help-10

[worthy cargo]

sqrt (3y^2-9x+4) isn't equal to what you wrote

[oblique nebula]

I kinda jumped through 20 steps to get in here so the rooting didn't happen on y

[worthy cargo]

so you didn't do

$$\sqrt{3y^2-9x+4}$$

[warm shale]

WOWWA

[oblique nebula]

Absolutely not, but from this perspective it looks easier

[worthy cargo]

$$\sqrt{3y^2-9x+4} \unequal \sqrt{3}y - 3x + 2$$

[oblique nebula]

I must have made a mistake somewhere :(

[warm shale]

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

[worthy cargo]

lol OK wtv

u get the point

anyways

once u solve for y

just plug it in directly to y^2

[oblique nebula]

Did this in step 2, the other 18 was just chaos

Well, I'll try over again

but thanks anyway

.close

[obtuse pebble]

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[worthy cargo]

@oblique nebula shouldn't be too hard if you need help I can walk you through it

[oblique nebula]

I'll try on my own

[obtuse pebble]

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[wicked ice]

how would i do these

[timid silo]

Do you know what asymptotes are

[wicked ice]

like the places where the line never meets

[royal basin]

given you know that each of these has equation xy = k

this is in fact piss easy

[hollow dove]

(just plug in the given points and solve for k)

[wicked ice]

so just 1 x 7 = k

oh

i thought it was something much harder

[hollow dove]

rectangular hyperbola are pretty nice

[wicked ice]

.close

[obtuse pebble]

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[woeful talon]

Question 13

B, c, d

Circle geo

I attempted b)

I got a+b = 90 but I don't think thats what they want

[hollow dove]

that seems reasonable to me

[cloud igloo]

Beautiful number 14

[woeful talon]

How would I proceed for C)

I wrote ce = 10-x

Surely this isn't right?

[hollow dove]

that's fine

[woeful talon]

ok, for d)

I) it's just xy/2

[hollow dove]

yeah

right angle triangle and all

[woeful talon]

For ii) I did (10-x)(10-y)/2

[hollow dove]

looks good

you can expand that and simplify it a bit I guess

[woeful talon]

(100-10x-10y+xy)/2

[hollow dove]

could write that as 50-5x-5y+xy/2

eh, it's not really much simpler

I think either form would be a perfectly fine answer

[woeful talon]

nvm I got

it

[woeful talon]

I have another question

[hollow dove]

,rotate

[woeful talon]

For a)

I did

8!/2!4!4!

For b) it's 2

And is C not the same as a?

[hollow dove]

C is slightly different, because we don't care about the ordering of the coins this time

all the ways in a) would count as a single way in c)

[woeful talon]

I'm still confused

[hollow dove]

so, HHHHTTTT and HTHTHTHT would count as the same way for part c)

because you have 4H and 4T either way, but they have different orderings, so they count as different for the purposes of a)

ah, wait

you can interpret it either way

and I think it actually does want ordering for part c (because the way it's used in part d implies you need ordering)

so, c is still different from a, because c doesn't require 4H and 4T

you could have 8H for part c)

[woeful talon]

What

[hollow dove]

c isn't the same as a

a wants the number of ways to order 4 heads and 4 tails

[woeful talon]

Yes

[hollow dove]

c doesn't require 4 heads and 4 tails

you could have 8 heads as a way in C

[woeful talon]

Oh

I get it

so is it just 8!/2!

[hollow dove]

I don't think so

each coin can either be head or tail, so you're making 8 binary choices

[woeful talon]

Is it just 2^8

Bruh

[hollow dove]

yeah

[woeful talon]

Ok for E)

I)

8!/5!3!2!

[worthy cargo]

wait

it's not for A I think

since H and T are indistinguishable

[woeful talon]

What's not for a

[worthy cargo]

2^8

oh

are u guys on c

[woeful talon]

Yeah a is 8!/2!4!4!

[hollow dove]

there are nCr(8,4) ways to place the 4 heads amongst the 8 places, with each arrangement also fully determining the tails

but it's symmetric in heads and tails, so divide by 2 (which agrees with 8!/2!4!4!)

[woeful talon]

So for E)

Is it this

8!/5!3!2!

Because U can arrange heads in 5 ways but it's the same and arrange tails 3 ways but it's the same and those are 2 objects

[hollow dove]

it wants a probability

so your answer should be between 0 and 1

[woeful talon]

Oh

so is it that the other way around

reciprocal of it

[hollow dove]

you've found the number of arrangements of 5 heads and 3 tails

part c) is the total number of arrangements

just divide the two

[woeful talon]

Oh

[worthy cargo]

ok sorry I don't mean to interrupt and i could totally be wrong but isn't A just 8!/4!4!

or 8 choose 4 , I don't really get why we are divisng by 2

this is the same question as how many 8 letter words can be formed with HHHHTTTT

[hollow dove]

if you choose the first 4 letters

you have HHHHTTTT

ah wait, I see what you mean

yeah, we count those differently

I was thinking as counting inverted permutations as the same

[woeful talon]

So you don't divide it by 2

But I don't understand

Why

[hollow dove]

how did you derive your solution to a)?

[woeful talon]

ok so

8!

