help-27

[restive elk]

I did

Uhh welp, I dont understand that part myself, since I guess its asking for one cycle for f(sin x)

And not sin x

[versed juniper]

i think the cycle is still 0 to 2 pi

or does it mean period

then its 0 to pi

[restive elk]

Any reasons on why...?

[versed juniper]

its still sin x

not sin 2x or anything

the cycle doesnt change just because you cubed it

[restive elk]

its sin x but cubic now

[restive elk]

Ooh, didnt knew that

thnx

but still

what after that?

[versed juniper]

integrate it?

[restive elk]

-8 pi

None of the options match!

[versed juniper]

let me guess

you integrated sin cubed -> -cos cubed

can you show your work

[restive elk]

Welp its a cubic function over a full period? So I didnt even need to do it? I gues?

[restive elk]

So sorry for that but I got no phone with me rn

I am on PC

[versed juniper]

,w integrate sin^3 x - 6 sin^2 x - 15 sin x - 1 from x=0 to 2pi

[woven radish]

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[restive elk]

I guess, it's correct

so.... what after this?

[versed juniper]

,w integrate sin^3 x - 6 sin^2 x - 15 sin x - 1 from x=0 to pi

[woven radish]

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Requested by k.tten

[versed juniper]

,w integrate | sin^3 x - 6 sin^2 x - 15 sin x - 1 | from x=0 to pi

[woven radish]

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[restive elk]

whatcha trying to do?

[versed juniper]

is it asking for unsigned area ?? what does cycle mean 😭

[restive elk]

cycle means when the sin x completes one rotation and then repeats itself

,w integrate | sin^3 x - 6 sin^2 x - 15 sin x - 1 | from x=0 to 2pi

[woven radish]

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[restive elk]

how do i make it show the answer in terms of pi?

[sudden blaze]

Divide by pi

The you have a × pi as an answer

[restive elk]

,w 57.4692/pi

[woven radish]

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[restive elk]

totally wrong ig

174/3

,w 174/3

[woven radish]

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[restive elk]

ok so this is the correct answer, according to gemini

but i dont understand how

[mystic scarab]

Because that's how you calculate the area 🤷‍♂️

The absolute value is there for taking into account that a normal integral would give you the signed area

[sudden blaze]

What was the original question

[mystic scarab]

#help-27 message

[sudden blaze]

Ah so the problem is that the answer doesn't match the given ones

[mystic scarab]

Yep

[sudden blaze]

Well only problem could be somewhere forming the correct polynomial,maybe smth went wrong

[mystic scarab]

Wait exactly

That polynomial doesn't go through the point (2, 47)

[sudden blaze]

Wow nice one

[mystic scarab]

It goes through (2, -47)

[sudden blaze]

💀💀💀

@restive elk

@mystic scarab I'll try to solve it myself later

[devout snow]

@restive elk Has your question been resolved?

[restive elk]

ye?

[sudden blaze]

We found your mistake

[restive elk]

What is it sir?

plss

[sudden blaze]

Your plinomial goes trough the point 2;-47 and not 2;47

[restive elk]

Didn't I mention that in the question itself?

[sudden blaze]

The question states it differtly than you said, so which one is correct?

[restive elk]

GPT had a hard time doing OCR... the inflection points are (2,-47)

[sudden blaze]

Ah ok

I'll try to solve it when I'm home from the gym, I'll DM you when I find smth

If you spot it before me, pls tell me

[restive elk]

pls do, tysm sirrrrr

[restive elk]

will do, tysmmmmmmmmmm

[devout snow]

@restive elk Has your question been resolved?

[devout snow]

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[oblique mulch]

Guys it says 72 is wrong but me and all the LLMs say it’s right? Help

[oblique mulch]

I solved the first formula for T and plugged into the second. Using 11.8 as delta X

[faint gorge]

i can see right through your hand

[oblique mulch]

Help

<@&286206848099549185>

[verbal vector]

3 sig figs

[verbal vector]

i got 72.6

didn’t round at all til the very end

if you wanna try that

[oblique mulch]

well im out of chances but thank you

[verbal vector]

alright

[oblique mulch]

.close

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[somber dawn]

for this i know the first part, but am stuck on finding the parametic

[wind mason]

Okay, do you have the line?

Or any info about it?

[somber dawn]

thats all that was given

[wind mason]

No, I mean have you derived anything

Like a point on it, or specific slope, etc

[somber dawn]

can't you turn 2x-y+3z =0 to <2,-1,3> and -x+2y+z=0 to <1,2,1>

then i found normal vector by doing cross multiplication

[cinder bobcat]

there's a very easy trick for this involving the normal vectors

[somber dawn]

please elaborate 🙏

[cinder bobcat]

the angle between the planes is the same as the angle between the normal vectors

[cinder bobcat]

also I think you meant <-1,2,1> here

[somber dawn]

correct, but i'm not confused about the angle part

the tricky part is the parametric

my teacher did this weird thing, am this is where I am confused at

[cinder bobcat]

to find the line of intersection, you need to solve the system of equations formed by the planes, which is what your teacher did

they solved for x and y in terms of z, and then they said z = k (usually I see z = t but k works too i guess)

[cinder bobcat]

this part is actually unnecessary

[somber dawn]

the question says to solve for the equation though.

[cinder bobcat]

saying x = (-7/3)t, y = (-5/3)t, and z = t is sufficient

[cinder bobcat]

the intersection of two planes forms a line, so you need the parametric equation of a line

which is technically 3 equations

[somber dawn]

thank you, i think this helped me better understand it

👍

since you're here could you please help me with this as well?

[devout snow]

@somber dawn Has your question been resolved?

[wind mason]

Ah, this is an application of $\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$, for $Ax+By+Cz+D=0$ a plane and $(x_0, y_0, z_0)$ a point

[woven radish]

;(

[wind mason]

The proof behind this used the normal vector to the plane going through the point

[devout snow]

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[devout snow]

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[restive river]

yo yo yo

[restive river]

@lyric hornet pov my teacher is a piece of shit and gives me a 90% even tho i got everything correct for smalls mistakes

[supple knot]

Don't open help channels to bitch and moan

.close

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[devout snow]

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[tough vigil]

1help

[tough vigil]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tough vigil]

question if y= a sin x + b cos x then (dy/dx)^2 + y^2 is a) a function of z or b) a constant

[tough vigil]

yeah

i think the ans is a function of x???

[pseudo basin]

what is z here

[tough vigil]

its was x

[pseudo basin]

btw just for future reference

you're supposed to just send your question directly

anyway, have you worked out what y'^2 + y^2 simplifies to?

