help-36

[tranquil pine]

The image of the identity operator $I: X \to X$ is the entire space $X$, so it is not a proper subset of $X$, no? am i tripping

[soft zealot]

♡LexQa♡

[desert mantle]

who cares if it is a proper subset

[tranquil pine]

Since it is not a proper subset, it is not possible for it to be closed as a subset of $X$

[soft zealot]

♡LexQa♡

[tranquil pine]

if my knowledge is correct

[tulip coyote]

Surely the whole space is closed, no?

[fallow wren]

no because the complement of X its the empty set that is open so X is closed

[desert mantle]

X is both open and closed in X. same for the empty set being both open and closed in X

[tranquil pine]

i see

well back to the original question, can we assume the statement is correct then?

[desert mantle]

why should we

just because of one example?

[tranquil pine]

well like u said, there was no provided counterexample

[signal knoll]

HELLO

[desert mantle]

just because I can't think of a counterexample atm does not mean there isn't one lol

[tranquil pine]

yeah fair enough

this channel stays open until then ig

I found this on stock exchange

https://math.stackexchange.com/questions/1087702/is-tx-a-banach-space

i guess thats it?

[grim nebula]

wait you're unmuted now

[tranquil pine]

yep i am back into the world snow

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

anyways bye

.close

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[dire terrace]

5. How spread are the college students’ weights to the mean?

[dire terrace]

is this variance or SD?

[prime oar]

Both

They both measure that

[dire terrace]

oh

3. How dispersed are the scores of the students from 100 respondents?
   i answered SD

is that also variance or only SD?

[prime oar]

Both measure that lmao

[dire terrace]

thanks!!

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[modern crypt]

$\frac{x^2my^n+1}^3{x^3m+1y^n}$

[soft zealot]

