help-33

[ionic owl]

I mean circumference is easy enough, you'd have 2pi * 5 = 10pi and 2pi * 2 = 4pi

[strange hazel]

you will use the radius

to find circumference

of each wheel

now

you r seeing the relation

between

linear and angular

well

in the ques

so

now you got steps

find

the angular velocity of each wheel using circumference

then put the omega in v=rw

and linear speed is same for both

[ionic owl]

Right so v = 5 * 1/60 approx 0.0833 and v = 2 * 1/60 =approx 0.03333?

[strange hazel]

well

you did wrong

thats not 1/60

ok

lemme finish my eating

i will send you solution

[marsh citrus]

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[ionic owl]

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[cloud surge]

.reopen

ey

[jaunty quartz]

i stuck in c) the airy equation getting bigger but in c it says its getting lower

[cloud surge]

Theorem; Any point equidistant from the end points of a line segment is on the right bisector of it.

Explain me, this theorem pls

[jaunty quartz]

how is it getting more more and smaller for next variable?

[cloud surge]

@tacit fjord

[tacit fjord]

bruh the fuck

Why are you randomly pinging me?

Don't do that thanks.

[marsh citrus]

@cloud surge Has your question been resolved?

[marsh citrus]

@cloud surge Has your question been resolved?

[marsh citrus]

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[tepid pond]

Rewrite as a single logarithim: 2 log(x) + 3 log(y) - log(z). I don't know how to approach this situation.

[jade iron]

use your log rules

[tepid pond]

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[junior sonnet]

I'm being asked to find the area of one pedel of the graph of the polar equation r = 4sin2theta

[junior sonnet]

but what I want to know is how do I know what the bounds for ONE petal are

in this case it's pi / 2 but I don't know why

I assume it's cuz you get sin pi when you plug that in

but in my opinion it makes sense for it to be pi /4

[still temple]

When the points where
[
\sin(2\theta) = 0
]

[elfin berry]

♡LexQa♡

[junior sonnet]

?

[still temple]

you are trying to find the integration bound of one leaf

[junior sonnet]

yes

[still temple]

The leaf will begin and end with r = 0

[junior sonnet]

oh I see

[still temple]

Which means
[
4\sin(2\theta) = 0
]

[junior sonnet]

sin (1) = 0

[elfin berry]

♡LexQa♡

[junior sonnet]

thus the leaf will go there

and the 4 for 4 sin (2 theta) implies 4 leafs?

[still temple]

Yes

[junior sonnet]

thank you!

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[marsh citrus]

Please don't occupy multiple help channels.

[agile radish]

can someone plz help me

by using this

[mystic minnow]

close two of the three rooms you have open

[agile radish]

i did

r they closed now?

[mystic minnow]

so what's the question?

[agile radish]

how to go from 1st step to second

[mystic minnow]

what theorem are they citing?

[mystic minnow]

x dy/dx = -(3x+y) which gives you
x dy = -(3x+y) dx, then add that back over

[agile radish]

so its just rearranging?

[marsh citrus]

@agile radish Has your question been resolved?

[agile radish]

can someone explain the part highlighted in red

as i dont undestand why F(y)=0

[marsh citrus]

@agile radish Has your question been resolved?

[marsh citrus]

@agile radish Has your question been resolved?

[marsh citrus]

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[marsh ice]

What is the formula for the nth term of a sequence?

[spare crypt]

The formula for the nth term of a sequence is often expressed as un = a + (n - 1)d, where a is the first term in the sequence, n is the position of the term in the sequence, and d is the common difference between successive terms in the sequence.

[bright jay]

You can easily google this

[marsh ice]

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[eternal tundra]

i don't think there is such an angle

https://en.wikipedia.org/wiki/Rotation_matrix

this is the matrix that rotates a vector counterclockwise by theta

[glossy wind]

right

[eternal tundra]

we are pretty much looking for an angle such that this matrix R equals your matrix A

so we need
cos(theta)=-0.5
cos(theta)=0.5

both of these are never true

[glossy wind]

hmmm

[hushed egret]

,w {-1/2, sqrt(3)/2} . {-sqrt(3)/2, 1/2}

[elfin berry]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 夢雪#2250

[hushed egret]

columns arent orthogonal

so not an orthogonal matrix (a rotation)

[eternal tundra]

oh, that is nice
didn't think about that

[glossy wind]

I do not quite understand

[hushed egret]

rotation matrices are orthogonal matrices

[glossy wind]

and the matrix given is not orthogonal so

it cant be rotated?

[hushed egret]

its not a rotation matrix

[glossy wind]

hmmm

Yeah sorry I still don't understand

may you please explain some more?

