help-13

[river topaz]

so y is 60

then + 75

[tepid blade]

Yeah

[river topaz]

135

• z

= 180

z is 45

how do we find x and w now?

[tepid blade]

Uh hold on

[cedar kiln]

@river topaz Has your question been resolved?

[cedar kiln]

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[bitter garden]

Déterminer les valeurs de x telles que : 3A=2(7-x) ; A<x²

[bitter garden]

<@&286206848099549185>

[olive crane]

yes

[bitter garden]

can u help

[cedar kiln]

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[empty river]

Ok. Today i asked for help, got some, but my head started hurting after a while. Now that it dosent hurt no more..

Could i get help trying to understand linear equations?
I have some tidbits of equalition rules, whatever happens on one side, happens on the other. And if theres mostly numbers without variables, we can have them on one side.

[empty river]

And i can solve the First question.

But honestly? The others my mind is kinda foggy

And since its kinda late. Id just like to get to figure out atleast something before going to bed

[hollow totem]

What do u need help with?

Part 1?

[empty river]

Part 1 and 2. Whatever is easier so i can get One part out of the way

Or is 3 even easier?

Again. This is a new world for me

[hollow totem]

let start with part 1

All you have to do, is to move the variable (x) to 1 side and all the number to the other side

[empty river]

Oh i know how to solve that one

2x + 7 = 4x - 13

2x -4x, and 4x - 4x

-2x + 7 = -13

-2x = 7 + 13.

-2x = 20.

-x = -10
X = 10

[hollow totem]

A small mistake here

[empty river]

:0?

[hollow totem]

But u get the idea

So u can solve for all question in part 1?

For part 2

I assume its a system of linear equation

One trick is the select one of the 2 equation, let choose -x+y=5

[empty river]

Oh no

No no.

I mean i solved it but it had so much guidance and assistance from someone else. I doubt i can solve it again alone

[hollow totem]

Then its a problem of practice

You just need to do it repeatedly

[empty river]

Ah. Okay. So skipping 1 for now.

How do we do part 2?

-x + y = 5

3x - 2y = -12?

[twilit escarp]

you want to make x or y disapear

lets try to make x disapear

[hollow totem]

@empty river

[twilit escarp]

multiply the first line by something as when we add the two lines together the x disapear and we only have y left

[hollow totem]

Here's a very detailed step by step solution, the point is for u to apply it to other solution

[empty river]

Hmmm

[hollow totem]

In Step 4, do exactly what u did in Part 1

[empty river]

Lets see..

3x - 2 (5+x)= -12

Uhhh...

[hollow totem]

Multiply the -2 term in, so we can get rid of the ()

[empty river]

3x -2 * 5 + x?

[hollow totem]

No

U need to multiply the -2 to each element inside the ()

[empty river]

Wait. I think i got something wrong

3x -2 * 5 * x?

[hollow totem]

You really need to improve ur arithmetic skill

[hollow totem]

No

[empty river]

Hmmm

3x -2 * 5 + -2 * x?

[hollow totem]

Bingo

Then solve for x

Easy

[empty river]

3x -2 * 5 + -2 * x.
2 times x Is 2x.
2 times 5 is 10.

3x - 10 - 2x = 12?

If i do the same thing i did the last time...

In the first task

3x - 2x. Is X.
And 2x minus 2x is 0.

X - 10 = 12?

X = 22?

...idk

..man it got quiet in here

....okay then

.close

[cedar kiln]

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[empty river]

.reopen

[cedar kiln]

✅

[empty river]

Nvm ima wait a few more minutes

Anyone?

..oh well.

.close

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[cedar kiln]

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[dense hull]

can someone help me get the other solutions ive got pi/18 so far

[dense hull]

<@&286206848099549185>

part b

ive already got one solution

just dont know how to get the other two

ty

<@&286206848099549185>

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[dense hull]

well by using the expression stated in part a, we can equate sinxtanx=sixtan(3x-pi/9)

solve for x

to get one solution

which is pi/18

but what are the others

[ruby tree]

you ve got sinxtanx=sinxtan(3x-pi/9) right?

[dense hull]

yes

and ive solved it down to x

and got one solution for x

but should be more graphically i think but ive forgotten how to do it

[ruby tree]

first you have solutions when sinx=0

[dense hull]

why its a tan graph im solving no?

[ruby tree]

how do you get rid of the sinx in the equation?

[dense hull]

ah i got rid of a solution didnt i

wait no im so confused

ive found the mark scheme

[ruby tree]

well in order to divide by sinx you need to know its not 0

[dense hull]

how do you know its no 0

sorry i hate trig

[ruby tree]

alr

lets say

you have

nvm idk

you have

sinxtanx=sinxtan(something)

bring it all in front

[dense hull]

bring what all in front

the bracket?

how

[ruby tree]

sinxtanx-sinxtan(something)=0

[dense hull]

ah

yeah go on

thats fine

oh

factorise ity

and for the bracket to equal 0

sinx has to be 0?

for one of them

[ruby tree]

so sinx(tanx-tan(something))=0

sinx=0 or tanx=tan(something) right?

[dense hull]

yeah

makes sense

according to the mark scheme theres anotherr solution as well

[ruby tree]

yeah tanx=tan(something) has more solutions

do you know how to solve tanx=tanθ

[desert flame]

hello,what is the question

[dense hull]

yeeh i did

i got x= pi/18

[ruby tree]

no like generally. what are the solutions

on a equation like tanx=tanθ

solve for x

[dense hull]

factor the tan?

