help-39

[lethal trout]

.close

[pearl pond]

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[pearl pond]

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[willow veldt]

help

[flint zenith]

ask your question

[willow veldt]

i need help solving these please

[flint zenith]

what have you tried

[willow veldt]

haven't really tried it yet, just been stuck and figured I could use some help lol

[flint zenith]

alr

you can factor the first one

that will cancel some stuff out

actually you dont even need to do that

plug in 0 for x and tell me what you get

[willow veldt]

ok one sec

i got 5

[flint zenith]

yea

so as the x approaches zero, the result of the function is equal to 5

now how about the next one

what happens when you plug in 0

[willow veldt]

would this be proper form for the first question ?

[flint zenith]

yup

I think

on the 2nd to last line do you still have a 4x?

can't really tell what that is

[willow veldt]

yes 4x

[flint zenith]

it should be 0

4(0)

not x

[willow veldt]

oh okok

its just x . like multiplied

[flint zenith]

other than that it's fine

but there is no more x

you're taking x to 0

so x becomes 0

[willow veldt]

should it be (0^2 - 4) (0-5) then ?

[flint zenith]

nope

$\frac{0^2 -25}{0^2 - 4(0) -5}$

[jolly parrot]

NahhFam

[willow veldt]

ohhh okok I see

[flint zenith]

yea

you're practically replacing every x with 0

[willow veldt]

alr i get that

that you Fam

thank*

[flint zenith]

np

for part b and c you will have to do some algebraic manipulation

[willow veldt]

could I try the other 2 and then can you check over it to see if i got it right?

[flint zenith]

because look at what happens when you plug in 0 for x in part b

yea

[willow veldt]

thanks

would this be correct for number 2 ?

or b i should say

[flint zenith]

yep

that's what I got

but why did u use L'Hopital's rule lol

[willow veldt]

a friend suggested it lol so idk

should I have used something else

in your opinion should I or should I not be using it for grade 12 calculus and vectors

[flint zenith]

it depends

have you learned L'Hopitals rule yet?

if you have then that's fine

but if you haven't I'd tried other ways

[willow veldt]

no we haven't learned it yet, I think we will next week

[flint zenith]

then yea don't use it

I'd multiply by ((3+x)(3-x))/((3+x)(3-x))

same as multiplying by 1

and distribute to see where it leads you

[willow veldt]

ohh okok

i see

lol

[flint zenith]

nope

lol

[jolly parrot]

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

[flint zenith]

$\frac{\frac{1}{(3+x)} - \frac{1}{(3-x)}} x \frac{(3+x)(3-x)}{(3+x)(3-x)}$

you get the idea

[willow veldt]

is it cool if you can show me ur answer and work for number 3, I didnt really know how to do it

[flint zenith]

Im gonna figure this out while you figure that out

[willow veldt]

yea lol

[flint zenith]

I will once you solve part b using this method

[willow veldt]

ok I will try it

[flint zenith]

there

[willow veldt]

okay

[jolly parrot]

NahhFam

[willow veldt]

ok i think i got it

[flint zenith]

show ur work

let's see

[willow veldt]

its uploading one sec

like that you mean?

[flint zenith]

yea

[willow veldt]

ayy okk

[flint zenith]

close enough xd

posting so I can see it easier

[willow veldt]

is for proper or is smth missing

[willow veldt]

okok

[flint zenith]

it's just formatted weird

idk

[willow veldt]

lol

[flint zenith]

I don't think the algebra is proper

[willow veldt]

meh close enough lol

[flint zenith]

it's w/e

so the last one

you need to multiply the top and bottom by the conjugate of $\sqrt{x+1} - 2$

[jolly parrot]

NahhFam

[flint zenith]

which is $\sqrt{x+1} + 2$

oops

[jolly parrot]

NahhFam

[willow veldt]

ok

i think I figured it out

is the final answer 1/24

[pearl pond]

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[fringe ice]

I do not understand b)

[fringe ice]

Probs got a wrong too but I can fix it myself

[pearl pond]

@fringe ice Has your question been resolved?

[bitter lodge]

In which direction is the bucket accelerating

[pearl pond]

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[alpine imp]

$\frac{dx}{x(y^2 - z^2)} = \frac{dy}{y(z^2 + x^2)} = \frac{dz}{z(x^2 + y^2)}$

[jolly parrot]

Leahalcyona

[alpine imp]

Find the first integrals

(maybe method of finding integrable combinations (properties of proportions))

[pearl pond]

@alpine imp Has your question been resolved?

[alpine imp]

<@&286206848099549185>

[pearl pond]

@alpine imp Has your question been resolved?

[austere mantle]

Of which equation exactly here

[alpine imp]

for example you can rewrite:

[jolly parrot]

Leahalcyona

$dx = x(y^2 - z^2)dt, \\
dy = y(z^2 + x^2)dt, \\
dz = z(x^2 + y^2)dt$

[alpine imp]

.

@austere mantle

[pearl pond]

@alpine imp Has your question been resolved?

[pearl pond]

@alpine imp Has your question been resolved?

[pearl pond]

@alpine imp Has your question been resolved?

[prisma elk]

no idea whats the question here @alpine imp

[pearl pond]

@alpine imp Has your question been resolved?

[buoyant linden]

@alpine imp what have you tried so far?

[alpine imp]

i tried and found some already, just wait cuz i can't send it rn, I'll be free in a few hours btw

[normal notch]

what's this supposed to be used for?

obv its a dif eq

looks like triality

in metric spaces

or something

hmm

[full relic]

<@&268886789983436800>

[stark owl]

Hello

can you guys help?

