help-23

[rough storm]

now note that for x->-inf we will have |x|=-x

so we'll have an extra -1 factor

with everything else the same, and multiplying it by -1, what will our HA be for x->-inf

[stiff vine]

Bruhh what

So I times 1 by -1

For other asymptote

[rough storm]

so our horizontal asymptotes will be -1 and 1 respectively!

we can verify this! :)

[stiff vine]

So I always time the answer by -1

For other asymptote?

[rough storm]

not necessarily

[stiff vine]

Then why do I do it here

[rough storm]

https://youtu.be/-PYebK8DKPc?t=3280

i've copied this video at the appropriate time that he goes over a similar problem

i hope he can explain the process better than I can

[safe radish]

@stiff vine Has your question been resolved?

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[grim basin]

Can someone please help with this problem

[grim basin]

I dont know how to approach it

[lean otter]

You should start by looking out for subset sequences

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[agile gust]

find all integer values of x and y that satisfy this inequality.

[agile gust]

$5x+3y<96$

[flat frigate]

meerpop

[agile gust]

i’m not sure how to begin this question

[plucky elk]

do you mean positivei ntegers

[agile gust]

yes thanks

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[wild hatch]

i dont understand what to put for union

[safe radish]

@wild hatch Has your question been resolved?

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@wild hatch Has your question been resolved?

[rough storm]

can you post the whole question

i.e. include part a

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[azure vine]

How do I solve these kind of questions, idk if this belongs here

[azure vine]

Do i draw up truth tables or is there an easier way to reason it

[marsh walrus]

sometimes you can see it

truth tables is pretty nuclear

an in between would be using identities to try to get one into the other, but it can be like searching in dark sometimes

so, its up to your style

wil the bot here do it i wonder

[azure vine]

alright it doesnt have an answer model so can you help

[stoic dune]

You don't necessarily have to write out an entire truth table, just need to find a single difference

[marsh walrus]

also just noticing that its curious one has r and the other doesnt

so if the second has a truth value that depends on r thats going to be a problem

damn i should read

[azure vine]

i got C

nuclear way 😆

[marsh walrus]

,w truth table (not (p equivalent q)) and p

,w truth table ((r implies p) implies (p and not q)) and p

[azure vine]

lemmee send a screen

shot

[marsh walrus]

here, look

5th and 6th rows here

thats gonna be a problem

[azure vine]

ftt and ftf?

[marsh walrus]

actually, the last 2 rows as well

because the truthiness depends on r

[marsh walrus]

but, theres no r in the first statement

lemme see

[azure vine]

never mind the (1) or -q that one was wrong

[flat frigate]

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[marsh walrus]

@azure vine looks like you got it

[azure vine]

Wowzers

Okay theres a time limit on this probably on a test so I really need to find another way

[azure vine]

elaborate on this pls :)

[marsh walrus]

well, you could have spotted that the truthiness of the second is going to depend on r, right?

[azure vine]

yes

[marsh walrus]

$(r \to p) \to (p \land \neg q)$

[flat frigate]

jan Niku

[azure vine]

yeah

[marsh walrus]

so

im not gonna pretend i know how to do these fast lol just trying to think

i mean if r is false, were done, the whole thing is true

[azure vine]

if r is false the first one is true

[marsh walrus]

oh, yea, sorry

[azure vine]

whole thing is true if p and -q is true (if r is false)

[marsh walrus]

thats ... q -> p right

$(r \to p) \to (p \to q)$

[azure vine]

uh wait

p and -q = - (-p v q)
- (p -> q)

[marsh walrus]

it is p to q right

im not crazy

[flat frigate]

jan Niku

[azure vine]

its the negation of p -> q

RIGHT?

[marsh walrus]

,w truth table (p and not q)

[flat frigate]

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[marsh walrus]

,w truth table (p implies q)

[flat frigate]

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[marsh walrus]

yes i misapplied demorgan

[azure vine]

Yep negan

negation

[marsh walrus]

$(r \to p) \to \neg (p \to q)$

[flat frigate]

jan Niku

[marsh walrus]

okay can we stare at this hard enough

[azure vine]

-(r->p) v p -> q
-(-r v p) v (-p v q)
r and -p v (-p v q)
r and (-p v -p and -p v q)
r and (-p v q)

,w truth table (r and (-p or q))

[flat frigate]

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Perhaps try rephrasing your question?
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Requested by danktagg

[marsh walrus]

man im so out of practice with these, sorry

it uses not, not -

[azure vine]

,w truth table (r and (not p or q))

[flat frigate]

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[azure vine]

,w truth table (r implies p) implies (not (p implies q))

[flat frigate]

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[azure vine]

F

[marsh walrus]

sorry im playing around with it

its really hard because you know like

the tables will work

but then they take so long

so you gamble on being clever

[azure vine]

shi i forgot negation on p -> q

Yeah

-(r->p) v -(p -> q)
-(-r v p) v -(-p v q)
(r and not p) or (p and not q)

,w truth table (r and not p) or (p and not q)

[flat frigate]

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[marsh walrus]

,w truth table not( (r implies p) and (p implies q) )

[flat frigate]

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[marsh walrus]

yea, i think this gives it

lemme make sure

[azure vine]

yes

never mind ill just use the dumb way lol

[marsh walrus]

sorry

[azure vine]

this is for my selection test for bachelor study

i have never had logic before

thanks for your time

[marsh walrus]

sorry i couldnt track it out

good luck

[azure vine]

(r and not p) or (p and not q) and p
needs to equal -(p <-> q) and p

for answer C

yeah nevermind

thanks

.close

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[shrewd solar]

hi

please dont waste channels

.close

ugh i cant do it

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[finite epoch]

I need some help

[finite epoch]

Bruh

[olive sonnet]

my fault

lmao

[finite epoch]

fizzie

Lol

[olive sonnet]

i lost the race 😭

[finite epoch]

Nice math prob tho

[plucky elk]

just ask already

[finite epoch]

@plucky elk

Un is a sequence

[plucky elk]

probably should translate if you want help

[finite epoch]

I mean

Okk

[devout shale]

Consider the sequence defined by …. Prove that …

[finite epoch]

Prove that -1≤Un≤2

Thanks mate

[finite epoch]

I tried substraction and was stuck with a Sin(nπ/2)

• 1-2n/n

@plucky elk you there?

