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[pliant estuary]

Is {3, x+13, x-13} a fundamental system for equation y'''-2y''=0?

[pliant estuary]

How can I do this type of question? I saw it once somewhere and they only checked if is linearly indpendent, so are there any next steps?

[vale wigeon]

{3, x+13, x-13} is not linearly independent

[pliant estuary]

wolfram says that it is

[vale wigeon]

show exact wolfram input & output?

as is, 1 (x+13) + (-1) (x-13) + (-26/3) (3) is 0...

[vale wigeon]

it's reading that as a set with a single vector in it

a single vector in R^3

[pliant estuary]

ohh, so just a difference in notation, that makes sense

well back to the question, so we know that it is linearly dependent, well whats the general process of checking whether it can be a fundamental system for equation?

[vale wigeon]

• are all of its members actually solutions to the DE
• is it linearly independent
• does it have as many functions in it as the order of the DE

[lone heart]

@pliant estuary Has your question been resolved?

[lone heart]

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[stark cosmos]

hello i have no idéa how to solve this question:

[stark cosmos]

3c)

can anyone show me/link how it is done?

[alpine sable]

is this a hamming matrix?

[stable pecan]

looks like it to me

but I have never studied information theory and error coding in a formal setting

so not sure

[vale wigeon]

@stark cosmos have you done anything with error-correcting codes before

[stark cosmos]

never before @vale wigeon

Yes it is a Hamming-matrix

a) has columnvectors that are different to each other and no one is equal to 0, thus it can correct 1 error

b) H*c = 0, thus it belongs to b

c) not even sure how to start

[vale wigeon]

hold on is that actually true tho

as in like

im not sure that implication holds lol

ok so wait do you like. know enough linear algebra and abstract algebra to not be intimidated at the phrase "a vector space over F_2" or is this gonna need to go down unlubricated

[stark cosmos]

i do not know what that means haha

[vale wigeon]

ok oof

ok so then do you know what "bitwise XOR" means at least

[stark cosmos]

still don't know im sorry

okay i just googled it

yes i know it

[vale wigeon]

bruh ok so you got thrust head-first into the deep end of the pool

wtf is your teacher thinking not teaching basic shit to you and then expecting you to know this like

okay

[stark cosmos]

i mean

they wrote like 2 pages for all this

[vale wigeon]

at least tell me you know your way around linear algebra in the context of real numbers, and understand how matrix multiplication works?

[stark cosmos]

and showed some proofs

yes ofc

[vale wigeon]

not gonna be strictly necessary but should help if you do

ok

great

yeah so ok

oh boy where do i begin

a code is a subset of the set of all bitstrings of a fixed length

[stark cosmos]

alright

[vale wigeon]

we call a code C linear if it's true for any strings s, t ∈ C that s XOR t also belongs to C

where XOR is that bitwise XOR i talked about

in a sense it is like addition except modulo 2

[stark cosmos]

yes okay

so 101 + 101 = 000, and 000 should belong to C?

something like that

[vale wigeon]

something like that yes

[stark cosmos]

yes

[vale wigeon]

we say that a code is able to correct 1 error if, for every string that differs from a codestring in exactly 1 position, we can tell definitively which codestring it is

[stark cosmos]

yes alright

[vale wigeon]

with a check matrix

uhhh

fuck

[stark cosmos]

so if C = 11 00 11, 00 11 00, 00 00 00, then 111 011 should be corrected to 11 00 11

[vale wigeon]

yes

hold on let me try to refresh myself on this because what im about to say might be convention dependent...

[stark cosmos]

no problemo ma'am

[vale wigeon]

fuck. can't find the book i was looking for

i think i remember what a check matrix is but ill be damned if i remember how to go from a check matrix to a generating matrix

[stark cosmos]

i think that kard(c) = 2^n
and len(c) - n = how many columns our matrix should have

then we just take len(c) - n c's and then we stack them over each other

M = [c1;c2;...;c(len(c) - n)]

after that we make sure that our matrix doesn't have replica columns by doing row reduction and then we are done I think

and ofc, no columns can be equal to a 0 vector

[lone heart]

@stark cosmos Has your question been resolved?

