help-13

[jolly remnant]

then you plug x and y in and get a map from t to a real number

it should simplify down to something *t + something else where t_0 <= t<= t_1

for any a,b,c with a<=b<=c and f linear then f(a)<=f(b)<=f(c)

this is because if f is lienar it can be written mt+k

a<=b<=c
ma<=mb<=mc or mc<=mb<=ma (depending on sign of m)
ma+k<=mb+k<=mc+k or mc+k<=mb+k<=ma+k

but this is exactly what you want

(replace the expression with f(a), f(b), f(c)

[cedar kiln]

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[modest nebula]

oh okay so f is either increasing or deacreasing, (say increasing), then f(t_0) <= f(t') <= f(t_1)

however f(t_0) > 0, f(t_1) > 0, but f(t') = 0, contradiction

[cedar kiln]

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[slim zinc]

I HAVE A QUESTION

[slim zinc]

woops

https://i.imgur.com/NSVvn52.png

substituting this u value (8)

https://i.imgur.com/yVw6ijz.png

i cant get to (9)

i think i might be doing something wrong

my next idea is to try and brute force it by using the product rule

i will picture what i worked out in a moment

[cedar kiln]

@slim zinc Has your question been resolved?

[slim zinc]

<@&286206848099549185>

the first one is what i worked out by substituting u

[slim zinc]

<@&286206848099549185>

[cedar kiln]

@slim zinc Has your question been resolved?

[cedar kiln]

@slim zinc Has your question been resolved?

[cedar kiln]

@slim zinc Has your question been resolved?

[cedar kiln]

@slim zinc Has your question been resolved?

[slim zinc]

<@&286206848099549185>

[feral coyote]

no one knows how to do this

[torpid sable]

There seems to be some missing information

[slim zinc]

i can give the whole paper but it requires knowledge of general relativity

and probably wont give anymore information than i gave

cause thats all it mentions of it

maybe ill take it to a pure physics discord and someone can make sense of it

[torpid sable]

but like where did that other P looking character come from. Not the rho, but the other one

Yeah, that might be your best bet

[slim zinc]

it was just showing that the density is explicitely a function of pressure

i dont think its relevantr

[torpid sable]

okay, that's what I was guessing, that it's just another function

[slim zinc]

its the only equation in the entire paper i cant follow

and it looks trivial

ill ask a physics discord

thanks

.close

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[crimson meteor]

Is it possible to simplify this further??

[crimson meteor]

(x+y)^2 = x^2 + 2xy + y^2

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[west island]

Hey I have trouble understanding the last two steps

[west island]

I/2 is supposed to be I^2/2 right??

Full problem + what I did

<@&286206848099549185>

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[crimson pike]

i proved that empty set is a subset of P(A) - P(B) already but im not sure how to continue

[wet fossil]

hmm for (ii) notice that the empty set belongs to both P(A) and P(B) and P(A) - P(B) equals all the subsets of A which are not subsets of B

[crimson pike]

yea so i got to the point where since empty set is in P(B) and empty set is in P(A) the difference would result in P(A) - P(B) not having the element empty set

but from b)i) i proved that empty set is always a subset of P(A) - P(B)

so i dont really get the distinction between empty set not being an element and it being a subset

[wet fossil]

The empty set is a element of P(P(A) - P(B))?
P(A) - P(B) is a set so the empty set is a subset of it but it's not a element of P(A) - P(B)

[crimson pike]

ohhh

okay i get it now

bruh its like i just got enlightened

TY so much man

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[crimson sedge]

how do you prove that:

x(y) = z(y) + t

(if originally x = z + (t/y) * y)

[tropic oxide]

this looks strange...

can you show the original problem please

[crimson sedge]

External angle = 180 - ((n-2)180)/n

someone has already helped me solve this in a simple way that does involve that the rule I mentioned above but I am still curious

[tropic oxide]

???

[crimson sedge]

n = the side of a polygon

[wanton grove]

how are these two related at all

[crimson sedge]

first

[wanton grove]

is this supposed to be the measure of an external angle of a n sided regular polygon?

or something else

[crimson sedge]

lemme first edit it bcz i typed something wrong

now,
E = 180 -((n-2)180)/n
En = 180n - ((n-2(180)/n * n
En = 180n - (180n - 360)
En = 180n (-180n + 360)
En = 180n - 180n + 360
En = 360
N = 360/E

I wanna now why En = 180n - ((n-2)180) instead of En = 180 - ((n-2)180)

.close

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[zinc mural]

$\frac{dM}{dt(n-aM)}=1$

[wraith dagger]

Xiszt

[zinc mural]

where n and a are constants

would you say this ode is seperable?

like could you intergrate both sides to dt and say its a seperable equation

[mental trail]

I mean technically you could integrate it in this form

have you tried integrating and doing the change of variables u = 1/(n-aM)?