[vagrant briar]

post your question in an unoccupied help channel

[pearl pond]

@alpine imp Has your question been resolved?

[worn bear]

Wdym find the first integrals?

Ive never heard of that

Wat class is this? @alpine imp

[midnight haven]

Hint: You can use Newton's second law of motion on both a and b

you will also need to do a bit of algebra here too

[humble lintel]

This question has gone completely off the rails, but if you multiply the equations by x, y, z respectively, you get ydy-zdz = xdx which you can integrate immediately.

[alpine imp]

Read my original question again and think about it

[humble lintel]

Maybe you should show some work rather than tell the person responding to you to think about it. This is your homework, not mine.

As I understand a first integral, e.g., a non-constant differentiable function that is locally constant on a solution to the PDE, that step leads you there.

[alpine imp]

Let me simplify things for you. Let's start by introducing the following:

(am not sure how true the statement about f_i \neq 0 is, tho)

@humble lintel

Are you listening?

[alpine imp]

And this property is invertible btw

Now, we can use the method of integrable combinations

to find the first integrals

According to the given condition, the system is implicitly defined by a symmetric form:

[jolly parrot]

Leahalcyona

[jolly parrot]

Leahalcyona

[alpine imp]

(have omitted some trivial explanatory steps)

[midnight haven]

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[worn bear]

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[rough stream]

.close

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[alpine imp]

look at the concept of the first integral in ode systems context please

[pearl pond]

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[quick venture]

Can somebody please help me with this

I've used IBP twice but I dont think I've done it correctly

[open rivet]

1. you forgot the bounds
2. no need for a second IBP

[quick venture]

Um

But

So I use the bounds for xsinx - cosx

[open rivet]

$\int_0^\pi x\cos(x)\dd x=[x\sin(x)]_0^\pi-\int_0^\pi\sin(x)\dd x$

[jolly parrot]

Bonk

[latent arrow]

What's the integral of sinx?

You don't need 2nd IBP to evaluate the integral of sinx

[pearl pond]

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[pearl pond]

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[dire basin]

why can we assume tension is proportional to (y_n+1 - y_n)/h

[pearl pond]

@dire basin Has your question been resolved?

[stuck knoll]

small angle approx?

[pearl pond]

@dire basin Has your question been resolved?

[dire basin]

isnt that like sin(theta) is approximately theta

is that what they're doing here

[stuck knoll]

is it not tan

am i bugging

[dire basin]

if ur evaluating the vertical component it'd be Tsin(theta)

[stuck knoll]

i was thinking between the string segment and the horizontal

vertical component of the tension is what causes the restoring force yeah

thats T sin theta

[dire basin]

oh right cause sin is approximately tan im dumb lmfao

actually why can we even assume the angle is small

[dire basin]

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[onyx spire]

how would I go about solving this?

[nimble lily]

so to work out where are are starting from: do you know what it means to be/not be O(f(n))?

[onyx spire]

Yes, it means g(n) grows at most as fast as f(n) for large n

[plain spade]

have you considered using an approximation for N! to help you get started?

[lilac jackal]

hmm i have a cool trick in mind

$\sum\frac{x^n}{n!}$ converges for all $x$

[jolly parrot]

𝓡𝓞𝓚𝓔𝓣𝓣𝓞─୨ৎ─❥ ♡ <𝟹❤

[lilac jackal]

(to e^x but you dont need to use that bit)

[plain spade]

That is more elegant that what i as going to suggest

[onyx spire]

So it approaches 0 which then means that that the C does not exist right

which means its not O(2^n)

[lilac jackal]

yes

in fact not O(x^n) for all x

[onyx spire]

okay thanks for the help

[lilac jackal]

no prob

[onyx spire]

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[woven oyster]

I think the answer are d,b,c can someone double check for me

[willow chasm]

yes

[woven oyster]

It's correct?

[pearl pond]

@woven oyster Has your question been resolved?

[latent arrow]

yes

[woven oyster]

Thanks

.close

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[trail swallow]

The answer for A says the direction of average acceleration is South 5 degrees West

But Im not sure how is that possible if you see the diagram

I've drawn vector v1 and v2, and the definition of average acceleration is change in v/change in t

change in v should be vector subtract of v2-v1

and ive drawn the subtraction on the right, and the result vector can't be South 5 degrees West, more like South X degrees North

How is my thinking or direction wrong?

Also the question assumes we start on the 1st quadrant at 0 degrees and travel clockwise

Also for context, I've calculated the degrees travelled after 40 seconds is 10.19 degrees clockwise

[little comet]

where is the drone after 40s?

[trail swallow]

it is travelled 10.19 degrees from 0 degree at quadrant 1

[little comet]

uhh is it similar to the diagram you drew

[trail swallow]

yes my diagram shows that

[little comet]

ok so how did you get this

[trail swallow]

Ok so the vector subtract applies to the velocity vector, which are perpendicular to the acceleartion vectors, so the vectors outside of the circle. I put them tail to tail, v1 and v2

[little comet]

ok i see what you're saying, you're doing v2 - v1 and getting a vector pointing up and to the right

[trail swallow]

yeh

lemme show

u

[feral sedge]

does the problem specify the direction at all?

