help-38

[worldly wing]

not exactly

but for simple things like this yeah it doesn't matter

[sudden drum]

okay thanks

im almost done with khan academy algebra 1

how do i make sure i remember that

[worldly wing]

as long as you don't mess up calculations, any manipulation using addition and multiplication you do ON BOTH SIDES will give equivalent equations

as long as you don't multiply both sides by 0

[high forge]

Bruh why did they use i? i is always the square root of -1 😂

[worldly wing]

not always, but yeah

that notation is cursed

it's algebra 1 though, so they don't even know what i is

[high forge]

Yeah 😂

[high forge]

Oh ok nevermind the imaginary stuff just comes to my mind when I see i

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[wanton wasp]

How do I solve this?

I'm a bit confused

[thorn copper]

k=cosec²x=1/sin²x

[wanton wasp]

yeah but then I can't just sqrt the whole thing

hold on

I somehow got that cot(x) = cos(x) * sqrt(k)

[thorn copper]

cotx=sqrt(cot²x)=cosx.sqrt(cot²x/cos²x)=cosx.sqrt(k) yup

[wanton wasp]

is the . to multiply?

[thorn copper]

yes

[wanton wasp]

Alright lemme see

[thorn copper]

what is cosx in terms of k

[wanton wasp]

its

um

hold on

sqrt(1-k)

?

wait no thats wrong

[thorn copper]

think simple

[wanton wasp]

sqrt[(k-1)/k]

is cos(x)

[thorn copper]

k=1/(1-cos²x)
1-cos²x=(1/k)
cos²x=1-(1/k)
cosx=sqrt[1-(1/k)]

[wanton wasp]

think you did a typo

[thorn copper]

where

[wanton wasp]

in the last line

oh wait nvm

you meant 1-(1/k)?

yeah ok

[thorn copper]

should be clearer

[wanton wasp]

yeah that's essentially what I put

[thorn copper]

just plug that here and i think you'll be done

[wanton wasp]

just put everything under the same denominator

[wanton wasp]

ok thanks!

let me try it

yeah I got to the right answer!

thanks!

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[shy nacelle]

Hey

[wanton rune]

!status

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3. I got an answer but I was told that it's wrong.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shy nacelle]

This formula will help 🙂

[wraith hinge]

1

[wanton rune]

multiply by conjugate of denominator in numerator and denominator. Then you can apply a substitution

this substitution will allow you to apply difference of cubes more clearly

[wraith hinge]

Hmmmm

Step one done

What will be the substitution

@wanton rune

[wanton rune]

$u=\sqrt[3]{x+27}-3$

[solid kiln]

🫎MooseyMooseMooser 🫎

[wanton rune]

then put everything in terms of u and take limits as u goes to 0. you can apply difference of cubes to denominator and get some cancelation between numerator and denominator

@wraith hinge

[wraith hinge]

Hmm

[wanton rune]

to put everything in terms of u, you must solve for x

this will be the difference of cubes. you will get on in the square root, but you can keep that a difference of cubes. what's important is applying difference of cubes to the x in the denominator (once its in terms of u)

[wraith hinge]

😩

Ok

I will give it a try

[wanton rune]

i believe in you :3

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[wraith hinge]

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[lyric flare]

Two workers had to mow a lawn. They agreed to each mow half of the lawn. The first worker started mowing 2hours and 16 minutes before the second worker. Till the 12;00 they had finished mowing 40% of the lawn, then they both took a 1,5hour brake and started working again. The first worker finished mowing his side of the lawn at 19;54 and the second worker finished at 20;10. When did each of the workers start mowing?

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[lyric flare]

<@&286206848099549185>

[frigid cloak]

dw man nobody answered me either.

[lyric flare]

😄

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@lyric flare Has your question been resolved?

[rigid pulsar]

Can you send a better picture of your working

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[sterile star]

Is this right?

[lament reef]

yes

[sterile star]

Is this one as well?

[uncut aspen]

you forgot about the t^2 lmao

and some parentheses to clear up your answer would be nice

[near vessel]

you have to use product rule for 5e^xsin x

[uncut aspen]

i think he did, but he just didn’t put parentheses around the answer

[sterile star]

I’m just trying to get done man

[near vessel]

i see

[dull island]

That’s in a different variable

[uncut aspen]

oops mb

[dull island]

That just disappears

[uncut aspen]

in that case, yea ur good with this one

[sterile star]

Or is this?

Or is it**

[dull island]

This is fine if you put parentheses around the sin+cos

[uncut aspen]

this one is good

[sterile star]

Okay

Thank you! I have like 6 more to do kms

[uncut aspen]

you got it!

trust!

[sterile star]

I’m losing sanity this 6 week course is killing me

[uncut aspen]

wait 6-week?

for all of calc 1?

[dull island]

Fellow Goodnotes user ?

[trim jolt]

@sterile star Has your question been resolved?

