help-13

[fringe mauve]

but we want y(x)

no y(t)

right?

have u delt with parametric ebfore?

[clear laurel]

No sorry

[clear laurel]

Yea its an x y graph so id assume so

[fringe mauve]

tahts ok

basically we need to get rid of t

no lucky for u sinc there is not air res

this si ez

[clear laurel]

Oh perfect 😋

[fringe mauve]

so we have x = 40cos(theta) t right?

we can just solve for t

and sub it in to our y frunction

to get a function with only y and x

[clear laurel]

No wait i dont have an x

At least i dont think so

[fringe mauve]

t = x/40cos(theta) no?

i just rearraged for t

[clear laurel]

I dont have both x and t i mean i tried finding t with t= v/a using the y values to find time for the whole trip of the projectile and got 1.035 but looking at calculators for this i dont think i got it right

[fringe mauve]

time for the whole trip?

why do u need that?

[clear laurel]

For some of the formulas

Right?

Or do i put random numbers for t

[fringe mauve]

i guess it could help

so u can find whre it lands

and also the nighest point

how did ufind the t

[clear laurel]

Using v(final) = v(initial) + at

And i think i was told that initial velocity for vertical is always 0 so it was v=at and rearanged to t= v/a

And a is the gravity which my teacher tells us we can use 10 instead of 9.8

[fringe mauve]

no

intial velcoity is not 0

[clear laurel]

Ah

[fringe mauve]

its ur 40sin(theta)

v(final) -= 0

at max height

[clear laurel]

Ohhh

Still i think id get the same answer just positive (explains my negative time lol)

Do i ignore the times that trajectory calculators gave me and use the one i have?

Like r those wrong or me

[fringe mauve]

what?

[clear laurel]

I was rlly lost so i found these calcs that like got me the crap but just to know if i was on the right track

[fringe mauve]

idk their probs right

but dw about those

u need 5 points

make all thsoe equations

in general terms

then just start subbing shit in

for those 5 points

do the beginning

the end

the highset point

and 2 point in between them

literally just solve for the projectilve motion

in terms of y and x

[clear laurel]

Begining wont count as a point but ill just do smth a bit later

So do i graph the velocities for x and y i have or the positions

[fringe mauve]

we know t = x/40cos(theat)

sub t into y = 40sin(theta)t - 1/2 g t^2

and simplify that

[fringe mauve]

no just position

[clear laurel]

I just not know what to sub in for t do i use the 1.035 i got?

[fringe mauve]

no

as in sub in t

replace all the t's

with that whole equation

x/40cos(theta)

and simplify'

[clear laurel]

Hold up sorru im trying to process

[fringe mauve]

when i say sub in t

have u done simultaneous eq

[clear laurel]

Yes i think so

[clear laurel]

Why is it cos and not sin?

[cedar kiln]

@clear laurel Has your question been resolved?

[fringe mauve]

leme hsow u

[clear laurel]

Bc i used cos for the x

[clear laurel]

Why would we use the v for y not x which is sin

[fringe mauve]

wdym

so v is ur 40

thats given in the q

were breaking it up to horiz and vertical components

[clear laurel]

No no i get that

[clear laurel]

Why wouldnt sin work here

[fringe mauve]

because

wehre did we get that from

we fromed an equation for the displacement

in the x direction

as a function of time

[clear laurel]

Ohhhh

[fringe mauve]

i simply

have u done simulataneous

equations

like

[clear laurel]

I think the y = at the start of the equation threw me off

[fringe mauve]

say u haev x + y =4

and xy = 6

solve for x and y

[clear laurel]

Im sorry i dont know that

[fringe mauve]

oh raelly?

[clear laurel]

I mean would we sub in 6/y for x

[fringe mauve]

yeah

pre much

and u solve for it

ur diong the same thing her

but for t

[clear laurel]

Ohhh

[fringe mauve]

try it on paper

and send it to me

[clear laurel]

Ok

[fringe mauve]

u should get a quadratic

[clear laurel]

What do i do from here

[cedar kiln]

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[wraith dagger]

Bean Man

[cedar kiln]

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[steep cedar]

hello, i have a question regarding the methodology of solving a problem. i understand the how, but i can't seem to fathom the why.

[steep cedar]

solving for d is apparently equal to this. but i have no idea why this works. d is part of the equation being divided by .06/12, yet the method for solving just separates d entirely and divides the rest to solve for d.
why is this possible?

how is it not the second equation = d/(.06/12)?

[upper laurel]

@steep cedar presumably you typed it wrong for them to get away with it

otherwise it would be d = (800000 * (.06/12) + 1) / (1 + .06/12)^(30 * 12)

[steep cedar]

i would have thought something similar, but surely enough the math checks out and how i have it shown to you is how my lecture is teaching it.

the answer to the equation as i typed it here is 796.4042012 and plugging that into the equation does return exactly 800000

[upper laurel]

and how sure are you that the d is inside the parentheses like that

[steep cedar]

that's exactly how the equation is done when you're solving for the variable that would be in 800000's place

i am sure, it's in my homework paper here

and the equation works

i just don't know why this is able to be done

i get the how, not the why

[upper laurel]

well its not

the equation you wrote has d ≈ 664

put the d on the outside, then you get d ≈ 796

recheck what you typed then try again

[upper laurel]

was not clear, reread

the equation you wrote has d ≈ 664

but the solution you provide has d ≈ 796

they are different

[steep cedar]

i am not seeing how you solve to get approximately 664 from that.

