help-0

[vale sapphire]

i'd like to see how a coin can both land on heads and tails in a single throw

[quaint pond]

yeah but like aren't you talking

about two different events

so as if in one of the events the coin lands on h and in the other one on t

[vale sapphire]

An event is another word for an outcome

We're only throwing our coin twice

We can't have the outcomes "the first throw is heads" and "the first throw is tails" at once

We're only doing two throws

nothing more

[quaint pond]

oh yeah

now i got it

[vale sapphire]

an event is just an outcome of a given experiment

If we cared about the results of doing the experiment twice then our universe would look different

it would be 4-tuples

Likewise, "both throws are heads" and "the first throw is tails" are incompatible events

[quaint pond]

yeah

[vale sapphire]

In set terms, {(H,H)} ∩ {(T,T),(T,H)} = ∅

now for probabilities

A probability is just a way to quantify how likely a given event/outcome is

[quaint pond]

yeah

[vale sapphire]

For a given universe, there may be multiple different probabilities

For example if you have magnetic coins that only land on heads, or unbalanced coins that land on heads more often than tails, the probability of each event will be different

[quaint pond]

yes

[vale sapphire]

But regardless of these different possible probabilities, they must all satisfy some basic properties

The fact that all probabilities lie between 0 and 1 isn't truly a property

it's a definition

0 and 1 are convenient numbers to work with

[quaint pond]

yeah like in cs 0 is false and 1 is true or?

[vale sapphire]

ye

But you could have chosen to have probabilities between 0 and 100, that's what you do whenever you talk about percentages

[quaint pond]

yes

[vale sapphire]

First property that makes sense is that if your universe is S, then P(S)=1

indeed, the outcome "any outcome will happen" is always verified

(there are some subtleties involved, but those only appear when dealing with infinite universes, so it doesn't matter here)

The second property is perhaps the most important

The probability of either one of two incompatible events happening is precisely the sum of the probabilities of each event

Using an example

The probability of "both throws are heads OR both throws are tails" is the probability of "both throws are heads" PLUS "both throws are tails"

[quaint pond]

yeah

that makes sense

[vale sapphire]

That's precisely what your definition says, with more barbaric notation

If you take a union of disjoint events, the probability of that union will be the sum of the probabilities of each event

[quaint pond]

yeah

[vale sapphire]

Here we have special notation for this

[glossy fulcrum]

can i ask a question

or channel busy

[quaint pond]

thanks systms btw

do you know any yoututbe channel or smt that

[vale sapphire]

We write $A \sqcup B$ to denote $A \cup B$ when A and B are disjoint

[ocean seal]

Syst3ms

[vale sapphire]

Honestly I didn't learn this from any youtube channel, just a good teacher

So unfortunately I don't have anything to recommend on that particular topic

Btw, the second point of your definition can be rewritten as $\mathbb{P}(\bigsqcup_{i=1}^\infty A_i) = \sum^\infty_{i=1} \mathbb{P}(A_i)$ with this notation

[ocean seal]

Syst3ms

[vale sapphire]

That notation helps tidy things up sometimes

[quaint pond]

yeah thanks

[sturdy wraith]

So hey I have been looking for a clarification. This is a question on Galois Theory in Abstract Algebra by Dummit and Foote, and I just want to know how the line in the given solution that goes $[\bar L: K]\mid [L_1:K]\ldots [L_n:K]$ came to be

[ocean seal]

AidenM27

[alpine sable]

hlo

[ashen axle]

May i get answer to this

<@&286206848099549185>

[velvet condor]

bruh

do you know Bézout's remainder theorem @ashen axle

let x^4-ax+b=(x^2-3x+2)Q(x)
you have x^4-ax+b=(x-2)(x-1)Q(x)
sub x=1 ->1-a+b=0
sub x=2->16-2a+b=0

[glass lichen]

that's typically just called remainder theorem

[velvet condor]

since no one is asking any problems let me give out some problems-

[glass lichen]

the channel is in use.

[velvet condor]

then ill just post it in another channel :/

[glass lichen]

Given you just responded to the problem I thought that would've been obvious

[alpine sable]

I don't think the channel is still in Use.

I have successfully Rendered a Sphere.
Now, I have coordinates of all the Points on the Sphere.
Currently, all the points on the Surface of the Sphere hold a Brightness Value of 255.
I can assign them a Value between 0 and 255, 255 being the brightest, and 0 being the darkest (null).

Now, if all the Points have a Value of 0, and I bring a Point sized Light Source on a horizontal axis, how do I calculate the brightness values (between 0 and 255) of all the Points on the Sphere.

[alpine sable]

I can deduce that the Greater the angle X, the Darker the Spot will be, but why's that?

[ionic jewel]

because it's further away

[alpine sable]

How do I determine or how can I cut circle so that contact point always stays in center line?

I dont get this step

[ocean seal]

Hhh

[alpine sable]

Why is there a 1/2

Maybe I should review the integration lessons

[pearl marlin]

$e^{x^2}= t\$ so $e^{x^2}. 2x dx = dt$ use this

[ocean seal]

learn4math

[alpine sable]

Wait where did you get the .2

[pearl marlin]

it will be $-2\int \frac{xe^{x^2}}{e^{x^2}.2x}dt$

[ionic jewel]

he might be using . as multiplication

[pearl marlin]

by derivative of power

[ionic jewel]

some confused places do that

[ocean seal]

learn4math

[alpine sable]

Im not sure im understanding

Sorry

Ill re watch the Integrations videos

Ty for your help though

[pearl marlin]

it will be $-2\int \frac{xt}{t.2x}dt$

check now

[ocean seal]

learn4math

[pearl marlin]

i put e^{x^2}=t , may this will help you

[alpine sable]

O yea i think i know about that substitution

But im not sure on why theres a fraction

Ty though

[pearl marlin]

okk you know that when you substitute some other variable you have to take intergration over that

[shell widget]

t= e^(x^2). Differenciate both sides with respect to x, you get dt/dx = e^(x^2) * 2x. We need to replace the "dx" in our integral, so dt/dx = e^(x^2) * 2x gives us dx = dt/[e^(x^2) * 2x]

[pearl marlin]

like when i put $e^{x^2}=t$\ it will give you $e^{x^2}. 2x dx=dt$ and from here $dx=\frac{dt}{e^{x^2}.2x}$

[ocean seal]

learn4math

[charred flint]

@alpine sable can you explain further?