[stark cosmos]

I mean

since it can only correct 1 error

i guess you can brute force to find it

just try first element = 0 and you will find Hc = 0

so i guess this is the answer

thanks @vale wigeon

/close

.close

[lone heart]

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[analog falcon]

why multiply that sum of sectors by 2

to find the area of sectors, u need the angle

can u show your work?

oh! yes yes

that is true

where did your work go wrong? can i see it?

no worries. Please take your time 🙂

not to multiply by 2

but apart from that, I think it's perfect

wait

oh I see where we're going wrong 🙊

let me perform some art

funny image

that small segment is not included in the sectors

yeah

now what shall we do

do u mean that this shaded segment is equal to the outer segment and thats how we can find it?

okay

woah, lemme have a look

Hmm, but the triangle and the green segment, both of them together will not give us red area alone. We need to still add one more segment (on the right side of AB in the diagram) to get the red area. And this segment is really just the same as the green segment

Oh I think I see

Yes, totally

Have you solved it this way?

yes I think

How will you find the segment's area?

perfect

I'll do the calculations too meanwhile you do them

[lone heart]

@peak lodge Has your question been resolved?

[analog falcon]

got anything?

I tried and was close (not perfect answer) but then realised a mistake I made, and retried, and now I'm further from perfection lol

Awesome 😎

can u show the work?

yeah

this is my funny work

you did it like
a - (b + c + d)
i did it like
a - b - c - d
so it's the same thing. I think I went wrong in the triangle's area part. Now who told me area of a equi trangle is √3/2 × side..................

this is perfect, just that in the last part, it should be + 1/2 √3 r²

then it'll fit well

yeah

[lone heart]

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[analog falcon]

.reopen

[lone heart]

✅

[analog falcon]

oh lol we're on the top now

okay

goodluck

Awesome, congrats!

I see. I had chapters related to circles but had relatively easier questions

yeah they are like that
its just a list of the key points that should be in your answer

welcome. I'm glad to help 🙂

.close

.close

[lone heart]

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[analog falcon]

oh wow i could close nice

[lone heart]

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[alpine sable]

hi, for this question: i just need help with the last one i cant seem to do it can someone help

[alpine sable]

for the last one i do b/SinB = to what?

[crisp falcon]

Proposed Arabic fourth average models

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[split heart]

Helli

[split heart]

I have a doubt in this question

[solid crane]

which?

[split heart]

14

Ima send till what i fid

Did

[solid crane]

which ques?

[split heart]

No. 14

[alpine sable]

,rccq

,rccw

[split heart]

I rationalised it but idk if it wasd necessary cuz i am not getting the image of what to do here

[solid crane]

hmm lemme check

[split heart]

I can give each no. A root

Where i can apply (a^m)^n for √50

So like √2 + (50^1/2)^1/2 +√15

Is that correct

Denominator also √5

[final viper]

you can further reduce them

[solid crane]

do you realise that a and b are reciprocals

[final viper]

sqrt(50) = 5root2

that makes 3 + 2root2 , and 3 - 2root2

(after dividing by 5)

[split heart]

Wait my last step i wrote wrong

It's not 2+√50

[split heart]

I find a bit hard understanding in text can u make it into mathematical form

[final viper]

$\frac{2\sqrt{25 * 2} + 15}{5} = \frac{10\sqrt{2} + 15}{5} = 2\sqrt{2} + 3$

[split heart]

Where did 10.'9.3

[ocean seal]

Oreo

[final viper]

sqrt 25 = 5

similar for b

just a - sign

[split heart]

Ohhhh

[final viper]

they are reciprocal of each other btw

[split heart]

?

[final viper]

a = 1/b

you'll notice once you rationalize

[split heart]

How is it uh reciprocal

New to irrational

[final viper]

well first try finding a and b

[split heart]

Cuz it is √5+√10

In b denominator is reversed

[final viper]

what b do you get after simplifying?

[split heart]

Wait i need to simplify like u said

[final viper]

it should be 3 -2root2

ye

[split heart]

Ok wait lemme write it

How did 16 become 3

[final viper]

thats 15

[split heart]

I mean 15

[split heart]

It says 2√2+3

[final viper]

because i divided by 5?

[split heart]

U alr divided it by 10 right

[final viper]

denom is 5

smh

[split heart]

Ye so u divided 10 with that 5

So denom becomes 2

I mean 1

[final viper]

i assume you know that (a+b)/c = a/c + b/c

[split heart]

Ye ye

[final viper]

so like 10root2/5 + 15/5

2root2 + 3

¯_(ツ)_/¯

[split heart]

Ohhhh

My bad

[final viper]

np

[split heart]

I was confused cuz then u should have cut √2 also

10 and √2 was single

My bad

[final viper]

all good

now find b

[split heart]

Ok lemme try b on my own

3-2√2?