[sterile star]

Yes

[sterile star]

Yessir 😄

Okay

This is everything

[dull island]

Use wolfram alpha to check

[sterile star]

I’m just gonna yolo it tbh

I’ve been working on this too damn long

.close

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[rose inlet]

,rccw

[digital girder]

Thanks

[rose inlet]

np

but i also have no clue what im lookin at lol

[digital girder]

I'm just wondering where the minus sign comes from

Simple mass spring system

[rose inlet]

that is

above my paygrade

[digital girder]

Ok I'll leave it here for someone to answer

Total energy of the system is kinetic energy + potential energy = 1/2 mx'' + 1/2kx

Which is constant so the derivative should equal 0

So then that means mx'' = -kx

But from F = ma, we get mx'' = kx

It's just a sign thing, I'm just wondering how to make the minus sign appear for a clean derivation

It might have something to do with d'Alembert's principle, or not

[bright quarry]

no it’s negative

[digital girder]

Oh

Yeah lol

[bright quarry]

the acceleration is opposite the direction of spring force

[digital girder]

But still

Why the minus

Hooke's law I guess

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[heady frigate]

A crate is dragged 22 m along level ground by a 90 N force applied at an angle of
40° to the ground. It is then dragged up a 6 m ramp onto a truck by the same force. If
the ramp is inclined at 25° to the ground, find the total work done.

[heady frigate]

i need this in the vector way

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@heady frigate Has your question been resolved?

[bright quarry]

do you know the formula for work

[heady frigate]

thanks

kinda

[bright quarry]

you’re welcome

what is it

you can ping helpers after 15 min btw

[heady frigate]

w = force dot distance

[bright quarry]

yes

so for the first part along level ground

what would the work be

[heady frigate]

w = |f||s|costheta

i get the first part

but the second part idk

[bright quarry]

did you try drawing a picture

[heady frigate]

yeah

is the angle 0 or 25

is the angle bewteen force and distance

[bright quarry]

well we assume the angle of 40 degrees to the ground would be the same

for the force

but now

it’s moving up an incline that is 25 degrees

above the ground

[heady frigate]

so theta is 25?

wt = w1 = w2

[heady frigate]

= 90(22)cos40 + 90(6)cos25

yeah mb

or is it 0

for angle

[bright quarry]

well i’d assume the displacement is vertical no?

like

[heady frigate]

huh

isnt the displacement on an angle

but the force is on the same angle too

25 degrees

[bright quarry]

hold on

i’ll draw a picture

for how i imagine it

[heady frigate]

ok

[bright quarry]

i think the angle should be zero

[heady frigate]

okay me too

i guess my teacher wrong

[bright quarry]

because the displacement and force should point in the same direction

along the incline

[heady frigate]

yeah

okay thanks

[bright quarry]

you’re welcome

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[still slate]

I'm trying to understand how to calculate the first simplicial homology group of the torus. If we let the 1-simplices of the torus be represented by a, b, and c, the kernel of the first boundary map will form the free abelian group whose set of generators is <a,b,c> and the image of the 2nd boundary map will form the free abelian group <a+b-c>. I know that for the purposes of computation, we can calculate H1=Ker(del1)/Im(del2) by establishing the relation a+b-c=0 and observing that Ker(del1) reduces to <a,b>. This means that H1 is isomorphic to Z2. However, I don't know how the cosets of H1 form a group that is isomorphic to Z2 from an algebraic perspective.
Can someone help me out please?

[still slate]

For reference, this is the labelling I'm using

!status

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2. I have begun but got stuck midway.
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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

[still slate]

2

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[drowsy ledge]

A ball begins falling from rest. horizontally, the ball is moving at a rate of 1.66m/s.
i'm trying to find how fast a constant gust of wind would need to be to keep the ball at a stand-still, assuming that the wind is horizontally moving in the opposing direction of the ball's movement.

air density = 1.225kg/m^3
ball radius = .254m
ball cross-section: .2026m^2
Fdrag = .5 * 1.225 * .5 * .2026 * 1.66 = .103N

this is as far as i've gotten, very stumped

[drowsy ledge]

any help would be greatly appreciated

[knotty quest]

viscosity given?

[drowsy ledge]

is that the drag coefficient?

[knotty quest]

ig

[drowsy ledge]

.5

[knotty quest]

okay

[drowsy ledge]

viscosity of the ball, you mean?

[knotty quest]

so what will be drag force?

[knotty quest]

air dude solids dont have tht

[drowsy ledge]

drag force should be .103N

oh word

[knotty quest]

y so?

im asking formula

ever heard 6pi(drag coefficient)rv?

[drowsy ledge]

i've not

[knotty quest]

ah ok

idk how explain then

[drowsy ledge]

i think we can ignore viscosity

it's not mentioned

[knotty quest]

drag coeffiecient mentioned?