[ruby tree]

no...

[dense hull]

god knows then

[ruby tree]

you didnt just remove the tans did you?

[dense hull]

arctan

[ruby tree]

huh?

[dense hull]

idk man

honestly

[ruby tree]

what did you do

[dense hull]

oh

i just removed them

it got me the right solution

[ruby tree]

oh alr

so

an equation like tanx=tany has infinite solutions: x= kpi+y

where k is 0, 1, 2 ,...

you cant get rid of tans as it will only give you one solution

[dense hull]

okay im following

[ruby tree]

alr so if you have tanx=tan(3x-pi/9) how would you solve it

[ruby tree]

(y is 3x-pi/9)

[dense hull]

still not sure lost me again

move everrything to one side?

[ruby tree]

well an equation like tanx=tany has solutions: x=kpi+y. Your equation is tanx=tan(3x-pi/9) which has solutions: x=...?

[dense hull]

pi/9

/3

pi/27?

[ruby tree]

are you telling me numbers?

[dense hull]

yeah

idk

im so confused sorry

[ruby tree]

alr lets go back a bit

[dense hull]

okay

[ruby tree]

do you know how to solve like sinx=sin(pi/4) for example?

[dense hull]

nope

oops

x=pi/4?

[ruby tree]

really? that first question seems a lot more complicated than that

[dense hull]

idk i just find this confusing

the rest of the paper has been easy

just this is baffling me

[ruby tree]

yes but that is just one solution. there are infinite. for example sin(3pi/4)=sin(pi/4)

for reference, what grade are you in so ik what i can say and what not

[dense hull]

so it depends on the domain

grade 13

uk

just before university

so bit of calculus trig algerbra

[ruby tree]

alr so

tanx=tany alr? thats the general form of the equation. you can subtitute anything in the place of y and solve for x

[dense hull]

yeah thats fine

[ruby tree]

the general solutions to this are x=kpi+y

now in every equation

you just need to subtitute the y that you have in the general solutions

for example

tanx=tan(pi/4) sooo x=kpi+pi/4

do you get this?

[dense hull]

kinda so the pi/4 is the difference

[ruby tree]

difference?

[dense hull]

sorry im thinking about it how its taught in physics

[ruby tree]

sure

[dense hull]

i can show u on a diagram if u would like

are u able to voicecall?

[ruby tree]

not really tbh i am not in the right place to do so

[dense hull]

okay no worries

ill draw it rq

[ruby tree]

so examples:

tanx=tan(pi/3)

x=kpi+pi/3

tanx=tan(pi/4)

x=kpi+pi/4

tanx=tan(pi/6)

x=kpi+pi/6

tanx=tan(pi/2)

x=kpi+pi/2

[dense hull]

is this kinda what u mean

[ruby tree]

no, not really tbh

what are these graphs

[dense hull]

tanx

sorry for poor drawing

[ruby tree]

ohh

wait

i mean kinda...?

you see tanx kind of repeats itself

over and over again

[dense hull]

yeah i do

but the thing im confused about

is that

by domain is between

0 and pi

[ruby tree]

thats not the concern here

[dense hull]

so it shouldnt repeat

oh shit

its the 3x

[ruby tree]

we are trying it more general

[dense hull]

compressing it

okay

yeah my bad

i think i understand

so

[ruby tree]

yes it doesnt.

[dense hull]

oh

how is there two solutions in the domain then

[ruby tree]

well it does but not infinitely

ignore the minus 1 this is generally the graph of it

[dense hull]

okay

so ur showing the number of oscillations in the domain?

[ruby tree]

that is not important tbh

[dense hull]

but i thought its the y value read of onto the other lines

for the x value

[ruby tree]

we are not on the question yet. we are trying to make you understand the general solution

[dense hull]

okay

[ruby tree]

so tanx=tany

y can be whatever

pi/3

pi/6

pi/8

whatever

ok?

[dense hull]

ok

[ruby tree]

for all y though

the solution for x is: kpi+y

in every case

you subtitute y for whatever it is

[dense hull]

yes

[ruby tree]

if it is tanx=tanpi/4 (y=pi/4) you do x=kpi+pi/4

do you get that?

[dense hull]

yeah youve subbed it in

[ruby tree]

alr

so if you have

tanx=tan(3x-pi/9) what is your y?

[dense hull]

pi/9

[ruby tree]

really?

tanx=tany

tanx=tan(3x-pi/9)

what is y

[dense hull]

3x-pi/9

[ruby tree]

great

so you can just subtitute it in x=kpi+y

[dense hull]

so kpi + 3x - pi/9

[ruby tree]

yes x= that

[dense hull]

okay

and thatll get me the other solutions generally

[ruby tree]

yes exactly

[dense hull]

brillaint

makes alot more sense

thank you so much

[ruby tree]

then you solve for x

x= kpi +3x-pi/9

so 2x=pi/9-kpi

x=pi/18-kpi/2

you see for k=0 you get the pi/18 that you found

but there are more solutions

for k=1...for k=2 etc

[dense hull]

where do i get the value of k from sorry

ive lost myself again

oh number infront of x

[ruby tree]

k doesnt have a value. it takes the values 0,1,2,3,4.....by putting k=0 you get a solution. by putting k=1 you get another solution etc. its what gives you the infinite solutions

[dense hull]

ohhhhhh

so i can just do that as many times as the x value fits into the domain?