Because we have 8 options at the start, 7 and 6 etc

But this doesnt account for the dupkicates

so there have to be 4 heads and 4 tails

You divide it by 4! Because U can arrange each of those 4 diff ways

Same for tails

[hollow dove]

yeah, we can just stop there (so it's just 8!/4!4!)

because if we swap the positions of heads and tails

then it does count as a different way in this question

I made the same mistake

[woeful talon]

oh

wait so the reason I divide by 4! Is right

Can you fully explain

[hollow dove]

yeah

8! is the number of ways to arrange the 8 coins

but we have 2 groups of 4 that are indistinguishable

[woeful talon]

Ok

[hollow dove]

so each group can permute amongst themselves in 4! ways

so we've overcounted by exactly 4! * 4!

because they can permute independently

[woeful talon]

Makes sense now

For E) vi)

Is it

2n!/n!(2n-n)!/2^2n

[hollow dove]

uhh

brackets, firstly

2n! is different from (2n)!

and also the stacked fraction is unclear

(2n)!/n!(2n-n)! is nCr(2n,n)

ah it's just rearranged from mine slightly

yes, they look like the same thing

you can rewrite (2n-n)! as just n!

and clean it up a bit more

[woeful talon]

Ok

Ty

U the best

[obtuse pebble]

@woeful talon Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[devout harness]

I have a question purely out of curiosity; is there a turing machine for every math conjecture/theorem such that the conjecture is true if and only if the associated turing machine halts, and such that proving wether or not the turing machine halts proves/disproves the conjecture? How would you prove or disprove this? I know that there is a turing machine that halts (or doesn't halt, not sure anymore) if and only if the goldbach conjecture is true, though I don't know what turing machine or the proof of that

[oblique nebula]

Well, that's a famous question Turing himself disproved like this:

Imagine a machine h, that can predict if a theorem is true or false, it makes the machine attached to it halt only if the theorem is false, and keeps going forever, if it is true.

Now imagine c , that copies the input and gives it as output

And lastly, imagine n , that negates the input (if it recieves a signal to halt from c , it keeps going and vice versa)

If you put these machines in the order h -> c -> n You get a more complex machine (I'll call it h+)

If you give h+ its own code, whatever h inside of it predicts, h+ will do the opposite.

Is there anything in my explanation that I can clarify?

[devout harness]

I think there's a misunderstanding - I know that there are turing machines of which you can't prove they halt or not and that there are conjectures you can't prove wether they are true or false. However, I meant to ask wether or not it is possible to, given a conjecture like the collatz conjecture or any other conjecture, prove of a turing machine that it halts if and only if the conjecture is true

[oblique nebula]

In my opinion, since you can't prove or disprove every statement that can be made (proven above), you can't prove or disprove that a yet unproved statement is provable.

I know it's kinda messy but it works

[polar fossil]

you can iterate over the computable numbers, and I'm tempted to say you can iterate over whatever structures are relevant for the problem, but I'm not sure

but we can talk about more than just the computable numbers

for collatz it's definitely possible, that shouldn't be hard to see

[oblique nebula]

How about twin primes?

[polar fossil]

yes easily

well

hmm

[devout harness]

how would you make a turing machine that halts if and only if the collatz conjecture is true? It's not trivial for me. Note that to my knowledge it's not proven that there aren't numbers that go off to infinity

[polar fossil]

yeah realizing now that's not as simple as i thought

if you could solve the halting problem in general then it would be possible but that's not what you asked

[oblique nebula]

I'm getting a bit lost in the terms here. Are you talking about the problem Swishfox asked or the "original one"?

[devout harness]

this person I just found phrased my question better than I did:https://cs.stackexchange.com/questions/66919/mathematical-conjectures-equivalent-to-the-halting-of-a-turing-machine

[polar fossil]

in this case i'm talking about solving the collatz conjecture

The twin prime conjecture is a Π2 statement, and so you can construct a TM with access to the halting oracle which halts iff the statement is false.
ayyy that's what i was theorizing as well hehe

[oblique nebula]

Thanks

[devout harness]

thinking about it a bit more, I can see how one would create a turing machine that halts if and only if the goldbach conjecture is false, but I don't yet understand wether or not its possible for all conjectures

[polar fossil]

sounds like no, from what that stack exchange link said

[obtuse pebble]

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[obtuse pebble]

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[brisk hamlet]

!status

[obtuse pebble]

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

[timid silo]

1

[tulip crypt]

Try expressing u_1 and u_2 as linear combinations of v_1 and v_2.

To do that, observe that it's solving a system of linear equations (once for u_1 and once for u_2).

[timid silo]

wdym? like setting $\begin{bmatrix} 1 \ 3 \end{bmatrix} = a\begin{bmatrix} 1 \ 0 \end{bmatrix} + b\begin{bmatrix} 1 \1 \end{bmatrix}$

[warm shale]

kyou

[timid silo]

and the same for the vector 1 4

[tulip crypt]

Yep.

[timid silo]

how does that help me construct the transition matrix?

[tulip crypt]

Wait wait oops it's from V to U.

So other way around.

[timid silo]

oh, I see

[tulip crypt]

Write v_1 in terms of u_1 and u_2, write v_2 in terms of u_1 and u_2.

[timid silo]

and then what do I do

once I have the solutions

[tulip crypt]

Note that $[\text{Id}]{V}^{U} = \begin{bmatrix} [v_1]{U} & [v_2]_{U}\end{bmatrix}$.

[warm shale]

TheUnTamed

[timid silo]

so it's a matrix

with columns equal to solutions

[tulip crypt]

Where $[v_i]_{U} = \begin{bmatrix} a_1\a_2\end{bmatrix}$ and $a_1u_1+a_2u_2 = v_i$.

Yep.

