help-41

[dusky moat]

Need help on 4

[lusty pine]

use simmilarity

[amber wasp]

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[hasty junco]

is the average growth rate and average growth speed the same

[leaden sonnet]

no

[hasty junco]

so which one do i need the derivative for

[leaden sonnet]

the speed

[pallid canopy]

https://www.youtube.com/watch?v=pfTTHx9kCHk

[leaden sonnet]

speed

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@atomic venture Has your question been resolved?

[cinder grail]

f(2)=13
f(-1)=-14

what

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@atomic venture Has your question been resolved?

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@atomic venture Has your question been resolved?

[pallid canopy]

.close

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[hallow rose]

just need help on getting a the formula for this ;/

[hallow rose]

its like sin / cos stuff

i forgor about

[remote sedge]

first determine whether you want to use sin or cos

[hallow rose]

cos

[remote sedge]

ok now you need to find the values of $a, b, c,$ and $d$ in the general form $a\cos(bx+c)+d$

[grizzled pagoda]

lpieleanu

[remote sedge]

first find $a$

[grizzled pagoda]

lpieleanu

[remote sedge]

it is the ampltitude

[hallow rose]

3

if im using 12 as the low

and 18 as the high

[remote sedge]

yes

[hallow rose]

ok

[remote sedge]

now, for b, use the fact that the period of the function is $\frac{2\pi}{b}$

[grizzled pagoda]

lpieleanu

[hallow rose]

ah

is the full period from 2,12 -> 26,12 or am i wrong

[remote sedge]

yeah thats right

[hallow rose]

so 2pi/24

pi/12

[remote sedge]

yeah

[remote sedge]

let's actually use $a\cos(b(x+c))+d$

[grizzled pagoda]

lpieleanu

[hallow rose]

ok so far 3 cos(pi/12(x+c))+ d

[remote sedge]

yeah

[hallow rose]

okok

[remote sedge]

now c is the phase shift

[hallow rose]

so shifted right 2

which would be x-2

then + 15

i think

for d

[remote sedge]

uh that way is possible

but remember that the normal function f(x)=cos(x) has a maximum at x=0

[remote sedge]

yeah

[remote sedge]

if you use x-2 then you would have to put a negative sign in front of the 3

[hallow rose]

ah

so

would it look like this

[remote sedge]

yeah

[hallow rose]

thanks v much

i think i got it

[remote sedge]

[hallow rose]

.close

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[worthy mountain]

could u differentiate this with the DI method

[honest dagger]

You technically didn't ping helpers, but if you meant to, please ping them only after you have not received help for at least 15 minutes

[worthy mountain]

mb

[honest dagger]

No biggie. Just informing you

[honest dagger]

What is DI method?

[worthy mountain]

https://www.youtube.com/watch?v=2I-_SV8cwsw

wat would be th ebsr way to approach it idk

integration by parts is so longg for it

[shadow stump]

they are the same method

[worthy mountain]

yea but

not rly

how would i

do that

for this

cuz itsonly

one thing

what do u think

is the best method

to approach

one like this
/

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[worthy mountain]

No

[worthy mountain]

.close

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[lilac halo]

arithmetic and geometric sequences are killing me, can anyone help me? i have no idea how to start solving this q lmao

[mint nacelle]

is this series given as arithmetic or geometric?

ah I think you don't need to know that

[lilac halo]

idk tbh

[mint nacelle]

first term = S2 - S1
second term = S3 - S2

[lilac halo]

OH I CAME ACROSS A Q SIMILAIR TO THIS TYY

i remember llol

[mint nacelle]

1 + k, 4 + 2k, 7 + 4k, 10 + 8k

interesting sequence

[mint nacelle]

but yeah there's enough of a pattern to see it, the answer is both

there's an arithmetic and a geometric part

[lilac halo]

the coef of k is times 2 but the normal number is arthematic

[mint nacelle]

yep the common ratio of the ks is 2

[lilac halo]

yeah

awesome

thanks

[mint nacelle]

np!

[lilac halo]

.close

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[tough mica]

I dont want to be a burden but can someone explain how they supposedly answer my question?

how did they got that W \ <w2,w3> line? I feel like they didnt cared to explain just jumped straight to the solution, I wanted to ask her in a comment how did she got there, but wanted to try to see for myself first

https://math.stackexchange.com/questions/5010357/find-a-in-mathbbr-and-a-subspace-mathbbw-such-that-mathbbs-oplus

[amber wasp]

@tough mica Has your question been resolved?

[tough mica]

,, \mathbb{W} \setminus \langle (2,3,1,-2), (1,-1,6,3) \rangle

[grizzled pagoda]

why is math so hard?

[amber wasp]

@tough mica Has your question been resolved?

[tough mica]

,, \mathbb{H} = {(x_1,x_2,x_3,x_4) : x_1 + x_2 - x_3+ 2x_4 = 0}

[grizzled pagoda]

why is math so hard?

[tough mica]

,, x_1 = -x_2 + x_3 - 2x_4 \ \begin{cases} x_1 = -y + z -2t \ x_2 = y \ x_3 = z \ x_4 = t \end{cases}

[grizzled pagoda]

why is math so hard?

