help-33

[whole linden]

it is

[quaint jungle]

oh then yes, i have 3 solutions

whats the next step?

[marsh citrus]

@quaint jungle Has your question been resolved?

[quaint jungle]

.close

[marsh citrus]

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[nova totem]

What have you tried

[karmic herald]

my first idea was to simplify but these numbers would be decimals if you applied the exponents on them

[nova totem]

Keep it as fractions

Simplify the denominator

[karmic herald]

oh

so 1/3 x 1/4

right

simplifies to 3^7/12

oh lol this question is way easier than I thought it was

thank you

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[indigo spade]

need to know which function these potent rows are

[dusky wing]

I am not clear on the question. But don't they both evaluate to 1/x, with the first one for |1 - x| < 1 and the second one for |1 + x| < 1

[indigo spade]

h

okk, how do you know how it is1/x?

[dusky wing]

1 + x + x^2 + ... = 1/(1-x) for all |x| < 1

So the first one becomes 1/(1-(1-x)) = 1/x
And the second one becomes -1(1-(1+x)) = -1/-x = 1/x

[indigo spade]

ok but this is just for the are where it converges right

[dusky wing]

Yes, the first one is only for |1 - x| < 1, which is the interval (0, 2)
And the second one is only for |1 + x| < 1, which is the interval (-2, 0)

[indigo spade]

is it possible to get a function for all x

[dusky wing]

Neither converge for all real numbers x

[indigo spade]

or would that just be a taylor ponynom

[dusky wing]

They are already taylor polynomials

[indigo spade]

yea but i mean like no matter if it converges

[dusky wing]

I am not sure what you mean. These sums diverge outside certain intervals

[indigo spade]

alright thanks

[marsh citrus]

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[cloud kettle]

Is it true to say that x is a subset of f(x) like domain being subset of range?

[devout mauve]

domain is generally not a subset of range

x and f(x) are elements, not sets, so neither is a subset of the other

[cloud kettle]

I see thank you for clearing this confusion of mine

Actually I was revising the composite function and I had written a note that domain of f(g(x)) is a subset of g(x)

Since what you said seem to be applied there do I have to cross that sentence from my note?

[devout mauve]

for $f\circ g$ to be defined, the range of $g$ needs to be a subset of the domain of $f$

[elfin berry]

Denascite

[devout mauve]

cause if g(x) wasn't in the domain of f, what should f(g(x)) mean

[cloud kettle]

I see

It is necessary for g(x) to be subset of x

Thank you

[devout mauve]

again, it makes no sense to say subset of x

x is an element

[cloud kettle]

right so it should range of g(x) to be subset of domain of f(x)

Am I right this time?

[devout mauve]

yes

[cloud kettle]

Ok I'll close the channel then

.close

[marsh citrus]

Channel closed

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[cloud kettle]

.reopen

[marsh citrus]

✅

[cloud kettle]

Is domain of f(x) and domain of g(x) would be same thing in f(g(x)) or different?

[devout mauve]

the domain of $f\circ g$ is the same as the domain of g

[elfin berry]

Denascite

[devout mauve]

cause you put the elements into g first

[cloud kettle]

Yeah

[devout mauve]

and the codomain of $f\circ g$ isthe same as the codomain of $f$

[elfin berry]

Denascite

[devout mauve]

cause they come out of f in the end

[cloud kettle]

Indeed

but if I say like arcsin(1/x) domain of 1/x would be R-{0} but domain of arcsin(1/x) would be something else

[cloud kettle]

Did I got you wrong here?

Because you said their domain same

[devout mauve]

you need to restrict the domain of 1/x so that it fits

when just writing formulas, the domains and codomains are often implicit

[cloud kettle]

But that's as f(x) not g(x) seperately

[devout mauve]

which is not great

I was talking about if $f$ and $g$ are defined as functions $f:A\to B, g:C\to D$

[elfin berry]

Denascite

[devout mauve]

I hope you have seen that notation before?

[cloud kettle]

Well it's 0 percent loaded. Let me refresh discord

[devout mauve]

f:A->B, g:C->D

[cloud kettle]

Yes

A is codomain and B is range

[devout mauve]

no

A is the domain of f, B is the codomain

the range of f is a subset of B

[cloud kettle]

Ah sorry

You're right

[devout mauve]

anyway, the point is when you write functions as formulas, stuff like arcsin(1/x) like you did, the sets A,B,C,D are not explicitly given

[cloud kettle]

I understand

[devout mauve]

only implicitly

and you have to choose them appropriately so that everything fits

[cloud kettle]

From what I know about implicit and explicit function is explicit is in terms of x only and implicit involves other variable

So I'm confused what you mean by implicity here

[devout mauve]

just like in the english sense

[cloud kettle]

Ohh

[devout mauve]

it's not explicitly given

[cloud kettle]

I understand

Thank you!