Now i gotta find √a and √b

Do i maake the roots into exponents?

Wait but then i have 2

In a and b

So 2*2^1/2?

Wait isn't that wrong somewhere

[split heart]

Are u still there

[final viper]

yup

now there is some crazy observations

do you know (a+b)^2

[split heart]

?

[final viper]

and (a-b)^2

expansion

[split heart]

Yes identity

[final viper]

leta take a first

[split heart]

a^2 + 2ab + b^2

[final viper]

3 + 2root2

[split heart]

Is this question average lvl or hars

[final viper]

its a good question

[split heart]

Cuz i learned irrational like 2-3 weeks back

[final viper]

above average

for someone who is new

actually hard

Lmao

[split heart]

I find these questions confusing cuz

[final viper]

but dw

[split heart]

I don't understand what to do

[final viper]

the point is

you cant do no thing with a root over 3 + 2root2

it just sucks

[split heart]

Wdym

[final viper]

i mean you cant rationzalize further

since denom = 1 now

[split heart]

Ye ye

[final viper]

so by observation you can compress 3 + 2root2 to (a+b)^2

the square and root will cancel

that is the reason

[split heart]

Oh how

[final viper]

well look at 2root2

it somehow resembles 2ab

no?

like both have 2

root2 can be a

1 can be b

[split heart]

Wait what

[final viper]

i mean

[split heart]

Oh ok ok

A and b

[final viper]

you gotta find the a and b

and produce their squares

to compress

so if we take a = root2

b = 1

2ab = 2root2

works well no

now we need a^2 and b^2

and what we have is 3

we have to split 3 in such a way that it produces a^2 and b^2

a being root2 , and b being 1

[split heart]

Wait wait

[final viper]

take your time

[split heart]

We are trying to make this

Into

[final viper]

the form of a^2 + 2ab + b^2

yes

[split heart]

a^2 +2ab + b^2

[final viper]

ye

correct

that is the motto

[split heart]

But we have no third digit

[final viper]

yes

thats why we split 3

[split heart]

Oh

Never did that in my life

[final viper]

in what 2 numbers should we split

so that we get a^2 + b^2 thing

remember

a = root2

b = 1

[split heart]

But how can we just split

3

[final viper]

break

[split heart]

We can't change the numbers right

[final viper]

into sum

of

two numbers

[split heart]

Oh

[final viper]

yeah lmao

[split heart]

So just change form

[final viper]

true

[split heart]

Can u like

[final viper]

what is a^2 and b^2

if a is root2 and b is 1

[split heart]

Use mathematical form when u use powers

Cuz it make me slower

[final viper]

oh ok

$3 + 2\sqrt{2} = 2 + 2\sqrt{2} + 1$

[ocean seal]

Oreo

[final viper]

does this ring bell

wait

[split heart]

Uh wait

[final viper]

$3 + 2\sqrt{2} = 2 + 2\sqrt{2} + 1 = (\sqrt{2})^2 + 2\sqrt{2} + 1^2$

[ocean seal]

Oreo

[final viper]

does this now ring bell

[split heart]

We made 3 into

2+1

[final viper]

true

[split heart]

But how did that two become √2 )^2

[final viper]

because

root2 times root2 = 2

[split heart]

Ohhhh

[final viper]

btw you can solve this ques without doing this compression

[split heart]

Oh

[final viper]

but that will be scary with roots

yeah lol

i mean you can

just put a and b

in the equation to prove

[split heart]

Is it easier or harder

[final viper]

idk you can try both

[final viper]

so like

[split heart]

Well i don't have much time

Maybe i will try it tmr

[final viper]

$3 + 2\sqrt{2} = 2 + 2\sqrt{2} + 1 = (\sqrt{2})^2 + 2\sqrt{2} + 1^2 = (\sqrt{2} + 1)^2$

[ocean seal]

Oreo

[split heart]

Cuz i gotta complete this and eat n slee

[final viper]

thats a

roota = (root2 + 1)

oh okay

you can try this later

[split heart]

But don't rush tho

[final viper]

sure

[split heart]

Cuz no use it someone just tells me the answer

[final viper]

true true

[split heart]

Oh ok i got it

I have never done splitting but i get the point

Ok now we compressed it so ur tellimg

Root of a just cancel that 2?

