help-0

[weak parcel]

I see what u mean

[prime glacier]

nah blud this stuff's in my physics final next week u got me questioning everything

[weak parcel]

LOL

how old are you

How would I find M? Would I do f=ma?

@prime glacier

[prime glacier]

16

[prime glacier]

well you have two equations

object Q is accelerating downwards so
mg - T = mg/4

because clearly mg > T

and object P is aceclerating upwards

so tension must be dominant over gravity

T - 4g = g

we already got that

so now we have
mg - T = mg/4
T - 4g = g

solve these equations to find the value of m

[weak parcel]

Wdym solve

Ur gonna have to be more transparent 😪

Like find tension

Substitute it

And stuff?

@prime glacier

[prime glacier]

nah

i mean

basically you can just add the two equations

in that case tension cancels out

so you're left with
mg - 4g = mg/4 + g

g obviosuly cancels out

and you can solve for m

[weak parcel]

So would this working not be valid ?

@prime glacier

[prime glacier]

hm wait hol up lemme check

it's valid but

where did the 1 in 196 go

@weak parcel

[weak parcel]

That’s a great question

So it’d be whatever 196/3

Divided by gravity

@prime glacier

I don’t have my call with me

Rn

[prime glacier]

ur what with u rn?

[prime glacier]

yeah

[weak parcel]

Calc

[prime glacier]

ah

[weak parcel]

Ok how would I do C?

Would it be same thing but just force pointing down

Like this

@prime glacier

[prime glacier]

what is C supposed to be lol

oh wait

the third one

nah i'm pretty sure ur supposed to actually measure it using the tension in the string

[weak parcel]

Wdym

[prime glacier]

Force exerted on pulley = Total tension

[weak parcel]

It asked in terms of g

[prime glacier]

• weight of the string but that's negligble here

[prime glacier]

right

T = 5g

[weak parcel]

Yeah so 10g?

Or 5g?

[prime glacier]

10 g

from what i remember

[weak parcel]

Tension is same through rope no?

Cos it’s light in extensible

[prime glacier]

it is but you're supposed to consider both the tensions

on both sides

at least that's how i've been doing these problems

[weak parcel]

Ok I’ve another question I’m not sure on

[prime glacier]

is it d?

[weak parcel]

Question 7

[prime glacier]

just equate the forces

[weak parcel]

It’s moving constant

In what direction tho

[prime glacier]

exactly

it's moving at a constant speed

so no external force acting on it

[weak parcel]

So they’re equal to each other?

[prime glacier]

yup

if the particle was accelerating the question would be more difficult

just equate the downoward and upward forces

and the forces acting to the left and right

[weak parcel]

2f +2g -22n = 3f-g-13n?

[prime glacier]

no no

not like that

13 N = 3F - G

and

[weak parcel]

Is it a simultaneous equation

[prime glacier]

22 = 2F + 2G

[weak parcel]

@prime glacier

[prime glacier]

by this do you mean equations in two variables

linear equations in two variables

i looked it up and yeah

simultaneous equations

[weak parcel]

Thanks

And what about question 8

I can’t visualise it

[prime glacier]

is that the entire question?

[weak parcel]

I need help visualising it

A diagram perhaps

[prime glacier]

particle of mass 4kg acted upon by a force of 8i + xj N. F acts at an angle of 135

i don't know how to solve this one tbh

could you send the entire question

without it cut out

[prime glacier]

oh i'm so stupid lol

[weak parcel]

Doesn’t 135 degrees have a gradient of 3x

[prime glacier]

tan 135 is -1

ok well

F acts at an angle of

135 deg, right?

and we have its vector form

if we have a particle moving along the xy plane

with the vector form

xi + yj

tan(theta) = y/x

can you try applying that to the first question

[weak parcel]

I’m so baffled

[prime glacier]

did you not do this stuff in class

it should be one of the first things you're learning

[weak parcel]

Nope

[prime glacier]

wow

hold on

i like to think of it as coordinates

imagine a point at xi + yj

that's like saying (x,y)

there's a height of y and and distance of x

let there be an angle theta

tan theta = y/x

welp i have to go sleep now it's 11:30

good luck on this stuff

[lone heart]

@weak parcel Has your question been resolved?