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[little comet]

im pretty sure it should be v1-v2

[trail swallow]

the question assumes we start on the 1st quadrant at 0 degrees and travel clockwise

[trail swallow]

this is def not true, our teacher taught us v2-v1

bc it's change in velocity so u use the later velocity

to subtract the initial

[feral sedge]

well then you labeled the vectors wrong

[little comet]

you labeled later v1 and initial v2 yeah

[trail swallow]

No i think ur corresponding those to the acceleartion vectors

[little comet]

its not really about the name of the vector, its about final - initial

[trail swallow]

the V1 is the straight line going to south

[little comet]

do you at least agree with this

[trail swallow]

look at this

V2 is the final velocity vector

[trail swallow]

yes

[little comet]

ok so whats the final and whats the initial here

and what do you get when you do the subtraction

[trail swallow]

it's this

v1 is the vertical one

v2 is the down left vector

[feral sedge]

are you 100% sure the rotation is clockwise?

[trail swallow]

yes

our teacher sent us this

[feral sedge]

hmm

seems like answer is wrong then

[trail swallow]

ehh, maybe idk

[feral sedge]

unless...

no

the angle around the circle traveled in 40s is small

[trail swallow]

its def 10.19

my answer matches the answer key value

just the direction is diff

my acceleration magnitude is correct

@feral sedge is the angle wrong moni

[feral sedge]

idk it should be north

[trail swallow]

East North?

[feral sedge]

no

[trail swallow]

?

[feral sedge]

west north

[trail swallow]

yeh

ok i cant

speak english

cause im korean

@feral sedge ty

[feral sedge]

np

[trail swallow]

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[trail swallow]

Cos-1(5/8)=51.32 degrees

The answer is wrong right?

why are all these answers off, C is 200/6.2=32.26

answer key seems wrong again?

[pearl pond]

@trail swallow Has your question been resolved?

[trail swallow]

<@&286206848099549185>

[latent arrow]

I agree with your answers

Just worked them out

The answer key is blatantly wrong, for part A even the units are wrong

It answers a different question than what was asked

[trail swallow]

alright thanks

.close

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[ivory shadow]

a

[ivory shadow]

which one is correct?

[light helm]

both

[ivory shadow]

when i tried this and its incorrect?

[light helm]

don't use braces

use normal ()

[ivory shadow]

i did

but still have that problem

[gray lantern]

you forgot to multiply the last part by 1/2

[pearl pond]

@ivory shadow Has your question been resolved?

[light helm]

show your work for this question

[ivory shadow]

i got it

thank you

.close

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[humble axle]

Im not sure if the answer is +inf or -inf

[rancid depot]

what is the limit of the factor on the left?

[humble axle]

What do you mean?

[rancid depot]

what is the limit of x^2+1

[humble axle]

Can you guide me step by step?

[rancid depot]

$\lim_{x \to \infty} (x^2+1)$

[jolly parrot]

LocalLunatic

[rancid depot]

do you know how to do this?

[humble axle]

Yeah, it's 1

X²/x²=1 then 1/x is cancelled

1/x²*

[rancid depot]

where did the /x^2 come from

[humble axle]

From the highest power

It will be divided to all

[rancid depot]

we are not doing limit of a fraction

[humble axle]

Limits with polynomials

That's what the teacher taught me

[rancid depot]

i suggest going over your notes to see when you need to divide by x to the highest power for a limit

[humble axle]

Can you explain?

I was sick when the teacher taught this lesson

[rancid depot]

https://www.youtube.com/watch?v=NmLljBAg82o&themeRefresh=1 look through some examples here

i cant explain a full lesson to you unprepared

[humble axle]

I know how to do it, maybe we have different ways

What could be your answer?

I'm just checking

[rancid depot]

the answer to $\lim_{x \to \infty} (x^2+1)$ is $\infty$

[jolly parrot]

LocalLunatic

[humble axle]

Where's the other binomial?

[rancid depot]

i was doing this in separate parts

$\lim_{x \to \infty} (2-x) = -\infty$

[jolly parrot]

LocalLunatic

[rancid depot]

then you multiply them together

[pearl pond]

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[strong nova]

i know that colA is an output thingy and nul A is an input thing, so p has something to do with the desired column length and q has something to do with the desired row length? How do i find these values

[cosmic charm]

how long is your column vectors

how many components does it have

[strong nova]

3

[cosmic charm]

so its a subspace of R how many?

[strong nova]

is that fr all i do

p=4 q=3??? or p=3 q=4?????

[cosmic charm]

well your col space is a subspace of Rq

and we know that your column space has 3 components

so q is?

[strong nova]

3????????????????????????????????????????????????

[cosmic charm]

yes

[strong nova]

i dont get this]

https://tenor.com/view/deadpan-thousand-yard-stare-funny-bruh-really-gif-2776150239966197556

thanks sir

.close

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[hearty flame]

i think my y bounds are wrong but im not sure why?

[pearl pond]

@hearty flame Has your question been resolved?

[brave sluice]

it's not right

the bound sqrt(36-y^2) is undefined when y is near -8 and near 8

[hearty flame]

should i break it up like this

[brave sluice]

that's better

you could also try using polar coordinates

[hearty flame]

ill just try this approach

not too familiar with polar coordinates

thank u

[brave sluice]

it will be simpler if you use symmetry

[pearl pond]

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[pearl pond]

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[fair eagle]

,rotate

[fair eagle]

Q6

,ritate

,rotatr

[jolly parrot]

Couldn't find a role matching tate!

Mar 4, 2025, 8:38 AM

[wet osprey]

,rccw

[fair eagle]

,rotate

[fair eagle]

Thanks

I used differentiation thingy and I don't know what to do now

I'm just stuck in general

[void grail]

Which exercise are you solving, 3?

[fair eagle]

Q6

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[pearl pond]

@fair eagle Has your question been resolved?

[leaden wadi]

What have you tried so far?

[fair eagle]

I did a now just finding b

[leaden wadi]

Given a remainder of -4 when divided by (x+1), what equation in terms of a,b,c, and e can you get?