[final viper]

yeah

square root of square cancels themselves

[split heart]

Damn

[final viper]

its like you are saying multiply 2 with itself and divide the result with itself

[split heart]

That thought process is epic

Cuz i got no image itself lol

[final viper]

lol

[split heart]

Lemme write the steps real qyick

Done for a

So a is

√2+1?

I mean √a

So b same and i get 0

The only change is minus

Hey I have a doubt in splitting

[final viper]

yessir

this time a - b whole square

[split heart]

I can only split and write it at the end or i can put it in between also???

[final viper]

any way

[split heart]

Oh

[final viper]

i mean

[split heart]

We wrote at end for this purpose

[final viper]

addition is commutative

[split heart]

True

But subtraction ain't

[final viper]

yes

so like

[split heart]

Can u give an examole

[final viper]

-2root2 thing

[split heart]

Where splitting is wrong

I got splitting for addition but for subtraction u can only write at end or also in between or anywhere

[final viper]

you gotto split so that it looks like

$a^2 - 2ab + b^2$

[ocean seal]

Oreo

[final viper]

here

(-2ab) can't loose its sign

thats the only concern

so like you can write it as

$a^2 + b^2 - 2ab$

[ocean seal]

Oreo

[final viper]

$-2ab + a^2 + b^2$

[ocean seal]

Oreo

[final viper]

all are same

[split heart]

How

[final viper]

its just terms in addition

lol

[split heart]

Subtraction ain't comutative

1-2 ain't 2-1

[final viper]

yes

but i didnt write 1-2 as 2-1

what i did was

1-2 = 1 + (-2)

now

[split heart]

U changed place

[final viper]

do you see the +

you CAN interchange 1 and (-2)

since + is commutative

[split heart]

Oh

I know

That

[final viper]

pog

[split heart]

But splitting is making me confuse basic stuff now hold u

Up

2--5 is 2+5

[final viper]

yes

[split heart]

2-5 is also 2+-5

[final viper]

yes

[split heart]

But i did not change place here

[final viper]

yes

its all valid

btw use a bracket to avoid confusion lol

2+(-5)

[split heart]

U went from a-2ab +b

To

a+b-2ab

[final viper]

yes

[split heart]

So u changed places

[final viper]

a + (-2ab) + b

i can change

about +

since

commutative

[split heart]

I did not get it hold up

[final viper]

a + b + (-2ab)

and then +(-stuff) = -stuff

[split heart]

Wait wait so ur telling

5-2+3 =5+3-2

[final viper]

yes

they same

[split heart]

Idk how i went dumb suddenly

My bad

[final viper]

its fine lmao

its just math mangling 🤓

[split heart]

When i learn smth hard i confuse basic sruff

[final viper]

yeah its normal

[split heart]

Yesterday happened when I learned negative base laws are different

Then why do they say subtraction is not commutative

Omg i am dumb

It was for lower classes

5-2 not equal to 2-5

But 5-2 is also -2+5

[final viper]

correct

[split heart]

My brain clicked now

[final viper]

nice

:D

[split heart]

I was thinking the stuff thought in lower classes

No wonder i got confused

[final viper]

i mean its right

[split heart]

So let's do b

B u 3-2√2

√a was

√2+1 right

[final viper]

yes

[split heart]

Lemme try b

?

No?

[subtle birch]

No

there is a -2ab term

[final viper]

we talkin about a

[subtle birch]

Yes

[final viper]

it has + 2ab

[subtle birch]

3-2*sqrt(2)

[final viper]

thats b

ungigachad moment /j

[split heart]

I thought what I did all went wrong

[subtle birch]

What is a?

[final viper]

3 + 2sqrt2

[subtle birch]

Okay then its correct

[split heart]

I gotta finish it now

[final viper]

imagine me pranking you for like half an hour

[split heart]

Hell.nah

So b is

2-2√2+ 1???

[final viper]

same procedure

can write 3 as 2 + 1

2 = root2 square

but this time -2ab

so a-b whole sq

yup

[split heart]

Wait what

b is

(√2-1)2

[final viper]

yes

then same shit

root cancels square

[split heart]

So √b is

√2-1

[final viper]

yes

now you are 90% done

just plug in a and b

in that equation to prove

[split heart]

I gotta find -2√ab now ahh

Plug in?