[drowsy ledge]

yeah it's .5

the Fdrag should calculate to .103N

if i calculated correctly

[knotty quest]

then u gotta consider it

[knotty quest]

how did u get that

u should basically balance drag force and mg

[drowsy ledge]

oh

[knotty quest]

by using this
in here, v is velocity

relative velocity

[drowsy ledge]

Fdrag = 1/2 * drag coefficient * cross-sectional area * air density * ball velocity

is how i got .103N

[knotty quest]

idk this formula lol

[drowsy ledge]

im assuming air resistance would need to be -1.66m/s but idk

not air resistance, but the force of wind

and idk how to find that against the ball

[trim jolt]

@drowsy ledge Has your question been resolved?

[drowsy ledge]

i think ive gotten somewhere yall

hold on

[drowsy ledge]

i calculated Fdrag wrong
Fdrag = .5 * 1.225 * .5 * .2026 * 1.66 = .171N

the ball falls for .713 seconds

horizontal velocity of 1.66m/s

required acceleration:
a = Δv / Δt ≈ -1.66 m/s / 0.713 s ≈ -2.33 m/s²

applying Newton's F = ma

(mass of ball) .45 * (a) -2.33 = -1.049

required wind speed:
-1.049 = 1/2 * (air density) 1.225 * v^2 * (drag coefficient) .5 * (cross-sectional area) .2026

thoughts please??

-sqrt(1/2 * 1.225 * .5 * .2026 + 1.049) = v

= -1.054

it doesnt make sense

the anguish

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[wraith hinge]

For this

[wraith hinge]

Why is it not $\vv H = \vc\varphi, H_0 e^{-jkR}$

[wraith hinge]

confusion

the only way i can see that happening is if you taylor expand $e^{-jkR}$ with [
e^{-jkR} = 1- jkR + \4{k^2R^2}2 + \dots
]
and assume $kR \ll 1$ and drop off second and up order terms

[wraith hinge]

ohh wait its the sum of both terms and they all contain e^jkR

so it cancels out

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[wraith hinge]

If a sequence converges to a value, then show that, all it's subsequences also converge to the same value.

[wraith hinge]

I have a proof for this I think

Consider we have a sequence $x_n$ I will denote it's subsequence as $x_n__k$

[solid kiln]

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

[kindred pier]

$x_{n_k}$

[solid kiln]

SWR

[wraith hinge]

sorry for the latex

im a noob

[kindred pier]

I was when I started. You learn by doing

[wraith hinge]

alright, so, we note first of all the subsequence $x_{n_k}$ can be written out as $x_{n_1}, x_{n_2}, ....$ such that $ 1 \le n_1 < n_2 < .... $

[solid kiln]

Arch

[wraith hinge]

Now, we note that, all $n_i$'s are basically positive integers.

[solid kiln]

Arch

[wraith hinge]

meh I'll just click a photo of what I've done on paper and send 💀

the basic idea is, there is going to be some q for all m, such that $n_q \ge m$

[solid kiln]

Arch

[wraith hinge]

then after that for all $k \ge q$ you get that they are ALL greater than m, for any given m, so you are able to say that for all values of epsilon, we have a value of q, and for all k more than that q, we have the inequality holding

[solid kiln]

Arch

[wraith hinge]

and then by definition we are done

I am not sure if what I've done is correct though

cause I am used to fakesolving

I would appreciate any comments, advice, and checking if this proof is valid

[trim jolt]

@wraith hinge Has your question been resolved?

[kindred pier]

Your rationale is fine. As for comprehensiveness of the proof, I'm too tired to read

But the idea you have is the right approach

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[austere totem]

is that attached image correct judging by the following question -- Boxes A and B contain 4 counters each. Box A contains 1 yellow and 3 orange counters and box B contains 3 yellow and 1 orange counter. A box is chosen at random and then a single counter is selected.

[hallow spruce]

seems fine

[austere totem]

thank you!!

enjoy the rest of your day!!

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[stuck holly]

the first doesnt converge but the 2nd?

[tall kernel]

Diverges

[stuck holly]

why

[tall kernel]

Limit test

Or other tests

[stuck holly]

what test

[tall kernel]

How did you determine the first series converges?

Oh I mean

Doesn't cnverg

[stuck holly]

limit test

[tall kernel]

Do the same for the second

[stuck holly]

theres no test to do for the 2nd

it doesnt converge and it doesnt go to infinite

it irregular

[wanton rune]

divergence test.

[stuck holly]

what test is that?

[wanton rune]

so if the limit is not 0...

[stuck holly]

prove it is not 0

[wanton rune]

??? sin(n) is a periodic function, its limit does not exist

[stuck holly]

so that applies to the rule?

the answer is that the sum is irregular?

[wanton rune]

its like (-1)^n in that it doesn't approach 0 as it goes to infinity

so sum (-1)^n diverges

[stuck holly]

diverges means it goes to infinity no?

[wanton rune]

no

[loud shell]

isn't sin oscillatory?