[fair eagle]

I already did part a

.close

[pearl pond]

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[pearl pond]

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[soft otter]

s

[jolly parrot]

nino

[soft otter]

oh it worked

testing the latex

but anyways im 90% sure my proof is correct

its a physics problem but im purely concerned with math

[rough forge]

i kinda don't get how you got there

[soft otter]

yeah i was kinda unsure

so basically i think dx/dt is

[
\frac{dx}{dt} = v \cos\theta \quad \text{(constant, no acceleration)}.
]

[jolly parrot]

nino

[soft otter]

and vertical dy/dt is v sin theta - gt

i mean taking the derivative of x(t) = cos , t gives that right?

[rough forge]

the derivatives and how you simplified it seems fine to me

i am questioning the substitution if done properly

[rough forge]

or did you consider gt-vsinθ-vsinθ = u-vsinθ

[soft otter]

hm idk looking at it it seems wrong

[rough forge]

,w Integrate[Sqrt[v^2-2 *g * t * v * sin(theta)+g^2t^2],t]

ugly

I will try to rewrite it with less constant chaos

[rough forge]

A quadratic under a square root, your best guess is to complete the square and apply a trig sub, I believe

[soft otter]

is that easier than the u-sub here?

[rough forge]

Let me see

[soft otter]

oh wait i realize

i use the trig identity to simplify it so they are the same

[rough forge]

This integral is geometrically speaking like the area under a half circle

[soft otter]

yeah that makes sense

[rough forge]

So you don'T have to actually evaluate it in that sense

I messed up a sign I think

[rough forge]

v^2+gt(...)

[soft otter]

Sorry I'll brb

[rough forge]

unpleasant

,w Integrate[Sqrt[a^2+x^2],x]

[jolly parrot]

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[rough forge]

Ok I see NOW how you got there

Seems it will simplify nicely due to the trig identity sin^2+cos^2=1

I would check with wolfram calculator

[soft otter]

Yup

So I think it works out

[pearl pond]

@soft otter Has your question been resolved?

[pearl pond]

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[white estuary]

it depends about what is your syllabus

i think this is more than enough

[devout drift]

Nice chart

[fair creek]

idk just keep perpendicular axes theorem in mind and you're fine

[shrewd mist]

Depends on your prof, thats probably more than enough. I have never bothered to learn anything other than 2 and 3, and I still sometimes mix them up.

You can actually derive them from the general formula

[fair creek]

calculus... 😔

can't escape it

[pearl pond]

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[warm osprey]

Can an orthogonal set be linearly dependent?
Consider S = {(0, 0), (1, 0)}
They are orthogonal as <(0,0), (1,0)> = 0 (using euclidean inner product)
But clearly they are linearly dependent

[west sapphire]

orthogonal nonzero vectors are linearly independent

but any set that includes the zero vector is linearly dependent

[warm osprey]

Okay, so suppose we get the question "Are the vectors in any orthogonal set always linearly independent", would the correct option be no because the zero vector basically

[west sapphire]

yes that would be correct unless they explicitly or implicitly said nonzero (e.g. if they said "orthonormal" that would implicitly mean nonzero)

[warm osprey]

Thank you

.close

[west sapphire]

yw

[pearl pond]

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[pearl pond]

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[midnight flicker]

The sum of the second and fourth terms of an increasing geometric progression is 45, and their product is 324. Determine the first term of this progression.

[fair creek]

it's just a

quadratic problem

a+d+a+3d=45

and

(a+d)(a+3d)=324

you get two solutions from this

[midnight flicker]

got you

but this is geometric progression

@fair creek

[white estuary]

since it is increasing, r>1

you know what the second and fourth term in gp is?

[fair creek]

i'm kinda

blind

sorry it's still the same

ar+ar^3=45

[white estuary]

lol

[fair creek]

a^2r^4=324

[white estuary]

and a^2 r^4 = 324

[fair creek]

it's still a quadratic problem

[white estuary]

yea

find r

from product

[fair creek]

hope this is enough nudge towards the solution

happy mathing!

[midnight flicker]

thank you

[white estuary]

did you solve it?

[pearl pond]

@midnight flicker Has your question been resolved?

[pearl pond]

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[worn goblet]

How

[worn goblet]

My friend said use substitution

[gray lantern]

is the top the original

[worn goblet]

Yes

[gray lantern]

what does the bottom say just to make sure

denominator

in the original

[gray lantern]

(just a bit difficult to read the denominator in the original)

[worn goblet]

Cosx-(5-4sinx)

Denominator

I was thinking

Taking x=pi/2 first

We get A = 1

If we don't open brackets

If we do we can't take anything common even if we make both sine or cosine

So by doing like this I got a=1

B=-4

[fickle schooner]

$$ \int \frac{1}{\cos x (5 - 4 \sin x)} $$

OR

$$ \int \frac{1}{\cos x - (5 - 4 \sin x)} $$

[jolly parrot]

StrangeQuarkAL

[worn goblet]

1st

[gray lantern]

bruh

[worn goblet]

And have to partially split it

[gray lantern]

the subtraction threw me off man 🙏

but yes u sub would work

[worn goblet]

How

[gray lantern]

nvm

[worn goblet]

How to split

<@&286206848099549185>

[pearl pond]

@worn goblet Has your question been resolved?

[pearl pond]

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[pearl pond]

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[pearl pond]

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[wise hedge]

I think I’m tweaking how do I find the point of intersection between two lines again….

[orchid pendant]

set them equal to each other

[wise hedge]

Do I solve for x?

