help-13

[sacred grail]

it's because expectation is linear

E(X1 + X2 + X3) = EX1 + EX2 + EX3

[solemn quail]

Here are some other perspectives to perhaps build instinct:
It's similar to the: here are 3 jacks and 1 king. Go around the table letting people pick and keep a card at random and the winner is the one who gets the king.
some people think that picking earlier is better because if you win, then the others don't even get to play
but 1/4
= 3/4 * 1/3
= 3/4 * 2/3 * 1/2
Another way to think about it is this:
pick the first ball but don't look at it. What's the EV of the second ball?

[sacred grail]

if the X1, X2, X3 represent the number on each ball here, they have identical distributions

[solid juniper]

[muted bear]

Linearity of expectation is so weird

[solid juniper]

linearity of expectation and good probability intuition are like the same thing

[sacred grail]

even if there is dependence structure between the random variables, when considering their expectation you can look at each quantity in isolation

[muted bear]

The two examples O dog gave are intuitive, but this one feels unintuitive

[sacred grail]

if you ignore the information about X1 and X3, X2 will behave just as it does when you consider the additional information

the important thing here is the marginal distributions of each random variable

[muted bear]

The high and lows cancel out

Ok yeah i think i get it

[sacred grail]

there's no cancellation per se, the distribution of X2 is what it is regardless of what X1 and X3 are

[muted bear]

Another way to think about it is to enumarate B1, B2,... ,B9

Then how you pick 3, there will be an equal amounts of cases with each B as X1, and so on

Then EV will take the average

You can assign each B value to be whatever since theure equal counts

[sacred grail]

expectation is integration and then there's no surprises

everyone knows integration is linear

[cedar kiln]

@civic coral Has your question been resolved?

[jaunty mural]

ik its brilliant, but this is the first place I read about this and immediately got it . Didn't really click before
https://brilliant.org/wiki/linearity-of-expectation/

[cedar kiln]

@civic coral Has your question been resolved?

[cedar kiln]

@civic coral Has your question been resolved?

[cedar kiln]

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[wraith dagger]

vicks

[mint nimbus]

I have no idea where to start

well I can try something but tbh I am not sure if it is even right

can anyone guide me

[cedar kiln]

@mint nimbus Has your question been resolved?

[mint nimbus]

nooo 😦

[cedar kiln]

@mint nimbus Has your question been resolved?

[mint nimbus]

no

<@&286206848099549185>

[cloud elm]

are there multiple choices

@mint nimbus

[mint nimbus]

nope

[cloud elm]

hmm

To calculate this probability we can use the complement probability

[mint nimbus]

yeah for one

1 - 1/n

for two no idea

[cloud elm]

To find the probability that the 1st bin is not empty, we subtract this probability from 1

sooo

1 - (1 - 1/n)^k.

no?

p(1st bin is not empty)

[mint nimbus]

that's what i was thinking too

but the whole thing raised to k

not 2nd term

but I'm not sure

[cloud elm]

you have the answer?

[mint nimbus]

nope

[cloud elm]

well i would go with those i provided as they are my final answers

[mint nimbus]

can you explain why you only raised the second term by k?

[cloud elm]

ummm

because we want to find the probability that the 1st bin is not empty after throwing all k balls into n bins

waiyt

Since each ball is thrown independently and uniformly at random into one of the n bins

the probability that any particular ball is not placed in the 1st bin is (n-1)/n.

Now

if you have k balls

and each has this probability of (n-1)/n of not being placed in the 1st bin

then the probability that all k balls

((n-1)/n)^k

[mint nimbus]

okay i see

[mint nimbus]

.close

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[jovial forum]

Hello, can anybody explain me this? I’m studying probability and the professor made an example and said.. let’s say we want to get all the particles position and velocity in the room, then he proceeded to represent the set of particles as such

[jovial forum]

N is number of particles

Why is it (…)^N and not (…)*N

Pls help me

[long swan]

That looks like a map of China

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[somber belfry]

help regarding metrics

Im looking at the metric $$d(x,y)=min{p(x,y), 1}$$
with $p(x,y) = |x-y|$.
for the triangle inequality there should be 4 cases if im not mistaken or not?
triangle equality being: $$d(x,z) <= d(x,y) + d(y,z)$$
$$ 1. p(x,z), p(x,y) and p(y,z) < 1$$
$$ 2. p(x,z) >= 1 and p(x,y) and p(y,z) < 0 $$
$$ 3. p(x,z) >= 1 and p(x,y) or p(y,z) >= 1 $$
$$ 4. p(x,z), p(x,y) and p(y,z) >= 1 $$

[wraith dagger]

Eichhorst

[cedar kiln]

@somber belfry Has your question been resolved?

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@somber belfry Has your question been resolved?

[somber belfry]

.close

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[twilit birch]

hey

[twilit birch]

is someone here

?

I have a string 1 2 3 4 5 6
How many arangements can be done to this string that no number in there can be to its initial position (1 can t be on position 1 2 can t be on position 2 and so on)

this is my problem

[buoyant latch]

I reckon finding the opposite is easier

How many ways would I still have numbers in their original spot

[twilit birch]

how come

[buoyant latch]

Idk just a hunch

[twilit birch]

and how would i do it

?

[buoyant latch]

Have you tried anything?