Zyme><SOL

$\frac{x^2my^n+1}^3{x^3m+1y^n}$
```Compilation error:```! Missing { inserted.
<to be read again> 
                   }
l.57 $\frac{x^2my^n+1}^
                       3{x^3m+1y^n}$
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)```

[final tangle]

can you take a pic of the problem

[modern crypt]

It's (x^2m y^n+1)^3/x^3m+1 y^n

[final tangle]

just take a pic pls

i suspect that there are multiple issues with notation

too many to be able to know what's intended

my best guess is

[modern crypt]

I'm uploading it rn

this is it

[final tangle]

ok

$\frac{(x^{2m}y^{n+1})^3}{x^{3m+1}y^n}$

[modern crypt]

i need to simplify using exponent law, it says answers should be expressed with positive exponents

[soft zealot]

ℝamonov

[final tangle]

what have you tried?

[modern crypt]

not much honestly, other than cubing what's in the numerator

[final tangle]

show me what you've done, whatever little it may be

[modern crypt]

so the new numerator would now be x^6m y^n+3

[final tangle]

that isn't right even assuming you have ()

[modern crypt]

oh

[final tangle]

how are you getting the y^(n+3)

[modern crypt]

I learned from previous questions that the exponents inside the bracket multiply with whatever exponent is outside the bracket

[final tangle]

yeh...

[modern crypt]

that was my best guess though, I didn't really know what I was doing

[final tangle]

you didn't multiply properly

[modern crypt]

oh

yeah probably, that rule worked when there was only a number as an exponent and not letters too

[final tangle]

the rule applies regardless of the number of terms in the power

you just need to multiply properly

[modern crypt]

so multiplying wouldn't be 3x2 and 1x3?

[final tangle]

${(a^{\text{this}})}^{\text{that}} = a^{\text{(this)}\cdot\text{(that)}}$

[soft zealot]

ℝamonov

[final tangle]

you should be multiplying 3 to the entire n+1,
not just whatever part of it you feel like

[modern crypt]

so 3n+3?

so if that's what you meant it would be (x^6m y^3n+3)/x^3m+1 y^n

I'd assume you'd have to then cancel out the terms

x^3m y^2n + 3 / x^ 1

that isn't the answer though

$\frac{(x^{6m}y^{3n+3})}{x^{3m+1}y^n}$

[final tangle]

that was just the first step

now apply the exponent law involving division

[soft zealot]

Zyme><SOL

[modern crypt]

ok

${(x^{3m-1}y^{2n+3})}$

x^1 = x

so I think we can remove that

or no, that's wrong

we'd have to move the 1 up

[soft zealot]

Zyme><SOL

[modern crypt]

this is the answer, last step I really just got the inference by looking at the answer

anyways, thanks

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[pulsar stirrup]

hi

[pulsar stirrup]

could anyone explain why i cant use my integral to calculate the surface area of the area outside the curve r(theta) = sin(theta)cos(theta) and inside r=0.5

my integral:

and the solutions:

[pulsar stirrup]

this is the graph btw

[final saddle]

@pulsar stirrup Has your question been resolved?

[midnight ginkgo]

$r(\theta) = \sin(\theta) \cos(\theta)$

[soft zealot]

barometer

[pulsar stirrup]

yes

[midnight ginkgo]

sorry i just needed to visualize it

what is this

[pulsar stirrup]

1/2 sin (2theta)

[pulsar stirrup]

<@&286206848099549185> or @midnight ginkgo 🙏

[midnight ginkgo]

Sorry I'm trying but my internet keeps going down

[pulsar stirrup]

oh shit okay np!:)

[midnight ginkgo]

Why do you have a 1/2

Oh nvm

Those are equivalent integrals

[pulsar stirrup]

its a formula for surface area for polar coor

[midnight ginkgo]

So you have the same as the answer key

[pulsar stirrup]

is it the same case here? cuz ive made another one nd i used another formula again

[midnight ginkgo]

maybe? i can't really tell what's going on

essentially the formula that you used in the first one is just a partially-worked out double integral

the answer-key writer just converted $\sin{\theta}\cos{\theta}$ to $\frac{1}{2}\sin{2\theta}$

[soft zealot]

barometer

[midnight ginkgo]

and also changed the theta bound to pi/4 and multiplied by two bc of symmetry

[pulsar stirrup]

yea

oh okay so theyre jus the same?

[midnight ginkgo]

yeah i think so

[pulsar stirrup]

Ah damn thank u ! so if this would similarly be the same too then?

(sorry for asking sm qs about these )

[midnight ginkgo]

what are these

[pulsar stirrup]

omg im so sorry hahah its 3 sin theta to 1+ sin theta

[midnight ginkgo]

okay lol

[midnight ginkgo]

i'm not sure since you're changing the bounds here

yeah no

the bounds on the second integral would be pi/2 to pi/6, not 0 to pi/2

[pulsar stirrup]

oh okay.. would you possibly know why i cant use these bounds in my integral? bcs i chose r going frm the circle to the cardioid , nd for theta i chose the bounds -pi/2 to pi/6 bcs thats where they intersect

[midnight ginkgo]

oh i misread that

should be -pi/2 to pi/6 for the bounds in that case i believe

oh i see why you're doing 0

[pulsar stirrup]

so then itll probably be my r bounds that r wrong for it to result in a diff answer?

[midnight ginkgo]

do you have the answer key?

[midnight ginkgo]

yeah

[pulsar stirrup]

oh fuck😭ithink i found where i messed up

[midnight ginkgo]

where

[pulsar stirrup]

green part

