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[elfin berry]

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[minor verge]

how do you do that with a 3x3 matrix? is there a formula?

[bitter solstice]

how did you do it for the 2x2 matrix? I'm curious now

[minor verge]

this is how we derrived the formula for a 2x2 matrix in class, do i just do the same thing for 3x3 instead

[bitter solstice]

right so what's different?\

do A-lambda I_n for the 3x3

find its determinant

it will be the above polynomial basically

[minor verge]

ok cool

tysm

[bitter solstice]

no worries!

it is more involved I get it

[marsh citrus]

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[marsh citrus]

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[rapid tangle]

Can someone help pls

[brittle shuttle]

isnt this the solution?

[marsh citrus]

@rapid tangle Has your question been resolved?

[marsh citrus]

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[ember flame]

is this the answer?

just cross checking

[ember flame]

B = ?

I did
(B^T C = D)(C^-1)
B^T = (DC^-1)^T
B = C^-1^T D^T

[stark turret]

your reasoning is correct

[marsh citrus]

@ember flame Has your question been resolved?

[marsh citrus]

@ember flame Has your question been resolved?

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[wraith vector]

Hello! I'm having trouble on this problem. I've attempted it but its incorrect.

[graceful owl]

What are the options in the dropdown?

[wraith vector]

these are the options

[visual snow]

first of all i love the pfp

[graceful owl]

The first one is reduced i believe

[visual snow]

the second one should actually be in echelon form

[wraith vector]

thank you !

mm

it still shows it as incorrect

but sadly this problem doesn't show which is incorrect

[static quarry]

check that last one again

[wraith vector]

oh its not echelon since theres no zero below the 1 ?

[static quarry]

right

[wraith vector]

would it be reduced echelon instead

[static quarry]

no, reduced echelon is echelon with additional properties

this isn't even echelon

[wraith vector]

Theres another problem thats similar. Are these fine before I submit it

[marsh citrus]

@wraith vector Has your question been resolved?

[wraith vector]

<@&286206848099549185>

It turned out to be wrong

but I'm not sure which are wrong

[ember flame]

Oh webwork

[ember flame]

I’m pretty sure second one is RREF

Free variables are x1 and x4

I’m assuming that you don’t have to do gauss jordan first if neither is part of the dropdown

So yeah, it’s probably just the second one

[marsh citrus]

@wraith vector Has your question been resolved?

[marsh citrus]

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[soft tangle]

Hello

[soft tangle]

Can I get help on this

This is my work

This is the question digitally

[low ruin]

Do you have a particular question

[soft tangle]

Yeah an I doing it right?

[lofty stirrup]

remember chain rule

[soft tangle]

I don’t know what to do i’m new to integration by parts

[lofty stirrup]

for derivative of u with respect to t

[still temple]

do you know tabular integration by parts

[soft tangle]

For the v i’m guessing?

[low ruin]

Basically for integration by parts you want to use the theorem. integral f(x)g(x) dx = integral f(x)dG(x)

[leaden monolith]

This is incredible

[low ruin]

From there you can directly apply the integration by parts formula

[soft tangle]

I know the formula

I just don’t know how to don’t how to do the problem

Because it is definite

[shadow thorn]

use integration by parts

[soft tangle]

I tried not sure if it’s correct

[shadow thorn]

let ln(5t) be u

[low ruin]

It does not matter if it is definite. The bounds are unchange

[soft tangle]

Yeah that’s what I did

[shadow thorn]

for the uv part, you have to evaluate it at 8 and 3 and subtract

basically ftc

t^4ln(5t)/4 at 8 - t^4ln(5t)/4 at 3

[low ruin]

You have not taken the derivative of ln(5t) correctly. Using chain rule you will get 1/t. Also it not clear what you are integrating with respect to. You shouldn't drop off the dt.

[shadow thorn]

8^4ln(40)/4 - 3^4ln(15)/4

and then at the end you have int 3 ^ 8 of t^2

with some 5 nearby

but then it goes away

you dont need to worry about 5 too much tbh cuz its basically a constant

derivative of ln(t) = derivative of ln(5t)

[soft tangle]

Ok I think i got it

Thanks

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[still temple]

what is the point of putting the eq for a line in symmetric form

[novel juniper]

Wdym symmetric form

[still temple]

sorry like the symmetric equation

where its like n = (a, b,c) x - x0 / a = y-y0/b etc

[novel juniper]

$\frac{x-x_1}{\sin\theta}=\frac{y-y_1}{\cos\theta}$?

[elfin berry]

Why am. I here

[novel juniper]

So the same thing but the trigonometric identities are replaced by a and b?

[still temple]

yeah i guess?

i mean n is the normal

[novel juniper]

https://math.stackexchange.com/questions/3012519/can-you-please-explain-what-the-symmetric-form-of-a-line-signifies

This may help

[still temple]

sorta like having it as rise over run?

[novel juniper]

Or are you talking about the form $n=xcos(\alpha) +ysin(\alpha)$?