[zinc mural]

I have intergrated it yeah

i said it counts as a seperable equation but my friend says a constant doesnt count

he says it needs to be a function

[mental trail]

huh?

a constant is a function

f(t) = C if he likes functions so much

$\int \frac{dM}{n-aM} = \int f(t)dt$ where $f(t) = 1$

[wraith dagger]

rafilou2003

[zinc mural]

hmm yeah

Just learnt

I can perform intergrating factor methon when its a constant co-efficient right?

or am i not allowed and have to use method of undetermined co-efficients

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[bleak juniper]

Can anyone help me with this??

[tropic oxide]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tropic oxide]

do you understand the notation, at least?

[bleak juniper]

yep.

[tropic oxide]

right

so then $\sum_{d_i \leq 6} (6-d_i) \geq \sum_{i=1}^n (6 - d_i)$ should be obvious, no?

[wraith dagger]

Ann

[bleak juniper]

yes this is clear

[tropic oxide]

right

so then the other half can be written as $6n - \sum_{i=1}^n d_i \overset?\geq 12$ can't it

[wraith dagger]

Ann

[tropic oxide]

btw can you remind me what "order" means for a graph

cause i forgor 💀

[bleak juniper]

no of vertices

[bleak juniper]

clear mam

[tropic oxide]

right so at least 3 vertices

oh, it matters that G is planar actually.

theres a bound on the edge count of a planar graph that's linear in n, isnt there?

[bleak juniper]

can you just tell me how to do it.
This is my assignment . I just want to be done with it

[bleak juniper]

didnt get it

[tropic oxide]

i remember that there was a theorem that went like "if G is a planar graph on n vertices and e edges then e ≤ (some linear function of n)"

but i do not remember what that linear function was

[sweet crest]

e=n(n-1)/2 if you pair all n

[cedar kiln]

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[inland valley]

how would I set this up as a system of linear equations when to find the distance from 2 points in 3D space is a non-linear process

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[inland valley]

.reopen

[cedar kiln]

✅

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@inland valley Has your question been resolved?

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[crimson sedge]

Having trouble on this problem

[crimson sedge]

So I think the answer to the first one is 1 right?

But like

There is an open circle there

At -1 where I circled here

I guess I just don't really understand how to do this problem

[cedar kiln]

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@crimson sedge Has your question been resolved?

[stable oyster]

thats not a multiple choice question

each of that is a question you have to do

oh nvm i misread your message

[stable oyster]

the open circle doesn't matter

the function still approaches 1 as x -> -1 from both sides

[stable oyster]

so you're right about this

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Oh ok

Sorry for the late response byw

Btw

[crimson sedge]

Can we work on part b for it pleaseM

?

So I think it says x approaching 2 from the left side

[stable oyster]

thats what the negative thing is yea

[crimson sedge]

So like

How would I do that

Would it be 0?

[stable oyster]

its what it sounds like

[crimson sedge]

But like I'm confused since there's 2 graphs

[stable oyster]

where?

[stable oyster]

just view it as one function with a bunch of disconnected parts

yeah its a bitwise function so just google that to see roughly what it is

[cedar kiln]

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[bright karma]

do i replace f(a) with 1

[bright karma]

@spice kraken you needed

[spice kraken]

use the formula below

$f'(a)=\lim_{x\to a}\frac{f(a+h)-f(a)}{h}$

[wraith dagger]

WhereWolf(ping if needed)

[spice kraken]

wait what

should be $h\to 0$

[wraith dagger]

WhereWolf(ping if needed)

[bright karma]

huh

i cant change what they give me

[spice kraken]

bruh there are no x in the limit

[bright karma]

wat

oh

um

@spice kraken did i make a mistake

[spice kraken]

bruh what are you doing

first

change the limit

second you plugged in all the wrong values

[bright karma]

WHAT

[bright karma]

help 😭

[spice kraken]

h=x-a

what does h approach when x approach a

[bright karma]

what

huh

approach approach?

[spice kraken]

.

[bright karma]

im so confused

what did i do wrong

wher did u go 😭

[spice kraken]

h=x-a

and x -> a

then h ->？

[bright karma]

x?

[spice kraken]

ｎｏ

[bright karma]

idk what type of logic puzzle is this

a?

[spice kraken]

no

[bright karma]

bothh???

[spice kraken]

.

.

h=a-a

[bright karma]

how am i supposed to know

h = 0?

[spice kraken]

yes

[bright karma]

so h -> 0?

[spice kraken]

calculate $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

[wraith dagger]

WhereWolf(ping if needed)

[spice kraken]

change the limit from $\lim_{x\to a} \text{to} \lim_{h\to 0}$ because we changed variables

[wraith dagger]

WhereWolf(ping if needed)

[spice kraken]

yep

what is f(a+h)

tip: it equals f(1+h)

[bright karma]

f(1)

[spice kraken]

no

[bright karma]

but h approaches 0

[spice kraken]

f(1+h) approaches f(1)

doesn't equal

[bright karma]

then how do i find h

[spice kraken]

???

you don't find h

[bright karma]

x;yuol9867545364tgh

[spice kraken]

plug in 1+h for 5x^2-3x+1

[bright karma]

5(1+h)^2-3(1+h)+1

5+5h^2-3-3h+1

5h^2-3h+3

factor now?