[upper laurel]

you are being inconsistent

the results are as I have shown you

are you aware what it means to have d on the outside?

[steep cedar]

then i'm not sure why this is working

[upper laurel]

you have d

on the outside

as you are showing

I will wait until you can identify what this means

it would help if you erased the outermost parentheses, by PEMDAS they are not required

you only need two pairs of parentheses, not five

[steep cedar]

you know, you are right, i did show the initial formula wrong. i see what you're pointing out. sorry about that. yes, the initial formula should look something like this instead.

but i still have the same problem

i just typed it wrong when i presented it initally

[upper laurel]

move the d one space to the left

(y/n) is this the same

[steep cedar]

i would say n

[upper laurel]

,,\frac{3\times5}{2}\ne3\times\frac52?

[wraith dagger]

mtt

[steep cedar]

i see, i understand that.

both are the same, yes

[steep cedar]

yeah using that principle i see how it translates to the equation.
my brain doesn't like it but i understand the correlation now.

thanks for the help, sorry for the initial confusion.

[upper laurel]

np

[steep cedar]

.close

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[eager robin]

Can someone explain for me the meaning of the definition, I am kinda getting parts of it but not every thing😅

[azure swift]

It’s the set of all possible outputs of f

If you were to input all possible values from the domain then you’d get all possible outputs

So set builder notation describes this by saying that your first object is what your set contains, i.e the f(x)’s and after the | you have your restriction

Which is that x is in A

That’s what the direct translation of the third set in your image is

The second one in your image (the one in the middle) is actually the exact same

[eager robin]

What does the A(f(x)=y) mean ?

[azure swift]

That’s just unfortunate spacing

They’re separate

So $(\exists x \in A),(f(x) = y)$

[wraith dagger]

Aslan

[azure swift]

But by convention we don’t write like that

But e.g. like this $\exists x \in A,(f(x) = y)$

[wraith dagger]

Aslan

[azure swift]

Do you know how to parse this with words?

[eager robin]

I am kinda new to it starting my first year in college and they sent to us a material to prepare for the lacture

So not really

[azure swift]

Is the notation totally new or are you familiar with them?

Have you gone through their definitions that is?

[eager robin]

I think its exist x that belongs to A, f(x) is equal to y

[azure swift]

Yup

One would usually also add “such that” where your comma is

[eager robin]

I have gone through it it is just complicated putting it together sometimes

[eager robin]

👍

[azure swift]

Try and always just say what each symbol means and combine it

The point of notation is just to avoid writing what u said since we’re lazy

[eager robin]

OK thanks for the help🙏

[azure swift]

Do you understand how the set in the middle corresponds to what I said earlier?

Would you like to get a feel for what it’s saying?

[eager robin]

I am kinda understand that you can definition the image of a functions in to different ways(sets), and that what they wrote basically

[azure swift]

The thing is the one in the middle is actually the exact one

The last one is sort of implicit

By convention leaving x without any quantifiers like that in a set usually just means existence but it’s totally not clear I think when starting out

[eager robin]

The first one is: y belong B when you have restrictions that exist and x that belong to A such that f(x) is equal to y
The other way is that the image of a functions is f(x) when you have the restrictions that x belongs to A

[azure swift]

The first one is saying that your element y in B is an output of f if there exists an x in A such that f(x) = y.

That is, pick any output of f, then that must mean there’s atleast one corresponding input that gets mapped to this output!

When they say y in B that’s part of the restriction btw

I’m interpreting your translation as “y is in B if … “ but that’s not what is being said here

[eager robin]

Wait so the restriction comes after I or before ?

[azure swift]

Yeah so ideally that’s how we should think about it. But really nothing is stopping you from directly writing that restriction before it aswell. As long as you’re clear that your object of interest (y in this case) is the one that the set consists of, which is whatever object that is before the | plus the constraints after this symbol.

So we’re saying the set consists of

all y in B such that …“some additional constraint”…

but this is just the same as saying

all y such that y is in B and …”some additional constraint”..

The such that corresponds to | in this case

[eager robin]

A ok i understand 👍

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[zinc herald]

Hello

[zinc herald]

Can anyone help me w domain and range

idk what to ask like I don’t understand anything I don’t have a specific question I js need help 😿

[twilit escarp]

Oh

[stable hull]

lmao

[zinc herald]

I understand the like f(3) replacing and then doing the backwards thing like ghg

then I get stuck at domain and range specifically and the graph stuff

[slate lintel]

i diagnose you with two Khan Academys and one Organic Chemistry Tutor

[zinc herald]

thank you

[twilit escarp]

"That will be 90$"

[lean lotus]

lmao

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[vale blade]

shouldn't it be a different answer?

[vale blade]

i let v(t)=xi+yj+zk
and then pi/2-arccos(z/45)=28, i get z=21.13 (this is good)
but then i do
arccos(x/45)=8, i get x=44.5621 (this is bad because x=39.35)

[cedar kiln]

@vale blade Has your question been resolved?

