help-38

[neon dirge]

for x e R

[sweet turret]

Wtf I'm so confused

[neon dirge]

(1 0 0) is an eigenvector for the eigenvalue 0

(0 0 1) is not an eigenvector

[sweet turret]

Ok

[neon dirge]

since M * (0 0 1) is not a multiple of (0 0 1)

[sweet turret]

U said any multiple of (1 0 0) maps to (0 0 0) ...how

[neon dirge]

M * (x * (1 0 0)) = x * (M * (1 0 0)) = x * (0 0 0) = (0 0 0)

[wooden plover]

M * (1 0 0) gives you the first column of M

Which is filled with 0s

[sweet turret]

But

M is (x y z) right

[wooden plover]

M is a matrix mate

[sweet turret]

So isnt it x + 0 + 0

[wooden plover]

We're talking about your matrix here

M is this

[sweet turret]

Oh wait

[sweet turret]

x + 0 + 0 =0

Which gives x=0

That's big brain

Thanks for help

I will juz do more practice

[neon dirge]

I'm a bit confused what you refer to with x + 0 + 0 = 0

but M * (x * (1 0 0)) = x * (M * (1 0 0)) = x * (0 0 0) = (0 0 0) describes why any multiple of (1 0 0) maps to the zero vector

likewise by arguing with the first column of M

[sweet turret]

Wait what about the last row

I still dgi

[neon dirge]

wat?

the last row of M is 0 0 0

[sweet turret]

Ya

[neon dirge]

yea

[sweet turret]

Why can't I use the same logic as x

[neon dirge]

wdym by that

[sweet turret]

How did u get 110

[neon dirge]

by multiplying M with (0 0 1)

[sweet turret]

Oh fk

I keep thinking M is the corresponding row

[neon dirge]

nono

M is the matrix

[sweet turret]

Okok

[neon dirge]

I also want to briefly point out how you could've checked your previous answer from the former task

[sweet turret]

Thanks for the help

[sweet turret]

Wut the

[neon dirge]

which would also give you the same result as your calc

I use y' synonymous to dy/dt btw

[wooden plover]

Yeah having the eigenvectors doesn't help very much in the first place tbf, you have a jordan block already

You still have to solve the system explicitly either way

[sweet turret]

Why did e-t get differentiated

[neon dirge]

here?

[sweet turret]

Ion think this in my syllabus

Yea

[neon dirge]

d/dt(e^-t) = -e^-t

I did that in reverse

[sweet turret]

Woah

[wooden plover]

You've seen integrating factors ?

[sweet turret]

Wtf how did u even think of reversing a differentiation

Yea

Oh

By using that

[wooden plover]

That's what lunatic did

[sweet turret]

[neon dirge]

nywys gl with further tasks and come back any time 🦩

[sweet turret]

Yea thanks again

Both yall

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[ripe valley]

Do we not have a formula for the sum of the first m odd numbers raised to the power k?

[ripe valley]

I am trying to derive a formula, by binomial expanding (2j-1)^k but uhh I don't see anything good

[ripe valley]

Generating function of this is e^z(e^2zm -1)/(e^2z -1)

[ripe valley]

But again, can't derive anything useful from here

[umbral dragon]

Maybe you could make use of just taking out the first m even numbers from faulhabers formula, since they have a common factor of 2^k?

[ripe valley]

Ehh yes that's a good idea let me try that

[ripe valley]

It worked, tyvm!

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[fair beacon]

can someone explain how you do this

[fair beacon]

i thought of doing something like this:

$\begin{pmatrix} 1 & 1 & 1 & 1 \ -1 & 1 & 1 & 2 \ 1 & 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} a \ b \ c \ d \end{pmatrix} = \begin{pmatrix} x \ y \ z \end{pmatrix}$

[solid kiln]

carburetor

[fair beacon]

and then solving for all x, y, z satisfying the equation

but i'm not sure how to proceed from here

also, $W + U = { w + u : w \in W, u \in U }$

[solid kiln]

carburetor

[acoustic flint]

What is LS(…)?

[dusty sleet]

I believe span

Or the usual ⟨ v1, v2 ⟩

[acoustic flint]

First time seeing that

[dusty sleet]

Me too

[acoustic flint]

Row reduce that matrix augmented with zero vector, if there is a single column without a pivot then it spans R^3

For the intersection one, just equate the linear combinations of each span

[trim jolt]

@fair beacon Has your question been resolved?

[fair beacon]

can you explain why this works?

i don't understand why you augment it with the zero vector

Ax = 0
if the RREF(A) has a single column without a pivot then there only exists the trivial solution x = 0

that would mean that the column vectors are independent right

but is that enough to say that it spans R^3?

[acoustic flint]

Three of the columns are linearly independent. That's not enough to say it spans R^3?

[fair beacon]

oh right got it

[acoustic flint]

Augmented because of the definition of LI

[fair beacon]

yeah i got it, you're finding solutions to Ax = 0 if i'm not wrong

[acoustic flint]

the point is to show that three of the four vectors are LI

Yeah

[fair beacon]

yep, i got that one thank you

i have one more question if you don't mind

[acoustic flint]

in this case, you don't need to actually find the set of solutions, it's more of a notation thing

sure

[fair beacon]

can we apply rank-nullity theorem here?

i was thinking of defining a transformation T : R^5 -> LS((1 1 1 1 1)) as T(x) = Ax then by rank-nullity dim(Image(t)) + dim(ker(T)) = 5 but rank(T) = 1 so dim(ker(T)) = 4?