[glacial hedge]

Could someone help explain the wedge product to me? It doesn't make that much sense right now

Why is the wedge product of dx and dx zero?

<@&286206848099549185>

[vale sapphire]

what do you mean by wedge product

[ionic jewel]

it's like a generalization of the cross product

i don't think I've ever used it but i have seen the term

[glacial hedge]

im very confused

they are bringing diffferential forms up in the last homework of multi var calc

[glass lichen]

so loosely speaking wedge acts like the cross product, but with differentials acting as vectors, and the cross product is 0 when the vectors are parallel

https://www.youtube.com/watch?v=PaWj0WxUxGg&list=PL22w63XsKjqzQZtDZO_9s2HEMRJnaOTX7&index=1

[midnight sundial]

have they established that the wedge product is alternating?

or anticommutative

[plush ermine]

I have a very basic question. Many operations have the property that O(cf(x)) = cO(f(x)), and the property that O(f(x) + g(x)) = O(f(x)) + O(g(x)). These operations include sigma summation, limits, derivatives, and integrals. Is there a name for this common property?

[glass lichen]

Linearity

specifically the 1st is sometimes called homogeneity and 2nd is additivity, but collectively we say O is linear

[plush ermine]

Ok. Thank you!

[tacit bridge]

calculate the broken fractions

[ionic jewel]

if you mean simplify, combine the fractions ont he top and the bottom, then vlip the bottom up

,w ((9/(x+2))-1/x)/(3/(10x)-1/(2x))

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[crystal cape]

how to solve this?

[glacial hedge]

try

expanding

(x-50)^2 and split up the sum into sum of x^2

and use the information from the first sum to remove the nastly things like -100x+2500

[crystal cape]

ok

[glacial hedge]

if you still need help i can walk it through

[twin pine]

Guys

[ocean seal]

andreask

[ionic jewel]

are you using the cross product symbol to mean multiplication?

[twin pine]

yes, I am not the best with TextEdit

If $f\left(x\right)=A\cdot cos\left(nx\right):+B\cdot sin\left(nx\right)$ show that
$$f:''\left(x\right)+n^2f\left(x\right)=0$$

[ocean seal]

andreask

[ionic jewel]

take the second derivative of f(x) and plug it in

it all cancels out

since the second derivative of sin(nx) or cos(nx) is -n^2sin(nx) or -n^2cos(nx) respectively

[twin pine]

My texbook says that $f:'\left(x\right)=-A\cdot n\cdot :sin\left(n\right):+B\cdot n\cdot :cos\left(nx\right)$

[ocean seal]

andreask

[glacial hedge]

thats the first derivative

[twin pine]

yep, is it the correct one?

[glacial hedge]

yes

no

wait

the -A * n * sin(n (missing the x))

its missing the x

[ionic jewel]

you can skip the first derivative

you know sin/cos/-sin/-cos is the 4 derivative cycle

so with two derivatives it'll cycle to the negative of the original

[glacial hedge]

@twin pine the textbook probably has a typo

[ionic jewel]

and each derivative pulls an n out to get the right thing

[glacial hedge]

because of chain rule, but you can do that in your head

[ionic jewel]

yea

[twin pine]

The only part which I do not get is that we do not know what A is so shouldn't it be the derivative of A (i.g A')

[ionic jewel]

no

a is a constant

so it just stays and does nothing when you take the derivative

[twin pine]

Ah I see

But how would I understand that it is a constant?

yeah, not the smartest question ever asked but forgive me...I am sleep deprived

[ionic jewel]

well

technically you dont

but it's not true if they aren't constants

and usually letters that aren't the variables are constants anyways

[glacial hedge]

f is a funciton of x not of a, so the only thing that can change (variable) is x

*x

[ionic jewel]

he's saying that A and B could be functions of x as well

you just assume they arent

they would probably say A(x) or B(x) if they were

[glacial hedge]

yes otherwise that would be very misleading for a textbook

[twin pine]

Okay, thanks a lot of guys!

[twin pine]

Trust me, that textbook is created to mislead

[mellow canyon]

hey, so how important would it be for me to learn the tan, cos, and sin trig values for common angles? like 0, 30, 45, 60, 90, etc . it feels like this is just memorization. is this something fundamental like times tables or is it not as important to learn?

[glacial hedge]

uh

i would just memorize the

45 45 90 triangles

and 30 60 90 triangles

because you can derivte the "easy" angles from those triangles

[glass lichen]

exact values*

[mellow canyon]

like the relationships between all the sides? for example, a leg is x, hypotenuse is sqrt2x, etc?

[glacial hedge]

yes

soh cah toa

[mellow canyon]

well I know those already. I mean a table showing what each function would output if you put in 30 degrees, 45 degrees etc

the textbook I'm self learning from has a mini chapter on this

[mellow canyon]

yes

[glacial hedge]

i guess I would try to memorize some of the sin and cos angles, but dont beet yourself over the head if you cant memorize all of them now, over time as you use them more you will start to memorize

[mellow canyon]

ok. I was just trying to gauge how important it is. for example, if you don't know your time tables that's kind of a staple. is this something super important or just something I should glance over once in a while and pick up along the way?

[ionic jewel]

i don't know them at all and I'm doing fine

[glacial hedge]

uhh not super important

[ionic jewel]

absolutely know the ones for 0, pi/2, pi 3pi/2 etc

the ones in the middle you can use a calc if you ever need em i guess

[glacial hedge]

^

[mellow canyon]

can pi/2 be an angle?

[ionic jewel]

it's the one pounting straight up

[mellow canyon]

well I gues it can, but is it denoted like that

[glacial hedge]

they are talking about radians

not degrees

[ionic jewel]

oh this guy uses degrees?

[mellow canyon]

yes I know this. I've just never seen an angle denoted in that way lol

[ionic jewel]

in that case make sure you know 0 90 180 270 etc

[mellow canyon]

so far yes. I'm by no means good at trig though lol

[glacial hedge]

?