[lone heart]

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[mighty cliff]

I made a drawing and wrote the formula V=1/3SH

S=S

but how to find the height for this formula

do not pay attention to the designations, they may be incorrect (I tried to solve it myself)

from the formula for the area of a rhombus I found a =sqrtS/sin alpha

that is AD, AB, DC etc

but I don’t understand how to find the height

[mighty cliff]

(edge is alpha and cut beta)

[lone heart]

@mighty cliff Has your question been resolved?

[mighty cliff]

<@&286206848099549185>

[hard panther]

Consider what will happen when you consider a 2d plane intersecting the original shape

If you draw a diagram with measurements on it, you should see some right triangles

You then use pythagoras to find the distance you're looking for

[mighty cliff]

?

I don't really understand what you mean

[hard panther]

Which distance is S

AF?

[mighty cliff]

S is the area of the rhombus
​

<@&286206848099549185>

[hard panther]

Sorry, got caught up in something

You know that because all the sides are aligned with the cut of B, That the rhombas has to be a square

Therefore the length of AD is S

You should be able to use pythagoras there to work out the height

Then you can just use the formula you gave

[mighty cliff]

what kind of rhombuses if there is only one at the base?

[hard panther]

It's a square

And you know its area

[mighty cliff]

and how can ad be S if AD is not an area

[alpine sable]

Still need help?

[hard panther]

Sorry

[mighty cliff]

yes

[hard panther]

Root s

Not s

[mighty cliff]

I don't understand anything in your reasoning

how can this be S

if formula S= a^2 sin alpha

a^2 = s/ sin alpha

a = sqrt s/sin alpha

and from these reasonings it cannot be S

[hard panther]

because you don't know alpha

AD is root S

therefore root s / 2 = a sin alpha

[mighty cliff]

and what will it give me

[hard panther]

You do the math and tell me

[mighty cliff]

a= S/sin alpha^2

.close

[lone heart]

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[hard panther]

You should be able to get an answer in terms of just S and A

[hard panther]

.close

[lone heart]

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[calm anchor]

Is this true?

[ivory pivot]

are they the same angle in the circunfercence? if so then yes thery are the same number

[tacit arch]

,w 4exp(i 3 pi/4) == 4exp(-ipi/2)

[ocean seal]

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Requested by rie.mann

[buoyant saddle]

-pi/2≠3pi/4

[calm anchor]

In that case, how do i know what is right?

-i4

”Write in polar form”

I cant use the formulas given because then i divide by 0

So i imagine it in the complex plane

And there the angle is straight down, therefore the angle theta is 3Pi/4 (which is the same angle as Pi/2)

<@&286206848099549185>

How do i write -4i in polar form?

[small talon]

Like using cis natation?

If you think of the problem using an agrand diagram, you can get the solution

[small talon]

What have you tried?

Let me walk you though it. Basically you can write -4i in the form a+bi
The magnitude of the imagingary part |b| is 4, so you know r=4.

SInce -b (which is the i component) lies on the negative part of imaginary axis (similar to y-axis), this is 270º or 3pi/2 radians.

Therefore -4i is 4cis(3pi/2)

[lone heart]

@calm anchor Has your question been resolved?

[calm anchor]

This is what i answered, however in the book, the answer is $4e^(-iPi/2)$

[ocean seal]

Tom

[lone heart]

@calm anchor Has your question been resolved?

[lone heart]

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[atomic harbor]

F(n,r) = $\frac{\sum\limits_{k=1}^{n-r+2} k\binom{n-k}{r-1}}{\sum\limits_{k=1}^{n-r+2} \binom{n-k}{r-1}}$

[ocean seal]

Normed

[atomic harbor]

This is what I've done so far. Is this correct? If so how do I simplify it to (n+1) over (r+1)?

[lone heart]

@atomic harbor Has your question been resolved?

[wintry panther]

Hi

Have you tried induction?

and I guess it should be n-r+1

The denominator equals to n choose r.

[lone heart]

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[fervent timber]

Is there a name for the following type of problem, and/or is there an algorithm I can use to solve it?

You are given a number of items, and a list of various groupings of those items. An item may be in multiple groupings. You are allowed to "mark" ONE item in each grouping. Your goal is to maximize the number of unique items marked.

[mortal trellis]

seems like an independent set problem in graph theory

make each item a vertex

connect two vertices if they are in the same group

then you want to pick vertices such that there are no edges between any vertices you pick

which is called an independent set

[fervent timber]

hmm

but an item can be in multiple groups

is there an algorithm to solve it?