[vale blade]

<@&286206848099549185>

[versed jasper]

i'm assuming the 8deg angle is towards the k-direction

like

[vale blade]

it says its towards the i direction tho

[versed jasper]

its hard to explain but
the 8 deg angle is like "upwards", like in the i-k plane

the 28deg one is the angle between the projection of the vector on the i-j plane with the i-direction

[vale blade]

hm?

this is the 28 degree one?

[versed jasper]

thats the 8 degree one
if you project the vector to the horizontal plane, the angle between that projection and i is 28deg

i think thats what it is

[vale blade]

but isn't 28 degrees the angle to the horizontal plane

that's an angle to a horizontal plane there?

[versed jasper]

i admit the wording is confusing

[cedar kiln]

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[vale blade]

https://tenor.com/view/im-not-finished-let-me-finish-shut-your-mouth-rude-gif-5681170

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[vale blade]

<@&286206848099549185>

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[ionic jolt]

Rational Expressions
Topic Name
Complex fraction made of sums involving rational expressions: Problem type 3

im having some trouble understanding the process on how to solve rational expressions but i cant seem to grasp how to tell what would be the LCD.

[long swan]

lcm of 5x and 10x ?

come on you can do it!

[cedar kiln]

@ionic jolt Has your question been resolved?

[ionic jolt]

i may have explained it wrong. when i tried to read the explanation, this comes up

and this is was as far as i got with it. it just stumps me

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[ionic jolt]

.reopen

[cedar kiln]

✅

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[wintry tiger]

This is Example 3 from section 2.4 "The precise definition of a limit" from Calculus ET by Stewart

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<@&286206848099549185>

[dire geode]

Are you trying to guess delta

Delta can depend on epsilon and the neighborhood of the value x is approaching

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[twilit abyss]

where does the /a go after the substitution

[digital cliff]

was there anything before this?

[digital cliff]

ah right

they just split it and considered it part of the constant C

took me longer than it should

[twilit abyss]

ah that actually makes sense, thanks!

.close

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[crimson sedge]

if you find the gradient vector for a graph

[crimson sedge]

is it equal to any scalar multiple of itself?

like $$ \nabla f(x,y) = \begin{pmatrix} 2x \ 2y \end{pmatrix} = \begin{pmatrix} x \ y \end{pmatrix} $$

[wraith dagger]

depresso

[hollow trail]

they are in the same direction (direction of biggest increase), but only one of them is the gradient vector

[crimson sedge]

i thought the gradient vector is just a vector field telling you the direction of biggest increase ?

[hollow trail]

the magnitude of the gradient tells you about how fast it is increasing in that direction

[crimson sedge]

oh i see

.close

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[warped robin]

How to verify this?

[dire geode]

Plug 13 into your starting equation

[warped robin]

Ok, then.

?

[dire geode]

That's it

[warped robin]

That's not going to verify

[dire geode]

Why do you think not

[warped robin]

Verify means if its correct or not by the same number.

Here's my example

[dire geode]

"by the same number" what does that mean

[warped robin]

In computing the final x: I just need to verify it if this is correct, so if the 2 sides (LHS, RHS) are the same, the final x is correct. But if the 2 sides are not the same, that means the final x is not the same.

[warped robin]

So, what is the verification of this

[cedar kiln]

@warped robin Has your question been resolved?

[warped robin]

<@&286206848099549185>

[frail yoke]

you have to equations that should equal one another and you found your x, to prove it as said previously you plug in your found x. Separate the two equations and prove

9=9 which means that your equations are equal

idk how else you want to prove this

[warped robin]

I think, there's wrong

13*2?

Also, it's still not the same equal LHS and RHS

Thanks for helping me but I need further more explanations.

[frail yoke]

wdym?

,w ((2*13)+1)/3

,w ((4*13)-7)/5

ahh

i see

forget i said anything

[wraith dagger]

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[frail yoke]

no nevermind i just wrote wrong

both are equal

[warped robin]

Ok thanks

.close

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[hard shard]

Can someone please help me with this

I'm really confused

[slender ginkgo]

radius = 16

• its a disk

what is ${r}$ is polar coordinate

[wraith dagger]

k

[hard shard]

It's like a kind of coordinate right

I have no idea how to solve this one

[thick forge]

circle of r = 16 at origin right

we want to imagine the area to the right y axis

what values of r and theta can cover this region

[hard shard]

Hmm I see

What happens after the graph?

<@&286206848099549185>

[thick forge]

what happens after the graph? what do you mean

[hard shard]

Like I'm confused on what I should do after

[thick forge]

Look at this, its a circle shaded to the right of the y axis

the shaded region is the thetas of, by looking at the unit circle, the area between pi/2 and 3pi/2

and if u imagine a bunch of half circles with radius of r=0 to r=16, it would cover the whole area

so its those bounds

[hard shard]

It it would be 0 and 16

[thick forge]

careful regarding the theta region, remember the unit circle as a function

[hard shard]

And then -pi/2

And pi/2

[hard shard]

Would that work?

[thick forge]

Yup :)

[hard shard]

Oh okay

Let me see

@thick forge do you know what to do for this one

i tried but it says its wrong

[thick forge]

if youre confused, first thing you should do is graph it, desmos

[hard shard]

How should I graph it?