[orchid pendant]

yeah

and then plug x into one of the equations for y

[wise hedge]

Ohhhh

Thank u

[orchid pendant]

thatll give you the coords

no problem

[wise hedge]

.close

[pearl pond]

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[frozen quiver]

hello, could someone help me on this problem. this lesson is called completing the square and my calcualtions arent the same as my teacher provided answer key. can someone help?

[rough forge]

what have you tried

[frozen quiver]

ill send my work

one second

[void gorge]

y = (x-2)^2-9

lol

[frozen quiver]

alright so this is my work

and this is the teacher ansswer sheet

[void gorge]

You have a mistake.

[rough forge]

your teacher is wrong

[frozen quiver]

wdym the teacher is wrong

[rough forge]

you started well last step is wrong tho

[void gorge]

This is correct answer.

[frozen quiver]

i just realized the extra plus bruh

[rough forge]

you can't just add a +4

you need to subtract also one, else you lose equality

[frozen quiver]

hm

[rough forge]

like you did

[rough forge]

then (x^2-4x+4) completes to (x-2)^2

[frozen quiver]

how this now

[rough forge]

yea basically

(x-2)(x-2) = (x-2)^2

[frozen quiver]

what about this one

hold up

i feel theres a mistake here too

[void gorge]

This is right.

[frozen quiver]

yeah i just re did my calculations

[rough forge]

your teacher still forgot to write +16

[frozen quiver]

why would there be + 16

[rough forge]

you cant just add terms as you wan

else you lose equality

[onyx granite]

Hi. Imagine you're adding 0.

You transform that 0 into +16-16=0.

To not modify the original expression.

[frozen quiver]

oh i see if my modified term b is negative i must add a positive equivalent to cancel it out?

[onyx granite]

English isn't my native idiom... What?

[rough forge]

when you add (b/2)² then you also need to subtract it

[frozen quiver]

ok yeah thats essentially what I said

[rough forge]

b = 8 you (b/2)² = 16 so you do +16 and -16

then you get

[frozen quiver]

and vice versa I assume

[rough forge]

x²-8x+16-20 not just x²-8x-20

[frozen quiver]

you know something ive notice is that the term I use to factor is almost always just half of B

[frozen quiver]

alright so heres what I got so far based off that

[rough forge]

the C term is always positive so switch it with +16

[frozen quiver]

wdym C term, 4 is the C term

16 is the modified B term

[rough forge]

ok lets call it differently

[frozen quiver]

X term

[rough forge]

you always work with the positive modified term

that is the +16

[frozen quiver]

the one inside or outside the parenthesies

[onyx granite]

(x+y)^2 = x^2 + 2xy + y^2, A square number is always positive.

What the flip.

[jolly parrot]

\begin{gather*}
y = (x^2-\underline{8}x+\underline{16})-16-4
\end{gather*}

[frozen quiver]

oh yeah these exist

[onyx granite]

Magic.

[frozen quiver]

oh I see
the Modified term is always positive
wait so wouldnt

be correct

[onyx granite]

That's correct. (I think)

Let my try that bot feature.

[rough forge]

no because there is no positive 16 in the first place

[frozen quiver]

I think we just wildly misunderstood it, she went straight for the answer and didnt bother writing the + 16 and the middle equasion

[jolly parrot]

AldaraLaRana

[rough forge]

basically yea, but again leaving out the +16 is still mathematically wrong

and you would and should get points off

so dont follow her

[frozen quiver]

i think she just didnt show her work

but the end answers correct right

[onyx granite]

Yup, is it.

[frozen quiver]

bam

alright one more question

[onyx granite]

bim.

[frozen quiver]

these have

no C

[onyx granite]

x^2 + 7x, 7 is our B term, the 2xy in the binomial square(x+y)^2.

We got x.

[frozen quiver]

huh

[onyx granite]

Then you solve for:

7x=2xy.

Got 7/2.

But that's only y, we want y^2.

We square up the 7/2.

49/4.

[frozen quiver]

what does the square have to do with anything

[onyx granite]

Ah...

[frozen quiver]

we didnt do ones without C

[onyx granite]

That because we are completing the square, so we want to our final answer to be in the form (x+y)^2.

[frozen quiver]

idk how she wants us to magically know it

hold up im gonna call the other guy over too rq

@rough forge

[onyx granite]

I'm bad at explaining, so yeah, it will be helpful.

[rough forge]

why are you pulling my cape

[frozen quiver]

i need thoust wisdom

[onyx granite]

That's me so fr.

[frozen quiver]

I have two questions without C and she didnt teach us how to do them

[onyx granite]

Oh... I have and idea on how to explain this, c = 0, solve in the same way.

,, y = x^2 + 7x + 0.

[jolly parrot]

AldaraLaRana

[rough forge]

oh these are even simpler

you need to factor right?

[frozen quiver]

yeah

[rough forge]

can you see a common factor

[frozen quiver]

X i guess

[rough forge]

yea

so factor that

[onyx granite]

I'm blind...

[frozen quiver]

this might be dumb but how would I factor 7x

[rough forge]

just x

[frozen quiver]

huh

then how would the equasion look

[rough forge]

ok

[frozen quiver]

yeah I see that

heres my teacher answers btw

[rough forge]

Oh

complete the square

not factoring

same thing

add and substract (7/2)²

your c is just 0 yea

[rough forge]

so this was the right thinking

[frozen quiver]

then whats 12.25

is that the modified B term

[rough forge]

,calc (7/2)^2

[jolly parrot]

Result:

12.25

[frozen quiver]

ok hold please

[rough forge]

yes

[frozen quiver]

oops thats supposed to be a -

[frozen quiver]

hows this

[rough forge]

good!

now recognize

[frozen quiver]

you mean factor?