[tough vigil]

yes

[pseudo basin]

and what did you get?

and show your work for it

[tough vigil]

it simplilfise to “(dydx​)2+y2=(acosx−bsinx)21​+(asinx+bcosx)2”

how to use latex??

[pseudo basin]

for a start, ^ for exponents...

but also you could post a screenshot or picture of the thing

[tough vigil]

(dydx​)^2+y^2=(acosx−bsinx)^-2​+(asinx+bcosx)2

[pseudo basin]

to the power of -2 ?

ok no

write this shit out on paper and send it here

also simplify it fully

[tough vigil]

i started from this

[pseudo basin]

wait what

dx/dy ??

did you not say dy/dx before?

why is dx/dy happening?

also $(a \sin(x) + b \cos(x))^2 \neq a^2 \sin^2(x) + b^2 \cos^2(x)$

[woven radish]

Ann

[tough vigil]

ooo yeah i get

.close

[devout snow]

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[pseudo basin]

??

what was this about?

[tough vigil]

i get it it will simplify to a^2+b^2=1 which indeed is a constant

option b

[pseudo basin]

since when was a^2 + b^2 equal to 1

dear lord you only stated like 1% of the problem to us 😭

[tough vigil]

[
\left( \frac{dy}{dx} \right)^2 + y^2 = a^2 + b^2
\quad \Leftarrow \quad \text{This is constant because it has no } x.
]

[woven radish]

BVB999

[tough vigil]

im sry its not 1 its aconstant beacuse no x and no trig function are left

not 1 im dumb i figured it out

\documentclass{article}
\usepackage{amsmath}
\begin{document}

Given:
[
y = a \sin x + b \cos x
]

Differentiate:
[
\frac{dy}{dx} = a \cos x - b \sin x
]

Square both:
[
\left( \frac{dy}{dx} \right)^2 = a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x
]
[
y^2 = a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x
]

Now, add them:
[
\left( \frac{dy}{dx} \right)^2 + y^2 =
(a^2 \cos^2 x - 2

[woven radish]

BVB999
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tough vigil]

Given:
[
y = a \sin x + b \cos x
]

Differentiate:
[
\frac{dy}{dx} = a \cos x - b \sin x
]

Square both:
[
\left( \frac{dy}{dx} \right)^2 = a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x
]
[
y^2 = a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x
]

Now, add them:
[
\left( \frac{dy}{dx} \right)^2 + y^2 =
(a^2 \cos^2 x - 2

[woven radish]

BVB999
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tough vigil]

[
y = a \sin x + b \cos x
]

[
\frac{dy}{dx} = a \cos x - b \sin x
]

[
\left( \frac{dy}{dx} \right)^2 = a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x
]
[
y^2 = a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x
]

[
\left( \frac{dy}{dx} \right)^2 + y^2 =
(a^2 \cos^2 x - 2

[woven radish]

BVB999
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tough vigil]

[
\left( \frac{dy}{dx} \right)^2 + y^2
= (a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x)

• (a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x)
  ]

[woven radish]

BVB999

[pseudo basin]

.reopen

[tough vigil]

[
\left( \frac{dy}{dx} \right)^2 + y^2 = a^2 + b^2
\quad \Leftarrow \quad \text{This is constant because it has no } x.
]

[devout snow]

✅

[tough vigil]

[
\left( \frac{dy}{dx} \right)^2 + y^2 = a^2 + b^2
\quad \Leftarrow \quad \text{This is constant because it has no } x.
]

[pseudo basin]

well, for what it's worth, you're right in that $y'^2 + y^2 = a^2 + b^2$

[woven radish]

Ann

[tough vigil]

so its a constat cause there is no x

is it my textbook answer says so???

???????????????????

[mystic scarab]

Correct

Actually, not because there are no x's

[tough vigil]

1 last question

[tough vigil]

yeah

[mystic scarab]

But rather because a and b are constants, hence a² + b² is a constant

[mystic scarab]

With this I mean that you could have had some y or y' (not in this exercise, though) on the RHS

[tough vigil]

the position of a particle in a straight line is given by y=3t^3 + t^2 +5 find the accelration at t=2

[mystic scarab]

Next time post different questions in different channels

So that we don't have too many messages to read through

[mystic scarab]

Do you know the definition of the acceleration?

[tough vigil]

yeah dv/dt

v=dy/dt

[mystic scarab]

Awesome

[tough vigil]

because y defines the positon

[mystic scarab]

So you have all the ingredients to calculate the acceleration

[tough vigil]

yeah i got 0.38 m/s^2

is it correct

and why do i get accelaration in square meter

[mystic scarab]

Nope

Show your calculations

[mystic scarab]

You don't in fact

Acceleration is in m/s²

[tough vigil]

[
\text{Velocity: } v = \frac{dy}{dt} = 9t^2 + 2t
]

[
\text{Acceleration: } a = \frac{dv}{dt} = \frac{d^2y}{dt^2} = 18t + 2
]

[woven radish]

BVB999

[tough vigil]

[
\text{At } t = 2: \quad a = 18 \times 2 + 2 = 36 + 2 = \boxed{38 \ \text{m/s}^2}
]

[woven radish]

BVB999

[tough vigil]

it is given in cm

[mystic scarab]

Yep correct

[tough vigil]

is it beacuse its the area under a triangle of accelration

[mystic scarab]

The positiony you mean?

[tough vigil]

yeah

[mystic scarab]

Ah alright

Then yeah it's 38 cm/s², so 0.38 m/s²

[tough vigil]

answer this

im confused beacuse of s^2

[mystic scarab]

It's a ratio of velocity over time

[tawny light]

you can think of dimensions right? its velocity/time

[tough vigil]

but why time is squared?

[tough vigil]

yeah

[mystic scarab]

This is meaningless, though

[tawny light]

velocity is m/s and you're dividing by another second. so m/s^2

[tough vigil]

oooooooo

yes

thnx bye chat

.close

[devout snow]

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[restive river]

c

[restive river]

yes

over n over

i got height 5

got the midpoint

everything

what do i do for c

no way u do some

5 distance

hence

3,4,5 triangle

cos that gives u 4

or u could check all 4 on the line from part b

but thats way too long

for 5 marks

[pseudo basin]

aight hold up

you did parts a and b right

can you give the equation for line l that you got

[restive river]

yes

[pseudo basin]

just for reference

[restive river]

8 6 23 0 x y −+=

ffs

wait

[pseudo basin]

reverse polish notation?