[plucky elk]

find the first 8 terms of sin(n * pi / 2) and see if you get a pattern

n = 1, 2, 3, ..., 8

[finite epoch]

sure

Why exactly 8

[plucky elk]

if you find a pattern before that you should stop

[finite epoch]

1,0,-1,0

@plucky elk

[plucky elk]

yea that looks right

so you proved -1 <= sin(n pi / 2) <= 1

[finite epoch]

Un = sin (n pi/2) + 1/n

@plucky elk what's next

[plucky elk]

try doing some algebra

[finite epoch]

wait

OHHHHH

n≥1

1/n≤1

sin(nπ/2) + 1/n ≤ 2

@plucky elk wait

[finite epoch]

How do I write this on paper

[plucky elk]

wot

[finite epoch]

uhh

I'm just gonna write for n=1,2,3,4,5,6,7,8

???

Writing is important right?

[devout shale]

Write

[finite epoch]

what

Write what

[devout shale]

Pronounce it

[finite epoch]

No like on paper

Know what, nvm

I got the point though thanks you're epic @plucky elk

.close

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[shrewd topaz]

@obtuse plover i cant find an equivalent for csc A

[ember loom]

mark everything thats already used

I think that will make it easier

[shrewd topaz]

well i have it laid out in front of me

i have six squares left

there isnt 1/sin anywhere

there is this but its right next to csc a which is what i need

like i have two cscAs but only one 1/sin

[solar hazel]

what is this

[shrewd topaz]

a puzzle where i have to rearrange the squares so the identities can be touching

[ember loom]

what about the right ones on the right corner. they seem to be missing

tbf none of them checks out

[shrewd topaz]

better crop

[safe radish]

@shrewd topaz Has your question been resolved?

[safe radish]

@shrewd topaz Has your question been resolved?

[shrewd topaz]

.close

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[midnight dust]

How would I even start this problem I don’t understand the question

[midnight dust]

Or what to do

This is optimization

[light shoal]

it doesn't say they have to be square or rectangular
how do you know that's the optimal shape

[midnight dust]

I don’t

I’m just assuming it’s going to be a square or rectangle because we’ve only done squares and rectangles in class

but we’ve never done a problem like this

[light shoal]

like it's at least plausible that drawing a circle and dividing it into three sectors would be better, but i haven't tried computing it

[midnight dust]

the best answer is probably a circle but I’m pretty sure my teacher is looking for rectangles

Assuming all 3 are equal and are squares or rectangles I need to find all 4 sides of one and then 3 for the others?

Not including the side that touches another pen

[light shoal]

yea i guess so,

and probably need to check whether three lined up in a row like that is best, or would some other arrangement save you some fence length

[midnight dust]

how many combinations are even possible

I’m guessing only 2

A line and an L shape

I don’t even know how to do a minimize problem she only taught maximizing

[light shoal]

yea seems those are the main possibilities

and then for a line, do you align them so they share the longer wall or the shorter one of the rectangle

so maybe 3 possibilities

[midnight dust]

Im just going to do a line this is worth like 5 percent of my grade with like 90 other homeworks this is a waste of time

Completion too

Do you think you could help with this one we’ve seen it before I just don’t understand it

[light shoal]

ok this one seems a bit more tractable

you have three dimensions let's call them L, W, H

you require L = 2W

and you require LWH = 36000

[midnight dust]

L is 2x and W is x

[light shoal]

yea

so now you have two variables

[midnight dust]

H can be y

[light shoal]

right

say (2x)(x)(y) = 36000

and then you have another equation for the surface area

[midnight dust]

Do I need to find y by doing substitution

[light shoal]

oh notice it says "open" box so it only has 5 sides

[midnight dust]

36,000/2x^2 = y?

[light shoal]

maybe first focus on finding a second equation

the one for the amount of material

but then yea

you would make that substitution so you just have one variable

and then minimize with respect to that variable

[midnight dust]

it’d be 4(xy) + 2x^2 for SA?

[light shoal]

let's see

two of the "vertical" sides would have area xy, the other two would have area 2xy, the bottom would have area 2x^2
so i calculate it as 2xy + 4xy + 2x^2

that's assuming the missing side is at the top

[midnight dust]

so now I find y?

The volume equation would probably be easier to get Y right

I don’t even think the first problem is possible

.close

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[midnight dust]

are you even there

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[ornate osprey]

If I have $3xy^2+2x^2y+8=0$ and need to find the point at $(-2,2)$, how would I product rule the first term?

[flat frigate]

Someone

Someone

Someone

Someone

[visual wren]

Did you learn implicit differentiation yet

[ornate osprey]

Yes, I am doing implicit differentiation

[visual wren]

With respect to y ?

[flat frigate]

Someone

[ornate osprey]

I'm confused on this part as I need to use the product rule. Do I put the 3x with both y and dy/dx?