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[lucid hinge]

prove this

[lucid hinge]

@devout mauve mr denascite

mr, mr

im disappointed mr

mr riemann

[main idol]

Stop pinging individual helpers

[lucid hinge]

[smoky plover]

Cant you just prove this with quadratic formula

[lucid hinge]

prove it

[main idol]

Is f a polynomial of degree n>1?

[smoky plover]

Oh I only considered quadratic oops

[lucid hinge]

yes

n>=1 would work right??

[smoky plover]

No

n=1 would only have 1 root

There’s a proof on Wikipedia

[lucid hinge]

no n>=0 would all work

[lucid hinge]

so it doesnt have the root in the form of a+sqrt b

so it doesnt matter

[smoky plover]

No it could have a root in the form a+sqrtb, but it only has one root so it can’t have a-sqrtb

[hushed egret]

@sharp vessel

[lucid hinge]

why can u explain

[sharp vessel]

ty

[lucid hinge]

no it still works

source: snow

yeh u cant have have a root of a+sqrt b for degree one

if coefficinets must be rational

redstone is right

[sharp vessel]

alright so deg(f) >= 2

[lucid hinge]

no it applies to deg(f)>=0

degrees 0,1 just don't have roots in the forms of a+sqrt b

[sharp vessel]

so?

[hushed egret]

b=0

[sharp vessel]

0 is a perfect square..?

wtf is going on

[hushed egret]

sure

then there arent any

[lucid hinge]

@hidden plaza talk

[sharp vessel]

there isn't a polynomial of degree 1 that satisfies the condition of the question

[elfin berry]

i love to boki

[hushed egret]

if if if if if if if

[sharp vessel]

therefore we know deg(f) >= 2

[lucid hinge]

[hidden plaza]

Q[sqrt(b)]

[sharp vessel]

am I missing something

no

pure

stop

[hushed egret]

Q[sqrt(b)]

[sharp vessel]

no pure

[hidden plaza]

[tight furnace]

yes n>=0 would work

[lucid hinge]

moni

[tight furnace]

anyone who doesnt think so needs to learn about things being vacuously true

[sharp vessel]

moni can u explain

how does a polynomial of degree 1 has a + sqrt(b) as a root?

[tight furnace]

it doesnt

thus all polynomials that do also have a - sqrt(b) as a root

[hushed egret]

therefore the theorem is true

[sharp vessel]

ok lmao

lets just consider degree 2 or more now

[hushed egret]

doesnt matter

[sage heath]

[sharp vessel]

(2) is easier to prove than (1)

a

i'll let u guys continue im bad

[sage heath]

so is here

[tight furnace]

2 isnt easier purely for the multiplicity reason

otherwise the proofs are identical pretty much

[hushed egret]

multiplicity is yuck

[lucid hinge]

ax+b=0, x=-b/a. Why can't -b/a = a+sqrt b?

[tight furnace]

????

[lucid hinge]

[hushed egret]

[tight furnace]

b and a are both rationals

-b/a is also a rational

[lucid hinge]

right

so what

is a+sqrt b not rationla

[tight furnace]

if b isn't a perfect square then its not rational

[lucid hinge]

ok ig

[lucid hinge]

wdym

[tight furnace]

the second question requires you to show it has the same multiplicity

the first one doesnt

[sage heath]

[lucid hinge]

right

[tight furnace]

since the proofs are otherwise pretty much identical that makes the first one easier

[lucid hinge]

how to prove

snow said binomial expansion

[hushed egret]

eh

probably dont need actually

[tight furnace]

sort of

[lucid hinge]

[tight furnace]

this is a relevant fact

[still temple]

lmfao i love how the whole gang from mg is here

[sage heath]

[lucid hinge]

[hushed egret]

[lucid hinge]

i previously thought this is a physics formula

[still temple]

pure lurking too

[lucid hinge]

[sage heath]

[sharp vessel]

chris

[sage heath]

it shud be renamed as purelurk

[lucid hinge]

so how

[sage heath]

[hidden plaza]

Redstone won’t let me talk

[sharp vessel]

whatt

[hidden plaza]

[sharp vessel]

oh

[sage heath]

[sharp vessel]

Q[sqrt(b)]

[lucid hinge]

[sharp vessel]

like is that really the proof u want to give to chris

does chris know what even that is

[lucid hinge]

q is rational

number

[sharp vessel]

a

[tight furnace]

its field theory

[hushed egret]

just prove the properties

[tight furnace]

lets not

[lucid hinge]

ok how to prove

[sage heath]

[sharp vessel]

fields ayy

[tight furnace]

ok imagine a polynomial such that f(a+sqrt(b)) = 0

[sharp vessel]

field extension

[main idol]

Boki got raided hard. Don't even know the question anymore

[lucid hinge]

@sage heath i challenge u with this question

[elfin berry]

i love to boki

[main idol]

Oh it's in the pins?