[twilit birch]

kinda

[crimson sedge]

show it

[buoyant latch]

Let’s see it

[twilit birch]

ok first i need to translate

[crimson sedge]

Frosst

what was the status command

[buoyant latch]

!status

[cedar kiln]

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

[crimson sedge]

I did ,status

ty

[twilit birch]

To solve the problem, we will use the principle of bijectivity, also known as the Inclusion-Exclusion Principle.

We have an initial string: 1 2 3 4 5 6.

In order for no element to be in its original position in the resulting strings, we must arrange the elements in such a way that we have no permutations (arrangements) where any element is the same position as in the original string.

We take each element and see how many ways we can put it in its original position. Next, we apply the Inclusion-Exclusion Principle to calculate the total number of possible arrangements that satisfy the required condition.

Item 1: We have 5 positions available to put 1 because it cannot be in its original position. We have 5 ways to arrange element 1.

Element 2: If 1 has taken position 2, then we have 4 positions available to put 2 (apart from its original position). If 1 took position 3, then we also have 4 positions available to put 2 (all apart from its original position). So in total we have 2 * 4 = 8 ways to arrange element 2.

Element 3: If 1 occupies position 3 and 2 occupies position 1, then we have 3 positions available to put 3 (apart from its original position). If 1 occupies position 3 and 2 occupies position 4, we also have 3 positions available to put 3 (besides its original position). If 1 occupies position 4 and 2 occupies position 1, we also have 3 positions available to put 3 (besides its original position). If 1 occupies position 4 and 2 occupies position 3, we also have 3 positions available to put 3 (besides its original position). So in total we have 4 * 3 = 12 ways to arrange element 3.

Element 4: Similarly, we will get 20 ways to arrange element 4.

Element 5: Similarly, we will get 30 ways to arrange element 5.

Element 6: Similarly, we get 30 ways to arrange element 6.

Finally, we add up the total number of ways we can arrange each element separately: 5 + 8 + 12 + 20 + 30 + 30 = 105.

Therefore, in 105 different ways, we can arrange the elements of the original string so that none of them is in its original position

[crimson sedge]

dude we wanted to see what you have tried

[buoyant latch]

Was what you tried “ask chatgpt”

[twilit birch]

i ve been trying to explain myself this

nah

[buoyant latch]

Because if it is, that’s not training your own thinking skills

And if that’s not what you tried

[twilit birch]

i got like 6 diff answers to that

[buoyant latch]

Then let’s see what you tried

Let’s see all of them

[crimson sedge]

Neb1, never look up the solutions or ask gpt for combinatorics and calculus.

[buoyant latch]

Or basic arithmetic

[crimson sedge]

@twilit birch if 1 is not allowed at the 1st position, how many possibilities does the 1st position have?

[twilit birch]

5! ?

[crimson sedge]

why 5!

[twilit birch]

oh just the first pos

[crimson sedge]

yeah

just the first pos

how many do you think are left if 1 isn't allowed?

[twilit birch]

we have 5 more digits so 5 more posibilities no?

[crimson sedge]

yeah

now the same on 2nd position

how many possibilities do we got if 2 isn't allowed?

[twilit birch]

5

[crimson sedge]

isn't this the same for all 6 positions?

[twilit birch]

no because i can t have idk like 1 3 2 4 5 6

because there ARE STILL 4 digits that are in their same exact pos

1 4 5 6 are in their initial position

[crimson sedge]

repetition is allowed or not?

as per the question

[twilit birch]

all of them need to be in different spots

repetition as in 6 6 6 1 2 3

no

[crimson sedge]

alr

[twilit birch]

just rearranging them

i found something about idk fixed point arrangements

or smth like this

i might think it s related

these are the answers that are given

so?

[crimson sedge]

back, had to go for a while

[twilit birch]

no problem man

[crimson sedge]

i translated, the question does not say anything about repetition

[twilit birch]

look i m not that good at these things with combination and all that but it s reffering to an array

it s a problem at a supposed comp science exam

and it s not the maths exa

exam

[crimson sedge]

I have not done combinatorics since a year but I'll try

[twilit birch]

ok

[crimson sedge]

exam?

[twilit birch]

it s an old subject

it s an exam for a uni

[crimson sedge]

kinda weird since I'm getting 5×5!

not sure where I'm making the mistake

[twilit birch]

at least you tried man

[crimson sedge]

oh it's same as 6!-5!

both are 600

[twilit birch]

how did u get here?

the same thing u told me earlier?

[crimson sedge]

no no that time I thought repetition is allowed

I confused myself

[twilit birch]

i know right?

[cedar kiln]

@twilit birch Has your question been resolved?

[cedar kiln]

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[random pivot]

,tex (sqrt(3) x sqrt(50)) / sqrt(6)

[wraith dagger]

Iver

[random pivot]

How do I simplify?

I already made sqrt(50) into 5sqrt(2)

Look at brackets or whatever they are called

Carefully, because I divded everything by sqrt(6)

[cosmic steppe]

sqrt(6) = sqrt(2) * sqrt(3) since sqrt(ab) = sqrt(a) * sqrt(b)

Use that to your advantage

[random pivot]

omg how i didint think of that

thank you

let me see if i can figure out now

,tex 5

[wraith dagger]

Iver

[random pivot]

@cosmic steppe ?

[cosmic steppe]

,w 150/6

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by umbraleviathan

[cosmic steppe]

Yeah

[random pivot]

what is this?