[pseudo basin]

ok right that's our l

[restive river]

yes

[pseudo basin]

do we have an equation for AB as well

[restive river]

no

we dont need it

just grad

gradient

[pseudo basin]

well ok

we can do a different route: CM = 5 defines a circle

take the eq of line l and that circle's equation and solve them as a system

[restive river]

what

whyd we do that

to find every point 5 units away midpoint

bit far fetched no

[pseudo basin]

i mean you did get that the height of the triangle is 5 right

so we gotta do it one way or another

use that info

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[restive river]

any help

[rare kernel]

with what

[undone chasm]

what’s ur question

[restive river]

this

i dont get their method

person above suggested a circle

but thats a bit long

[rare kernel]

send the question again please

[restive river]

but it would work

[rare kernel]

theres actually a formula for area of triangle

[restive river]

c

[restive river]

well yh

[restive river]

ur joking

😭

what in the hell

[rare kernel]

joking? u just have to plug in the points

u already have the area

[restive river]

bro ive never seen this IN MY LFIE

[rare kernel]

ok then leave it

[restive river]

😭

bro

[rare kernel]

stop crying

[restive river]

i feel like my skls so far behind

everytime i ask for help

[crystal fiber]

No worries.

[restive river]

everyone somehow has some unique genius simplified method

[devout snow]

@restive river Has your question been resolved?

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[valid iron]

I am not sure where to start with this...

[valid iron]

we have to find Sn here

[wispy oyster]

perhaps some identity of cos^n(kπ/n)

in terms of another sum

[valid iron]

euler?

[wispy oyster]

yes

[mystic scarab]

Well, I would simply do it with area = 1/2 base × height 🤷‍♂️

[valid iron]

how should i deal with the n power on cos

like de morvie had something to do with powers but im not familirized with it yet...

wait i think i got it

is the answer 0

[wispy oyster]

no

you need the identity

$\cos^n\theta
=\frac1{2^n}\sum_{j=0}^n\binom nj,e^{,i(n-2j)\theta}$

[woven radish]

mia

[wispy oyster]

stems from de moivre

plug this into the sum so you get a nested sum

then you will realize inner sums are either 0 or n depending on j

im not sure why you would attempt this question if you werent too familiar with de moivre

[devout snow]

@valid iron Has your question been resolved?

[devout snow]

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[storm raptor]

Is there a way to bound $\frac{e^z}{z^2 + 1}$ on $|z| < 2$

[woven radish]

dyxn

[storm raptor]

Cant seem to get anything to work

[woven radish]

dyxn

[storm raptor]

:/ doesn't work

oh wait

uh

the function isn't bounded on the disc

whoops 💀

.close

[devout snow]

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[storm raptor]

.reopen

[devout snow]

✅

[storm raptor]

The constraint is on |z| = 2 mb

[raven leaf]

is this part of a bigger complex analysis question

[arctic field]

|z^2 + 1| >= |z|^2 - 1 = 3

[raven leaf]

I don't agree with the second equality - where did you get that from?

[arctic field]

oh 3

close enough

[raven leaf]

Oh, it's an equality constraint

got it

[devout snow]

@storm raptor Has your question been resolved?

[storm raptor]

yeah I figured this out lol, thanks

[storm raptor]

bound an integral

but that's all I needed, so thanks

.close

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[queen narwhal]

Okay. So. I just randomly thought of this problem, and I genuinely do not know what the answer is, so I can't validate it by anything other than a proof. Also I barely understand what I'm asking, and no this is not for homework, this is purely for casual/fun interest.

So, I am standing somewhere in multi-planar space - somewhere which can be located on a complex plane. North of the origin is the positive imaginary, south of the origin is negative imaginary, east of the origin is positive real, west is negative real.

I'm trying to figure out where I'm standing in relation to the origin. North of me is positive imaginary, but... to my east is negative imaginary, to my south is negative real, and to my west is positive real. How is this possible, and where am I standing?

Valid answers are:

• That's not fking possible bro
• It IS possible and here is your position relative to the origin

[gray summit]

I don't think that's possible

[queen narwhal]

Why not, can you prove it

[gray summit]

Idk

I'm just saying it based off of common sense, I'm neither good with proofs nor with graphs

[queen narwhal]

I mean that there could be multiple planes in this space, and that my position may** not be** on the complex plane. I’m actually not sure how many planes there are, but there are definitely at least three (X, Y, Z)

Yes

Probably… sorry I’m a baby at math so my terminology may not be very precise 😄 Would visuals help?

I guess what I’m imagining is it’s possible that maybe the planes have been wrapped around to form a sphere, making the planes no longer infinite

Oof well give me a second because I have to redraw my visual, apparently I merged all the layers and failed to save a copy of the unmerged drawing...

For a little more context by the way, I'm describing metaphysical objects in a little project I'm working on. So the rules definitely are a tiny bit permissive, mostly because I'm still trying to figure out what all the rules are 😄 But there are some rules that I know for sure are fixed

Okay here we go

look at them colors

So, the black dot is the origin right, and the colors represent the planes. Obviously we know that imaginary and real numbers can overlap each other and stuff, but the general point is the yellow is the positive imaginary, red is positive real, green is negative real, blue is negative imaginary

So somewhere along the way while I'm God or whatever making these fancy numeric systems, I dropped my universe making tool down some mysterious wormhole

And it landed somewhere where the planes look like this

yes

I'm trying to figure out how to explain without giving more context than necessary lemme think

It's actually really hard to explain without sounding really mythological, so I'll just do that lmao

So we start with the first image here right
#help-27 message

The "origin" is a location, actually. It's a mythological location that only a bunch of aliens know how to access.

The second image is a compass, which points towards the origin. You only know what the colors mean, and that when the compass points in that color space, it means that's the direction you have to go in to get closer to the origin. So if the compass points to the yellow, you need to move along the positive imaginary line to get closer.

But as you can see, the cardinality is twisted in a weird way. Does that suggest that the compass is oriented wrong, or that you are oriented in a strange way in relation to the origin?

There are project related reasons that they are specifically the real and imaginary plane, which I could explain but idk if it would help make the solving of it any easier

Hmm

I suppose the one contextual detail that may possibly help is: the aliens specifically do not want you to find the origin. They know that if you find where North, South, East, and West is of the origin (aka if you have the first image) you'd be able to find it. There is an alien guarding one of these paths and hiding it from plain sight so that you never find it.