[sage heath]

complex conjugate root

[tight furnace]

show conjugates of roots are also roots essentially

[sage heath]

[sage heath]

yea

[lucid hinge]

bruh

thats literally the question

[hushed egret]

you only need to prove like

[tight furnace]

i was telling riemann yes

[lucid hinge]

so

how

[tight furnace]

imagine a polynomial such that f(a+sqrt(b)) = 0

well

see if you can write f(a-sqrt(b)) in terms of f(a+sqrt(b))

[lucid hinge]

yeh i get that

product too

not only sum works like that

[hushed egret]

lol you cant see the space

[lucid hinge]

stop copying what i said smh

[hushed egret]

and then you're done

[lucid hinge]

[sharp vessel]

not copying if u write it in latex

[lucid hinge]

?????

[hushed egret]

cuz you just conjugate the whole thing

[lucid hinge]

wtf

[upper comet]

Use $\overline{az + bz^2 + ...} = a\overline{z} + b\overline{z^2} + ...$

[lucid hinge]

wtf lol

i gotta switch classroom

[upper comet]

Euh you can see what I mean

[sharp vessel]

snow whats ur macro for conjugate

oh ok

[sage heath]

gone

[sharp vessel]

\conj

[lucid hinge]

u usre

[sharp vessel]

yes they are sure

[elfin berry]

Castroploiin¹

[lucid hinge]

what is the line above f mean

above the function

f

[hushed egret]

conjugation

[upper comet]

Conjugate

[sharp vessel]

conjugate

[lucid hinge]

switch classroom

i know

[sharp vessel]

[lucid hinge]

BRUHudsfkjd

1 sec

[hidden plaza]

I mean that’s field theory

[sharp vessel]

conjugate is highschool stuff

[sage heath]

[sage heath]

is it qft

[hushed egret]

[main idol]

Does this follow from Vieta?

[sharp vessel]

no

[main idol]

The second equal sign

[sharp vessel]

from properties of conjugate

[hushed egret]

it follows from both of the things i posted before

[sharp vessel]

the polynomial is just a sum

[hushed egret]

cuz f is polynomial you just shove the conjugate through

[main idol]

Oh this

[sharp vessel]

so u conjugate each term

[hushed egret]

ye

[sharp vessel]

and then the coefficients stay the same cuz they are real

[hushed egret]

rational*

[sharp vessel]

(and not fake)

[sharp vessel]

oh right

[hushed egret]

isfake

[sharp vessel]

this isn't complex

ye

they aren't fake

[hushed egret]

isreal

[sharp vessel]

what does that have to do with me being dumb

[hidden plaza]

In disguise

[main idol]

I poop Jerusalem cuz my shit isreal

[hushed egret]

nothing

[sharp vessel]

oh i see the pun

bruh

chris does it all the time wtf am i thinking

[sharp vessel]

ok no

thanks but no

[lucid hinge]

im back

[hushed egret]

hi back

[sage heath]

where front

[lucid hinge]

ok so like

@sharp vessel if u conjugate the roots, that's equal to conjugating the entire

polynomial

[sharp vessel]

ye

if its easier for u

then think about the polynomial

and plug in a + sqrt(b)

and equate to 0

you know thats true

now, conjugate both sides

conjugate of 0 is 0

and on the left, the only thing that changes are the roots

like

ye

[lucid hinge]

i see

[sharp vessel]

and then boom boom boom u proved

little square

QED

sign it

publish it

[sharp vessel]

$\filledsquare$

[elfin berry]

RedstonePlayz09
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp vessel]

dottedsquare

linedsquare

u need to add macro for this snow

[lucid hinge]

nice i get it

thx guys

.close

[marsh citrus]

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[sage heath]

[marsh citrus]

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[still temple]

In the puzzle sudoku, one fills in a 9 by 9 square grid with numbers from the set 1,2,3,4,5,6,7,8,9, with exactly one occurrence of each of these numbers in each row, one in each column, and one in each of the nine 3 by 3 boxes. Suppose that you are constructing such a puzzle, and have filled in the upper left 3 by 3 box completely. How many possible arrangements of numbers remain for the upper right 3 by 3 box?

[still temple]

I dont even know where to begin

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

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[still temple]

.reopen

[marsh citrus]

✅

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[shell trellis]

dont get this

lemme show my work where im at

[ionic stirrup]

you made a typo in the second line

should be -2a^3 - 4a^2 - 2a^2 - 4a

then factor from there

[shell trellis]

anyone? why does that become a 2 from a 3?