[mellow canyon]

I always see stuff in degrees. it's simple enough to just convert them right?

[glacial hedge]

yes

[mellow canyon]

I always just convert if I need to

[ionic jewel]

yes but once you finish whatever class ur in you never use degrees again

haven't used em in years

[glacial hedge]

well when you get to calculus radians become much more usefull than degrees

[mellow canyon]

oh alright. so when a angle is denoted in radians, do you just input that in the place of degrees in a calculator?

[mellow canyon]

radians are based off the actual circumference of the cirlce, aren't they?

[glacial hedge]

put your calculator into radians

*radians mode

[mellow canyon]

I know that, but you would still input it as sin(pi/2)=

right?

[glacial hedge]

yes, 1 radian = 1 arc length wrapped around circle

[glacial hedge]

yes

[mellow canyon]

oh alright, that's simple enough. thank you guys. one more thing before I go

what's the thing with memorizing various values among the unit circle? it seems similar to a table, is it just the same thing represented visually with more angles

[glacial hedge]

yes, easier to visualize most of the time

cos is the x component of the point on circle

sin is the y component of the point on circle

[mellow canyon]

is that the same concept? why would I want to know all the angle measures in a unit circle

[mellow canyon]

point on circle? can you explain this a bit more

[glacial hedge]

yes its the same concept, its the same reason for memorizign the quadratic formula or any other thing

[glacial hedge]

wowe thats hard to read

[mellow canyon]

I haven't memorized the quadratic formula lol. there's usually a different way to do it, and if you can't you can just look it up. unless I have a test on it in the future, I don't plan on

[mellow canyon]

ok, so in here you've got a standard angle

[ionic jewel]

i have the quadratic formula memorized

it's very handy

[glacial hedge]

sheeesh fancyy!

[mellow canyon]

for what exactly?

[ionic jewel]

quadratics

they come up a lot

[mellow canyon]

80% of quadratics can be done without it

maybe more

[glacial hedge]

no

[ionic jewel]

no

[mellow canyon]

and you can always look up the formula if you need it

[ionic jewel]

only because your teacher gives you ones that can be solved without it

[glacial hedge]

no NOn ONonononnono, the problems that the teachers create... most can be done by factoring

[ionic jewel]

the vast majority (read infinite) of them you can't factor your way

[glacial hedge]

but in real life your gonna be working with 134.384238776x^2 + 0.3423x -154 = 0 or somethign really ugly

[mellow canyon]

why not complete the square, factor, rearrange them etc

[mellow canyon]

wtf

[glacial hedge]

You cannot factor some things

[ionic jewel]

because polynomials outside of your algebra classes don't look nice

[mellow canyon]

well that's a headache you compute anyways. ok so let's say that most applications require it, why can't you just look up the formula?

[glacial hedge]

yes

[ionic jewel]

you could, but it just takes time

[glacial hedge]

but why would you want to look up every single time

[ionic jewel]

no reason not to just spend 15 minutes and know it

[mellow canyon]

I don't normally memorize things unless I use them a lot

[ionic jewel]

you will use it a lot

[mellow canyon]

because I use it...like 5 times so far outside of class lol

[glacial hedge]

just wait till future math classes, so many things to memorize

[mellow canyon]

which classes or real life applications?

[glacial hedge]

hmm.... almost every stem class

[mellow canyon]

I'm in 8th grade, so I don't know if I'll need that. if I do, I'll memorize it

[ionic jewel]

to be fair it's relatively easy to understand the proofs of almost all of the hard theorems, but you don't have time to rederive it on an exam

[mellow canyon]

I just don't see why I would need it, since I am taking honors geo and then trig after that. so I'm gonna memorize something I don't need for 2+ years??

[ionic jewel]

i don't think I've taken a single math class since algebra that i haven't used quadratic eqn

[glacial hedge]

lmao that would be a terrible test taking strategy.. hmmm Im gonna check my answer by deriving and proving said formula

[ionic jewel]

you don't plan on making it to calc?

[mellow canyon]

obv but that's in more than two years. I doubt I'll remember it that well. it's easier to just memorize it beofre a class that requires it

[ionic jewel]

i mean think of it like the Pythagorean theorem. you almost certainly have that one memorized, it's just something you have to know

you could look it up every time but you don't need to

[mellow canyon]

because I need it for most of my classes. I don't need QF in the near future

[glacial hedge]

@ionic jewel do u know sameer?

[ionic jewel]

no

is that a person?

[glacial hedge]

sameer deez nutz on ur forehead

[ionic jewel]

[glacial hedge]

[mellow canyon]

intellectual comedy

[ionic jewel]

you do what you want but it's very nice to know, for everything

[mellow canyon]

I guess. Thanks for the help lol. I'm teaching myself trig so I don't relaly have a teacher to ask random stuff

[ionic jewel]

you will be expected to know it in general, they will throw quadratics on a calc exam or whatever and not tell you beforehand

[hoary cipher]

Hey

[twilit ibex]

Hello Bunny the Duck if you have time, please check Q-5 🙂

[hoary cipher]

Is anyone available for a stupid logarithms question?

[twilit ibex]

Maybe, show us 🙂

[glacial hedge]

yes

[hoary cipher]

On 117

[glacial hedge]

which one

[hoary cipher]

117

I get x=10

But the answers say 10, 1

[glacial hedge]

@hoary cipher your mising somethign obvious

what is 1 to any power

[hoary cipher]

1 why

[glacial hedge]

when you put in x=1

you get

$1^(log(1)+1)=1^3$

[ocean seal]

Elonmosqito96

[glacial hedge]

wow thats messed up

[hoary cipher]

But why would I put x=1 😖 ?

[glacial hedge]

when u see

x^(blah) = x^(blah)

you know that x=1 is always going to be an answer regardless of blah

[hoary cipher]

Mmm I think they have missed it in my text book

Thank you !

[glacial hedge]

yw

[hoary cipher]

So whenever it happens to place 1?

[glacial hedge]

?