[hushed egret]

rotation matrices are orthogonal matrices

so if its not an orthogonal matrix then its not a rotation matrix

[glossy wind]

okay well yeah

thats the part

I dont understand

[hushed egret]

which part

[glossy wind]

if its not an orthogonal matrix then its not a rotation matrix

[hushed egret]

a => b so not b => not a

?

[glossy wind]

wut

[hushed egret]

all oranges are fruits so if its not a fruit its not an orange

?

[glossy wind]

okay

So I suppose

thats just a rule

it has to abide by for it to be able

[hushed egret]

its just logic

[glossy wind]

to be rotated

[hushed egret]

well orthogonal matrices and rotation matrices are the same thing

like

if you rotate the axes

theyre gonna stay orthogonal

[glossy wind]

yeah

!solved

.close

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[light swift]

How please

I literally have no clue how to do this

[rotund violet]

um whats true of a diagonalizable matrix?

actually im busy rn srry wait for somebody else to help

[light swift]

i got not clue besides PAP^-1 = something

ok np

I dont know the rules to check I cant find them

[candid pendant]

Just quick one, have you looked at this?

https://math.stackexchange.com/questions/132552/showing-a-matrix-is-not-diagonalizable

[light swift]

no

I didnt understand what the guy said

I got my teachers work and i cant understand

[candid pendant]

What did your teacher do?

[light swift]

he's writing in the runic language or something

[candid pendant]

Nope his working is great!

So do you understand how to read eigenvalues off a diagonal matrix?

[light swift]

read?

i got the eigen values

-1 and 2 right?

but lamda 2 and 3 is -1

[candid pendant]

Better question might be which part of his proof do you not get?

[light swift]

So first a traignular matrix just has the empty area making a triangle right?

[candid pendant]

Particularly, a triangular matrix has all value below the diagonal = 0

[light swift]

So because it's triangle means that the diagonal is lamda?

[candid pendant]

or atleast a lower triangular

yes it means that the eigenvalues are the values on the diagonal

lambda is the typical symbol used for eigenvalues

[light swift]

okay why does he switch to putting lamda infront now?

i know sorry i just want to type it faster

[candid pendant]

hahaha no worries im being pedantic

so do you know what equation an eigenvalue and corresponding eigenvector satisfies?

[light swift]

i have no clue what that means. I just know how to find the eigen value and vector

are you talking about the PAP^-1?

[candid pendant]

So an eigenvalue $\lambda$ satisfies the equation $Av=A\lambda$ for a corresponding eigenvector $v$

[elfin berry]

todadqa

[candid pendant]

where A is the matrix btw

have you seen that before?

[light swift]

yes i think

[candid pendant]

just the definition of an eigenvalue

okay well what happens is

[light swift]

yeah ive seen it

[candid pendant]

it is rearrange to

$(A-\lambda I)v=0$

[elfin berry]

todadqa

[candid pendant]

and we are looking for v not the zero vector

[light swift]

yes

[candid pendant]

so now we are looking for v that satisfies this equation right?

lets denote v as $(x_1,x_2,x_3)^T$

[elfin berry]

todadqa

[candid pendant]

then all he has done is solve the above equation

[light swift]

uh

[candid pendant]

Sorry to be clear, we write v as the column vector version of $(x_1,x_2,x_3)$

[elfin berry]

todadqa

[candid pendant]

So if you multiply $(A-\lambda I)v$

[elfin berry]

todadqa

[candid pendant]

then you get the matrix he shows (he row reduces to basically just simplify the systems of equations)

which helps you find a general version of the eigenvector v

does that make sense so far?

[light swift]

I see he only does the part

oh wait

ohhh

A is infront

[candid pendant]

:

)

[light swift]

so the right part switches

I got it

wait now what do i do

[candid pendant]

So if you write down $(A-\lambda I)v$

[elfin berry]

todadqa

[candid pendant]

have a look at what you have

you will end up with a column vector of three equations

[light swift]

this is all set to the zero vector right?

[candid pendant]

yes!

exactly 🙂

and then you end up with 3 simultaneous equations (i.e. a system of equations)

[light swift]

yeah so u get the eigen vectors

3 vectors? 2 being the same

[candid pendant]

erm not quite

[light swift]

because two eigen values are the same

[candid pendant]

just because the eigenvalues are the same, doesn't mean you can only find 2 the same

for example, if you set t=1 and then t=2, then they aren't the same vector right?

[light swift]

yeah :(

[candid pendant]

the key thing is you can't find linearly independent eigenvectors

which in the case of two vectors, just means they aren't a multiple of eachother

in the case of multiple, it means that one vector is not a linear combination of the others

[light swift]

alright

so basically if they arent linearly indendent it's not diagonizable?