[visual wren]

$3xy^{2}\frac{dy}{dx}+6xy$

[flat frigate]

javier

[ornate osprey]

what?

that is just not right

[visual wren]

Yea i don’t get your question mb

[faint raptor]

remember that y^2 becomes 2y and not just y

[ornate osprey]

right

but still dunno what i do with the 3x :/

$\frac{d}{dx}(3xy^2)$ is what I am doing

[flat frigate]

Someone

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

[ornate osprey]

I don't know what I do with the 3x at the very bottom

[visual wren]

Wdym find the point

What is the original question

[ornate osprey]

I literally cannot find the point with that 3x(y) there

[ornate osprey]

this

[flat frigate]

Someone

[faint raptor]

classic xy situation

[ornate osprey]

Sorry

[flat frigate]

Someone

[ornate osprey]

.solved

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[dapper hamlet]

How to expand this

[hot thistle]

im not sure this has a nice expansion

specifically the i * 4^i

[dapper hamlet]

Well I cheated using wolfram and it’s not supposed to

But I need to know how to do it

[hot thistle]

the point is that chances are you did something you shouldnt have

assuming you agreed that this doesnt have a nice expansion

in which case show all the work and the original qn so others can help

[dapper hamlet]

Ignore the last lines

But i computed the original equation into wolfram as well as my 2nd final line and the end answer is the same

So chances are I’m supposed to expand that

[stoic dune]

It looks like you're trying to apply master theorem, but that can't typically find better results than asymptotic growth afaik

[dapper hamlet]

This is called master theorem?

[stoic dune]

Well, you can start with Σ a^i, and take the derivative in terms of a. This gets you a formula for Σ ia^i

[safe radish]

@dapper hamlet Has your question been resolved?

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[red mountain]

determine implied domain of sqrt"1-x"-4y=-5

[red mountain]

i dont know what to do

theyve already solved the first part being f(x)=(sqrt"1-x"+5)/4

[icy lance]

whats the domain of sqrt(x)

[red mountain]

f(x)=(sqrt"1-x"+5)/4

[icy lance]

?

[red mountain]

oh

i dont know how to find domain

[icy lance]

it requires knowledge of some parent functions

here said function is sqrt(x)

,w graph y=sqrt(x)

[flat frigate]

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[icy lance]

what do you notice

[red mountain]

im not sure

[icy lance]

hm okay, can i do sqrt(-2)

[red mountain]

sure?

[icy lance]

can i?

in the real numbers?

[red mountain]

oh no

you cant

[icy lance]

can i square root any negative?

[red mountain]

no

[icy lance]

can i do 0

[red mountain]

no

[icy lance]

you sure

[red mountain]

now im not lol

[icy lance]

sqrt(0) is just 0

not problems there

the domain of sqrt(x) is x>=0

[icy lance]

you have a sqrt(1-x) therefore what restriction is 1-x under

[red mountain]

idk this is confusing the hell outta me

[icy lance]

if its sqrt(1-x), can 1-x be negative?

[red mountain]

yes because itll make it positive

double negative

[icy lance]

i didnt say x

i said can (1-x) be negative

[red mountain]

then no

[icy lance]

so 1-x>=0

1>=x

that is your domain

[red mountain]

it wants it in interval notation

[icy lance]

(-infty,1] then

[red mountain]

i still dont reallt get it tho

[icy lance]

when you look at domain youre seeing what x can be for the function to actually be defined in whatever number system youre in

you have a sqrt(1-x) and we established that the input of a square root cant be negative

therefore 1-x>=0

has to be otherwise the root isnt defined in R

which means x<=1

[red mountain]

are you just replacing the 0 wuith the 1

flipping the signs

[icy lance]

im not replacing things

i just added x to each side

[red mountain]

okay

thanks

.close

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[safe radish]

@indigo sail Has your question been resolved?

[safe radish]

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[vivid mortar]

https://cdn.discordapp.com/attachments/839368000746684436/1212629579311620176/IMG_4798.jpg?ex=65f2884d&is=65e0134d&hm=fd9df725cc07bd0d89d12a94641ea3a4ef76dbf60d620f713fbc5b84cbb32da1&

[vivid mortar]

Hi, does anyone know how to do c here?

its supposed to be very easy but i cant figure out the trick

[plucky elk]

use z score

[vivid mortar]

as in z = ( x-mu)/sigma?

[plucky elk]

right

[vivid mortar]

im just unsure how to apply that here

[plucky elk]

https://www.khanacademy.org/math/statistics-probability/modeling-distributions-of-data/z-scores/v/ck12-org-normal-distribution-problems-z-score

[vivid mortar]

Thank you, I watched the video but Im still unsure how to apply that to this problem, would you mind showing me how you'd set up the z-score equation?

I keep getting k = (50+4k-50)/4 which doesn't seem to lead anywhere

[safe radish]

@vivid mortar Has your question been resolved?

[plucky elk]

how does that use z score ?

you need 2 of them. one for the left inequality and one for the right

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@vivid mortar Has your question been resolved?

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[dusk fog]

how did step 1 get to step 2?

I got the part where -2 cos 20 sin (80 -x) turns to - (sin(100 - x) + sin(60 -x)), but idk about the right side

[median vigil]

,, \sin(\ang{90}-x) = \cos(x)

[flat frigate]

cloud

[dusk fog]

oh damn

didn't think of that.

w8

yeah that checks out, tysm

.close

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[dense wadi]

Is there any tricks with u substitution

[dense wadi]

lets say this

how do i know which term to become u

ik its given here but lets say a random problem

[restive niche]

Sometimes u subs can be hard to spot, its just a matter of practice

In this case it is noticing that the derivative of sqrt(x)=1/(2sqrt(x))

There is no "general method" to always know what u sub to make

[median vigil]

it mostly comes down to experience. things to look out for:

• functions inside other functions
• functions on the denominator
• functions which have a their derivative multiplied by the rest of the integral
  it's still basically just judgement and trial and error

[safe radish]

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[tiny heart]

??