ididnt take into account

im so sorry

[midnight ginkgo]

oh word

[pulsar stirrup]

tysm for checking the first integral tho !

[midnight ginkgo]

yeah

honestly i can't remember on the second one 😭 been a minute since i did that

no wait

[pulsar stirrup]

huh

[midnight ginkgo]

no sorry

i think you were right with this one

[pulsar stirrup]

yea no i just double checked too

it should be right… right?

cuz were jus substracting

the green

[midnight ginkgo]

mmm

i think that should be covered when you do the radial bounds

do you know what the answer is supposed to be?

[pulsar stirrup]

yes wait

oh oop nvm they only gave the integral in the answer key

[midnight ginkgo]

that's what you had though

so slay

[pulsar stirrup]

nono i put it in red there cuz i thoight mine wasnt equal to that😭

[midnight ginkgo]

oh it's not

[pulsar stirrup]

💔

no

not mine,,

equal to each other that idk

[midnight ginkgo]

this is confounding me for no reason

OHHHH

nevermind

[pulsar stirrup]

YEA?

oh

[midnight ginkgo]

they literally just did the area of the cardioid minus the area of the circle for that intersection

i think you can't do it with double integration because the angles aren't the same at the intersection points

[pulsar stirrup]

huh why not

[midnight ginkgo]

lemme make a little diagram

so at A, the 1 + sin(theta) is at the angle -pi/2, but the circle is at the angle 0. so even though they intersect, it's not at the same angle

[pulsar stirrup]

oooooooooooooooohhhhhhhhhhhh

thanku sm that was confusing me way too much than it should

[midnight ginkgo]

me too

i think once you learn double integrals it's hard not to want to just use them all the time lol

[pulsar stirrup]

yeahh exactly😭

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[golden sundial]

What number should come after 1, -3, 5, -7, 9, -11, 13?

[broken moon]

-15?

[slender oyster]

stay on your own channel only

[strange shore]

#❓how-to-get-help

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@golden sundial Has your question been resolved?

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[tranquil pine]

just a quick notation question, but is there any major difference between those two signs?

[tranquil pine]

the first is what i use usually for approximations, but the second i also see from time to time and i vaguely remember some sort of difference

i dont know their particular names to look them up haha

[jagged seal]

https://math.stackexchange.com/questions/864606/difference-between-≈-≃-and-≅

[sharp forge]

I’ve seen the second one used for isomorphism

[tranquil pine]

yeah same

[tranquil pine]

okay ty

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[sleek edge]

me back

[sleek edge]

hey so in here wont there be infinite solutions?

i hav made this desmos graph, wait lemme share

https://www.desmos.com/calculator/ebu9uhhzdx

what they've asked for is when they dont meet right?

[rare cradle]

Yea should be infinitely many solutions

Bc you can make n arbitrarily large

[sleek edge]

indeed

wait unless they're asking for when they intersect

[rare cradle]

Maybe it was a typo

[sleek edge]

oh no

yea maybe

there are lots of them

wait but then too answers dont match

in the key answer is given 4

[final saddle]

@sleek edge Has your question been resolved?

[sleek edge]

oh hey moni

[muted prairie]

so a circle centered around 3-2i with radius n/4

hello

[sleek edge]

yea and another at 2-3i with 1/n radius

[muted prairie]

vs a circle centered around 2-3i with radius 1/n

[sleek edge]

[muted prairie]

so it seems like they'd intersect when n/4 + 1/n < sqrt(2), right

[sleek edge]

[muted prairie]

or no greater than

[sleek edge]

yea

[muted prairie]

but the set should be naturals n such that n/4 + 1/n < sqrt(2)

which should be finite

[sleek edge]

[muted prairie]

n=1,2,3,4 checks out

[sleek edge]

hm?

wot how did u get it tho

i mean

[muted prairie]

Their radii are 1/n and n/4

[sleek edge]

graph says otherwise tho

yea no i understood u

but

desmos

[muted prairie]

ah hmm

ok

[sleek edge]

what did i do wrong

[muted prairie]

problem makers made the same mistakr I did then ig

[sleek edge]

wheres mistake tho

[muted prairie]

if the conditions were like |z-3+2i|<n/4

then it would work my way

[muted prairie]

Assuming sum of radii ≥ distance guarantees intersection

[sleek edge]

[muted prairie]

what's also needed, and probably forgotten: difference of radii ≤ distance

[sleek edge]

hm i dont get it

oh

ohh

ohhhh

icici

ahh

so its wrong q?

i mean

infinite sol

[muted prairie]

ye

[sleek edge]

right?

[muted prairie]

should be

[sleek edge]

i mean even otherwise we still do get infinite solutions no?

[muted prairie]

nah then you'd get only solutions to n/4+1/n < sqrt(2) as solutions

Those are the cases in which the 2 circles are outside each other

[sleek edge]

oh

icic

thanks moni

[muted prairie]

np

[sleek edge]

.close

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[wet hornet]

May I have some help getting started on this one?

[wet hornet]

I cannot figure out where to even begin lol

[vital crag]

https://www.cuemath.com/geometry/arc-length/