[lunar magnet]

because 6y will multiply with y^2

6y will multiply with every term between parentheses

[marsh citrus]

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[short wadi]

hey guys so simple question, how do you calculate the distance between a vector and a subspace? I know how to do it from vector to vector, but the question 4b) asks me the distance between the vector x and subspace W..

[marble glacier]

https://www3.nd.edu/~dyer/Teaching/ex3_soln.pdf

First page

[marsh citrus]

@short wadi Has your question been resolved?

[short wadi]

thank you

a lot

.close

[marsh citrus]

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[short wadi]

.reopen

[marsh citrus]

✅

[short wadi]

I didnt know y hat was always the vector i would use to find the distance to x

.close

[marsh citrus]

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[flat grail]

Hi! Could anyone help me with an elastic collision question please, I've done part a

[flat grail]

This is the workings for part a (might be helpful)

I figured doing where gamma is deflection might help, but I got no where

[marsh citrus]

@flat grail Has your question been resolved?

[marsh citrus]

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[naive oasis]

hi

[naive oasis]

i need help in a question

(2x^2+3y^2) = length , (x^2 - 2y^2\3) = breadth find area of rectangle

[stoic saddle]

what does that backslash mean?

[naive oasis]

i got answer as

[slow thicket]

Show work.

[naive oasis]

divided by

[naive oasis]

sure

[stoic saddle]

that's usually written with a forward slash.

i believe i've pointed this out to you before.

[naive oasis]

mam and sir plz wait imma show you guys my work okay?

[stoic saddle]

maybe i misremember.

[slow thicket]

Hi, I need help with a question. It states the following: let $2x^2+3y^2$ denote the length and let $x^2-\frac{2y^2}{3}$ denote the breadth.

I need to find the area of a rectangle. Here is what I’ve done so far:
(Add work)

I’m stuck on:

(Add what you’re stuck on/ don’t understand).

I’d love getting help.

Thanks in advance.

[elfin berry]

Heisenbug

[slow thicket]

Use such a structure. It will make it more accessible and people would much rather answer such questions.

[naive oasis]

i just wanted to clarify that is this answer is correct or have i done wrong

[stoic saddle]

could be simplified further, but no mistakes thus far.

[naive oasis]

if i had done wrong plz let me know helpers

[naive oasis]

does that mean this answer is correct?

[stoic saddle]

no, it means that you aren't done.

[slow thicket]

It means that you have no mistakes, yet you need to keep on going.

[naive oasis]

idont know how to simplify this more can you help me

[naive oasis]

thank you heisenburg for letting me know the meaning of the sentence

[stoic saddle]

$-\frac{4}{3} x^2 y^2 + 3x^2 y^2 = ; ?$

[elfin berry]

Ann

[naive oasis]

x^2 y^2/3

[stoic saddle]

incorrect

surely you would not argue that $-\frac{4}{3} + 3 = \frac{1}{3}$?

[elfin berry]

Ann

[naive oasis]

uh

nope

[stoic saddle]

then why do you claim that $-\frac{4}{3} x^2y^2 + 3x^2y^2 = \frac{1}{3}x^2y^2$?

[elfin berry]

Ann

[naive oasis]

sorry leamme correct

its

13 x^2 y^2/3

[stoic saddle]

-4/3 + 3 = ?

[naive oasis]

5

5/3

[stoic saddle]

yes, correct. -4/3 + 3 = 5/3.

[naive oasis]

yes

so

sorry

i didnt saw "-" sign in the question thats why i said 13

[stoic saddle]

and then you also failed to see it in my message

or you saw it but chose not to ask me where it came from

[naive oasis]

im sorry

btw

thanks ann for helping me

.close

[marsh citrus]

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[potent dove]

Hello Guys can someone help me with this problem?
At 12 noon, ship A travels due west at 100 knots. After 4 hours, ship
B leaves the same port on the direction N30◦W at 80 knots. How
fast is the distance between the two ships changing at 8 P.M. in the
same day?

[potent dove]

Hi

[marsh citrus]

@potent dove Has your question been resolved?

[marsh citrus]

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[potent dove]

.reopen

[marsh citrus]

✅

[potent dove]

<@&286206848099549185>

[brittle vessel]

draw a picture, it will help

[proud ice]

what have you tried

where are you stuck

[marsh citrus]

@potent dove Has your question been resolved?

[upper comet]

Find an expression for the distance between them

[marsh citrus]

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[karmic pilot]

Apparently I'm unclear as to what a rational function is. Could someone help clarify this for me? I posted a screenshot of a quiz, but don't worry this is not a graded quiz (I can show further proof of this if necessary). This is a "quiz 0" meant to be used as a refresher for my calc course.