[acoustic flint]

Why is T mapping from R^5 to <1,1,1,1,1>?

that can't be defined as Ax since A maps R^5 -> R^2

[fair beacon]

oh i see

so the dim(Image(t)) is at most 2

dim(ker(T)) = 3 or 4

[acoustic flint]

yeah

which <1,1,1,1,1> clearly isn't

[fair beacon]

yeah right

so the answer is no right?

[acoustic flint]

mhmm

[fair beacon]

and also is there a way of doing this without using rank-nullity?

[acoustic flint]

rank-nullity is the best to reason about it

Im not sure of a way that isn't implicitly using rank-nullity

[fair beacon]

oh alright

because this problem set was given before the theorem was introduced to us

[acoustic flint]

I think now that you put rank-nullity in my head it' s all i can think about

let me see

I mean ultimately you know you are going from R^5 to R^2 so you will be losing at least 3 dimensions

[fair beacon]

yeah

[acoustic flint]

but again, saying that is pretty much leading to rank-nullity

[fair beacon]

i have a tutorial on tuesday where they'll discuss this pset and if they show a method that doesn't involve rank-nullity directly i will let you know 👍

thank you very much

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[tardy ledge]

how do i find the surface area of this leaf?

[tardy ledge]

its a special leaf called string-of-pearls and im trying to find its surface area but the problem is the formula is obv 4pir^2 but if i measure the distance from center to the rightmost edge (from this angle) i will be measuring the arc length

so hwo can i find the surface area given that

[woven python]

assume it spherical and measure its diameter would be the simplest way to go about it🤔

wait is it like flat or a bit spherical

[tardy ledge]

its almost spherical

but how do i find diameter

if im only able to measure the arc lengths

[tardy ledge]

yes thesee

[west sleet]

Looks like grapes fr

[woven python]

easiest would be using a ruler on the edges

like vernier callipers

[west sleet]

Idk maybe you could do surface area of revolution

[tardy ledge]

is there a mathematical way to do it? this is part of a larger thing im doing where im trying to use calculsu to find the surface areas of leaves and see if its correlated ot the tree heights

[boreal apex]

this is more of an ellipsoid. Try to find the length of the semi-major, semi-minor, and intermediate semi-axis radii and calculate the area from that

[tardy ledge]

oh yeah

[tardy ledge]

i have it graphed in geogebra

could you help me

[west sleet]

Mean there's a lot of estimates you could do ig

[woven python]

depending on how much you want to estimate, this would also work

[tardy ledge]

i have it kinda centered around the origin here

[tardy ledge]

no this wouldn't work

cuz i want to use calculus for it as much as possible

[west sleet]

Then just surface area of revolution then

[tardy ledge]

yeah okok

so

[west sleet]

I mean that makes a lot of a assumptions

But ye

[west sleet]

Maybe you'd want a different angle

[tardy ledge]

its assuming its a perfect sphere right

[west sleet]

No not that it's a perfect sphere

[woven python]

applied mathematics in a nutshell

me thinks

[west sleet]

You can maybe use an ellipse

Gl bro

[tardy ledge]

😭

do you think its hard

[west sleet]

As aero said it's more of an ellipsoid

[west sleet]

Time consuming ish

[tardy ledge]

yeah lol actually its part of a larger project im doing

[west sleet]

Why you doing this anyway

[tardy ledge]

for my hs i need to have an "internal investigation" on something we like applying the math we learned lol

[woven python]

just

[tardy ledge]

so i like biology so i thought this would be cool

[woven python]

assume it spherical vro 🙏

[tardy ledge]

lmaoo

i think ima do elipse though

[woven python]

sure

[tardy ledge]

js to be accurate as possible and not oversimplify

[west sleet]

But that'd be too eas-

[tardy ledge]

cuz i gain marks for making assumptions as long as i back it up

[west sleet]

Also these def aren't spheres

[woven python]

just so you know that surface area of an ellipsoid is... a bit difficult than a sphere

[tardy ledge]

if i have this, then for surface area of elipsiloid what could i do

[west sleet]

Just a teeny bit

It's fine

What could go wro-

[tardy ledge]

i would mark out the points on the upper side of the x axis that are like the "edges" of the leaf right?

[woven python]

vs

[west sleet]

I mea-

[woven python]

4pi*r^2

[tardy ledge]

this is perfect

💀

i knew my ia was going to be too easy lmao

i acc did another leaf before and that was a lot of pages

so this is going to be enough to set up the rest of the lab i feel like

[woven python]

i mean, uh, good luck i guess

[tardy ledge]

thank you

but first

could i also get some help on where to start

[woven python]

it's such a small leaf too assuming it spherical wouldn't change the values that much-

unless you want to be rigorous...

[west sleet]

I mean what do you even need this surface area for anyway

What is the purpose of calculating this

[spice fulcrum]

solid area of rev

i think

[west sleet]

Covered

[spice fulcrum]

ah

unlucky

[west sleet]

Rip

[tardy ledge]

well my research question was "does the surface area of a leaf correlate to the height of the tree" since in my biology textbook it said taller trees typically compete for sunlight by having larger leaves so it would explain the adaptation

but obv its not true for all cases

and thats kind of waht im trying to see

[west sleet]

I mean in that case ig you might need to be kinda exact

[tardy ledge]

yeah

i did the banana leaf before

[spice fulcrum]

complex equations always leads to more marks though right 💀

[tardy ledge]

it used matrices and i had a lot to explain with polynomial interpolation and vandermonde matrix

[tardy ledge]

hopefully

😭

[spice fulcrum]

thats awesome

[tardy ledge]

but also rn could i get some pointers on where to start with the surface area of revolution stuff

assuming its an ellipse

whats like the first thing i should do once i have it in geogebra

is the first step to create an equation that is like (x)^2 + (y)^2 = 1

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@tardy ledge Has your question been resolved?