[wanton rune]

yes

[stuck holly]

so if it dont converge is always diverge?

[wanton rune]

diverge can mean bounce between values, like -1+1-1+1... bounces between values constantly, but note the partial sums don't get closer to any particular value

[wanton rune]

if the sum doesn't coverge, it diverges yes

if the sequence inside does not go to zero, the sum will not be convergent

[lofty walrus]

i think you owe him a proof that sin(n) doesn't go to 0

[stuck holly]

no i know it doesnt go only to 0

i just thought diverge means necessarily go to infinite

or a fixed amount

[loud shell]

actually the sum just bounces between -1 and 1 and hence its not convergent but oscillatory

[stuck holly]

i didnt know that irregular solutions are diverge too

[stuck holly]

yeah but is oscillatory by definition divergent?

[loud shell]

nah

[lofty walrus]

depends on your definition on divergence

[loud shell]

divergent and oscillatory are different things

[stuck holly]

ight

ill just say irregular then

[loud shell]

yeah

[stuck holly]

can u help me with another real quick bro?

i have a doubt

[loud shell]

yea

[stuck holly]

thx

what method can i use for this

[lofty walrus]

what have you tried?

[stuck holly]

thats what im unsure ab

should i do ax/ax+1?

idk if it helps at all tho

i dont see limits to compare it to neither

[loud shell]

uhm...

[acoustic flint]

Ratio test is a_{n+1} / a_n btw

[stuck holly]

mb

[loud shell]

use this test

[stuck holly]

if its limit goes to 0 tbh, its a irregular form

[loud shell]

irregular form?

[acoustic flint]

He means indeterminate form

[stuck holly]

ye

[loud shell]

yea

im getting this limit 1 actually

hence divergent

[stuck holly]

why u get 1

[loud shell]

just apply limit

[frail drift]

yeah the limit is 1

[stuck holly]

but the limit is indeterminate how yall getting result?

[loud shell]

$\lim_{x \to \inf} \left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)}=1$

[solid kiln]

Taki

[stuck holly]

just the x is ^

[loud shell]

ahh.. $1=1^{\frac{1}{x}}$

[solid kiln]

Taki

[loud shell]

for any x

[stuck holly]

k

but

here says inf^0 is indeterminate

how u get result out of indeterminate

isnt indeterminate mean that u cant get result unless u use tricks

[loud shell]

thats why you use limits

[acoustic flint]

1/inf = 0, 0^0 = 1

[stuck holly]

what

[loud shell]

not necessary 0^0=1

[stuck holly]

we dont have 1/inf

[loud shell]

somethimes 0^0=0 too (in limits)

[stuck holly]

yeah but u do stuff

u just straight said that inf^0=1

[acoustic flint]

Why’s that

[loud shell]

0^0 is indeterminant too

[loud shell]

i never said that

[stuck holly]

herer

[loud shell]

where is inf^0 there?

[stuck holly]

x^(1/x)

[loud shell]

bruh

thats (1/x)^(1/x)

[stuck holly]

isnt that the same tho

1 stays 1

[loud shell]

no

tf..

are you even calculating limits yourself?

follow the procedure to find limits

you'll get 1

[stuck holly]

ight btw

if u have value with - and +

u usually do the module yeah?

but if the module diverge

then u check if ax<ax+1

yeah?

do u do this too

[loud shell]

:?

what?

[frail drift]

$\lim_{x->\infty} \frac{1}{n^{1/n}} = \lim_{x->\infty} \frac{1}{e^{log(n^{1/n})}} = \lim_{x->\infty} \frac{1}{e^{\frac{1}{n}*log(n)}} = \frac{1}{e^0} = 1$

[solid kiln]

smygalz

[stuck holly]

ay thx

[frail drift]

that's another way to calculate that limit

[main sigil]

hmm the variables are kinda wrong in the latex

[stuck holly]

Doesnt this mean inf^1/x always 1?

[frail drift]

oh sorry instead of x->inf it should be n->inf

[main sigil]

What about (x^x)^(1/x), this is technically inf^(1/inf)

as x -> inf

[stuck holly]

Uhm

Yah

Okok i understand

[frail drift]

every time you have an exponential use this e^log trick

it helps

[stuck holly]

Thx

[stuck holly]

@frail drift do u follow this procedure too?

[lofty walrus]

does this work because e^x is continuous?

[stuck holly]

No its cuz u have e^0

[frail drift]

i don't know what you're talkin about

[lofty walrus]

continuous at 0 i guess

[stuck holly]

Like when u have sum of (-1)^x

U try with module

And if not try to see if ax<ax+1

[frail drift]

the first thing you do is the limit

[frail drift]

i guess it works because x = e^log(x)

if x > 0 obv

[stuck holly]

Ye but do u do the rest

As i do

Im asking cuz I want to understand 1 thing

[lofty walrus]

i think you can't pass the limit through a function like that if the function isn't continuous

[stuck holly]

Leibiniz

I remembered

[frail drift]

yes also

[stuck holly]

Name is leibiniz

[frail drift]

yeah use leibniz rule

[stuck holly]

Ok so

If ax<ax+1 isnt true

Like the sinx example

Then what u get?