[warm shale]

TheUnTamed

[timid silo]

ok I'll solve it right now, and try to see if it's correct

thank you

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[sinful eagle]

I would like help on dealing with this

[harsh summit]

Divide 14/7 and 5/15

[sinful eagle]

Why is that possible

[harsh summit]

And 8/4

[harsh summit]

Since its multiplication bro

[sinful eagle]

But they are sepperate

[nocturne minnow]

$\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$

[warm shale]

dldh06

[sinful eagle]

Ya i know but

🤨

[nocturne minnow]

For multiplication of fractions, you multiply across the numerator and denominator

So $\frac{1}{2} \times \frac{6}{7} \times \frac{14}{5} = \frac{1 \times 6 \times 14}{2 \times 7 \times 5}$

[warm shale]

dldh06

[sinful eagle]

Sorry i thought of something else

I know this

Just confused with the many other assigments i Did

Thanks you

.close

[obtuse pebble]

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[true summit]

what is the question asking me to do

[true summit]

like what is a standard form of a differential equation?

that?

[obtuse pebble]

@true summit Has your question been resolved?

[tardy epoch]

,rotate

[tardy epoch]

Find y'' and y'

Then find a linear combination of them that's equal 0

[obtuse pebble]

@true summit Has your question been resolved?

[true summit]

so is this alright?

[true summit]

<@&286206848099549185> im sorry to disturb you all , but i kinda need this

[obtuse pebble]

@true summit Has your question been resolved?

[tardy epoch]

No

compare y'' and y

Don't think y' needs to be in your DE

[obtuse pebble]

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[fluid snow]

not A and not B*

[hollow dove]

there are a lot you could list out

like A and B is the same as B and A

(commutativity law)

are there any more specific ones you'd like?

[fluid snow]

A course in logic I'll take next year will cover some laws of predicate logic, it says, though they didn't give us any further information yet

[fluid snow]

Those are the obvious ones, are there any that are not so obvious?

[hollow dove]

I can just send a list, if you'd like

most of the named "laws" are obvious though

[fluid snow]

oh, sure

[fluid snow]

Thanks. Is that from a website?

[hollow dove]

it's from a reference book

you can find it in my bio (page 26-27)

[fluid snow]

Alright

Thanks

.close

[obtuse pebble]

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[tall burrow]

hi

[tall burrow]

can someone help me with this

[fathom flicker]

, rotate

[tall burrow]

like how to approach it

sorry

i shouldve rotated it

[fathom flicker]

no worries!

[obtuse pebble]

@tall burrow Has your question been resolved?

[tall burrow]

hello?

[fathom flicker]

make sure to react with the ❌ to the bot

or else the channel will timeout and close

[tall burrow]

okay

um

[hazy elk]

you need to think of it logically

so you have 100 total, with 3 categories:
people who
only have a laptop, 40
only have an ipad, 20
and both, 10

so here's how i would approach the top middle

the only people that would qualify are people that have an ipad, but no laptop, so the answer must be 20

it's pretty simple

[tall burrow]

?

answer to what

im trying to first like

make sense of how i would complete the table

is anyone there?

[tall burrow]

okay.. nvm then

is this right at least?

[tall burrow]

is this right

[obtuse pebble]

@tall burrow Has your question been resolved?

[tall burrow]

almost !

how would i approach these last questions

[brazen viper]

So you have the percentage of laptop owners and non-laptop owners who own iPads. Are these percentages the same (indicating no relationship), or are they different (indicating some sort of relationship)?

College statistics course only: ||(If this were a proper statistics course, you might have to show that the difference was not due to sampling randomness, but an underlying difference in the distributions, but reading the questions I don't think this is the case for your class.)||

@tall burrow

[obtuse pebble]

@tall burrow Has your question been resolved?

[tall burrow]

the percentages are not the same

[brazen viper]

(indicating some sort of relationship)

@tall burrow ^

[fathom flicker]

@grizzled shore

[obtuse pebble]

@tall burrow Has your question been resolved?

[obtuse pebble]

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[hidden eagle]

help pls, the problem: Jika P(x) dibagi oleh (x^2+1) dan (x^2+3) sisanya berturut-turut adalah (2x-3) dan (4x-5) maka tentukan sisanya jika P(x) dibagi oleh (x^2+1)(x^2+3).

[hidden eagle]

so its basicly says

if p(x) is divided by (x^2+1) then the remainder will be (2x-3)

and if p(x) is divided by (x^2+3) then the remainder will be (4x-5)

so what will be the remainder if p(x) is divided by (x^2+1)(x^2+3)

?

i am stuck because both (x^2+1) and (x^2+3)

is an imagener

meaning the determinan is lower then 0

[obtuse pebble]

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[fossil spoke]

these are the answers

[warm shale]

jstn.

[fossil spoke]

<@&286206848099549185>

[obtuse pebble]

@fossil spoke Has your question been resolved?

[fossil spoke]

???

[obtuse pebble]

@fossil spoke Has your question been resolved?

[fossil spoke]

.close

[obtuse pebble]

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[crisp elk]

Let A1, B1, and C1 be the midpoints of the sides BC, AC, and AB of triangle ABC, and A2, B2, and C2 be the midpoints of the altitudes drawn from vertices A, B, and C respectively. Prove that the lines A1A2, B1B2, and C1C2 are concurrent.

[crisp elk]

I know that we are supposed to use ceva in A1B1C1

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[reef gorge]

yo i need some help with 19.b

im kinda bad at simultaneous equations so if someone could help me

[harsh remnant]

Which equation do you wanna start with?

Upper or bottom?