[tough mica]

,, \mathbb{H} = {(x_1,x_2,x_3,x_4) : x_1 + x_2 - x_3+ 2x_4 = 0} \ \mathbb{H} = {(-y + z -2t, y, z , t) : y, z , t \in \mathbb{R}}

[grizzled pagoda]

why is math so hard?
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tough mica]

,, \mathbb{H} = {(x_1,x_2,x_3,x_4) : x_1 + x_2 - x_3+ 2x_4 = 0} \ x_1 = -x_2 + x_3 - 2x_4 \ \begin{cases} x_1 = -y + z -2t \ x_2 = y \ x_3 = z \ x_4 = t \end{cases} \ \mathbb{H} = {(-y + z -2t, y, z , t) : y, z , t \in \mathbb{R}}

[grizzled pagoda]

why is math so hard?

[jolly matrix]

what exacrly is unclear

about that step

[tough mica]

I just said, she jumped to the solution right away

without explaining the algebra behind it

[jolly matrix]

hmmm

i can try to explain it

[jolly matrix]

so you basisically strat with a larger space W and youre removing (subtracting) the span of vectors w2 and w3 getting the complement of <w2,w3> in W

thats like saying "take W and remove everything that can be made from w2 and w3

[tough mica]

what I wrote here was a mistake

I meant $\mathbb{H} \setminus \langle (2,3,1,-2), (1,-1,6,3) \rangle$

[grizzled pagoda]

why is math so hard?

[tough mica]

idk why I wrote W

[tough mica]

yes but replace the W by H

[jolly matrix]

yes

sry

[tough mica]

thing is, she gave me some set builders without explaining the algebra behind

she said, "some calculations lead to"

[jolly matrix]

lol

but she is very good

when i solved it 2,3,1,-2 is a better choice for w1 than i got

i got -2,-3,-1,2 which is technically correct but

shit at the same time

[jolly matrix]

btw could i answer your question or didnt it help?

[tough mica]

you just explained the notation

[jolly matrix]

yes

[jolly matrix]

ill try to write it down

[tough mica]

.close

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[stuck kindle]

let $S_n$ be the symmetric group and $N \subseteq S_n$ a normal subgroup. If $N$ contains a transposition $\tau$, why is that $N = S_n$?

[grizzled pagoda]

hanno

[stuck kindle]

I know that $\pi N = N \pi$ because N is normal

[grizzled pagoda]

hanno

[stuck kindle]

but what can I do with this? And what to do with the transposition?

[indigo cloud]

if pi is arbitrary then pi tau is in pi N, so it also has to be in N pi

write out what that means

[amber wasp]

@stuck kindle Has your question been resolved?

[stuck kindle]

so for any $\pi$, there exists a $\nu$ in N, so that $\pi \tau = \nu \pi$?

[grizzled pagoda]

hanno

[jolly matrix]

N contains a transposition τ and N is normal in n thennn N must contain all conjugates of τ since transpositions generate N must equal Sn

[stuck kindle]

I don't get that last point

I can see now how any $\pi \tau \pi^{-1}$ is in N

[grizzled pagoda]

hanno

[stuck kindle]

but how would I have to choose pi to get any transposition of S_n?

[jolly matrix]

2 million bucks and I'll tell you

Let's say we have τ=(1 2) in our normal subgroup N

[indigo cloud]

how does in general pi(a1,a2,...,an)pi^-1 for some cycle look like

[jolly matrix]

Then you aenerate anyy transposition
to get any transposition (i j), we can conjugate τ=(1 2) by a permutation π that Maps 1 to i
Maps 2 to j
Then π(1 2)π^-1 =(i j)

[jolly matrix]

I love your pfp

Anyway

To get (2 3) choose π that sends 1→2 and 2→3
To get (1 3) choose π that sends 1→1 and 2→3
And so on

since N is normalall these conjugates must be in N giving us every possible transposition

[stuck kindle]

oh wait so we take pi = (1<->i, 2<->j) right? Then pi (1<->2) pi^-1 = (1<->i, 2<->j)(1<->2)(1<->i, 2<->j) = (1->i, 2->j, i->2, j->1)(1<->i, 2<->j) = (1->1, 2->2, i->j, j->i) = (i<->j)

[jolly matrix]

Good ? Or you still have questions

[stuck kindle]

I see

[jolly matrix]

How did you write that

But yes exaz

Exactly

You got it

[stuck kindle]

and so we can transform tau into any transposition we want

that's a cool idea

[jolly matrix]

I know

You're welcome

[amber wasp]

@stuck kindle Has your question been resolved?

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[swift spoke]

how can i solve this

[split sail]

use pythagoras's theorem

x² + (x + 1)² = (x + 4)²

then go from there

[swift spoke]

k thanks

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[amber wasp]

Please don't occupy multiple help channels.

[pallid canopy]

...

.close

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[empty canyon]

$$\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{1}{x^{2}-2x+1}$$

[grizzled pagoda]

SilentALume

[empty canyon]

What are the solutions?

[final briar]

gimme a second

ill try and help here

[glad pulsar]

did u try anything?