oh

[spice kraken]

we have $\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0}\frac{(5h^2-3h+3)-3}{h}$

[wraith dagger]

WhereWolf(ping if needed)

[bright karma]

5h^2-3h/h

5h-3

so -3?

bcuz h approaches 0

[spice kraken]

good

[bright karma]

wait

how do we know f(a) = 3

[bright karma]

i thought f(a) = 1

[spice kraken]

??

a is 1

f(1) is 3

[bright karma]

howd -f(1) turn into -3

um

[spice kraken]

..

f(1) =3

-f(1) = -3

3

[bright karma]

how do we know that

OH

bcuz we plug in 1 into the equation

[spice kraken]

yes

[bright karma]

which gives 3

[bright karma]

ok so heres my work

alr tyyyy

[spice kraken]

you forgot to write f'(1)

[bright karma]

where

[spice kraken]

line 6

[bright karma]

where in line 6

if i need to write it in line 6 i need to write it in every line after?

[spice kraken]

left most

[spice kraken]

no

[bright karma]

f'(1) = [(5....

[spice kraken]

yep

[bright karma]

does that look better

[spice kraken]

look good

[bright karma]

oke ty.

.close

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[turbid spoke]

help

[turbid spoke]

.close

[cedar kiln]

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[worldly bluff]

Got wrong answer

[worldly bluff]

Not sure where I messed up

[spice kraken]

what's the correct answer?

[worldly bluff]

x=ab/(a-b-c)

[thorn hollow]

I think your answer will reduce to that if you factor

Maybe not -c hmm

[worldly bluff]

this is my answer written more clearly

[thorn hollow]

Oh you’ve got an overall negative yeah this works

[worldly bluff]

isnt this all i can factor

how does that help me?

[thorn hollow]

You can factor the denominator by grouping

c(a+b) + (b+a)(b-a)

To (a+b)(c+b-a)

[thorn hollow]

Difference of squares on the right

[worldly bluff]

what is that

[thorn hollow]

a^2-b^2=(a+b)(a-b)

[thorn hollow]

And you can factor (a+b) out then cancel it

[worldly bluff]

that is insane wtf

thanks

.close

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[small igloo]

I need help on solving this equation csc^6 10º - cot^6 10º - 3csc^2 10º x cot^2 10º

[small igloo]

how do i use the bot thingy

[carmine whale]

U mean calculator ?

If u r talking about Wolfram alpha

,w csc^6 10º - cot^6 10º - 3csc^2 10º × cot^2 10º

[small igloo]

not that

the bot that uploads the equation as a png

[carmine whale]

Oh

$\csc^6 10º - \cot^6 10º - 3 \csc^2 10º \cdot \cot^2 10º$

[small igloo]

yeah that

[wraith dagger]

•Iᴛᴀᴅᴏʀɪ•

[thorn hollow]

$\csc^6 10^\circ - \cot^6 10 ^\circ - 3 \csc^2 10 ^\circ \cdot \cot^2 10 ^\circ$

[wraith dagger]

Alti

[small igloo]

that's better

I know to solve this equation I need to use the Pytaghorean identities

I believe I should be dividing the equation by sin^2 first

actually hm

[cedar kiln]

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[ornate crescent]

How exactly would I go about this problem? I imagine I would integrate from 0 to 6, but other than that I'm very confused on what f(x) would be here.

[muted bear]

What is the function for a circle (or really a semicircle)

[ornate crescent]

pi(r^2)?

why a semi-circle?

[muted bear]

The function, not the fprmula for area

[ornate crescent]

I'm unsure.

[muted bear]

does $\sqrt{x^2-r^2}$ look familiar

[wraith dagger]

GarlicB

[ornate crescent]

It does not.

[muted bear]

I forgot the y=

ok how about $x^2+y^2=r^2$

[wraith dagger]

GarlicB

[ornate crescent]

That does look more familiar, yes.

[muted bear]

try to isolate y

[ornate crescent]

I'm not sure this is something I would say I've worked with before though.

I see how that would become the function you created earlier.

[muted bear]

What is r in this problem?

[ornate crescent]

is this the function for a circle?

[muted bear]

Function for a semicircle

Because of the squareroot, we have to choose one of either positive or negatibe

In this case, i chose positive

So we can cheat by multiplying by 2

[ornate crescent]

Ah that makes sense.

[muted bear]

So then do you know how to use the cross section formula in tjis case?

[ornate crescent]

I'm a bit confused about that as it says the slices are squares.

Or rather, unsure how to incorporate this information with the rest of the problem.

[muted bear]

ok lets look at one slice at say x=1

Im not gonna simplify anything

we have the length of the line segment with endpoints on the circle is $2\sqrt{9-1}$

[wraith dagger]

GarlicB

[muted bear]

Excuse my idiocy, this formula is for a hyperbola

The r and x should be switched for the semicircle

[ornate crescent]

so r^2-x^2 under the root?

[muted bear]

So $2\sqrt{9-1}$ is the side length of the square

[wraith dagger]

GarlicB

[muted bear]

the area of said square is 4(9-1)

[ornate crescent]

Wait how do I know that that is the side length of the square?