The compass has been constructed by humans, and absolutely definitely works - and we can assume that it always does reliably point to the origin - but because of this alien and its "mysterious ability to relate to the metaphysical world", what will be unclear is whether we are in fact getting any closer to the origin, or if we're just going in circles forever

lore reasons
(I do have a more satisfactory answer, but basically lore reasons)

Well, that is actually helpful, because basically what I could conclude from this is one of two (possibly one in the same) thing:

• If and only if the compass always does point to the origin reliably, then the way the real world relates to the origin makes the compass very deceptive. This is because the aforementioned "guardian alien" is manipulating the real world so that humans with the tools available literally cannot physically find the origin, it would be impossible
• The compass is mapped the way it is because that directionality maps to something concrete in the real world - orienting it any differently would make the compass not work in the real world. That suggests the compass needs to be able to map something almost "super real" in order to arrive at the origin, another deception this alien would be capable of

So to explain a little more fully: there are three worlds in this project I'm making, one is "reality", one is sorta like a super-reality, platonic, metaphysical place. "Heaven" basically. It's assumed that math rules over all the worlds - so anything that can be calculated with math is true for both Heaven and Reality, but just because it's not true in reality doesn't mean it's not true in Heaven

There is a reason I'm using the complex plane in this specific example, but I won't go into that 😄

This?

oh wow i didn't expect that to be invisible

It almost feels like the compass has to be sort of clock-ish to actually work correctly

Basically two different hands, one which cares about imaginary, one which cares about real - but they still might have some challenges pointing any particular way

I'll give it more thought and see what I can do with it

Thanks for the help ❤️

(do I have to mark this channel's discussion as finished or something)

.close

[devout snow]

Channel closed

Closed by @queen narwhal

Use .reopen if this was a mistake.

[queen narwhal]

o

that worked

[devout snow]

Available help channel!

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[dark dawn]

Hi I need help with 2ii using the hence method

[dark dawn]

The left side is what I did for part i

<@&286206848099549185>

[potent tusk]

hence method?

[dark dawn]

like by using part i answer

I got stuck again

[sudden blaze]

Hence means like therefore

[sudden blaze]

.

[dark dawn]

Ya

I can't solve it

<@&286206848099549185>

[devout snow]

@dark dawn Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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[devout snow]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[broken wagon]

Hi, I'm a little unsure why this is wrong. I was assuming to be independent you needed to sum to 0

[broken wagon]

And dependent was if it didn't so I interpreted it as "if same direction, then dependent and if it has a counteracting direction then independent"

[devout snow]

@broken wagon Has your question been resolved?

[broken wagon]

Oh that reminds me, it's been 15 minutes

<@&286206848099549185> It's not a rush so please help others if needed and assist at your connivence, thanks in advance

Just ping me when you do

[devout snow]

@broken wagon Has your question been resolved?

[devout snow]

@broken wagon Has your question been resolved?

[devout snow]

@broken wagon Has your question been resolved?

[formal delta]

Need help?

[broken wagon]

Well if possible yeah, I'm just unsure what I did wrong

[formal delta]

Top row (left to right):
1. First field: Uniform field pointing right no rotation
Answer: Path independent
2. Second field: Diagonal lines, consistent direction no curl
Answer: Path independent
3. Third field: Vertical vectors only, evenly spaced no curl
Answer: Path independent

Bottom row (left to right):
4. First field: Clear circular flow rotation present
Answer: Path dependent
5. Second field: Another rotational/circular field rotation present
Answer: Path dependent
6. Third field: All vectors horizontal, uniform no rotation
Answer: Path independent
7. Fourth field: Horizontal field again, uniform no rotation
Answer: Path independent

@broken wagon

[broken wagon]

Oh I see, sorry my eyes aren't too good but I can't recall a 7th field

Unless you mean the 7 attempts lol

[potent tusk]

???

[formal delta]

What

[broken wagon]

As in, there are only 6 graphs there

[formal delta]

Bro there aren’t seven graphs ik

Read the thing

I was counting differently

[broken wagon]

Alright, thanks

[formal delta]

I got I

U*

1. Path independent
   2. Path independent
   3. Path independent
   4. Path dependent
   5. Path dependent
   6. Path independent

[broken wagon]

Okay, I apreciiate this, thank you

[formal delta]

Did that help

Top row (left to right):
1. Graph 1: Vectors point uniformly to the right → no curl
Answer: Path independent
2. Graph 2: Vectors diagonally increasing → no visible circulation
Answer: Path independent
3. Graph 3: Vectors point straight up → uniform, no rotation
Answer: Path independent

⸻

Bottom row (left to right):
4. Graph 4: Obvious circular rotation (counterclockwise) → has curl
Answer: Path dependent
5. Graph 5: Also shows a rotating pattern → curl present
Answer: Path dependent
6. Graph 6: Horizontal vectors, no swirl or circulation → straight and uniform
Answer: Path independent

[potent tusk]

So tell me how the 5th graph has any rotation

I really think this is ai

[broken wagon]

It did

[formal delta]

Bruh

I’m getting on the airplane mb bro

[formal delta]

Just help him i can’t rn

[potent tusk]

<@&268886789983436800>

likely not malicious but blatantly wrong

[modest dagger]

If you can't answer somebodies question don't spam llm junk at them.

[winter torrent]

chatgpt is not a cheat code for reality, and we do not need you to run it for us.
do not post output from chatgpt in this server ever again (unless specifically requested).

[potent tusk]

did you uhh get it right?

[broken wagon]

Well I was trying to be polite but that's what I put it initially

So at least one of it was wrong anyway unfortunatly

[potent tusk]

here is what i think it is

[broken wagon]

That wasn't it unfortunatly but I really appreciate the assistance

.close

[devout snow]

Channel closed

Closed by @broken wagon

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stark shuttle]

Hello

[stark shuttle]

When distributing union or intersection why doesn't it switch like from like 4th line to the 5th line?

Comparing it to the 3rd to the 4th line

.close

[devout snow]

Channel closed

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[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[zenith spoke]

Use a double integral to find the area of :

[zenith spoke]

so from the wording of the question it is clear that 3cosθ=<r=<1+cosθ and not the reverse order

my main problem is with determining the bounds of θ

[lyric hornet]

the bounds of theta can be found by finding theta such that (1+\cos\theta=3\cos\theta)

[woven radish]

PajamaMamaLlama

[zenith spoke]

$1+\cos\theta=3\cos\theta\implies\cos\theta=\frac 12\implies\theta=\frac{\pi}3+2k\pi,\ k\in\mathbb{Z}$

[woven radish]

pirateking0723

[zenith spoke]

i guess that i am fine if i just consider 0=<θ=<2π so that the solution of this equation is θ=π/3 right ?

but then what about the other bound

[lyric hornet]

there's another solution to cos(theta)=1/2 👀

[zenith spoke]

ah -π/3

[lyric hornet]

which is coterminal to 5pi/3 yep

then ofc setting up (\iint_{R}dA) is pretty easy

[woven radish]

PajamaMamaLlama

[zenith spoke]

so $\int_{\frac{\pi}3}^{\frac{5\pi}3}\int_{3\cos\theta}^{1+\cos\theta}r\dd r\dd\theta$?