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@tardy ledge Has your question been resolved?

[tardy ledge]

now that I have the equation how can I calculate the surface area?

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[hallow bluff]

Can anyone help me, I’m stuck.

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@hallow bluff Has your question been resolved?

[hallow bluff]

<@&286206848099549185>

[cedar elk]

Draw heights from L to AB and AD

Then label some angles

@hallow bluff

[hallow bluff]

Okay, will do.

Anything after that?

[cedar elk]

Notice something

[hallow bluff]

How exactly do I draw the lines? Any specific point on AB and AD?

[cedar elk]

<CBP = <LHP

Where H is foot from L to AB

[hallow bluff]

okay I see it now.

So how do we make it so <LPA = <CPB?

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[native gull]

i need some help here

[red mountain]

,rccw

[red mountain]

$\lim_{x\to 2} \left(2x\sqrt{\frac{x}{x-2}}-x\right)$

[solid kiln]

Bonk

[red mountain]

is this the question?

[frosty slate]

LHopital?

[native gull]

can you help me

[red mountain]

!noans

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[red mountain]

!nosols

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[red mountain]

fk, wrogn one

[small glen]

OH

i was just about to explain it

should i delete?

[red mountain]

no its alright

just keep it in mind for next time

[small glen]

use simple algebra in the beginning in order to use L Hopital, and then be a bit careful with the derivatives. next once its in proper form you can substitute x for 2

[small glen]

will do

[native gull]

what are you doin' in the 2 step

[small glen]

multiplying 2x and square root of x

[native gull]

i don't understand

[small glen]

erm

wait

take it step by step

multiply this first

[native gull]

(2x* sqrt(x))/sqrt(x-2)

what's you think

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[dusky pike]

me gots a question

[dusky pike]

if i wanted to turn a number from one base to another with a matrix

what could it look like

say I wanted to input a number in base 10 and get out a number in base two or something

[ionic pendant]

i'm not aware that that's something possible to do by matrix multiplication

[dusky pike]

yeah

[ionic pendant]

usually base conversion is done by taking the quotient and remainder a bunch of times

[dusky pike]

i was pondering how to solve a problem

it was around the lines of having a special base that is based on a function of n

and you could only multiply each place by one or zero

and i was trying to see if you can represent any number using that base

whatever

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[fleet notch]

Why is abs necessary for the ln x^2 + 25 term? Isn't x^2 + 25 always positive?

[fleet notch]

(given x in R)

[lone basin]

The absolute value is not necessary for ln(x^2 + 5)

[fleet notch]

(also I already caught the mistake don't worry. I just don't have the corrected answer as a screenshot)

[lone basin]

If you have $\int \frac 1u \dd{u}$, you can simplify to $\ln(u)$ if $u > 0$

[solid kiln]

King Leo

[fleet notch]

So for functions that are always positive, abs isn't necessary for ln?

Thanks!

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[glad pumice]

Can someone please help me on this question:

[glad pumice]

I know BD = 6

and EC=28

and I think triangle ADE is an isosceles triangle

and EC is the base of the quadrilateral BCDE, same as AE is the base of triangle ADE

idk what to do next

<@&286206848099549185>

height of triangle ADE is square root of {312}

area is $14\sqrt {312}$

[solid kiln]

RamenNoodleMan21

[glad pumice]

how am i supposed to calculate the area of shape BCED

<@&286206848099549185>

i was thinking you could simplify the quadrilateral into simpler shapes

helllp

it;s my last question please help

nevermind i just solved it

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[past widget]

Did you finish it?

It'd be something like:
[ADE]/[ABC] = (AD•AE)/(AB•AC)

So [ADE]/[BCED] = (14•19)/(25•42 - 14•19) = 19/(75 - 19) = 19/56

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[limpid dawn]

status

[low cloud]

i dont know what that means

[slender shard]

How far along are you in solving the problem?

!status

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1. I don't know where to begin.
2. I have begun but got stuck midway.
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4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
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7. None of the above

[low cloud]

3

[limpid dawn]

!show

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

[slender shard]

Can you show your work?

[slender shard]

e^ax shouldn't change.

The coefficient comes down, but the exponent does not change.

[low cloud]

so it says 5x

[slender shard]

e^5x -> 5e^5x -> 25e^5x -> ...

[low cloud]

okay i see where i went wrong thx

i should be able to do it now

[low cloud]

actually how do i know when to stop if they both just continue going forever

[slender shard]

So this integral is interesting because it is cyclical.

e^x doesn't change however the trig function cycles back to itself after four steps.

And the only difference will be the coefficient.

So you will get a multiple of the original integral.

[low cloud]

okay that makes sense

[slender shard]

Which you can move to the left-hand side(LHS).

It will look something like this.

[low cloud]

do you mean like this?

[slender shard]

One moment, I'll show you.

Doh, made a mistake.

The fourth term is the same as the original integral with only a difference in the coefficient.