[elfin berry]

Why am. I here

[novel juniper]

If this is what you're talking about then it allows you to easily find the distance between the origin and line which can make solving problems easier

[still temple]

no im not talking about anything related to trigonometry

[novel juniper]

I'm not very familiar with the form you're using, mind defining all the symbols?

[novel juniper]

Ah, lines in 3D sorry, thought you were talking about lines in 2D

Ok so a,b,c are the direction cosines of the x coordinate, y coordinate and z coordinate respectively

That allows you to find the inclination of the line with the axes with relative ease

[still temple]

sorry ! i should have clarified

[still temple]

hmm we never learned about this in class, do you think u could send a video explaining it

[novel juniper]

https://youtu.be/ZoCxUV893fo?si=Yq43L7VcpJidMhT3

Sorry a,b,c are direction ratios not cosines

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

thanks, i appreciate you

.close

[marsh citrus]

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[waxen remnant]

is this legitimate

[waxen remnant]

trying to do trig sub for the integral

with $x=\tan \theta$

[elfin berry]

znsn

[waxen remnant]

and when im going through i got $\int \frac {tan\theta \cdot \tan^{-1}(\tan\theta) d\theta} {\left(1+x^2\right)^2}$

[elfin berry]

znsn

[fervent rampart]

you certainly can, although i'm not sure how useful it would be

[sand fable]

x = tan(theta) is a nice way of doing this integral

[waxen remnant]

i did get a result that works

i just want to see if the method is legitimate, because

arctan(tan(theta)) isnt always theta

[fervent rampart]

you can restrict the domain of theta to just -pi/2 to pi/2 (which is what we usually do for trig subs anyway)

[waxen remnant]

oh yeah you are right

🤦‍♂️

thanks

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[sinful fog]

Help

[sinful fog]

Hiii can someone teach me how to compare key features of graphs? Like domain range end behaviors etc

[marsh citrus]

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[marsh citrus]

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[dusky ledge]

Not sure how to do this one? if anyone could hlelp me out

[wise musk]

Just differentiate the entire thing wrt x

Meaning d/dx

[dusky ledge]

i dont really know how to start it

[marsh citrus]

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[still temple]

I need help with cosines.

[marsh citrus]

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[still temple]

.close

[fickle stream]

Hello, I'm trying to find the value of pi with only using a square and a circle and apparently, things don't seem great with my work, can anyone guide me how to do that?

[rotund pike]

This channel isn't yet ready for new people (it takes a bit after the channel is closed in case the channel was closed by mistake).

[marsh citrus]

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[cedar glade]

how do i simplify to get to the second equation?

[cedar glade]

<@&286206848099549185>

[marsh citrus]

@cedar glade Has your question been resolved?

[cedar glade]

My tries lead to large unsolvable terms

[marsh citrus]

@cedar glade Has your question been resolved?

[cedar glade]

<@&286206848099549185>

[marsh citrus]

@cedar glade Has your question been resolved?

[open relic]

Hello. Notice that the vector is
(y'(t)*(-A), x'(t)*A)

where A is that factor multiplying x'(t) in the second component

find the modulus using those terms

[marsh citrus]

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[dire root]

I am stuck on 3.b.

[dire root]

I need help getting started

[shadow warren]

first graph the function

so you can get a good visual of it

[dire root]

x= -1 and x=2

[shadow warren]

that’d be the roots yes

[dire root]

Maximum turning point

[shadow warren]

did you graph the function?

[dire root]

is it like that?

[shadow warren]

yep

now we can see the area that is bounded yes?

[dire root]

yes

[shadow warren]

alright so what would be the integral here

[dire root]

integral of 2x+2-x^2-x

[shadow warren]

from?

what are the bounds

[dire root]

2 and -1

[shadow warren]

yep

[dire root]

i got 21/2

[shadow warren]

i got 4.5 instead

what did you input as your integral?

[dire root]

i did it again and got 7/2

[shadow warren]

that’s still a bit off

[dire root]

i put x+2-x^2

[shadow warren]

yep and the anti derivative would be?

[dire root]

x^2/2 + 2x - x^3/3

[shadow warren]

yep

now plug in

[dire root]

2 and -1

[shadow warren]

yep

[dire root]

ahh

9/2

i got it

[shadow warren]

alright good

[dire root]

thank you

.close

[marsh citrus]

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[viscid tulip]

ABCD is a square where DM = MC. it is also known that BN is perpendicular to AM. prove that triangle BCN is an isosceles triangle

[marsh citrus]

@viscid tulip Has your question been resolved?

[viscid tulip]

<@&286206848099549185>

[night mica]

what you have tried?

[severe dawn]

yes!

[viscid tulip]

i have seen that triangle DAM is similar to NBA
i called DM = x from there AB = 2x and so on expressing each side i can with x. using the similarity i gained ratio and input the x values but it doesnt seem to help

my main thing was to define NB and NC with X but i cant find NC

[viscid tulip]

there is proly a simpler way im just not seeing

[night mica]

right

[viscid tulip]

proly angels

but i cant find it

MNCB is cyclic quadi

if that helps

[night mica]

do you think that the congruent sides are NB and NC?