[reef gorge]

lets start with upper

[harsh remnant]

Great

Let's put y on the LHS

[reef gorge]

whats a LHS

[harsh remnant]

Move 2y to the left, which it already is in the left side

And the rest to the RHS

LHS = Left hand side

[reef gorge]

ah ok

[harsh remnant]

RHS = Right hand side

[reef gorge]

okay so i write

2y = 5x+1 = 0

right?

[harsh remnant]

Why are there two equal signs?

[reef gorge]

oh my bad

2y = 5x+1+0

so wouldnt it just be 2y = 5x+1?

[zinc sage]

hi

[reef gorge]

hi

[reef gorge]

whats that bro

[zinc sage]

can anyone explain this

i need help

[harsh remnant]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[harsh remnant]

Get your channel

[reef gorge]

so is it 2y = 5x+1?

[harsh remnant]

Still incorrect

What are you doing by moving to the left?

[reef gorge]

wdym?

i just move the 2y to the left

[harsh remnant]

No, you're substracting the same value on each side

[harsh remnant]

This is a different equation

[reef gorge]

why would i be subtracting?

if its 5x + 2y + 1 = 0

and i move the 2y to the left

[harsh remnant]

I'm not literally talking about the word "move" here

[reef gorge]

then what do i do?

[harsh remnant]

Shouldn't it be more like 2y = - 5x - 1?

[reef gorge]

why would it suddenly become negative?

i don't get it

[harsh remnant]

There may be a better way to explain

But it's just math

[obtuse pebble]

@reef gorge Has your question been resolved?

[azure willow]

subtract (5x + 1) from both sides of the equation

by subtracting from both sides they remain equal and the equation still holds

and this gives the form 2y = -5x -1

[obtuse pebble]

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[reef gorge]

.close

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[raven halo]

Hey guys I am having trouble solving this for convergence or divergence

I thought that I could say that this is like geometric sequence

But I don’t think I can show that 1/ln(n) is less or more than one

Or I could use a direct comparison text where $$ln(n) < smaller Than sqrt(n)$$

[warm shale]

Rasul

[wet nebula]

You have the right intuition, but you can do even better by looking out what happens when n > e^2, for example

[raven halo]

Wait so n>e^2 always works right

So I could use it in a comparison test

[wet nebula]

When n>e^2, you have that 1/ln(n) < 1/2

so [1/ln(n)]^n < (1/2)^n for all n

[raven halo]

Owww and (1/2)^n is convergent and my series will be convergent too

[wet nebula]

Yes

[raven halo]

Thank u very much:)

[wet nebula]

You're welcome 🙂

[obtuse pebble]

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[grizzled raven]

I got
C(X) = 6000 x

[obtuse pebble]

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[tame oasis]

Hye everyonee, do you what the red one means? is it position vector = vector x unit vector?

[hot sonnet]

Hello. It's
Position vector = length of vector x unit vector

[obtuse pebble]

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[timid silo]

please help me solve this question :

Suppose there are n+1 people in a city, all of whom have distinct amount of money in
their account (no two person have same amount of money). Prove that there is two people
among them, whose wealth difference can be evenly distributed among n people.

I know I need to use pigeonhole principle, but still can't solve it

[forest sinew]

hrm? whats the restriction on division of the wealth difference

[timid silo]

I don't think there are any. i don't get your question perhaps

[forest sinew]

i mean why cant you just divide any number by n

[timid silo]

oh. cuz they are not multiple of n, i guess. like not mn where m is an integer . is that what you were asking?

[frosty river]

Consider the differences. There are n differences

Well no

There are more differences than you need

[timid silo]

yeah

[frosty river]

So even better for pigeon

[timid silo]

ok. then?

let me tell you how far i have gotten :

consider the differences between consecutive amounts of money:
d 1 =a 2−a 1
​d2=a3−a2
⋮
dn =an+1−an
​
We want to show that there are two people whose wealth difference can be evenly distributed among n people. This means that there exist i & j such that dj is a multiple of di, and therefore, dj can be distributed among n people with each receiving di

Now I have problem going ahead

[drifting wraith]

it works with n=2, all 3 differences can't be odd

no clue what to do next

[timid silo]

could you explain a bit more pls?

[drifting wraith]

ok i get it

yeah there are (n+1)c2 differences, way more than you need

look at what happens with n=2, then n=3, it's easy

on god

[timid silo]

ok let me try

@drifting wraith still can't crack it

[drifting wraith]

suppose n=2, so we want to find a difference that's even

we have 3 differences, suppose the first one is odd, and the second is odd, then the third one will be even

that's mostly it

[timid silo]

but how do u prove it when n is more?

[drifting wraith]

it literally keeps working the same way

a
a+2k+1
a+2j+1

once we have 2 odd differences

the difference of these 2 people is even: 2(k-j)

if you have 4 people, to make a difference that's divisible by 3 you still want to find 2 differences that have the same remainder

maybe the first difference is 3k+1, and the second is 3k+2, but the third will have to coincide with one of those, or else just be divisible by 3 already

[timid silo]

ok i am getting it a bit. let me see

[drifting wraith]

i was overcomplicating it, thinking about (n+1 )c2 differences

[timid silo]

In what is this math used for?

[drifting wraith]

you keep comparing with the poorest guy, there will be n differences, that's enough

[timid silo]

i don't understand

[timid silo]

its a bit of combination permutation

[drifting wraith]

well it's the same thing i explained

just you can actually designate the poorest person as a

[timid silo]

kindly explain again. why should i only keep comparing with the poorest guy

[timid silo]

What type of work uses permutation?