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[dull comet]

Can someone help me with this one: Let {fn} be the Fibonacci numbers. Prove by induction that every third Fibonacci number is an even number. (Hint: Assume f3k is an even number.)

[teal kayak]

Where are you stuck? Show work

[dull comet]

Im struggelig to formulate a proof for the odd numbers and whole numbers

because the pattern is: odd, odd, whole

But honestly idk what to do next

[teal kayak]

I think this doesn't need any cases

use the property of fibonacci numbers

$f_{3k+3} = ?$

[soft zealot]

numbpy

[teal kayak]

use fibonacci property

[dull comet]

like this?

[dull comet]

.

[teal kayak]

why do you have 3(h+3) in index?

It should be 3(h+1) = 3h+3 right?

[dull comet]

because the task assumes that f_3k is an even number

f_3k

[teal kayak]

yes, you assumed it to be true for P(n) for induction hypothesis

Then you want it to be true for P(n+1)

[dull comet]

yup

so i chage it to: f_3(k+1)

[teal kayak]

or P(k+1) which means f_{3(k+1)}

[dull comet]

yup

[teal kayak]

yess change it

[dull comet]

so it is: f_3k + 3

right?

Im srry but i do not quite understand what you mean with "change it" in this instance

[dull comet]

.

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[teal kayak]

.reopen

[final saddle]

✅

[dull comet]

?

[teal kayak]

I meant in the picture

Anyway

Use the Fibonacci property on f_(3k+3) twice

[dull comet]

What do you mean with FIbonacci property

like the number?

[teal kayak]

Yeah the fact that f_(n+2) = f_(n+1) + f_n

[teal kayak]

Dudeeeee, nah why did you separate + 3 from the index 😭

[dull comet]

because is thought you mean that is should switch it with the fibo number

anyways

[teal kayak]

No, meant like f_(3k+3) = f_(3k+2) + f_(3k+1)

[teal kayak]

I think you are confused with the indexing

[teal kayak]

$f_{3(k+1)} = f_{3k + 3}$

[soft zealot]

numbpy

[teal kayak]

This is wrong

You have 3k+3 NOT 3(k+3) = 3k+9

Just work with f_(3k+3) remove the brackets. They are confusing you

[teal kayak]

Why do you have 3k+9?? Look at what you have written. It's clearly wrong

Like let k=0 then what you have written means

f9 = f6 + f3

[dull comet]

but it is: f_9 = f_6 + f_3

isnt that right?

[teal kayak]

Nope

f_9 = f_8 + f_7

[dull comet]

hmmm

so f_3(k+1) + f_3(k) = f_3(k+2)?

[teal kayak]

No

Why are you using brackets?

[dull comet]

because i asumed that f_3k = a whole number

[teal kayak]

You mean an even number

[dull comet]

yes

which means that f_3(k+1) should also be an even number

[teal kayak]

Yeaaah and you just need to worry about f_3(k+1)

Which is f_(3k+3)

[dull comet]

yup

[teal kayak]

Just work with that

[dull comet]

so should i add that with f_k?

because i need to prove that every third fibo number is even

[teal kayak]

Yeah, that's why we are using induction

THE INDUCTION IS ON k in 3k

Not on 3k itself

[dull comet]

but what can i do with: f_3k+3 + f_3k ?

[teal kayak]

Why do you have + f_3k?

[dull comet]

because that is our asumption

and to find the third number we need to ad the to before that

[teal kayak]

Huh? Our assumption is that f_3k is even right

[dull comet]

yes

[teal kayak]

Absolutely no need to do that

Just start with f_3k+3

Expand it using Fibonacci property

[dull comet]

so

f_3k+3 + f_3k+6 = f_3(k+3) ?

[teal kayak]

How did you even come up with that?

Do you know the Fibonacci property?

Write it

[dull comet]

f(3k+3) = f(3k+2) + f_(3k+1)

[teal kayak]

Yess, finally

Now, apply the Fibonacci property on the first term again

ie on f(3k+2)

[dull comet]

how can i do that? I understand that the fibo property is: f_n-1 +f_n-2 but how can i apply it on the first term

[teal kayak]

Just remember this, reduce the index by 1

[dull comet]

do you mean: f(3k +2) = f_(3k+1) + f_3k

[teal kayak]

Yep

[teal kayak]

Put this back here

What do you get?

[dull comet]

f(3k+3) = f_(3k+1) + f_3k + f_(3k+1)

[teal kayak]

Simplify a little

[dull comet]

can you simplify it. Everything is written with indexes?

[teal kayak]

You can add something

[dull comet]

no way

f(3k+3) = 2(f_(3k+1) ) +f_3k

[dull comet]

or did i mess up again?

[teal kayak]

Do you realise you have completed the proof almost

[dull comet]

to be honest no, but

[teal kayak]

See the right hand side

[dull comet]

yes

[teal kayak]

Does it look even?

[dull comet]

yes because we asumed that f_3k was even. And the first thing is multiplied with 2, which means it must be even

so have i done it now?

[teal kayak]

Yes cause the sum of two even numbers is also even

[dull comet]

correct

[teal kayak]

So, f_(3k+3) is even

This completes the induction

[dull comet]

nice. Thx for your patience and help

[teal kayak]

Sure

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[silver cedar]

hi!

[silver cedar]

I need help with part (b)

this is physics by the way

[versed crater]

if you didn't know there is a physics discord

[silver cedar]

oh I tried it there but they said I’m not allowed to post assignment questions

[stoic canyon]

oh

[final saddle]

@silver cedar Has your question been resolved?

[grim nebula]

it should just be the difference of those two lengths no?