[final viper]

put value

[final viper]

ab is just (A-B)(A+B)

if you notice

A^2 - B^2 identity

[split heart]

??

[final viper]

i mean

[split heart]

So √2+1)√2-1

[final viper]

ab = $(\sqrt{2}+1)(\sqrt{2} - 1)$

right

[ocean seal]

Oreo

[split heart]

U can use properly

Property

[final viper]

isnt it (A+B)(A-B)

yesh

so its all ez

now

you can do it

[split heart]

wait so

2-1

[final viper]

ye

[split heart]

1

√1 is 1

[final viper]

yes

[split heart]

Damn

Lemme write steps hold u

Up

[final viper]

what steps are you even writing

[split heart]

All the things u tell

[final viper]

i think it is better to know why we are doing

[split heart]

To solve the answer

[final viper]

than how we are doing

because the how can be produced

alr alr

its okay

[split heart]

Wait wait

So it's -2*1

So -2?

[final viper]

ye

and

[split heart]

-2√ab is -2

[final viper]

now roota - rootb

[final viper]

yes

[split heart]

Hold up

Now i do the final step?

No wait wrong

We did wrong

We used the value of √a and √b

[final viper]

what

just subtract them

roota - rootb

whats wrong?

[split heart]

The question is √ ab

No no wait

Inside -2√ab

I am dumb

I compressed a and b

[final viper]

ah

[split heart]

Still wrong

We used root of a and b

So tehcnically wrong

[final viper]

what

is technically wrong

[split heart]

-2√ab

[final viper]

its -2

no?

[split heart]

A is

Different

√a is √2+1

[final viper]

yes

[split heart]

It says ab

Not.√ab

[final viper]

the question says -2rootab

[split heart]

Ye

[final viper]

the root is right there

$\sqrt{a} * \sqrt{b} = \sqrt{ab}$ btw

[ocean seal]

Oreo

[split heart]

Yes

[final viper]

so its all good

go ahead

[split heart]

No no

Then i don't need an extra root in my step

[final viper]

yes

[split heart]

So i just have to cut that root?

[final viper]

you already have root on a and b

what cut

[split heart]

So i wrote it wrong

[final viper]

you are directly writing rootab

[split heart]

Watch my step

[final viper]

multiplying them

[split heart]

I wrote root on top of root a and b

[final viper]

yeah

that should be the reason

[split heart]

So that's wrong

[final viper]

you dont need that

its extra

[split heart]

My bad

[final viper]

no worries

[split heart]

All good?

[final viper]

yes

[split heart]

Now final step right?

[final viper]

yes

also dont go adding another -ve sign in front of -2rootab lol

[split heart]

√2+1-(√2-1)-(-2)

[final viper]

since the -ve is already there

i knew you would do that

Lmfao

[split heart]

Why not loo

Lol

We can write it as (-

+-

[final viper]

you already have -2root(ab)

[split heart]

I wrote to avoid confusion lager

[final viper]

now you just have to add it

[split heart]

Later

Wdym

[final viper]

i mean

the double -ve makes it +

which is +2rootab

[split heart]

Ok

My has

Bad

[final viper]

which is not what we wanted

[split heart]

But in question it is -2√ab

[final viper]

yes

you already know -2rootab

[split heart]

Ima make it plus no problem

[final viper]

you just put "-2" in place of it

[split heart]

Ohhh damn

A big mistake

Ty ima make it plus

[final viper]

yeah that happens

thats why its advisable to put everything into one equation

than breaking in parts

but thats just an advice

[split heart]

Can u write the last step cuz i feel very sleepy

I mean i wrote it righr

It's correct right

[final viper]

yeah

√2+1-(√2-1)+(-2)

[split heart]

Lemme write n send

[final viper]

its done

open brackets

all stuff cancels

√2+1-√2+1-2 = 0

[split heart]

I know that

The text says that

Proving is the problem

Lol

[final viper]

thats pretty much it

its proved once you write that

[final viper]

looks good to me 👌

[split heart]

Les goooo

Oreo the clutch god

[final viper]

lesgo

was this your homework btw

[split heart]

Tehcnically no

[final viper]

coaching tuition stuff

[split heart]