[viscid tulip]

now that ur saying

maybe its BN and BC or BC and NC

tbh

seems way more reasonable that it has something with the side BC

thats where the square qualities come in good

so i guess its gotta be something with side BC

either NB or NC

[dire root]

Its okay i got it

[severe dawn]

yippee

[viscid tulip]

was thinking of calling angle NBC a and NMC = 180 - a

somehow maybe proving BNMC is a kite

look there is also Pythagoras in many triangles

maybe expressing NB and NM with X

as we have BC = 2x and MC = x

[harsh zinc]

use this

[viscid tulip]

how can that help

other than BM being a diameter

and NBC + NMC = 180

[harsh zinc]

did you notice that <BNC = <BMC?

[viscid tulip]

good sight, they are both on BC if we draw the circle. so lets let BNC = BMC = a to work like this?

[harsh zinc]

sure

[viscid tulip]

might be easier to draw the circle

[harsh zinc]

then, how can we use the fact that DM = MC?

[viscid tulip]

great

MC = 0.5 BC

then a must be 60

as if its a 30 60 90 triangle

[harsh zinc]

no, it doesn't have these angles

[viscid tulip]

?

MCB is a right triangle

oh yeah

my mistake

is supposed to be MB

not BC

if i went on saying that

[harsh zinc]

okay, what can you say about triangles ADM and BCM?

[viscid tulip]

well they are congruent

[cedar glade]

THANK YOU SO MUCH BRO i got it now much love man🙏🙏🙏🙏🙏🙏

[viscid tulip]

then ABM is isosceles

[harsh zinc]

yes, so <AMD = <BMC, right?

[viscid tulip]

yes

<MBA = 180 -2a

then both the small ones are a

[harsh zinc]

now you should use again that BNMC is cyclic

[viscid tulip]

how can i use it

[harsh zinc]

well what properties of it you know?

[viscid tulip]

opposite angels sum up to 180

of the quad

[harsh zinc]

right

[viscid tulip]

NBM + NMB = 90?

i dont see something that helps

[harsh zinc]

yes but that's not what you need

[viscid tulip]

i need BCN

[harsh zinc]

so, <NMC + <NBC = 180

[viscid tulip]

true

[harsh zinc]

you know NMC, right?

[viscid tulip]

NMC is 180 - a

NBC is then a

so NBM = 2a - 90?

[harsh zinc]

back to this

180 - a + <NBC = 180

so you have <NBC = <CNB

[viscid tulip]

oh i see

thanks

if we say AN = m
how can i express the perimeter of triangle BCN using m?

maybe proporions

thanks for helping

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[craggy lance]

Hello

[craggy lance]

I have one simple question

What is this type of question called?

I would like to practice them more

But I don’t know what to search

Many thanks if anybody knows

[marsh citrus]

@craggy lance Has your question been resolved?

[craggy lance]

<@&286206848099549185>

[shrewd forge]

algebraic simplification x in terms of n

[graceful elbow]

Probably a logarithmic problem, about finding x in terms of n.

[craggy lance]

Thanks guys

[craggy lance]

Nothing came up with this

[craggy lance]

Will try this

Couldn’t find anything either

Any other ideas on what to search?

[marsh citrus]

@craggy lance Has your question been resolved?

[dim schooner]

$hint:\(\sqrt{27})^{4n+6}\
=(\sqrt{27})^{2(2n+3)}\
=\Big((\sqrt{27})^{2}\Big)^{2n+3}\
=\Big(27\Big)^{2n+3}\
=\Big(3*3^2\Big)^{2n+3}$

[elfin berry]

nico

[dim schooner]

are you still here?

[still temple]

@craggy lance

[craggy lance]

Yes

[craggy lance]

I know

I want the name

[dim schooner]

the name of what ?

[craggy lance]

The type of question

What are they called

I want to solve more of them

[dim schooner]

i think they are just called exponential functions

found this for exercices https://www.math-exercises.com/functions/exponential-function

[craggy lance]

Thanks

So much

.close

[marsh citrus]

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[little condor]

How do we tell if the answer we get in lim x->∞ questions is finalized?

[marsh citrus]

@little condor Has your question been resolved?

[marsh citrus]

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[quaint sky]

let me try this

(a^2-6)((20/a+5)(-4))

is this (a^2-6)(-80/a-20)

.close

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[plucky jolt]

How do I solve equations with log?

[plucky jolt]

I figured it out so close if you see

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[trail flint]

ok I think I might have this now but no fully sure

This seems like its it

but I don't really get the values

[mint gust]

!status

[marsh citrus]

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

[trail flint]

They are to small for me to really see

[trail flint]

3

0.16692906

[mint gust]

Are you working out the integral?

[mint gust]

Or are you doing something else

[mint gust]

What is your question

[trail flint]

that's the integral

worked out

I don't know if that's right to start

[mint gust]

Are you working for the area of y or just the integral

[trail flint]

This is the full question

I need to find the total area of that region

so I set up a function for the upper half of the region and a function for the lower half

to integrate then add

the upper half I deemed was y * 1/y - 0.145

[mint gust]

Riemann sum is it?