[cosmic steppe]

Just checking

[random pivot]

150???

ok

[cosmic steppe]

3 * 50

[random pivot]

but why 3x

oh

removed all

sqrt

[random pivot]

smart method

ty

[cosmic steppe]

your fraction is equivalent to sqrt(150/6)

[random pivot]

ye

😁

.close

ty

[cedar kiln]

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[scenic chasm]

Hi.
Just small question about binary system
So their is 4 type , binary and octal (8) and decimal (10) and hexadecimal (16)

So my question is about this (5) and (4) for example what they call it , and also I can do calculate In it using only fhe devide method on it? (I mean for example for types above their is others methods to convert)

[median holly]

thats base 5 and base 4

yesyou can find it by the division method

[scenic chasm]

So their is only devision method for base like This?

[median holly]

i cant say

i know only that method too

although im comfortable with powers and use those

[scenic chasm]

Oh ok

[cedar kiln]

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[warm tree]

why are the two steps marked wrong, but the answer is correct

[granite knoll]

because w = x^2 - 36, not w = x^2

[warm tree]

im writing it in terms of w

oh

i see

thanks

.clsoe

.close

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[silver osprey]

Conservation of mass related question I guess,
I'm trying to find the concentration of a drug within 5L of blood, where it is being filtered out at a rate proportional to its concentration.
The initial dose is 200mg at 4mg/s, and is followed by a continuous administration at a rate of 0.25mg/s, which balances the rate at which it is filtered out.
It's assumed that the level of blood is constant, and the concentration stays constant during the 0.25mg/s phase, im just wondering how to find that concentration.

Sorry if there is not enough information, this is just driving me up a wall

[silver osprey]

This is for an intro mathematical modelling course if that helps at all

[cedar kiln]

@silver osprey Has your question been resolved?

[silver osprey]

<@&286206848099549185> when you get a chance

[verbal compass]

ooh thats actually an intresting question

[silver osprey]

its for homework and alot of my classmates are going with 40mg/L but my thought process is that it would start filtering the second its administered so it would be abit less once the full 200mg dose is done

[verbal compass]

concentration is measured in mg/L?

[silver osprey]

ye

a rough diagram that to visualise my thought process i did but that’s about as far as i’ve gone

could never really wrap my head around conservation of mass equations in modelling

[verbal compass]

i got 39.95 😭

wtf

[silver osprey]

could you show me how you got to that? im confused as to how to approach it

[verbal compass]

ok so 200 mg at 4 mg/s is administered, which would take 50 seconds, right?

lets model everything as a function of time, so C(t) = concentration at a time

f(t) = rate of filtering as a function of time

f(t) = k*C(t) (via proportionality)

now lets work out C(t)

C(t) = (mg of drug)/5

to find the mg of the drug at any point, we know were getting +4 mg every second, but losing f(t) mg every second

so then mg of drug = 4t - f(t)

so C(t) = (4t - f(t))/5

and f(t) = k*(4t - f(t))/5

rearrange it to get f(t) = 4t/(5/k + 1)

so now we have f(t) in k and 1

[silver osprey]

ooh ok

[verbal compass]

ok i overcomplicated this alot

once you get C(t) = (4t - f(t) )/5
thats all u need to find

ignore all the proportionality stuff

we know f(50) = 0.25

so when t = 50, f(t) = 0.25 (cuz adding 0.25 balances it)

[verbal compass]

sub that here

you get C(50) = 39.95

[silver osprey]

you are

beautiful cute and amazing

[silver osprey]

tysm

.close

[cedar kiln]

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[silver osprey]

.reopen

[cedar kiln]

✅

[silver osprey]

actually sorry lmao

how would i find K moving forward?

also since f(t) is a rate does subbing it into C(t) = (4t - f(t) )/5 as 0.25 work?

sorry just trying to wrap my head around it

lmao nice bot

<@&268886789983436800>

[verbal compass]

yes you can sub f(t) = 0.25 and t = 50 (once all 200 mg have been used)
this is how i found C(t) at t= 50

but that wont help you find k

[silver osprey]

mm ok

[verbal compass]

for k you need to consider the proportionality

you have a function for C(t) in terms of f(t) and t

[silver osprey]

but is 0.25 not whats being removed at t=50 not what has been removed in total?

[verbal compass]

write f(t) = k*C(t) , so then you have an equation in f(t), t and k. then sub known values of t and f(t) (like 50 and 0.25)

[verbal compass]

after the 200 mg is injected after 50 seconds, 0.25 mg/sec is administered. so its not a total, but a rate per second. since it perfectly counteracts the rate of filtration, we can assume f(t) for that period = 0.25, so whatever it being removed is being perfectly accounted for to keep concentration constant

[silver osprey]

ahhhh ok sorry i get it

its 5am for me

[verbal compass]

oof

you should get some sleep

come back to this in the morning

ping me if you need any more help

[silver osprey]

yeah, ive got the class for it at 1pm so i was just trying to crunch it out before then, its been a hectic weekend

thankyou again for your help

you're amazing

.close

[cedar kiln]

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[verbal compass]

any time

[cedar kiln]

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[chrome flax]

Does this look correct so far?

[sage forge]

Usually one would just substitute 3sin u = x

[chrome flax]

U have to use a suitable hyperbolic substitution

[cedar kiln]

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[west turret]

how do i solve this?