[drifting wraith]

a
a+2k+1
a+2j+1

when i wrote this, it was like vaguely metaphorically right, and now i realized it's literally true

[timid silo]

ask in #discussion

[timid silo]

ok

so...

[drifting wraith]

you compare the poorest person with other people, and n differences will be enough

either two differences will have the same remainder, show how that's sufficient
or they will all be different, that's sufficient for some other reason

[timid silo]

ok oh

i got it

thank you bro

@drifting wraith

[drifting wraith]

👍

i'm still thinking about differences you can prove it with just raw amounts, that has to be the intended solution

with n+1 people, you never have the case where all remainders are different

[obtuse pebble]

@timid silo Has your question been resolved?

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[trim trail]

Isn't the series in (a) the harmonic series? (after cancellation with $m!$)

So doesn't the sum diverge (to infinity)?

[warm shale]

StatisticalCat

[frank monolith]

in the second term, (m+2)(m+1) will be left in denominator

for third (m+3)...

how is this harmonic?

[trim trail]

ohhhh yeah ur right sorry

[trim trail]

ok so I got it can u lmk if i am right?

basically $m\geq 1$, so at most, the sum is $\frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +... = e-2 < 1$

[warm shale]

StatisticalCat

[obtuse pebble]

@trim trail Has your question been resolved?

[trim trail]

<@&286206848099549185>

[trim trail]

yo can someone help plz

[tardy epoch]

Yes this is right

[trim trail]

thanks!!

any hints on how to go about b

[trim trail]

I get that the series never converges to 1 and e is 2 plus that series

is just stating this enough for part (b)?

[tardy epoch]

No

if e were rational then it can be written as a fraction of two integers

Arrive at a contradiction using this assumption

[trim trail]

ok lemme try

[trim trail]

is this satisfactory

i showed that 2 < e < 3

[timid silo]

@drifting wraith what do you think? i tried this. is this ok?

[trim trail]

yo wrong chat i think

[tardy epoch]

No

Do you know what rational means

[trim trail]

yeah i got my mistake im working on the full proof 😓

I forgot that the sequence could converge my bad

[obtuse pebble]

@trim trail Has your question been resolved?

[obtuse pebble]

@trim trail Has your question been resolved?

[obtuse pebble]

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[forest glade]

im lost on how to solve this one

[native inlet]

When doing surface area of 3D shapes, consider breaking the shape up into elementary parts :) this is merely a hexagon and 6 triangles

[forest glade]

so the triangles

12*4 for half a triangle

dang

[native inlet]

not for half...

[forest glade]

oh the whole thing

[native inlet]

now we have six of those triangles so we yield 12*4*6 :)

do you understand that part?

[forest glade]

mhm

[native inlet]

so, how do we get the area of that hexagon?

[forest glade]

we have the

radius?

[native inlet]

not quite, we have something different

[forest glade]

we have half of its length

[native inlet]

that 6m is called an apothem

[forest glade]

ooooo

yes

[native inlet]

the radius is from the center to a vertex

[forest glade]

i completely forgot i just did a whole thing on the apothem 🤦‍♀️

[native inlet]

the apothem goes from the center to the side to form a right angle :)

ok so the formula for the area of a polygon is $A=\frac{ap}{2}$

[warm shale]

MrFancy

[native inlet]

very simple, we just need the apothem and perimeter

[forest glade]

8m

8*6

[native inlet]

8 isn't the perimeter, it's one side length

[forest glade]

yeah sorry thats not what i meant

48

[native inlet]

not quite

we still have to divide by 2

[forest glade]

wait why

[native inlet]

^ the formula :)

[forest glade]

oh oh

you are a step ahead of me

6*48/2

is the area

of just the polygon

[native inlet]

wait yes

im sorry that was my bad

[forest glade]

144 is the polygon area

or not

[native inlet]

im sorry I'm confusing myself

it is im certain

sorry my brain just like absoluetly did not brain there

[forest glade]

me most of the time

432m

• 432m^2

[native inlet]

there we go :)

[forest glade]

hooray!

you are such a good teacher MrFancy

thank you!!

.close

[obtuse pebble]

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[buoyant ingot]

.reopen

[agile fiber]

The question I need help with is
"Parameterize the following...
a) x + y = 2
b) 3x + 2y - z = 2"
I'm having trouble finding out where to start (don't remember parameterization)

[obtuse pebble]

@agile fiber Has your question been resolved?

[timid silo]

hi

do you understand where the equations of the corner come from?

[agile fiber]

I do not, not fully sure what im looking at. I haven't done this kind of math in a few years i kinda need a refresher

[timid silo]

do you remember how to add vectors? head-to-tails?

and dot product

[agile fiber]

if i saw a numerical representation i would probably understand it a bit better, but i dont recall how to do those off the top of my head

for a little more context this was an assignment for the first day of Linear Algebra

[timid silo]

first day... uhm

[agile fiber]

is this more something i would learn later on?

[late stump]

are you familiar with vectors

[agile fiber]

yes

[timid silo]

you already know this?

[agile fiber]

what's the line at the top saying exactly?

I think i get it i just havent done this kinda stuff in like 2 years

the theta represents the angle between the two vectors right

[timid silo]

yeah

so

[agile fiber]

oh oh i see, i found it in my notes

[timid silo]

if the angle betwen two vectors are 90 degrees, then a dot b = |a||b|cos(90) = 0

[agile fiber]

sorry im flipping back through my notebooks from a few years ago

yeah

[timid silo]

okay, the idea is that you could say you already know 1 point of the line

then you move to another point that is on the line

that would give you a vector

if you multiply that vector by any scalar that vector wouldn't change its direction, so you still on the line

[agile fiber]

alright i see

now when it comes to parameterization, what do i do to parameterize the equations?