[ruby tree]

yes yess

[dense hull]

cheers mate

.close

[cedar kiln]

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[ruby tree]

see ya

[cedar kiln]

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[crimson sedge]

Monica has 3 dice which the sum of oposite numbers is always 7. She positions them side by side so that the numbers touching are the same, getting a 3 digit number in the superior sides. Example: the number 436 can be obtained. How many different numbers can she get?

[crimson sedge]

Could someone explain to me why 6.4.4 = 96 isn't the answer?

[undone halo]

Wait

where () is the side number

actually

[crimson sedge]

hmm

I just thought: well you have 6 options for the first numbers, 4 for the second and 4 for the third

since when u choose the first number in the side, you automatically exclude 2 for the second

[undone halo]

i just have a hunch there might be some duplicates

[granite knoll]

how about the ordering of the numbers? is that valid

(FYI: "dice" is the plural already)

[undone halo]

Is vertical dice reading allowed

Like
4
3
6

[crimson sedge]

no

[crimson sedge]

wdym

[granite knoll]

was just trying to think if the ordering of the numbers matters. e.g. 436 vs 346. but maybe you already account for that with 6 * 4 * 4

[crimson sedge]

oh ok

yeah it already counts but somehow there's more numbers you can get

I don't understand where is the mistake with 6.4.4. Really bugging me rn

[granite knoll]

you can try asking in #probability-statistics , there are people good at combinatorics problems there

[crimson sedge]

thank you

.close

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[cedar kiln]

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[analog wraith]

2. Solve the following using small angle approximations:
   (i) cosθ + sinθ + tanθ = 0

[analog wraith]

Answering this question

I got to θ = 2

I havbe the questions but not the answers, can someone confirm if I got it right?

My calculations look like this

[mental trail]

O^2 - 4O + 2 is definitely not the expression below

[analog wraith]

ffs

im dumb asf

give me 2seconds

🤣

[analog wraith]

Could I factorise out theta from the 3rd line?

something like

θ(-θ + 4) = 2

or is this a situation I should use the quadratic formula?

[sharp pike]

You should definitely use the quadratic formula for that

[analog wraith]

okay, ill use quadratic formula, ty

so would the answer be -2+√2 or -2-√2

[sharp pike]

How did you get that quadratic equation

[analog wraith]

-b+-√b^2-4ac / 2a

[sharp pike]

No lol I meant theta^2-4theta+2=0

[analog wraith]

this is my full working

[sharp pike]

what identities did you use

[analog wraith]

This is the question

sin = theta, tan = theta, cos = 1 - theta^2/2

[sharp pike]

um no

how did you get that?

[analog wraith]

that's the identities for small angle approximation no?

[sharp pike]

oh

I mean I graphed this on desmos and the angles are not very small

[twilit escarp]

even when theta approaches 0 ?

[sharp pike]

Well when theta is 0, the sum is 1

and the graph is continuous close to theta=0

[analog wraith]

Can someone else try to solve:
Solve the following using small angle approximations:
(i) cosθ + sinθ + tanθ = 0

to see what they get so I can see where I went wrong

[cedar kiln]

@analog wraith Has your question been resolved?

[analog wraith]

<@&286206848099549185>

[static whale]

these are right

so you get 2theta + 1 - theta^2/2 = 0
which rearranges to ||theta^2 - 4theta - 2 = 0||
quadratic formula gives theta = ||(4 +- sqrt(16 + 8))/2 = 2 +- sqrt(6)||
we discard ||the + solution|| because ||it is too large|| and we get theta = ||2 - sqrt(6) approx -0.45||

[analog wraith]

so I got it right? 😄

well

I got the solution but didn't specify whcih one

o wait

nvm

rip

got it wrong xD

okay, thank you 😄

[static whale]

wait sorry i messed up the algebra a bit!

[analog wraith]

thanks alot ❤️

[static whale]

nws

i fixed it in the message

[analog wraith]

thank you, sorry, I had already seen the edit, that's why I said thanks 😄

!close

.close

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[turbid sphinx]

I have no idea how to solve this one and feel as though I can learn from the answer: Let's say you have a bag with a blue and a green marble. You pick one at random. If it's a blue marble you put it back in the bag, then you put another blue marble into the bag. If it's a green marble you do the same but with a green marble. If you do this a very large number of times, what are the odds that the bag will tend toward having an even split between blue and green marbles?

[turbid sphinx]

I know that the more one color is picked the more likely it is to be picked the next time, so there's a good chance the same color gets picked the first few times then those odds just run away and it becomes effectively all blue or all green, but the case can exist where the 1/3 chance of picking green second after the first marble was blue, evening it out. If things keep being even - which by law of large numbers they ought to be more likely to - then you wind up with about a 50/50 split of blue and green marbles.
I would still consider it a tendency toward evenness if there are 10000 marbles and 5003 of them are green.

I know that if the number of marbles in the bag goes up evenly between blue and green the odds it gets thrown off get infinitesimally smaller; the odds have to converge somewhere, but where? And how do I calculate it?