[candid pendant]

yes

for a matrix to be diagonalisable, you need to find 3 eigenvectors which are linearly independent

(i THINK)

Cus then if its diagonalisable, you can set P (from your PAP^-1 = D) to the matrix which has each of those eigenvectors in each column

Have you got notes on this topic btw?

[light swift]

Nope

[candid pendant]

weird haha

[light swift]

Yea online classes be like that

thank you for the help

[candid pendant]

nw

does that make sense tho

[light swift]

it makes sense more

I get everything until the conclusion now

[candid pendant]

okey which bit of the conclusion?

[light swift]

I cant even read it

[candid pendant]

is it the fact that if there aren't 3 linearly independent eigenvectors, then it isn't diagonalisable?

maybe i can read it cus ik what they mean haha

[light swift]

Okay ik imma sound really stupid. But what does linear independent mean again...

[candid pendant]

no thats not stupid at all

so the formal definition might be confusing for you

not sure what level of math you are doing lol

but simplest way

is that if you have vectors $v_1, v_2, ..., v_n$

[elfin berry]

todadqa

[candid pendant]

then these are linearly independent as long as for every $v_i$, it is not a linear combination of the other vectors

[elfin berry]

todadqa

[candid pendant]

just in case, a linear combination is just a scalar sum e.g.

$2v_1 + 3v_3$

[elfin berry]

todadqa

[light swift]

I remember u just write a linear compination that equals zero or soemthing

[candid pendant]

yhhh

[light swift]

I hate this class

[candid pendant]

the formal definition

is that the only solution to a linear combination of those vectors = 0

is the trivial solution (where all coefficients, of each vector, are equal to zero)

[light swift]

So basically you need the same amount of linearly independent eigenvectors to span the space

[candid pendant]

lol its a pretty satisfying topic tbh

[light swift]

i liked calc better :(

[candid pendant]

the fact that you can find 3 linearly independent eigenvectors, then out pops this P

and then calculating PAP^-1 gives you EXACTLY the matrix with just the eigenvectors on the diagonal

[candid pendant]

basically yh

[light swift]

I can d this too?

wait imma stick with the first way

I get it now

thank you so much

I hope you get lots of cookies and goodness

goodnight :)

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[paper canopy]

is there a way i can get (1-x^(n+2))/(1-x) to = x^(n+1)

[paper canopy]

im mid process in proving this

[whole hazel]

@paper canopy use sum of a GP

[paper canopy]

huh what?

[whole hazel]

geometric progression

[paper canopy]

not sure what that is

[whole hazel]

um

like 2+4+8+16+32+64+...

is a GP

[paper canopy]

ok

so how does the sum help

my teacher told me to do proofs through using n+1

and solving the equation so it equals the next sequence

so i have x^n..... x^(n+1) = ( 1-x^(n+1) )/(1-x) + x^(n+1)

( 1-x^(n+1) )/(1-x) + ( (1-x)x^(n+1) )/(1-x)

( (1-x^(n+1)) + ( (1-x)x^(n+1) ) )/(1-x)

reducing that down to what my question was originally

is there a way i can get ( 1-x^(n+2) ) /(1-x) to = x^(n+1)

[paper canopy]

so how does the sum help

[whole hazel]

@paper canopy what topic of math is this under

induction?

[paper canopy]

yes

[whole hazel]

oh shit

ok

$x^n..... x^(n+1) = ( 1-x^(n+1) )/(1-x) + x^(n+1)
( 1-x^(n+1) )/(1-x) + ( (1-x)x^(n+1) )/(1-x)
( (1-x^(n+1)) + ( (1-x)x^(n+1) ) )/(1-x) $

$x^n..... x^(n+1) = ( 1-x^(n+1) )/(1-x) + x^(n+1)

( 1-x^(n+1) )/(1-x) + ( (1-x)x^(n+1) )/(1-x)

( (1-x^(n+1)) + ( (1-x)x^(n+1) ) )/(1-x)$

whatever

[paper canopy]

$( (1-x^(n+1)) + ( (1-x)x^(n+1) ) )/(1-x)$

[elfin berry]

sorinmarkauv

[whole hazel]

lets go through it together ok

[paper canopy]

close

[whole hazel]

ok so the assumption is

for n=k

assume $1+x+x^2+...+x^k=\frac{1-x^{k+1}}{1-x}$

[elfin berry]

ed

[paper canopy]

yup

[whole hazel]

now we need to prove $1+x+x^2+...+x^k+x^{k+1}=\frac{1-x^{k+2}}{1-x}$

[elfin berry]

ed

[paper canopy]

oh well i have that

why is it x^k+2

i already proved that

[whole hazel]

because we need to prove for n=k+1

[paper canopy]

i thought for some reason im supposed to get to x^k+1

[whole hazel]

no....