That was using the formula of the circle right?

[muted bear]

Yes

[ornate crescent]

Which is how we obtained this equation initially.

So, by squaring that, that just gives us a square?

[muted bear]

Yes

[ornate crescent]

This sound ridiculous to type out but

[muted bear]

Its good to type it out

[ornate crescent]

Interesting. Okay.

[muted bear]

Help it make sense

[ornate crescent]

Yeah.

I mean "so by making it a square makes it a square" feels very silly to ask. lol.

But, okay cool. that is a good information to have.

[muted bear]

So in general, to find the volume when the cross section is squares is the formula (let me type out the tex)

$\int_a^b(f(x))^2dx$

[wraith dagger]

GarlicB

[muted bear]

So you know your bounds and the function then plug and chug

[ornate crescent]

and f(x) here is the formula of a circle. which can be derived with y^2+x^2=r^2

Correct?

and 2*int from 0 to 3?

[muted bear]

Yeah

[ornate crescent]

all in all this is 2int 0 to 3 (2sqrt9-1)^2

squared.

[muted bear]

$2\int_0^3 (2\sqrt{9-x^2})^2dx$

[wraith dagger]

GarlicB

[ornate crescent]

Okay so I'm a little confused on where to find x here.

in the function of a circle, where would x come from?

thinking of this on a 2d plane makes sense.

[muted bear]

recall x^2+y^2=r^2

y=sqrt(r^2-x^2)

[ornate crescent]

but with the given information I'm not following.

[muted bear]

y=f(x)

f(x)=sqrt(r^2-x^2)

[ornate crescent]

yeah that makes sense.

I think I'm just unsure where I would find x with the information given.

[muted bear]

What do you mean by find x?

[ornate crescent]

Oh you know what.

I'm smooth braining hard.

lol

nvm.

This does make sense.

Thank you so much! This actually fills in a lot of gaps I was having with some of the more recent material.

[muted bear]

Youre very welcome!

[ornate crescent]

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[late wadi]

hi, i just have a quick question about this problem:

i initally say that x is NOT in the set of anbnc (because it is in the complement)

after this, if i was to split anbnc into something like

x is not in the set of a, x is not in the set of b, x is not in the set of c

are they connected by ands? or by ors?

[late wadi]

i see that the formal definition of an intersection is A ∩ B = {x ∣ x ∈ A ∧ x ∈ B}.
but does this switch when x is NOT in the set?

[flint plinth]

if it's not in A∩B∩C, then it's not in all of them, so there's at least one that it's not in

[late wadi]

hmm

[flint plinth]

so it's not in A, or it's not in B, or it's not in C

[late wadi]

OHHH

that makes so much more sense

duh

tysm

[flint plinth]

sure

[late wadi]

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[flint whale]

Not sure if this is a jump discontinuity or a infinite discontinuity.

[royal loom]

infinite discontinuities only require one side of the limit to approach infinity

jump discontinuities require both sides to approach finite values, yet that these finite values are not equal to eachother

[flint whale]

So it would be a non removable infinite discontinuity?

[flint plinth]

if the limits on each side are different, it's not a removable discontinuity

[flint whale]

Oh ok thanks

So it would be the same for this as well then?

[royal loom]

idk what do u think

[flint whale]

yes

also

would you consider a piecewise function

discontinuous outside of its domain?

[royal loom]

@solid juniper

you would consider it undefined

[flint plinth]

does it make sense to talk about continuity or discontinuity outside of the domain?

the definition of continuity at a point requires the function to be defined at that point

[solid juniper]

[cedar kiln]

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[odd sierra]

wat im doing wronog 😭

[modern compass]

are you converting the .83 into seconds?

[odd sierra]

I guess

That is what i am stuck

[modern compass]

multiply .83 by 60 to get the number of seconds

[odd sierra]

49.80?

so if i put 50 maybe it work

[modern compass]

seems reasonable, yeah.

[odd sierra]

Perfect it work

Thnx

.close

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[tropic oxide]

,rccw

[uneven siren]

10. I get it that the answer is b

But the mathematical proof is something like Vct= Vc-Vt

Just wanna know why its minus here

[tropic oxide]

imagine you're on the train

compared to a stationary observer, all speeds as observed by you are shifted by the same vector

your own speed is v_t to the stationary observer (bc you're on the train), but 0 to you

so that shift is -v_t

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[somber talon]

wouldn't the answer be $5?

the answer booklet says $6

300cm of wire means 3m of wire

$2 for every wire, means, $2 for 1 metre

now we have 2m left

another $2 spent

1m left

spent $4

oh I got it ty

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[dreamy zenith]

if i calculate the direction of a vector thats in quadrant 4 and it comes out as -16.11 can i write the direction as 343.89 degrees instead of -16.11 degrees ?

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[stray otter]

how is this even logical?

[stray otter]

these 2 parts are the same thing isn't it?