<@&286206848099549185>

[stuck crystal]

technically there are infinite possibility for it to be equal, blue majority, and green majority

so unless there a limit, the possibility would require a huge deal of effort to find or its just infinity

sicne there are infinity combination

[turbid sphinx]

well i was thinking there are three possibilities over enough time, not infinite:
~ either one of the two colors gets picked a few consecutive times (the odds of one color getting picked increase with the amount of marbles of that color in the bag) and that causes that color to take over completely or
~ the two colors stay relatively even, which gets more likely as it continues to tend to happen

i doubt the possibility of staying at some other ratio, as the bigger color could easily take over before the number gets too large to change, but we'd have to run simulations for that

[fair geyser]

even if you do it the opposite way, add the color you didn't pick

you get decreasing probability that it's even split the longer it goes

i think

[turbid sphinx]

i was under the impression that the longer it goes the law of large numbers keeps it around the specific point wherever it is

[fair geyser]

that's alos true, but i mean exactly equal

anyway i don't have anything to say

so i;m just googling so far

[turbid sphinx]

yeah but it'll never be exactly equal when the number of marbles in the bag is odd, and even if they're not equal if the proportion they're not equal by is statistically insignificant (i.e. 5003 out of 10000) then it would probably just as easily equalize given a few more iterations

[fair geyser]

so you're asking what's the probability that it will be split at least once

i don;t know what i thought you were asking

is that correct though?

[turbid sphinx]

...

I'm asking two things:
First: the accuracy of my conjecture that a system like this will almost always end up in one of the three states i mentioned
Second: given the first conjecture the odds it ends up in the middle state

[fair geyser]

but what does it mean to "end up"

[turbid sphinx]

the status of the contents of the bag after an arbitrarily large number of iterations

[fair geyser]

it will not be evenly split

even if you add the color you didn't pick back in

i don't understand, it's fine if you can't explain, i almost never can explain anything

[turbid sphinx]

so you start with a bag with a blue marble and a green marble

you pick one at random

let's say blue

you put a second blue marble in the bag

so now you have two blue and one green in the bag

[fair geyser]

yeas but that's more complicated

[turbid sphinx]

that was my question in the first place!!!

[fair geyser]

so i'm trying to understand it on the scenario where it's the other color that gets added

[turbid sphinx]

that always ends up even!!!

[fair geyser]

that doesn't make sense

if you name the number of steps, it's not 100%

if you increase the number it becomes smaller

at least that's my conjecture

if you don;t name the number then it doesn't "end up"

[turbid sphinx]

what's not 100%? what becomes smaller as which number increases?

am i being trolled?

[fair geyser]

after 10th step there are 12 marbles
the probability is <1 that they are evenly split
after 12th step there are 14 marbles
the probability is less than it was after 10th

i'm kinda demanding an explanation, it's trolling in that sense, that you were supposed to be the asker

but i'm acting like it's me

i genuinely don't know i;m not pretending

[turbid sphinx]

If you have three marbles (2b 1g) the odds of picking the green marble are 1/3, but doing so resets the odds to 50/50, then you pick an unbalance and with say 3g 2b the odds of them evening out is 2/5, then 3/7, then 4/9, then 5/11, then 6/13, as it edges infinitely closer to 1/2

By law of large numbers, the more marbles are in the bag, the less likely a massive swing is!

[granite knoll]

how do the odds get "reset"?

[turbid sphinx]

it's the difference between a 2/5 becoming a 2/6 and a 500/1001 becoming a 500/1002

[turbid sphinx]

the number of each kind of marble returns to completely even

[granite knoll]

but you still have to account for the probability of "getting there". 1/2 the first draw is blue, times 1/3 the second draw is green, times 1/2 the next is blue, etc. right?

well, they're not independent so I don't think you can just multiply them

[turbid sphinx]

cut the 1/2 parts; the unbalance is the same whether you pick blue or green

[stuck crystal]

we could concluded it as binomial distribution

i am just not sure how to do it

[fair geyser]

if I look at the interval near the middle that scales with the number of steps, that seems to be true, it's more rarely outside with more steps

[stuck crystal]

because it does seems like that the number of ways that it ends in same number if marbles for each color gets smaller amd smaller each time

like those pachinko phenomenon

for each draw, there are two possible outcome, or intersection

and at the bottom, there should only be 3 results

for odd numbers of marble there have to be only one

and 3 for even

[turbid sphinx]

what are you even talking about
there are two options for when the odds aren't equal (regularize or diverge further) and one when they are (put another marble in)

[fair geyser]

@turbid sphinx so you're saying that suppose it stayed even after 6 steps, avoiding the runaway that we would expect

the next 6 steps have better chances to repeat this

because now our definition of close to even expanded

[turbid sphinx]

yes they would

[fair geyser]

with more marbles

so the speed of running away is smaller

[turbid sphinx]

speed of potential runaway is smaller therefore greater odds of self-regulation

[fair geyser]

okay

well I don;t understand why there's tendency to self regulate, you're saying there's like an attractor in the middle

and it's weaker than the sides

but then it becomes very strong, even stronger than the sides, as long as we stay near by chance, we will eventualyl stay near by law

[turbid sphinx]

i'm not saying there's attraction in the middle; i'm saying as the number of iterations goes up the pull from the sides gets weaker

[granite knoll]

but that's assuming that the "right" sequence of marbles were pulled earlier on. isn't it?

[fair geyser]

aha

[turbid sphinx]

those are the odds i'm asking about

[granite knoll]

because if you pull a handful of blues in a row (at the start), it's less and less likely they'll end up even

[turbid sphinx]

exactly

[fair geyser]

like if there's literally a massive body that attracts you, but it only pulls you on average, and due to variance maybe you end up further away

[turbid sphinx]

i'm saying due to the volatility of initial conditions the odds that it ends up any way other than those three states is lower than any of those three specific states

[fair geyser]

but then it's easier to get even further away

you have 100 keys, and you're trying to find the right one for the lock
you try them all in a random sequence
which key is the "average" the one that's most likely to fit?

i'm not sure if it's relevant at all, sorry in advance

[turbid sphinx]

I am fed up with the three of you. We are no closer to the solution than the initial conjecture (which has not changed from the initial proposition). I will no longer tolerate your waffling; it is borderline trolling. Contribute or leave.