[mortal trellis]

which just means its connected to a lot of other items

there are some algorithms

but the problem is np complete iirc. which basically means its hard to compute

[fervent timber]

hmm maybe this is the XY problem
would it help if I told you the specific problem I'm trying to solve?

[mortal trellis]

cant hurt

[alpine sable]

how to find the derivative of 3^sqrt(x)

[lone heart]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fervent timber]

it's a variant of this problem
https://www.youtube.com/watch?v=6hVPNONm7xw&t=403s&ab_channel=GoingNull

I want to figure out how many people you would need for N hat types, but under the constraint that you can't see the hat type of the person to the right of you as well

[fervent timber]

and I basically generalized the question to the above

assume that each possible scenario is an "item", and each grouping is specifically the list of scenarios that are assumed from your perspective

[mortal trellis]

I dont wanna watch a video right now

[fervent timber]

yeah fair enough

pretty sure I described it earlier in words

I'll try finding it

N people sit in a circle. each person wears a hat that can be of N different colors (multiple people can have the same color). Each person can only see the hat color of the other N-1 people.

Everyone must guess their own color hat at the same time. the group wins if at least 1 person guesses their hat color correctly. what strategy can they use to guarantee they win every time? (The solution is that there is a strategy btw).

This is the variant: You have N colors and a group of people but the twist is that each person can't see the person directly to the right of them (so they can only see K-2 others in a group of K people). The goal is to figure out if a strategy is possible for some value of N or K.

actually maybe this is more complicated than I thought

maybe I should think about it further and ask again later

[mortal trellis]

these prisoner hat problems are always difficult

[fervent timber]

yeah it feels like that lol

the problem in the video has a cool answer if you haven't seen it before btw
there's a special way to solve it that makes it way easier

I made up the variant myself so there's no solution I can look at unfortunately. And the "special way" doesn't work on my variant.

anyway, maybe I shouldn't be thinking of this problem rn and focus on other stuff

so I'll close this channel for now

.close

[lone heart]

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[devout lake]

@vale crag referring to what we were talking about earlier

[devout lake]

in that game where you play RPS against each other at [1/5, 3/5, 1/5] vs [1/5, 3/5, 1/5] respectively

[vale crag]

1/5 you mean

[devout lake]

that is obviously a nash equilbrium and the expected value for the players is 0

yeah****

but you change it to [1/3, 1/3, 1/3] vs [1/5, 3/5, 1/5] and its still 0

does that mean theres 3 equilbria for those 3 combinations?

[vale crag]

"[1/3, 1/3, 1/3] vs [1/5, 3/5, 1/5]" this is not an equilibrium as I argued earlier

cause the second player can change their strategy to increase their gains

[devout lake]

oh

makes sense

[vale crag]

(the playing full rocks one)

[devout lake]

i hella confused myself bc I didnt think it was one but since it was equal

thx lol

.close

[lone heart]

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[lone heart]

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[shut plinth]

im kinda confused how they got from the first line to the second line

[long axle]

is the first line:

$85+10 \log_{10} {32}$

[ocean seal]

Stephen

[long axle]

do u guys use base 10?

@shut plinth

lmk ur thoughts

if the second line is true, it would necessitate that 10log_10 (32) = 15

which...doesnt look all that good

[dire trail]

bro this is wrong, the answer is around 100.05 but it isn't 100

[unreal meteor]

actually it’s 100.0515 🤓

[shut plinth]

ya

[shut plinth]

ya

[shut plinth]

uhh

[shut plinth]

how?

wait

[dire trail]

bro the calculation aren't true

[shut plinth]

do you do

$$log32/log10$$

+85?

[ocean seal]

cat

[dire trail]

that may be true in engineering study but not in math

[shut plinth]

huh

um i got 86.5

oop nvm i did it wrong

[long axle]

$10 \log_{10} {32} = 10 \log_{10} {2^5} = 50 \log_{10} {2}$\

[shut plinth]

i got 100.05

[ocean seal]

Stephen

[shut plinth]

isnt it just log32^10

and + 85

[long axle]

?

[long axle]

yea

but thats still not 100

[shut plinth]

the ten becomes the exponent

[shut plinth]

oh ok i think he rounded it

[dire trail]

100.05 = 100 ...