[west turret]

<@&286206848099549185>

[fluid talon]

4pi* r^2 = 356 * 2

yo uhave the equation

[west turret]

its 2pi r ^2

because its a hemisphere

[fluid talon]

yeah same thing

i multiplied 356 by 2

[west turret]

so what do i do

[fluid talon]

they gave you the surface area of it

[west turret]

yeah

[fluid talon]

so they're equal

[west turret]

i need to find the diameter

[fluid talon]

first you find the radius, just multiply it by 2 to get the diameter

[west turret]

how do i get the radius though

thats what im asking

[fluid talon]

you have the "build" an equation

[west turret]

how

[fluid talon]

you have the formula for surface area of a hemisphere, and you have the value of it

so you just make them equal

[west turret]

356= 2pi r^2 + pi r^2

[fluid talon]

2pi r^2 = 356

356= 2pi r^2 + pi r^2

this

why did you add the second part?

[west turret]

ok so we divide it by 2pi?

[fluid talon]

yep

[west turret]

accident

[fluid talon]

and find the square root

then multiply the radius by 2 to get the value of the diameter

[west turret]

radius^2 = 356/2pi

what from here?

[fluid talon]

what is the inverse operation of exponent?

it is the root

right?

[west turret]

yh

so square root

so radius = square root of 356/2pi

[fluid talon]

yep

[west turret]

so when i put in in my cacl

[fluid talon]

but you want to find the diameter, not the radius

[west turret]

and i get the answer

should i press 'Ans' then times that by two for the radius

[fluid talon]

diameter = 2 * radius

[fluid talon]

exactly

[west turret]

or should i round the radius to 1dp before multiplying it

the answer is wrong

[cedar kiln]

@west turret Has your question been resolved?

[cedar kiln]

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[pulsar beacon]

Can someone help idk if I’m doing it right

[crimson sedge]

8-1+4*2 all over 10

So 15/10

Or 1 and a half

Just sudstitute what you found G(2) and G(-1) to be

[pulsar beacon]

so part 1 and 2 are right

[crimson sedge]

Yup

[pulsar beacon]

and for part 3 i do this

[crimson sedge]

Yes

[pulsar beacon]

ok hold up

[crimson sedge]

Do it in steps

$\frac{(2)^3-(-1)^2+4(2)}{10}$

[wraith dagger]

HadarS

[crimson sedge]

$\frac{8-(1)+8}{10}$

[wraith dagger]

HadarS

[crimson sedge]

$\frac{15}{10}$

[wraith dagger]

HadarS

[pulsar beacon]

3/2

[crimson sedge]

$1\frac{1}{2}$

[wraith dagger]

HadarS

[crimson sedge]

Yes

[pulsar beacon]

ohhhh ok

so i had ogtten it right the f

[crimson sedge]

I love this Tex bot

[pulsar beacon]

ty for the help

[crimson sedge]

Np

Recommend to your friends

[pulsar beacon]

how

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[late rapids]

What does it mean for a sequence to be increasing and bounded

[late rapids]

It’s in reference to this explanation in the bottom

[dire geode]

increasing means the terms are getting bigger

bounded means there's a number that's greater than all terms in the sequence

[late rapids]

Okay I think I understand

It’s referring to a_n right

[dire geode]

a_n is the sequence yes

[late rapids]

Oh okay thank you have a nice day

.close

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[edgy swan]

Can someone help me figure out where I messed up?

It said the answer is supposed to be -5ln|x+2| + 5ln|x+1| + 4/(x+1) + C

So I am close, I just don’t know where the 5ln|x+1| comes from, and I don’t know how they get rid of the x

[dire geode]

fyi this channel's gonna close

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[hollow ember]

I'm on number 2, and this is what I have so far

I don't think the set spans R^2...

but I'm not sure

I haven't learned how to figure it out with like linear dependence or something yet

I can say (1, 0) where 1=1+2(0)+2(0)

and a=1, b= 0, c=0

so then

y=-1 -2(0) + 3(0)

that doesn't equal 0

so (1, 0) doesn't work there

but I dont know if it can work with another combination of a, b, c

so idk if that's a valid counterexample to say the set doesn't span R^2

[cedar kiln]

@hollow ember Has your question been resolved?

[cedar kiln]

@hollow ember Has your question been resolved?

[cedar kiln]

@hollow ember Has your question been resolved?

[crimson sedge]

the sets elements arent all linearly independent

(2,-2) is a multiple of (1,-1)

therefore they can be written as linear combinations of eachother

[hollow ember]

So does that mean the set doesn’t span R^2?

[cedar kiln]

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[feral moon]

Little confused where to go from here, is n= 2^n?

[cedar kiln]

@feral moon Has your question been resolved?

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@feral moon Has your question been resolved?

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@feral moon Has your question been resolved?

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[upbeat fern]

I would've done volume - surface area but the surface area is bigger so

I don't know what to do

[sinful crane]

The answer is A

a 5 by 5 cube is painted black on the surface

means the inner 3 by 3 is not painted in black

You simply just substract the side by 2 to get the side of the inner cube

[upbeat fern]

Oh yeaaaaa

That makes sense

Okay ty

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[edgy laurel]

does anyone know the equation of a skewed quadratic? or a similar function

[crimson delta]

wdym

[cedar kiln]

@edgy laurel Has your question been resolved?

[edgy laurel]

i’m trying to find a function that can model increasing velocity and decreasing velocity, but it can’t be a quadratic. it needs to increase quickly, reach a max and then decrease at a slower rate

[south tundra]

Something like 1 - x^4 perhaps?