[limpid holly]

W approach

[tiny heart]

stop everyone be sending that image 💀😭😭 lol

anyways

so

i did

aplha + aplha + beta

= -b/a aka 3

which is

2 aplha + beta = 3

thats eq one

and then i did

[pure ibex]

wait wait what do you mean by alphas and betas

what exactly are you doing here

because 3 is not a root

[tiny heart]

yeah its not but its -b/a

cuz sums are equal to -b.a

-b/a

wait

question 17 be

BTw

[pure ibex]

sir this is a cubic function

[tiny heart]

....????????

i thought it didnt matter tho

[pure ibex]

ive never seen someone do it that way but what was your answer?

I cant read that

[tiny heart]

idk how to get the answer 💀😭

[tiny heart]

sorry yeah my handwriting gets messy

[tiny heart]

my school teacher approached the question this way

well another question

similar to this

[pure ibex]

was it quadratic?

[tiny heart]

na

ill show pic

[pure ibex]

ahhh I see what this is

ok so to answer your question

2a + b = 3

a^2 + 2ab = 0

a^2b = -4

b = -a/2

-3a/2 = -3

a = 2

b = -1

2, 2, -1

[tiny heart]

yea -1

ok ty

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[frozen marlin]

hmmm

well, we know what two of the digits must be

@small epoch is it a specific number of digits

...

thinks

well it must be exceedingly small, i believe

brb

[drowsy citrus]

There must be some contraint.

Is it for ALL numbers ?

Nah nah.

ALL numbers as in, two digit, three digit, four digits.....

Shit.

Make different cases, then.

[frozen marlin]

wait hang on

i think i got smth

[drowsy citrus]

What ?

If you have a question to upload, please upload it seperately.

[frozen marlin]

$n=\sum_{n=0}^{n_{d}-1}10^{n}d_{n}$

@drowsy citrus bro atleast read the chat b4 coming in

wait hang on one mistake

$n=\sum_{n=0}^{n_{d}-1}10^{n}d_{n+1}$

[flat frigate]

ren

[frozen marlin]

d_n is the nth digit; n is the number in question, and n_d is the number of digits.

now given n mod 4 = 0, what is the probability of (d_1 + d_2... d_n_d) mod 4 = 0

yes

[buoyant pond]

i think it is sigma 2•3^(n-1) n is the number of digits

yea that i do

[drowsy citrus]

Nope. What bout 164 ?

[frozen marlin]

given $\left(\sum_{n=0}^{n_{d}-1}10^{n}d_{n+1}\right)\operatorname{mod}4=0$, what is the probability of $\left(\sum_{n=1}^{n_{d}}d_{n}\right)\operatorname{mod}4=0$

[flat frigate]

ren

[frozen marlin]

my god that is long

lmfaoooo

wait
let's expand the series, maybe.

$\left(1\cdot d_{1}+10\cdot d_{2}+10^{2}\cdot d_{3}\ ...\ 10^{n_{d}-1}\cdot d_{n_{d}}\right)\operatorname{mod}\ 4\ =0$

[flat frigate]

ren

[buoyant pond]

the digits would be either of 0 4 8

[frozen marlin]

probability of $\left(d_{1}+d_{2}+d_{3}\ ...\ d_{n_{d}}\right)\operatorname{mod}4=0$

[buoyant pond]

nothing else

[flat frigate]

ren

[buoyant pond]

just simple pnc

[frozen marlin]

@buoyant pond what?

[buoyant pond]

i think its right tho

Oh okay

[drowsy citrus]

This is what, the sum of digits divisible by 4 ?

[buoyant pond]

Rip

[frozen marlin]

@small epoch checked my expansions yet??

[drowsy citrus]

Also, what is this and how did this thing show up ?

[frozen marlin]

it's... stuff

why did bro remove my emojis

@drowsy citrus why did you want to remove it?

[drowsy citrus]

Woow !! That explained a lot !

[drowsy citrus]

What ?

[frozen marlin]

please be respectful

@small epoch got anything? imma brb

okay

imma be back in a bit. ping if u got anything useful

...

thanks, mods

:D

wait try and use combinations

or combinatorics

whatever

OH im an idiot

one second

just find out

the probability of a number's digits

summing to a multiple of 4

and multiply that by 0.25

u didnt get anything?

sighs

imma just say sum mod 4 when i wanna say that the sum of the digits is div by 4 i cant type that much\

for 1 digit numbers, prob of sum mod 4 = 0.2

oh.

it should though

because that throws us off

but ok

nah not really

just messing with ya

we dont need to inc 0

for two digit numbers (including 1 digit numbers):

for 4--04, 13, 22, 31, 40
for 8--08, 17, 26, 35, 44, 53, 62, 71, 80
for 12--39, 48, 57, 66, 75, 84, 93
for 16--79, 88, 97

the digits add to a multiple of 4

24/100

??

i know

sigh

you know what

imma work this out

and DM it to you

nah lol this is a pretty awesome question

i need to go somewhere, i'll message you if i get anything

i mean i think so

idk whatever u want

[safe radish]

@small epoch Has your question been resolved?

[safe radish]

@small epoch Has your question been resolved?

[craggy sparrow]

What is the question?

[safe radish]

@small epoch Has your question been resolved?

[cunning escarp]

Yes, what is the question good sir?

[slender pendant]

Is that the original question?