[frail drift]

but the sinx series is not an alternating series

[stuck holly]

Its negative and positive

Its alternating just like (-1)^x

[frail drift]

no because (-1)^x alternates at every step

you have -1, +1, -1, +1

instead sin(n) is not alternating for every next N

[stuck holly]

Okok

[frail drift]

so you can say that sum does not converge

[stuck holly]

U would yet diverge anyway right? When leibiniz

[frail drift]

this sum oscillates it's not convegent nor divergent

[stuck holly]

No not sinx

I mean when used Leibniz properly

ax>ax+1 mean diverge?

[frail drift]

ax>ax+1 this means converge

idk if ax < ax+1 if it means divergent

[stuck holly]

Thx

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[maiden crane]

Let $T$ be a subset of the integers from $1$ to $209$, where $|T| = 11$. Is it possible for all non-empty subsets of $T$ to have distinct sums of their elements?

[solid kiln]

exams (back may 30)

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[frail drift]

If |T|=11 you know that the number of non-empty subsets is 2^(11)-1 = 2047

ok?

let x be the minimum integer of a set T
You would have that the minimum sum any subset of T is x and the maximum sum could be x+200+201+202+203+204+205+206+207+208+209=2045+x

So you have 2046 possible sums but you have 2047 possible sets so at least two sets have the same sum (pigeonhole principle)

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[maiden crane]

o alright thank you

[maiden crane]

.reopen

So you're saying it's not possible that all non-empty subsets of T have distinct sums of their elements?

@frail drift

[frail drift]

yes

[maiden crane]

But why did you make x not equal to 199 @frail drift

Shouldn't that be 199 to get the maximum subset sum?

[frail drift]

yes but in that case you would have that the min sum is 199

so the difference between max sum and min sum is 2046

ok so i proved it when the min value is 199 and i want to generalize it for every possible choice of 11 numbers

[maiden crane]

But why do you have 2046 possible sums, where did that number come from @frail drift

Are we counting x to be x = 1?

[frail drift]

oh sorry

if the minimum sum is x and the max sum is 2045+x

you have (2045+x - x +1) possible sums

i actually got this solution from here
https://math.stackexchange.com/questions/1170537/pigeonhole-question-about-distinct-sums

maybe it can help

[maiden crane]

Still really confused, how they came up with this set of numbers, seems like they just chose whatever they wanted @frail drift

They picked some sum of numbers that was -1 less than the total different sums (128)

[frail drift]

yeah that answer is not clear

look the one after

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[maiden crane]

alright well i'll use that link to guide me, thanks for that @frail drift

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[radiant crow]

can someone pls explain me this? thank you

[raw magnet]

hi so the first part is trivial

bc it's symmetrical equation

you can interchange a, b, and c

so interchange them s.t a <= b<= c

you got that part?

[radiant crow]

am back

wait i'll go rea

d

[radiant crow]

yes

[raw magnet]

well

since a<=b

we can assert this is bigger than a/(b+c)

trivially

since b<=c

this is bigger than

[radiant crow]

thanks

alot

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[simple jackal]

can someone verify my expression for the diagonal entry of an orthogonally diagonalizable matrix (symmetric) A? also A is psd

[marble wharf]

well if you just want to show A_ii >=0, its enough to consider e_i^T A e_i

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[simple jackal]

clever

thanks!

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[ancient edge]

How does that become that?

[ancient edge]

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[vernal cosmos]

Hi, how can I solve this for h? green line is tangent to circle - c is known (or x can be known then c is unknown, doesn't matter), d is radius also known, alpha can be between 0 to 70 deg,

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[vernal cosmos]

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[rancid perch]

I know that x = log10(2), but how can I show the rest of the work to derive that

[proper kernel]

use that $\log(a^b)=b\log a$

[solid kiln]

mtt

[rancid perch]

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[hidden elk]

hey all! is there an easy way to determine trhe inverse of larger matrices?

[hidden elk]

finding the minors is very tedious and also the algorithm is very difficult to remeber

[ionic pendant]

using gaussian elimination, augmenting with the identity matrix [ \begin{amatrix}{1} A & I \end{amatrix} \xrightarrow{\text{RREF}} \begin{amatrix}{1} I & A^{-1} \end{amatrix} ]

[solid kiln]

cloud

[ionic pendant]

note that if A is not row reducible to the identity matrix then it is not invertible

[hidden elk]

gotcha

i just saw a youtube video which helps understand thew algorithm portion soo much better

just imagine covering the column and row and boom thats your minor then solve

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[spark aspen]

Can anyone help me with my situationel math problem for sec 5 math its on trigometry

[kindred pier]

!da2a

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[spark aspen]

]

anyone can help me do this ?