Like just the bounded by thing

[hard shard]

Hmm I see

What would the next part be?

I just find a hard time finding the next part

[hard shard]

I know the first part would be 0 and root 15 for sure

A little bit unsure about the other 2 points

[thick forge]

i agree

the other two points will be on the left side of the red line

[hard shard]

do you know how to find the other 2 points

[thick forge]

so its the shaded area left of the red line

[thick forge]

unit circle

[hard shard]

would it be 3pi/4

[hard shard]

hm

and 5pi/4?

not sure if thats right

[thick forge]

3 pi/4 is for y=-x

so not that

[hard shard]

yeah

oh

[thick forge]

this will be one of them

top of green area is bounded by this purple line

whats the theta for that

[hard shard]

of the dot in the middle?

would it just be theta

[thick forge]

no

its the left side of the x axis

[hard shard]

would it just be pi

[thick forge]

yup

[hard shard]

sorry that took so long

so it would be 0, root 15,

then 5pi/4, pi?

or would it be switched around

[thick forge]

switched around, bc pi<5pi/4

[thick forge]

heres the specific region you were finding btw

[hard shard]

ah i see

hm okay

[thick forge]

i graphed it by graphing the two given functions

then created a circle thats bounded by the red line on the x axis, and 0 on the y axis

[hard shard]

and then the other 2 would be just 0 and root 15 right

could you help with another please

[thick forge]

You can do it yourself :)

[thick forge]

im confident in u

[hard shard]

can you help guide through

[thick forge]

https://www.desmos.com/calculator

graph the function of x

[hard shard]

okay

i got 0 and 3 for the first 2 points @thick forge

[thick forge]

okay

[hard shard]

would th eother 2 points be similar to the unit circle

like would it be pi/2

and 3 pi/2

[thick forge]

yes theta will always be the range of the region expressed on the unit circle

[hard shard]

ah i see

so would it just be 0 , 3

and then pi/2

and then 3 pi/2

i get confused about the order

do you know how this one would work

i got that other one

[thick forge]

the same as every other one with one exception

[hard shard]

whats that?

[thick forge]

note that tiny circle in the middle

see how its not shaded

[hard shard]

yes

its just blank

[thick forge]

what will that change about our range of radiuses for the shaded region

[hard shard]

oh wait

so will the other 2 points be from that area

so would it be 0

and 5pi/3?

[thick forge]

Jay

read the question

[hard shard]

oh wait

[thick forge]

it word for word tells you the answer for radiuses and thetas

[hard shard]

7pi/4

bruh im so slow

[thick forge]

You're okay lol

just please

have some confidence

You can 110% do these

[hard shard]

im just scared to get it wrong

[thick forge]

I understand

[hard shard]

ive been stuck on these for hours

[thick forge]

I will happily check your answers

[hard shard]

so if i get even 1 wrong it drops all my progress

[thick forge]

but think of a way to solve these, confidently attempt, and I can check

[hard shard]

this software is so cooked

wait so for the one above

should i put 1, 9

and then 5p1/3

and then 7p1

[thick forge]

1 and 9 are correct

but those last 2 are the area NOT shaded

[hard shard]

i meant 7p1/4

but yeah

[thick forge]

u want the sahded region

[hard shard]

so just 0

[thick forge]

yup

to 7pi/4

[hard shard]

ah i see

yeah

thanks a lot for the help

.close

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[fringe oasis]

hi, im a bit confused how this equation goes from 2* root2* root13 to root 13?

[bleak viper]

,, \frac{8\sqrt{2}}{2\sqrt{2}} = 4

[wraith dagger]

Astar777

[fringe oasis]

ohhh okay

thank you!

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[coral jasper]

I should solve it by getting triangle of zeros in the matrix

[steel canopy]

Have you try substitution yet?

[coral jasper]

i tried wait i will show you

this only gives me two answers -1 and zero
the answer should be 0 for inf sol
and 1 and -1 to no sol

<@&286206848099549185>

[steel canopy]

I am pretty sure for i) a can't equal 1,-1, 0 but it can any other values. You double check with them

[cedar kiln]

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[plain holly]

how

it doesnt separate, cant make it into a homogenous function, cant make it into a lde it isnt an bernoulis form either

not sure how to proceed

[pastel vault]

$\frac{dy}{dx} = 1 + \frac{xe^y}{e^x}$ is a linear first-order DE

[wraith dagger]

south's secret twin brother

[pastel vault]

wait no it isn't

but if you try subbing in y = ln z for example

[plain holly]

i tried making substitutions couldnt find anything suitable

[pastel vault]

$\frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx} = e^y \frac{dy}{dx} = z \frac{dy}{dx}$

so $\frac{dz}{dx} = z + x \cdot z \cdot e^{-x}$

[wraith dagger]

south's secret twin brother

[pastel vault]

now this is linear first order

[cedar kiln]

@plain holly Has your question been resolved?

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[obtuse elbow]

lets have r = tau wandering if the its du= 1-dr or du= -dr, if its the first would that mean that the expression below is missing a component.

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[carmine otter]

can anyone explain the first step?

how did e appear ?