[rough forge]

no recognize the square

[rough forge]

yea

well done!!

[frozen quiver]

oh so same thing

ig my teacher just wants it simplified

[rough forge]

ye

completing the square

[frozen quiver]

still wonder how she got that

hmm

oh well

I think im good but ill keep the channel open for 15 mins in case i get confused on something

[rough forge]

yes this is wrong

(x-2)²-1 = x²-4x+4-1 = x²-4x+3

thats what happens when you just add terms

[hasty edge]

That should be (x-2)^2-||9||

[pearl pond]

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[frosty hatch]

I have an elementary set theory question regarding a definition I can’t understand,
The definition is that of the product of a family of sets

The latter being defined (for a family Xi (i in I) as
{ F | F defined in I and for all i within I F(i) is in Xi }
Where the F’s are functional graphs

To exemplify this I used this simple family indexed by {0,1}
Thus this gives me the set of graphs with said property (written at the bottom)
However I don’t see how this expression is equivalent to the basic two set Cartesian product which is {(3,5),(3,6),(4,5),(4,6)} (we should only be accounting for the second coordinates of the graphs)

[pearl pond]

@frosty hatch Has your question been resolved?

[pearl pond]

@frosty hatch Has your question been resolved?

[frozen quiver]

<@&286206848099549185>

[deft orchid]

i don't get how this definition is supposed to work

what does "F is defined in I" mean and where exactly does F come from, and what does F(i) mean

[merry carbon]

And I wish someone pinged <@&268886789983436800> on you - oh wait...

[deft orchid]

now as it stands it seems like I is both index set for (X)i and domain for F's, if that's the case then what set do functions F come from

[frosty hatch]

F’s go from I to the unions of Xi’s with the condition F(i) belongs to Xi for all i

[deft orchid]

okay, i think i get it

so in my understanding the definitions don't really produce the same objects

but objects produced by both definitions are isomorphic (so there is bijection between them), which allows us to treat them as the same object

specifically, we can associate each function graph F from (1...n) to some set X with n-tuple of elements from X

@frosty hatch does that make sense?

[pearl pond]

@frosty hatch Has your question been resolved?

[frosty hatch]

That does make a lot of sense considering the graph definition comes with that of a function -from the product to the set of sets- of a "projection" which for a certain i gives: F —> F(i)
I suppose that’s the bijection (if we extend all of the i projections to a common function over the whole thing) which then allows us to simply consider the graphs as their second coordinates
Thank you

[pearl pond]

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[brazen vector]

so by a previous theorem it suffices to show that f(V) is open in Y for every open set V in X

so what i thought about is the following: Let $V\subset X$ be open in $X$, then $\forall p\in V\ \exists\delta>0\ \text{such that}\ x\in X\ \text{and}\ d_X(x,p)<\delta\implies x\in V$. now $f:X\mapsto Y$ is continuous on $X$ and so $f$ is continuous on $V$. Thus $\forall\varepsilon >0\ \exists\delta>0\ \text{such that}\ d_X(x,p)<\delta\implies d_Y(f(x),f(p))<\varepsilon$ and $x\in V$ so that $f(x)\in f(V)$. Thus $f(V)$ is open in $Y$

d_X and d_Y denote the metrics on X and Y respectively

is this correct? or is it missing something

well i think it has a problem

since i didnt use the compactness of X

[south inlet]

I don’t get how this proves that f(V) is open

Also as a quick side question, do you know what a homeomorphism is?

[brazen vector]

i am not sure too maybe i thought that this proves that f(V) is open while it is not. But i thought that this proves f(V) is open since i just found a neighborhood of f(p) which is a subset of f(V) for every open set V no ?

so i proved that for any p in V , there is a neighborhood of p in V such that for each x in V, f(x) in f(V) where these f(x) form a neighborhood of f(p), didnt I?

[brazen vector]

no

i wont encounter this here since i am studying real analysis not point set topology and the author didnt mention homeomorphisms in the chapter that he presented the concepts needed from point set topology in

[south inlet]

I think you just showed that all points at a max distance delta from p are mapped to points at a max distance epsilon away from f(p)

Which is directly from the definition of continuity

[brazen vector]

yes but arent all points f(x) elements of f(V)?

[south inlet]

Yes

[brazen vector]

if so , then doesnt that form a neighborhood of f(p) in f(V) ?

which means that f(V) is open no ?

because this is the case for any p in V

[south inlet]

But I don’t think that necessarily means that ALL points in the neighbourhood of f(p) are image points of f

Or at least I don’t think you have proven that

[brazen vector]

ah i see your point

true

[south inlet]

You showed all image points under f of x at a distance less than epsilon from p is mapped to the ball of radius epsilon from f(p), you need to kind of show the converse, that all points in the ball of radius epsilon from f(p) are image points of f restricted to V

[brazen vector]

yes you are right , thats what i should do because it is directly applying the definition of continuity to f^(-1)(x) on f(V)

so i need to prove that if $p\in V$ then $\forall\varepsilon>0\ \exists\delta>0\ \text{such that}\ \ d_Y(y,f(p))<\varepsilon\implies d_X(x,p)<\delta\ \forall y\in Y\ \text{and}\ x\in V$

right?

[pearl pond]

@brazen vector Has your question been resolved?

[pearl pond]

@brazen vector Has your question been resolved?

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[jade moss]

need help

[jade moss]

i got 22/6 but i think im wrong

.close

[pearl pond]

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[spiral pivot]

@jade moss figured it out?