[velvet escarp]

how many digits does the number have? is that known or not given?

[frozen marlin]

not given

it's for all numbers, i believe

[velvet escarp]

oh ok

[frozen marlin]

@small epoch sorry didn't get time to solve it

well, we have 4 helpers on the case

shrugs

[slender pendant]

My guess would be like 25%. Let me test that , and see if i can prove it

[cunning escarp]

Oh lord, I have been trying but not a constant pattern

We have true, true, false, true

i don't know how to solve this

In case 0, we have yes, case of 4, yes again, case 8 yes again, but in case of 12 we have a no, and then in case of 16 we have yes and so on

Would you mind if I use python to find answer?

Ok

[drowsy citrus]

Guys rather than that, can we try finding, all the numbers divisible by 4 and whose sum is divisible by 4, case-wise ?

[cunning escarp]

Number of true outcomes: 62493
Number of false outcomes: 187507

Probability is 0.33328355741385657

Ok, we have 0, 4, 8, 16, 40, 44, 48 etc.

My bad

Maybe we can try again

[slender pendant]

Generally speaking 25% of all numbers have sum divisible by 4

[cunning escarp]

I have tried it on 100 million numbers and here's the result, if you want more information plz contact me via discord direct messages

Number of true outcomes: 6249991
Number of false outcomes: 18750009
0.3333326933336405

The sequence now is; 0, 4, 8, 32, 40, 48...

Wrote it by mistake

I don't know

It's more like 1/3

probability

I must ask, are you asking for probability or possibility

because both are different

Ok, I get it. I got confused. My bad. I though you were asking for possibility

Then it is around 25% only

[slender pendant]

You can prove that it doesnt matter n mod 4

But then you just prove that all numbers have 25% of chance that their sum is divisible by 4

You can prove that whether number is divisible by 4 or has remainder 1. The probability is same

[cunning escarp]

But, that proof is no longer used because it lacked mathematical back, and axiom it is based on was proved incorrect

[slender pendant]

No i mean you prove it from scratch

Yoh dont assume that its true for numbers divisible by 4

[cunning escarp]

You're correct, my apologies for the oversight. Let's correct the approach.

Let's denote the number as ( \overline{abcd} ), where ( a, b, c, d ) are digits. Given that the number is divisible by 4, ( cd ) must be divisible by 4.

Now, let's consider the sum of the digits of the number: ( a + b + c + d ).

We know that ( cd ) is divisible by 4, so ( c ) and ( d ) must form one of the pairs {00, 04, 08, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96}. For each pair, there are 10 possibilities for ( a ) and ( b ) (0 through 9). Among these 100 possibilities, we need to count how many of them result in a sum divisible by 4.

Let's go through each case:

• For every 4th multiple of 4, the sum of the digits will be divisible by 4.
• For every 4th multiple of 4 plus 2, the sum of the digits will also be divisible by 4.

This is because if ( cd ) is divisible by 4, then the sum of the digits is also divisible by 4 regardless of ( a ) and ( b ). Therefore, 25% of numbers divisible by 4 also have a sum of digits divisible by 4. Thus, the probability is 1.

[flat frigate]

KavyaSahai

[slender pendant]

I mean it looks like that "number is divisible by 4" is just a distraction

[cunning escarp]

Is your help forum solved?

@small epoch

What is it?

I'll try my best

[slender pendant]

Isnt this just PIE

[cunning escarp]

The probability that the sum of digits of a leap year is also a leap year between 0 and 10000 is: 0.25

How?

good sir?

[slender pendant]

P(It is divisible by 4) - P(it is divisible by 100) + P(it is divisible by 400)

[cunning escarp]

It seems like, bigger the dataset you have, more closer the probability is to 1/4.
The probability that the sum of digits of a leap year is also a leap year between 2000 and 2024 is: 0.29
The probability that the sum of digits of a leap year is also a leap year between 0 and 2024 is: 0.26
The probability that the sum of digits of a leap year is also a leap year between 0 and 10000 is: 0.25

[slender pendant]

You calculate all the probs

Wait is the mini problem finished?

[cunning escarp]

I had the same question for my Ph.D., and I did my hand on more than 10,000 leap years

So, I am well-versed with this question

[slender pendant]

I mean i got an idea thats why im asking

[cunning escarp]

The probability that a randomly chosen year is divisible by 4 is ( P(\text{divisible by 4}) = \frac{1}{4} ), because 1 out of every 4 years is a leap year.

The probability that a year is divisible by 100 is ( P(\text{divisible by 100}) = \frac{1}{100} ), because only 1 out of every 100 years is not a leap year (divisible by 100 but not by 400).

The probability that a year is divisible by 400 is ( P(\text{divisible by 400}) = \frac{1}{400} ), because only 1 out of every 400 years is a leap year despite being divisible by 100.

Therefore, the expression for the probability that a randomly chosen year is a leap year using the leap year rules is:

[ P(\text{Leap year}) = P(\text{divisible by 4}) - P(\text{divisible by 100}) + P(\text{divisible by 400}) ]

[ = \frac{1}{4} - \frac{1}{100} + \frac{1}{400} ]

[ = \frac{100}{400} - \frac{4}{400} + \frac{1}{400} ]

[ = \frac{97}{400} ]

So, the probability that a randomly chosen year is a leap year using the leap year rules is ( \frac{97}{400} ).

Which is around 0.25 only

[flat frigate]

KavyaSahai

[slender pendant]

Ok so prob that sum of digits of leap year is divisible by 4 is 1/4.