[empty canyon]

Yes

[glad pulsar]

show

[empty canyon]

I got one solution but the other I have no idea

[glad pulsar]

what did u do

[final briar]

yeah

[empty canyon]

$$\left(\frac{3}{2}-\frac{\sqrt{5}}{2},\phi+1\right)$$

[grizzled pagoda]

SilentALume

[empty canyon]

This is one of them

However the other solution I have no idea about

[final briar]

yeah sorry i cant help out here i think

[empty canyon]

I found it

$$\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3},\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)$$

[grizzled pagoda]

SilentALume

[empty canyon]

$$y=\left(\left(\frac{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\left(\frac{1+\sqrt{5}}{2}\right)+1\right)}{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)}\sum_{k=1}^{x}1+\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\frac{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\left(\frac{1+\sqrt{5}}{2}\right)+1\right)}{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)}\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)\right)-\sum_{n=1}^{\left(\frac{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\left(\frac{1+\sqrt{5}}{2}\right)+1\right)}{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)}\sum_{k=1}^{x}1+\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\frac{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\left(\frac{1+\sqrt{5}}{2}\right)+1\right)}{\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}}{3}+\frac{\sqrt[3]{\frac{25}{2}+\frac{3\sqrt{69}}{2}}}{3}\right)-\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)}\left(\frac{2}{3}+\frac{\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}$$

[grizzled pagoda]

SilentALume
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[empty canyon]

Bruh

[amber wasp]

@empty canyon Has your question been resolved?

[amber wasp]

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[glass wraith]

I don't even know what the fuck this means

I'm only given a bunch of functions in fx

[amber wasp]

@glass wraith Has your question been resolved?

[pallid canopy]

anymore context?

[glass wraith]

I do have functions

like 1/x plus 3

[amber wasp]

@glass wraith Has your question been resolved?

[amber wasp]

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[tepid estuary]

can someone help? this is my last question i dont wanna get it wrong 😭 (498 questions answered)

[honest dagger]

!nosols

[amber wasp]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[gloomy ferry]

yep, mb

[split sail]

Are the measures of any of the angles provided?

[tepid estuary]

Nope

[split sail]

Oh is it saying less than 10?

[honest dagger]

I'd start by ruling out angles one at a time

[split sail]

This is why we label with variables but anyway

[honest dagger]

yeah it threw me off at first also

[tepid estuary]

7 is greater im pretty sure

one of the greater angles

[honest dagger]

It looks like it would always have to be, yeah, but I'm trying to figure out if there's a way to know definitively

[tepid estuary]

it kinda just expects us to use the theorem

which is so confusing

because it throws me off when theres no measures

[honest dagger]

Which theorem?

[tepid estuary]

exterior angle theorem

[gloomy ferry]

I tried rationalising by looking at the horizontal line and having a perpendicular intersecting the angle 9

[tepid estuary]

hmm

[tepid estuary]

this is kinda what ive been doing

ohhh i got it

.close

[amber wasp]

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[nocturne plaza]

hi. :) I got two dif answers, 32x^7-(6/x^3) and 12x^8-5+(6/(4x^2)). I think the first one might be right, but I'm not sure if either are xD. Could someone check? :o

[thorny terrace]

u dont need us

search derivative-calculator.com into google

type in ur function

and let it differentiate

[nocturne plaza]

I tried Symbolab and it did it wayyy more complicatedly, but I'll try that source :)

[thorny terrace]

u can even press the green checkmark to see if ur answer is equivalent

[weak zinc]

(if you do want us to check, can you explain how you got each answer you found? )

[thorny terrace]

sorry its .net not .com

[nocturne plaza]

oki they did it very weirdly, but they said my answer is equivalent so yay :D

[thorny terrace]

yes ur first answer is correct

ur second one isnt

[nocturne plaza]

yeah I thought so, I did the derivative and simplification kinda backwards

I'm super confused on what happened here, but it says it's equivalent so I'm probs just reading it wrong 😅

[nocturne plaza]

I would really appreciate it, thank you. :) Lemme know if this is legible, haha. (4x^12/x^4)-(5x^4/x^4)+(3x^2/x^4), so I just split up the sections here. Then I simplified it all to 4x^8-5+(3/x^2). Then I just did the derivative of all three parts.

so I got 32x^7-(6/x^3). Realizing that the 2nd one is def wrong tho, woops :')

OHHH wait omg, I see what the calculator did actually :O

how do I always realize after asking @~@

[weak zinc]

It's like that sometimes getting to think out what you're doing and explain it to someone else can do it for you

[nocturne plaza]

haha very true.

Alright, I really appreciate it. Thank you everybody :)

For the help and for the website :D

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[devout ledge]

first I multiplied top and bottom by (-96)^(4/5)

so

,, \frac{3^{\frac15} \cdot (-96)^{\frac45}}{-96}

[grizzled pagoda]

smeagol

[devout ledge]

I don't know how to get like bases though

<@&286206848099549185>

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[devout ledge]

no

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[oblique oak]

heres the solution

but my question is how could i find P

[shadow stump]

the diagonal matrix uses the eigenvalues and P uses the eigenvectors

[oblique oak]

ohh

but thats only for diagnolizable matrices right?

[shadow stump]

well diagonalization is what they did. if it's not diagonalizable then finding large powers will be difficult

[oblique oak]

but you can do that for any matrix cant you

[shadow stump]

not all matrices are diagonalizable, no

[cunning birch]

if your matrix weren't diagonalizable, polynomial division can be an alternative:

Let $A = \begin{pmatrix}2&-1\1&4\end{pmatrix}$, find $A^{100}$

[grizzled pagoda]

rafilou is not not born in 2003

[oblique oak]

what is polynomial division?