[glacial hedge]

what does this mean

[hoary cipher]

I mean

Isn't there a rule that works like that?:

[glacial hedge]

yes

[glacial hedge]

a must be greater than 1 because 1^(blah) = 1

[hoary cipher]

So isn't that a problem to place x=1?

[glacial hedge]

?

[sterile beacon]

Can someone help me with this exercise?

it says "Calculate"

b is 7

and a is 8

[warped phoenix]

how do i do this?

y = x-385000/38

do i switch y and x

channel in use

[bleak harbor]

My bad

[warped phoenix]

actually, i know this is an inverse function

and i know how to solve it

but how do I identify that?

I only know what to do here because the unit im working on is literally on special functions

but if i was given this question out of the blue, id substitute T for 1000, divide d by both sides, yadda yadda yadda i'd get the wrong answer

[worthy hatch]

y is directly proportional to x^2.

when x has a certain value, y=6.
find the value of y when x is doubled.

pls help me with this qn ty

[warped phoenix]

channel in use

[worthy hatch]

thanks so much in avance

oh oaky thanks

[warped phoenix]

how do i do this?
y = x-385000/38
do i switch y and x

actually, i know this is an inverse function
and i know how to solve it
but how do I identify that?
I only know what to do here because the unit im working on is literally on special functions
but if i was given this question out of the blue, id substitute T for 1000, divide d by both sides, yadda yadda yadda i'd get the wrong answer

<@&286206848099549185>

anyone?

[distant otter]

can you not just isolate

[warped phoenix]

wdym

[distant otter]

1000 = (D-..)/38

isolate d

[warped phoenix]

well wouldnt it be 1000(D) = D-385000/38

And then you divide both sides by D

or 1000*

to get D by itself

[distant otter]

1000*38+385000=D

why you dividing by D

[warped phoenix]

i meant divide by 1000 srry

[distant otter]

I'd just multiply by 38

on both sides

then you can add 385000 on both sides to get D

[warped phoenix]

well wouldnt that be 423000(D) = D

[distant otter]

[warped phoenix]

here

[distant otter]

okay so you have

1000*38 = D-385000
1000*38+385000 = D
D=1000*38+385000

aren't you done now?

[distant otter]

uhh what is 1000(D)?

is it 1000 multiplied by D

[warped phoenix]

1000 is T, so the LHS of the equation is T(D), so wouldn't it be 1000(D)?

[distant otter]

oh...

lmao

no sorry that's not how functions work

[warped phoenix]

ohh

[distant otter]

When you write it like that you're writing 1000 multiplied with D

[warped phoenix]

ah you substituted 1000 for the entirety of T(D)

[distant otter]

in this case, T(D) has the value 1000

yes

T(D) is 1000

T(D) is a number

[crude rose]

T(D) is just the function T with an argument of D

not T times D

[warped phoenix]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh right 🤦‍♂️

sorry lol

[distant otter]

you better be

no just kidding

😛

[warped phoenix]

😆

ok well ig now i know i dont need to make an inverse equation to solve this

[distant otter]

well...

that's actually what you're doing

when you isolate D

[warped phoenix]

:000000

[distant otter]

D(T)=T*38+385000

that's the inverse function

[warped phoenix]

wait why would it be that and not D = T(D)-385000/38

isnt making an inverse just switching the things

so you switchT(D)'s place with D

[distant otter]

the inverse function is defined as D(T) where D is is distance as a function of time

the function you were given was T(D), where time was a function of distance

[warped phoenix]

ahhhhhhh

so the inverse is not a switching of the function and the variable's place in the equation

but a switching of themselves

because the opposite of T(D) = D(T), and the opposite of D is T here

assuming what i said was in agreement with your statement, then yeah i think i got it

[distant otter]

uh lmao

f(x)=5x+1 is the same as x(f)=5f+1

if that's what you're saying

[topaz scaffold]

The notation is a bit weird

[distant otter]

I'm just replying to this

but yea the notation is weird

[topaz scaffold]

It's more like x=5f(x)+1

[hybrid grove]

wow you guys doing high level math

[topaz scaffold]

No?

[hybrid grove]

wdym

[topaz scaffold]

Ig high level depends on what level you are at

[hybrid grove]

definitely looks high level

[topaz scaffold]

So different people have different ideas for what high level is

[distant otter]

this is like low level US high school mathematics

[topaz scaffold]

Yea

[hybrid grove]

bruh im in 9 th grade

[topaz scaffold]

This is around your level then

Maybe a bit higher

[hybrid grove]

ye havent learned that yet

but ok

[warped phoenix]

wait so these arent inverse then 😦

[distant otter]

no lol

an inverse function is not the same as switching the notation

5x+1=3 is the same equation as 5t+1=3

where x has been replaced by a t

[warped phoenix]

right

[distant otter]

but it doesn't matter if it's an x or t, it's just a symbol we use to represent a number

[topaz scaffold]

Yep

Usually the notation for inverse functions is $f^{-1}(x)$

[ocean seal]

RipeOrange

[topaz scaffold]

Keep in mind that it's not actually being taken to the -1st power, it's just the notation

[distant otter]

an inverse function, g, to a function f, say defined on the real numbers so f:R -> R is a function that is both the function's left and right inverse, so f(g(x))=x and g(f(x)) = x

[warped phoenix]

@distant otter so then how is D(T) = T an inverse of T(D) = D?

if simply switching the notation doesnt change anything

[glacial hedge]

Is inverse of a matrix distrubtive i.e (A+B)^-1 = A^-1 + B^-1

[oak chasm]

@glacial hedge No, as the sum can be invertible while A and B can be noninvertible.

[glacial hedge]

Ohh.. i see ty

[oak chasm]

No problem.

[glacial hedge]

@oak chasm Sorry to ask again how would I go about solving this?

[oak chasm]

@glacial hedge Well, you have three invertible matrixes. If you multiply them together, they stay invertible.

So, which of those is a product of some of those three matrixes?

[glacial hedge]

A, D, C?

So b is the answer?

[oak chasm]

OK, how did you get D?

Oh, I see A(A + B).

Yes, that's right.

[glacial hedge]

Ayy!! thank you, i forgot that adding 2 invertible matrices doesnt necessarily make an invertible matrix

[glass lichen]

yeah B is the only one not a product of invertible matrices

[oak chasm]

You're welcome.