[crimson sedge]

Well, yeah, kinda

The line on the right is also the extra block peeping out

1s

[stray otter]

oh

I see

yeah that makes sense

[sweet crest]

some cubes have all except one side transparent

[stray otter]

I got it now

thx

.close

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[green vapor]

Hi, I am writing the proof of the triangle inequality but I am stuck on this part and not sure how to progress. I did the other two cases where a < 0, b <0 and a>=0,b>=0. Not sure how to finish this one.

[green vapor]

<@&286206848099549185>

[sweet crest]

what's |a| when a<0?

[green vapor]

-a

[sweet crest]

so where is the problem?

[green vapor]

I just dont know how to finish the proof

I need to get it to |a+b| on the rhs

[sweet crest]

so you need to prove b-a>=|a+b| a<0, b>=0?

[green vapor]

i am trying to prove |a+b| <=|a|+|b|

[sweet crest]

well, you already proved |a|+|b|=b-a a<0, b=>0

[green vapor]

yea

i dont know how to move forward to finish it

[sweet crest]

look at cases |a|>b and |a|<=b

here #help-13 message

[green vapor]

wdym

wouldnt i need to create subcases where a+b>=0, a+b<0

[sweet crest]

that's identical

cos a<0 => |a|=-a

[green vapor]

yeah

[sweet crest]

so look at those two

[green vapor]

ok

honestly idk man

i have been working since yesterday morning

im running on 3 hours of sleep

and this assignment is due in 2 hours

[sweet crest]

time scheduling problem

[green vapor]

perhaps

i forgot this assignment existed until 2 days ago

so it was unfortunate

[green vapor]

Wdym tho

do i create those subcases?

and even if i do how would i write that

[sweet crest]

suppose a+b<0 then solve

[green vapor]

ok

ill do both and ill ask for confirmation

[sweet crest]

*prove that it holds

[green vapor]

gotcha

[cedar kiln]

@green vapor Has your question been resolved?

[green vapor]

@sweet crest

Is this good?

changed it a bit

but i think thats all?

@sweet crest do you think this is good?

if so i think im done

[sweet crest]

yeh, looks messy but ok

[green vapor]

ok sick

[sweet crest]

0>=-b

[green vapor]

wdym

[sweet crest]

Since b>=0

[green vapor]

yea

oh

i see

[sweet crest]

there is a simpler way btw

[green vapor]

you mean like this?

[green vapor]

well too late for that now

wait that makes no sense

why is it 0>=-b

[sweet crest]

-a<=|a|<=a same for b, add them up -(a+b)<=|a+b|<=a+b and a<=|a|

So |a+b|<=a+b<=|a+b|

wops

[green vapor]

i dont get it

is my proof correct though?

[sweet crest]

-|a|<=a<=|a|

[green vapor]

or is thre still something wrong?

[green vapor]

yeah i learned that before

[sweet crest]

-(|a|+|b|)<=a+b<=|a|+|b|

[green vapor]

thats the simpler way right?

welp nothing i can do about it now

as long as my current proof is ok im fine

[sweet crest]

then you prove |b|<|a| <=> -|a|<=b<=|a|

yeah, should be fine

[green vapor]

ok thank you

[sweet crest]

except <= and <

[green vapor]

wdym?

did I mess up the usage?

where?

[sweet crest]

Since b>=0 b>0

[green vapor]

this?

or this

[sweet crest]

both

[green vapor]

oh

whats the right way then

i dont get how the first one is wrong

oh

its supposed to be >=-b right

[sweet crest]

yeh second one is ok

[green vapor]

yeah i see where i messed up on the first one

[green vapor]

ok

im gonna hand it in now

i should start doing this from an earlier date now

finally done

my god

[cedar kiln]

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[floral pagoda]

How can I show that matrix A is the root of the equation?

[uneven matrix]

plug it in and calculate

[floral pagoda]

I need to show it on paper, any suggestions?

also west coast>>>

[uneven matrix]

my suggestion is to deal with it

its not as bad as u think

[uneven matrix]

idek bro im just down bad and hit play on lana del ray spotify

[floral pagoda]

yeah I understand, I just started learning matrix

so im kinda confused where to even start

[uneven matrix]

oh just calculate A*A - 6A + detA * E

and itll end up being tthe 0 matrix

[floral pagoda]

okay thank youuu

what does E mean tho?

[uneven matrix]

the identity matrix

| 1 0 |
| 0 1 |

[floral pagoda]

ah ok thanks

appreciate you spending your time explaining

[uneven matrix]

np

[cedar kiln]

@floral pagoda Has your question been resolved?

[floral pagoda]

ok I got to this part, what should my next step be?

[uneven matrix]

you have a mistake

you calculated -6A and then did -(-6A) --> +6A

[floral pagoda]

oh bruh

I'll fix it

[floral pagoda]

thanks for pointing it out, should be good now

[uneven matrix]

once again you have a mistake, you copied A^2 incorrectly

[floral pagoda]

ok, so its 0 instead of -36, fixed👍

what should my next step be?

[uneven matrix]

now youd multiply E by the determinant of A just like you did with -6A

and add it again

[floral pagoda]

🫡

[uneven matrix]

yep

[floral pagoda]

I am very gratefull for your help. How may I prove that the matrix is the square of the equation?