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[river apex]

Hey

[river apex]

I need help figuring out how to calculate the possibilities of combinations in a game.

There is this game called mastermind

I don't need to explain the rules but here is what I want to calculate

There are 6 possible colors, but I'm trying to figure out all the possible combinations of any 4 of them

There can be duplicates such as red red blue yellow or yellow yellow yellow yellow etc

How do I calculate this?

[burnt mulch]

Ngl I’d just do casework:

• all 4 are the same
• 3 are the same and the fourth is different
• 2 are one color and 2 are another color
• all four are different

[river apex]

I don't know how to do that

[burnt mulch]

Are you familiar with combinations?

[river apex]

Not really. It's something I was interested in learning when I saw the gane

Game

Is there any way you could just show me how to do these and then I can learn from there?

[burnt mulch]

https://brilliant.org/wiki/combinations/

[river apex]

There are 6 colors and I want all the combinations of any 4

[burnt mulch]

This is basically what you’re looking for

[river apex]

I'll read that. I will. But I'm at the hospital rn and I'm just curious how to do this one lol

Would you mind showing how to calculate the combinations of this game

Thanks though I'll read that later

[jagged fiber]

would it just be 6^4?

[burnt mulch]

Wait

[jagged fiber]

ok im actually gonna fail how is it not 6^4

[burnt mulch]

Does the order in which you pick the colors matter

Ex. Is YRYR the same as YYRR

(Yellow and red)

[burnt mulch]

Repetition

[jagged fiber]

if the answer is no would it be 6C4

and if the answer is yes 6P4?

[burnt mulch]

You can repeat colors…

[jagged fiber]

omg im so dumb

💀

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[neat herald]

guys real quick question is the instantaneous and average rate of change the same thing

[kindred pier]

No

[neat herald]

how is it different?

[kindred pier]

One is instantaneous and the other is average. Sounds dumb but that's kinda the whole thing

If rate of change is constant, then they're the same. Otherwise, no

[neat herald]

whats the difference between instantaneous and average tho

i know that the average is (y2+y1)/2 over (x1+x2)/2

[hallow kite]

if i have a position-time graph, the average rate velocity would be the slope of the line between the two points you’re interested in looking at

that line is called the secant line, it requires you to define some time interval to take the average over

instantaneous velocity is the velocity at a particular moment in time, this would be the slope of the tangent line to the position time curve at that point

[neat herald]

give me an example of instantaneous

[ionic pendant]

let's say you go in a car, then your speed is the rate of change of your position over time. your average speed is measured across the entire trip. for example if you go 80 km in 2 hours you went an average of 40 km/h. but your instantaneous speed is what you read on your speedometer, which may be higher or lower than the average speed

[neat herald]

so for example if you stop for a red light, your instantaneous speed would be o km/h but your average speed would still be 40 km/h?

[ionic pendant]

yes

and your average speed depends on what interval you measure on. for example if you went 50 km in the first hour, then your average speed that hour is 50 km/h, but your average speed over the entire trip is still 40 km/h

[neat herald]

ok thanks!

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[latent axle]

What am I supposed to do for part b exactly?

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@latent axle Has your question been resolved?

[tranquil widget]

write x in terms of y for y=x^2

then taking that expression, integrate wrt y (dy)

integrate such that yoou get the area above the curve and line and below the curve and line

because theyve mentioned that this y=c line divides these areas into two halves

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[simple anvil]

i need help with this simple physics question on work

i swear my answer is right but idk why it's saying it's not

[deft plaza]

wd=Fxd so wd on 20N block=20x75x10^-2

not sure if thats right

[simple anvil]

whats wd?

[analog shale]

(you don't use "x" for multiplication)

[wraith hinge]

work done

[simple anvil]

i got 9 by doing 12 * 0.75

because i assumed the normal force and gravity both have a work of 0

since the formula is Fdcos(theta)

theta being the angle between the displacement and the force

so the angles are 90 and 270 both of which are 0

so it cancels out all the vertical work

so the only work could be the work from the tension

[wraith hinge]

[wraith hinge]

how did you get displacement ?

as 0.75 ?

[simple anvil]

it's given in the problem

it says the block moves 75 cm to the right

so i divided by 100 to get it in nmeters

[wraith hinge]

did you solve for the accel of the system?

[simple anvil]

nah i didnt think i needed to

bc i took F as just the tension force

which would be the weight of the block hanging off the table

but apparently not since i got the problem wrong

[wraith hinge]

i doubt that

in the work done formula

force

would be mass times accelration

mass would be 20/g and accel would be equal to accel of the system

g is gravity

get it ?

[simple anvil]

why wouldnt F be the tension force though?

since it's the only force acting on the block

that has a non zero work

[wraith hinge]

lets say there is a block still on the air

neglect air resis

what would be the work done by gravity if the block moves horizontally downward by 10m

[wraith hinge]

draw the F.B.D of the 20N block

what would you get the tension as ?