[karmic pilot]

I thought a rational function would be anything that could be expressed as a fraction. Am I wrong in thinking the only ones that wouldn't fit this are the ones with square roots or fractional exponents

[late geode]

have you tried googling it

[karmic pilot]

Yes

[upper comet]

[late geode]

and what came up when you looking it up

[karmic pilot]

I spent 10 minutes reading about it and I still can't come up with an answer

[late geode]

copy what's written word for word

[karmic pilot]

In mathematics, a rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials. The coefficients of the polynomials need not be rational numbers; they may be taken in any field K

[upper comet]

both the numerator and the denominator are polynomials.

[late geode]

and if you dont know what a polynomial is, you'd need to look that up

[karmic pilot]

everything I selected seems to apply

[upper comet]

b. can be simplified
c. Just a polynomial
h. |x| doesn't count

[obsidian shard]

nope, at the very minimum A is wrong

[late geode]

you've also missed one that you should've checked

[karmic pilot]

yea I saw b can be simplified, but 1/1 is still a fraction

[upper comet]

A is fine I think

[late geode]

A is wrong

[obsidian shard]

How is A fine?

[karmic pilot]

I didn't know that absolute value doesn't count though

[late geode]

and if you dont know what a polynomial is, you'd need to look that up

[upper comet]

Does not simplifying it make it not a rational function?

For a.

[karmic pilot]

I mean I've watched hour long videos on what a polynomial is and read about it, apparently I tsill don't get it so.

[obsidian shard]

I don't think simplification is allowed

[upper comet]

b would count then

[karmic pilot]

Google says even a function f(x) = 1 is considered to be a rational function

because it can be expressed as 1/1

[upper comet]

It's probably interpreting "rational" differently then

By definition, 'rational functions' are functions that are a ratio of two polynomials

[karmic pilot]

I am still reading stuff and can't get the answer right

I thought it might be c, f, g

or b,c,f,g

but entering those doesn't work

[upper comet]

Why c?

[upper comet]

^

5 doesn't count as a polynomial generically

[karmic pilot]

This is what I get when I "just google it"

[upper comet]

Constant terms are constant * x^0; their degree is 0

[karmic pilot]

This doesn't count because the denominator has a negative exponent

This maybe counts? but can just be simplified to 1

[upper comet]

"In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and positive-integer powers of variables."

[karmic pilot]

This doesn't count because denominator is a constant

[upper comet]

There's always going to be conflicting definitions, but I have rarely seen constant terms pass as polynomials when it matters

[karmic pilot]

This doesn't count because the denominator has a fractional exponent

This doesn't count because of the sqrt

[late geode]

constants are polynomials too

[upper comet]

Why not?

[late geode]

don't discriminate just because there's no variable

[upper comet]

My bad

[karmic pilot]

the sqrt(3*x^2) counts?\

[upper comet]

It's √3 * x^2

[late geode]

the root only applies to the 3

[karmic pilot]

Oh

Maybe that's the answer I've been missing then

[late geode]

as that's where the bar stops

[upper comet]

If math is typeset like that, you can be sure in assuming the root ends where the bar does

Handwriting is different obviously

[karmic pilot]

These were the correct answers

The constant values counted as polynomials

and as a rational function as well apparently

Thanks for the help, I think while the varying definitions of rational and polynomial functions confused me, the big problem I had was thinking that one expression was sqrt(3x^2)

.close

[marsh citrus]

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[still temple]

hi, this is a past paper from my uni's group course, not homework

it's mainly the first part i'm confused on

wait can't i just

cauchy

that's embarassing

[mystic minnow]

you probably can. You then have to prove that theorem.

[still temple]

yeah, thanks

can someone help me show direct productness?

i can't seem to prove that all elements of G can be expressed as <x>H

[marsh citrus]

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[marsh citrus]

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[lucid hinge]

how is this simplified

just the Tn step

q* the fraction

.close

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[ivory laurel]

I can’t seem to find a website explaining what addition is in predicate logic

[ivory laurel]

I haven’t come across ii before

[still temple]

the xor?

$\oplus$

[elfin berry]

hibyehibye

[ivory laurel]

Yes

[still temple]

its same as logical OR but 1 and 1 is 0

its exclusive

[ivory laurel]

So it’s the exclusive or?

[still temple]

yeah

[ivory laurel]

Essentially or without the true and true

[still temple]

mhm

00 0
01 1
10 1
11 0

[marsh citrus]

@ivory laurel Has your question been resolved?

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[wraith gull]

Can I get some help on what to do next on this circle geometry

[wraith gull]

So I know I could find the equation of the normal but I would be missing the y-int

[marsh citrus]

@wraith gull Has your question been resolved?

[marsh citrus]

@wraith gull Has your question been resolved?

[wraith gull]

No you wanker

.close

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[flat aspen]

Where did I go wrong in my working for, $\frac{1-x}{2}=\frac{3-x}{4}+5x$?

[elfin berry]

kangaroo rat

[main idol]

where did your second equation come from?

1/2 turned into a 2?