[cunning birch]

we find that the minimal polynomial of $A$ is $(\lambda-3)^2$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

so the idea

is to do the following polynomial division:

$\lambda^{100} = (\lambda-3)^2Q(\lambda) + a\lambda + b$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

now we have to find a and b

[cunning birch]

basically euclidian division

with polynomials

[cunning birch]

we now solve for a and b

plug in $\lambda = 3$: $3a + b = 3^{100}$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

differentiate this relation:

$100\lambda^{99} = 2(\lambda-3)Q(\lambda) + (\lambda-3)^2Q'(\lambda) + a$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

plug in $\lambda = 3$: $a = 100 \times 3^{99}$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

so $b = -99\times 3^{100}$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

(because of this)

[cunning birch]

now we can "replace" lambda by A

which means $A^{100} = aA + bI$

[grizzled pagoda]

rafilou is not not born in 2003

[cunning birch]

feel free to say if there are steps that are unclear for you @oblique oak

[oblique oak]

ohh this makes sense

thank you!!

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[viscid plaza]

when applying the first derivative test

[viscid plaza]

do i solve for critical points and singular points or just criticals

[rich latch]

by critical point you mean where the derivative does not exist?

[viscid plaza]

i mean where the derivative equals 0

singular points r where it does not exist

[rich latch]

sorry singular point

[viscid plaza]

yeah

im pretty sure it would

[rich latch]

when you apply first derivative test, I think the singular points won't have a derivative.

[viscid plaza]

i dont think that would change whether or not the function would have a local max or min at that point though

[rich latch]

Yeah, if there is no derivative and the left hand derivative is positive but right hand is negative that would be a local max, the other way around for local min

so if you want global max/min, you'd have to get all the critical points, and singular points that are singular even though the function is continous there (like a sharp point) as well as other discontinuties

[viscid plaza]

mhm

thank you

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[craggy flame]

how would i show that sigma is bijective? ive only proven that matrices are bijective

[keen pawn]

Matrices aren't necessarily bijective

every linear transformations including non-bijective ones can be reprsented using a matrix

[craggy flame]

yeah i meant like showing that certain ones are

[craggy flame]

how would i go about showing that this is a matrix

[unborn plaza]

why do i feel like sigma isn't surjective?

[keen pawn]

it isn't necessarily one-one IMO $e^x+1 + e^x-1 = e^x+2+e^x-2$

[grizzled pagoda]

ƒ(Why am. I here)=I don't Know

[unborn plaza]

i'm not sure what you're showing there

[keen pawn]

nvm, misinterpretd the question

[unborn plaza]

but for example, let f(x)=0
then there is no g with sigma(g) = sigma(f) +1

[craggy flame]

where does g come from in this problem

[unborn plaza]

"there is no g with sigma(g) = sigma(f) +1"
this is a way of saying "sigma(f)+1 is not in the image of sigma"

in other words sigma(f)+1 is never produced by sigma for any input

[rapid peak]

i.e. the antiderivative of 0 is in the image of sigma. Call it C.
Then there is no function who's antiderivative is C+1, so C+1 is not in the image of sigma.

[unborn plaza]

yes

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@craggy flame Has your question been resolved?

[craggy flame]

so its not surjective??

[unborn plaza]

it sure seems that way to me
C^inf(R) is smooth functions R->R right?

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[lime jasper]

Where do i put the angles

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[tiny sonnet]

not exactly a problem i need help with just need help finding a video i can learn this material from

[tiny sonnet]

im not too sure what i shuold search either

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[hardy sierra]

Hi, im extremely lost on what I am doing wrong in this problem.

[hardy sierra]

My understanding is that I should use the top graph for x, and bottom graph for y cordinates and then draw the points. I do that, and I seem to only get partial credit

im wondering am I missing points then?

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[median kettle]

indefinite integral
this question is answer is
x^10-3x^4-5x+c am i correct? Or
x^10-3x^4+c-5

[split sail]

the 2nd one

-5 is outside of the integral

[median kettle]

x^10-3x^4+c-5

[split sail]

yes

[median kettle]

what about this question

[split sail]

you're integrating only t⁷

the other terms are outside the integral

[median kettle]

its supposed to be t^8/8+11t^3+4t^2+c or t^8/8+c+33t^2+8t

[split sail]

the 2nd one

[median kettle]

thank you.

[split sail]

read this

[median kettle]

yeah i got it

[split sail]

np <3

[median kettle]

thanks :>

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[median kettle]

determine f(x) given f''(x) = 8x^3-12x+3 is my answer correct? i feel theres something wrong.

[split sail]

,rotate

[split sail]

-12x²/2 is not -3x²

[median kettle]

☠️

no way

[small pendant]

you cannot take out the c form f'(x)

in addition to the other mistake you made

[median kettle]

wym

[small pendant]

you got 2x^4-3x^2+3x+c after you simplified the first integration, and then you took the c out

you cannot take the c out

you integrate the c, and get cx

and then you add another constant

[median kettle]

oh

cx+c

[small pendant]

make them c_1 and c_2 because they can be different c

[median kettle]

c1x+c2

[split sail]

is this correct?!

the 1st integration

[small pendant]

yes

[split sail]

is that a 3 or a 2?

man nvm

[median kettle]

thank you :>

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[rancid aurora]

Question 6

[rancid aurora]

question 6

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[midnight karma]

Hi everybody,
Hope everyone's doing well. I've a dataset with timestamped XYZ values for basketball in terms of euler to define its rotation. Given that I've XYZ how can I identify the axis of rotation of the ball and graphify this in terms of base 0 space.
eg
Basketball_X= 8.01
Basketball_Y=-1.23
Basketball_Z=0.56

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[lime verge]

do i use this formula

to solve this?