[cedar kiln]

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[gloomy prism]

y is inversely proportional to the square root of x.
It is known that y = 64 for a particular value of x.
Find the value of y when this value of x is increased by 300%.

[gloomy prism]

i only know the formula is y=k(sqroot x)

[tropic oxide]

no

[spice kraken]

inversely proportional

[tropic oxide]

it's $y = \frac{k}{\sqrt{x}}$

[wraith dagger]

Ann

[gloomy prism]

ohh my bad

i still dont know how to solve the question tho

[tropic oxide]

do you know what "increased by 300%" means?

[gloomy prism]

multiplied by 4?

[tropic oxide]

yes

what happens to y when x gets multiplied by 4?

[gloomy prism]

is it divided by 4

[tropic oxide]

no

y is inversely proportional to the square root of x.

not to x itself.

[gloomy prism]

huh

what should i do

[tropic oxide]

$\frac{k}{\sqrt{4x}}$

simplify this

[wraith dagger]

Ann

[gloomy prism]

k/2 sqrt x?

so y divided by 2?

[tropic oxide]

yes

[gloomy prism]

ohh tysm the answer is 32 right

thanks

.close

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[jagged trail]

$$Prove that \sum_{k=0}^{n}(-1)^k $$
$$\begin{smallmatrix}
n \
k$$

[cedar kiln]

Please don't occupy multiple help channels.

[jagged trail]

$$Prove that \Sigma_{k=0}^{n}(-1)^k \binom n k= 0$$

[crimson delta]

$\binom n k$

[jagged trail]

thanks

[wraith dagger]

Denascite

[crimson delta]

and \sum instead of \Sigma

[tropic oxide]

Prove that $\sum_{k=0}^n (-1)^k \binom{n}{k} = 0$.

[wraith dagger]

Ann

[tropic oxide]

here, fixed the badtex for you.

[crimson delta]

and dont put the text inside the $

[tropic oxide]

!status

[cedar kiln]

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

[jagged trail]

thankyouu

[tropic oxide]

do you know the binomial theorem?

[jagged trail]

ye

I'm just absolutely terrible with proving stuff and have no idea how to even begin with such things

[tropic oxide]

ok can you state the binomial theorem for me

[jagged trail]

$\binom n k = \frac{n!}{k!(n-k)!}$

[wraith dagger]

Feevo

[tropic oxide]

that's not the binomial theorem

that's just the formula for binomial coefficients

[jagged trail]

OH so
$(x + y)^n = \sum_{k=0}^n \binom n k x^{n-k}y^k$

I think

[tropic oxide]

yes this

er

a and b, or x and y?

pick one

[jagged trail]

oh whooops

[wraith dagger]

Feevo

[tropic oxide]

there we go

now, what values can you pick for x and y to make the RHS of this into $\sum_{k=0}^n \binom{n}{k} (-1)^k$?

[wraith dagger]

Ann

[jagged trail]

so I basically have to find $x^{n-k}y^k=(-1)^k$

[wraith dagger]

Feevo

[tropic oxide]

cringe wording but sure...

[jagged trail]

why is it cringe ;-;

I'm legit so bad at math idfk how

[tropic oxide]

x and y can be whatever you want

what could you make them into so that x^(n-k) y^k becomes (-1)^k?

for the love of god do not see this as equation-solving.

[jagged trail]

if x is 0 and y is -1 ig

[buoyant latch]

You’re telling me (0 - 1)ⁿ = 0?

[jagged trail]

not me forgetting that this is multiplication

1 and -2, for instance then

no

idfk

[tropic oxide]

x=1, y=-1

[jagged trail]

i am dumb

[buoyant latch]

You just aren’t as experienced

[neat dune]

or both 😛 (but it's too early to tell if u r dumb )

[jagged trail]

.close

[cedar kiln]

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[cedar kiln]

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[dense lichen]

i'm confused with 7b

i think what i did is wrong because i'm double counting stuff and also every time you add a man, it creates more gaps

so are there 16x17x18x19 ways to organize the remaining 4? instead of 16^4

and i don't know how to solve the double counting problem :( am i apporaching it wrong?

[cedar kiln]

@dense lichen Has your question been resolved?

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@dense lichen Has your question been resolved?

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[mossy halo]

Corroll me plese

[mossy halo]

2,7 1 FALSE

b ) True c) fasle d) true ,e) false, f)true g) true

[broken mist]

,rccw

[mossy halo]

is correct?

[mental trail]

For b)

Is the empty set one of the elements of {0}?

[mossy halo]

2,8 a)true b) true c) false d) true e) true f)false

empty set is a subset of {0}

not a element

[mental trail]

So 2.7 b is ?

[mossy halo]

True

i think

[mental trail]

What does $\emptyset \in {0}$ mean ?

[wraith dagger]

rafilou2003

[mossy halo]

So , its false

then

[mental trail]

Yes

Also what does $\subset$ mean for you?

[wraith dagger]

rafilou2003

[mental trail]

Is there a difference with $\subseteq$?

[wraith dagger]

rafilou2003

[mossy halo]

okay, there is a difference

give me a sec

in short terms,\subsetteg is like =

so , if we have 2 sets like A = {1,2} and B = {1} B is subset of A but not ⊆

if B was {1,2} then i can use ⊆ symbol

[mossy halo]

correct?