[trail flint]

no this is shell method

[mint gust]

Ok

[trail flint]

finding volume of an object creating by rotating around x-axis

created*

[mint gust]

Why do you think your answer is wrong then

Remember, the left and right endpoints are given by 1/x

Subbing in the value, I would think 1/7 is around 0.1428

Instead of 0.145

Maybe that’s why you were wrong?

[trail flint]

yeah I think so

the integral is right

the values seem wrong though

are those 0.04 increments?

[mint gust]

Your integral is 1 unit wide and at least 1/7 unit tall

So you should be looking for an answer between 1/6 and 1/7

[trail flint]

I got this

as one of them

[mint gust]

This looks a little too high, 1/6= 0.16666

Try and rewrite your limits of integrals

[trail flint]

I don't get it

I come to this

is this right?

[marsh citrus]

@trail flint Has your question been resolved?

[trail flint]

Shell method is confusing the hell out of me

https://cdn.discordapp.com/attachments/907022703230345237/1204598312196382821/image.png?ex=65d55099&is=65c2db99&hm=97c5420459aaaa835a96c618705132a8d926430e83dec127adbb1161b7511218&

I still don't really get this

here is a help video for reference

[marsh citrus]

@trail flint Has your question been resolved?

[trail flint]

<@&286206848099549185>

[marsh citrus]

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[trail flint]

.close

[marsh citrus]

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[marsh citrus]

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[marble blade]

Where is the -x coming from

I understand uptil the second step

Where did the 10 -x = 3 come from?

[low ruin]

So this comes from basic substraction

To go from the 2nd line to 3rd line you subtract x on both sides

[marble blade]

Thank you

.close

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[still temple]

Can somebody help me with the case 2? I didn't understand at all.

[main idol]

Which equation first do you not understand

[still temple]

In case -2 all the four equations. Basically the entire case 2

Like how they equate alpha = beta²-2 and all. The entire logic i didn't understand or the concept

[main idol]

alpha = beta^2 - 2 comes from the given information that alpha^2-2 is also a root if alpha is a root

[still temple]

Why did they do "AND" the alpha = beta²-2 and beta = alpha²-2 ?? How they equating 4 roots?

[main idol]

Those aren't 4 distinct roots

Two equations, two roots

[still temple]

could you explain it elaborately? what's the purpose of those cases since this question has been bothering me and there's no explicate solution on internet.

[marsh citrus]

@still temple Has your question been resolved?

[main idol]

The purpose is to find what a and b are

[marsh citrus]

@still temple Has your question been resolved?

[topaz bronze]

some please help me https://hips.hearstapps.com/hmg-prod/images/math2-1569526052.jpg?crop=1xw:1xh;center,top&resize=980:*

[still temple]

I know the purpose but equating beta² = alpha² -1 I couldn't understand

[main idol]

Where's that equation

[main idol]

You mean this one?

I alpha and beta are distinct roots so their other roots are also distinct, but there are only two roots so beta's other root must be alpha

And vice versa for alpha's other root

You just need to realize there are only two distinct roots for the case alpha and beta distinct

[still temple]

so you are saying root alpha itself has roots?

[main idol]

No

[still temple]

Could you explain the entire case -2? when alpha ≠ beta. How did they arrive with all those possibilities? That's the part i didn't get it

[marsh citrus]

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[main idol]

alpha = beta^2 - 2 follows from just 2 facts and 1 case. alpha and beta are roots. If alpha is a root, so is alpha^2 -2. You're in the case alpha is not beta

[main idol]

I've been explaining that the entire time

Please read

[marsh citrus]

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[still temple]

Got it got it.

[marsh citrus]

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[opaque robin]

when i tried in rectangular i got the spiral to be

(0,0),(5,5),(2.5,12.5),(0,12.5),(-1.25,11.25),(-1.25,10)

[opaque robin]

Please help me devise a spiral function for this

"All this started when i took a isosceles triangle and the split it into half and then extended the other side creating another congruent triangle and it went on to create this beautiful spiral like shape (similar to the golden ratio spiral but not the same(i am unsure))"

[still temple]

i just realized that there's a desmos feature

https://www.youtube.com/watch?v=7bE4Rt5Iuqc

[opaque robin]

@still temple

but this is for either linear, quadratic or n polnomial format

[still temple]

nvm i tried this, im getting absurd results

[opaque robin]

ye

u should beacuse we don't have general formula of sprial curve

do we?

[still temple]

yeah ig

[opaque robin]

but it is like polar right?

[still temple]

$r=ae^{b\theta}$

[elfin berry]

deltaG

[opaque robin]

polar

ye

[still temple]

yup

[opaque robin]

not rectangular

hmm

we can convert this into

[still temple]

oh

[opaque robin]

rectangular

[still temple]

that one is

$r=a\theta$ ig

[elfin berry]

deltaG

[opaque robin]

even this is polar

...

but in this r is constant too

[opaque robin]

<@&286206848099549185>

[still temple]

wait those are straight lines?