[crimson sedge]

e^ln(x) = x

[carmine otter]

ok tq

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[tough vigil]

im sorry if this is a poorly written question:
for what real number k does (kx)! > x^x as x approaches infinity?

[lyric narwhal]

my guess would be that no such value of k exists

x^x grows much faster than a factorial function

actually no nvm

[tough vigil]

that was my guess too but when i see the infinite sum of n^n/(2n)! it converges

[lyric narwhal]

it does

do you know about stirling's approximation?

[tough vigil]

it means that the denominator grows faster right

[tough vigil]

yes i do

[lyric narwhal]

it does

[lyric narwhal]

so we have $(kx)!\approx\sqrt{2\pi kx}\left(\frac{kx}{e}\right)^{kx}$ as $x\to\infty$

[wraith dagger]

kheerii

[tough vigil]

i had done some numerical analysis, it shows that when k<1,1 it diverges for about 20ish iterations but when k>1.15 it seems to converge

[lyric narwhal]

we should be able to get an equation for the exact value of k where it changes behavior

i think any value of k>1 would converge actually

[tough vigil]

it might, maybe 20 iterations wasnt enough

is stirling's approx more accurate as x approaches infinity?

[lyric narwhal]

this means $\frac{(kx)!}{x^x}\approx\sqrt{2\pi k} x^{(k-1)x+\frac1{2}}\left(\frac{k}{e}\right)^{kx}$

[wraith dagger]

kheerii

[lyric narwhal]

which clearly diverges for k>1

and converges for k<=1

[lyric narwhal]

yes

[tough vigil]

i can see that

[lyric narwhal]

stirling's approximation is just the first term of a series

there are upper and lower bounds too

[tough vigil]

yeah they do

i think i need to do some more iterations

[lyric narwhal]

it's actually $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1{n}\right)\right)$

[wraith dagger]

kheerii

[tough vigil]

yeah ive read that you can make the approximation more accurate by adding more terms

but i dont really know how does the big o notation work

[lyric narwhal]

there's a whole series

[lyric narwhal]

it just means that the second term of the series is some constant * 1/n

[tough vigil]

ah i see

[lyric narwhal]

more precisely, $$\lim_{n\to\infty}\frac{\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}-1}{\frac1{n}}=c$$ where $c$ is a non-zero finite real number

[wraith dagger]

kheerii

[tough vigil]

if the stirling's approximation is good enough as x approaches infinity, is it safe to substitute the factorial term with stirling's approximation here?

[lyric narwhal]

in most cases you should be fine

if there's some sort of cancellation though you should probably consider higher order terms too

[tough vigil]

can you give me an example?

[lyric narwhal]

it's similar to how assuming sinx=x is fine for certain limits but not fine for others

[lyric narwhal]

take the limit I gave above

[lyric narwhal]

this one

if you just take the first order approximation you'll just end up with 0

but it's not 0

because there are higher order terms

[tough vigil]

ah that makes sensee

i think i will use the approx term for now as it is good enough. thanks for your help!

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[crimson sedge]

I need a helper

[valid egret]

Maybe

[cedar kiln]

@crimson sedge Has your question been resolved?

[pastel vault]

they're timed out

.close

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[junior stratus]

Is this still simplifiable?

[dreamy void]

you could technically rationalize the denominator

[junior stratus]

I mean it told me to find its value but after that its aight?

[dreamy void]

I'd say yea

[junior stratus]

Alright thanks

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[eager hemlock]

Is there any quick way to compare these two?

[wraith dagger]

ren

[eager hemlock]

But how does this help?

[cedar kiln]

@eager hemlock Has your question been resolved?

[mental trail]

Even though sin and cos are a bit related, there is no easy way to compare the two because of the decreasing function 1/(1+x^2)

My guess would be that the sin integral is bigger because it's positive on the parts that are "bigger"

other than that I see no easy link

[potent fractal]

Subtract one from another and use the fact that $\sin(x)-\cos(x) = \sqrt{2}\sin(x - \frac{\pi}{4})$

[wraith dagger]

EQUENOS

[mental trail]

and then?

[potent fractal]

And then I thought of writing some inequalities for $\frac{\sin(x-\frac{\pi}{4})}{1+x^2}$ but it doesn't seem too fast, yeah

[wraith dagger]

EQUENOS

[cedar kiln]

@eager hemlock Has your question been resolved?

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[slender ginkgo]

what have u done

so far

[cedar kiln]

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

[crimson sedge]

erm

so

so my thinking process was

there are 9 people in total
if a women is in front and last thats
W.......W
so we get 2! as we can change the position of the women and 7! to change the people/dots

so wont that be 2!*7!

but the answer is

7!x 2 x3

[slender ginkgo]

there are 9 people in total

[crimson sedge]

yes

[slender ginkgo]

here's my thinking

[crimson sedge]

y 3?

[slender ginkgo]

3 women

pick one of 3 as the first step

[crimson sedge]

wont one women be counted in the group of 7?

[crimson sedge]

ohh

that makes sense

so ur taking one women

[slender ginkgo]

${\underbrace{\binom{3}{2}}_{\text{Pick 3 out of 2 women}} \cdot 2!}$

[wraith dagger]

k

[crimson sedge]

mmh

2!??