[flat aspen]

idk

thanks

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[hard ivy]

is there a way i can cancel out a (x+6)^6

[raw sandal]

Let the consecutive angles be AOB, BOC and COD, such that the largest of them is the angle AOB and the smallest is the angle COD. If m<AOB - m<COD = 28⁰, the measure of the angle formed by the bisectors of the angles AOD and BOC is:

[main idol]

!help

[marsh citrus]

Please read #❓how-to-get-help

[main idol]

Factor it out from the two terms

[hard ivy]

wdym

how would that look

[dusk coral]

$$6(x+6)(x+6)^6+3(x+6)^6$$

[elfin berry]

Bestower

[dusk coral]

can you take it from here?

[hard ivy]

hopefully

.close

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[hazy idol]

$\rm{Var}[X+Y]=\rm{Var}[X]+\rm{Var}[Y]+2\rm{Cov}(X,Y)$ no?

[spark hollow]

Got that but does the minus change anything?

[elfin berry]

SkyTwX

[hazy idol]

What about $\rm{Var}[aX]$? Choose a=-1

[elfin berry]

SkyTwX

[spark hollow]

Ah the variance is the same true

So I just use that twice?

Var(X+Y-Z)=Var(X+Y)+Var(-Z)

[marsh citrus]

@spark hollow Has your question been resolved?

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[lucid hinge]

[marble glacier]

Ooof

I’m guessing that means no ln?

[ionic lava]

ax² + bx + c = (x - 1)² 🤔

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@lucid hinge Has your question been resolved?

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[nova dew]

Can anyone explain me what directional derivative is?

[nova dew]

I know the formulas and the definitions but I don't understand it from the core

[marsh citrus]

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[marsh citrus]

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[still temple]

how do I solve this

[fathom ridge]

You can write any straight line in the form of y = mx + c

Just chuck in a few values as per your convenience and solve for m and c

[still temple]

I don't know how to do that.

.close

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[fathom ridge]

[marsh citrus]

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[somber forum]

Can someone help me with a proof by induction. I am An orientation of a graph G=(V,E) is any directed graph G'=(V,E') arising by replacing each edge {u,v}∈E either by the directed edge (u,v) or by the directed edge (v,u).
Prove by induction that for every planar graph there is an orientation such that each vertex has at most five outgoing edges. I already proved that a undirected graph can have at most degree 5 to be planar but i am not sure how to get to the directed graph and how to make the prove by induction.

[marsh citrus]

@somber forum Has your question been resolved?

[somber forum]

<@&286206848099549185>

[marsh citrus]

@somber forum Has your question been resolved?

[marsh citrus]

@somber forum Has your question been resolved?

[marsh citrus]

@somber forum Has your question been resolved?

[marsh citrus]

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[gray shale]

there's nothing that i didn't get with limited development
for instance we know the development of sqrt(i +x)

[gray shale]

,w taylor sqrt(1+x) in 0

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Max.cc#5208

[gray shale]

to get the devoplment at the point 2 instead of 1 we do :

sqrt(2 + h) with h = x -2
so sqrt(2(1 + h/2))

so sqrt(2) * sqrt(1 + h/2) and so can develop the right expression (we just replace x by h/2)

but why does it work

i means sqrt(2 +h) = sqrt(0) so ....

[marsh citrus]

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[sick nest]

how do I solve this

[distant peak]

implicit differentiation i believe

[sick nest]

y^2=2x+4-x^2

y=sqrt(2x+4-x^2)

dy/dx*y=sqrt(2x+4-x^2)

dy/dx*y=1/2(2x+4-x^2)^-1/2

[distant peak]

$2x + 2y\frac{dy}{dx} = 2$

first derivative

[sick nest]

dy/dx*y=1/2sqrt(2x+4-x^2)

[elfin berry]

Ｈｅｒｅｌｓ

[sick nest]

2x+2ydy/dx=2

[distant peak]

?

[sick nest]

how about the d^2y/dx^2

thing

[cobalt sentinel]

U have to first isolate dy/dx from this, then differentiate the whole thing 1 more time

[sick nest]

2.2y(dy/dx)=2-2x
3.dy/dx=2-2x/2y

[distant peak]

you differentiate again both side

without forgetting chain rule and product role

[sick nest]

4.dy/dx=((-2)(2y))-((2-2x)(2y*dy/dx)/2y^2

then i'd do the product rule

I mean

I wouldn't

[elfin berry]

Stephen

[sick nest]

how about for d^2y/dx^2

[cobalt sentinel]

Now use quotient rule

[sick nest]

okie

[cobalt sentinel]

Do u see how I got this

[sick nest]

by getting rid of the 2

factoring out the 2

[cobalt sentinel]

Yea

[sick nest]

-1(y))-((1-x)(y)/y^2

-y-y+yx/y^2

[cobalt sentinel]