Although I suppose you could do the same with the second iteration. 🤔

[low cloud]

okay ya i see that its the same

and that is what signals me that i can stop and have the answer

[slender shard]

Yes.

e^x * cosine or sine will be like that.

[low cloud]

it says that this is still wrong

[zinc ginkgo]

Try simplifying

[low cloud]

i dont know what that would change?

[slender shard]

One moment, let me make an easier example.

So note that I went through two iterations of the DI method. You can write the third line as an integral which happens to be the same as the original integral except that it is negative.

[low cloud]

okay so for my question would i just multiply by 2401/125 in stead of dividing by 2?

[slender shard]

On the third iteration of the DI method, you would make the fourth line into an integral which is similar to the original integral with only a difference of a coefficient.

[low cloud]

ugg im so confused

[slender shard]

Ok, let me expand it then.

[low cloud]

sorry i dont know why this isnt clicking i feel like it souldnt be this hard for me to understand

[slender shard]

It's all good. We're here to help you understand. 😉

[low cloud]

wait why does the last box get two lines to it

[slender shard]

I chose to stop there because if you move horizontally across, you are making an integral.

And that integral at that iteration of the DI method happens to be a similar to the original integral with only a difference of a coefficient, -125/343.

I suppose its one of those things you have to go through the proof of integration by parts to understand why that is the case.

The DI method is just integration by parts in a tabular form afterall.

[low cloud]

okay so you make an integral on the last step, why?

[low cloud]

ya but the examples we did all went to 0 and thats how i knew when to stop

which i guess is why he gave us a hw question like this to practice but it still sucks

[slender shard]

Because for this integral, both parts will continue on forever if you choose to keep going on. We rely on the fact that on the fourth iteration, it happens to repeat so to speak.

e^5x will be just a difference of a coefficient. sin(7x) will just alternate between +- cos(7x) and +-sin(7x).

But on that third iteration, it makes an integral which is just a difference of a coefficient, -C * I.

[low cloud]

where the "I" is what the integral of the original function would be?

[slender shard]

Yes.

[low cloud]

okay cool

[slender shard]

And because it is just -c I, you can add it to the left-hand side.

And to solve for I, you can just divide both sides by 468/343.

[low cloud]

where did 468 come from?

[slender shard]

I + 125/343 I

343/343 I + 125/343 I

[low cloud]

oh got it

also why can u turn both integrals into "I" when one is sin and the other is cos

[slender shard]

I messed up. 😛

Pretend I did a fourth iteration.

https://tenor.com/view/fml-sylvester-cat-annoyed-head-bang-gif-13045781

Let me fix that.

[low cloud]

okay so it should be 625/2401e^5xsin7x

[slender shard]

I think I got that right now. 🥴

As you can see, even when you "know" what you're doing you can still make mistakes. 😛

Although now that I think about it, you can stop at the second iteration. 😛

[low cloud]

omg your going to make me cry

[low cloud]

hahah your all good

[slender shard]

My apologies, I'm making this all kinds of complicated. 😄

[low cloud]

🥹 its okay

[slender shard]

But you see how it is cyclical and the original integral reappears?

[low cloud]

ya

[zinc ginkgo]

,w int exp(5x)sin(7x)

[solid kiln]

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[slender shard]

I swear to god constants are my immortal enemy.

[low cloud]

is that one just simplified or is there something wrong in my work?

[slender shard]

Wolfram's answer is just simplified.

[low cloud]

okay sick i was really scared that there was another mistake and i actually would have cried this time

[slender shard]

This is really a unique case with e^x and a cosine or sine function that is cyclical.

I don't think there are any other examples in which this will occur, but I'm probably mistaken.

[low cloud]

ya i hope that i dont have to do this again

thanks tho

[slender shard]

Oh, you'll probably see this on a test.

Just know that if you see e^ax * cos(x) or e^ax *sin(x), you'll want to use this.

[low cloud]

great

well at least now im 99% sure i know how to do it

[slender shard]

Better than 0%.

[low cloud]

thx again

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[wraith hinge]

Hi guys

[empty orchid]

Hm?

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@wraith hinge Has your question been resolved?

[west sleet]

你好

@wraith hinge

What do you want to ask cuh

@wraith hinge this is the place to ask

[wraith hinge]

ohhhh

kk

[west sleet]

Gah damn bruh

So what's the question

[wraith hinge]

I was asking if this is the place where math lovers are ?

😅

[empty orchid]

...

[wraith hinge]

cuz I am one too and I like to learn from people

only this

[final kiln]

nah we all hate math :)

[wraith hinge]

sorry

no please be honset, I really want to learn from you guys if this is teh case

[west sleet]

Is English your first language

[wraith hinge]

uhh yes, please don't be mean I am just asking

[empty orchid]

This is a math server. Either:

1. You like helping people with math
2. You want to learn math
3. You like to watch others do math (weird but valid)

[west sleet]

Then you can read the #❓how-to-get-help

And #rules to get an intro to this server

[wraith hinge]

Okay Thanks, I will be right back

[eternal nimbus]

4. can’t do math, don’t love math, only monkey

[wraith hinge]

So I've got this question, and I am getting the answer as 90degrees, can anyone verify it?

#help-38

[eternal nimbus]

Area = (pi)r^2, circumference = 2pi(r)

Personally I’d solve for radius then use either area or circumference to find the angle

[wraith hinge]

I gues its about the sector of the circle, right? so like the area will be: theta/360 * pi*r^2 and perimeter of the sector will be: theta/360 *2 * pi * r + 2r. Isn't that so?