actually not quite

should be these two

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@silver cedar Has your question been resolved?

[silver cedar]

hi!

yes I did path difference between incident - path difference between reflected

[grim nebula]

yeah you should be able to get those lengths with trig

[supple mantle]

Path diff. 1 should be like: dcos(40°) right?

[silver cedar]

yeah I got dsin(50°) but it’s the same I think

[supple mantle]

Wait that's not it

[silver cedar]

oh

I thought they were equivalent

[supple mantle]

d is gonna have to be divided into a big section and a smaller section

So that you can use right-angle trig

[silver cedar]

does this seem okay?

[supple mantle]

Oh you got d already

[silver cedar]

oh sorry! I forgot to mention I got d as 10000nm

[final saddle]

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[sleek edge]

hey there

[sleek edge]

so in here what does largest interval mean

[final saddle]

@sleek edge Has your question been resolved?

[grim nebula]

possible values of g(3)?

so min max

and everything in between?

[sleek edge]

max cant be 2 tho

[grim nebula]

,w 1 + 1

[sleek edge]

[grim nebula]

wait

[sleek edge]

[soft zealot]

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Requested by 夢雪#2250

[grim nebula]

[sleek edge]

no but 2 cant be

[grim nebula]

yeah

[sleek edge]

because we need it conti

[grim nebula]

also can't be 1/3

[sleek edge]

the guy who set this question didnt consider this?

[grim nebula]

maybe it's just

like

others miss some values

[sleek edge]

[grim nebula]

only b covers everything

[sleek edge]

bruh

[grim nebula]

[sleek edge]

why not give (-∞ ,∞ )

[grim nebula]

[sleek edge]

hey wait i hav another one

[grim nebula]

[-∞ ,∞ ]

[sleek edge]

\cci -\y,\y

[grim nebula]

[sleek edge]

why is it not loading

the pdf

[grim nebula]

,tex \def\cci#1,#2 {[#1, #2]} \let\y\infty $\cci -\y,\y $

[grim nebula]

[sleek edge]

what if we cheat a bit

lets say

p(x) = x⁵+x⁴+x³

i mean we don’t really know much about f n g

[grim nebula]

there are no x^(3k+2) terms

in P

[sleek edge]

[grim nebula]

but being a multiple of x^2 + x + 1 necessitates

[sleek edge]

why do u say so

[grim nebula]

or does it

it doesn't

[sleek edge]

hm?

[grim nebula]

because f and g are poly

so you only have cubic

and one above cubic

so 3k and 3k+1

[sleek edge]

oh hm

icic

[grim nebula]

so you need 3k+2 to cancel

[sleek edge]

[grim nebula]

shopping rn so im doing this in my head

[sleek edge]

u havnt bought me christmas presents yet

[grim nebula]

it's gotta be 0 right?

because otherwise the 3k+2 never cancel

[sleek edge]

idk

[pale oracle]

$1/3 \le g(1) \le 1$

[grim nebula]

like

[soft zealot]

Disorganized

[sleek edge]

[grim nebula]

like you can only ever cancel something in the middle

[pale oracle]

by FTC....

[grim nebula]

when you do the multiplication

but never on the outside

[pale oracle]

This integral is equal to g(3) - g(0)

[sleek edge]

oh wait it’s another question

[sleek edge]

^^

[grim nebula]

no okay it might not be 0

but the coeffs have to add to 0

[sleek edge]

what if we multiply n divide by (x-1)

[grim nebula]

[pale oracle]

.rotate

[sleek edge]

so its x³-1 in denom

[pale oracle]

.rotate

[sleek edge]

,rotate

[sleek edge]

@pale oracle that question is no longer in discussion rn

[grim nebula]

[pale oracle]

Oh

[grim nebula]

like

write P(x) = (x^2 + x + 1)Q(x)

the coeffs of Q have to add to 0

[sleek edge]

[grim nebula]

cuz the 3k+2 terms have to die

so like

you look at windows of 3

[sleek edge]

[grim nebula]

and the 3k+2 term looks like a+b+c

[sleek edge]

[grim nebula]

like

if Q(x) = a + bx + cx^2

and you do the multiplication

[sleek edge]

oh

icic

[grim nebula]

the x^2 term is gonna be a+b+c coeff

[sleek edge]

oh okay i get it now

thanks

[grim nebula]

[sleek edge]

.close

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[white rock]

how do i do this

[tired walrus]

which part?

[white rock]

i cant even do the first

[tired walrus]

okay well let's go through this step by step

[white rock]

ok

[tired walrus]

first off

i think it will be good to write down the coordinates of all named points

A, B, C, D, P, Q

[white rock]

ok

[tired walrus]

this will help us for all parts

so let's just get this done so we can refer to it later

would you like to try doing this on your own or should i give you pointers?

[white rock]

A: -60,21

B: 40,26

[tired walrus]

don't omit the parentheses

[white rock]

C: (-60,18)

D: (40,8)

P: (-60, 40)

Q: (40,40)

O: (0,0)

[tired walrus]

okay, let's see

okay, yeah, all your coordinates are correct

now for part a

you are asked to find the equation of line AB

you know the coordinates of A and B

[white rock]

find gradient

[tired walrus]

do you know how to write the equation of a line through two points with known coordinates?

finding the gradient is one step along the way

[white rock]

y=1/20x+b

do i just select a random y value?

[tired walrus]

what do you mean by "a random y value"...?

[white rock]

o

nvm

i do (y-y1)=m(x-x1)

[tired walrus]

so can you write out the equation?