[timid silo]

you need to represent any point of the line with 1 parameter, for example, a vector whose head is exactly at that point and have one parameter

asides forming a vector with 2 points of the line, you still need to orientate it with respect to the origin, in general is the point (0, 0)

the equation above is just the result of adding two vectors, head-to-tail, and lets why this only depends on t

1. we already know the points p0, p1, to make a vector that has the same direction of the line

2. you just need to multiply by a number (say t) to make that vector larger or smaller, to reach the other points of the line

[agile fiber]

I see

I think i get it enough to take care of it. Ill check up with my professor in the morning to make sure

Thank you

[timid silo]

just an example to compare

[agile fiber]

Appreciated

[timid silo]

for a plane its different

well, it could be easily in a different way

av1 + bv2 is called linear combination, intuitively, think of the x-y plane, thats... well a plane. Anyways, the point is, you can take a times vector v1 + b times vector v2 and fulfill the plane if they have different directions

now our plane is determinated by the same logic, the vector from p0 to p1, and the vector from p0 to p2

i hope this gives you an intuition

[obtuse pebble]

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[molten plaza]

What is the primary operator here?

[nocturne minnow]

,rotate

[molten plaza]

I am confused if its the second conjunction or the second to last implication

I mean the primary/main operator

<@&286206848099549185>

[molten plaza]

Can you help me with my question?

[limber quartz]

"it"?

so you have statements A, B, C, D, S

[molten plaza]

Im getting confused as the rule says there is supposed to be only onr primary operator

[limber quartz]

what rule

[molten plaza]

But to me it looks like there are 2

In a propositional form

[limber quartz]

I see a "then", an "and", a "not", a "or", ...

[molten plaza]

So the question im asking is what is the main/primary operator here

[limber quartz]

I have no idea about that

but you can solve this thing by filling in your truth table

I would make a spreadsheet

[molten plaza]

Can you? That would be very helpfull

[limber quartz]

you should make the spreadsheet

it's dumb-as-rocks easy

you need 2^n rows, where n is the number of statements

you have |{A,B,C,D,S}| = 5

2^5 = 32

I'll show you the spreadsheet setup, hang on

so the first column is statement A. Half of the 32 rows are True (1), the other half is False (0)

the second column is statement B. half of the statements correspoding to statement A being true, are true, the other half false; half the statements corresponding to A being false are true, the other half false.

and so on.

A -> 16 true, 16 false
B -> 8 true, 8 false, 8 true, 8 false
C -> 4 true, 4 false, 4 true, 4 false, 4 true, 4 false, 4 true, 4 false
D -> 2 true, 2 false, 2 true, 2 false, 2 true, 2 false, 2 true, 2 false, 2 true, 2 false, 2 true, 2 false, 2 true, 2 false, 2 true, 2 false
S -> alternate T/F

@molten plaza

now you just make some columns that will evaluate operations on A, B, C, D, S with increasing complexity until you have the entire compound statement

and just see what rows result in '0' (false)

https://sites.millersville.edu/bikenaga/math-proof/truth-tables/truth-tables.html

[obtuse pebble]

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[hazy sky]

hey

Graphically determine the order of magnitude of the mean value over the interval [-3,-1] of the function f

how am i supposed to know that looking at the graph ?

[obtuse pebble]

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[tardy epoch]

fyi channel's gonna close

[hazy sky]

ah

.close

ill do it again

whatever

[teal idol]

hello

Im from Peru

[hazy sky]

i think i fcked the channel @tardy epoch

.close

[obtuse pebble]

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[spark jungle]

can someone help me with chemistry

[spark jungle]

there is no chemistry server

Please

and this unit is unit conversions

and thats math

like 99%

[late stump]

there’s a chemistry server in #old-network but you can post it here, maybe someone will know

[errant lark]

[spark jungle]

do u know unit conversions

lemme send question

how do i do 6

[errant lark]

,rotate

[teal idol]

help please

[obtuse pebble]

Please read #❓how-to-get-help

[errant lark]

Open your own help channel.

[errant lark]

Which one @spark jungle

[spark jungle]

number 6 if anyone can help

[teal idol]

sorry

Im new

[spark jungle]

6

https://media.discordapp.net/attachments/903430332324405288/1143031481984356474/1116326425088241714.png?width=895&height=671

[errant lark]

Okay. Firstly, what are grams and Kilograms unit of ?

[spark jungle]

Wdym

unit of?

[errant lark]

What do you measure in kilograms?

Like, if you have a length to measure, you use metres or centimetres or sth else.

[spark jungle]

weight

[errant lark]

Uhhh... Common misconception. Weight is actually in newtons, or a unit called Kg-weight which is equivalent to 9.81 newtons.

[spark jungle]

oh

[errant lark]

Regardless, let me tell you that mass is measured in Kilograms or grams or such.

[spark jungle]

ok

[errant lark]

What about mL ?

[spark jungle]

not sure

[errant lark]

That stands for millilitre if you dunnno.

[spark jungle]

liters?

ye

[errant lark]

Yeah. It's similar to litres. Just smaller. So, we measure volume in it.