[empty orchid]

,rccw

[wraith hinge]

yep

[empty orchid]

Yes.

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[wraith hinge]

uhhh you guys get the question?

[empty orchid]

Although I do not know about that +2r part.

[empty orchid]

Yes.

[wraith hinge]

see the kength of the arc has the formula: theta/360 * 2 * pi * r, cool

and when we need to find the total perimeter of the sector

then we need to add the extra two radii of the sector

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[wraith hinge]

making the perimeter of the sector theta/360 * 2 * pi * r +2r

[wraith hinge]

am I not right?

[eternal nimbus]

I’m confused 🤔

[wraith hinge]

wait a second

[eternal nimbus]

No need for +2r,

[exotic pine]

need

[eternal nimbus]

2pi(r) = circumference, theta/360 equals percentage of circle

Unless I’m misunderstanding something

[exotic pine]

the sector also has its "legs"

[eternal nimbus]

I like this method, the theta/360 factors out and you get this system of equations

[zinc ginkgo]

Looks right
https://www.cuemath.com/measurement/perimeter-of-a-sector/

[exotic pine]

the red?

[wraith hinge]

here

[zinc ginkgo]

2/360 = 1/180

[eternal nimbus]

You’re right, my bad, completely forgot about the legs

[eternal nimbus]

Mu bad 🙏🙏🙏

[empty orchid]

Okay.

[west sleet]

Yea it's just a simultaneous equation

[empty orchid]

So now we have a system of equations, in terms of theta and r.

[wraith hinge]

can you checkout my solution?

also when I found the value of beta, then the radius or the circle was coming to be 6.5, and when solving the equation using the radius the value of theta wasn't matching the value of any of the options given in the options, so I took radius to be 7cm

#help-38

pleade help guys

<@&286206848099549185>

????

[eternal nimbus]

There doesn’t seem to be a mistake, I don’t have a pen or paper with me right now, so I’m not 100% sure

What angle did you get, you can just plug back in to do a quick check

I think there should be two solutions because it’s a quadratic, one small circle with large angle and one big circle with small angle.

[wraith hinge]

yes this is a quadratic equation

and

I am getting radius = 6.5 for the second value of the equation

but when substituing this value in the equation the answer doens't match with any of the options given in the question

so taking r = 7, we get theta = 90degrees

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[midnight vessel]

It's for calculating the number of spectral lines during transition from $n_2^{\text{th}}$ to $n_1^{\text{th}}$ orbit.
If I need to go from $n_2=4$ to $n_1=1$, total number of transitions would be$\$
$4\to3:4\to2:4\to1\3\to2:3\to1\2\to1\\$

So, the formula would be $\binom{n_2-n_1+1}{2}$. But I don't understand how. I came up with an alternate solution

Number of spectral lines = $\sum_{k=0}^{n_2}{(n_2-n_1-k)}\\$
But the first one looks much simpler to implement.

[solid kiln]

GoldBarley

[midnight vessel]

How do you get to the first formula?

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@midnight vessel Has your question been resolved?

[silver glade]

because there's one transition for each pair of orbits

n2-n1+1 is the number of orbits from n1 to n2

[midnight vessel]

Oh, and the 2 represents the transition from x->y?

[silver glade]

you're looking for pairs

yeah

[midnight vessel]

Oh, yea
Out of n2-n1+1, im selecting a pair of two.

Thanks for the help!

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[solemn hound]

What have you tried

[royal swan]

i dont where to start

[solemn hound]

Do you understand what the question is asking

[royal swan]

yes

[solemn hound]

Yea so I’ll just say it here so we’re on the same page it’s asking you at what time t is the distance from M to N the smallest

Try to make a function that is equal to the length between M and N

[royal swan]

i know

[solemn hound]

Like for a given t what is the length between M and N

[royal swan]

if its on the same side

i could come up something

but they're on the opposite side

[frosty slate]

Hmm

[solemn hound]

Okay so ur struggling with making that function I see

[frosty slate]

Use the cosine law

[royal swan]

ohhhhh

thank you

let me try that

[frosty slate]

lol

[royal swan]

is it t=16?

@frosty slate @solemn hound

[frosty slate]

idk

[solemn hound]

I’m too lazy to check

Show ur work

[frosty slate]

same

[solemn hound]

And I’ll check

[frosty slate]

lol

[royal swan]

my answer match the provided answer

[solemn hound]

Gg

[royal swan]

thank u 2

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[halcyon estuary]

Help.

[halcyon estuary]

gotta find the K value that holds X^-10

I chose k = 5

[lone basin]

The coefficient of x^(-10)?

[halcyon estuary]

Yeah.

I suppose that's how its' called.

[halcyon estuary]

binomial theorem, yeah

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[elfin cedar]

I must have a fundamental misunderstanding of cylindrical shells because I keep getting things wrong. Bounds are $$ y=\sqrt{x}+2 $$, y = 2, x = 4. Revolve around x axis. Inverse of the function is (y-2)^2, which should be the height, bounds are 2 and 4. Since there is no offset, radius should just be y. Integrate with respect to y.

[solid kiln]

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

[limpid dawn]

Can you show your work

[elfin cedar]

$$ \pi\int_{0}^{4}\left(\left(\sqrt{x}+2\right)^{2}-4\ \right)dx $$ this I believe should be correct and it evaluates to about 92.