in whatever form?

[white rock]

x-20y+480

=0

[tired walrus]

,calc -60 - 20*21 + 480

[soft zealot]

Result:

0

[tired walrus]

ok yeah that checks out

it will be more convenient for us to have it in slope-intercept form. can you convert it to that?

[white rock]

20y=x+480

y=x/20+24

[tired walrus]

ok, good.

that does it for part a.

are you able to continue for parts b, c and d?

[white rock]

how would i find the depth of r?

is its jsut 40

?

wait no

i need to find or

isee

would or just be x=0

so you sub it into ab

and find the depth like thaat?

[tired walrus]

wording...

[white rock]

OR

is x=0

AB

is y=x/20+24

[white rock]

sub x=0 into y

y=24

[tired walrus]

i've highlighted in red the depth they ask you to find.

[white rock]

40-24=16

so its 16m?

[tired walrus]

yes

i don't like the way you worded this but your calculations and answer are correct.

[white rock]

part c i have no idea

how to find a curve

[tired walrus]

you are told to assume that your curve has equation y = ax^n with a and n yet unknown.

you know that the curve passes through A and B and you know the coordinates of both

[white rock]

so what would i do with that info?

[tired walrus]

what are the coordinates of A?

oh

wait hold on

my bad

the curve passes through C and D, not A and B

[white rock]

yes

[tired walrus]

the process doesnt change, only the numbers do

so, what are the coordinates of C?

[white rock]

-60,18

[tired walrus]

(-60, 18) yes

ok, can you write down as an equation the statement "The curve y = ax^n passes through the point (-60, 18)"?

[white rock]

ok

would i sub in the values into the equation?

[tired walrus]

yes that is exactly what you would do.

[white rock]

so 18=-a60^n

and 8=a40^n

[tired walrus]

bad notation and also wrong for one of these

(-60)^n is not equal to -(60^n)

[white rock]

o

[tired walrus]

a * (-60)^n = 18 and a * 40^n = 8

[white rock]

ok

[tired walrus]

do you see how to proceed from here

[white rock]

simultaneous equation?

[tired walrus]

... sure

[white rock]

a=18/(-60)^n

a=8/40^n

[tired walrus]

typo and missed pair of parentheses, also inefficient approach

better to go (-60)^n/40^n = 18/8

from which you will get n = 2

[white rock]

how do i get it into that form?

[tired walrus]

divide one of these by the other

gtg

[white rock]

alr

i got it thanks

.close

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[winged gorge]

for 1(a), to find the turning point whats the method to use?

is it completing the square?

[crimson token]

yeah

[winged gorge]

ok my turning point is (2.5,4)

is it correct?

[crimson token]

,w vertex of y=(x-1)(x-4)

[soft zealot]

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[crimson token]

no unfortunately, you have the y coordinate incorrect

[winged gorge]

why?

[crimson token]

What did you get when you completed the square?

[winged gorge]

oh i get it

[crimson token]

[winged gorge]

ok thanks

[final saddle]

@winged gorge Has your question been resolved?

[final saddle]

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[winged gorge]

.reopen

[final saddle]

✅

[winged gorge]

hi

is my turning point for 1(b)

(-1,-4) ?

[crimson token]

,w turning point of (x+3)(x-1)

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Sneaky#0928

[crimson token]

in future you can go to wolframalpha.com and ask him directly

but yes that is correct

[final saddle]

@winged gorge Has your question been resolved?

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[distant quest]

hello can someone help me to do that ? I didn’t understand anything

[winter hazel]

^rotate

[marsh temple]

,rotate ccw

[marsh temple]

[manic herald]

partial sum formula for a geometric series?

[final saddle]

@distant quest Has your question been resolved?

[winged gorge]

okay

[final saddle]

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[toxic tree]

https://gyazo.com/b610c82443eec1617503df1741070694.png

[toxic tree]

Do I really need to make a 4 atom truth table or is there another way to do this

[vale moat]

you could use a truth table, or logical equivalence laws

[final saddle]

@toxic tree Has your question been resolved?

[hushed sonnet]

Yup logical equivalence lass would be the way to go or you'll be writing a truth table of size 2⁴

Distributive law should get you a long way

[toxic tree]

thanks guys!!

❤️

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[mellow cedar]

$\lim_{x \to 0} \frac{x}{ \frac{-x}{1+x}}$

[soft zealot]

Mehdi_Moulati

[mellow cedar]

like this ?

[severe dawn]

just use division sign and brackets

[rare cradle]

@upbeat jacinth what are your pronouns?

[soft zealot]

Mehdi_Moulati

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@upbeat jacinth Has your question been resolved?

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[dark loom]

i need a hint with this example. i need to simplify this:

[final saddle]

@dark loom Has your question been resolved?

[dark loom]

<@&286206848099549185>

somebody helps me

[carmine kite]

what does : represent in this context?

[dark loom]

it's division

[carmine kite]

never seen it like that interesting…