[nocturne minnow]

I think he was questioning the spelling of liters, cause you spelt it litres

[spark jungle]

it doesnt really matter

[errant lark]

I don't think so. Lol

[spark jungle]

i just need help on how to do this

[errant lark]

Ah. Well, anyway, i prefer British ones.

[spark jungle]

ok

[errant lark]

Now, You need to convert from $\frac {g}{mL}$ to $\frac{Kg}{L}$.

[warm shale]

Enemagneto

[nocturne minnow]

It's just converting, using dimensional analysis, this video could help a lot https://www.youtube.com/watch?v=vWovdihteqs&ab_channel=TheOrganicChemistryTutor

[spark jungle]

ye

[spark jungle]

i'll watch it in a bit thx

[errant lark]

Yeah. That also works.

[spark jungle]

ok

how do i convert

[errant lark]

Perhaps try watching first and then ask if you don't get it?

[polar fossil]

10

[errant lark]

😶

Well, 9.81 Newtons to be more precise. Lol

[nocturne minnow]

Or 2.2e-03 Kips

[spark jungle]

@errant lark thats videos different

cuz if i multiply

i get to variablews

2

idk what to do

[nocturne minnow]

What do you mean you get two variables?

[spark jungle]

if I multiple

9.11g/Ml

[nocturne minnow]

The concept is the same, you're converting between units

[spark jungle]

i dont understand

[nocturne minnow]

What don't you understand?

[spark jungle]

how to conver to kg/L

[nocturne minnow]

The video is showing you have to convert from one unit to a different one, that's the same exact thing you are doing

[spark jungle]

its 16 minutes long

[nocturne minnow]

Okay, and?

[spark jungle]

i dont have enough time

[nocturne minnow]

If you don't the concept, you should learn it to understand it

[tardy epoch]

then make time

Or watch enough where it answers your problem

[spark jungle]

i procrastinated for the whole day

[tardy epoch]

Skip, 2x speed, etc.

[nocturne minnow]

If you had the time to sit in a help channel for 15 minutes, you have the time to watch a 16 minute video

[spark jungle]

@nocturne minnow

9.11g/mL x 1kg/1000g

if both grams cancel

[nocturne minnow]

Good

[spark jungle]

is it 9.11/ml x 1kg/1000

[nocturne minnow]

Now you need mL to L

[spark jungle]

?

or do 9.11 and 1000 disappear

[nocturne minnow]

Why would it disappear?

[spark jungle]

i forgot how to cross multiply

[spark jungle]

so like this

and than

[nocturne minnow]

You don't cross multiply

[spark jungle]

What

but they cancel right?

[nocturne minnow]

I don't know what you're confused about, it means what it means, you don't cross multiply

[nocturne minnow]

9.11 and 1000 don't cancel out, the units do

[spark jungle]

ohhh

9.11kg/1l

no

wait

[nocturne minnow]

You handled g to kg with this, now you need to convert mL to L, how would you add on to this expression for mL to L?

[spark jungle]

there

[nocturne minnow]

Why is over 1 L?

When you started with mL?

[spark jungle]

9.11g/mL x 1kg/1000g x 1000ml/1l

grams and ml cancel

[nocturne minnow]

Good

Now simplify that

Do the math

[spark jungle]

9.11kg / 1l

[nocturne minnow]

Oh that was your answer

[nocturne minnow]

Yes this is right

[spark jungle]

do u know what a mole is

https://media.discordapp.net/attachments/903430332324405288/1143031481984356474/1116326425088241714.png

9.24x10^24 particiles to moles

[nocturne minnow]

A unit of measurement for amount of substance

[spark jungle]

do u know how i would start with 8

perhaps

[nocturne minnow]

1 mol = 6.02x10^23 particles

Use that conversion

[spark jungle]

Damn

thats

ok

[nocturne minnow]

If you recognize the term "Avogadro's number", that's what you're using

[spark jungle]

so particles cancels out

easily

and its just 1 mol x 9.24x10^24

oh wait

divided by 6.02 x 10^23

[nocturne minnow]

Yes that's better

[spark jungle]

how do i simplify that

because I also need 3 significant figures

[nocturne minnow]

You can use a calculator, right?

[spark jungle]

i mean

why not

[nocturne minnow]

So use it

[obtuse pebble]

@spark jungle Has your question been resolved?

[obtuse pebble]

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[sage dagger]

I need help with implicitly differentiating tan(x/y) = x + y

[sage dagger]

I have the steps I’m up to hold on

[royal basin]

oh ugly.

[sage dagger]

It is

it doesn’t look very pleasant

I think ive done something wrong possibly

dunno where tho

Oh wait

Last line should be -xy^-2

I forgot to write the -2 exponent

(Still don’t know tho)

[royal basin]

$\frac{1}{\cos^2(x/y)} \cdot \frac{y - xy'}{y^2} = 1 + y'$ is how i would have written it

[warm shale]

Ann

[sage dagger]

For which part?

[royal basin]

first line

[sage dagger]

Oh well

I’m too deep into this to go back

how can I salvage what I have I know it’s doable

[royal basin]

not a thing

you can always backtrack and erase your work

[sage dagger]

please I’ve rewritten this so many times already

[sage dagger]

how can I reach the answer from what I’ve got

[royal basin]

i am not gonna answer that in case there is an earlier mistake that i missed.

[rotund isle]

HEY HW R U GUYS DOIN

[sage dagger]

How can I factor out the dy/dx

assuming I have -dy/dx on the LHS and the rest is equal to 1

[royal basin]

hello, these channels are for math help.
for casual chatter, please go to #discussion or #chill.

add xy sec^2(x/y) dy/dx and subtract 1 from both sides

[sage dagger]

you mean -xy^(-2)? not just xy

[ornate moon]

Hello!