[tawny hill]

yes it is the frobenius method

[lime verge]

so stuff like this

always work?

[tawny hill]

yes

[lime verge]

can i always use this formula to do similar problems?

[tawny hill]

its not formula

[lime verge]

cuz i see nmy teacher

with answers with sines and cosines

okok can u help me

solve it

plz

PLZ my fellow brother

[tawny hill]

its just general expression of series

[lime verge]

idk how to attempt it

[tawny hill]

plug the solution in the equation

[lime verge]

okok

[tawny hill]

and found recurrence relation on cn

[lime verge]

so what goes inside what? what do i put on x and what do i put on yn?

[tawny hill]

just substitute de y,y' and y''

what year of school are you in?

[lime verge]

2nd

u

u said substitute y' and y''

[tawny hill]

yes

[lime verge]

to what serieis?

thats a yt vid i was watching

so would it be the coefficents?
so y = -x -1
so i plug in that for serieis?

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[eternal summit]

How to do written division?

[solar gust]

Written division you mean the thing like this
L
|

[eternal summit]

?

How to divide

[gloomy tide]

it would be good if you had an example problem to work on

[solar gust]

Probably better to watch a video if it is general method

[eternal summit]

no. teach me

[dull pike]

No

[eternal summit]

YEASSS

EEEEEE

E

[solar gust]

But nobody will xd its not made for this

[dull pike]

Stop trolling please.

[gloomy tide]

this isn't how to be popular

[dull pike]

.close

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[solar gust]

He took the responsability 🫡

[primal holly]

ariana grenade

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[eternal summit]

A train traveled 700 meters at a constant speed in 50 seconds. While maintaining the same constant speed, the same train traveled a distance equal to its length in 15 seconds.

Calculate the length of this train. Record the calculations.

[amber wasp]

Please don't occupy multiple help channels.

[solar gust]

Close the previous one

[eternal summit]

This is the triangle?

so v*t

,calc 700/50

[grizzled pagoda]

Result:

14

[solar gust]

50sec = 700
15 sec = ?

[eternal summit]

,Calc 14*15

[grizzled pagoda]

Result:

210

[eternal summit]

210 m

[solar gust]

Seems fine

[eternal summit]

@solar gust

No way I got 1 point for that in 2023 exam

…..

That is so easy

I was 15 there

@solar gust

[solar gust]

Can you not ping

[eternal summit]

….

.close

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[eternal summit]

It’s 2?

[celest cove]

if that j is irrelevant, then yeah

otherwise, no

[eternal summit]

It’s not relevant

[pastel star]

sqrt(81) - sqrt(49)
9 - 7
2

[wind jewel]

yes right

[eternal summit]

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[tribal stone]

how do i change the units to gy/s

[tribal stone]

mb its electrons/gs

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[turbid pulsar]

Is it true for all n that no more than n vectors can be independent in r^n space

ie no more than 15 vectors in r^15 space

[cunning birch]

a linearly independent family in R^n can only have at most n vectors, yes

[turbid pulsar]

ok cool

tyty @cunning birch

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[regal swift]

need help with part b

i dont get how they computed b0

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[inner prawn]

translation: prove that the congruence always has a solution to every natural number m

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[simple drum]

Let $\mathbb{Z}$ the set of integers. \
Signaling: $\mathbb{Z}^-={x \in \mathbb{Z} | x<0} and \mathbb{Z}^+={x \in \mathbb{Z} | x>0}$ \
Prove by using diagonalization that this set is uncountable: \
${X \subseteq \mathbb{Z} | X \subseteq \mathbb{Z}^+ or X \subseteq \mathbb{Z}^-}$

[grizzled pagoda]

prograce

[simple drum]

Call the set X and assume that X is a countable set therefore by definition there exists an injective function from X to $\mathbb{N}$ , and from a theorem there exists a surjective function $f : \mathbb{N} \rightarrow X.$ Indicate $f(n)=K_n$, define a new set S like this: \
$S = {x^n | x^n \notin K_n \wedge x \in \mathbb{Z}^+}$ \
From the definition of S, $\forall n \in \mathbb{N} : x \in \mathbb{Z}^+ therefore S \in X$ , $\forall n \in \mathbb{N} : if x^n \in K_n then x^n \notin S and if x^n \notin K_n then x^n \in S therefore S \neq K_n$ \ We reached contradiction (that f is surjective) therefore the set is uncountable

[grizzled pagoda]

prograce

[simple drum]

Can someone check my work

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[modest chasm]

can someone explain me this?

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[modest chasm]

<@&286206848099549185>

[silk fossil]

Thinking

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[summer bear]

Did i do the right steps to solve for x? (Ping me)

[thick shale]

,rccw

[thick shale]

No

First step itself is inaccurate

[grizzled pagoda]

Pro_Hecker

Pro_Hecker

[thick shale]

@summer bear

[summer bear]

And then how you solve got x?

[ionic terrace]

move the terms without sqrt to right side; square the whole thing again after

your first objective is to get an equation without sqrt, before trying to find x.

[grizzled pagoda]

Wild123

[ionic terrace]

This is a theoretical example, to show you the pattern of solving..

[summer bear]

So 4 - squrt x+1 = squrt 2x+5?

[ionic terrace]

no; first step is correcting 2nd line, like Pro_Hecker said.