[uneven matrix]

you mean root?

you just did

[floral pagoda]

yes the root

[uneven matrix]

say p(x) = x^2 -6x + |x|E then a root means that p(A) = 0

and you just showed that p(A) = 0

[floral pagoda]

ohhh ok thanks a lot man!

you saved me so much time

[uneven matrix]

no problem!

[floral pagoda]

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[tired violet]

I need to derive a closed-form expression for this summation.
I tried using partial fractions but at last i obtained some harmonic numbers which do not have any known closed-form expressions.
How can I solve this?

[neat dune]

Mind if u show what u get from partial fractions

I sorta had to work with this sorta thing to find $\sum_{n= 1}^\infty \frac{1}{\sum_{m= 1}^n m^2}$ recently

Anyways if ur not taking to infinity and there isn't any telescoping going on there probably isn't any pretty closed form

Just a bunch of Euler constant and digamma function ish u can find by inputting into wolframalpha

However at least for my problem, when you take the series it turned out to be 6(3 - 2log(4)) which was somewhat pretty

I kinda did it in a complicated manner but you might have an easier time using the aymptotics of harmonic numbers

[wraith dagger]

992qqoloy

[tired violet]

this is from wolframalpha lol

[crimson sedge]

this is correct i think

[neat dune]

oh yeah that's probably the prettiest you can get , but if you take to infinity there's telescoping that cancels out most terms but the first 4 from the sum 1/(4(j-2))

[crimson sedge]

its not that difficult

yu[

yup

just expand and this will come

i tried a different approach tho

1/j^-4=1/(j-2)(j+2) and then plug values from 3 to n and terms will start getting cancelled and ull be left with 8 terms

which on simplifying will give the answers i suppose

[neat dune]

my thing had partial fractions on a cubic cus sum of squares is (n(2n + 1)(n+1)/6 so that's why it was harder I guess

[crimson sedge]

cube?

[neat dune]

yeah cus 3 factors of first degree

[crimson sedge]

first as much as i remember square is n(2n+1)(n+1)/6

[neat dune]

It partial fractioned out to like 1/n + 1/(n + 1) - 4/(2n +1)

[crimson sedge]

cubic is (n(n+1)/2)^2

[neat dune]

no im saying the closed Form of sum of squares is a cubic poly

[crimson sedge]

oh yeah

yeah

[neat dune]

I don't wanna try to figure out what the sum of the reciprocal of the sum of cubes is

[cedar kiln]

@tired violet Has your question been resolved?

[neat dune]

But yeh it was really fun cus then you could split into like (1/n - 2/(2n + 1)) + (1/(n+1) - 2/(2n + 1)) and then with some slight correction (hence the +3) you'd have the negative of partial sums of the alternating harmonic series, minus the 2nd half of the partial sum of the harmonic series. Then take to infinity, so ln(2) for the alternating, and u could calculate the other value by rewriting it as a Riemann sum and then taking the integral from 1 to 2 of 1/x, which was also ln(2)

And that's how u got 6(3 - 4ln(2))

[cedar kiln]

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[hollow igloo]

I need help

[hollow igloo]

Do I do

7x+28+x-9=180

?

[crimson sedge]

29?

[hollow igloo]

oh mb i meant 28

[crimson sedge]

then yes

[hollow igloo]

x=20

what do i do next?

[crimson sedge]

interior angle of the polygon? is this the full question

[hollow igloo]

yup

[crimson sedge]

ah ok so ABC is a part of a regular polygon

and you've found x

[crimson sedge]

what would be the angle then

[hollow igloo]

i got 168

[crimson sedge]

yes

[hollow igloo]

not sure how i am meant to get the number of sides

[crimson sedge]

you divide the exterior angle by 360

do you see the exterior angle

[hollow igloo]

x-8?

[crimson sedge]

yes

let me try to draw for a simpler example

[hollow igloo]

okay thanks

[crimson sedge]

ok

say we have an equilateral triangle (all sides equal to 60 degrees)

[crimson sedge]

what would be the exterior angle

[hollow igloo]

120

[crimson sedge]

using that do you see how we can get the number of sides

[hollow igloo]

to be honest

nope

[crimson sedge]

.

[hollow igloo]

so 360/120?

[crimson sedge]

what do you get

[hollow igloo]

3

oooooo

[crimson sedge]

see

[hollow igloo]

so 360/x-8?

[crimson sedge]

and what is x-8

[hollow igloo]

12

[crimson sedge]

so how many sides

[hollow igloo]

30 💀

[crimson sedge]

yes

[hollow igloo]

30 sided polygon 💀

thanks so much

[crimson sedge]

np np

[hollow igloo]

is trianle ABC right angle?

[crimson sedge]

how can you tell?