[fair geyser]

but there's no one else

we're entertaining you while you wait for the fourth person

[turbid sphinx]

I genuinely hope you find yourself in a dark place and the people you turn to for help mock your plight. Go away.

[fair geyser]

no we won;t discuss why that would be good cwatson, that's definitely off topic

[turbid sphinx]

<@&268886789983436800> these guys are 100% trolls

[fair geyser]

you;re confusing one kind of disrespect for another

[turbid sphinx]

I just wanted help with a question

[slate lintel]

um. not sure exactly what's going on here but it seems like frownyfrog you should dip out atp

anyway a random walk goes back to the origin infinitely many times

and i think that's what you're describing

[turbid sphinx]

yeah, and i'm wondering the odds that runaway doesn't occur under the initial conditions

[slate lintel]

0

[turbid sphinx]

0?

how so?

[slate lintel]

well the probability that it never returns to equal is 0

it won't stay there

[turbid sphinx]

i thought if it perchance manages to get to like a 10/10 split out of 20 then the odds of it becoming 15/15 out of 30 or a 50/50 out of 100 become much higher

[slate lintel]

probably is

[turbid sphinx]

if you're saying that there are infinite possibilities over an infinite timespan, what of the odds of divergence becoming infinitely less likely over that time?

[slate lintel]

uhhhh

https://mth49601spring2010.wordpress.com/2010/02/04/lecture-10-a-1d-random-walk-visits-the-origin-infinitely-often/

idk, this looks digestible if you work at it

there's also the wikipedia version https://en.wikipedia.org/wiki/Random_walk

i mean yeah if you get off to a bad start it'll take longer to recover

but eventually you will

might be after the sun explodes though

[turbid sphinx]

yeah but random walks have equal probablility of the options, whereas the question at hand involves diminishing odds

[slate lintel]

ok lemme reread the question hold on

in my mind the fastest way to answer this is to simulate it

it does seem unstable

you can probably use recurrence relationships to analyse it?

[turbid sphinx]

idk recurrence relationships

[slim jasper]

I dabbled in a somewhat related problem recently -- not quite the same, though, and it's not immediately clear whether the techniques there are still useful when the growth rates are the same. https://mathoverflow.net/questions/470050/a-random-urn-problem-do-the-faster-duplicating-balls-always-dominate

[slate lintel]

that's the same problem isn't it?

[turbid sphinx]

This seems to imply that the system is deterministic

wait lemme check that article rq

[slate lintel]

oh i see the difference now

[turbid sphinx]

that one has the black balls being more likely than the white ones by putting two more in as opposed to just one

[slate lintel]

ye

[slim jasper]

Right.

[turbid sphinx]

found it https://en.wikipedia.org/wiki/Pólya_urn_model

[slim jasper]

Good find.

[turbid sphinx]

Also, when there's two numbers in parentheses one above the other like the (n/n.1) in this:

what does that mean?

(this expression is in the Polya Urn article)

[slim jasper]

That's a binomial coefficient.

[turbid sphinx]

just looked it up i recognize the formula

isn't that the one that applies to combinations sampling without replacement?

[slim jasper]

In the case with one white and one black ball initially (x=y=1), the whole expression seems to work out to a claim that after n steps, every possible count of black balls is equally likely -- which sounds so strange that I may be misinterpreting something.

Hmm, no, actually that does seem to make some intuitive sense.

[turbid sphinx]

i worked it out myself

it... evens out given x, y = 1???

[slim jasper]

Suppose that after 99 draws there's an exactly 1% chance of there being 1 black ball, 2 black balls, ..., or 100 black balls. What is the probability of 37 black balls one draw later? That can happen either if there were 36 before and we drew black (with a probability of 36/101) or there were 37 balls before and we drew white (with a probability of 64/101), so the total probability is ((36+64)/101) * (1/100) = 1/101.

[slim jasper]

Yes. Fairly surprising but it seems to check out.

[turbid sphinx]

i was under the impression that given 6 balls seeing 3b 3w was more likely than seeing 1b 5w

[slim jasper]

Yeah, that's what I would have expected intuitively too. actually I had intuitively expected that outcomes in the middle would be less likely than lopsided ones.

We could probably have worked out the right answer for the x=y=1 case without consulting Wikipedia first, if only we'd had the courage to compute the probabilities for the first few steps by brute force, and then notice the pattern ...

[turbid sphinx]

but the odds of landing purely on one side is 1/2*2/3*...*n-1/n => 1/n

[slim jasper]

Oh, that should in itself have been a clue.

[turbid sphinx]

the odds of landing in the middle every time is 1/3*2/5*...[(n+1)/2]/n (given n is odd)

how does that tie in?

OH

[slim jasper]

Oh right, if you want to know about how the entire history looks, instead of just the distribution of the final counts, then we don't have an answer yet. The "exchangeability" section of the Wikipedia article looks like it could help there, though.

[turbid sphinx]

i jujst realized that staying in the middle is just one possibility out of many which would lead to the same end

[slim jasper]

Indeed.