Can u just do it on paper or latex and send

[sick nest]

yeah

[cobalt sentinel]

Don’t forget to do the implicit differentiation chain rule

[cobalt sentinel]

U messed up at the y•(dy/dx) part, the derivative of y is not that

[sick nest]

what is it

[cobalt sentinel]

Think abt it, what’s the derivative of ‘y’ by itself

[sick nest]

y

[cobalt sentinel]

No

What’s the derivative of x by itself

[sick nest]

1

[cobalt sentinel]

So why not the same for y

[sick nest]

good point

so im now here

[cobalt sentinel]

No, u forgot the dy/dx lol

[sick nest]

😭

[cobalt sentinel]

Derivative of y is 1, but u have to chain rule

[sick nest]

one sec ill fix it

[cobalt sentinel]

Check ur signs in numerator

[sick nest]

right sides supposed to be a plus

x

[cobalt sentinel]

Yep

[sick nest]

then what do I do

[cobalt sentinel]

Now set that equal to d^2y/dx^2

And for the dy/dx’s, sub back in our original value of dy/dx

[sick nest]

?

[sick nest]

.

this one right

[elfin berry]

Stephen

[cobalt sentinel]

This is what u got originally right

[sick nest]

yeah

then I distributed the x in

and then brought the y's down

[cobalt sentinel]

How did u get y^4 in the denom

[sick nest]

I brought the y's down lol

[cobalt sentinel]

The denominator of the numerator’s fraction should be a ‘y’

Not a y^2

[sick nest]

oh what I should've did

is combined those

and then rbought the y down

[cobalt sentinel]

Also, u forgot to give the -y term in the numerator a y denominator

[sick nest]

because they aren't seperate

y's

[sick nest]

wdym

[cobalt sentinel]

U can only bring the y down if every term in the numerator has a y denominator

Otherwise it’s incorrect

[sick nest]

ooh

should I make it to where it has a common denominator

on the top

in the numerator

[cobalt sentinel]

Yes

[sick nest]

ik how to do that I think

its supposed to be -y^2

mb

am I able to get rid of the -y^2 in the numerator

yeah I did

and by doing that I got

-1+2x-x^2/y

[cobalt sentinel]

Wat

No

Those fractions in the numerator aren’t being multiplied

Get everything to have one common denominator

Like combine the fractions

[cobalt sentinel]

From here

[cobalt sentinel]

👍

[sick nest]

then what do I do

[cobalt sentinel]

Nothing much u can do, other than factor those last 3 terms in the numerator

Doesn’t rlly matter tho

[sick nest]

these are the options

[cobalt sentinel]

Ok yea so u have to factor

Immediately which ones can u cross out?

From the answer choices

[sick nest]

top one

[cobalt sentinel]

And

Hint: denominator

[sick nest]

b

[cobalt sentinel]

And

[sick nest]

and c

[cobalt sentinel]

Ye

So down to d and e

[sick nest]

I mean e

mb

[cobalt sentinel]

Now just factor these last 3 terms in the numerator

[sick nest]

how would I do that without factoring y

as well

[cobalt sentinel]

Just don’t include the y

[sick nest]

you can do that?

[cobalt sentinel]

Just factor the quadratic x

Yea

Why not? It’s not affecting anything else

[sick nest]

because they aren't their own nomial

[cobalt sentinel]

They are

[sick nest]

so you said factor out x

-1+2-x

-1-x

[cobalt sentinel]

No, I mean like

[sick nest]

1-x

?

[cobalt sentinel]

Same way u would factor x^2 + 8x + 15

Like that

Oh ok now I see what u were thinking, this is what I meant^

[sick nest]

its oriented oddly lol

-x^2+2x-1

then I factor out the -

im assuming

-(x^2-2x+1)

-((x+2)(x-1))

[cobalt sentinel]

Nope

One thing, notice how in the answer choices, the y^2 in the numerator is positive, thus, try factoring out a -1 from the whole numerator

[sick nest]

(x+2)(x-1)?

[cobalt sentinel]

No

[sick nest]

I can't factor x out of it all so what do you want me to do 😭

I mean I can but it would look weird

[cobalt sentinel]

No I’m saying to factor the x^2-2x+1

[sick nest]

yeah that's what i've been factoring

[cobalt sentinel]

That’s not the right answer

[sick nest]

so it should multiply the equal 1 not -2

so (x-1)(x-1)

is the ans

so it would be -[y^2(x-1)^2]/y^3

-y^2-(x-1)^2/y^3

[fickle gust]

https://tenor.com/view/among-us-sus-yhk-among-twerk-among-us-twerk-gif-23335803

[sick nest]

Breh

[cobalt sentinel]

Well ur answer says (1-x)^2 so something’s off

[sick nest]

u right

can I flip it?