[wraith hinge]

ignor ethis

[simple anvil]

oh

would tension be m1a = m2g?

[wraith hinge]

what?

[simple anvil]

idk

im having a brain fart

[wraith hinge]

did you do this ?

[simple anvil]

yes

[wraith hinge]

what is the equation

[wraith hinge]

that you have got ?

[simple anvil]

Ft = m1a

[wraith hinge]

man ignore the vertical forces

[simple anvil]

lol

[wraith hinge]

[simple anvil]

since tension is going to the right it's posisitve

so just = ma

ye

i see

[wraith hinge]

so did you undeerstand why your answer is wrong?

[simple anvil]

i do

how do i use the second block to help me find the acceleration?

[wraith hinge]

draw the FBD of the second block

[simple anvil]

i cant set the tension forces equal to each other right

cause the system is moving and not in equilibrium

[wraith hinge]

❌

tension along the same string is always equal

[simple anvil]

oh shoot

nice

so a = m2g/m1

sweet

[wraith hinge]

:/

try again

it is wrong

[simple anvil]

oh what

the tension of the left block is

m1a

and the tension of the right block

[wraith hinge]

indicate the accelration of the block too

[simple anvil]

is m2g

[simple anvil]

accelerating down

[wraith hinge]

yes

[wraith hinge]

wrong

draw the FBD again this time showing the accel

[simple anvil]

like this you mean?

sorry for being slow btw

physics is my weakpoint

[wraith hinge]

yes now

what is the equation you achieve from this

[simple anvil]

Ft - Fg?

idk im lost

or are you jsut trying to say that g should be -g since a is down?

[wraith hinge]

no

[wraith hinge]

since the block is moving downward

it should be the oppsite

Fg-Ft

because tension is upward and g is downward

[simple anvil]

oh bc Fg is larger than Ft?

since Fg is the "leader"

[wraith hinge]

yes

[simple anvil]

Fg is only equal to Ft in this scenario if the system was in equilibrium?

since theyd balance each other?

[wraith hinge]

yes

but here it isn't

[simple anvil]

ye

[wraith hinge]

so what would you equate it to ?

[simple anvil]

ok so Fnet of the bottom block is

Fg-Ft

in the y direction

Fnet in the x direction of the top block

is ma

so u eqate those two?

equate

[wraith hinge]

just think about the second block for now

[simple anvil]

oh then

Fg-Ft = m2a

[wraith hinge]

Fg - Ft = m2a

[simple anvil]

ye

[wraith hinge]

yes.

[simple anvil]

and a is the same a as the first block

since the blocks are attached and move as one?

[wraith hinge]

yes

from the first equaiton

ft would be ?

Ft

[simple anvil]

m1a

[wraith hinge]

sub in second equation

[simple anvil]

Ft for seconb lock is

Fg - m2a

so m1a = Fg - m2a

[wraith hinge]

[simple anvil]

m1a = m2g - m2a

and i plug in from there and solve for a

[wraith hinge]

yes

[simple anvil]

then once i have a, i do m1a * displacement

and thats it?

[wraith hinge]

[simple anvil]

ok imma give it a try

[wraith hinge]

👍

[simple anvil]

holy shit it worked

the problem seemed so simple on the surface

and i guess it kind of was

but i didnt think there was going to be that many steps

thanks a lot bro

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[orchid wagon]

lasso crop 😭

[main sigil]

Hmm why is it cropped?

Is it a test?

/ exam

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[steel oyster]

!status

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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

[livid girder]

this question appeared in previous year question paper

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[limpid shoal]

how do you solve from here?

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[limpid shoal]

so its 45.8?

nah its correct

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[split chasm]

is there more context?

what issue do you have with the work her

why is there an issue with p^* being 1

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[wraith hinge]

<steinmetz>
So i dont understand why when converting between the time and frequency domains the angular frequency is oftentimes neglected? Take for example: [
\6vt = V_m\6\cos{\omega t + \varphi} = \6\RRR{V_me^{j\8{\omega t + \varphi}}}\Iff V = V_me^{j\varphi}
]
$\omega$ is inevitably lost by going in the $\Implies$ direction, and returning by the $\Impliedby$ direction will not tell us what $\omega$ is

[wraith hinge]

ok im just going to assume this is purely notational

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[pliant cliff]

hi

[plush igloo]

you can calculate all the possible arrangements and then subtract from it all the arrangements where 2 S's are next to one another

7 letters gives us 7! possible ways of arranging them

if we consider 2 S's to be in a group, there are 5 letters and 1 group we need to arrange and that gives us 6! possible ways of arranging

7! - 6! = 6!6

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[civic thorn]

How can I prove $\frac{1}{x}>0$ on the interval $(0, \inf)$

[solid kiln]

dingypine

[marble wharf]

depends on what you can use

[main sigil]

x is positive, so you can multiply both sides of the equation by x ig, but yeah, it depends on what you can use