[opaque robin]

ye that is one property of this

straight line

s

it's just absurd thought of mine on paper which seems to devise something cool and new in math

idk

[raven olive]

I mean converting to polar isn’t that difficult x = rcos(theta)
And y = rsin(theta)

[opaque robin]

but that doesn't solve the problem

[opaque robin]

@raven olive

[raven olive]

The last coordinate seems to be off

[opaque robin]

naw it is fine till there

after that i wasn't able to find

like it was too small

[raven olive]

0,5 after that?

[opaque robin]

no

[raven olive]

Why not

[opaque robin]

no how

[opaque robin]

read this

[raven olive]

I did 0,5 follows the pattern

[opaque robin]

ok did you get a function

[raven olive]

No time rn but I’ll come up with one later

[opaque robin]

ok

close.

[raven olive]

It’s .close

I’ll contact you later

[marsh citrus]

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[marsh citrus]

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[trail badge]

Hi ! I need help please
f(x)= x^4-8x^2+2
f'(x)= 4x^3-16x

1. Study the variations of f and draw up your variation table
2. Deduce the local extrema of f
3. Provide guidance of f(x) as x ∈ [ -2 ; 1 ]
   My teacher told me to factorise f' but I really don't know how to do it.

[valid cape]

set f'(x) = 0 to find the roots

factorising is for finding the roots

[trail badge]

ok but I don't really know how to do it can you explain me please ? (I'm a beginner in maths 😬 )

[valid cape]

take the 4x out

$4x^3-16x=0$

[elfin berry]

FungusDesu

[valid cape]

after that you should get an equation under the form of AB = 0

which means A = 0 or B = 0

[trail badge]

ok thanks i'm gonna try that

[marsh citrus]

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[grizzled locust]

how do you do this

[still temple]

do you know newton lebnitz theorem?

leibnitz*

$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt = a'(x)f(a(x))-b'(x)f(b(x))$

[elfin berry]

deltaG

[marsh citrus]

@grizzled locust Has your question been resolved?

[grizzled locust]

no i have not heard of it looks like the quotation formula

[still temple]

quotation formula?

[grizzled locust]

quadration

i used deposit formula to solve it

not sure if it is correct translation

[still temple]

its definitely not the correct translation

[grizzled locust]

but i rewrote it with d [f(t)] x 1 and then F(x) - F(1) and then F(1) goes away

i can take a photo

[still temple]

yeah that works too ig

[grizzled locust]

but i dont understand how the derivitive of cos can be cos

[still temple]

send your work

[grizzled locust]

red says i name the cos(t^2) as f(t)

[still temple]

the answer is correct, i dont understand the process due to the language :(

[grizzled locust]

wait i will rewrite it

[marsh citrus]

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[grizzled locust]

.reopen

[marsh citrus]

✅

[grizzled locust]

.close

[marsh citrus]

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[marsh citrus]

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[proud zealot]

can somone help me with this

i found normal vector norm to be sqrt(4x^2 + 4y^2 +1)

using polar coordinates i get r*sqrt(4r^2 +1)

im not sure for the raidus though

is it sqrt(2)?

i tired it but didn't find the right number

nvm forgot to add the circle

sorry

.close

[marsh citrus]

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[proud zealot]

.reopen

[marsh citrus]

✅

[proud zealot]

actully no its still wrong

[marsh citrus]

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[void elm]

!status

[marsh citrus]

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

[void elm]

2

Here, K is a generalized eigenspace.

I have let v be in the intersection and let m and n be the smallest natural numbers such that (λI - S)^m(v) = 0 and (μI- T)^n(v) = 0

Then I take the difference to get (λI - S)^m - (μI - T)^n = 0

From there, I assume without loss of generality that m>n, and rewrite the same thing m instead of n as the exponent of (μI - T)

And then I used binomial expansion

After that, I don’t know what to do; I can distribute v back into the sum and apply each operator in the sum to it, but I can’t see what good that does for me

Alternatively I could attempt to rearrange terms somehow, but I don’t see a way to do that either, so now I’m stuck and need a bit of help

[marsh citrus]

@void elm Has your question been resolved?

[marsh citrus]

@void elm Has your question been resolved?

[marsh citrus]

@void elm Has your question been resolved?

[marsh citrus]

@void elm Has your question been resolved?

[marsh citrus]

@void elm Has your question been resolved?

[void elm]

<@&286206848099549185>

[blazing pulsar]

You have a vector v in the intersection of the LHS, you want to show it must be an element of the RHS. i.e., there exists a k such that (S+T - (\lambda + \mu)I)^k v = 0

Now expand using the binomial theorem or whatever and apply your assumptions as needed

You can try to keep going down your path if you want, but you should probably add instead of subtract them. Then your binomial expansion will be messier because you have to keep track of the m and n's but maybe that is preferable.

[marsh citrus]

@void elm Has your question been resolved?

[void elm]

expand the RHS?

[void elm]

I don’t see how this helps me…

[nova gale]

To prove the statement, let $v \in K(\lambda, S) \cap K(\mu, T)$. This means that $Sv = \lambda v$ and $Tv = \mu v$. We want to show that $(S+T)v = (\lambda + \mu)v$.

