find x and y int then connect them boom line

if you want to have it accurate you use the slope.

rise/run rise goes up and run goes across

[mossy sandal]

oh...

you just

put the points into the equation

and solve for c

so the point at b is

[timber gale]

c or b

[mossy sandal]

uh

any point that passes through the line

[timber gale]

yes you can always rearrange the formula or equation to find your missing points

[mossy sandal]

as long as we have the slope we can put the points

[timber gale]

YES B IS THE Y intercept

m is the gradient

thats all you need for a linear equation

[mossy sandal]

4 = 1 + c

ok c is 3

the y intercept is at 3

so its y = -2x + 3

wait

it was 6 in the answer key

wtff

[timber gale]

cuz it is 6??

i think your confusing yourself at this point

[mossy sandal]

i am

ok i'll just explain whats going on

inside my head

in y = mx + c

to find c

you can substitute the values of x and y of any given point on the line

and the slope is -2 as we derived

b is (1,4)

[timber gale]

..

stop

stop

[mossy sandal]

aaaaaaa

[timber gale]

let me explain

to find y intercept

wait let me think fo ra moment

[mossy sandal]

💀

[timber gale]

yeah ik

yeah nah

@mossy sandal

pls do your algebra again

so the point is (1,4)

equation is y=-2x+b ( we found m/slope yes slot it in )

now add in the points

[mossy sandal]

oh...

i took x + b

completely forgot about -2x

[timber gale]

4=-2(1)+b

[mossy sandal]

RAHHSF

43G

[timber gale]

b=6

bruh

cmon man

[mossy sandal]

i dont practice maths

i just get distracted every time i try to practice

and then i always make mistaktes

[timber gale]

better start practicving maths or youll end up like me

[mossy sandal]

like this

[timber gale]

just 5 days ago i had no clue how to draw a linear graph aswell lol

while my ass is doing functions

[mossy sandal]

well atleast i knew that

💀

alr thanks then

[timber gale]

but i had to grind like 20 chapters or so of math questions.

[mossy sandal]

its worth it ig

[timber gale]

and i have an exam in aproximately 7 hours

.

[mossy sandal]

go study

dont help me no more

[timber gale]

but im here

yeah ik lol

i kinda confused myself aswell

[mossy sandal]

alr ill close it

[timber gale]

aight ciao

[mossy sandal]

.close

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[keen lotus]

how to prove that DA=DB when C,H,D in 1 line

[keen lotus]

triangle ABC has medians BF, AE intersect at H. CH intersects AB at D. Prove DA=DB

[vocal sleet]

@keen lotus Has your question been resolved?

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[timber gale]

umm faf

[timber gale]

@fluid obsidian sorry to bother you if your not bothered doing it its fine

taken channel soz

[gilded widget]

is it the same question

[timber gale]

but howd you get the 4/5

yes

[gilded widget]

send it again

[timber gale]

siince im trying to reproduce what he did

[gilded widget]

what was the help number before

[timber gale]

https://cdn.discordapp.com/attachments/486841106298961930/1101142820720480316/734794827077648586.png

[gilded widget]

do you mean the 4/2

[fluid obsidian]

Which 4/5?

[timber gale]

yes

[timber gale]

umm sprry bad memory

[gilded widget]

you wanna take the total amount

minus the ones that are less

[fluid obsidian]

Nice pfp

[gilded widget]

thanks

[timber gale]

what so if its 40000 its 4

[gilded widget]

?

[fluid obsidian]

Where is it from?

[gilded widget]

its called rabbids

rabbids invasion

[timber gale]

yeah

[fluid obsidian]

Ah alright

[timber gale]

so how does one get 4?

[fluid obsidian]

We want a number less than 40000 so the number will start with 3 right

[timber gale]

ah

[fluid obsidian]

And the remaining 4 digits are 4,4,5,6

Which can be arranged in how many ways?

[timber gale]

4!

[fluid obsidian]

/2! As two 4s

[timber gale]

yep

so if we apply that to this question

[gilded widget]

is the ! necessary

[timber gale]

so it would be 5! - 3!