[lyric hornet]

5pi/6?? 👀

[woven radish]

pirateking0723

[lyric hornet]

now that can be evaluated using Calc II knowledge :)

[zenith spoke]

well i checked wolfram to get a number quickly so that i can compare it with the given solution

although it is relatively easy to evaluate it by hand

but then it turns out that the answer is different

[lyric hornet]

🤔

strange

,w int pi/3 to 5pi/3 of int 3cosx to (1+cosx) of ydydx

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xxmrfancyu2xx

[zenith spoke]

answer key says π/4

i have a detailed solution

but i didnt read it

[lyric hornet]

ah wait, bounds on theta should be from pi/3 to 5pi/3

[zenith spoke]

did you mean to swap them by any chance ?

[lyric hornet]

oh wait yeah lol

[zenith spoke]

because what you wrote here is the same as what you wrote to wolfram

[lyric hornet]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[zenith spoke]

so the integral now would give 2π

[lyric hornet]

but yeah that's weird

[zenith spoke]

which is positive but still not the desired answer

[lyric hornet]

how did the answer sheet get pi/4?

[zenith spoke]

here

[lyric hornet]

I am actually so confused what ur prof is doing here

[zenith spoke]

actually this is from stewart's book

well to me it seems that he reached these bounds just after drawing the figure

and didnt figure them out beforehand

[mortal crypt]

[lyric hornet]

really? huh

[zenith spoke]

yea

[lyric hornet]

yeah idk man im sorry I may have overlooked something but I have no clue what I did wrong

[zenith spoke]

dw man no need to be sorry

instead , tysm for your help

[mortal crypt]

so uh

about that

seperating the integrals was correct i thinjk

[zenith spoke]

why to separate it into 2

and how to get the bounds π and π/2

[mortal crypt]

For the first part of the cardioud and the circle

in the upper right quadrant

then taking the area betwen the two ends up just as the normal integral

however, not the the upper right quadrant

as the r = 3cos(Theta) ends up going negative at that point

[zenith spoke]

wdym by the 2 ends

[mortal crypt]

ill send a picture

[zenith spoke]

the ones in red ?

[zenith spoke]

alright

[finite crater]

what is solution please

[zenith spoke]

#❓how-to-get-help

[topaz crag]

Malika, ce forum est pris, il faut en ouvrir un nouveau

[mortal crypt]

If we take the inegral to be the same as before, then the 2nd part ends up encompassing not just the small bit between the cardioud and the circle, but because the circle has a negativbe radius, we end up counting that negative area as well

its easier to translate into regular graph form

https://www.desmos.com/calculator/ukygon0m7t

We want the area between the green line, and the circle. That would be whatever area is greater than the black line and less than the green curve for the first part. But when the black line goes below 0 after pi/2. Then the integral ends up counting all that negative area

[zenith spoke]

so i want the area between the green and the black curves but then i will need to split into 2 integrals before and after θ=π/2 and flip the sign of the integral after π/2?

[mortal crypt]

You will have to modify the lower bound of the inner integral to go to the origin instead of the circle

[zenith spoke]

ah wait

after π/2 the area is no longer between the 2 curves but instead between the black curve and the x-axis, ie y=0

[mortal crypt]

yep

[zenith spoke]

thats why the second integral will have lower bound 0 and upper bound 3cosθ (flipped the bounds so that the integral is positive)

[mortal crypt]

the upp bound will still be the 1+cos(theta) as the outer edge is the heart and not the circle

?

[zenith spoke]

well, the idea is clear as the figure is in front of me rn

but i need to figure everything out without any reference to the figure

so that i can solve a question like this without needing a diagram

[mortal crypt]

drawing stuff is a ok

[zenith spoke]

ik but i think it is troublesome to draw sometimes thats why i usually avoid it

but well maybe i understand a bit without referring to the figure too

the need to split is that 3cosθ<0 for π/2<θ<π

but we dont want to have 3cosθ<0 since this is a radius

is this the correct reason or does this just happen to lead to the result of splitting by chance

[mortal crypt]

For pi/2 < theta < pi, A line from 1 + cos theta to 3 cos theta ends up crossin the circle at 2 points and not just 1 like before

And for pi/2 < theta < pi, the second point that the line crosses is the one at the origin

So any line in between the origin and the heart (1+costheta) is going to be the area instead

https://www.desmos.com/calculator/zzcpbrfjnw

[devout snow]

@zenith spoke Has your question been resolved?

[mortal crypt]

We want the area between the heart and the circle. So we can think of adding up infinitely many boxes between them. However, because 3cos pi/2 ends up at the origin, that means that the box between 3cos and 1 + cos only needs to go as far as the origin, r = 0 for that point

if r = 0 at pi/2 in this case. Then that means the box used for area will only ever need to go down to that r = 0 for any radius

If I instead took the box from 1+ cos theta down to 3cos theta, then the box will go past the the point at the origin and go through the circle to the other side of it, counting both the area we want as well as the area of the circle as well

as the point of the origin is also a point on the circle

[zenith spoke]

ohhhh i see. This helped me get the idea too

but why does he just stop at θ=π

is it because he only wants to find the area of only one of the 2 bounded regions

and then use symmetry ?

[mortal crypt]

when theta = pi, then the heart thing radius is 0

the 2 small areas are the same, so rather than setting up a second integral that would be almost the same, we can just multiply by 2

[zenith spoke]

yea i agree with this

[mortal crypt]

hold up, im latexing some stuff, 1 min

got it

you got an integral set up?