First, we compute $(S+T)v = Sv + Tv = \lambda v + \mu v = (\lambda + \mu)v$.

Since $v$ was arbitrary, this shows that $K(\lambda, S) \cap K(\mu, T) \subseteq K(\lambda+\mu, S+T)$.

[elfin berry]

Flamey

[void elm]

K is a generalized eigenspace though

so (T - μI)^n (v) = 0 and (λI - S)^m (v) = 0 for some m,n

[marsh citrus]

@void elm Has your question been resolved?

[void elm]

<@&286206848099549185>

reposting

[cinder ore]

Replace V with 17.6 divided by the formula sir.

[void elm]

huh?

[cinder ore]

REPLACE V WITH 17.6 DIVIDED BY THE FORMULA SIR.

DO YOU HEAR ME NOW?

SIR?

[void elm]

?

[void elm]

.close

[marsh citrus]

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[robust dune]

a

[north crater]

what kind of question is this

is this a joke

[marsh citrus]

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[earnest mica]

I've been trying to solve this question but I keep getting the wrong answer. I thought that integrating 1/4x^2+1 should give tan^-1(2x)?

[white willow]

What's the first step you do?

[earnest mica]

Since it's an improper fraction, I divided 2x^2 by 4x^2+1 and got 0.5 - 0.5/ 4x^2+1

oh my gosh i just checked the formula sheet i was doing it wrong i figured it out sorry for wasting your time

.close

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[fickle compass]

Hi

[fickle compass]

The length of the latus rectum of the parabola $25[(x-1)^2 + (y-1)^2] = (3x+4y-2)^2$ is

[elfin berry]

❄ѕησωƒℓαкє❄

[marsh citrus]

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[fickle compass]

bro-

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[velvet lava]

How are these graph transformations the same?

[shell ether]

If it's absolute value?

[fleet fossil]

If f is symmetrical across the y axis then f(x) = f(-x). Does that help?

[velvet lava]

But isn’t -x a transformation in itself

Reflection in the y axis

So is 5-x two transformations. A reflection and a translation

[fleet fossil]

Well f(x) isn’t a graph, if that’s what you’re getting at

It’s the value you get when you put in x

So f(x-5) isn’t changing anything about f, it’s not translating it or anything. It’s just the value of f at the point x-5

[velvet lava]

Let’s say f(x) is a graph

Like x^2 for example

What would the transformations be

[fleet fossil]

If I understood transformation correctly (not familiar with that term as I’m not a native speaker) a transformation would be -x^2+5

So g(x)=-x^2+5 would be a transformation

[velvet lava]

Like how would it make move

Up or down , left or right

[fleet fossil]

How would what make the graph move

[topaz lagoon]

translation refers to left right up down movement

with your question, (x-5) and (5-x)

[velvet lava]

Yes pls

[topaz lagoon]

(5-x) can be simplified to (-(x+5))

yes?

[velvet lava]

No

That would be -5-x

[fleet fossil]

The x is already negative

[topaz lagoon]

oh nvm

[fleet fossil]

Two negatives cancel out

[fleet fossil]

You mixed up the sign in the second one

It was minus

In the original question

So just spelling mistake

[topaz lagoon]

-(x-5) and (x-5) are same graph. There's no translation, but there is a reflection over x axis

sorry didn't see that @fleet fossil

[velvet lava]

So it would give the same value

If the graph is symmetrical

[topaz lagoon]

value??

[velvet lava]

Sorry shape

And Position

[topaz lagoon]

yes same shape

position same too

just a reflection in x axis

[shell ether]

not y?

[fleet fossil]

f(x-5) isn’t a shape though. That’s what I find so confusing about how functions are taught in high school. It’s just not correct

[topaz lagoon]

you can see it both ways, actually

[topaz lagoon]

that's true

[velvet lava]

Ok thanks everyone

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[whole minnow]

Problem with functions

A clock manufacturer can make a particular clock with a cost of $150 per unit.
It is estimated, that if the selling price of the clock is ''x'' , then the number of clocks that is sold in a week is [ 1250-x ]

a) Express the weekly utility of the manufacturer as a function of the selling price of each clock. Specify the domain in which this expression makes sense. Sketch the graph of the utility function.

[whole minnow]

ok so I know that the utility is earnings-cost

I understand this function too but I don't understand how to express the utility as a function

the domain in this one should look like this too 👍

<@&286206848099549185>

[marsh citrus]

@whole minnow Has your question been resolved?

[marsh citrus]

@whole minnow Has your question been resolved?

[marsh citrus]

@whole minnow Has your question been resolved?

[whole minnow]

<@&286206848099549185>

[marsh citrus]

@whole minnow Has your question been resolved?

[whole minnow]

<@&286206848099549185>

[marsh citrus]

@whole minnow Has your question been resolved?

[marsh citrus]

@whole minnow Has your question been resolved?