Now we get the other 2

[frozen marlin]

this is STILL GOING

WHAT

[slender pendant]

What is the prob that number divisible by 4 is divisible by 100

[frozen marlin]

I WENT FOR A CLASS AND CAME BACK

AND THIS IS STILL GOING

[slender pendant]

Its 1 / 25

[frozen marlin]

don't you mean 0.04

aka 1/25

ok

yep

[slender pendant]

Yes my bad

[frozen marlin]

dedication

[slender pendant]

Now that its divisible by 400

[frozen marlin]

feel like he was using GPT

[cunning escarp]

def is_leap_year(year):
if year % 4 == 0:
if year % 100 != 0 or year % 400 == 0:
return True
return False

def sum_of_digits(n):
return sum(int(digit) for digit in str(n))

def probability_sum_leap_is_leap(start_year, end_year):
count_leap = 0
count_sum_leap = 0
for year in range(start_year, end_year + 1):
if is_leap_year(year):
count_leap += 1
if is_leap_year(sum_of_digits(year)):
count_sum_leap += 1
return count_sum_leap / count_leap if count_leap != 0 else 0

start_year = 0
end_year = 10000

probability = probability_sum_leap_is_leap(start_year, end_year)
print(f"The probability that the sum of digits of a leap year is also a leap year between {start_year} and {end_year} is: {probability:.2f}")

If this code is visible to you, which i used

[frozen marlin]

dam

[cunning escarp]

It is clearly evident that, it is 0.25% only.

If it is a leap for example, 2020. Then the digits should add upto 4 which is also a leap year.

[frozen marlin]

yep it's 0.25

[slender pendant]

Its like

[frozen marlin]

hang on

[slender pendant]

,calc 0.25 - 1/25 + 1/25 * 1/4

[flat frigate]

Result:

0.22

[slender pendant]

0.22?

[frozen marlin]

then the probability of a number's digits adding to a mult. of 4 is 1/16

[slender pendant]

Does it make sense

[frozen marlin]

PEMDAS/BODMAS

?

[cunning escarp]

I mean the larger the dataset we use, more we approach it

[slender pendant]

Bot did it how i wanted it to be done ig

Do you know the ans?

[cunning escarp]

Like if I use the dataset, 0 to 100, probabilities would be different, but if I used 0 to 10,000 probability is closer to 0.25 and when I use the dataset 10 100,000,000, the probability is 25.0000000000068%

The probability that the sum of digits of a leap year is also a leap year between 0 and 100 is: 0.36

[slender pendant]

Then ig i have to write a code in a fast enough programming language to test it today

[frozen marlin]

wait wait wait
oh my fucking god
i think there's a very simple solution; i might be wrong.

as we approach inf for the dataset

[slender pendant]

Im goona go write some rust brb

[cunning escarp]

No, you're getting me, wrong, that is the probability of getting leap year if the digits are also a leap year

[frozen marlin]

the sum loops constantly, from 1 to 9*n_D

AND

since one in every four numbers

is a multiple of 4

one in every four numbers

I SWEAR TO GOD IM AN IDIOT

THIS DOESN'T EVEN NEED ALL THIS

IT JUST NEEDS INTUITION

[cunning escarp]

I used python, because, I know python only, and It was 0.25 only, for 0 to 100 million years, the probability was 25.000000068%

If you can check the python code, it is attached below

def is_leap_year(year):
if year % 4 == 0:
if year % 100 != 0 or year % 400 == 0:
return True
return False

def sum_of_digits(n):
return sum(int(digit) for digit in str(n))

def probability_sum_leap_is_leap(start_year, end_year):
count_leap = 0
count_sum_leap = 0
for year in range(start_year, end_year + 1):
if is_leap_year(year):
count_leap += 1
if is_leap_year(sum_of_digits(year)):
count_sum_leap += 1
return count_sum_leap / count_leap if count_leap != 0 else 0

start_year = 0
end_year = 100

probability = probability_sum_leap_is_leap(start_year, end_year)
print(f"The probability that the sum of digits of a leap year is also a leap year between {start_year} and {end_year} is: {probability:.2f}")

If there's any error, plz let me know

Maybe they would be using dataset like 0 to 100 where it is 0.36

[slender pendant]

I wrote the first problem in python too. But to test it as best as possible we need a big datset

[cunning escarp]

A 100 million is big enough dataset I guess

[frozen marlin]

by the way

@cunning escarp

this is off

because @small epoch doesn't include 0 as a multiple of 4

so whatever

[cunning escarp]

Oh

That makes sense

Lets make a small adjustment

Still, The probability that the sum of digits of a leap year is also a leap year between 1 and 100000 is: 0.25

[frozen marlin]

no

it is not 0.25

good sir

[cunning escarp]

What would be it then?

[frozen marlin]

no idea

[cunning escarp]

https://tenor.com/view/you-have-no-idea-gif-27149353

[frozen marlin]

me rn

nah just messing with ya

it's probably still 0.25

[cunning escarp]

Okay

[frozen marlin]

xD

i mean

4

1111

2222

2020

2002

just off the top of my head

1124

oh yea lmfao

forgot about that

2020, 4, 1124

664

[cunning escarp]

https://colab.research.google.com/drive/1yTHnrfW17V2JB62zZBYNAXMP6EfC9maV?usp=sharing

This is code, if anyone wants to mess with it, they are free to mess with it to make it better, i guess

Can you send me book or website link?

In which the question is mentioned

Okay

[slender pendant]

For first 100_000_000 leap years its true

I mean its 25%

Approx

[safe radish]

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[forest lance]

Hello

[forest lance]

I need help

How to find x

[lean otter]

do you know trigonometry

[forest lance]

Yes

[pastel vortex]

You know the side lengths
Trig functions can relate side lengths to angles, remember SOHCAHTOA?