2

[zenith spoke]

take your time

[mortal crypt]

I got the integrals, Also I made the ones for if I didn't use symmstry

$A = 2(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{3\cos\theta}^{1+\cos\theta}rdrd\theta} + \int_{\frac{\pi}{2}}^{\pi}\int_{0}^{1+\cos\theta}rdrd\theta}) = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{3\cos\theta}^{1+\cos\theta}rdrd\theta} + \int_{\frac{\pi}{2}}^{\pi}\int_{0}^{1+\cos\theta}rdrd\theta + \int_{\pi}^{\frac{3\pi}{2}}\int_{0}^{1+\cos\theta}rdrd\theta + \int_{\frac{3\pi}{2}}^{\frac{5\pi}{3}}\int_{3\cos\theta}^{1+\cos\theta}rdrd\theta$

oop

[woven radish]

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

[mortal crypt]

lowkey, I do not want to really calculate 4 double integrals

[zenith spoke]

no you dont need to

[mortal crypt]

so symmetry it is

🙂

[zenith spoke]

yea sure

[mortal crypt]

1 + cos(pi) = 1 + -1 = 0

[zenith spoke]

you dont even need to evaluate that

i only wanted to know how to set up the integral

the calculation is left as an exercise to the author

[mortal crypt]

[zenith spoke]

i still want to know how to set up the integral purely algebraicly (idk if this is a word lol), ie without depending on the figure in any way

but i will try that alone

i wont take more of your time and i now understand why the integral was set up like this thanks to you

[mortal crypt]

[zenith spoke]

tysm for your help and time

have a great day/night

.close

[devout snow]

Channel closed

Closed by @zenith spoke

Use .reopen if this was a mistake.

[zenith spoke]

.reopen?

[mortal crypt]

sure?

.reopen

i think u have to

[zenith spoke]

ah no i was asking if you wanted to add something

[mortal crypt]

I just wanted to say, that you should be totally fien with drawing things. At some point you will be drawing things in your head or on paper, especially more geometric arguments like these

I can go through my recent notes, and it's not just equations

its a few drawings as well

[zenith spoke]

the thing is that i am not used to draw these sorts of shapes so i definitely need a bit of time to draw a figure like that

which i dont have in exams for example

[zenith spoke]

yea sometimes i imagine the figure

[mortal crypt]

Here's an exmple of something I drew, even if vague

you dont need to care about the other math, but drawing that pentagon really ends up important intuitively

[zenith spoke]

but i am not sure how to draw a graph of the sort given in this question (r=1+cosθ in this case)

[mortal crypt]

ah I see, that would come down to kinda already knowing what 1+costheta looks like

[zenith spoke]

yea i get what you mean

[zenith spoke]

exactly

it is much easier to imagine it if you already know what it looks like

[mortal crypt]

drawing it in x,y plot and then transferring it to polar or vice versa is a great way to change the perspective and gain insight

like, If I only looked at the xy plot and not the r theta version, there's a chance I might not have caught the trick with the origin

[zenith spoke]

isnt r=1+cosθ something painful in crtesian coordinates

[mortal crypt]

sqrt(xx+yy) = 1 + x/sqrt(xx+yy)

It ends up being somewhat okay, not but pretty

what i mean is

giogn from

y = 1 + cos(x), to r = 1 + cos(theta)

[zenith spoke]

yea exactly

[zenith spoke]

ohhh

[mortal crypt]

you can think of it as taking the xy plot, and wrapping it aroudn the origin

also stretching the higher the y point is

[zenith spoke]

hmmm i see

maybe i have to admit that sometimes a sketch will make life easier when setting up multiple integrals

but i just hate to draw a figure for some reason

anyway tysm again

[mortal crypt]

yw 🙂

short crappy drawing

of 1 + cos theta

i am eating paella rn

afk

[zenith spoke]

.reopen

[devout snow]

✅

[zenith spoke]

i reopened it so that we can continue leisurely without being concerned about the channel closing suddenly

[zenith spoke]

you flipped 2π and 0 so that you get the same exact figure as that above instead of a reflection of it right ?

but why does just rotating one half like that give the desired figure

and why does it give a reflection of the desired figure if you just rotate without flipping first

[mortal crypt]

It's not really rotating, but rather stretching around in a circle

And If I stretched first without flipping, then the part on the right side of the graph, the 2pi, would end up on the x axis. But instead of the theta increasing when going counter clock wise, it would be decreasing

If I draw at line at x = 2pi, then its originally vertical, however, after the stretching, it would end up horizontal. So both the left and right parts end up stretching, its just that the left half is stretching and rotated by 3pi/2 while the right part is stretched and rotated by -pi/2

[zenith spoke]

but i dont see any difference, because if you look at it as stretching around in a circle then you wont be changing the radius that you are going in from the original radius right ? at least thats what you are doing here as far as i can see

[mortal crypt]

I say stretching because although it would be the same radius from the origin, if we take a y=1 line and a y= 2 line. the distance between (0,1) and (pi,1) aroudn the circle ends up being pi. whereas if I took the points (0,2) and (pi,2) from the y = 2 line and stretched it around the circle, then the distance around the circle of radius 2 would be 2pi for and not pi like the first line

If I instead took the graph y = 0, and looked at the points (0,0) and (pi,0), the distance between them is pi. And then when stretched around the origin, they end up at the same point

so the distance between (0,0) and (pi,0) -> (r = 0, theta = 0) and (r = 0, theta = pi) ends up being 0. so compresssed

,rccw

[mortal crypt]

the distance gets distorted when transforming functions from a y = f(x) to a r = f(theta) setup

lowkey really interesting math that goes on behind the scenes here

consider 2 points a, and b on the y = 2 graph

And we consider the distance between them along that y = 2 line

Let a = (0,0), and let b = (pi,0).

Then the distance between them along that line is just pi

However, If we then move that to the r = 2 space.

And take the distance along the r = 2 line. we end up with something else

As these 2 points are halfway across the circle. and we are only allowed to move along the r = 2 function. Then we calculate half the circimcfrance. So D' = C/2 = 2pir / 2. And as r = 2. that means that the new distance between these two points along the line r = 2 is D' = 2pi, where as D = pi

So as distance is not the same in this transformation. I would say stretching rather than just rotation

You can alternatively think of this as calculating the length of the line from A to B

u there?

[zenith spoke]

by this sentence you mean to take the "usual" distance between (0,2) and (pi,2)

[mortal crypt]

yep

it might be better to think of it in terms of a line between those 2 points ctually lel

[zenith spoke]

ohhh i see your point

yea if you look at it like this then it is reasonable to call that stretching

[zenith spoke]

well you can consider the arc between these 2 points in the r=2 space as a line between them

because that is actually what a line in this space look like right ?