[solid kiln]

Fragment

[limpid dawn]

this doesn't look like shell method

[elfin cedar]

I don't have my phone with me right now, but I am pretty sure I am not making an algebra error here. y - 2 = sqrtx, (y - 2)^2 = x

[elfin cedar]

It's not, it's the washer method.

[limpid dawn]

ok I see

[elfin cedar]

I set up another method to compare.

The bounds I have are correct, correct variable to integrate with respect to as well.

[limpid dawn]

well that's your region basically

[elfin cedar]

yes

[limpid dawn]

your outer radius should be what

oh nvm

you set it up

ok but what does the solution say?

[elfin cedar]

I do not have a solution sheet, I just know that it is not correct.

Am I getting the radius wrong?

[limpid dawn]

hmm why not

[elfin cedar]

2pi * int r * h dy should be the form

[limpid dawn]

that's shell method

[elfin cedar]

yes

[limpid dawn]

oh i see you tested both

and got different results

[elfin cedar]

I was comparing the evaluation yes

but yes I don't know whether I am incorrectly finding h or r

[limpid dawn]

what were your bounds

[elfin cedar]

2 and 4

they are correct

[limpid dawn]

yes

[elfin cedar]

the bounds aren't relevant to r are they?

[limpid dawn]

so h should be h = (4-(y-2)^2)

[elfin cedar]

ugh

thank you

[limpid dawn]

and r should be just y

[elfin cedar]

I'm braindead

[limpid dawn]

,w Integrate[2pi * y * (4-(y-2)^2),{y,2,4}]

[solid kiln]

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[elfin cedar]

yeah that fits the result

for the other method

thanks

[limpid dawn]

yep

[elfin cedar]

I guess I just have to pay attention to what's actually getting revolved

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[wraith hinge]

Can i get some help with a pattern in my game?

[wraith hinge]

So in my game there are harvesters and planters with areas like this

⬜️⬜️⬜️⬜️⬜️
⬜️⬜️⬜️⬜️⬜️
⬜️⬜️⬛️⬜️⬜️
⬜️⬜️⬜️⬜️⬜️
⬜⬜⬜⬜️⬜️

What would be the most efficient layout of both harvesters and planters

Sorry if this a dumb question

[austere matrix]

gang what.

[wraith hinge]

Nvm

I thought this would be easy for some math nerds

[exotic pine]

well maybe cause we dont have any context??

[zinc ginkgo]

Clearly it's minesweeper

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[real herald]

Could i have shown this by factoring out the 4 instead? as 4n is always even too?

[real herald]

also could I have done this with a direct proof? Did i need a counterpos

By this i mean a = 2n, then a^2 = (2n)^2 = 4(n^2) which is always even

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[real herald]

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[tight wind]

can somoneone help me w how to start solving

[tight wind]

(asking for total vertical distance)

[lone basin]

Assuming t -> infinity?

[tight wind]

yes

[lone basin]

Ok lets consruct a series

Assume n is the amount of bounces that have passed

[tight wind]

$\sum{n=1}^{\infty} (2/3)^n$?

[tight wind]

oh shoot

[lone basin]

$\sum_{n=1}^{\infty} (2/3)^n$?

[solid kiln]

King Leo

[tight wind]

Yes thats what i meant

[lone basin]

,w sum (2/3)^n from n = 0 to 1

[solid kiln]

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[lone basin]

You should start at n = 0, because the ball technically falls 18 feet before the first bounce

[tight wind]

$\sum_{n=0}^{\infty} 18 \cdot (\frac{2}{3})^n$

[solid kiln]

alwaysonce

[tight wind]

would this represent it

Forgot the 18 the first time

[lone basin]

Yes

So do you know how to find where this series converges to?

[tight wind]

18 / (1-(2/3)) = 54?

[lone basin]

✅

[zinc ginkgo]

Are you missing a factor of 2 somewhere

[lone basin]

?

[tight wind]

yay ty

[lone basin]

Wait

Maybe

[tight wind]

Oh

Where

oops yeah i put it in and 54 is wrong

Oh like bc we have to add it bouncing up and bouncing down

[tight wind]

Do u think 2(54) - 18 would cover it

[lone basin]

One sec

,calc (5/3)/(1 - 2/3) * 18

[solid kiln]

Result:

90

[tight wind]

wait also when i do this what should my notation look like

Like is the geometric series formula thing supposed to be s_n = a / (1-r)

i forgot if its s_n or smth else

[lone basin]

$\mathrm S_n$ refers to a limited sum

[solid kiln]

King Leo

[lone basin]

You should just use $\mathrm S = \frac{a_k}{1 - r}$

[solid kiln]

King Leo

[lone basin]

$\sum_{n = k}^\infty r^n$ converges to $\mathrm S = \frac{a_k}{1 - r}$

[solid kiln]

King Leo

[lone basin]

@tight wind to help construct the proper sequence, can you find how much the ball has travelled between

• [0, 1]
• [1, 2]
• [2, 3]
• [n, n + 1]

[tight wind]

assuming the sequence is r^n Is a_k always equal to a_0

[lone basin]

You dont know that, because the summation might start at n = 5 or whatever other number

[tight wind]

ohh

$\sum_{n = k}^\infty a \cdot r^n$ converges to $\mathrm S = \frac{a_k}{1 - r}$ writingto visualize

[solid kiln]

alwaysonce

[lone basin]

(in this formula, you can technically replace a_k with r^k

[lone basin]

Btw a is a sequence, while a_1 is a number

[tight wind]