you assume true for n=k

then show with the assumption that its true for n=k+1

so the term on the RHS is $x^{k+2}$

[elfin berry]

ed

[paper canopy]

ok well it didnt go the correct direction

but here you can see i had 2n-1= n^2

and i proved by adding n+1 or k+1 to the right side

and got the solution to look like the original formula

n^2 just this time it was (n+1)^2 where n+1 is some element of Z

so i can just add 1 to the n and try to prove it? i think i just answered my own question.

i did

i didnt realize thats how i was supposed to plug it in to "solve" for

[paper canopy]

i got this already so i just didnt realize that was what i was trying to prove

thanks!

[whole hazel]

yw

!

[paper canopy]

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[fading kiln]

can someone pls help me with this problem? I keep getting it wrong

[sharp mortar]

Doesn't look like a product solution

[fading kiln]

I don't know what to say. I solved for X and Y and multiplied them. Then used the initial values to get the constants

[marsh citrus]

@fading kiln Has your question been resolved?

[fading kiln]

@sharp mortar do you know what i need to do instead?

i figured it out i just had some positive and negative signs mixed up

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[ashen crown]

Hello there I have a question regarding synthetic division. Why do you need to put the remainder over what was divided. The answer I get is 2x^2+2x+1 with a remainder of 10. But that remainder needs to be added as 10/(x-4) I am just trying to wrap my head around what that means exactly.

[ashen crown]

10/(x-4)

[whole hazel]

when you divide 38 by 5

we say 7 remaimder 3

but we really mean $7+\frac{3}{5}$

[elfin berry]

ed

[whole hazel]

because $\frac{38}{5}=7+\frac{3}{5}$

[elfin berry]

ed

[ashen crown]

ohhhhh

Okay

So basically it could be anything then

If it was pi

It would just be over pi

If you dividing by pi rather

[whole hazel]

yea

does that answer your question

[ashen crown]

Yes and thank you! You have been most helpful

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[marsh citrus]

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[prime mountain]

i know this is pretty simple but i keep on getting these wrong, some help in the right direction would be appreciated.

for #13 i need to round to 6 decimal places

[main idol]

,w sum j=5 to 15 1/(j^2+4) * 1.0

[elfin berry]

Results provided by WolframAlpha

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Requested by riemann#2529

[main idol]

Well that definitely looks less than 1

Show in your calculator what you're typing

[prime mountain]

my calculator

i have a ti-83 and it doesnt have summation notation

so im just using google

0.144340 could it be this?

[main idol]

Use Wolfram instead

[prime mountain]

could you doublecheck cuz this is my last try on this question

[main idol]

Maybe?

Just copy that fraction

[prime mountain]

i cant copy/paste

0.1443405718

this is what im getting

she wants me to round to the 6th decimal place

what would that be?

[main idol]

Then type it out

[prime mountain]

i did

whats the rounded final answer?

[main idol]

This should be right as long as you round correctly

[marsh citrus]

@prime mountain Has your question been resolved?

[marsh citrus]

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[still temple]

How we plot the sinusaloide function asin(bx+c)+d ?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[tropic zenith]

Find the Maclaurin expansion of order 2 of the following functions. Express the residual term as x^n B(x) with suitable n and B(x) bounded in a neighborhood of x=0: f(x)= 1/1+x

[tropic zenith]

$f(x)=\frac{1}{1+x}$

[elfin berry]

Yousssef

[tropic zenith]

$f'(x)=-\frac{1}{(1+x)^2}$

[elfin berry]

Yousssef

[tropic zenith]

$f''(x)=\frac{2}{(1+x)^3}$

[elfin berry]

Yousssef

[tropic zenith]

$f'''(x)=-\frac{6}{(1+x)^4}$

[elfin berry]

Yousssef

[tropic zenith]

$f(x)=\frac{1}{1+x}$ $f'(x)=-\frac{1}{(1+x)^2}$ $f''(x)=\frac{2}{(1+x)^3}$ $f'''(x)=-\frac{6}{(1+x)^4}$

[elfin berry]

Yousssef

[tropic zenith]

$f(0)=1$ $f'(0)= -1$ $f''(0)= 2$ $f'''(Bx)= -\frac{6}{(1+Bx)^4}$

[elfin berry]

Yousssef

[tropic zenith]

$f(x)=\frac{1}{1+x}$ $f'(x)=-\frac{1}{(1+x)^2}$ $f''(x)=\frac{2}{(1+x)^3}$ $f'''(x)=-\frac{6}{(1+x)^4}$ => $f(0)=1$ $f'(0)= -1$ $f''(0)= 2$ $f'''(Bx)= -\frac{6}{(1+Bx)^4}$

[elfin berry]

Yousssef

[tropic zenith]

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[covert onyx]

integral is split

i dont understand the signs

[still temple]

why not?