You have $\sqrt{2x+5}+\sqrt{x+1}=4$

[grizzled pagoda]

Wild123

[ionic terrace]

you square it

get 1 sqrt + some terms

[summer bear]

Then ?

I do what u said?

[ionic terrace]

Yes.

Then you move the "some terms without sqrt" to right side, and square again. Even if they have x in them.

[summer bear]

Ohhh

I get it now

Thanks dude

[ionic terrace]

You're welcome. Can send your progress while you work on it.

[summer bear]

Im not home rn cant

[ionic terrace]

Remember to square correctly, with the (a+b)^2 formula

oh alright.

[summer bear]

.close

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[inner prawn]

translation: prove that the congruence always has a solution for every natural number m

[indigo cloud]

factor

ok nvm doesnt quite work. but at least covers a lot of cases. hmm

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.reopen

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✅

[inner prawn]

<@&286206848099549185>

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[violet blaze]

so we want to show (2x+1)(3x+1) = 0 , always has a solution mod m

first, if m is coprime to 2 or 3, than we are done

so

let us take m = 6k

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[simple drum]

Prove that $a_{n}=n^{((-1)^n)}$ is not bounded

[grizzled pagoda]

prograce

[simple drum]

By definitikn

[fluid wren]

!status

[amber wasp]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[simple drum]

I started by trying to use the negation of the defintion that for every M in R there exists n such that |a_n|>M but I don't know how to continue

[fluid wren]

A suggestion for starting writing a proof would be to first think about why you consider the statement to be true

[simple drum]

Well, I thought of it because if n is even then it's going plus infinity but if n is odd then it goes to minus infinity but I don't know how to translate that into the boundness definition that there will never be a One number that is bigger than all numbers you know ?

[fluid wren]

So given M, you need to find some even number n such that |a_n| will be larger than M

Well firstly a_n = n when n is even

And |a_n| > M becomes equivalent to n > M

So you just need an even integer greater than M

You can construct one using ceil or floor of M

[simple drum]

Okay

Can I say it's 2M?

[fluid wren]

2M may not be an integer

[simple drum]

Oh true

2 ceil of M is an even integer?

[indigo cloud]

(its quite handy to prove that in general you only need to consider natural M in the def of (un)boundedness)

[simple drum]

I can say, for every real number there exists a natural number bigger than it, therefore I can choose a natural M ?

[indigo cloud]

yes

[simple drum]

Oh can I also choose 2[M]?

That's three different answers now lmao

[proven vapor]

any even number larger than M works

[indigo cloud]

you should make sure that M is positive btw

[simple drum]

Oki

Ty all

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[shrewd elm]

pls don't troll in the help channels

.close

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[leaden sonnet]

<@&268886789983436800>

(In case they delete)

[gloomy tide]

we have message logs dw (but thank you)

[leaden sonnet]

Oh ok

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[tough mica]

the last condition is saying that
dim(R^4)=dim(S+T) + dim(<(1,2,2,1)>)
4 = dim(S+T) + 1
dim(S+T)=3 from dimension counting right?
also: dim(S+T)=dim(S)+dim(T)-dim(SnT) from the theorem of dimensions of finite vector spaces
right?
dim(S+T)=dim(S)+dim(T)-dim(SnT)
3 = dim(S)+dim(T)-dim(SnT)
but from first condition we know dim(S)=dim(T)
3=2dim(S)-dim(SnT)

@limpid sun

<@&286206848099549185> @limpid sun @limpid sun

[upper raptor]

!volunteers

[amber wasp]

Helpers are just people volunteering their time to help you. Be polite and patient.

[upper raptor]

Also wait 15 minutes

[tough mica]

15 minutes have passed brother

[tough mica]

also he is the one who asked for my progress

[upper raptor]

Oh I read the time as 12:37 and not 27, I am very very sorry

[tough mica]

is ok, let me compute the orthogonal complement of H

H : 2x3 - x4 = 0

x4 = 2x3

(x1,x2,x3,x4) = (x1,x2,x3,2x3)

(x1,x2,x3,2x3) = x1(1,0,0,0) + x2(0,1,0,0) + x3(0,0,1,2)

H = <(1,0,0,0),(0,1,0,0),(0,0,1,2)>

now that we got a basis for H let me find the orthogonal complement of H

v = (x1,x2,x3,x4)

i) v . (1,0,0,0) = 0

ii) v . (0,1,0,0) = 0

iii) v . (0,0,1,2) = 0

i) x1 = 0
ii) x2 = 0
iii) x3 + 2x4 = 0

x3 = -2x4

(x1,x2,x3,x4) = (0,0,-2x4, x4)

(0,0,-2x4,x4) = x4(0,0,-2,1)

H perp = <(0,0,-2,1)>

since we know $\mathbb{S} \cap \mathbb{T} = \mathbb{H}^{\perp}$

[grizzled pagoda]

938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

we know dim(SnT) = dim(Hperp)

dim(SnT) = dim(<(0,0,-2,1)>)

dim(SnT) = 1

3=2dim(S)-dim(SnT)

3 = 2 dim(S) - 1

4 = 2 dim(S)

dim(S) = 2

dim(T) = 2

right?

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[timber dome]

hi guys, could someone help me with this exercise?