[mortal crypt]

yep

[zenith spoke]

i suppose that such transformations are studied in topology ?

idk maybe i am wrong since i am not familiar with topology

i barely have a vague idea of it

[mortal crypt]

I haven't done topo yet, but I know this sort of thing is doen in differential geometry

[zenith spoke]

ohhh

[mortal crypt]

which was what the formula from before was a part of

[zenith spoke]

i see thats nice

[mortal crypt]

The idea of distance, and how that changes is a pretty big part of it

[zenith spoke]

what are you studying these days

[mortal crypt]

I think I've put it on the back burner for now, as my source material and most source materials online don't really have a great answer for my question on it

It ended up being a question of "why am I using the numbers in this way? what does this intuitively mean?"

So I've gone back to doing analysis which I should've done a while ago

[zenith spoke]

ohhhh

so rn you are studying analysis ?

[mortal crypt]

yep

[zenith spoke]

real analysis ? or you've already done that?

[mortal crypt]

real analysis, my rigor was really lacking early on in my math, and I'm paying the price for it

-1/12 if you know you know

so I'm building from the ground up to get better at that

[severe echo]

really?

im taking real

analysis rn

[zenith spoke]

i didnt get this ngl lol

[severe echo]

do you have any advice for how i should go about the class

[zenith spoke]

ah i see yea math gets too rigorous as get deeper

[mortal crypt]

https://www.youtube.com/watch?v=w-I6XTVZXww&pp=ygUSbnVtYmVycGhpbGUgLSAxLzEy this thing

[zenith spoke]

i am studying real analysis too

[zenith spoke]

ah yea i know this

[mortal crypt]

it threw me off so badly i was lowkey crank

for a while

[zenith spoke]

rearrangement

[severe echo]

nice

[zenith spoke]

you cant play as you like with any infinite sum

[mortal crypt]

fortunately I found this sereis

[zenith spoke]

there have to be some conditions for you to be able to rearrange the terms as you like

(absolute convergence)

[mortal crypt]

$\sum_{n=1}^{\infty}{\frac{1}{n^2+1}}$

[woven radish]

Yeatte

[mortal crypt]

this really helped me out

[zenith spoke]

i am not sure if i can say something helpful

[mortal crypt]

^

[zenith spoke]

but i recommend that you go through the proofs of the propositions/theorems by yourself

so pick up the theorems of the course

try to prove them yourself

[severe echo]

oh ok

nah cuz lowkey my school is ass

[zenith spoke]

check if your proofs are correct/ if they dont exactly match that doesnt mean that you are wrong so ask on this server for example or something like that

[zenith spoke]

how did it help

it helped you know when you can rearrange and when you cant?

[severe echo]

yeah this server seems like a really good resource

[zenith spoke]

ohh i see

[severe echo]

yeah mickey mouse clubhouse university bottom 10 stem school

[zenith spoke]

rn my university is trash too lol but i will try to get into a better one

that being said, it doesnt mean anything for the only thing that matters is your work to understand the topic

[mortal crypt]

before, I though that I could do whatever I wanted with any series, after rearranging, I ended up getting 1/2 + "zeta(-2)" + "zeta(-4)" + .... assuming I could do the things I thought I could and so i would get 0 for each term except the first. But a sanity check told me otherwise. that I needed to be careful

as 1/nn+1 def convergeed to more than 1/2

[zenith spoke]

ohhh

now i see

[mortal crypt]

so I knew the only step that was wrong was assuming finite values for the zeta 'series' and manipulating those without worry

[zenith spoke]

yea it gets tricky with infinite series

[mortal crypt]

I also accidentally foudn the solution to the suum anyway lel

[zenith spoke]

one can easily get something totally wrong just by assuming that usual manipulations for finite sums work just fine for infinite ones

[zenith spoke]

btw are you a math major ?

[mortal crypt]

I am a bakery worker

[zenith spoke]

oh thats nice

[mortal crypt]

I'm not in college for various reasons, so i study on my free time

[zenith spoke]

everyone has their hardships/reasons for not doing something

i hope you can bypass all of your problems

and btw you dont need college to be knowledgable

well you are the best example

you were just teaching me

[mortal crypt]

[zenith spoke]

nice to see that you are interested in math lol

math is too fun ngl

[mortal crypt]

I took a break from it for a while, but I couldn't stay away

[zenith spoke]

hahahahaha

yea feel free to take breaks every now and then

[mortal crypt]

I finished up some ideas from hs for physics and it ended up leading me down a rabbit whole all the way to diff geo

[zenith spoke]

ohhh

[mortal crypt]

electromag is really nice

[zenith spoke]

in my opinion most of physics is really nice

[mortal crypt]

ignore air resistence

[severe echo]

this might be happening cuz like infinity kind of wraps around

well not really but

[zenith spoke]

there was a point in time where i was choosing between pure math and theoretical physics

[zenith spoke]

the part that i hate in hs

[severe echo]

you know how in computer science once you get to the end of the signed digits and you keep increasing you actually end up wrapping around

[zenith spoke]

well not too long ago since i just started university some months ago lol

[mortal crypt]

I should prob go back to college hm

[zenith spoke]

well you know whats strange about series like this , you can literally get any number that you want

[mortal crypt]

I heard bout that ye

[zenith spoke]

if you can then i encourage you to

i think that maybe it is better to close this now that we are done from everything (probably) since it is a help channel

nice to meet you yeatte

[mortal crypt]

aye, and to you too

[zenith spoke]

we will probably have more encounters in the future since we are both in this server xD

i am looking forward to it

[mortal crypt]

🙂

[zenith spoke]

have a great day

.close

[devout snow]

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[devout snow]

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[muted sapphire]

I need help

[muted sapphire]

How to prove the first one

[full cloak]

alright, so you can prove it like this

think about it

anytime you draw a line between two points, what are you doing in terms of their odd/even degrees

[devout snow]

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[fathom hound]

i need help

[topaz axle]

where's it from?

[fathom hound]

a game on roblox

easter egg hunt

[topaz axle]

green is middle

because rule 3 says, you need to do VGB or BGV, so green is not 1 or 5
but if it's 2 or 4, then rule 4 doesn't allow it

hm

so like, PBGVG?

[fathom hound]

ill try that

it worked

thanks

[topaz axle]

it doesn't make sense, there's only 4 colors, so I don't even have to use pink

alright

[fathom hound]

.close

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[strange nimbus]

OK, the easiest way will be a Venn diagram.

Have you done this sort of thing with one before?

[daring depot]

Hmmm, nah.

I have done probability before, but not question like this where there is incomplete data

[strange nimbus]

OK, draw a Venn diagram with three circles.

One circle for discrete math, one for statistics, one for linear algebra.

[daring depot]

Okay, then?