[turbid sphinx]

given 5 balls the odds are equal given there are 4 possibilities, two are calculated to be 1/4 and the other 2 have intuitively equal odds

weird but i get it.

Thanks a lot!!!

before i close, i'd like to ask if you're gonna do anything about someone asking a question and 3 people crowding in all being equally, deliberately unhelpful so nobody else has to deal with that

[void sand]

I'm not exactly sure they were being deliberately unhelpful...

[turbid sphinx]

you sure this isn't a troll response?

[void sand]

hm, I suppose it's not exactly helpful

[turbid sphinx]

schrodinger's a-hole, not joking until you're called out on it

[void sand]

I think this was needlessly aggressive though.

[turbid sphinx]

anyway i blocked all three

[void sand]

that being said, I do think that perhaps they should've disengaged earlier

@turbid sphinx anyways, if you're done here, you should close the channel

[turbid sphinx]

thanks

.close

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[versed kayak]

hi

[versed kayak]

In how many ways can 5 different toys be packed in 3 identical boxes such that no box
is empty, if any of the boxes may hold all of the toys?

can i know how to do this by beggars method/stars and bars method ?

[amber cargo]

Recurrence relation

[blazing zephyr]

this might be wrong, but my method is that assume all the boxes have 1 toys in them, so there are 2 possible configurations left, toys that are in the same box and toys that are in diffrent boxes, so 2×5!/3!, ×5! to permute the toys and /3! to remove the permuration of boxes

tho like i said, this is pribably wrong

yeah its wrong

https://brilliant.org/wiki/distinct-objects-into-identical-bins/#:~:text=Distinct objects into identical bins is a problem in combinatorics,which objects are grouped together.

[versed kayak]

what is S

my approach was like

we can write

x1 + x2 + x3 = 5\

since no box can be empty

[blazing zephyr]

the boxes are distinct no?

[versed kayak]

x1 = t1+1 , x2 = t2+1 , x3 = t3+1

[versed kayak]

cant we use this if the boxes are distinct ?

[blazing zephyr]

sorry i meant the boxes are identical

[versed kayak]

yea but here x1 x2 x3 representa the number of balls in boxes

[blazing zephyr]

also im pretty sure your method works for when the toys are identical and the bins are distinct

[versed kayak]

okk this is reverse situation ig

[blazing zephyr]

since your solution does care about the permutation of the bins (ie x1 x2 x3) but doesent care about the toys (which is the value of it

[versed kayak]

yea yea

got it

[cedar kiln]

@versed kayak Has your question been resolved?

[versed kayak]

<@&286206848099549185>

[blazing zephyr]

.

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[strong jacinth]

how do i do question 2F

[blazing zephyr]

$$\frac{1}{\sqrt{x}}=\frac{1}{x^{\frac{1}{2}}}=x^{-\frac{1}{2}}$$

[wraith dagger]

Skill_Issue

[strong jacinth]

yeq

[strong jacinth]

sorry for the late response

so then i got

-1/2 x ^-3/2

<@&286206848099549185>

[crimson sedge]

hola

[strong jacinth]

hi

[cedar kiln]

@strong jacinth Has your question been resolved?

[crimson sedge]

speak Spanish

[strong jacinth]

no

its good i figured it out thsnks

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[crimson sedge]

Well, I don't speak much English, nor do I know much, so to speak.

[austere ember]

quieres ayuda?

[crimson sedge]

si, no entiendo ingles

jaja

[austere ember]

ah

soy un hablante nativo de ingles (pienso que son las palabras correctas)

pero estudie espanol por cuatro anos

probablamente hay una canal especificamente para ayuda en espanol

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[rose cliff]

b ii

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[rose cliff]

b ii

[cedar kiln]

@rose cliff Has your question been resolved?

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@rose cliff Has your question been resolved?

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@rose cliff Has your question been resolved?

[latent current]

Where are you experiencing the difficulty? @rose cliff

[rose cliff]

I tried solving the question and couldnt get 1 + root7 at all

[latent current]

How did you start? Can you show me your approach?

[latent current]

Difference of the 2 radii is not the length/width of the rectangle

If you notice carefully, the bigger circle is not touching the rectangle at the far edge

[rose cliff]

Oh right

how would you get the difference then?

[latent current]

Hint: Pythagorean theorem

Lets just start from here

[rose cliff]

i dont see it

[rose cliff]

Hm ok

sorry but I still dont get how they get the inequality could you just show me?

[latent current]

AJ^2 + JI^2 <= AI^2 for the rectangle to be inscribable?

If AI was a tad bit shorter, the bigger circle would cut the rectangle

So $AJ^2 + JI^2 <= AI^2$

@rose cliff let me know if you get it

[wraith dagger]

fukwerint

[rose cliff]

Yea i get that

[latent current]

So just plug in the values

Let small radius be = r

Then bigger radius is 2r

(By previous part)

So
AJ = 3 + r
JI = 3
AI = 2r

Just plug in and solve the inequality

If you have any confusion you can ask right away

[rose cliff]

Nice i got it

that was a good explanation ty

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[agile ether]

whats wrong? did I miss something?

[livid hound]

yes, first step

how are you getting
4x^3 + 16/x^2

[agile ether]

(x²)² = x⁴ right? then 4(x^4-3) = 4x³

[livid hound]

where's 4(x^4-3) coming from
and that isn't 4x^3 either

i don't understand what you're trying to do

[agile ether]

idk how to explain to english:(

x⁴ so the power multiply with x become 4x and the power (-) by 1

[livid hound]

oh, you're trying to differentiate?