[cobalt sentinel]

Nah we can’t do that

Lemme check smth

[sick nest]

kek

[cobalt sentinel]

Ok even derivative calculator says we’re correct from this point, double check to see if we missed any signs anywhere

Wait a second

[cobalt sentinel]

Do u see how this became that^

I’ll brb gimme 2 mins, see if u can figure it out

[sick nest]

ok

[cobalt sentinel]

U understand?

[sick nest]

got rid of a y

and put it on the bottom

I mean

got rid of the y on the top

and divided it

to the bottom

[cobalt sentinel]

No, they just factored out the dy/dx from those two terms

dy/dx = y’ btw

[sick nest]

makes since

[cobalt sentinel]

Ok u understand?

[sick nest]

im just trying to think of where we went wrong

with it

I mean which answer choice

[cobalt sentinel]

Real quick, just start from here

Remember it’s supposed to be a +x

[sick nest]

ok

[cobalt sentinel]

So [-y + dy/dx(x-1) ]/y^2

Yes?

[sick nest]

yes

[cobalt sentinel]

Now just sub in our dy/dx

[sick nest]

ok

[cobalt sentinel]

Becomes [-y + (1-x)(x-1)/y]/y^2

[sick nest]

[-y+x-1/y(x-1)/y^2

yep

we take out the -?

[cobalt sentinel]

Which is [-y - (1-x)(1-x)/y]/y^2

U see what I did?

[sick nest]

yeah

[cobalt sentinel]

Now see if u can go from there

[sick nest]

-y-(1-x)^2/y^3

?

[cobalt sentinel]

-y^2 first term

Remember everything common denominator

[sick nest]

yeah

-y^2-(1-x)^2/y^3

[cobalt sentinel]

Now see ur answer choice?

[sick nest]

d

?

coolio

thx thx

for your time

[cobalt sentinel]

Np

[sick nest]

.close

[marsh citrus]

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[ashen flicker]

How did they get first step

?

[lucid hinge]

they squared both sides

to get rid of the absolute signs

[ashen flicker]

I dont understand

why would squaring get rid

lol im dumb

[lucid hinge]

cause say

|-6|

if u square a negative number

it becomes positive

if u square a positive number

it also becomes positive

[ashen flicker]

oh ok

ah

.close

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[ashen flicker]

thank you

[lucid hinge]

@devout mauve ur not welcoming tho

sir denaste

[marsh citrus]

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[lucid hinge]

can i get help in using the methods in pic 2 to solve example 4?

||@devout mauve ||sir

[north sparrow]

so, have you gotten some sample values?

[marsh citrus]

@lucid hinge Has your question been resolved?

[marsh citrus]

@lucid hinge Has your question been resolved?

[lucid hinge]

napping will be back

[marsh citrus]

@lucid hinge Has your question been resolved?

[lucid hinge]

still napping

[marsh citrus]

@lucid hinge Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[karmic herald]

forgor how to do this 💀

[karmic herald]

nvm bruhs too long

.close

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[karmic herald]

nvm bruh didn't know how to do it

[karmic herald]

.reopen

[cobalt sentinel]

U know how to cross simplify?

[karmic herald]

yup

[cobalt sentinel]

Do that

[karmic herald]

would it be where it's like a/b x a/b = ab = ab?

[cobalt sentinel]

Not sure what u mean

U can simplify every term in those fractions

[karmic herald]

not so sure if I understand

[cobalt sentinel]

So like for example

[elfin berry]

Stephen

Stephen

[cobalt sentinel]

U see how that happened?

[karmic herald]

10/6 simplified is 2/2?

[cobalt sentinel]

No

I cross simplified

The 6 and the 3 became 2 and 1

The 10 and 5 became 2 and 1

You could just multiply the numerator and denominator right off the bat but you’d have to do simplifying after rather than before

If u don’t wanna do this method, then just multiply and simplify after

[karmic herald]

would it be cross multiplying or straight multiplying

in this scenario

so like multiplying denom with denom

num with num

[cobalt sentinel]

I mean yea if ur going to simplify later

[karmic herald]

so in this case it'd be $\frac{12xy^2}{24xy}$?

[elfin berry]

Zyme><SOL

[cobalt sentinel]

Yes

[karmic herald]

which simplifies to $\frac{2xy}{4}$?

[elfin berry]

Zyme><SOL

[cobalt sentinel]

No

x/x is what

[karmic herald]

1, but 2-1 is also 1

we have a squared, not sure if that applies only to the y

[cobalt sentinel]

Yes only to the y

[elfin berry]

Stephen

[karmic herald]

so it'd actually be $\frac{12y}{24}$

[cobalt sentinel]

No

[elfin berry]

Zyme><SOL

[karmic herald]

forgot to remove the x

[cobalt sentinel]

Yep that’s correct