[buoyant latch]

B is also ⊆ A

If B ⊂ A then B ⊆ A

It’s like < and <=

If b < a then b <= a

[mossy halo]

ok

okay, exercies are correct?

[buoyant latch]

Wait what the empty set is in every set

[mossy halo]

empty set is subset of every set

[buoyant latch]

No

The set containing the empty set is a subset of every set

Think about it

[mossy halo]

2.7 B is an example . So, that expression false?

[buoyant latch]

${0}\cap{1} = \emptyset$

[mossy halo]

i don't know how to read that

[buoyant latch]

Do you agree

[wraith dagger]

Frosst

[mossy halo]

no

[buoyant latch]

Then what should the intersection be

[mossy halo]

okay i agree

there isn't any number

okay, i going to do next exercies

[mental trail]

?????

No. The set containing the empty set, for example, is not a subset of the empty set

The empty set itself is a subset of every set

[buoyant latch]

Is the empty set not in the empty set

[mental trail]

The empty set is not an element of the empty set

The empty set has NO elements

[buoyant latch]

Oh

[mental trail]

"If A is a set, every element of the empty set is an element of A". This statement is true because the empty set has no elements

So for any set A, $\emptyset \subseteq A$

[wraith dagger]

rafilou2003

[buoyant latch]

Ah if I select nothing from A I get the subset the empty set

And unions and intersects give sub/supersets not elements

[mental trail]

With what was said about the difference between $\subset$ and $\subseteq$, it would be nice to check 2.7 f) and g) again

[wraith dagger]

rafilou2003

[buoyant latch]

That’s my bad

I also got confused

[mental trail]

And check 2.7 e) as well

[mossy halo]

2

It is Okay ?

a ⊆ C

[mental trail]

What is okay?

[mossy halo]

my proof

[mental trail]

That A is a subset of A ?

[mossy halo]

A ⊆ C

[mental trail]

There is no proof here

You just stated "If A ⊆ B and B ⊆ C, then A ⊆ C"

Or are you talking about what's under

[mossy halo]

okay, how to write it in normallway

what's under

[mental trail]

Are you trying to prove that A ⊆ A ???

[mossy halo]

I am trying to proof A ⊆ C

[mental trail]

You're trying to prove {1,2} ⊆ {1,2}

[mossy halo]

yeah , i just use an example of sets

[mental trail]

Ok, what's the definition of "A ⊆ C"?

[mossy halo]

by exercise is A ⊆ B, B ⊆ C, and I need to proof that A ⊆ C

[mental trail]

Ok

[mental trail]

.

[mossy halo]

A is subset or equal to C

[mental trail]

And what does that mean?

[mossy halo]

dont understand a question

[mental trail]

What does "A is a subset of C" mean?

[mossy halo]

It mean , that they have same elements

[mental trail]

They have the same elements ?

So A = C ?

[mossy halo]

if B is subset or equal to C and A is subset or equal to C then A is equal or subset of C

i don't how to write it in normall way

[mental trail]

What is the definition of "A is a subset of B" ? How do you know, if I give you two sets A and B, that one is a subset of the other?

[mossy halo]

because it's written in exercise

[mental trail]

????

Ok well here's something for you :

If A = {1,2,3} and B = {1,2}, is one a subset of the other? And why?

[mossy halo]

B is subset of A because A have numbers{1,2} and B have numbers{1,2} so by deffenition of subset subset is a part of a given set

[buoyant latch]

What does it mean to be “a part of a given set”

[mossy halo]

it means that they have similiar elements

[buoyant latch]

Ah, say that more clearly

What has what elements

What do you mean similar

[mossy halo]

set B has set A elements in it

[buoyant latch]

What does “has” mean

[mossy halo]

contain

[buoyant latch]

We’re trying to define contain here

How do you know {1,2} is a subset A

And why did you check {1, 2}

[mossy halo]

i just create 2 random sets

[buoyant latch]

Why didn’t you choose {ball, apples}

[mossy halo]

what?

[buoyant latch]

Why randomly choose sets

[mossy halo]

i really don't understand what you want to hear

[buoyant latch]

Why did you choose {1,2} to see if A has this

Where did this {1,2} come from

[mossy halo]

from my head

okay forget about {1,2}

[buoyant latch]

Then how can you reason that B is a subset of A

You never mentioned B

[mossy halo]

by exercice . A ⊆ B B ⊆ C. I need to proof that A ⊆ C.

that all exercise

[buoyant latch]

It’s the same problem

What does ⊆ mean

[mossy halo]

subset or equal

[buoyant latch]

What does it mean to be equal

[mossy halo]

that sets have same elements in it

or have same value

[mental trail]

Yes, A = B means that "Elements of A are elements of B AND elements of B are elements of A"

[buoyant latch]

$\forall x\in A(x\in A \iff x\in B)$ means $A = B$

[wraith dagger]

Frosst

[mental trail]

So now, from this definition of A = B, can you guess how A ⊆ B is defined ?

$\forall x(x\in A \iff x\in B)$ means $A = B$

[wraith dagger]

rafilou2003

[mossy halo]

$\forall x(x\in A \iff x\in B)$ means $A ⊆ B$

[buoyant latch]

Ah that’s enough as well

[wraith dagger]

Sali_12

[buoyant latch]

No, that would be equals

[mental trail]

We just said that this is equals

[buoyant latch]

I mean he’s not wrong

But like

[mental trail]

"Means" is not "implies"

[mossy halo]

okay, but u need to proof that A ⊆ C

[buoyant latch]

Mhmm I suppose it’d mean is defined as

[mental trail]

Yes, but how can we prove it when we dont know what it means?