[pliant siren]

ok

so you have a function for the amount of clocks he sells, lets call it a(p), a for amount, p for price

now you need to sketch the graph of the weekly utility utility

to do that, you first need to find an expression for it

we already know how many people will buy our clock at a given price, a(p)

and to know how much we earn, we need to multiply this amount with the price, so we have a(p) * p

now from that, we need to subtract the cost per clock

[whole minnow]

Alright

[pliant siren]

we can get that by saying p * 150

so we have a(p) * p - 150p

now put in a(p)

(1250-p) * p - 150p

now distribute the p into the brackets

1250p - p^2 - 150p

= 1100p - p^2

and there you have it

do you understand the concept, @whole minnow

[whole minnow]

Thanks a lot yeah I got it

[pliant siren]

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[whole minnow]

Ill close later because this problem has a) b) c) d) e) etc

[pliant siren]

k

[whole minnow]

But they might be easy now

[pliant siren]

if you get stuck, ping me

[whole minnow]

Thank you

[pliant siren]

youre welcome

[pliant siren]

@whole minnow are you done?

[whole minnow]

Yeah just finished

The others were easy once I had the utility function

Thanks friend

.close

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[sullen ice]

What is this used for?

[stray hound]

this is called the poisson distribution. if there is an average number of occurrences of an event, you can use it to find the probability of k occurrences

[gleaming pecan]

this is the probability distribution function of a random variable with Poisson distribution X and its expected value and variance

[stray hound]

for example, if a Meteor strikes the earth an average of once every 10 years, lambda = 10. then P(X = k) is the probability that a Meteor strikes the earth k times

in 10 years*

[spark otter]

uh I don't think so

the expectation wouldn't add up, this is for lambda = 1

but if on average, 10 meteors strike the earth during a period of 100 years, then lambda = 10

and so the probability that k meteors strike the earth during a period of 100 years is P(X=k)

[marsh citrus]

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[drowsy bronze]

Can someone help with problem b

[drowsy bronze]

.close

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[dense wigeon]

Heya, what's the best strategy for integrating this? I'm struggling

[marsh citrus]

@dense wigeon Has your question been resolved?

[main idol]

Have you learned Integration by parts

[dense wigeon]

Yes

[main idol]

do that then on your last expression

[marsh citrus]

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[dim zodiac]

Two parallel lines are lines that increases/decreases at the same rate. And you're given a point, then it's to find the constant.

What kind of function are you finding?

Exactly, an equation to a line that is parallel to y=-4x+1. How does a line function look like?

This is the formula for a line at a specific point. A line function, linear function, has its form as y = mx + b

Due to that the line is parallel, the line we're finding is something in y = -4x + b

You know a point where the line goes trough, use it.

[mystic minnow]

That works, parallel lines have the same slope so you can pull the slop from the given line.
You can then plug your point into y1 and x1 and simplify your equation to standard form y = mx + b

You can also do what Good said, and use y = mx + b, you know m, and plug your point into x and y to find b.

[marsh citrus]

@still temple Has your question been resolved?

[mystic minnow]

yep

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[pseudo kayak]

So to find the limit of this...i multiplied the numerator nd denominator with that

As n goes to infinity..what will the next step be?

Like factoring out sqrt(n)??

[proud ice]

sounds reasonable

[pseudo kayak]

But the thing is..im a bit confused like...sqrt(sqrt(n))?

[proud ice]

Are you asking how to factor $\sqrt{n}$ out of $\sqrt{n+\sqrt{n}}$?

[elfin berry]

SWR

[pseudo kayak]

Ya like aftr this?

[proud ice]

this isn't quite correct

[pseudo kayak]

In wolframalpha the answer is 1/2..idk how

[velvet wharf]

1/2

you typed it wrong prob

its infity

[pseudo kayak]

I factored it like(sqrt(

[velvet wharf]

oh - sqrt n

[proud ice]

the irony

[pseudo kayak]

i thought its infinity too but these alpha🥲

[proud ice]

If you don't know how to factor, you can solve for $x$ in $\sqrt{n+\sqrt{n}}=\sqrt{n(1+x)}$ to get the factor

[elfin berry]

SWR

[pseudo kayak]

Like sqrt(sqrt(n)(sqrt(n)+1)) +sqrt(n)

[mystic minnow]

you can also divide everything in the numerator and denominator by root(n)

[pseudo kayak]

Ahh okkk

Yes now its 1/2..wow

Thanks🥹

.close

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[vague leaf]

Hey, Need to learn this. this is an example

[vague leaf]

what im wondering is

why is there a extra row for division on the left?

[harsh falcon]

technically, we first apply g(x)=-x, then h(x)=x^2, then k(x)=x+1
so in total f(x)=k(h(g(x)))=(-x)^2+1

but since we are squaring, that part is not really needed, as you noticed

[vague leaf]

so

sorry i might need that simplified

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@vague leaf Has your question been resolved?

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[lyric kelp]

why is C not defined in this context

[dreamy trench]

It says enter "an" antiderivative

Just one of them

[lyric kelp]

so when its 1 theres no C?

[fervent yarrow]

You should enter any antiderivative of f

[still temple]

You can let C = 0

There are infinitely many functions that are antiderivatives of $\frac{-18}{\sqrt{1-x^2}}$

[elfin berry]

45

[still temple]

They all differ by a constant

[still temple]

or 1, or 69, or 420, or pi

[fervent yarrow]

69 is an interesting number

[still temple]

so sad that 42069 isnʻt prime

😭

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[livid maple]

Can anybody please explain what "superlayer" for this tensor is?