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[young seal]

pls help

[pallid canopy]

Did you try to find two equations for d and n, where d is number of dimes and n is number of nickles

[young seal]

yes

I am struggling

[quick ridge]

what did you get so far

[quick ridge]

how much is a nickel worth

and how much is a dime worth

[young seal]

0.05

[young seal]

0.1

[quick ridge]

right so how can we use this to make an equation including the 1.75

[young seal]

0.05x + 0.1 (25-x)= 1.75

idk

[quick ridge]

great

now just solve for x

[young seal]

so confused

[quick ridge]

apparently not

[young seal]

hmm

0.05x + 2.5-0.1x = 1.75

-0.05x = -0.75

x = 15

10 dimes

15 nickes?

[quick ridge]

x was nickels, yes

you’re just a chill guy fr

[young seal]

also this is pretty confusing

[young seal]

bruh

[quick ridge]

did you draw a picture

[quick ridge]

bruh

[young seal]

[young seal]

[quick ridge]

draw one

[young seal]

one sec

[young seal]

Like that?

@quick ridge

[quick ridge]

yea

[young seal]

What’s next

I don’t think I have enough clues

It’s so hard

Help

[quick ridge]

you realize that part of learning math is taking time to think it through for yourself right

[young seal]

Let me try

[quick ridge]

start by labeling

[young seal]

Btw if I try to find the side length I use rad on the calculator right?

[quick ridge]

you can use degrees

just be in the correct mode

[young seal]

This?

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[upper ember]

find the difference between the maximum and minimum value of the function

[rich latch]

what do you have in mind so far?

[frank flint]

tan(f(x))=7

[night crater]

yeah

just solved it to be that

actually not exactly

there will be periods where they are both negative

[frank flint]

tan(arctan((2x-1)/(x+2)))=?

it's not relative about negative

[night crater]

it is tho, because arctan gives a negative value is the thing is negative

[rich latch]

true.

[night crater]

when -2<x<1/3, both of them are negative

and the fix I think is simple, I think this is just arctan(pos/neg 7)

[frank flint]

tan(arctan((2x-1)/(x+2)))=?

[night crater]

wait no

[frank flint]

tan(a+b)=?

[night crater]

tan a+ tanb/1-tanatanb

tan(arctan((2x-1)/(x+2)))=2x-1/x+2

[frank flint]

same solve about that

last, tan(f(x))=?

[night crater]

oh yeah right

it is arctan 7

[frank flint]

clear?

[night crater]

oh yeah right it is just arctan 7 in a pos and negative value lmao

mb XD

[upper ember]

definitely not true, checking with the answer

[night crater]

I have the graph lol

[upper ember]

I understand that the answer to the problem will be 0?

[night crater]

why 0?

[upper ember]

ah, I misunderstood you, I calculated arc(7) as the maximum and minimum value at once

[night crater]

oh lol

np

[upper ember]

What does the graph say about the maximum and minimum of a function?

[frank flint]

x!=-2 && x!=1/3 f(x)=arctan(7)

f(-2)=-pi/2 + arctan(1/-7)

[upper ember]

By the way, do you understand well the inequalities that are solved through derivatives?

[night crater]

I am not the right person for calculus lol

[frank flint]

arctan(7)-(-pi/2+arctan(-1/7))

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[upper ember]

.close

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[unkempt echo]

Pentagon ABCDE, M, N, P, Q is the midpoint of BC, AE, AB, DE. Proves that if MN cross PQ at the midpoint of PQ then MN//CD.

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[tough mica]

\textbf{1.- } Let $\mathbf{S} = \langle (2, 3, 1, 1), (1, 1, 1, 0) \rangle$, $\mathbf{T} = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 + x_3 + x_4 = 0; ; 2x_1 - x_2 + x_4 = 0 }$ and $\mathbf{W} = \langle (2, 1, 2, 1) \rangle$. Define, if possible, a linear transformation $f: \mathbb{R}^4 \to \mathbb{R}^4$ such that:

[
f(\mathbf{S}) = \langle (0, -1, 0, 1) \rangle, \quad f(\mathbf{T}) = \langle (-1, 0, 1, 0) \rangle, \quad \text{and} \quad \mathrm{Im}(f) = \mathbf{W}^\perp.
]

[grizzled pagoda]

938c2cc0dcc05f2b68c4287040cfcf71

[tough mica]

hopefully i translated correctly

[small pendant]

um

im confused on the domain of F

am i trolling or is $S\not\in\mathbb{R}^4$?

[grizzled pagoda]

Arnavutköy

[tough mica]

S is a subspace of R4

[small pendant]

oh the subspace generated by (2,3,1,1) and (1,1,,1,0) I understand

[tough mica]

yeah?

[small pendant]

ok can you come up with the respresentation of the vectors which generate T

[tough mica]

a basis for T?

you just need to solve the system of two equations

,w rref {{1,0,1,1,0},{2,-1,0,1,0}}

[grizzled pagoda]

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[tough mica]

x1 = - x3 - x4

x2 = -2x3 - x4

(x1,x2,x3,x4) = (-x3-x4, -2x3 -x4 , x3,x4 ) = x3(-1,-2,1,0) + x4(-1,-1,0,1)

T = <(-1,-2,1,0),(-1,-1,0,1)>

[small pendant]

ok

so now if we can just show that the image of the generators lie within the corresponding images

we should be done

[tough mica]

we havent even touched the last condition

[small pendant]

sorry we are done with the first two conditions then

[tough mica]

do you know the dimension of the kernel and dimension of the image ?

[small pendant]

oh you are correct, thats the way

the dimension of the image is $3$. do you understand why?