[carmine lion]

One card is drawn from a standard pack of cards and kept on a table. A second card is drawn and placed beside it on the table. What is the probability that 2nd card is from a different suit to the first

isn't it 3/4

[crude rose]

different suit?

[tough hatch]

not really, from the wording of the problem it seems that the cards are drawn without replacement

[woeful pulsar]

are they drawn from the same deck

[carmine lion]

oh wait

is it 39/51

[carmine lion]

yes

[crude rose]

oh suit is symbol

[woeful pulsar]

yeah that's it, there are 51 cards left

[crude rose]

ohhh

[carmine lion]

lol i struggle in probability

its not as intuitive as other stuff like calculus

to me at least

[pulsar fractal]

what???

2 arcsec??

[oak chasm]

Let's see.

,w derivative of arccsc(x)

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[oak chasm]

@pulsar fractal Looks liek the right one is arccsc.

[pulsar fractal]

thats so weird

[oak chasm]

,w derivative of arcsec(x)

[pulsar fractal]

how theres a glaring mistake

[glacial hedge]

they probably had a typo

[ocean seal]

Results provided by WolframAlpha

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

Requested by Chai T. Rex#7323

[pulsar fractal]

yea they were probably copying off and they left the lhs unchanged

ah they fixed it in their latest edition

[pulsar fractal]

set them equal and solve

[warped phoenix]

how do i solve this correctly? I got the right answer just by plugging different values in as a x (started from 1) until I got they equaled each other

oh ok hold on

[pulsar fractal]

easy quadratic since the 1086s cancel out

[carmine lion]

ugh

i h ave another probability

question

[warped phoenix]

2x^2 - 10x = 0?

[pulsar fractal]

yes

[shrewd otter]

You set f(x) = g(x) and find x.

Your approach was quite naive and brutish.

[warped phoenix]

😦

[carmine lion]

A box contains 3 red marbles and 7 white marbles. A marble is drawn from the box and a marble of the other colour is then put in the box. A second marble is drawn from box.

how do i find P(both marbles are white | both marbles are same colour)?

[shrewd otter]

At the beginning, what is the probability of drawing a white marble?

[carmine lion]

7/10

[shrewd otter]

Okay, and then how many marbles remain?

[carmine lion]

wdym

oh

[shrewd otter]

How many are left in the box?

[carmine lion]

9, but u add other coloured marble

[shrewd otter]

Hold your horses.

Of which, how many are red?

[carmine lion]

uh 4?

since u add 1

[shrewd otter]

I said hold your horses.

We haven't added it yet.

So 4 is incorrect.

[carmine lion]

ok then 3

[shrewd otter]

No, there can't be fractional red marbles.

Okay.

[carmine lion]

i dont get what ur question was

[shrewd otter]

And how many are white?

[carmine lion]

uh 6

[shrewd otter]

So we have 3 red and 6 white marbles.

Since we pulled a white, we add a red.

How many red marbles are there now?

[carmine lion]

yes i already have the whole tree drawn :(

[shrewd otter]

So now there are 4 red and 6 white marbles.

What is the probability of drawing a white marble?

[carmine lion]

6/10

mate i've already drawn the whole tree diagram

[warped phoenix]

how do i know to do this btw? like what in the question prompted you to set them equal to each other

[pulsar fractal]

because they want to know when the populations are equal

[carmine lion]

they have same population

[shrewd otter]

The question asks you to find when they are equal.

[carmine lion]

so that must mean the functions are equal

[warped phoenix]

ohhhhhhhhh

righttttttttttt

🤦‍♂️

TY!!!!

[shrewd otter]

So what's the problem?

[carmine lion]

that question

at the bottom

P(both marbles are white | both marbles are same colour)

[shrewd otter]

What's the formula for P(A | B)?

[carmine lion]

$\frac{P(A\cap B)}{P(B)}$

[ocean seal]

abe

[shrewd otter]

Have you found P(B)?

[carmine lion]

idk how to

[carmine lion]

so i guess thats why im stuck

btw is the answer 42/47

[shrewd otter]

I really don't know, honestly.

Finding P(B) seems tricky.

[carmine lion]

wouldn't it just be the sum of white and red

[shrewd otter]

As B is itself dependent on something else.

[carmine lion]

because those are the only 2 options

where u get double

(42+5)/100

then u divide P(WW) by that?

idk if P(WW)/P(WW OR RR)

is the answer

[shrewd otter]

The chance of pulling two of the same colour depends on which colour you pulled first.

I don't know if it's the sum.

It might be.

[carmine lion]

should i ping helpers

[glacial hedge]

i think you shoudl post it in the probability section

#probability-statistics

[carmine lion]

ok

[strong furnace]

it is

[carmine lion]

it is?

[carmine lion]

so 42/47?

[strong furnace]

I did not do the computation

[carmine lion]

ok

but thank yhou

[strong furnace]

42/48 ?

[carmine lion]

whaa

[strong furnace]

3/10*2/10

for RR

so 42/(42+6)

[carmine lion]

oh rihgt

my bad

careless error

[glacial hedge]

Is this true?

[shrewd otter]

It seems true. I don't know how to prove it.

[glacial hedge]

yeah thats what i thought

[alpine sable]

HELLOING, does anyone know how i can find the semi-major axis while only knowing the radius of the periapsis an speed at that point, i thought itd be a=GM/v^2r but i get a decimal meter which does not spark joy, mr Google said its a = ((GM/4pi^2) T^2)^1/3 only

uwu

pls ping me

[glacial hedge]

not a physics channel 😦

[alpine sable]

wheres the physics channel...

[glacial hedge]

its called Mathematics not Physicmatics xD

[alpine sable]

but its math.

[glacial hedge]

physics != math

[alpine sable]

whats ! meant to be

[shrewd otter]

!= is how programmers say "not equal to".

[alpine sable]

oh

why not

[glacial hedge]

someone else wanna explain why math isnt physics?

[shrewd otter]

$$physics \subset maths$$

[alpine sable]

[ocean seal]

Hurubon

[alpine sable]

but im 14

you can learn it uwu

[strong furnace]

explain how you got that?