[covert onyx]

the negative are pulled out right?

making the adding integral negative

[still temple]

yes

[covert onyx]

ty

im overthinking

[still temple]

np

[covert onyx]

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[pale path]

whats 1 + 1

[lilac bane]

When the impostor is troll

[pale path]

[lilac bane]

[pale path]

bro fr

i need it for my exams

[lilac bane]

Provide a photo of your problem

[lilac bane]

[pale path]

alr i figured it out

its 3

ty :)

.close

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[covert onyx]

i was understanding how partial fractions worked until i came across this, the whole denom just bein Ax+B throws me off

[covert onyx]

i firght thought see if denom can be factored which it cant

so im thinking if we cant factor denom and its proper of course we put the whole thing under Ax+B or Dx+E for example

???

[proven timber]

Since it can't be factored, you can't use partial fractions

[covert onyx]

oh ok ty

.close

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[wicked hearth]

Hi

[wicked hearth]

I need to prove if this is true or false

and show why

And I know that is false, but I need to show why I say it's false

x ; y ϵ R

<@&286206848099549185>

[hollow quest]

Show any singular counterexample and you have proven that its false

[wicked hearth]

That´s a good point

May I use numbers instead of x and y ?

[hollow quest]

Yep

Choose any x and y where it doesnt hold

Then thats enough proof

Now as for why you might need to think a bit more about it, but it will have something to do with whether x and y have the same signs or not

[wicked hearth]

yes, I can see it

if Y is negative

the right side turns in a addition

and left side in a rest

[hollow quest]

Note that if x is also negative then right side turns back into subtraction

-x-(-y)=y-x so when x and y are both positive or negative it will hold, if they are not the same sign then the inequality does not hold. And obviously if any or both of them are 0 then it clearly holds

[wicked hearth]

awesome!

thx

.close

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[still temple]

hmmm im unsure with this one

[still temple]

do i do binomial expansion of x^25?

[stoic saddle]

no

you can't expand x^25 anyway it's just x^25

but you might want to expand (x + x^2 + ... + x^10)^5 instead

[still temple]

yeah thats what i meant, i ended up getting 5631

okay throwing a random number at you is not going to magically make you tell me if it is right or wrong

lemme just

send steps

oh wait all of my steps are wrong

i am assuming a 6-sided dice whats wrong with me

okay i think i got it, it is 831,204

thanks

.close

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[ionic owl]

I'm not clear on what this is asking

[ionic owl]

Is it saying to find the point that I've drawn in there in red?

Directly below where it was on the second sighting?

[marsh citrus]

@ionic owl Has your question been resolved?

[marsh citrus]

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[junior atlas]

Hi can someone help me? Based on the given, the sequence is decreasing. But when I tried to test it using monotonic sequence theorem, the answer is that the sequence is increasing. Im so confuseddd

[junior atlas]

Is my general formula wrong?

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[pale meteor]

can someone help me

solve this

[dry spindle]

thales' theorem might help

then derived proportions

[spark berry]

No, just do 70-10

KJ = KS + SJ

SJ = KJ - KS

[dry spindle]

i was talking about tj

[spark berry]

Oh didn't know it is called thale's theorem

[dry spindle]

plus they want lt

and tj

thales's theorem

yeah

and fundamental simmilarity theorem is an alternative

if you wanna find the paralel's length

[spark berry]

I just think of these as similar triangles and that's it

[dry spindle]

thales' theorem doesnt really involve simmilar

its just a mix of proportions between the side lengths formed by the paralel

and not the paralel itself

[spark berry]

I just create 2 triangles inside of my head

The bigger one and the small one

[dry spindle]

thats kinda fundamental simmilarity theorem

thales' states:
if ST||KL, then SJ/KS = TJ/LT, among other combinations

in a triangle

[spark berry]

Then you know as long as the line segments inside the triangle are parallel, then they are equal by via transversal lines

That's how I think of this problem most of the time

[dry spindle]

could u exemplify

kl =/= st

[spark berry]

I mean, line ts and line kl must be parallel, it literally says in the given

So via transversal lines

you know

[dry spindle]

i dont

[pale meteor]

I just need the equation and I could solve it, I did these other ones just don't know how to setup this equation

[dry spindle]

.

[pale meteor]

ty

[marsh citrus]

@pale meteor Has your question been resolved?

[pale meteor]

@dry spindle

[dry spindle]

?