[gilded widget]

uhh i believe so

wait hold on

[timber gale]

5!-3!=114 book said 48

whats with 48 nowadays

[gilded widget]

wait yeah i think i know

uhhh gimme sec let me try to put it in a sentence correctly

so total amount of arrangements is confirmed 5!

we can confirm that much

[timber gale]

yes

[gilded widget]

if 3, theres 4! arrangements

if 4 theres 4! arrrangements

if 5 theres 4! arrangements

ahh

[timber gale]

nani

[gilded widget]

so basically

[timber gale]

is that a ahh understanding or ahh screaming

[gilded widget]

it would be 5! - 4!*3 i think

[timber gale]

whered *3 come from

[gilded widget]

yep it is

[timber gale]

ah those 3

[gilded widget]

the if statements came from

if X is placed at the front

then there are 4! arrangements

and 3 things fill the spot for X

yeah 48 makes sense

[gilded widget]

anything else?

[timber gale]

if you dont mind me asking this

if its a less than question why are we doing 3,4 and 5

[gilded widget]

we did it the long way i believe

[timber gale]

i swear permutations theres either long way or logic i dont know anymore

[gilded widget]

what i just did was taking the total amount of arrangements

and minusing the ones larger than 30000

[timber gale]

yes

[gilded widget]

to get all the ones less than 30000

[timber gale]

I see

and then you had the if statements alright let me try and soak these into my brain

[gilded widget]

we could have easily just done

2* 4!

im so dumb

[timber gale]

well

ive done

2x4x3x2x1

=48

[gilded widget]

yeah thats right

thats 2*4!

[timber gale]

ah yep

[gilded widget]

just ignore everything i said before

[timber gale]

but it just doesnt work on duplicates ig

[gilded widget]

uhh

[timber gale]

or i just straight up ignore duplicates at this point

[fluid obsidian]

Which question are you asking?

[gilded widget]

12345

arranged below 30000

[fluid obsidian]

You have a nice pfp

[timber gale]

we were tryna figure out the way you were doing it and reproduce it

[gilded widget]

lmao thanks

[timber gale]

I feel like ive heard this convo before

[fluid obsidian]

That will be 3×4!

[gilded widget]

i should have just stuck to 2*4!

[gilded widget]

wut

less than 30000

not more than

[fluid obsidian]

It's 5!-2(4!) or 3(4!)

[timber gale]

48 is the answer tho

[gilded widget]

but thats for more than 30000

we want less than 30000

[fluid obsidian]

Oh is it

[gilded widget]

which is just like reciprocated numbers

[fluid obsidian]

Yeah my bad

[gilded widget]

that you said

[gilded widget]

ok is that all

[timber gale]

yes

[gilded widget]

nice

[fluid obsidian]

It's 5!-3(4!) or 2(4!)

[gilded widget]

yea

[timber gale]

but i have more questions to go through and permutations is really broad so im gonna suffer for a while so...im gonna go suffer now.

conclusion permutations just aint it

[fluid obsidian]

Do you know about rabbid invasion?

[gilded widget]

hm

[timber gale]

yes, its a cartoon

[fluid obsidian]

The thing in Meehee pfp is rabbid

[timber gale]

yes?

you aussie aswell meehee?

.close

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[fluid obsidian]

Yeah

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[hearty dragon]

How do i do this question

[hearty dragon]

It gave me this

Is it right?

[timber gale]

your image or paste didnt work

[timber gale]

im not gonna lie man, i dont know how to figure that question out but you need to do working out.

[hearty dragon]

Its number 12

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[vast shale]

PGF question

i know that the coefficient of the power t is the probability x = that number

for example 1/6 t ^2 means that the probability x = 2 is 1/6

i answered 0 here since no there is no coefficient of power of 3

gave me wrong answer

[plain minnow]

This is a property of the PGF when expressed as a polynomial (sometimes from a Taylor expansion). It is not the generalization.

We have that for a PGF $G_{X}(t)$, with $X$ being a discrete random variable, that $\mathbb{P} (X = k) = \frac{G^{(k)}_{X} (0)}{k!}$.

[twin meteor]

Mikkel

[vast shale]

triple differentiate, plug in 0 all over 3 factorial

?

i dont understand this

[plain minnow]

Yeah

[vast shale]

it is what it is thank you

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[ruby relic]

How many bitstrings of length eight either begin with 00 or end with
101?

[vast shale]

where are you facing difficulty in this problem?