[agile ether]

yes

[livid hound]

still wrong,

you're trying to do multiple things at a time, some partly
making this a complete mess to interpret

there are two main ways to go about this

[agile ether]

so i need to do it step by step?

[livid hound]

yes

and make it clear when you're actually taking derivatives

just having
y =
and =
indicates that everything will be equal to y

method 1: power chain rule immediately
method 2: first expand (just expand, don't make any attempt to differentiate in this expansion step)
then differentiate using power rule

[agile ether]

i will try

@livid hound sorry can I multiply x² with 4x^-1

so 4x^2(-1)?

[livid hound]

whys the (-1) positioned like that

[agile ether]

4x^-2?

[livid hound]

no, that's not how exponent multiplication works

[agile ether]

i mean (2(-1))

[livid hound]

still not good

[agile ether]

oh i forgot

if x²(x²) = x^(2+2)

then what about x²(4x^-1)?

[livid hound]

a^b * a^c = a^(b+c)

same, just add the powers

[agile ether]

ik this rule

ohhh

not multiply?

[livid hound]

no

just add

[agile ether]

so x²(4x^-1) = 4x^(2-1) = 4x

[livid hound]

yes

[agile ether]

tytyty

i forgot about this lol

i confused bcause i remember do the 9^2 = 3^2(2)

@livid hound wait is 4x^-1(4x^-1) = 8x^-1-1 or 4x^-1-1?

[livid hound]

neither

[agile ether]

then

[livid hound]

4 * 4 = ?

[agile ether]

oh lmaoo sorry

16x^-2 ?

[livid hound]

yeh

[agile ether]

i forgot about index thing 😭🙏

i think its right this time

ty! @livid hound

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[true rover]

i need help with this abomination

[ruby tree]

alr

[true rover]

:>

[ruby tree]

starting from (a)?

[true rover]

yea

[ruby tree]

ok so do you know limits, derivatives and stuff like that or nah?

so i know what to use or not

[true rover]

i know

[ruby tree]

alr so the instanteneous rate of change of g at 2.5 would be the limit (x--->2.5) of (g(x)-g(2.5))/(x-2.5))

[true rover]

yea

[ruby tree]

alr moving on to b

[true rover]

wait

how did u get (g(x)-g(2.5))/(x-2.5))

[ruby tree]

thats the rate of change of a function at a point b

g(x)-g(b)/x-b

idk if you know this tho

[true rover]

yea i don't

so i use that if its instant?

[ruby tree]

how do you usually calculate a rate of change

[ruby tree]

yes for instant

[true rover]

oh

[ruby tree]

but the limit of that

not just the fraction

lim(x--->b) of g(x)-g(b)/x-b

but if you dont know this maybe you shouldnt at this point

lets see another way

[true rover]

yea

[ruby tree]

if you look at the graph of g you can that between 2 and 3 the rate of change stays the same. its like a straight line

[true rover]

yea it does

[ruby tree]

then the instaneneous rate of change at any point will be the same

[true rover]

at any point between 2 and 3?

[ruby tree]

yes.

so like 2.5 , 2.6 , 2.7 its the same

[true rover]

yea

[ruby tree]

do you know what the average rate of change will be from 2 to 3?

[true rover]

uh. the slope?

[ruby tree]

yep

and how do you find that

[true rover]

y1-y2/x1-x2

[ruby tree]

yes and in this case?

whats y1 y2 etc

[true rover]

0.5?

i mean the slope

[ruby tree]

um no

[true rover]

y1 is 4 y2 is 1?

[ruby tree]

well yes

x1 x2?

[true rover]

2 and 3

[ruby tree]

so 4-1/2-3

which is...

[true rover]

3/-1

[ruby tree]

so -3

[true rover]

yea

[ruby tree]

so thats the change of rate at any point between 2 and 3

the different quotient the exercise wants is g(2)-g(3)/2-3

which is -3

[true rover]

yea

[ruby tree]

so moving on?

[true rover]

wait thats all?

[ruby tree]

i mean yeah thats what it asks for

in (a)

[true rover]

ok

[ruby tree]

alr so (b)

we need the limit(x-->3) of g(f(x))

we will use a substitution

do you know how?

[true rover]

uh.

idk

[ruby tree]

alr we will set u=f(x)

we need to see what number u is approaching when x is approaching 3

in other words

we need the lim(x-->3) of u so the lim(x-->3) f(x)

[true rover]

uh where do we

oh its from the question

[ruby tree]

hm what?

[true rover]

the x->3

[ruby tree]

yeah it says so

[ruby tree]

alr so what is this

[true rover]

ye.

[ruby tree]

no i am asking like what is lim(x-->3) f(x)

[true rover]

uh there is a hole?

[ruby tree]

that means that f doesnt take a value there. but the limit still exists. like what value is f approaching when we are approaching 3

[true rover]

3.1?

[ruby tree]

huh?

the values of f are in the y axis

[true rover]

oh

2

[ruby tree]

yes.

so the limit we want

which is

lim (x-->3) g(f(x)) will be the limit (u-->2) g(u)

[true rover]

we delete f completely?

[ruby tree]

yes because we subtituted f(x) with u (f(x)=u) and we changed the number that u is approaching to 2)

[true rover]

what about h(x)?

oh

its that

ok

where did we get "u" from

[ruby tree]

we did not get it from anywhere we just set f(x) to be u

f(x)=u

u is simply f(x)

[true rover]

oh

i wonder if my teacher will understand that.

while correcting the papers

[ruby tree]

well has she/he taught you how to calculate limits like that

with substitution?