[hollow igloo]

im just guessing tbh

it looks like a right angle triangle

but is there any way i can be sure

[crimson sedge]

bad

clue: pythagoras

check if it works

[hollow igloo]

👍

ur very smart

thanks again

[cedar kiln]

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[warm tree]

really confused by what im doing wrong here, i've attempted variations of this problem. I feel like I'm applying the product rule correctly, so ig there's an algebra mistake somewhere or something

just saw it, didn't carry the half for some reason

.close

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[warm tree]

.reopen

[cedar kiln]

✅

[dire geode]

Simplify the function before finding the derivative

[warm tree]

.clsoe

.close

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[pseudo hinge]

hi, I would like help with these please

[pseudo hinge]

i really couldnt get my head around this one

pls ping me if you respond (sorry ik thats rude)

[paper robin]

for the first one start by completing the cone and getting its volume

[pseudo hinge]

how would i complete the cone

[paper robin]

like draw the whole cone

[pseudo hinge]

sure

but then, how would i know the height

(sorry if im asking really dumb questions)

[paper robin]

similarity of triangles

[paper robin]

oh no they're not dumb dw

[pseudo hinge]

right, i don't know anything about that

okay, thats fine

i'll study similarity of triangles

can you tell me something about the second one

[cedar kiln]

@pseudo hinge Has your question been resolved?

[pseudo hinge]

<@&286206848099549185>

sorry for the ping

[pseudo hinge]

but can someone help me with these

[sweet crest]

https://discord.com/channels/268882317391429632/1152983813903876187

w8, not the same

[gilded sonnet]

Well for the first one just count the volume

The formula is V=\dfrac{\pi h}{3}({r}{1}^{2}+{{r}{1}}{{r}{2}}+{r}{2}^{2})

$V=\dfrac{\pi h}{3}({r}{1}^{2}+{{r}{1}}{{r}{2}}+{r}{2}^{2})

idk how to use the bots o nthis server

It's just this

@pseudo hinge

[cedar kiln]

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[ornate wadi]

Trying one more time, as I couldn't grasp from the last explanation before a bot closed the channel.

[ornate wadi]

I'm a math teacher in a small school, but trained in English--I say that to point out that while I'm passionate about math, I'm not extensively formally trained. That said, I'm motivated to pass on to students why certain patterns/methods work, not just that they do. I need to make sense of it for myself before I can do the same for others, obviously. Considering the photo/snip below:

https://imgur.com/a/S32fYFD

...how do I explain to students, mathematically, why I can simplify before the multiplication, instead of multiplying "straight across" the fraction, and being left with a very awkward fraction to reduce to lowest terms.

[cinder venture]

They are divisible by each other

[ornate wadi]

Advice from previous respondant that I couldn't quite make heads or tails of how it helps break things down: do the explanation of fraction multiplication in the opposite direction to factor 21/44 into 7/11 and 3/4
multiplication is associative
multiply 11/7 with 7/11 to get 1

[ornate wadi]

I get that multiplication is commutative, but isn't there division at play as well?

[cinder venture]

That’s now what I meant

You’re asking why you can divide those fractions, simplifying them before multiplying no?

[ornate wadi]

Yeah, specifically the numerator of one fraction by the denominator of another, before doing the multiplication?

[cinder venture]

Do you agree that if you look at 21 and 7 they both have something they can be divided by?

[ornate wadi]

Yes, 7

and 11 and 44 by 11

[cinder venture]

Yeah, so you can simplify before multiplication because the values have terms they can be divided by

[ornate wadi]

Coming from a different angle, the muliplication at play is essentially multiplying the result of two division problems (fractions) together. Written like this: (-11/7)*(-21/44). I can't rationalize how I can "reduce" the terms of the "division" problems when they're in different parts of the equation.

I recognize I'm not being terribly clear. I understand if you can't make heads or tails or my "roadblock"

[wraith dagger]

Dyssrupt

[junior dome]

now you can simplify.

hope that shows why this works.

[ornate wadi]

It shows me that it works, but *why * am I able to swap denominators, mathematically? Perhaps it's easier to break down/explain using the multiplication of the fractions (2/3) * (15/16). I've tried working out why this concept works using those simpler numbers, with no success.

[cedar kiln]

@ornate wadi Has your question been resolved?

[worn kraken]

Perhaps it would be clearer in a general form, where you have $\frac{a}{b}\cdot \frac{c}{d}$. That is equal to $a \cdot (1/b) \cdot c \cdot (1/d)$. \ Because multiplication is communative, we may rearrange it to get $a \cdot (1/d) \cdot c \cdot (1/b)$, which is equal to $\frac{a}{d} \cdot \frac{c}{b}$.

[wraith dagger]

Calculustache

[cedar kiln]

@ornate wadi Has your question been resolved?

[ornate wadi]

@worn kraken that was so very helpful! Thanks! For my Grade 8 students, that's probably a bit much, but it helped me make sense of it!

I honestly spent the last hour in a circular loop with ChatGPT trying to explain that he commutative property applied to the operation in the below picture, without breaking down how!