[forest lance]

Yes

[pastel vortex]

So you can set some trig function of x equal to one side over the other
And use inverse trig functions to solve for x

[forest lance]

Like this? X=Tan^-1 12/15

[forest lance]

But how would I find the actually angle in terms of degree

[pastel vortex]

I'm pretty sure 15 isn't adj

And as to how to get the angle, it's basically a guessing game
For 12/15, it won't be any neat angle, so you can either leave it in terms of inverse trig or get an approximation

[forest lance]

I did that but it was incorrect the answer was 39⁰ and I was confused on how

[pastel vortex]

I'm confused too

[vapid cypress]

tan^-1(12/15) = 37.5 degrees approx

standard angle ig

they approx it to 39

[forest lance]

Ok thanks

But is it possible without a scientific calculator

[pastel vortex]

37.5 to 39?

[timber moss]

,w arctan(12/15)

[vapid cypress]

idk close enough

[vapid cypress]

oof

[flat frigate]

Results provided by WolframAlpha

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

Requested by triaxyz

[timber moss]

.

38.66 rounded to 39

[pastel vortex]

Oh
Well that makes sense pf

[forest lance]

Well thanks

But I have a test on this tomorow is writing tan^-1 12/15 good enough

[timber moss]

yes especially if you cant use calculator

[forest lance]

Because I'm in 9th grade and don't understand all those calculations

.close

[safe radish]

Channel closed

Closed by @forest lance

Use .reopen if this was a mistake.

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[normal torrent]

are these affine planes?

[normal torrent]

cus i dont think so

and i forgot how to make a proof with logic

so could i get help with that pls. you dont have to spoon feed me, just help me a bit

[safe radish]

@normal torrent Has your question been resolved?

[normal torrent]

😦

<@&286206848099549185>

[safe radish]

@normal torrent Has your question been resolved?

[normal torrent]

😦

[safe radish]

@normal torrent Has your question been resolved?

[covert yoke]

@normal torrent so, I know the definition of an affine plane, but I have no clue whatsoever what those doodles are supposed to represent.

My supposition is that b) is supposed to represent the projective plane, with the weird line that jumps from left to right being the line at infinity

Which if that is true, that would make a) the affine plane.

But as it stands, I am much more likely to interpret those images as graphs than as planes without context.

[normal torrent]

Sorry

The question was if they qualified as affine planes

And they said that lines are differentiated based on "drawimg style"

Dotted lines, straight

Etc

Being curved didnt matter

[covert yoke]

Well, I will definitely need to know what a dotted line vs a solid line means

And what do the points mean for that matter

[normal torrent]

Wait lemme geer on my pc

hi

[normal torrent]

its just how it is drawn

you see in a) there is a X shape going from corner to corner

thats a different style from those going from side to side

and those going up and down

points represent the intersection of two "lines"

and if no point is there, there is no intersection

now my problem with this, is that on c) its just a point

how can u have an intersection of lines, with no lines

using the axioms, i dont think i can say these are affine planes

i think i just answered my own question

thanks bye

.close

[safe radish]

Channel closed

Closed by @normal torrent

Use .reopen if this was a mistake.

[normal torrent]

.reopen

[safe radish]

✅

[normal torrent]

i actually dont know

as ive never had anything about affine planes or projection in school

and definitely not incidents geometry

[covert yoke]

so like

are they asking if any of these can exist on an affine plane, or if any of these ARE affine planes?

[normal torrent]

ARE

if so i think they arent

but im so unsure

[covert yoke]

whether or not they are affine planes depends very strongly on what everything in these images actually means

[normal torrent]

lines just mean lines, and if a line is drawn in the same style, it is part of that very same line. in other words a lin can be discontinuous but still be connected.

intersections between lines are symbolised with a point

[covert yoke]

???

[normal torrent]

idk what else to say

[covert yoke]

ok so

[normal torrent]

idk what else it could mean

[covert yoke]

I thought I remembered b having a T junction where three of the same style line came together

which would be literally impossible if the interpretation was as you mentioned

[normal torrent]

so is this wrong

[covert yoke]

I don't know

[normal torrent]

bc thats what the problem said

[covert yoke]

but I looked up and I misremembered

there's a junction where 4 of the same line comes together

which is ambiguous

it's an intersection of a line with itself, but it could be in one of two configurations

that being said, that's impossible in an affine plane.

[normal torrent]

i just thought of a "line" here as more of a family. where the drawing style signifies which family it comes from, and therefore makes it a line

[covert yoke]

so b cannot be contained in an affine plane

I don't know what "family" means in this context.

[normal torrent]

idk bro, thats just how i visualized it

like it said, ive never even come across this in school

[covert yoke]

so, if we run with what you said, then a, b, and c cannot be affine planes in and of themselves.

a and b, cannot be embedded inside of an affine plane

[normal torrent]

ok let me send the problem

[covert yoke]

but c can be embedded inside of an affine plane.

[normal torrent]

(På matematik camp kan vi ikke finde ud af at tegne, så linjer behøver ikke være lige - vi adskiller linjer med tegnestil i stedet. Hvis to linjestykker har samme stil og rammer samme punkt, så er de dele af den samme linje. Linjer skærer kun hinanden, hvis der eksplicit er sat en prik ved deres skæringspunkt. De eneste punkter i systemet, er dem, som er markeret med prikker).

I incidensgeometri er et affint plan en mængde af punkter og en mængde af linjer, som opfylder følgende krav:

For hver 2 forskellige punkter findes der præcis én linje, som går gennem begge.