[pale meteor]

did i do it worng

?

bro yall cant just reply yes or no holy shit

.close

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[exotic vine]

Hi could someone help me with this exercise? Is in spanish tho

[fleet dagger]

$x^2 - y^2 = x - y$

[elfin berry]

Doggo

[fleet dagger]

you can use difference of squares

$(x - y)(x + y) = x - y$

[elfin berry]

Doggo

[fleet dagger]

if x - y = 0, you're done

if x - y isn't 0, then divide both sides by x - y and you get x + y = 0

so R is basically x = +-y

[lethal cloak]

no

[fleet dagger]

wait I'm confusing myself again right

yeah on the second part

[lethal cloak]

$(x)=(1-y)$

[hushed egret]

(x-y)/(x-y) = 1

[elfin berry]

villarrubi

[fleet dagger]

yeah

[lethal cloak]

and from here idk how to continue

we are stucked here @exotic vine and me

[hushed egret]

you dont actually need all this algebraic manipulation

[exotic vine]

But the exercise said that we need to demostrate the that is a equivalence relationship and we dont know how to do that

[lethal cloak]

yeah, we need to find if its simetric, transitive and reflexive

[hushed egret]

check the axioms

reflexive so xRx should be fairly straightforward

[lethal cloak]

but its not xRx

its xRy

[hushed egret]

this is checking reflexivity

[glass silo]

Check the definition of an equivalence relation

[exotic vine]

If we could understand it we wouldnt be here asking for help 🥲

[exotic vine]

And how do we check the other two things?

[hushed egret]

well symmetric is if xRy then yRx

[exotic vine]

so it would be checing if $y^2 -x^2=y-x$ ?

[elfin berry]

__lidiaaaa__02

[exotic vine]

checking*

[hushed egret]

which is like

theres nothing to check

its just true

[exotic vine]

so i just write that in my notebook and it would be okay? whitout checking anything?

i mean we just assume that its correct?

[hushed egret]

well

theyre the same equation essentially

you just multiply both sides by -1

you can write like

[exotic vine]

okay

so we have checked simetry and reflexivity

how do we do transitivity?

[hushed egret]

well do you know the statement for transitivity

[exotic vine]

I know that for example if we have the set {1,2} and {2,3} we need to have {1,3} for it to have transitivity

[hushed egret]

like

what does the transitivity axiom actually state that the relation should satisfy

[lilac creek]

It states that if two quantities are both equal to a third quantity, then they are equal to each other

[hushed egret]

[lilac creek]

xD

[exotic vine]

dont mind him

[hushed egret]

transitivity is if xRy and yRz then xRz

[lethal cloak]

it seems someone studied theory

[lilac creek]

thats just the same imo

[exotic vine]

yes, so how do we demostrate in this exercise that?

[hushed egret]

no it isnt

[lilac creek]

also i copied it from google

[hushed egret]

well its not equivalent

[exotic vine]

and how do i know that its correct?

like how i demostrate it?

[hushed egret]

well

notice that the sum of these two equations is exactly what you're looking for

[exotic vine]

oh i see

[hushed egret]

so you're done

[exotic vine]

and one last thing

it said in the exercise that we need to find the classes?

what does it mean with that?

[hushed egret]

the equivalence classes?

[hushed egret]

then you'd go and do something like this probably

[exotic vine]

i dont get it

[hushed egret]

like

you need to identify all the elements which are "equal" to each other under the equivalence relation

[hushed egret]

so like from this analysis that doggo did

[marsh citrus]

@exotic vine Has your question been resolved?

[exotic vine]

oh okay i get it

thank you

[hushed egret]

you'll probably have to identify all the sets tho

[exotic vine]

but it will be infinite sets right?

[hushed egret]

not too sure about that

needs to be checked

the relation looks kinda sus

[exotic vine]

well ill just ask my teacher tomorrow

thank you

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[hushed egret]

[marsh citrus]

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[still temple]

T is the solid tetrahedron with vertices (0,0,0), (1,0,0),(1,1,0) and (1,0,1)

Can someone please tell me how to get equation of this tetrahedron?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

<@&286206848099549185>

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@still temple Has your question been resolved?

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@still temple Has your question been resolved?

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[supple badger]

Heyy, anybody that can help me with this? Find eigenvalues and their eigenvectors of operator f = span..., R^3, which is defined as: ...

[supple badger]

I know that I have to start with finding the linear map, but I forgot how to do it

[marsh citrus]

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[supple badger]

<@&286206848099549185>

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[supple badger]

<@&286206848099549185>

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[sacred socket]

Annuity Due And Simple Annuity
Anyone give me this problem and solving example?

[bright jay]

Google problems to do

[sacred socket]

Idk

I can't find one

Idk if correct or not

[bright jay]

Then google a problem, do it yourself, then post what you did, someone one here can check

[sacred socket]

@bright jay
Can you help me find one?