[alpine sable]

ahemm

[glacial hedge]

also @alpine sable u can use latex for the equations

[strong furnace]

show your work and then some1 can help

[rocky dock]

im kinda stuck on an a system of equations problem using elimination

[glacial hedge]

mhm?

[alpine sable]

idk if you can get it

[glacial hedge]

have u tried using a augmented matrix and putting in echelon form?

[rocky dock]

no i added rows 1 and 3

and got
-8z = -32
z = 4

however, thats where i get stuck

[glacial hedge]

mhm

[strong furnace]

just describe your idea

not the computation

[glacial hedge]

then plug in z

into all equations

[strong furnace]

I ccan't read that anyway

[alpine sable]

i used algebra manipulation stuff but its wrong apparently

[strong furnace]

bro just what law did you use? and how?

[rocky dock]

but wouldnt i have to use row 2 first and eliminate a variable again?

[alpine sable]

idk what law, i just moved the stuff like my weeb teacher teached me

[glacial hedge]

bruh

no one can really help that...

[strong furnace]

how is any1 supposed to help you ?

[alpine sable]

with the power of fwendship

[glacial hedge]

@alpine sable do u know sameer?

[alpine sable]

whos sameer

[strong furnace]

go read about kepler's laws( basically conservation of angular momentum due to central force )

[glacial hedge]

sameer deez nutz on ur forehead

[alpine sable]

[strong furnace]

applied on gravitation

[alpine sable]

i did, didnt get it

[rocky dock]

elon did you see what i said?

[alpine sable]

o-o

[glacial hedge]

@rocky dock maybe i misesd it want to go to channel 1 cuz this ones being used rn

[strong furnace]

have you done basic rotational mechanics and conservation of angular momentum?

[rocky dock]

ok, thanks

[alpine sable]

kindaa?

i only know the equation of centripetal force and gravity of 2 objects

[vestal hawk]

hey guys how does e^(4lnx) cancel out to x^4

[strong furnace]

kepler's law is just the application of that on the gravitational force

[alpine sable]

uhu

i thought this wasnt a physics channel

i have been decieved

and backstabbed

and maybe bamboozled

[strong furnace]

quite possibly bamboozled*

[alpine sable]

so

[crude rose]

e^lnx = x so you have x^4

[alpine sable]

wtf do i do

so

you say that i cant do anything with knowing somethings speed at periapsis

[strong furnace]

give it another read idk how any1 can help you with how little we have to work with here , I am not going to derive the whole thing for you here there's probably something on the internet that can help you

[alpine sable]

like the semi major axis and consequently the apoapsis

mm

A

i need to put the phone number to be in the physics server

not pog

[vestal hawk]

OOOH THX

[glacial hedge]

could someone walk me through this? I got the only solution is 0 but oviously thats wrong

<@&286206848099549185>

[vague coral]

Shouldnt you solve AX = O ?

[glacial hedge]

thats what i did

and got only x=9

*x=0

[vague coral]

that's weird

let me try

[glacial hedge]

@scenic patrolrel

nvm

i just can do matrix stuuf properyly

[mortal void]

you can say me dumb but im getting so confuse on how -3x-2^2=12 not -12?

[glacial hedge]

??

u mean why is x not -12

?

[mortal void]

that x is multiply sign

[glacial hedge]

oh

[vague coral]

[mortal void]

sorry cause i forgot the way to put the sign

[glacial hedge]

the negatives cancel

[vague coral]

when you multiply two negative numbers, you got a positive number

[glacial hedge]

^

[mortal void]

yeah

but

[vague coral]

dont ask why

[mortal void]

-3 x (-2^2)= -3 x 4 = -12

[glacial hedge]

Another question how do i do this rather than to just plug in random values or solv efor the augmented matrix?

[mortal void]

is what im thinking

[glacial hedge]

-2^2 != (-2)^2

[mortal void]

is what im thinking

wait what lol

i was doing it wrong way this whole time then omg

[vague coral]

I dont even know what Col is (I'm not used with english notation yet)

[glacial hedge]

@vague coral how do i do this

[vague coral]

@glacial hedge what is Col ?

[mortal void]

if its -2^2 then i was putting value 4 and if its 2^3 then i was putting value in negative which is -8

its what i was thinking this whole time omg

[glacial hedge]

Col means subspace of columns of A

like the column space

or maybe can u walk me thorugh this? its pretty much the same method...

[vague coral]

$c_1 \begin{pmatrix} 1 \2\ -3 \end{pmatrix} + c_2 \begin{pmatrix} 2 \ 0 \ -4 \end{pmatrix}= \begin{pmatrix} c_1 + 2c_2 \ 2c_1 \ -3c_1 -4c_2 \end{pmatrix}$

[ocean seal]

Herels

[glacial hedge]

oh so i have to solve that 😦

[vague coral]

Well yes and no

[fierce musk]

Problem: Suppose a large number of particles are bouncing back and forth between x=0 and x=1, except that at each endpoint some escape. Let r be the fraction reflected each time; then (1-r) is the fraction escaping. Suppose that the particles start at x=0 heading toward x=1; eventually all particles will escape. Write an infinite series for the fraction which escape at x=1 and a series for those which escape at x=0. Sum both series. What is the largest fraction of the partocles which can escape at x = 0 (Remember that r must be between 0 and 1).

I got answers P/(1+r), r/(r+1) and 1/2P

Could someone confirm my answers please?

Here are my workings

[charred flint]

don't think you need P there, the fraction that escape is between 0 and 1

[fierce musk]

hmm

but r

represents the number of reflected particles

which is between 0 and 1

The way I interepret the last part of the question is that they are asking for the largest fraction of particles which escape in total at x=0

So thats why I put the series at x=0 over P

[charred flint]

the fractions are right and everything

[fierce musk]

so total that escape at x=0 over total particles present

[charred flint]

you mean P to be a the number of particles right?