Oh

wth

Ugh this unit is so conufsing

I thought a was just a constant

[tight wind]

Like isn't a always 18 here

[lone basin]

No, $a_0 = 18$

[solid kiln]

King Leo

[tight wind]

18 meters, 18(2/3) + 18(2/3), 12(2/3) + 12(2/3), uhh

[lone basin]

Try to separate it differently

What youre doing will be confusing, because from h = 18 to h = 0, the ball technically only performs a half bounce

So i would recommend like this:

Calculate red distance, calculate green distance

[tight wind]

Red distance is 30 and green distance is 20

[lone basin]

So whats the ratio

[tight wind]

Then what is a_2

[tight wind]

3:2

[lone basin]

In the sequence $a_n = 18 \cdot \qty(\frac 23)^n$, $a_2 = 8$

[solid kiln]

King Leo

[lone basin]

It should be (next distance) / (previous distance)

[tight wind]

When ur finding where this series converges to using a_k/(1-r) why am i using 18 as a when 18 is just a_0 and not a_k

What is k here

[lone basin]

k is the start of the summation

[tight wind]

2/3

[lone basin]

$\sum_{n = \textcolor{red}k}^\infty r^n$ converges to $\mathrm S = \frac{a_\textcolor{red}k}{1 - r}$

[solid kiln]

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

[tight wind]

OHHHHH

[lone basin]

And what is the first term of the sequence

Wait one sec

[tight wind]

If n were to start at 2 here, then to find where it converges would it be 8/(1-(2/3))

[lone basin]

✅

[tight wind]

umm

is it not 18

im cooked for my test😭

[lone basin]

18 is this blue distance

But you havent reached n = 1 yet

[tight wind]

How come 1 isnt the vertex

Oml

[lone basin]

#help-38 message (tell me if you need me to explain further)

[tight wind]

I think im misunderstanding what a bounce is 💀

Ok i see

then the first term is 30

[lone basin]

So you have:
$$\sum_{n = 0}^\infty 30 \qty(\frac 23)^n$$

[solid kiln]

King Leo

[lone basin]

Can you find what that converges to?

[tight wind]

Oh yes 90

Tgats teh right answer than k u

[lone basin]

W spelling, but np

[tight wind]

LMAO

.close

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[lone basin]

(btw you technically never needed the summation. You could have just use a0 = 30 and r = 2/3 to invoke the S = ... formula)

[tight wind]

Oh ok true

[zinc ginkgo]

This was also right. Just do 2 * 18 (2/3) + 2 * 18(2/3)^2 +...

And factor out a 2

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[tight wind]

like i couldve done -18 + 2$\sum_{n=0}^{\infty} 18 \cdot (\frac{2}{3})^n$

[solid kiln]

alwaysonce

[tight wind]

?

[lone basin]

,w -18 + \sum_{n = 0}^{\infty} (18 * (2/3)^n)

[solid kiln]

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[tight wind]

Ok um

[lone basin]

Something messed up

[tight wind]

,w -18 + \sum_{n = 1}^{\infty} (18 * (2/3)^n)

[solid kiln]

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[tight wind]

nvm

[lone basin]

!done ?

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[tight wind]

Earlier when i ddi 2(54) - 18 it worked though

[lone basin]

If the answer is 90, how did 2(54) + 2 work

[tight wind]

Sry brainfart edited

[lone basin]

You need to ask riemann about that 🤷‍♂️

[zinc ginkgo]

This is correct

[tight wind]

Oh u typed it wrong here

,w -18 + 2 * \sum_{n=0}^{\infty} 18 \cdot (\frac{2}{3})^n

[solid kiln]

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Requested by a1waysonce

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.close

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[crimson halo]

I need help when explaining how both c angles of the triangle are congruent, what would be the reason but mostly why?

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[rigid orbit]

The box is 48 cm, if the diameter ratio between large and small circle is 2:1, what is the minimum area of the circle indside the box?

[rigid orbit]

π(48 -x)², then find the X vertex

X vertex= 48, then (48-48)²

I sum up all of the area of circle but couldn't get the result

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[mighty lynx]

i did answer the first step of the question which is to get the inverse function but i don't know how to get the domain and range

[wraith hinge]

domain will be x\in R since the denominator will never be 0, range will be y\in [-3,3) since there exist a horizontal asymptote y = 3, and since the numerator can be 3(x^2-1), where x^2-1 always bigger or equal to -1, the lower bound is -3.

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[wheat stirrup]

Solve for $x = \frac{a\pi}{b}+k2\pi,k\in\mathbb{Z},\frac{a}{b}$ in its simplest form: $2\sin{x}\sin2x=3-\sqrt{3}\sin{x}$

[solid kiln]

nhknyugn

[ripe valley]

!status

[trim jolt]

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

[wheat stirrup]

Step 2

[ripe valley]

What did ya try

?

[wheat stirrup]

I tried applying some trigonometric formulas but then I got stuck...
My first attempt was to try converting these into a single variable t=sin(x)

$4\sin^2x\sqrt{1-\sin^2x}=3-\sqrt{3}\sin{x}$

[solid kiln]

nhknyugn

[ripe valley]

Not that bad to solve

[wheat stirrup]

Do I have to check the case if \cos{x} is negative => = \sqrt{1-\sin^2x}?

[ripe valley]

No i believe

[ripe valley]

Square both sides, u have a kinda nice quartic

[wheat stirrup]

$16\sin^4{x}(1-\sin^2{x})=9-6\sqrt{3}\sin{x}+3\sin^2x$?