[bright jay]

Nope, that's what google is for

[sacred socket]

Please?

[bright jay]

You can google on your own

[sacred socket]

I can't even see it

If I Google it

Pretty please help me?

[viral lily]

[sacred socket]

Help?

: (

<@&286206848099549185>

[marsh citrus]

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.

[bright jay]

Guess mods muted you

.close

[marsh citrus]

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[light kindle]

need help on the second box

[late hare]

bruh

so where are you stuck at exactly

doing 0.06*50?

+0.04*15

[light kindle]

like

the part where it says

0.04x

but that isnt correct

so im lost

i know i have to write a eqution

but i dont know how

or where to start

[late hare]

first empty box

is literally 0.04$

[light kindle]

it said 0.06x

[late hare]

0.04x

[light kindle]

each

[late hare]

after 50

it's 0.04

[marsh citrus]

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[marsh citrus]

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[frosty galleon]

How does this work

[leaden monolith]

What grade are you in

[static quarry]

that second question is amazing, if that's all there is

[dusk coral]

you call yourself an advanced math student?

solve for W, rn

[dusk coral]

is their more info provided for question 6, if not then y=r-h

[static quarry]

well W is obviously a subspace because what the hell else would you use W for 😀

[dusk coral]

assumptions assumptions

W is a variable

that has an obvious value to actually competent mathematicians such as myself

I am saddened you can't see it

[frosty galleon]

So like all the method is is to move around r and h and make is negative and stuff

[dusk coral]

ya, but is that equation really all thats provided

[frosty galleon]

Yes

[dusk coral]

not much you can do with those 3 variables by themselves

oh intresting

there is quite literally an infinite number of possibilities for each variable

[frosty galleon]

Ok thanks bro

So like just to make sure

Ok yes

Right

[marsh citrus]

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[cold mica]

https://cdn.discordapp.com/attachments/922932003370836021/1052095868771254343/image.png I am stuck as to how to use proportions here to find the sides

[cold mica]

So far I have gone up and down trying to use proportions connecting the angles

but I cannot figure it out

[old dawn]

hmm it feels like we are missing some context here?

what happened.

channels flashed again

[cold mica]

Oh um

so I have this problem here

https://cdn.discordapp.com/attachments/922932003370836021/1052095868771254343/image.png

I am directed to use

proportions

to solve but

I am confused how to use them

ive tried to use proportions but i just confuse myself more

[whole hazel]

do you understand which sides are proportional one another?

[cold mica]

no I do not

my teacher is just like read the book and youll figure it out

and im just lost 😭

[whole hazel]

KL is proportional to LM

JL to LN

JK to MN

[cold mica]

ok and then I just align and cross multiply correct?

to find x and y?

[whole hazel]

sed

yes

[cold mica]

ok so I got

x = 35

and y = 18

I hope thats right

[marsh citrus]

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[agile abyss]

can someone explain

how to do both of these problems

[whole hazel]

for question 10 try writing f(x) without any exponents

[agile abyss]

would that just simplify

to

be

a+x

[whole hazel]

no it doesn't

[agile abyss]

then how do you write it without any exponents

(1/a+1/x)^-1

then

what would you do from there

[whole hazel]

$f(x)=(\frac{1}{a}+\frac{1}{x})^-1$

$f(x)=(\frac{1}{a}+\frac{1}{x})^{-1}$

[elfin berry]

ed

[whole hazel]

$f(x)=(\frac{x+a}{ax})^{-1}=\frac{ax}{x+a}$

[elfin berry]

ed

[agile abyss]

how does

1/a + 1/x become x+a / ax

ohh

nvm

common denominator

and then the -1

flips it

ok

then what

[whole hazel]

$\frac{1}{a}+\frac{1}{x}=\frac{x}{ax}+\frac{a}{ax}$

[elfin berry]

ed

[agile abyss]

now what?

[whole hazel]

now theres an vertical asymptote at x=-a

a*

[agile abyss]

what

which one

at x = -a

or x = a

[whole hazel]

-a

[agile abyss]

how do you know htat

[whole hazel]

oh the a* was to correct grammar 🤣

[agile abyss]

ax / x + a

[whole hazel]

x cannot be -a as the denominator would be 0

then it would be indeterminate

i think thats the right word

[agile abyss]

ohh I see

so that would be /0

[whole hazel]

i think undefined is the correct word

[agile abyss]

which is a verticle asymptote

then

whats the other one

or how do you get the other one

@whole hazel

the answer is x = -a and x = a

how would you get x = a

[whole hazel]

you mean y=a

[agile abyss]

sorry

y=a

how do you get that

[whole hazel]

for horizontal asymptotes take the limit as x approaches infinity

[agile abyss]

does it matter

positive

or negative?