$\forall x(x\in A \Longrightarrow x\in B)$ means $A ⊆ B$

[wraith dagger]

rafilou2003

[mental trail]

Every element of A IS an element of B. Only this, and not the reciprocal, means that A is a subset of B

[mossy halo]

okay

[mental trail]

So now

[mossy halo]

then i can write $\forall x(x\in A \Longrightarrow x\in C)$ means $A ⊆ C$

[wraith dagger]

Sali_12

[mental trail]

Yes

So proving A ⊆ C means proving the statement to the left

How do we prove a statement that starts with "forall" ?

[mossy halo]

okay, but how to write answear if i have a test

first i need to write

$\forall x(x\in A \Longrightarrow x\in B)$ means $A ⊆ B$

[wraith dagger]

Sali_12

[mossy halo]

then

$\forall x(x\in A \Longrightarrow x\in C)$ means $A ⊆ C$

like that?

[mental trail]

?

You wrote the same thing twice

[wraith dagger]

Sali_12

[mental trail]

No that's not it, you skipped a step

Anyways

[mental trail]

We want to prove this statement

The proof of a statement that starts with "forall" will start with "Let..."

It will ALWAYS be the case

So

"Let x be an element."

Then we have to prove that $x\in A \Longrightarrow x\in C$

[wraith dagger]

rafilou2003

[mental trail]

So what do we write first?

[mossy halo]

proof that A ⊆ B

[mental trail]

No, we know that A ⊆ B, that's in our hypotheses

So we don’t have to prove it

We have to prove that "Statement 1" implies "Statement 2"

How do we write the proof of that?

[mossy halo]

let x E A .

since A ⊆ B , then x E B

since B ⊆ C and x E B , then x E C

since x we take randomly from A , we show that x E C , we proved that A ⊆ C

is that okay?

[mental trail]

yes

[cedar kiln]

@mossy halo Has your question been resolved?

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[cedar kiln]

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[meager pike]

Ive set c = omega * e^(alpha t)

got the dervative of that and set it equal to DlapaceC + alpha c

idk what else to do from here

i also have tried taking the dergative of domega/dt

oh wait NEVERMIND

by doing it that way i see that dthea/dt = dlapacanC e^-alphat

i can plug that back in and i get the solution i think 🙂

.close

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[bleak monolith]

How I make this

[primal socket]

What do you need help with?

[bleak monolith]

how can I calculate with a?

[primal socket]

The first example in a?

[bleak monolith]

wait

[primal socket]

Waiting

[meager pike]

always ready

[primal socket]

I don't understand textbooks that give a subtask a that has 3 examples in it

[primal socket]

Forgot a closing bracket there

[bleak monolith]

Oh ok

[meager pike]

this is gonna be (2 sqrt(5) - sqrt(3)) * (2 sqrt(5) - sqrt(3))

when you square brackets like this you foil out

[primal socket]

Yeah, otherwise only the sqrt(3) is getting squared

[primal socket]

Have you learned about binomial expansion formulas yet?

[bleak monolith]

Yeah

[primal socket]

Well, then. You've got a square of difference

$(a-b)^2$

[wraith dagger]

USS-Enterprise

[primal socket]

How do we expand this

[bleak monolith]

ahh

[primal socket]

Yeah, always try to spot things

Think of $2\sqrt{5} = a \land \sqrt{3} = b$

[wraith dagger]

USS-Enterprise

[bleak monolith]

Is this right yet

Well is kinda like that

[primal socket]

I have to be real with you, this example is none of the 6 you posted above

[meager pike]

LMAO

[primal socket]

It however is correct, yes.

[bleak monolith]

I have this in extra page

[meager pike]

thats the way you solve these problems though

[bleak monolith]

Sorry

[primal socket]

No problem 🙃

[meager pike]

you expand it out, do the multiplican and simplfly

[bleak monolith]

Ok

.close

[cedar kiln]

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[dire geode]

probably some half angle identity

,tex .half angle

[wraith dagger]

riemann

[cedar kiln]

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[crystal mauve]

I have to prove using the axioms that x < y <=> -x > -y
How to do it? I have no ideas 😭

[dire geode]

well use the axioms

copy them here from your book if you don't know them by heart

[crystal mauve]

but i don’t really know which ones i should use

[dire geode]

try

[crystal mauve]

that’s the only one assignment that i don’t have solution to

[dire geode]

you shouldn't need a solution to do this if you're taking a class that requires you to prove staetments, just axioms

[crimson delta]

how would you do this if you didnt have to use axioms explicitly

[crystal mauve]

i have to use them in this example

[dire geode]

...