My practise book

[final viper]

ah

[split heart]

But dis too tough compared to my text

[final viper]

yes

[split heart]

It's like text i can speedrun in 1 hr

[final viper]

you in 10th standard i suppose

[split heart]

No actually

[final viper]

or 9th

[split heart]

Yep

[final viper]

ah

[split heart]

But like a person told it's too hard to

Solve

For an average 9th yr old

[final viper]

thats pretty nice if you understood all that

[split heart]

9th grader*

[final viper]

because its hard enough for the 10thies too

[split heart]

I mean school text is pretty easy for me

[final viper]

lol

[split heart]

I did understand every thing

But i need questions to what I did remember

[final viper]

ic

[split heart]

Idk if the application lvl of this comes i will solfe

But i remember every step u told

Yesterday I learned a big big exponent question

And like i understood it very properly

So i can do a similar question now

[final viper]

thats good

[split heart]

This is a bit tricky cuz

Compression may vary

Tmr i will learn the other method idk if u will be the same person

Bye

[final viper]

yeah takes time

its just the way of looking things

but reverse

cya

if you are done you can .close

[split heart]

I hope to meet u again

[final viper]

sure

[split heart]

Good explanation 10/10 fr

For me atleast

[final viper]

Lmao tq

i got this jee advance thing in like 2 days holy shit

i should get going

[split heart]

Bye

.close

[lone heart]

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[hybrid meadow]

how to find interior and closure of sets made of functions?

[lone heart]

@hybrid meadow Has your question been resolved?

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[grave matrix]

"Show A |-> A^2 for A in SL(2,C) is not surjective on SL(2,C)"

I already have that if B in SL(2,C) is diagonalisable then there is A in SL(2,C) with A^2 = B.
Do I just look for a non-diagonalisable SL(2,C)-matrix?

[lone heart]

@grave matrix Has your question been resolved?

[alpine sable]

no, the converse might not hold - you would need to check that seperately

[grave matrix]

Calling this squaring map T, I need to show that im(T) != SL(2,C)

Which is to say there is some SL(2,C)-matrix that isn't the square of any SL(2,C)-matrix...?

[alpine sable]

yes

[grave matrix]

Now, I've already just shown that (diagonalisable matrix) ∈ im(T)

So a desired matrix is not diagonalisable

Meaning it has only one eigenvalue?

[lone heart]

@grave matrix Has your question been resolved?

[lone heart]

@grave matrix Has your question been resolved?

[lone heart]

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[brittle slate]

Can someone solve this for me pls

[lone heart]

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

[brittle slate]

1

[rustic coral]

This is going to sound very contrived, but consider $\int^{b}_{a} \frac{f(x)}{f(x)+f(a+b-x)} \dd{x}$

[ocean seal]

Civil Service Pigeon

[lone heart]

@brittle slate Has your question been resolved?

[lone heart]

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[silver portal]

I need help

[silver portal]

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[icy charm]

Nn siri cach plus wla xi wakala khasek a la carte ou sf

[silver portal]

Huh

[slender quiver]

whats the problem

[silver portal]

The triangles

Wasnt in order sorry

I need to know how to solve these for my test

<@&286206848099549185>

[wary stream]

https://www.youtube.com/watch?v=YiFwvAFk-xs&ab_channel=TheOrganicChemistryTutor

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@silver portal Has your question been resolved?

[lone heart]

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[hardy trench]

I have exam for uni entrance tomorrow and have essentially done 0 work on conic sections, one of the possible subjects. Am I on the right track with this?
Assuming a cone formed from the line z=ax, T in origo. Call the axis of symmetry S. Call the angle between S and the conic line (l:ax) v. Call the angle between the "cutting" plane and the axis of symmetry θ. Call the curve formed by intersecting points of the "cutting plane" and the cone the "curve of intersection"
Start of explanation:
Assume a circle is formed by the curve of intersection. If so, then the z-coordinate of all points on the curve of intersection must be constant. in this case θ is obviously = 0 deg. Assume circle had focal points A and B in its center. If B is moved "up" the z-axis, i.e. l, an ellipse will form. This is obviously true for all 90 deg> θ>v (v is the angle between S and l) . When the angle reaches θ=v, then the plane must be parallel to l, meaning B is at infinity. The parabola is an ellipse with one focal point at infinity.