[slender ginkgo]

this is another way of thinking about it

[crimson sedge]

pls tell me that

[slender ginkgo]

method a: step-by-step

1. pick one out of three women for the left-most spot
2. pick one out of two women for the right-most spot
3. arrange the other 7

method b: picking

1. lock the 2 spots for these two women
2. pick 2 out of 3 women and place them those 2 locked spots
3. multiply by 2! cuz there are 2! ways those locked spots can be switched
4. arrange the other 7

method a is easier

but method b is more mathy imo

heres computation for method b

[crimson sedge]

wait

[slender ginkgo]

${(\binom{3}{2}\cdot 2!) \times (\binom{7}{7} \cdot 7! )}$

[wraith dagger]

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

[crimson sedge]

that makes sm sense

thank you so much

so ur basically doing 3! * 7!

the answer comes out right

would thta be okay?

[slender ginkgo]

Should be 3P2 7P7

[crimson sedge]

mmh ic

thank you

[cedar kiln]

@crimson sedge Has your question been resolved?

[past wave]

I think your problem is resolved so I'll close this

.close

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[eager hemlock]

how to determine if this is convergent or not?

[vagrant elbow]

=< 1/x^p

[eager hemlock]

so it's convergent when p>1?

[mental trail]

when p > 1, it converges. What about the other cases?

[eager hemlock]

when p=1, it div

[mental trail]

why?

[eager hemlock]

1/x div

ooh no

[mental trail]

is that what we have though?

[eager hemlock]

wait

even if 1/x div, we can't say a series < 1/x div

[mental trail]

yup

[eager hemlock]

hmmmmmmmmmmm

[mental trail]

do you think it's gonna be abs convergent?

(p <= 1)

[eager hemlock]

wait I forgot the behaviour of 1/x^p when p<=1

[mental trail]

let's just say 1/x^p > 1/x

and think about |sin(x)|/x instead

maybe it can be useful

dividing the integral into a sum of smaller integrals

for example

say a = 2pi I don't know

It doesn't matter which value a has since the integral of a continuous function on a closed bounded interval is finite

and doesn't affect convergence

so a = 2pi

we start by integrating from 2pi to 3pi

then from 3pi to 4pi

etc...

summing all those integrals would give us the integral from a to infinity, would you agree?

[eager hemlock]

lemme think

[wraith dagger]

Allophane

[eager hemlock]

Do you mean like this?

[mental trail]

yes

[eager hemlock]

but why it doesn't affect convergence

[mental trail]

starting from 2pi instead of another number?

what is the integral from a to 2pi of |sin(x)|/x?

[eager hemlock]

0

[mental trail]

no

but it's finite, right?

[eager hemlock]

1/pi

[mental trail]

don't try to find a specific value xdd

[wraith dagger]

Allophane

[eager hemlock]

damn I don't know

[mental trail]

I told you

we don't care about the exact value

but is it finite? infinite?

[eager hemlock]

finite

[mental trail]

yes, why?

[eager hemlock]

The sine function is continuous and bounded and x is in the finite range

[mental trail]

more concise

|sin(x)|/x is continuous

on the closed interval [a,2pi] (or [2pi,a])

[eager hemlock]

so if a function is continuous on a closed interval, then its sum is finite on this interval?

[mental trail]

its integral

and closed and BOUNDED interval

[eager hemlock]

Noted

[mental trail]

anyways let's indeed look at the plot of |sin(x)|/x to get some inspiration

[mental trail]

since we now established that convergence/divergence of the integral from 2pi to infinity

is the same problem as convergence/divergence of the sum

[eager hemlock]

but here the interval is not closed and bounded

[mental trail]

so we can't say anything about the integral

[mental trail]

we only applied the argument here

to switch the beginning point from any point 'a'

to the well chosen point 2pi

(Chasles property)

[eager hemlock]

so we don't know the situation of other cases?

[mental trail]

we're about to

since we now established that convergence/divergence of the integral from 2pi to infinity
is the same problem as convergence/divergence of the sum
let's now look at the plot of |sin(x)|/x to get some inspiration

each term in the infinite sum

is the area under one of those bumps

that get progressively smaller

pretty easy to show why the bumps get smaller and smaller in area right?

[eager hemlock]

x is getting larger

[mental trail]

you kinda get the point but the proof is just about comparison

|sin(x)| is pi-periodic

so $\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx$

same as $\int_{(n-1)\pi}^{n\pi}\frac{|\sin(x+\pi)|}{x+\pi}dx$

[wraith dagger]

rafilou is not not born in 2003

rafilou is not not born in 2003

[mental trail]

I just did x -> x - pi shift

(u sub)

so same as $\int_{(n-1)\pi}^{n\pi}\frac{|\sin(x)|}{x+\pi}dx$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

which is smaller than $\int_{(n-1)\pi}^{n\pi}\frac{|\sin(x)|}{x}dx$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

since $x +\pi > x$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

so case closed, the area of each bump is decreasing

does the area of each bump go to 0 as n goes to infinity? yes/no and can you show why/why not?