[robust kraken]

ye

use cos(x)^2+sin(x)^2=1

[jade moss]

im not there yet lol

i just use pythagerom therom to find the opposite

[spiral pivot]

This is what I was about to suggest, so excellent work

[pearl pond]

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[soft otter]

The length hanging down as a function of time and the speed when the chain loses contact with the table are derived using Newton's second law and energy conservation.

Length as a function of time:
The differential equation governing the motion is ( y''(t) = \frac{g}{\ell} y(t) ). Solving this with initial conditions ( y(0) = y_0 ) and ( y'(0) = 0 ) gives:
[
y(t) = y_0 \cosh\left( \sqrt{\frac{g}{\ell}} , t \right)
]

Speed when losing contact:
Using energy conservation, the speed ( v ) when ( y = \ell ) is:
[
v = \sqrt{ \frac{g(\ell^2 - y_0^2)}{\ell} }
]

Final Answer:
The length hanging down is (\boxed{y(t) = y_0 \cosh\left( \sqrt{\dfrac{g}{\ell}} , t \right)}) and the speed when contact is lost is (\boxed{v = \sqrt{\dfrac{g(\ell^2 - y_0^2)}{\ell}}}).

[jolly parrot]

nino

[soft otter]

To verify the solution, we systematically apply Newton's second law and energy conservation:

---

1. Deriving ( y(t) ) via Newton’s Second Law

Force Analysis:

• The gravitational force on the hanging portion at time ( t ):
  [
  F = \mu y(t) g
  ]
• Total mass of the chain: ( \mu \ell ).
• Newton’s second law (( F = ma )):
  [
  \mu y(t) g = \mu \ell \cdot y''(t) \implies y''(t) = \frac{g}{\ell} y(t)
  ]

Solving the Differential Equation:

• The equation ( y''(t) = \frac{g}{\ell} y(t) ) is a linear ODE. Its general solution is:
  [
  y(t) = A \cosh\left( \sqrt{\frac{g}{\ell}} t \right) + B \sinh\left( \sqrt{\frac{g}{\ell}} t \right)
  ]

Applying Initial Conditions:

• At ( t = 0 ), ( y(0) = y_0 ):
  [
  y_0 = A \cosh(0) + B \sinh(0) \implies A = y_0
  ]
• Differentiate ( y(t) ):
  [
  y'(t) = A \sqrt{\frac{g}{\ell}} \sinh\left( \sqrt{\frac{g}{\ell}} t \right) + B \sqrt{\frac{g}{\ell}} \cosh\left( \sqrt{\frac{g}{\ell}} t \right)
  ]
• At ( t = 0 ), ( y'(0) = 0 ):
  [
  0 = B \sqrt{\frac{g}{\ell}} \implies B = 0
  ]

Final Expression for ( y(t) ):
[
y(t) = y_0 \cosh\left( \sqrt{\frac{g}{\ell}} t \right)
]

---

2. Speed When the Chain Loses Contact (( y = \ell )) via Energy Conservation

Initial Energy:

• Initial potential energy (center of mass of hanging chain):
  [
  U_{\text{initial}} = -\frac{1}{2} \mu g y_0^2
  ]
• Initial kinetic energy: ( 0 ).

Final Energy:

• Final potential energy (entire chain hanging):
  [
  U_{\text{final}} = -\frac{1}{2} \mu g \ell^2
  ]
• Final kinetic energy:
  [
  KE = \frac{1}{2} \mu \ell v^2
  ]

Energy Conservation:
[
U_{\text{initial}} + KE_{\text{initial}} = U_{\text{final}} + KE_{\text{final}}
]
[
-\frac{1}{2} \mu g y_0^2 = -\frac{1}{2} \mu g \ell^2 + \frac{1}{2} \mu \ell v^2
]

Solving for ( v ):
[
\frac{1}{2} \mu \ell v^2 = \frac{1}{2} \mu g (\ell^2 - y_0^2) \implies v^2 = \frac{g(\ell^2 - y_0^2)}{\ell}
]
[
v = \sqrt{\frac{g(\ell^2 - y_0^2)}{\ell}}
]

---

3. Dimensional and Physical Consistency

• Hyperbolic Function: The argument ( \sqrt{\frac{g}{\ell}} t ) is dimensionless, as required.
• Velocity: Units of ( \sqrt{\frac{g(\ell^2 - y_0^2)}{\ell}} ) simplify to ( \sqrt{g \ell} ), which is dimensionally correct.
• Edge Cases:
  • If ( y_0 \to \ell ), ( v \to 0 ) (no motion if chain is already fully hanging).
  • If ( y_0 = 0 ), ( v = \sqrt{g \ell} ), matching free-fall intuition.

---

Final Answers:
[
\boxed{y(t) = y_0 \cosh\left( \sqrt{\dfrac{g}{\ell}} , t \right)}
]
[
\boxed{v = \sqrt{\dfrac{g(\ell^2 - y_0^2)}{\ell}}}
]

not sure if this is right

[jolly parrot]

nino
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nino
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[pearl pond]

@soft otter Has your question been resolved?

[pearl pond]

@soft otter Has your question been resolved?

[soft otter]

can i please get help just like going over this

i just wanna look at the math

[pearl pond]

@soft otter Has your question been resolved?

[pearl pond]

@soft otter Has your question been resolved?

[pearl pond]

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[pearl pond]

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[wraith badger]

how can i solve for x and y in y+1=x^4 and (x+1)^3 = y

[toxic lichen]

that looks like a rather ugly 4th degree equation. where did you get it?