[ruby relic]

how exactly do I find the common numbers

[rugged orchid]

Does either mean or or xor

[ruby relic]

or

[vast shale]

what do you mean by common here?

[ruby relic]

we need to subtract ones with 00 and 101 together due to double counting right

[vast shale]

yep good

so you need to remove the count of strings starting with 00 and ending with 101

you are facing problem in this?

[ruby relic]

so is it 2^6+2^5-2^3?

[vast shale]

yes

correct!

[ruby relic]

no

I searched it on internet (to double check) and multiple website have written something else

[rugged orchid]

What numbers in 8 bits start with 00 and end with 101

[ruby relic]

How many bit strings of length eight either begin with 00 or end with 101?
But there are also ⌊20/6⌋ = 3 that are divisible by both 2 and 3, so the total is 10 + 6 − 3 = 13. (c) How many bitstrings of length eight either begin with 00 or end with 101? Solution: There are 26 that begin with 00, 25 that end with 101, and 23 that start with 00 and end with 101.

https://freshtakepublishers.com/how-many-bit-strings-of-length-8-either-start-with-01-or-end-with-01/

[vast shale]

ohh I get it

[ruby relic]

hello?

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[ruby relic]

<@&286206848099549185>

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[wintry walrus]

Looking for help with question 6.e not sure how to solve with a number in front of the squared term

[vast shale]

factor out a 2 from all the terms first

[wintry walrus]

-14 I mean

[vast shale]

not just the d^2 term

factor out 2 from every term of the quadratic

2(d^2-4d+1)

then complete the square of the inside

[wintry walrus]

How do I do that?

[vast shale]

the same way as you did the earlier parts

,tex .cts

[twin meteor]

Toby

[wintry walrus]

[vast shale]

complete the square of d^2-4d+1

[wintry walrus]

How?

[vast shale]

how did you do part a?

[wintry walrus]

6.a?

I just haft the 2 x and x ^2

[vast shale]

do the same thing with x^2-4x+1

[wintry walrus]

(d-2)^2 - 3

[vast shale]

yeah

d^2-4d+1 = (d-2)^2 - 3

now multiply everything by 2 to get the final answer

[outer warren]

not like that

exponentiation takes priority over multiplication

you just leave that term as
$2(d-2)^2$

[twin meteor]

ℝamonov

[wintry walrus]

And is that the answer ?

[outer warren]

no

you didn't read what I said

now the factor of two has disappeared from the first term

[wintry walrus]

That was an accident

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[fast quarry]

I dont understand how w^-1 = x

[sage burrow]

are you comfortable with everything on this table

i.e what i presume you have already deduced

[fast quarry]

yeah yeah

[sage burrow]

definition of inverse:
an element a has an inverse 'a^-1' such that

a * a^-1 = e, a^-1 * a = e

[fast quarry]

its just the w^-1 = x is not making sence to me

[sage burrow]

here we have a=w

[fast quarry]

OH

[sage burrow]

what element 'w^-1' of the elements (s,t,u,v,w,x,y,z) satisfies this

your latin square should tell you

(note e=s, here)

[fast quarry]

i completely forgot about this axiom lol

[sage burrow]

no worries lol

hope it makes sense now

[fast quarry]

thanks yeah

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[glass galleon]

am i right wit A here?

<@&286206848099549185>

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[outer kite]

The following initial value problem has to be solved
I have no idea how to solve that, can somebody please help 🙂

[outer kite]

$x^{'}(t) = ln(t)*e^{-x(t)}, x = x(t), x(1) = 2$

[twin meteor]

Jonas

[flat whale]

,w solve dx/dt = log(t) e^(-x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by riemann#2529

[flat whale]

probably try separation of variables

[outer kite]

alright, I try

[vast shale]

ks

[outer kite]

Got it. THX a lot!

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[graceful fox]

$$\forall n \in \mathbb{N}^*, \int{0}^{1} \frac{x^n}{1 + x^n} dx = \frac{\ln(2)}{n} - \frac{1}{n} \int{0}^{1} \ln(1 + x^n) dx$$

[twin meteor]

koji1001

[graceful fox]

how to show this

[orchid yarrow]

Do you know about by parts?

integ by parts*

[graceful fox]

I tried

[orchid yarrow]

Let's try to make (x^n-1)x in numerator

does that clicks something?