[crystal mauve]

i have four examples to do and 3/4 are done

[dire geode]

dena is giving you a helpful hint

[crystal mauve]

that’s the last one

[dire geode]

and i asked you to post your axioms

and yet you've done neither

what do you want from helpers

[crystal mauve]

i would multiply them by -1

[crimson delta]

ok. now post your axioms

[crystal mauve]

i have 13 of them

out of 14 as far as i know

notes are in polish (sorry 😅)

[dire geode]

try using the axioms that have multiplication and -x or -y

[crystal mauve]

x + (-x) = 0 is the only one with -x

[crimson delta]

I would instead try to do it only with addition

[crystal mauve]

no idea

no results

[crimson delta]

so, you need at least one axiom about < which you should use. if I say "only addition", which one will it be

[crystal mauve]

12

[crimson delta]

aka "add on both sides". what could be useful to add here

[crystal mauve]

the opposite of one of the factors

[crimson delta]

ok, lets say -x

what do you get

[crystal mauve]

0 < y - x and 0 > x - y

[crimson delta]

only do it to the left inequality

the one we start with

we want to do something to it

and end with the other one

ok, so 0 < y-x

where we used axiom 4

now, what else could we add

[crystal mauve]

yup

[crystal mauve]

0

[crimson delta]

well adding zero does not that much

[crystal mauve]

-y

that’s the solution

omg boss

thank you so much

[crimson delta]

youre welcome

[sharp creek]

Can someone help me to solve an integral ?

[cedar kiln]

@crystal mauve Has your question been resolved?

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[fiery whale]

For two n-vectors a,b, is a x transpose(b) necessarily invertible? trying to show that C = B + a x transpose(b) is invertable, given B is invertable

[crimson delta]

not even close

a * b^T has rank 1

[fiery whale]

?

its an n*n matrix

a=nx1, b^T = 1xn, so a * b^T is n*n

[crimson delta]

yes

do you know what rank of a matrix means

[fiery whale]

oh well yeah

leading 1s

[crimson delta]

all the rows of a*b^T are multiples of b^T

so if you do row reduction, you'll just get a single nonzero row

[fiery whale]

oh, didnt know that

so definetly not invertible

[crimson delta]

your claim is not true in general. you need to know something about a and b in relation to B

[fiery whale]

only other condition is 1+b^T * B^-1 * g =/= 0

but dont see how that helps

[crimson delta]

well at least thats exactly the correct condition

try solving (B+ab^T)x=y for x

let s = b^Tx, solve it in terms of s

[fiery whale]

x = B^-1 (y-as)

[cedar kiln]

@fiery whale Has your question been resolved?

[cedar kiln]

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[cyan void]

how do i get started

[cedar kiln]

@cyan void Has your question been resolved?

[cedar kiln]

@cyan void Has your question been resolved?

[cyan void]

@slate niche

<@&286206848099549185>

[fleet bloom]

Stop spamming helper

s

where do you deed help in

[cyan void]

above

[late matrix]

You just need the value of the angle DXG? Because with the value of DXF, the value of FZG; and G is the incenter of the triangle XYZ, I understand that it would get its value of the angle DXG.

[cyan void]

hm

[cyan void]

Would it be 36

[late matrix]

The value of DXG it would be 46°

[cyan void]

still confused

[late matrix]

Wait

[fleet bloom]

<@&286206848099549185> for @cyan void

it was lylz

[cyan void]

OH

ITS HALF

[late matrix]

Do you understand that now?

It seems strange to me that it's just like that, because the other data would be filler xd

[late matrix]

this is the definition of the incenter, with this and with the value of the DXF angle, it would already be

[cyan void]

That makes sense

[cyan void]

it not 96

i thought it was centriod

would it be 68

[late matrix]

this is another problem?

[cyan void]

ya

the other one is right

[late matrix]

ok ok

[cyan void]

did you figure it out

[late matrix]

I'm looking at how to solve it, for now I've gotten DE = 128.

[cyan void]

[late matrix]

DE = 128; EY = 60; DW = 68

[cyan void]

yay

[cyan void]

i can only figure out the 90

[late matrix]

<HBC = 31°

[cyan void]

ahh 31 + 59 = 90

[late matrix]

yes

[late matrix]

I'm not entirely sure, because theory always has its between the lines, but I think they would be 1, 3, 4

[cyan void]

ah it was these 2

[late matrix]

Yes, LN is not a segment bisector, i was wrong about that.

[late matrix]

¿1?, 2, 3, 4

[cyan void]

ahh

[late matrix]

Ahh i was wrong about 3 haha

[cyan void]

running low on tries

[late matrix]

It's confusing to me that it's in English haha but it was still a matter of reading it better

[cyan void]

ohh

i could try translating it

[late matrix]

1, 3, 5, 6

[cyan void]

W

correct

[late matrix]

6

[late matrix]

6

[cyan void]

😭

[late matrix]

why the emoji?

[cyan void]

painful

i dont like math

[late matrix]

y = -74; z = 90; x = -34

[late matrix]

but you need to study, you may not like it simply because you have not found a pleasant or comfortable way to study

[cyan void]

i can do the other parts just not geometry and proofs

[late matrix]

and why not geometry and proofs?

[cyan void]

i forget all the vocabulary

[cyan void]

like this

like angle bisector

[cyan void]

all dis stuff

[late matrix]

A matter of studying it little by little, the same thing happens in algebra, trigonometry, etc.

[cyan void]

my brain gets fried after all this

the other algebra stuff is a bit more understandable but this is a different level

[cyan void]

i finished basically all my hw besides proofs

[late matrix]

well

[late matrix]

oh ok ok, the same thing happens to me with geometry

[cyan void]

do you know how to do proofs

😭

[bleak monolith]

You forgot say closed

[cyan void]

And that bc and ad is given

I Only know ac and ac is reflexive

@bleak monolith

[cedar kiln]

@cyan void Has your question been resolved?

[cedar kiln]

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[robust glen]

Unsure what this is asking