[marsh citrus]

Please don't occupy multiple help channels.

[livid maple]

<@&286206848099549185>

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[still temple]

Hey i just need some clarification for beginning vector diagrams

[still temple]

Can someone quickly explain the difference between wind blowing from and at a bearing of an angle

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[sturdy acorn]

Bruh what did I do wrong

at the point 1, 0 isn't the derivative equal to 1?

Meaning that the slope at that point is POSITIVE. and it remains POSITIVE no matter the y - so as x increases, so does y

meaning that the domain is 0 to infinity, no????

Like what kind of brain dead error am I making, I feel like this problem should be trivial at its hardest

[marsh citrus]

@sturdy acorn Has your question been resolved?

[hushed vector]

Try (1, infinity)

It's
y(1)=0
Which means, x=1,y=0
And since we're talking about domain... We'll see x

@sturdy acorn
tell me if it's wrong...

[sturdy acorn]

ah that didn't work

I figured out what is up tho, thank you!

.close

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[fresh osprey]

How did my teacher get the degree 135 for the plane?

[fresh osprey]

i guess i dont really understand how to resonate it with the unit circle

vector navigations ^

[marsh citrus]

@fresh osprey Has your question been resolved?

[marsh citrus]

@fresh osprey Has your question been resolved?

[low ruin]

Northwest is 135 degrees from the positive x-axis. Is this what you are asking?

[fresh osprey]

yeah I don’t get it like

like what is he subtracting sometning am I not seeing something

[low ruin]

So if you don't know what you are asking I can't help

[fresh osprey]

oh

[low ruin]

Dp you have any specific questions

[fresh osprey]

T

I don’t understand ur answe

[low ruin]

So is there any particular thing you don't understand?

[fresh osprey]

yeah the 135 degrees

basically the first steps of the process from the beginning

[low ruin]

Okay imagine a compass

[fresh osprey]

ok

[low ruin]

At 0 degrees we are doing North

[fresh osprey]

ok

[low ruin]

We rotate 45 degrees counter-clockwise that is NW

correct?

[fresh osprey]

sorry was doing something

Why 45

[low ruin]

West is 90 degrees counter clockwise

If we go Northwest then we are going between North and West. As result we have 45 degrees

[fresh osprey]

ok

so

it’s a 454590 triangle

[low ruin]

Now we want to determine the angle from the positive x-axis

[low ruin]

not yet

[fresh osprey]

ok

[low ruin]

So what is it?

[fresh osprey]

The angle?

[low ruin]

the angle ye

*yes

[fresh osprey]

I’m not sure

[low ruin]

If we rotate from the positive x-axis by 90 degrees we get North correct. Because positive angle are counter-clockwise and negative angles clockwise by definition

Are you still there

I will give you the answer. We get 135 degrees. Since we want to find the x-coordinate, and y-coordinate we use the unit circle definition cos(theta) = x and sin(theta) =y. As a result we get 300cos(135), 300sin(135)

[marsh citrus]

@fresh osprey Has your question been resolved?

[north raptor]

open

/reopen

[marsh citrus]

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[still temple]

Did I do number 1 correctly?

[still temple]

I feel like I did something wrong idk why

Maybe the greater than sign is not correct

[dim zodiac]

Is this a test?

[valid cape]

the graph seems to be correct

[still temple]

no its not

[still temple]

It’s hw

it doesnt have -3 as a root

[valid cape]

albeit a bit ugly

[still temple]

-3 is x intercept

[valid cape]

but he does know the general concept

[still temple]

Is the set notation correct?

That’s the main thing

[valid cape]

if its asking >0, find the part of the graph thats above y=0

and thats your answer

[valid cape]

no

[still temple]

Oh shit

What did I do wrong now

[still temple]

What’s the correct set notation?

[valid cape]

for example if you want to express $x>0$, you do $${x \in \mathbb{R} | x>0}$$

[still temple]

Oh shittt

Wait

[elfin berry]

FungusDesu

[valid cape]

now im shit at latex

but

[still temple]

Is the lil part at the bottom count as well?

Since it’s technically above the x int

[valid cape]

- wikipedia

[still temple]

This paet

Does that count?

[valid cape]

no, its below y=0

[still temple]

My teacher doesn’t make us write it like dat

[still temple]

The HA is 1/3

[valid cape]

yes, its still above y=0 so it counts

[still temple]

Bruhhh

[valid cape]

your graph is not the best looking

so yeah

it should look somewhat like this

[still temple]

So it’s x>0 or x<1/3?

My graph looks crap the

Tbh

[valid cape]

no

[still temple]

What

Ohhh mb

X<-3

?

[valid cape]

you see this redline here

thats your roots

[still temple]

Yes

[valid cape]

and...?

[still temple]

X>0

[valid cape]

correct

and theres your answer

[still temple]

Ok Ty

My graph looked really weird

.closr