[solid kiln]

nhknyugn

[ripe valley]

Wait actually nvm

[wheat stirrup]

Yeah this is where I'm stuck

I don't know how to simplify it any further :P

[ripe valley]

Probably not the right approach then

[ripe valley]

Divide by cos^3 x maybe?

Nvm the 3..

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[echo crystal]

anyone who can help with a physics problem help 8 pls

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@wheat stirrup Has your question been resolved?

[magic eagle]

wb using 2sin(a)sin(b)

or using sin2x=2sinxcosx

[wheat stirrup]

I tried that

[magic eagle]

oh mb

[magic eagle]

got any ideas?

[wheat stirrup]

Nope

[solid jewel]

write 2sinxsin2x as cosx-cos3x

take the sin term to the LHS

cosx+√3sinx-cos3x = 3

the maximum value of the function on the LHS is 3 and this occurs when cosx+√3sinx = 2 and cos3x = -1 simultaneously

from the first equation you will get the condition x = (6n+1)𝜋/3 and from the second x = (2m+1)𝜋/3

take the intersection and you will be left with the solution set x = (6m+1)𝜋/3, m∈ℤ

@wheat stirrup

[wheat stirrup]

I don't understand this part. How did you conclude cosx+√3sinx = 2 and cos3x = -1

[solid jewel]

the maximum value of cos+√3sinx = 2

and the minimum value of cos3x = -1

only for these values would the equality hold

you have to find all the points of maxima of the function on the left

cause it varies from [-3,3] and the RHS is 3

[solid jewel]

global maxima that is

not local

[barren acorn]

Take m=0 => x = 60°
I plugged it into original equation, the equality doesn't holds

[solid jewel]

2(3/4)=3-3/2

it holds

[magic eagle]

i tried plotting a graph

x comes out nearly as pi/3

[solid jewel]

ye the solution set is correct

[magic eagle]

oh, in desmos the graph was showing an error idk y

hen i equated them

btw which app/website is it?

[solid jewel]

desmos

[magic eagle]

huh

[magic eagle]

alr

oh u simplified it

i just got lazy and sent that given eqn lol

[solid jewel]

i just took it all to one side

no simplification

[wheat stirrup]

Is it a coincidence that the RHS is the maximum value of the LHS?

[solid jewel]

its most prob intentional

[wheat stirrup]

Yeah I guess so

[solid jewel]

cause otherwise it would be much harder to solve

[wheat stirrup]

Alright thanks ;)

.close

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[solid jewel]

np

[floral hollow]

.close

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[amber heart]

wtf

[dusk summit]

so this is my question

and what i have so far

i just need help on what i need to do next

[amber heart]

Ur doing it wrong

[dusk summit]

is my work right so far

[dusk summit]

how

[amber heart]

Ur working out is not right

[dusk summit]

whats wrong?

[amber heart]

P x a x D

[dusk summit]

lemme open another channel

.close

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[feral mist]

Help I dont know how they got this

[limpid dawn]

You look at the right matrix and start 1 maps to 2

then you look in the left matrix for the 2, 2 maps to 4

so the result is 1 maps to 4

and you do that each time

2 maps to 4 on the right
on the left 4 maps to 1
so 2 maps to 1 is the result

[feral mist]

Ohhh

Thank you so much

.close

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[runic spade]

Translate:

3. Given: A...D₁ is a cube with an edge length equal to √72 cm.
   Find: the distance between the lines AA₁ and BD₁

4. Given: ABCD is a rectangle, MB ⊥ (ABCD), AB = 60, BC = 80.
   Find: the distance between the lines MB and AC

I need help not with solving the tasks themselves, but with drawing for these tasks

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[runic spade]

<@&286206848099549185>

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[runic spade]

<@&286206848099549185>

[frosty slate]

edge length?

@runic spade

Is that Russian?

[runic spade]

Yes

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[sturdy skiff]

how would one decompose $\frac{x^4+1}{x^3+2x^2+x}$

[solid kiln]

haveaniceday12

[wraith hinge]

oh

fun

[sturdy skiff]

it's definitely not doing
$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+1}$

oops

[wraith hinge]

do you know

polynomial diviion

division

[solid kiln]

haveaniceday12

[sturdy skiff]

yes, but we've only been given one example and we didn't use that

but i skipped lecture so maybe i missed that

how would one use polynomial long division for this

[wraith hinge]

u are close

tbh polynomial division is the easiest way

[sturdy skiff]

i know that the general formula is some function + remainder/original denom

[wraith hinge]

take the leading term(x^4)

divide by leading term of denominaor

(x^3)

[sturdy skiff]

yes

[wraith hinge]

multiply x by entire divisor

substract the original dividend

then repeat the process

[sturdy skiff]

ik how to use polynomial long division

i've done it before, i am quite familiar

[wraith hinge]

then what's the problem you are facing?

[sturdy skiff]

in this case, i don't know what will be what when i am done

[wraith hinge]

explain

[sturdy skiff]

like when i get my final result, what will that actually be in terms of the decomposition

well acc i think i know

but what will be the remainder

this is what we were told

so using polynomial long division, i guess that the thing we get after dividing is q(x)

[stable crescent]

whats the question?

[sturdy skiff]

but what is ax^2 + bx + c in this case

[stable crescent]

^

[sturdy skiff]

the question is right in front of you idk what i can say

[stable crescent]

where