[whole hazel]

$\lim_x\to \infty f(x)=\lim_x\to \infty \frac{ax}{x+a}$

shit

$\lim_{x\to \infty} f(x)=\lim_{x\to \infty} \frac{ax}{x+a}$

[elfin berry]

ed

[whole hazel]

there we go

[agile abyss]

bro im lost 💀

[whole hazel]

$=\lim_{x\to \infty} \frac{a}{1+\frac{a}{x}}$

[elfin berry]

ed

[agile abyss]

how does this help get verticle asymptote

[whole hazel]

$=\frac{a}{1}$

[elfin berry]

ed

[whole hazel]

so the horizontal asymptote is y= a

[agile abyss]

how did you go from this

[agile abyss]

to thi

[whole hazel]

i divided both sides of the fraction by x

[agile abyss]

would you multiply by 1/x on both top and bottom

[whole hazel]

^

that is correct

[agile abyss]

ok

so from this

a/1

how do you get y = a

[whole hazel]

because the limit as x approaches infinity = a

[agile abyss]

ohh

I see

im stupid

i forgot

a /1

= a

[agile abyss]

now how do you do 11

[tight furnace]

Wow that's a cursed problem

Is this jee or something

[agile abyss]

jee?

[tight furnace]

difficult Indian math exam

[whole hazel]

you just assume the roots are 1 and -7

im pretty sure

thats what you're meant to do

[tight furnace]

I know it for this combination of basic calc concepts but in very non-formulaic ways

[agile abyss]

how did you get -7

as one of the roots

[whole hazel]

1+7=8

[agile abyss]

I assume you got 1 from x apporoching 1

[tight furnace]

alternatively

[whole hazel]

nothing to do with that

it has to do with x-1 on the bottom of the fraction

[tight furnace]

you need the numerator to be 0 for the limit to exist at all

[whole hazel]

^

yes

so x-1 must be a factor

of the numerator

[agile abyss]

ok

then what

[tight furnace]

That imposes a restriction on a and b

If you want you can write a in terms of b from that

But anyways

[whole hazel]

you can handle this im gonna die to lin alg

[tight furnace]

Only 1 unique value of b will make the limit equal to 8

[agile abyss]

but would there be

multiple

cause you can also change a

[whole hazel]

no i think theres exactly one a and b

[tight furnace]

But you already have a restriction between a and b because the numerator has to be 0 at x=1

Deciding b will determine a that way, just for the limit to even exist at all

[agile abyss]

I think my best option is just pray this won't be on the test

[tight furnace]

you could write the numerator as (x-1)(x-c) if that would help

c being another variable introduced just for the purpose of being easier to solve for

[marsh citrus]

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[tight furnace]

,w integral from -2 to 1 of xe^(-x^2)

[elfin berry]

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Requested by monikanicity#0205

[glossy wind]

wait did they drop a negative?

[tight furnace]

Looks good to me

No they switched the subtraction order in the numerator as well

[glossy wind]

oops I see indeed

thanks

.close

[marsh citrus]

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[tight furnace]

np

[marsh citrus]

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[coral turtle]

can someone explain how to factor this

[distant peak]

if both terms have a factor in common, you can factor by this factor

[still temple]

You can only really factor that a

[coral turtle]

ye ik u can do it by a but how like i dont understand this lool

[distant peak]

rt

[coral turtle]

thx bro ik what that shit means

[still temple]

[
a(x-16)^2 - ay^2 \iff a((x-16)^2 - y^2)
]

[elfin berry]

♡LexQa♡

[still temple]

They're just trying to help :)

[coral turtle]

yeah man this factoring stuff just doesnt make sense it goes over my head

[merry terrace]

Let (x-16)= z

az²-ay²
= (√az)²-(√ay)²
Now put it into (a+b)(a-b)=a²-b² then when you get the result put the value of z in

[still temple]

Yeah I mean just do difference of squares here if you want that lol

[coral turtle]

thx 4 the help but i have no idea the process in why and how u just took the a out like that

.close

[marsh citrus]

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[mystic lark]

guys is this correct?

[mystic lark]

or wrong?

🙏

please I'm not sure

I got dy2/dx2 on symbolab

and its 2-4x

but

I'm not sure about the constants

that I got

[glass silo]

,w y’’=2-4x, y(1)=3, y(0)=2

[elfin berry]

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Requested by chartbit#3847

[mystic lark]

nice

[glass silo]

Looks right to me 😉

[marsh citrus]

@mystic lark Has your question been resolved?

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[barren coyote]

the first one I think must also be