Bruh

I couldnt see ur questrion sorry

[wary stream]

Find the characteristic polynomial , do you know how?

Have you looked up the formula?

Btw the reason you can use .close and .reopen is because you gave yourself the helper role

$λI_n$ is just the identity matrix with lambda (which is just a variable) down the diagonal

[ocean seal]

dldh06

[wary stream]

So you do A - $λI_n$

[ocean seal]

dldh06

[wary stream]

Then find the det

[lone heart]

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[quartz lance]

Hey I need help finding the domain of this piecewise function

[quartz lance]

I am watching a youtube video explaining that the domain is any value the x can take, I understand that the first case x + 3 can only take values greater or equal to -2 and must be less than 1

However, since x is not a denominator where we cannot divide by 0, nor is it inside of a root, where we cannot root a negative number, shouldn't any real number work?

I tried inputing (-infinity, infinity), but it rejected the answer. I don't see why the restriction should matter if I input 1 for x, which is part of the restriction, but 1 + 3 = 4, there is no problem for that

[vapid shuttle]

any real number could work, but this piecewise function restricts the domain

start with the case of -x+2 where x>1

does this have any domain restrictions, inside the boundary of x>1?

[quartz lance]

so x can be 2, 3, 4 ...

it cannot be 1 or 0 or negative

as in it says that, but nothing is really stopping it right? like dividing by 0

[vapid shuttle]

you might be confusing real numbers with natural numbers btw

[quartz lance]

are real numbers any numbers that are not irrational?

[vapid shuttle]

real numbers contain the irrational numbers

[quartz lance]

oh ok

[vapid shuttle]

but anyways

[vapid shuttle]

can you answer this for me?

[quartz lance]

(1, infinity)

that's the bound

[vapid shuttle]

yes or no

does it have domain restrictions?

[quartz lance]

it does have a restriction

[vapid shuttle]

being?

[quartz lance]

it cannot be 1 or less than 1

[vapid shuttle]

yes okay good

[quartz lance]

anything that is 1 or less than 1 cannot work

[vapid shuttle]

and that is the only one (for that piece of the function)

great

so now move on to the next piece

the next piece tells you that

x can equal 1

and when it does, the output is 8

[quartz lance]

so it is in the function

[vapid shuttle]

so now the only part of our function that is currently restricted is that it can't be less than 1?

right?

[quartz lance]

correct

[vapid shuttle]

before we move onto the next and final piece

and this final piece, tels us that x can be less than 1

but it must be greater than or equal to -2

[vapid shuttle]

so this, is our final and only restriction on the entire function

[quartz lance]

so it be 0, -1 or -2 but not -3

[vapid shuttle]

sure. but again, it could also be -1.1, or -sqrt(2), or -8/6, etc.. other numbers that aren't counting numbers

the part that matters is that they must be greater than or equal to -2.

[quartz lance]

right

[-2, 1] U [1, ∞)

[vapid shuttle]

no

can't x=1?

[quartz lance]

oh wait oops

1 and -2 are part of the domain

so I have to use []

[vapid shuttle]

yes

but

you don't unionize something at the same value on both sides, that doesn't make sense

you would just write

[-2, inf)

[quartz lance]

ah ok so if it includes on both ends like 1 in this case I shouldn't write it

[vapid shuttle]

right, because normally the union is meant to join together two disconnected intervals

[jagged raptor]

its not wrong but its better not to

[vapid shuttle]

and here, the intervals aren't disconnected if they both include 1 like that

you typically use it for something like [-1, 2] U [3, 8]

[quartz lance]

thanks for explaning, I was trying to figure out why it can't take -infinity and infinity because technically it doesn't break any math rules

[vapid shuttle]

yea, the functions themselves were nice in this problem for you, so that theoretically you each piece would have a domain of all the real numbers

but since the piece wise function restricted the domain, you gotta follow the rules

[quartz lance]

right that's what I wasn't sure about

I thought oh it doesn't have a fraction, so I don't really have a rule I can't break

anyways thanks for explaining!

.close

[lone heart]

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[lost flax]

Hi can someone explain to me what I did wrong? Ty

[lost flax]

-54 instead of -32?

like this

yeah that was it lol

.close

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[lost flax]

i have no idea how to go about this, someone pls guide me ty!