[fierce musk]

yes

P is the total number of particles in the beginning yes

[charred flint]

The wording just wants a fraction, not a number

like if it asked "what percentage of particles escaped at x=1"

you'd give like 50% or something, not a particle number

[fierce musk]

yes i understand that

but

[charred flint]

all the math stuff is right

[fierce musk]

oh alright

but

1/2P

how can i write that without P

to express the largest fraction of particles that can escape at x=0

if it were 1P/2

that would be 1/2 of P

but im getting P as part of the denominator

[charred flint]

umm I assumed you meant P times 1/2

[fierce musk]

oh

no

[charred flint]

oh I didn't see the page where you put P in the denominator uhhh

[fierce musk]

r / r + 1 is what i got for the particles that escape at x=0

[charred flint]

there's a problem there, dividing by a (number of particles) makes the units really strange

yea the answers I have are 1/(1+r), r/(r+1) and 1/2

[fierce musk]

woah how did you get 1/1+r

wwait

nvm

they said all the particles eventually escape

so thats why i assumed that

I couldnt get a series properly for the x=0

so thats where i invoked my assumption

hold on

ill show you something

P( 1 - r )( r^ (2n-1) )

this is what i got for x=1 as the particles escaping at that stage

but you see i cant transpose 2n-1 into the form k( n - 1 ) where k is some real number

and if i cant do that

i cant use it in the formula for the sum of the series

ah

yeah true

hmm

@iron stag stop

@fair crater So how would you write P( 1 - r )( r^ (2n-1) ) in series form?

[alpine sable]

<@&268886789983436800>

[fierce musk]

ahhh

silly me

yes thank you sir @fair crater

@charred flint thanks for the help too

[analog lintel]

guys i have a question if anyone is willing to help atm

I have one more question for my assignment but im stupid and forgot how to do it

want me to copy and paste the question its nothing hard its grade 11

[fierce musk]

Holdup, Plurmant, if you arent busy, i'd like to review over my problem again with you as iam unsure with one of my answers @charred flint

sorry bluey

go ahead

[analog lintel]

lol no worries

The profit of a skateboard company can be modelled by the function P(x) = – 14x2

• 133x – 63 where
  p(x) is the profit in thousands of dollars and x is the number of skateboards sold, also in thousands.
  When will the company break even, and when will it be profitable?

this is the question also sorry for interrupting u before

[ionic jewel]

the company breaks even when profit is 0

[analog lintel]

i think i have to show my work

[ionic jewel]

for x >= 0

yes you do

you have to solve for x, by setting P(x) = 0

it's profitable when the profit is positive, or P(x) > 0

[analog lintel]

BUNNY I LOVE U

ty

im really dumb lol ty again

[charred flint]

just ask here morf

[fierce musk]

oh ok

yeah man could you showu how you got 1/1+r for one of the answers?

I got P/1+r for one of them

which i noticed wasnt in your answer set

[charred flint]

yea there's no reason to have P

the problem doesn't even mention P

[fierce musk]

no P is a variable i made

P represents the number of particles at the start

[charred flint]

yea just as a general tip if they don't mention a variable you won't have to invent a name for it

[fierce musk]

yes i just realised that

thanks

[charred flint]

the question asks for a fraction instead of a number of particles

[fierce musk]

yeah i understand that now so i know im wrong somewhere

but i can trace my error to the first series I got

the series for x=0

just would like to know how you set up your series at x=0

[charred flint]

you can reuse the series for x=1

x=0 looks like x=1 except each term is multiplied by r

like the 2nd bounce is r times the 1st bounce

the 4th is r times the 3rd

[fierce musk]

sorry i meant series for x=1

[analog lintel]

@ionic jewel bunny sorry to bug you but would the equation be -14(0)+133(0)-63? just double checking with you.

[charred flint]

gotcha

[ionic jewel]

no

[charred flint]

so it takes 2 bounces to go back to x=1

[ionic jewel]

you want to solve -14x^2+133x-63 = 0

[charred flint]

and for the first time it's 1-r of the particles

so the series looks like (1-r) + (1-r)r^2 + (1-r)r^4 +...

[ionic jewel]

then take the positive solution (since x has to be greater than 0)

[charred flint]

(1-r)*[1+r^2+r^4+...] = (1-r)/(1-r^2) by geo series

[analog lintel]

ty bunny, again, Im stupid lol ty

[charred flint]

and factoring gives you 1/(1+r)

[fierce musk]

thank you @charred flint

[undone herald]

can someone help me understand what exactly l is here?

or what it represents

I can barely follow this proof, and what l is is just confusing me

(this is from this document fyi https://math.uchicago.edu/~may/REU2015/REUPapers/Lytle.pdf)

[indigo jetty]

l is a certain value, which in this context is the limit of x_n and z_n as n -> infinity

[undone herald]

so it's saying as x_n and z_n approach infinity, they will converge on a certain value

[indigo jetty]

yup

and this value is l

[undone herald]

and if y_n is a value between these two values, it will too converge on l?

[indigo jetty]

yeah

[undone herald]

the second line of the proof, it's saying that, for values of y_n greater than some value N, y_n will have converged?

as y_n -l is equal to 0?

[indigo jetty]

basically it's saying y_n converges to l

[undone herald]

ohhh

I think I'm starting to get it

[indigo jetty]

that whole line is basically the formal definition of a limit

[undone herald]

so in the next part, it's saying that we know that x_n converges on l, and so does z_n

what does max(N1, N2) mean?

sorry I'm in hs atm, so I really don't know much lmao

[indigo jetty]

it's basically given two numbers N1 and N2, pick the larger of the two

[undone herald]

right ok, thanks

[indigo jetty]

👍

[undone herald]

yeah, that actually helps explain limits better

[terse cradle]

just a quickie 😉

[shell widget]

@terse cradle CD = 10cm. Then use pythagorus' theorem

[terse cradle]

ohhhhh so I dont use both sides

@shell widget what about this one just tryna get my head around it

[shell widget]

same pythagorus' theorem to find BD, where AD is the perpendicular bisector of CB

[terse cradle]

i did 8^2+14^2 but it wasnt the right answer

I feell ike im missing something lmao

[shell widget]

thats not the correct theorem

14^2 = 8^2 + (BD)^2

find BD

[terse cradle]

so i use that formula to find BC

[shell widget]

BD first

then multiply by 2 to get BC

[terse cradle]

wheres BD tho Ive done no study on this 😭

Could u quickly run through the whole thing like just formula wise and I should get it

[shell widget]

Do u see the imaginary line going down from A to BC?