[wraith badger]

lemme show the original question

[toxic lichen]

hmm

ok yeah unfortunately the ugly quartic is there still

[wraith badger]

apparently the solution is pretty simple and direct

😭

[toxic lichen]

x^4 - x^3 - 3x^2 - 3x - 2 = 0

is what this reduces to

[toxic lichen]

says who?

[wraith badger]

the teacher

i think you do something with the logs

and dont rewrite them as equations

also a classmate found a loophole in the question

[dapper tinsel]

do anyone knows hyperbola

[wraith badger]

so you can do it in a few steps

🗿

[toxic lichen]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[toxic lichen]

!da2a

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[wraith badger]

WAIT

im so sorry

[toxic lichen]

and what would that be

[wraith badger]

i thought the paper was calculator

[autumn trellis]

looks like -1 is a sol?

[wraith badger]

non coalcaultor*

[toxic lichen]

these reduce to an ugly cubic without a rational root

[wraith badger]

but its calculator

[toxic lichen]

yeah so that lets you knock the degree down to 3

x^3-2x^2-x-2=0

good luck solving that lol

[autumn trellis]

google cardanos method

[toxic lichen]

ok and what school do you think teaches that

[wraith badger]

that looks painful

[autumn trellis]

thats why i said google

[toxic lichen]

don't try to be a smartass in help channels @autumn trellis

[wraith badger]

its okkk im pretty sure i was just supposed to plot the two equations

[toxic lichen]

that ain't it chief

[wraith badger]

Can somebody explain b?

This is the answer to a) btw if it has to be used in b)

NVM I GOT IT

[pearl pond]

@wraith badger Has your question been resolved?

[pearl pond]

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[rapid wagon]

Prove/Disprove: if $a_n$ converges to a finite limit then $\lim_{n \to \infty}(\frac{a_n}{n})^n=0 \
Proof: $a_n$ converges therefore $a_n$ is bounded. let's look at the expression $(\frac{a_n}{n})$. $\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} \frac{1}{n} \cdot a_n = 0$ (a series converging to zero multiplied by a bounded series, converges to zero) using the definition of limit let's choose $\epsilon = 1$ there exists N such that for all $n \geq N$ we get $|\frac{a_n}{n}| < 1$ therefore $|\frac{a_n}{n}|^n$ is a geometric series with |r|<1 therefore $\lim_{n \to \infty}(\frac{a_n}{n})^n = 0$

[jolly parrot]

prograce
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[pearl pond]

@rapid wagon Has your question been resolved?

[rapid wagon]

Can I have someone check my proof?

<@&286206848099549185>

[tropic saddle]

its not a geometric series

there is no common ratio

the a_n/n could be changing quite "randomly"

you want to get rid of the a_n. you cant control it

you already know its bounded

call the bound a

work with that

[rapid wagon]

Ok

Idk how to use a here..

[proper dove]

Epsilon delta

[pearl pond]

@rapid wagon Has your question been resolved?

[tropic saddle]

|a_n|<=a, so therefore |a_n/n|<= ?

[rapid wagon]

|a/n| and then I can say |a/n| converges to zero

Or a/n

Yea

.close

[pearl pond]

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[thorn shoal]

we don't have the answers for this

but can i know what the new height is for this question?

i got 0.78 and idk why but im quite certain that im wrong ://

[light helm]

show your work

[thorn shoal]

so for a

the volume is like 6.48

so the missing part is 23.52

then

let the ratio be v/3

where v is the new radius

1/3 * (v/3)^3 * 3^2 * 10 = 23.52

then i got like v = 2.77

then the new height = 0.779

[sage hawk]

How do you calculate the volume without the radius of the liquid's surface?

[light helm]

all your calculations should be exact in terms of pi

[thorn shoal]

nah

but pi cancels out on both sides

[thorn shoal]

set ratio

cuz its similar triangles shit

[light helm]

how did you get 6.48 for the volume

[thorn shoal]

in terms of pi that one is

the part b calculations is all just cancelled out tho

[light helm]

but how did you get it

[thorn shoal]

cuz ratios

the height ratio

6/10

so the radius is in this ratio

so radius is 1.8

[sage hawk]

,w 1/3 * 1.8^2 * 6

So the volume is 6.48pi

[thorn shoal]

?

wdym lol

,w 1/3 * (1.8)^2 * 6

[jolly parrot]

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Requested by rex4878

[thorn shoal]

pir^2h

[sage hawk]

Just forget that

[light helm]

ok, so you should be writing the pi for those values

in terms of pi doesn't mean you don't write pi

[thorn shoal]

ye for part a it is

but for part b it isn't

and i just cancelled out pi

[light helm]

volume would be 6.48pi,
and missing/empty part would be 23.52pi
i don't understand your work after that,

[thorn shoal]

ye

ok so

uhh

cuz the ratio of radii

is equal to ratio of heights

so i let the radii of the empty part of the cone in part b be v

then

1/3 * pi * (v/3)^3 * 3^2 * 10 = 23.52

pi

[light helm]

ok

,w solve 1/3 * (r/3)^3 * 3^2 * 10 = 23.52

[jolly parrot]

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[light helm]

,calc 3/5 * cbrt(2) * 7 ^(2/3)

[jolly parrot]

Result:

2.7662617752351

[light helm]

that'll be the length of the red line

apply the ratio between h and r to get the height

and subtract that from 10

[thorn shoal]

mhm

[thorn shoal]

ye

so then i got the ans of 0.779

[light helm]

seems like that was what you did,
your work just wasn't clear

[thorn shoal]

is that correct?

ah

ok thanks then

+close

.close

[pearl pond]

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[pearl pond]

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[white estuary]

help

[white estuary]

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.

my working till now