(note in advance that just because the general term of a series goes to 0 does not indicate if the series converge/diverge)

(but we will use this info later)

[eager hemlock]

I guess yes

the denominator will be infty ig so the bump will go 0

[mental trail]

I'm gonna need more rigorous than that

to show something (non-negative) goes to 0

you can bound it (upper bound it) by something that goes to 0

[eager hemlock]

<=|sinx|/x...?

[mental trail]

?

[eager hemlock]

hmmm i don't know ig

[mental trail]

how do we generally bound |sin(x)|

[eager hemlock]

<=x

[mental trail]

uhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

that's useful when we're close to x = 0

when we're making x -> infinity, do you think this will be useful?

[eager hemlock]

<=1

[mental trail]

yes

you were kinda close tho

but

|sin(x)|/x <= 1/x

so what is the integral of the upper bound

i'm sure integrating 1/x isn't a big hustle

[eager hemlock]

lnx|^infty_a

[mental trail]

ok so

huh

[mental trail]

what are the bounds of the integral on one bump?

[eager hemlock]

lnx|^npi_(n-1)pi

[mental trail]

alright

does that go to 0 as n->infinity?

[eager hemlock]

hmm no

[mental trail]

can you show why/why not then?

[wraith dagger]

Allophane

[eager hemlock]

ln(n-1/n)

uhh seems like yes it goes to 0

[mental trail]

ok yes

n/(n-1) = 1 + 1/(n-1)

1/(n-1) goes to 0

so ln(n/(n-1)) goes to ln(1) = 0

So far, we now know the area of each bump decreases over each iteration, and goes to 0

[mental trail]

again we remind this

Just because the general term of a series goes to 0 doesn't mean the series converges or diverges

[eager hemlock]

hmmm then why did we do this

[mental trail]

it will be useful later

what kind of series for example has the general term go to 0 but the series go to infinity?

an example

[eager hemlock]

1/n

[mental trail]

yep correct

so

I won't go too much into detail because that's not what we want to prove

but $\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

behaves a lot like $\frac 1n$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

the idea is this:

$\frac 1x\geq \frac{1}{(n+1)\pi}$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

oops yeah I gotta change a little something

We can show in fact that a lower bound is $\frac{\sqrt 2}{4(n+1)}$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

by using this

and using $|\sin(x)| \geq \frac{\sqrt 2}{2}$ when $x\in [n\pi + \frac \pi 4, n\pi + \frac{3\pi}{4}]$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

so anyways, our series is bigger than constant * harmonic series

so the original integral doesn't absolutely converge

@eager hemlock so do you think you know where I'm going with this?

[eager hemlock]

I'm literally already lost now

[mental trail]

anyways let's just say

we know the area of each bump decreases over each iteration, and goes to 0

but the sum of the (positive) areas of the bumps diverges

so the original integral isn't absolutely convergent

is there a way for it to still converge?

[cedar kiln]

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[nocturne compass]

Take n binary codes, $c_1, c_2, \ldots, c_n$ of length $k<=2^n$. Show that there is at most $\lceil{\frac{n^2}{4}}\rceil$ pairs of codes such that the hamming distance between them is one.

[wraith dagger]

bigpufik

[nocturne compass]

<@&286206848099549185>

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[upper stirrup]

What's the probability of flipping 2 heads in a row before flipping a tails and then a heads?

[upper stirrup]

not sure how to solve it algebraically/if my answer is right

so by symmetry, the probability of flipping tails then heads before heads then tails is 1/2

so the question is the same as P(flipping 2 heads in a row before heads then tails)

and that is 1/2 because after the first heads, either tails or heads decides it

so 1/2 is the final answer

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[fast basin]

How do I transpose this to solve for V1? This is my first one with an exponent on it.

[dreamy void]

you can take the 1.35th root on both sides

[fast basin]

oof I wouldn't even know how to do that

would there be a function for that on my calculator? I only see sq rt

and 3 sq rt

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[blissful fossil]

I need help with this. What I have here is a parametrized vector function for light ray reflections being represented as a vector field. The light ray reflection field is the function R_{R}(X,Y). It is composed of three functions, R_x, R_y, and R_z. What I wish to do is scale each and every vector so that way they all reach the focal point of the surface, when the paramter t = 1. So to do this, I multiply it by a scaling factor of mu. But mu is a function of x and y as well. So I take the divergence of the scaled reflection field and set it equal to 0. But then I get stuck

[blissful fossil]

Is this the correct way of going about solving for the scaling function mu?

Or is there another better way?

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[eternal pollen]

What did I do wrong? Explain in first principles please

[dusk goblet]

-(3x+1) = -3x - 1

you did -(3x+1) = -3x + 1

[eternal pollen]

Aaah

I forgot about the parentheses

Thank you

[dusk goblet]

you’re welcome

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[tacit citrus]

Hi, so I have a question about related to Combinatorial calculus. So, I have 7 different balls and I want to put them in 3 different boxes, but in a way that all of the 3 boxes have at least 1 ball.

Where Im getting confused is when I do for example groups of (4,2,1) and when i do: C(7,4)xC(3,2), should I multiply by 3 factorail since its 3 different boxes?

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[tacit citrus]

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[elfin onyx]

What do you mean by C(7,4)xC(3,2)

[tacit citrus]

imagine

i have the seven balls

and i want one of the boxes to have 4 balls, the other 2 and the other 1

and since

the balls are all different

i thought i had to do C(7,4)x C(3,2)

to create the "groups"

[elfin onyx]

You mean

$ C_n^k $

[tacit citrus]

?