[graceful fox]

owww

i think i got it

let me write it

yea did it

thank you

[orchid yarrow]

No worries

[graceful fox]

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[hearty dragon]

How do you do 12

[hearty dragon]

I know its permutation

But when i did it it gave me a huge number

So im not sure if i messed up or not

[wraith venture]

Should you count the permutations ?

[hearty dragon]

I did

For permutations you divide the toal amount over the amount you need

So for number 12 it would be 17/10

Then you would do a formula

[wraith venture]

You did, but should you ?

[hearty dragon]

I think yeah

[wraith venture]

Is every permutation alphabetically ordered ?

[hearty dragon]

Yeah

[wraith venture]

Is it though ?

[hearty dragon]

What do you mean?

[wraith venture]

"they must be arranged alphabetically"

[hearty dragon]

Thats why im using permutation

Since they have to be arranged alphabetically

[wraith venture]

But you shouldn't

Because you should not be allowing every permutation

You should only be allowing one

The alphabetically ordered one

[hearty dragon]

?

Were organizing books

[wraith venture]

Yes

Once you picked 10 books, there's only 1 way of arranging them

[hearty dragon]

But what if were choosing 10 different books each time

So that would create more combinations

[wraith venture]

But still only 1 permutation per 10 books

So it's 17C10

[hearty dragon]

Wouldnt it be 17p10

Since the order you do it is important

[wraith venture]

What does 17P10 represent

[wraith venture]

It's precisely because it is that it's C instead of P

[hearty dragon]

oh now i get it

I have to find the combinations of which 10 books im choosing

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[dry rover]

suppose h(z) = f(z) + g(z)

if |f(z)| is always greater than |g(z)| for all z in a simple closed contour, in which h,f,g are holomorphic.

since |f(z)| > |g(z)| >= 0 for all z in the interior of the contour

so |f(z)| must always be larger than 0?

then |h(z)| will always be bigger than 0 in the interior of the contour?

So, there are no 0s? at all?

i suppose if there is a case where |f(z)| is larger than |g(z)| in some annulus, and |g(z)| is larger than |f(z)| in the smaller circle (that together with the annulus, makes up the region h(z) is being considered in, then there might be 0s then?

but in case 1? h(z) is never 0?

[flat whale]

so |f(z)| must always be larger than 0?
this is true regardless of g(z)

then h(z) will always be bigger than 0 in the interior of the contour?
did you mean |h(z)| here?

[dry rover]

yes

thank you for pointing that out, edited

[dry rover]

yes, so erm, sorry for ping, by rouche's theorem, |h(z)| will always be greater than 0, i.e there are no 0s in a circle for when |f(z)| is always greater than |g(z)|?

wait, doesn't need to be a circle

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[lofty spindle]

I do not get question 6 (vectors)

[lofty spindle]

can't seem to solve it graphically (cant construct it)

[brittle lion]

you can assume the direction of one vector first

suppose a is facing to the positive-x direction

[lofty spindle]

I understand that but how do I connect that vector addition indicated

[brittle lion]

when you add two vectors, let's say a and b, you put the 'tail' of vector b on the 'tip' of vector a

[lofty spindle]

I understand but is mine considered incorrect

Mine is following that rule though

The tail of vector b starts like what you have did

But the letters are switched

[brittle lion]

the angle between a and b should be 45°. Here, it's 90°

[brittle lion]

oh i see

[lofty spindle]

good point

but how do i graph the vector addition?

[brittle lion]

the graph should look kinda like this

there's two method you can use, wait let me draw it

[lofty spindle]

tq

im going to draw the parallelogram

did i do it right?

[brittle lion]

in your diagram, the vector b and c is swapped

[lofty spindle]

im confused

why must the letters be in that specific way

because the vectors are still from tail to head relative to A

but the letters are just swapped

[brittle lion]

you have to read the information provided by the problem

[lofty spindle]

is there a rule that im unaware of?

[brittle lion]

the angle between a and b should be 45°, not 90°

[lofty spindle]

you're right

let me fix that

[brittle lion]

alright

[lofty spindle]

@brittle lion

i got to here so far

is this the vector that im trying to find?

[brittle lion]

yes!

[lofty spindle]

perfect

[brittle lion]

keep in mind that since they are unit vectors, the length are all 1

[lofty spindle]

got it

question

which angle do i have to use

i am confused about that

is it the angle between -2b and 3c?