[strange nimbus]

OK, now we work from the center out.

How many take all three? Write that in the middle.

[daring depot]

Okay, it's 1

[strange nimbus]

OK, that was the overlap of three circles. Now we do the overlaps of two circles.

How many take statistics and linear algebra?

[daring depot]

It's 3

[strange nimbus]

OK, so the overlap of S and L circles has two parts in the Venn diagram.

If you lightly shade it in, you'll see that.

So, we have those two parts of the Venn diagram adding up to 3.

So, something + 1 = 3.

Write that something in the unfilled-in part of the overlap of S and L.

[daring depot]

Okay, that's 2

[strange nimbus]

OK, so write that in. Then, we look at the overlap of L and D.

How many people take both linear algebra and discrete math?

[daring depot]

I don't know, the data is not included in the question

[strange nimbus]

OK, so there are two parts of the Venn diagram for this, so write x in the unfilled in part.

What about D and S? How many people take discrete math and statistics?

[daring depot]

It's also not included...

[strange nimbus]

OK, so write in y for the unfilled in part.

Now let's do the circles themselves.

How many people take statistics?

[daring depot]

11

[strange nimbus]

OK, so the four parts of the S circle should add to 11.

So, what will the unfilled in part be?

[daring depot]

8 - y?

[strange nimbus]

OK, good.

What about the L circle's unfilled in part?

[daring depot]

19 - x

[strange nimbus]

OK, what about D?

[daring depot]

28 - x - y

[strange nimbus]

OK.

Now, we do all the parts.

All the parts added together add up to what?

[daring depot]

50

[strange nimbus]

OK, so add up all the parts, set it equal to 50, and simplify the equation as much as you can.

[daring depot]

I get x = -y + 8

[strange nimbus]

OK, so x + y = 8.

Then, I guess you can simplify the D alone part of the Venn diagram.

I can't see any more simplifications.

Oh, wait.

Let's see.

Never mind.

OK, from here, you can do part a, but I don't see a way to get b and c.

[daring depot]

So, does this mean there is no answer for b and c?

[strange nimbus]

There might be a way to get another equation with different coefficients for x and y when you simplify it so that you can solve simultaneous equations, but other than that, I can't see a way.

Right now we have x + y = 8.

From the five statements they give, all except the discrete mathematics one lead to no variables when simplified, and the discrete mathematics just gives us x + y = 8.

I suppose you can replace y with 8 - x.

Hmm.

It might be even worse than that, since the first sentence implies that it might not be all 50 who take at least one of the three courses ("Out of 50 students in a campus, there are several students..."). That can't refer to students that take all three, since the one student who does that isn't several students.

[devout snow]

@daring depot Has your question been resolved?

[strange nimbus]

I'd recommend answering b and c with something like: "If we take x as the number of students who take discrete math and linear algebra without taking statistics, the answer is whatever/50."

x could also be written as |D n L n S'|.

[daring depot]

Ohh okay

It should be a good approach for that

Thanks for the help

[devout snow]

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[strange nimbus]

You're welcome.

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[strange nimbus]

Oh wait! |(D u L) n S'|.

[strange nimbus]

.close

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[restive river]

Need help with this geometry problem

[restive river]

.close

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[thin geode]

,rotate

[thin geode]

Pls only give very subtle hints

I need to use 4k-2 and the factorials in some way, probably by multiplying

[devout snow]

@thin geode Has your question been resolved?

[thin geode]

i got it

multiply 4k-2 with k-1

factor out the 2 and put it into the k-1

makes 2k-2

[devout snow]

@thin geode Has your question been resolved?

[real mason]

look at the factorials
see if you can rearrange(4n-2) * (2n-3)!/(n-2)! to be (2(n+1)-3)!/((n+1)-2)!
if you do that you're done

try expanding the RHS

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[final forum]

Someone please dm me if they are willing to help me with my pre cal hw it’s just 2 questions

[vital sedge]

Just ask them here?

[final forum]

I’m desperate istg

[vital sedge]

Just send the questions

[wispy oyster]

ask your question here

[soft umbra]

Send your question here

[final forum]

These 2 questions

44 and 45

[runic grove]

set x + 5 = 0 and y = 0, those values of x and y are your centers

this ellipse is basically shifted

[final forum]

I honesylu don’t know what’s going on 😭

[runic grove]

all the properties remain the same, but all the points on the ellipse have been shifted 5 points in the negative direction

okay let me graph this

[wispy oyster]

in standard ellipse form
$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$
the center is (h,k)

[woven radish]

mia

[runic grove]

this is a normal ellipse, with it's center being (0, 0)

now that is the same ellipse but it's shifted 5 units in the negative x direction

[final forum]

This is a sample question we did in class

[runic grove]

all the points, the focii, the end points are just shifted 5 units in the negative x direction

[runic grove]

that's all there is to it

[final forum]

I’m still so confused 😭

[wispy oyster]

https://youtu.be/lmBcss4NJuI?si=a_SO4SQDpksKmktK

[devout snow]

@final forum Has your question been resolved?

[final forum]

Not yet

[final forum]

It’s not close to the question ☹️

[cedar bison]

https://youtu.be/OPSCKXXvWiM?si=NWmt0e5g1b5RuFXv

Ig that's exactly what r u looking for

[cedar bison]

This guy is the best
Physics, calculus, chemistry

[devout snow]

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[devout snow]

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[winter meadow]

I’m really struggling to understand maclaurin and Taylor polynomials does anyone have some resources I could use ?

[cinder bobcat]

khan academy is always a good place to start

[winter meadow]

It always comes back to khan academy doesn’t it

Frick this chungus life

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[opaque talon]

when ur constructing the table

[opaque talon]

how dyu know the categories to use

this question is so long man

[pseudo basin]

it would be better with a venn diagram tbh

3-way tables are kinda tricky

[opaque talon]

ah they use a table in the mark scheme

Ok i'll try that

even that sounds kinda tricky

ah

how many bubbles should there be

i mean circles in the venn diagram

idek

[devout snow]

@opaque talon Has your question been resolved?

[opaque talon]

for iv

on the denominator

i get an extra 0.15

cuz not S is 0.35+0.15

and not T is 0.15 + 0.15

idk

[primal ferry]

look at it in terms of the overall area

this is the part signified by S' union T'

this should rule out the extra 0.15

[opaque talon]

ahhh

okok

thx

[primal ferry]

np

[opaque talon]

idk if u have any tips for the venn diagram here as well? 😅

[primal ferry]

oh that wasnt answered yet?