Givet en linje l og et punkt P, som ikke ligger på l, findes der præcis én linje, som går gennem P og ikke skærer l.

Der findes 4 punkter, således at der ikke er 3 af dem, som ligger på samme linje.

Afgør om nedstående figure er affine planer.

wait lemme translate

sry

(At math camp we can't figure out how to draw, so lines don't have to be straight - we separate lines by drawing style instead. If two line segments have the same style and hit the same point, then they are parts of the same line. Lines intersect only each other if a dot is explicitly placed at their point of intersection (the only points in the system are those marked with dots).

[covert yoke]

(At math camp we can't figure out how to draw, so lines don't have to be straight - we separate lines by drawing style instead. If two line segments have the same style and hit the same point, then they are parts of the same line. Lines intersect only each other if a dot is explicitly placed at their point of intersection (the only points in the system are those marked with dots).

In incidence geometry, an affine plane is a set of points and a set of lines that satisfy the following requirements:

For every 2 distinct points there is exactly one line that passes through both.

Given a line l and a point P that does not lie on l, there is exactly one line that passes through P and does not intersect l.

There are 4 points such that no 3 of them lie on the same line.

Determine whether the figures below are affine planes.

[normal torrent]

yes thanks

given these axioms, they must not be affine planes

no?

[covert yoke]

so being very pedantic here. None of these are affine planes. Because they're just finite collections of lines and points

[normal torrent]

is incidence geometry really so nitty gritty

i havent studied it, but it seems very simple

its just if lines intersect, with no regard to length or angles right?

[covert yoke]

you might have more luck in #point-set-topology perhaps. I'm not entirely sure I have the appropriate background for this.

[normal torrent]

ok

thanks

have a good day

.lcseo

.close

[safe radish]

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[safe radish]

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[shy steeple]

<@&286206848099549185> right>

?

dang it i was wrong

[teal linden]

What do you notice about the two triangles?

In terms of, similarity.

[shy steeple]

both right

[teal linden]

Yes, but they also share the same angle l

[shy steeple]

ye

[teal linden]

Now, with that in mind, how can you relate these two triangles?

what two sides can you setup a relationship?

[shy steeple]

j and i

[subtle python]

Do you know Thales theorem ?

[shy steeple]

no

[teal linden]

i is an angle. I wanted you to see that because that a testament that they are similar triangles.

[shy steeple]

i mean i understand the process but when i calculate it i get some number barely off

[teal linden]

To solve this, you need 1 side from the big triangle that corresponds to the smaller side in the smaller triangle

[shy steeple]

is the answer 9.5

or 9.6

[teal linden]

Think of that answer

[shy steeple]

ohhh im dumb

yea i js got it wrong

[teal linden]

your asked the find n

[shy steeple]

imma do this later lmao im about to go insane

[teal linden]

here

you see that in the smaller triangle, your given that JH is 24 yards, and HI is 27.5 yards

[shy steeple]

i got it wrong

so like

[teal linden]

and i am here to help

give me a sec

so you have two triangles

what side does JH correspond to?

(basically, what side is JH similar to on the big triangle)

[shy steeple]

uhhhh

KL

[teal linden]

Good

now

IH corresponds to what side on the big triangle?

or where it says 27.5 yards

[shy steeple]

uhhh

hl

[teal linden]

sorry if my wording threw you off, but IH corresponds to the LI

or in other words, IH on the small triangle is similar to LI on the big triangle

see what i mean?

[safe radish]

@shy steeple Has your question been resolved?

[safe radish]

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

2sets

[lean otter]

Universal set: {1,2,3,4,5}

A={1,2,3} , B?

[tight void]

unfortunately you have to give me more info

[lean otter]

so, since the left overs in the U set, are 2, B will also have 2 digits?

or can it have 5 digits

[tight void]

is B equal to A complement?

[lean otter]

yes

thats

what im asking

fr

[lean otter]

??

[tight void]

I'm sorry I'm a lil confused on your question

okay so walk me through this

we know that the universal set is {1,2,3,4,5}

[lean otter]

yes

[tight void]

and A is equal to {1,2,3}

[lean otter]

yes

[tight void]

now we don't know what B is right

[lean otter]

yes

[tight void]

now what's the question

it would be great if you can share the original problem statement

[lean otter]

will B be A complement
like will it be, B={4,5}?

[lean otter]

its my question

[tight void]

B doesn't have to be A complement

[lean otter]

im confused about the sets and universal sets

[tight void]

B can be anything

[lean otter]

oh

[tight void]

B can be A

[lean otter]

truee

ok so

im confused cause of a question

[tight void]

mhm

[lean otter]

yo

ok so

[lean otter]

salam, tumhe urdu atti he?

[lean otter]

ok so in here, 32 students represents the elements of the universal set right?

[lean otter]

Wa ʿalaykumu s-salam (and upon you be peace) HAN ATI HA

[lean otter]

abhe yaar, agar uski samaj nehi arehi angrezi mein, mein sumjha sakti hu ke wo kiya kera he

agar urdu angrezi se ehsan he

?

[lean otter]

ahaha yar english theek ha\

teekh he, bus upar se lagraha tha ke aapko shaid puri tarha sumhaj nehi arehi

marzi tumhari

[lean otter]

yar im just confused abt the universal set in a ven diagram

[lean otter]

yahan pe

ye sawal mera

hmm

[tight void]

how many numbers are between 41 and 50

[lean otter]

9

ik how to solve it but

[tight void]

there is 10 lol count it on ur fingers

[lean otter]

im just confused about the universal set

[lean otter]

OK BUT

WHY ARE YOU ASKING THAT

ok so in here, 32 students represents the elements of the universal set right?