.close

[cedar kiln]

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[cedar kiln]

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[dreamy zenith]

$4cm+10cm/s^2$

[wraith dagger]

putridplanet

[vestal bear]

what

[lucid topaz]

no

[dreamy zenith]

if im finding the displacement how do i solve this

[vestal bear]

show the entire question

[dreamy zenith]

do i just get rid of s^2 because displacement is distance

[vestal bear]

no

show the entire question

[vestal bear]

is the s^-2 just there because of the t^2

[dreamy zenith]

no

[vestal bear]

this is a weird vector

you can start by plugging in t=2 and normalizing it

[dreamy zenith]

how did the s^2 disappear

[vestal bear]

the t^2 cancels it out

[dreamy zenith]

but you plug in 2 for t to get 4

which makes 2.5 10

[vestal bear]

t^2 has units of s^2

[dreamy zenith]

?

[vestal bear]

t is seconds no

so t^2 is seconds squared

[dreamy zenith]

ok

ok

[vestal bear]

s^2/s^2=1

this can also be turned into paramterics if you know what those are

I'm more familiar with that

[cedar kiln]

@dreamy zenith Has your question been resolved?

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[opal crystal]

How do I do this without graphing it?

[thorn lotus]

Are you familiar with translations, scaling, etc.?

[opal crystal]

I am not

[thorn lotus]

ok. you're familiar with y = x^2, no?

[opal crystal]

Yes, I am

[thorn lotus]

Good. Translation just means moving it around the graph

so if you have (x-a)^2, that means you're moving y = x^2 to the right by 'a' units

so if you have (x-1)^2, that's x^2 moved to the right by 1

[opal crystal]

Would (x+2)^2 mean it's moving one to the left?

[thorn lotus]

Yes! Exactly!

[opal crystal]

Ahh so it's backwards ha, interesting

[thorn lotus]

Now let's do up down

[opal crystal]

I meant x+1*

But same difference

[thorn lotus]

Yeah I got what you meant

[opal crystal]

That was my next question

Is that what the '-1' is?

[thorn lotus]

hmm?

what do you mean

[plain umbra]

Yes, the -1 shifts the whole parabola one unit down

[opal crystal]

In the equation here, would the -1 be the up/down?

[thorn lotus]

Ah, then yes

[opal crystal]

Okay, that makes sense

[thorn lotus]

The 8 scales the y =x^2, making it thinner or wider

Also

the sign of that 8 dictates where it opens

[opal crystal]

How exactly does it do that?

[thorn lotus]

if it's positive, the parabola opens up, negative means it opens down

Imagine y = -x^2

what does that look like

,w plot -x^2

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by toomanycooks_

[opal crystal]

Parabola

[thorn lotus]

It opens downwards because of the minus sign

[opal crystal]

Yes

[thorn lotus]

if it's positive, it opens upwards

I do want to note though that for all of this to work, your parabola has to be in the form $a (x-h)^2 + k$

[wraith dagger]

TooManyCooks

[thorn lotus]

Anyway, so where does your parabola open?

up or down?

[opal crystal]

Up, since hte 8 is positive

[thorn lotus]

Good. Now the vertex is the determined by the left/right and up/down translations you did

in this case, where is the new vertex

[opal crystal]

(2,-1)

[thorn lotus]

excellent

can you tell me what are the options for that last box?

[opal crystal]

Minimum and maximum

[thorn lotus]

ah ok

so what is f(2)

[opal crystal]

-1 i believe

[thorn lotus]

right

so, knowing that the parabola opens upwards, would that be a minimum or maximum?

[opal crystal]

So minimum since it's opening upwards

[thorn lotus]

Yes, exactly

[opal crystal]

I'm just confused on the 8 now. I know it affects the upwards/downward, but how can I figure out exactly how it affects how thin/thick it is?

[thorn lotus]

ok, let's look at the plot then

,w plot y = x^2

[wraith dagger]

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Upgrade to WolframAlpha Pro!

Requested by toomanycooks_

[thorn lotus]

,w plot y = 8x^2

[wraith dagger]

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Requested by toomanycooks_

[thorn lotus]

Look at the y-values

8x^2 rises much faster

[plain umbra]

Wolfram did a poor job at explaining it, it changed the axis scale lol

[thorn lotus]

Lemme make a plot

there

i hope that makes it clear

[opal crystal]

Ahhh yes, that makes it a lot more clear

So positive = thinner
negative = wider
right?

[thorn lotus]

no

[opal crystal]

Oh

[thorn lotus]

positive or negative jsut tells you whether it opens up or down

[opal crystal]

Ohhh right

[thorn lotus]

if a>1, it's thinner

if 0<a<1, it's wider

so if a = 1/2

that's wider

[opal crystal]

I see, that makes sense. I forgot about how the negative affects it here lol

[thorn lotus]

There

might be a better way to visualize

[opal crystal]

Sweet, thank you all for the help! That cleared things up a lot

[thorn lotus]

👍

[opal crystal]

Yes! That's perfect, it helps seeing it on a graph

You all have a good one:)

.close

[cedar kiln]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[flat sparrow]

how'd they get 1/4?

[surreal cave]

$\left(\frac{f}{g}\right)(-1)=\frac{f(-1)}{g(-1)}$

[wraith dagger]

MrFancy

[flat sparrow]

?

[rain drift]

What is f(-1)? What is g(-1)?

On the graph I mean. What are the values?

[flat sparrow]

Ah i see great

thanky ou

how abourt b and a