[weary wyvern]

Isn't that the whole question

[lost flax]

oop haha

[proven leaf]

lemme retry that rq 😅

[weary wyvern]

Just use taylors theorem

[lost flax]

i do have the expansion x^n/n! thing

[proven leaf]

not quite x^n

[lost flax]

i think the cn thing is confusing me lol

[proven leaf]

that's y=e^x about x=0

[lost flax]

oh

[proven leaf]

first do you know the general Taylor series?

[lost flax]

this?

ive been using this

[weary wyvern]

Yes now a=3

[lost flax]

oh, thats all?

[proven leaf]

that's what the a term means :))

[lost flax]

what is the cn thing supposed to mean?

ohh

[proven leaf]

the nth coefficient

@weary wyvern why'd you downvote your own response? 😅

[lost flax]

so its just asking me for the first 5 terms?

haha

[weary wyvern]

To keep things balanced

[proven leaf]

not terms coefficients of those terms

[lost flax]

oh?

[willow patio]

LOL

[proven leaf]

yin-yang, Taoist I see? xD

[lost flax]

what r the coefficientds

like what is it referring ot i mean

[proven leaf]

c_n is the nth coefficient the term in front of the (x-3)^n

[lost flax]

oh i see

so i would write only the coefficients

[proven leaf]

which is $\frac{f^{(n)}(a)}{n!}$

[ocean seal]

XxMrFancyu2xX

[proven leaf]

yep :))

[lost flax]

ah i see

i will try it

[lost flax]

is the factorial included in the coefficient?

[proven leaf]

ofc it's what makes the Taylor Series work!

[lost flax]

okayy

and e^x no matter how many times u derive it is e^x right

[proven leaf]

yep! :))

[lost flax]

okayy

[proven leaf]

it's the special function

[surreal meadow]

||so is 0||

[lost flax]

yay cool!

thank you guys sm!

.close

[lone heart]

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[proven abyss]

Anyone know which option this is?

[lone heart]

@proven abyss Has your question been resolved?

[lone heart]

@proven abyss Has your question been resolved?

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

[lone heart]

@proven abyss Has your question been resolved?

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[heady finch]

how do I do this question

[heady finch]

I've already done
A = (a+b)/2
A' = (b+c)/2
G = √ab
G' = √bc

Dunno how to proceed with that though

[lone heart]

@heady finch Has your question been resolved?

[heady finch]

<@&286206848099549185>

.close

[lone heart]

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[opaque kelp]

So here I got my tq correct 21.7/21.651 but afterwards I'm wrong not sure why

[opaque kelp]

I did 180-45-90 to find angle r=45

Then I did 180-45-60-45=30

Since I need to find pt aka the adjacent side and ik the opposite side =25cm

I did tangen30°=opp/adj=25/PT=25.651/PT which PT=44.4 at the end

But the ans is 37.5 which I'm not sure what I did wrongly

[random gorge]

,rotate

[random gorge]

this is wrong

[opaque kelp]

my working or?

[random gorge]

tan 30 = opp/adj = TQ / PT

TQ isnt 25

[opaque kelp]

no tq is 25.651

[random gorge]

its 25 (sqrt3)/2

[opaque kelp]

oh

[random gorge]

but its surely lees than 25

[opaque kelp]

oh wait

its 21.7

[opaque kelp]

its 21.7

but my teacher says i have to do it in 5 sf value

which is 21.651

[random gorge]

yeah right

[opaque kelp]

but i dont understand where i am wrong

not sure why my PT is wrong

[random gorge]

but i suggest you to write it in this form and at the end convert sqrt 3 to 1.7

[opaque kelp]

oh hmm ok

[random gorge]

solve it again with 21.7

[opaque kelp]

alright

its 37.6

but the ans key says 37.5

er idk wat happening....

[random gorge]

because you converted sqrt 3

[opaque kelp]

i didint square root anything tho?

i just use the calc not rlly sure whats the value of tangen

since i am allowed to use it

[random gorge]

tangent 30 = (sqrt 3) / 3

use this

[opaque kelp]

ohjhh

i did it wronly...

i used tq as 25

mb mb sorry-_- yea it correct now

er sorry for disturbing u:)

ty tho

.close

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[lone heart]

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[fathom harness]

Lofi girl back to doing homework

[fathom harness]

.close

[lone heart]

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[tidal fog]

some1 plz

[vale wigeon]

which one(s) are troubling you?

[tidal fog]

all of them basicly