[elfin onyx]

/( C_n^k )/

[tacit citrus]

yhea i dont understand that

its basically combinations, i think that is the name

[elfin onyx]

Yeah

[elfin onyx]

I just dont know why it didnt turn into latex text but anyways

[tacit citrus]

oh okok

but yhea my question is

if I have to mutiply by 3!, in that situation

[elfin onyx]

We definitly should care about the position and the arrangement of the balls cuz as you said they are not the same

[tacit citrus]

so i have to multiply by 3! then, right?

[elfin onyx]

Is every single balls different from all the others?

[tacit citrus]

yhea yhea

all different

[elfin onyx]

I dont think that this expression is correct C(7,4)x C(3,2)

[tacit citrus]

why not?

[elfin onyx]

I actually dont understand why you choosed exactly this expression

[tacit citrus]

Cause I wanted to create 2 groups, one with 4 balls and the other with 2 balls

[unreal shale]

maybe you could think it in another way

[elfin onyx]

You only tought about this exact arrangement

[unreal shale]

What I'm thinking is, what do you need to do to have at least one ball in every box?

[tacit citrus]

Yh only that one

[tacit citrus]

But I just want to think about this particularly arrangement

[unreal shale]

ooh for 4,2,1 groups?

[tacit citrus]

Yes yes

[unreal shale]

I think you should consider the orders of the boxes since 4,2,1 wouldn't be the same as 2,1,4

[tacit citrus]

Yes

But I should multiply by 3 factorial right?

[unreal shale]

yeah

[tacit citrus]

Because in other arrangement where is like: (3,2,2)

When I do C(7,3)xC(4,2)xC(2,2)

[unreal shale]

or (5,1,1)

[tacit citrus]

Where I would have to multiply by 3factorial divided by 2 factorial right? Like this 3!\2!

[unreal shale]

that i wouldn't be sure about

because we said the boxes were different

that's why we said we had to multiply by 3! before

[tacit citrus]

Yes but when I C(4,2) multiplied by C(2,2) an order is already imposed there that is why I divided by 2!

[unreal shale]

because even though two boxes have the same ammount of elements, they are not the same

let me think through it

[tacit citrus]

It would be something like this

[unreal shale]

I think there is a lot of ambiguity

[unreal shale]

because we could give names to the boxes

like box a, b and c

[tacit citrus]

Yes yes

[unreal shale]

so (3,2,2) would mean a has 3, b has 2 and c has 2

now, since they have names

a,b,c ≠ a,c,b in any cases, because the boxes are independant of the balls.

I think I'm not explaining myself clearly

[unreal shale]

let's call this A= (3a, 2b, 2c)

the arrangement A is never the same as the arrangement B = (3a, 2c, 2b), even if both boxes a had chosen the same balls, box c in B had chosen the same balls as b in A, and box b in B chosen the same balls as c in A.

because of that, we wouldn't need to divide the 2!; we only care about the order of the boxes, not their contents

[unreal shale]

so with this you would be taking account of all the possible (5,2,2), but taking out all cases where b and c are exchanged

have i explained myself better now?

[tacit citrus]

but i saw an exercise that said when i do groups with the same ammount of elements an order would be imposed already

i can send it to you

[unreal shale]

let me see it

maybe we can work it out together

[tacit citrus]

but it is not in english

portuguese

[unreal shale]

I speak Spanish and English

maybe I can get something out of it

[tacit citrus]

there it is

[unreal shale]

if not I'll just use a translator don't worry

yes, it is correct, in this case have to divide because if we read closely it says that the five envelopes are equal

so if we name envelopes a, b, c, d, and e

we would have that (a,b,c,d,e) = (b,a,e,c,d)

because of that we also do not multiply by 5!

[tacit citrus]

yhea

but the example i said

i did 3!/2! because i have 3 groups and i divided by 2! because of the multiplication C(4,2) C(2,2)

[tacit citrus]

in this exercise they divide by 3! when the groups have the same amount of elements, but if you see when they do groups of different amout of elements like: C(8,3) x C(5,2) they dont divide by 2!

[unreal shale]

yes, let me guide you in why I think you should divide in one case and not the other

in the case with the envelopes, we had been told that the envelopes were the same

[tacit citrus]

man do you accept call?

i think it would be easier to explain

[unreal shale]

so when computing the cases, we had to be careful with those with the same ammount of photos; that is because a case with (2a,2b,2c,1d,1e) would be considered the same one as a case like (2b,2c,2a,1d,1e) or (2a,2d,2e,1b,1c): So, to take all those repeated cases out, we divide by 3!

[tacit citrus]

yhea

but for example

[unreal shale]

can't right now, my internet is bad and i'm riding back home 😔

[tacit citrus]

(3b,2c,1a) wouldnt be the same as (3a,2c,1b)?

because when they do groups

with different amount of elements

[unreal shale]

it wouldn't be the same in the case of the boxes, because they told us they were different

[tacit citrus]

they dont divide

[unreal shale]

so it's not the same if we have a in the first position, as if we have b in the first position