[civic thorn]

for context this is the actual question

so i took the derivative of ln(x)

since if it's increasing f'(x) should be > 0

second shape therom just says f'(x) > 0 then f is increasing on the interval

[main sigil]

Do you even need to prove that 1/x > 0 then? Isn't it obvious enough

[civic thorn]

i honestly dont know but just incase a similar problem shows up like this on an exam i should probably know how to

thats why i was asking how to prove 1/x > 0

[main sigil]

ig that you'd either have to go to the definition of 1/x, or just use the property that you can multiply it by positive number (x) and it wont change

so you could do 1/x * x > 0 * x

and 1 > 0

[raw magnet]

well say it's' not then, 1/x > 0; take the reciprocal and then x < 0

idk

contradiction

[main sigil]

hmm

[raw magnet]

wait

[main sigil]

reciprocal

1/0

[raw magnet]

the other way around

sorry

[main sigil]

multiplying by x is sufficient here i believe

[raw magnet]

yeah

[civic thorn]

alright i see

[main sigil]

It requires just the fact that every inequality is equivalent to itself multiplied by a positive real number, that 1 > 0 and 1/x * x = 1

1st is accepted and proven together with introduction of inequalities for reals
2nd is obvious
3rd is either simple algebra, or the fact that 1/x is multiplicative inverse of x (for non-zero x)

[civic thorn]

alright i see, thank you guys! i appreciate it

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[simple jackal]

what does the down arrow mean

[simple jackal]

is it y approching x from the left

[stark bison]

No, from the right

[simple jackal]

oh yes youre right

thanks

[stark bison]

The arrow indicates that y is approaching x from upwards and upwards is the positive vertical direction, just like the right is positive horizontal side (conventionally)

[spiral kettle]

i guess it would be confusing if they named it “right-continuity” and the down arrow meant “from the left” lol

[simple jackal]

i’ve only seen notation using a + or -

[spiral kettle]

not anymore!

[simple jackal]

yea

v cool

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[wraith hinge]

I figured by having those two points I could come up with a vector

or a line joining those two points

but I'm trying to come up with a perpendicular to that line

is there a way to do this?

I know if it was for example, a 2 coordinate vector

I could use the slope formula rise/run and find the negative reciprocal to get the perpendicular

[ionic pendant]

in higher dimensions it becomes more useful to use the dot product

[wraith hinge]

but what would we dot it with?

[blissful ridge]

use orthogonality ?

[ionic pendant]

any vector in the plane can be written as pointing from (-1, 1, 0) to some point (x, y, z)

[wraith hinge]

I don't understand what you're suggesting here

are you saying we should dot the first direction vector with an unknown vector xyz?

[ionic pendant]

we should dot the direction vector you found to be perpendicular to the plane, with a direction vector pointing from a known point in the plane (-1, 1, 0) to an unknown/arbitrary point in the plane (x,y,z)

[wraith hinge]

so then we create a new direction vector which includes the points (-1,1,0) and (x,y,z)

let me try that

$B= (x,y,z)-(-1,1,0)$

[solid kiln]

Remlis

[wraith hinge]

$B=(x+1,y-1,z)$

[solid kiln]

Remlis

[wraith hinge]

and then you suggest we dot this with the 1st direction vector?

[ionic pendant]

yes

[wraith hinge]

alrighty

$(x+1,y-1,z)\cdot (2,-4,-1)$

[solid kiln]

Remlis

[wraith hinge]

$= 2x+2-4y+4-z$

[ionic pendant]

the dot product returns a scalar

[wraith hinge]

oh right

[solid kiln]

Remlis

[ionic pendant]

and what should that be equal to if the two vectors are perpendicular?

[wraith hinge]

0

$2x+2-4y+4-z=0$

[solid kiln]

Remlis

[wraith hinge]

$2x-4y-z+6=0$

wait a minute

we already have our

cartesian equation

[solid kiln]

Remlis

[wraith hinge]

alright perfect, thank you

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[marsh burrow]

Is the answer that I got here correct?

[marsh burrow]

for this exercise

[marsh burrow]

iphone 15 pro max taking pictures like a damn nokia

[delicate bobcat]

,w 2y(x+1)y'=4+y^2, y(0) = sqrt(5)

[solid kiln]

Results provided by WolframAlpha

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Requested by modus7591

[marsh burrow]

ok so no.

can you help me fix it maybe? please

[delicate bobcat]

Let's see

It should be

[marsh burrow]

why that? where does my y(x) go?

[delicate bobcat]

ah typo 2y obviosuly

[marsh burrow]

ok, i got here so far, and then i tried a change of variable

[delicate bobcat]

so it seems you don't have 2 there?

[marsh burrow]

oh, right

[delicate bobcat]

ok let's proceed

that's little important

but in fact you could see that:
(y^2 + 4)' = 2y