[true rover]

yes.

i am just too dum to remmber.

i gave up on it after i saw how confusing the questions are.

so uh continue?

[ruby tree]

sorry i ve got some things irl. i can continue a bit later

[true rover]

ok

[ruby tree]

alr i am back

so what is the lim (x-->2) g(x)

[true rover]

uh

4?

[ruby tree]

no.

[true rover]

wat

oh

-1?

i just realized it wont go on the exam, i dont need to be worried about it.

thx for help tho

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[bleak pike]

simultaneous equation, 80 = ax^6 10 = ax^3 answer for x = 2. how?

[upper ruin]

Divide the two equations side by side

[bleak pike]

can you elaborate

[upper ruin]

$$\color{red}{80 = ax^6}$$

[wraith dagger]

Alberto Z.

[upper ruin]

$$\color{green}{10 = ax^3}$$

[wraith dagger]

Alberto Z.

[upper ruin]

$$\frac{\color{red}{80}}{\color{green}{10}} = \frac{\color{red}{a x^6}}{\color{green}{ax^3}}$$

[wraith dagger]

Alberto Z.

[upper ruin]

$$\Rightarrow 8 = x^3 \Rightarrow x = \sqrt[3]{8} \Rightarrow x = 2$$

[wraith dagger]

Alberto Z.

[bleak pike]

ok thank you that cleared it up for me

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[west path]

how do i start solving this equation?

[wicked stag]

Arc tan = arctg?

[west path]

yeah

[wicked stag]

2tan^(-1)x = tan^(-1) 2x/(1-x²)

Use this formula to evaluate 2 tan^(-1) 1/2

[west path]

😨

let me write it down

im not getting it

2tan to the (-1)x?

[wicked stag]

wait

Im sending formula

[west path]

yes pls

[wicked stag]

next

[west path]

this is tg(a+a)

[west path]

the same formula

[wicked stag]

Use thia formula

[wicked stag]

like same but not same

[wicked stag]

This is how this formula is derived

Did you solve ?

[west path]

im trying to understand this

[wicked stag]

Ok

[west path]

holy

i think i get it

[wicked stag]

Cool

Now try solving it

[west path]

i am

is hella long

[wicked stag]

Umm

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[ancient helm]

i can't find this don't even know how to start

[cedar kiln]

Please don't occupy multiple help channels.

@ancient helm Has your question been resolved?

[hazy shore]

Have you considered what implications can stem from x <= 1/2 etc?
For example, 2x <= ...

[cedar kiln]

@ancient helm Has your question been resolved?

[cedar kiln]

@ancient helm Has your question been resolved?

[ancient helm]

anyone?

[cedar kiln]

@ancient helm Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[forest coral]

can someone help with this

[cedar kiln]

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

[forest coral]

i have no idea what im doing 😭

1

like ik the cartesian eqn formula is Ax+By+Cz+D=0

and ik (A,B,C) is normal to the plane

so i tried using the other plane eqn they gave me

(ABC)=(2,-1,1)

and a point to get the eqn

but idk it didnt work

textbook says this is the ans

[nimble mountain]

u can find the plane uniquely from two lines right

[forest coral]

yeas

[nimble mountain]

the first line is the one they gave

the second is the line from the two points given

[forest coral]

wdym first line is the one they gave

[nimble mountain]

that plane

[forest coral]

does it not matter thats its perpendicular?

[nimble mountain]

the vector (2,-1,1) is perpendicular to the plane

like the plane is defined by that perpendicular line in this case

[forest coral]

wait i think im getting somewhere

give me 5s

ok erm i got closer to the answer but

i got the wrong signs

[nimble mountain]

show ur work

[forest coral]

uhhh i took the two vectors and took the cross product and got the theoretical normal

and then used apoint

[nimble mountain]

u did it right

its the same

just multiply throughout with -1 to get it in the same form

but its the same

[forest coral]

OHHHHHHHHHHHH

[nimble mountain]

🍞

[forest coral]

its just the x should be positive thing

THANK U!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[nimble mountain]

🍞

[forest coral]

.close

[cedar kiln]

Channel closed

Closed by @forest coral

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[shadow shard]

How do i do 8? I dont know where to start? I dont want the answer just a start. thanks if you can help 😊

[warped coyote]

Notice how both terms have a tan

[shadow shard]

Why is it always so simple 😭

.close

[cedar kiln]

Channel closed

Closed by @shadow shard

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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

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

[hoary stone]

how do i do this

[cedar kiln]

@hoary stone Has your question been resolved?

[hoary stone]

<@&286206848099549185>

[modern cairn]

why dont you just put in random values of x

@hoary stone

[hoary stone]

dawg how

[modern cairn]

in the question they asked us to use technology so does that mean we can use a graphing calculator

[hoary stone]

yeah

ive been using desmos and im lost

[modern cairn]

oh ok

so do you know that a logarithm of a negative number does not exist?

[hoary stone]

new question

[modern cairn]

,rotate

[modern cairn]

what part of this question you are struggling this?

is it that you are having difficulties in understanding the graph?

[hoary stone]

the whole thing, plotting the points

[modern cairn]

ok

so this is the graph

ill just explain it to you

do you know that a logarithm can never take a negative value as an input?

[hoary stone]

no

[modern cairn]

thats one of the fundamental properties of log

you know this?

[hoary stone]

no