[dark loom]

i tried, but get only this

<@&286206848099549185> ?

[final saddle]

@dark loom Has your question been resolved?

[high zenith]

<@&286206848099549185>

[dark loom]

.close

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[dull rampart]

Please help me with geometry homework so I can get an A

[dull rampart]

I wasn’t here the days we went over this and the second one is a quiz correction

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Channel closed

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[wary grove]

huh

[supple mantle]

@inner charm what's with opening and closing channels?

[inner charm]

im dont open this!

its about 30 minutes

[supple mantle]

[inner charm]

[supple mantle]

[inner charm]

im in lofi girl channel dont disturb

[supple mantle]

Your name is on the channel

[inner charm]

（；´д｀）ゞ

[final saddle]

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[rustic musk]

Can you show your work

[rose yew]

uhh

can i explain

[rustic musk]

Like take a picture

[rose yew]

ohh

1 sec

@rustic musk

[rustic musk]

Look at this step

[rose yew]

yes?

-9x1=-9

-9x-1=+9

[rustic musk]

ohh hwhoops

Your right

You missed the parenthese though

ok give me a moment I'm looking through the rest

Did you solve sin(x) = -1/9?

[rose yew]

arcsin(-1/9)

=-.1113

which isnt in the answer range

[0,2pi)

[rustic musk]

note pi-arcsin(-1/9) is in the answer range

and sin(pi-arcsin(-1/9)) = -1/9

[rose yew]

wait so it would be arcsin(-1/9)

and then the answer of that pi - answer?

[rustic musk]

But I'm a little confused too because that's more than pi/2

[rose yew]

yup

[rustic musk]

lemme pull up desmos

[echo needle]

so?

[rustic musk]

The answer of pi/2 was marked wrong

[rose yew]

okay

[echo needle]

it wants the decimal i think

[rose yew]

so its pi - (arcsin^1(-1/9))

what about the first

arcsin(1)

got it

ty

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[open rain]

it says R is an equivalence relation

and x is a non empty set

show that for elements out of X either this option or that option is true

[tulip coyote]

Show that equivalence classes are either identical or disjoint

[open rain]

ohh that's what it means

| | are equivalence classes

[tulip coyote]

Well, assume they have a common element and use properties of equivalence relations

[tulip coyote]

Yea $[x] := { y \in X : y R x }$ usually for $R$ being the equivalence relation

[soft zealot]

chartbit

[tulip coyote]

As per here, assume $z$ is such that $z\in [x]$ and $z\in [y]$

[soft zealot]

chartbit

[open rain]

ok in my book it is only covered by 2 sentences

wait what is this example about?

@tulip coyote

is it to teach me how equivalnce classes work?

i am confused af

it says to either show they are the same or that they are disjunct

[tulip coyote]

Yep basically, they're telling you to show here that two equivalence classes are either identical or different

[open rain]

oh

[tulip coyote]

To do that, if we have this, we have that $zRx$ and $zRy$

[soft zealot]

chartbit

[tulip coyote]

Can you say something about x and y now?

[open rain]

i guess we had an equivalenz relation

i don't get where you got the z from

[tulip coyote]

Let them have a common element

[open rain]

but ok for R it means it's reflexive, symmetrical and transitive

[tulip coyote]

What we aim to show is that if they do, then we're forced to have the equivalency classes being the same

[open rain]

hmm ok i tried to watch videos on those classes can't find anything useful

you took a z out of the class [x] and [y]?

[tulip coyote]

Basically equivalency classes are the set of all elements that are equivalent to each other

[open rain]

and then compared the z R x and z R y?

[tulip coyote]

So if an equivalence relation is a way of saying things are the "same", an equivalency class is the set of each things that are the "same"

[open rain]

does this mean everything in that set is the same? like {1,1,1,1,1,1}?

[tulip coyote]

Not quite, e.g. let's take the relation $R$ on $\bZ$ to be $xRy$ if $(x-y)$ is divisible by 2

[soft zealot]

chartbit

[tulip coyote]

Then the equivalence classes basically turn into $[0] = {\ldots, -2, 0, 2, \ldots}$ (the even integers) and $[1] = {\ldots, -3, -1, 1, 3, \ldots }$ (the odd integers)

[soft zealot]

chartbit

[open rain]

ohhh

and the thing on the left is the class?

i have seen that one before yeah

[tulip coyote]

You can choose any other odd or even integer to represent each equivalence class as appropriate

[open rain]

ok wow this makes things even more complicated now -_-

[tulip coyote]

e.g. You could choose $[69]$ to represent the odd integers or $[420]$ to represent the even integers

[soft zealot]

chartbit

[open rain]

cuz in that example [0]and [1] would be disjunc

[tulip coyote]

Yep that's kinda the point of this exercise 😉

[open rain]

but in my exercise we don't know what R is

[tulip coyote]

Take any equivalence relation, and then you want to show that it'll hold

[open rain]

we have no info about R

[tulip coyote]

You don't need to know what the relation R is, only that it is an equivalence relation

[tulip coyote]