[royal basin]

no, i mean from the last line, add the leftmost term to both sides

[ornate moon]

Can someone help me wth my math?

[sage dagger]

Oh right

[frank monolith]

he has a error in last line

[royal basin]

@ornate moon please open your own channel for that. read #❓how-to-get-help for instructions

[sage dagger]

where

[frank monolith]

the y^-2 which you already mentioned

[sage dagger]

Yeah I fixed that tho just haven’t sent new sc

[royal basin]

@frank monolith mind taking over?

[frank monolith]

um okay

[timid silo]

because everybody b talking im too then

[frank monolith]

#discussion and #chill

[sage dagger]

so now I have this

[frank monolith]

yeah

take dy/dx common

because you need to find it

[sage dagger]

Okay so I have dy/dx[xy^-2 • sec^2(x/y)]

[frank monolith]

and shift the +1 to LHS

[sage dagger]

Alright yeah got that

So then I divide by my terms inside dy/dx right

[frank monolith]

yup

and if you want to simplify further, you can convert y^-1 to 1/y and take LCM

[sage dagger]

Sweet

I wanna check if what I have can be turned into this

,w implicit derivative tan(x/y) = x + y

[warm shale]

Results provided by WolframAlpha

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

Requested by waterrbeam

[sage dagger]

Hm

[frank monolith]

where's the 1 at in the denominator?

you forgot the 1

[sage dagger]

Wait where does 1 come from

[frank monolith]

after taking dy/dx common

the 1 already present was shifted to LHS

but there was another

[frank monolith]

.

[sage dagger]

There was?

[frank monolith]

take dy/dx common again

[sage dagger]

This is my factored one which doesn’t have a 1

[frank monolith]

it is wrong

[sage dagger]

Oh

Oh there would be a +1 in there right

[frank monolith]

yes

[sage dagger]

Alright we pretend there’s a +1 on the denominator now is that correct?

[frank monolith]

yes

[frank monolith]

.

[sage dagger]

And that’s how they got this right

[frank monolith]

yes

[sage dagger]

Okay thanks for all your help

.close

[obtuse pebble]

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[obtuse pebble]

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[haughty hemlock]

I am stuck on this assignment question on probability. I know how to find the pdf for single random variable and other simple stuff, but for multiple random variable, i don't understand the concept of Jacobian transformation factor(when & how to use it). If someone can help me understand, that'd be really helpful.

[obtuse pebble]

@haughty hemlock Has your question been resolved?

[obtuse pebble]

@haughty hemlock Has your question been resolved?

[obtuse pebble]

@haughty hemlock Has your question been resolved?

[obtuse pebble]

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[alpine gull]

can anyone help me prove that the limit of this succession tends to -infinity?

[harsh remnant]

Where's the limit?

[alpine gull]

the limit of An with n->+infinity to study its behavior

[trim portal]

I don't feel like that tends to -infinity

[sage geode]

Consider the subsequences with odd- and even- numbered terms

If they converge, are the limits equal?

[alpine gull]

its a question with 4 possible answers
a) it doesnt have a limit
b) it has maximum but not a minimum (this is the right one)
c) it has maximum and minimum
d) it's not bounded from below

also i just realized i thought it had to diverge to -infinity since it's supposed to have a maximum but not a minimum, but im not sure if thats true

[sage geode]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[sage geode]

[trim portal]

The sketch was incorrect, I confused maximum and minimum with boundedness

and yet it converges

the sketch is a bit awful, but you know what I mean

[sage geode]

Hm? I don't think that's how the graph looks like

Wait

[twilit knot]

Anyway it's possible to have a sequence unbounded from below but doesn't diverge to -inf

[trim portal]

This sketch isnt supposed to model your function, it's just example of function with maximum but not a minimum

[alpine gull]

does this mean the question is against the rules? D: it's not from homework, its an exercitation to prepare for an exam

The limits of the subsequences seem to be equal unless I messed something up

[sage geode]

That rules concerns helpers, not helpees

[twilit knot]

Nah I typed something I shouldn't have dw

[trim portal]

Yes, they are. It indeed converges to 0

[sage geode]

Yeah so the sequence converges to 0

[alpine gull]

after proving that it converges to 0, how do I get to proving that it has a maximum but not a minimum?

[sage geode]

I guess show that infimum is not contained in the sequence

[alpine gull]

i got it, tysm =)

.close

[obtuse pebble]

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[wicked ice]

what does like the sideways m mean

[royal basin]

that is a summation symbol and it is an uppercase greek sigma

[wicked ice]

what does it mean

[twilit knot]

That you add up the expression which comes after, indexed by i in this case

[wicked ice]

so say i was given this how would i do it

[obtuse pebble]

@wicked ice Has your question been resolved?

[wicked ice]

<@&286206848099549185>

[quiet crystal]

x bar means arithimetic mean

which you can calculate by the original method

[wicked ice]

what about standard deviation

.close

[obtuse pebble]

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[obsidian quail]

What do I do with the e^sin 20°

[sweet cobalt]

Are you allowed to use a calculator?

[obsidian quail]

Yea

[sweet cobalt]

Well e^sin(20) and cos(20) are both constants, so you can just calculate the RHS

[obsidian quail]

ohh so i can just treat that expression as the sine ratio

got it

[sweet cobalt]

Yes :)

[obsidian quail]

.close

[obtuse pebble]

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