[grizzled pagoda]

Arnavutköy

[tough mica]

yeah, is three

because W has dimension 1, its orthogonal complement will have dimension 3

dim(W) + dim(W perp) = dim(R⁴)

since they are complementary subspaces

[small pendant]

yes

[tough mica]

so the dimension of the kernel is 1, do you understand why?

do you happen to know rank nullity theorem?

[small pendant]

yes, i do understand why

[tough mica]

good

Im(f) = Wperp

and W perp has three vectors inside, dim(W perp) = 3

dim(Im(f)) = 3

dim(Ker(f)) = 1

we need to map one of our vectors to the vector zero

the thing is: to uniquely define a linear transformation, idk if you know this but

[small pendant]

ok i claim that the kernel must be in both $S$ and $T$

[grizzled pagoda]

Arnavutköy

[tough mica]

why is that

[small pendant]

okay. suppose the kernel isn't in $S$.

[grizzled pagoda]

Arnavutköy

[small pendant]

Then, do you agree that $f(2,3,1,1)$, and $f(1,1,1,0)$ are linearly independent?

[grizzled pagoda]

Arnavutköy

[tough mica]

f(2,3,1,1) and f(1,1,1,0) should be linearly dependent or no?

we are being told f(S) = <(0,-1,0,1)>

f(S) = <(0,-1,0,1)>
f(<(2,3,1,1),(1,1,1,0)>) = <f(0,-1,0,1)>

[small pendant]

yes, thats the contradiction

if the kernel isn't in $S$, that implies $f(2,3,1,1)$ and $f(1,1,1,0)$ are linearly independent, when we know that they are linearly dependent

[grizzled pagoda]

Arnavutköy

[tough mica]

fr

same for T, then

[small pendant]

yeah

so the kernel is in $S\cap T$, right?

[grizzled pagoda]

Arnavutköy

[tough mica]

yeah

[small pendant]

and what is $S\cap T$?

[grizzled pagoda]

Arnavutköy

[tough mica]

lets find the intersection

S = <(2,3,1,1),(1,1,1,0)>
T = {x in R4 | x1 + x3 + x4 = 0; 2x1 -x2 + x4 = 0}

a(2,3,1,1) + b(1,1,1,0) = (2a+b,3a+b,a+b,a)

(x1,x2,x3,x4) = (2a+b,3a+b,a+b,a)

i) (2a+b)+(a+b) + a = 0 ==> 2a + b + a + b + a = 0 ==> 4a + 2b = 0 ==> 2a = -b
ii) 2(2a+b)-(3a+b)+a = 0 ==> 4a + 2b - 3a -b + a= 0 ==> 2a +b = 0 ==> 2a = -b

[small pendant]

,w det {{-1,-2,1,0},{-1,-1,0,1},{2,3,1,1},{1,1,1,0}}

[grizzled pagoda]

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[tough mica]

they have an intersection

[small pendant]

wait did we find the kernel though

yeah, this intersection will have dimension $1$ namely

[grizzled pagoda]

Arnavutköy

[small pendant]

so we know it is the kernel

[tough mica]

we can map the vector that generates SnT to (0,0,0,0)

[small pendant]

yes

[tough mica]

i) (2a+b)+(a+b) + a = 0 ==> 2a + b + a + b + a = 0 ==> 4a + 2b = 0 ==> 2a = -b
ii) 2(2a+b)-(3a+b)+a = 0 ==> 4a + 2b - 3a -b + a= 0 ==> 2a +b = 0 ==> 2a = -b

now we go back to the generic form of S and plug in the linear dependency

in

(2a+b,3a+b,a+b,a)

b = -2a

(2a-2a,3a-2a,a-2a,a)

(0,a,-a,a)

SnT = <(0,1,-1,1)>

Ker(f) = <(0,1,-1,1)>

[small pendant]

okay

[tough mica]

f(0,1,-1,1) = (0,0,0,0)

[small pendant]

yes

[tough mica]

to uniquely define a linear transformation we need to define it in a basis of its domain

for example , f : R⁴ -> R⁴

[small pendant]

correct

[tough mica]

we are looking for the basis of R4 of

<v1,v2,v3,(0,1,-1,1)>

[small pendant]

so we already have the set $(1,1,1,0), (0,1,-1,1), (-1,-1,0,1)$, right?

[grizzled pagoda]

Arnavutköy

[small pendant]

we want to add a fourth one such that it forms a basis

[tough mica]

T = <(-1,-2,1,0),(-1,-1,0,1)>

S = <(2,3,1,1),(1,1,1,0)>

[tough mica]

unsure

[small pendant]

2nd and third form basis for T, while first and third form basis for S

im just doing this to keep it simple

because we can then assign values for $f(1,1,1,0)$, $f(0,1,-1,1)$, and $f(-1,-1,0,1)$

[grizzled pagoda]

Arnavutköy

[tough mica]

wait a second

f(S) = <(0,-1,0,1)>
f(T) = <(-1,0,1,0)>

[small pendant]

yes

[tough mica]

I am trying to see how you got that basis

{(1,1,10, (0,1,-1,1), (-1,-1,0,1)}

[small pendant]

because we know from your calculations that $(0,1,-1,1)$ lies in the bases of $S$ and $T$

[grizzled pagoda]

Arnavutköy

[small pendant]

then we add other vecotrs in $S$ and $T$ linearly independent to $(0,1,-1,1)$ to get bases for $S$ and $T$

[grizzled pagoda]

Arnavutköy