What properties of an equivalence relation can you use here?

[open rain]

ok i get that is the answer but why

[tulip coyote]

See these ones you said

[tulip coyote]

Why do the classes have to be disjoint or identical, or?

[open rain]

i mean i can't know

[tulip coyote]

https://tenor.com/view/yes-you-can-you-can-do-it-you-got-it-gabriel-iglesias-gabe-iglesias-gif-17555406

[open rain]

because you can enter different numbers or x or y

[tulip coyote]

Equivalence relations $\sim$ are such that:

$a\sim a$,

$a\sim b$ if and only if $b\sim a$, and

if $a\sim b$ and $b\sim c$, then $a\sim c$

[open rain]

yeah i get all of those

[soft zealot]

chartbit

[open rain]

but what changes if i show that for z R x or z R y

[tulip coyote]

You have that this is already true. Hint: use the second one on the first relation, then you can use the third one

What happens then?

[open rain]

z R x, x R z

[tulip coyote]

and then...?

(swap them around and you'll see something!)

Actually wait hang on no

[open rain]

x R z, z R y, x R y?

[tulip coyote]

Yes this!

[open rain]

hmmm

[tulip coyote]

You have that x is related to y, which is what we want

Do you see why that will force the equivalence classes to be identical?

[open rain]

but in this case it would be wrong right?

hmm

[tulip coyote]

Hmmm, not really, they're disjoint, which is what we said, they're either identical or disjoint

[open rain]

but in our case how do we know we can get that z ?

what guarantees that we get that z out of x and out of y

[tulip coyote]

What we said is that if two equivalency classes have a common element, then they must be the same

[tulip coyote]

We don't know that one exists, but as above, if it does, then they're the same

[open rain]

ah ok so it means we are not done?

we only say if that's the case then they are identical otherwise they are disjoint

[tulip coyote]

We pretty much are, provided you explain this here!

Just one element in common is enough to force them to be the same

E.g. consider the classes $[420]$ and $[0]$ from what I said before

[soft zealot]

chartbit

[open rain]

ah ok this was still just theoretical though right? we can't say anything further than that because we have limited info

[tulip coyote]

Pretty much, we've just shown what they said to, that two equivalency classes must either be the same or disjoint

[open rain]

ahh ok

[tulip coyote]

These two equivalence classes have elements in common

[open rain]

you are right

[tulip coyote]

But e.g. $[69]$ and $[420]$ don't under my relation beforehand

[soft zealot]

chartbit

[open rain]

i understand what they wanted to show with that task now

👍

[tulip coyote]

Nice nice, does that all make sense now?

[open rain]

yeah it does

but still i am not sure if i would have found that way to proof it myself

but i guess i should always check those definitions first

[tulip coyote]

Yea a good place to start would be to work with what you know, like the definitions of equivalency classes and equivalence relations here

Then see if you can use those to get anywhere, a lot of proofs are like that

There are some which are more complicated but something like this is just definition finessing (there is a subtle point I guess about why "one element in common is enough to force them to be the same" is enough, but meh )

[open rain]

yeah i think spamming examples will make me better in this

i noticed i understand things better by doing examples rather than just rewatching the lecture

thx dude 👍

[tulip coyote]

No worries, my pleasure

[final saddle]

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[weary mist]

hi, for integrations with improper fractions, is it possible to use only partial fractions instead of long division method for all functions?

[final saddle]

@weary mist Has your question been resolved?

[unreal anvil]

hello @weary mist
i am assuming you are talking about functions in the form of
P(x)/Q(x) where deg(P)>=deg(Q)
am i correct?

[weary mist]

yes

[unreal anvil]

well, i do always use partial fractions fashioned method to tackle such problems

but I'm not sure the name for it

e.g.
(x²+1)/(x+1)²

i would simply do
(x²+2x+1-2x)/(x+1)²

in stead of doing
(x²+1)/(x+1)==A+B/(x+1)+C/(x+1)²

[weary mist]

what about these type of function
(2x^3 + 4x^2 + 5) / (2x^2 + 1)

[unreal anvil]

oh, i will push them one by one , well, actually it's more like long division

for your example

[weary mist]

so technically i would still need to use long division

damn

[unreal anvil]

yea, but you can kust manipulate them this way

or

[unreal anvil]

this

I'll show both methods for your example

(2x³+x-x+4x²+2+3)/(2x²+1)
= ((2x³+x)-x+(4x²+2)+3) )/(2x²+1)
= x + 2 + (-x+3)/(2x²+1)

and

(2x³+4x²+5)/(2x²+1) == Ax + B + (Cx + D)/(2x^2+1)

[weary mist]

ah

[unreal anvil]

and solve for A,B,C and D

[weary mist]

thankyou sire

bless

.close

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[final saddle]

@south zealot Has your question been resolved?

[echo mason]

Try drawing a diagram.
Things will become much clearer.

[lethal totem]

What is a major segment of a circle

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