[brittle lion]

My interpretation to the problem is that you are told to calculate the magnitude analytically and then use the diagram to verify

so..

do you know how to break down a vector into it's x and y components?

[lofty spindle]

no

i didnt learn that from my teacher

[brittle lion]

oh okay let's find a different way

[lofty spindle]

in fact we were never shown how to do these type of questions

[brittle lion]

oh shoot

[lofty spindle]

i can show you the question we did in class

[brittle lion]

don't worry I'll help you

[lofty spindle]

this is the question that we did in class today

thanks man

what i meant to say is that we never learned how to do it with 3 vectors

[brittle lion]

okay i think we'll use cosine rule

[lofty spindle]

kk

[brittle lion]

yeah

[lofty spindle]

oh

wait

angle between -2b and 3c would just be 45 degrees

[brittle lion]

yeah that's correct

[lofty spindle]

ok that's easy then

just use sin rule

[brittle lion]

okay

I'm gonna use my laptop

I'll show you another method to solve it

[lofty spindle]

kk

thanks

something's not adding up

the answer is supposed to be 1.64

[brittle lion]

okay let's see

oh i know

[brittle lion]

you should have connected the tip of 3c with the tail of a

[lofty spindle]

for the addition vector?

[brittle lion]

yes

[lofty spindle]

ok

[brittle lion]

like this

[lofty spindle]

give me 5 m

ill redraw it

[brittle lion]

alright

I think we should calculate a - 2b first

[lofty spindle]

would just be 2 no?

for magnitude

[brittle lion]

I'm not sure though

[lofty spindle]

hm

im just going to move on to another question

im wasting too much time on this question

[brittle lion]

alright sorry for not being able to give much help

[lofty spindle]

ay man

thanks for trying

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[lofty spindle]

close

no idea how to close it

ill just wait till it time outs

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[vast shale]

Hey

[vast shale]

Find the equation of the trigonometric graph

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[craggy helm]

How does it know the interval?

[craggy helm]

does it assume it's between 0 and 1 because it wasn't given or what? I'm confused

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[flat whale]

Ask whatever website that's from

[craggy helm]

oh so it's the site

ok gotcha

.close

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[craggy helm]

Can somebody explain me this?

[craggy helm]

I'm confused.

[thin vale]

2y^2 is y^2+y^2 right?

so you have y^2+y^2=y^2+4

subtract y^2 from both sides of this equation

and you get y^2=4

[craggy helm]

omh

I'm seriously losing my mind over the simplest things

Sorry! Such a dumb question! Thank you!

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[dry rover]

so erm, just to continue from earlier, since my question wasn't fully answered

for rouche's theorem, given h(z) = f(z) + g(z)

where |f(z)| > |g(z)| for all z in the interior of some closed contour

so minimum of |g(z)| is 0, and thus, |f(z)| must be greater than |g(z)| >= 0

then, |h(z)| will always be greater than 0 then?

so h(z) does not have any zeros?

for all z in the interior of said contour.

h,f,g are holomorphic in the interior of the contour

it sounds simple, but i am trying to understand rouche's theorem in a concrete way, starting with "simple" examples.

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[dry rover]

for rouche's theorem, given h(z) = f(z) + g(z)
where |f(z)| > |g(z)| for all z in the interior of some closed contour

so minimum of |g(z)| is 0, and thus, |f(z)| must be greater than |g(z)| >= 0

then, |h(z)| will always be greater than 0 then?
so h(z) does not have any zeros? since it has the same number of 0s as f(z) which is never 0

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[dry rover]

.reopen

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✅

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@dry rover Has your question been resolved?

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Channel closed

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[hidden badger]

how do i do (b)

i dont know what i did wrong

.close

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[prisma viper]

can anyone help me with b

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[prisma viper]

2

[strange crater]

show what you started doing

[potent anvil]

think about whether A affects the probability of R

[prisma viper]

yh, it does

[potent anvil]

so if two events affect each other, are they dependent or independent

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@prisma viper Has your question been resolved?

[prisma viper]

independant

oh no, dependant

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@prisma viper Has your question been resolved?

[heavy yoke]

so if you choose box one does that change the probability of a red ball getting drawn?