D is the point where that line touches BC

And it is the midpoint of BC

[terse cradle]

so would that just be 8?

[shell widget]

That would be AD

AD = 8, AB = 14, BD = unknown

using pythagorus' theorem, 14^2 = 8^2 + (BD)^2, find BD

Then multiply by 2 to find BC

[terse cradle]

would BD be 16.12?

[shell widget]

,w solve x^2 = 14^2 - 8^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by Sup?#7960

[shell widget]

uh

[terse cradle]

ohhhhh -

[shell widget]

11.489

We cant take -2sqrt(33) since length is not negative

[terse cradle]

got that

ohhh

[shell widget]

2sqrt(33) is approximately 11.489

better to keep it as 2sqrt(33)

[terse cradle]

so what formula do i use now

I tried doing the other one it said no solution

[shell widget]

just multiply by 2

[terse cradle]

oh so I just x2 11.48

[shell widget]

2sqrt(33) * 2 = 4sqrt(33) = BC

[terse cradle]

Why is this making no sense to me

[oak chasm]

@terse cradle Does the Pythagorean theorem make sense as far as finding a side length of a right triangle when you have the other two side lengths?

[terse cradle]

yep

[oak chasm]

OK, and so you have a right triangle on the right side of the diagram.

So Pythagorean theorem gets you the bottom of that half of the big triangle.

Does that part make sense?

[terse cradle]

yep

[oak chasm]

OK, now the thing to go beyond that is to know that when you divide an isosceles triangle into two right triangles, the right angle hits exactly in the middle of the unequal side.

Does that part make sense?

[terse cradle]

yep

[oak chasm]

OK, so if it hits exactly in the middle, the bottom part of the right side is the same length as the bottom part of the left side.

Does that make sense?

[terse cradle]

yes

[oak chasm]

OK, so the bottom part of the right side is half of the whole bottom.

So, you double it.

So, the steps are to get the bottom half of the right side using the Pythagorean theorem.

Then double it.

Does that make sense?

[terse cradle]

somewhat

[oak chasm]

What are you stuck on?

[terse cradle]

like how do we find the whole length of CB

[oak chasm]

OK, so we find the length of half of it.

The right half.

[terse cradle]

would it just be 11.48

for one half

[oak chasm]

Yes, that's right.

[terse cradle]

then do times it by 2

[oak chasm]

Then since that's half the length, you double it.

Right.

[terse cradle]

Ohhhhhhhhh

SOrry im a slow learner 😭

[oak chasm]

Oh, that's OK.

[terse cradle]

could we do one more

[oak chasm]

Sure.

[terse cradle]

Then I think I should have the hang of it

[oak chasm]

OK, so the triangle OAB (not drawn) is isosceles.

Let's call the middle of the chord C.

So, AC is half of AB.

So, we have a right triangle AOC.

We know OA. We know AC, What's OC?

Does that make sense so far?

[terse cradle]

OA is 14 yeah

[oak chasm]

OB is also 14 because it's the center to a point on the circle.

So it's the radius.

So, isosceles.

OA = OB.

[terse cradle]

oh ok I see

So AB is 4

AC is 2?

[oak chasm]

Right.

[terse cradle]

So I do 14^2 - 2^2

[oak chasm]

Yes, that's right.

[terse cradle]

and then I should have the right answer

[oak chasm]

14^2 - 2^2 = OC^2

Well, almost.

You have to square root that.

[terse cradle]

yeah ofc

I got 13.85 and rounded it to 13.9

[oak chasm]

OK, do they want rounded answers or exact?

[terse cradle]

rounded

[oak chasm]

Oh, OK.

[terse cradle]

Ok last one if u want and then Ill stop bothering u @oak chasm

i know what to do if I have two values

[oak chasm]

OK, so you have three sides of a right triangle.

Write the Pythagorean theorem equation.

[terse cradle]

c^2+b^2+a^2?

[oak chasm]

No, a^2 + b^2 = c^2.

What are a, b, and c here?

[terse cradle]

a = 18

b = x, c = x?

[oak chasm]

No, a is across from the right angle.

Sorry 18 is.

[terse cradle]

ooh so its c

= 18

[oak chasm]

So, 18 is the hypotenuse, which goes by itself.

Right.

So, x^2 + x^2 = 18^2.

Does that make sense?

[terse cradle]

Yes

[oak chasm]

Do you know how to go from here with algebra?

[terse cradle]

like 50% just give me a quick run down

[oak chasm]

OK, so combine like terms on the left side.

What do you get?

[terse cradle]

well each triangle is 90 degrees

dont we have to use that

do determine one of the sides

[oak chasm]

We are with the Pythagorean theorem.

That requires a 90 degree angle.

So, we have x^2 + x^2 = 18^2.

Now we use algebra.

We're done with the trigonometry.

So, we combine like terms on the left.

What's x^2 + x^2?

[terse cradle]

I forgot

wait

2?

[oak chasm]

Almost.

[terse cradle]

since x = to 1?

[oak chasm]

No, we don't know what x is yet.

What's x + x?

[terse cradle]

x^1

[oak chasm]

No, it's 2x.

[terse cradle]

x^2

2x

[oak chasm]

x times x is x^2.

[terse cradle]

yeah tahts right

[oak chasm]

But x + x is 2x.

[terse cradle]

omg

[oak chasm]

So, x^2 + x^2 = 2x^2.

Same idea.

[terse cradle]

Oh ok

[oak chasm]

So, we have:
x^2 + x^2 = 18^2
2x^2 = 18^2

Now, what do we do next to get x closer to being by itself?

Using just algebra from before you learned triangles.

[terse cradle]

no idea

[oak chasm]

OK, so we look at what side has x on it.

It's the left side.

Then, we look at how to calculate it if we knew x.

To do that, we do the power first.

x^2, so we square it.

Once we do that, we multiply it by 2.

Do you remember that?

From order of operations.

Like if you have 2 times 5^2, you do the 5^2 first to get 25, then you multiply by 2.