[long axle]

ugly

[dire trail]

bro studying engineering

[shut plinth]

💀

ok i got one more quesiton

does this = 6?

[dire trail]

nop

never mind

[long axle]

dont give answer

[shut plinth]

huh

what 10?

[long axle]

$a^{b+c} = a^b a^c$

[ocean seal]

Stephen

[long axle]

do that first

[shut plinth]

OH

2^log_2(5)*2^log_2(2)

[long axle]

simplify all that

[shut plinth]

5*2=10

tyty

.close

[lone heart]

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[neon frigate]

heloooooo

[neon frigate]

im willing to learn

[vapid shuttle]

what relationship do you see between y and x

[neon frigate]

what do you mean by relantion ship

[agile grove]

mk so what is the base formula for a linear equation?

[vapid shuttle]

like, do they look related to eachother

[neon frigate]

what just happened

[vapid shuttle]

anything in common between 1 and 2? or 2 and 4?

[neon frigate]

0,0

[agile grove]

Formula we have to consider all the x values

[neon frigate]

that they are multiplied by 2

[vapid shuttle]

great you're basically done

that is the rule

[agile grove]

okay ima give a hint

[vapid shuttle]

y is x multiplied by 2

[agile grove]

y = mx + b

[vapid shuttle]

right?

[agile grove]

does that ring a bell?

[neon frigate]

umm

not really

i know its like

y= something

[vapid shuttle]

Apicito

earlier you told me what the rule was

can you restate it

you already got it

[neon frigate]

y=2x

[vapid shuttle]

great

boom

done

[neon frigate]

ohhh

ok

is this one

different

or its similar

[vapid shuttle]

it is similar

not the same answer

but the same process

let's compare x's and y's

5,6,7,8...

1,2,3,4,...

are they growing at the same rate?

[neon frigate]

yes

they are like

4

[vapid shuttle]

so every time x goes up 1

y goes up how many

[neon frigate]

4

[vapid shuttle]

no

[neon frigate]

wait

[vapid shuttle]

when x goes from 5 to 6, y goes up how many?

[neon frigate]

y goes up 1 time

[vapid shuttle]

yes

so y grows at the same rate as x

[neon frigate]

yeah

[vapid shuttle]

we would say y=x. But this doesn't work right? because clearly like 5 isn't equal to 1 and so on

so it is like x got a head start

[neon frigate]

yeah

[vapid shuttle]

a head start by how many?

[neon frigate]

4

four

[vapid shuttle]

yes

so what should the rule be

[neon frigate]

so i put

y=x+4?

[vapid shuttle]

perfect

well

no

sorry

[neon frigate]

i was happy for a sec

[vapid shuttle]

this is giving y the head-start

you're very close

[lofty creek]

Its close

[neon frigate]

wait how do i give x a head start

[vapid shuttle]

well this is saying, y is 4 more than x

you want x to be 4 more than y

right

so just retry writing that

[neon frigate]

y=x-4?

[vapid shuttle]

yes

[neon frigate]

ohh

i get it now

[vapid shuttle]

good work

[neon frigate]

this one is the same

procces

or different

[vapid shuttle]

same process, slightly harder

do you want to work on that one?

[neon frigate]

yeah

so then i can just do it by myself

[vapid shuttle]

Okay sure

every time x goes up by 1, how many does y go up?

[neon frigate]

2

[vapid shuttle]

yes

so we think, y=2x

but then we look

[neon frigate]

didnt they both get like a headstart

[vapid shuttle]

sort of yes

[neon frigate]

but y more

[vapid shuttle]

initially we are thinking it should be y=2x, but we know we have to adjust this to make it true

right?

so try the first value

[neon frigate]

yes

[vapid shuttle]

we think that 17=2(4)

17=8

not quite

but what do we need to do to make it true?

[neon frigate]

i dont know

[vapid shuttle]

add how many to 8

will make it equal 17

[neon frigate]

9

[vapid shuttle]

mhm

so maybe the rule is not y=2x

but is instead what?

[neon frigate]

3

how is that sus

[vapid shuttle]

remember what we did last time

we kept the rate

the y=2x

and we added or subtracted something

to account for the headstart

now we need to do the same thing here

the rate of change is y=2x

you need to add or substract something to make the rule correct though

[neon frigate]

so do we do like

y=2x+9

[vapid shuttle]

that is correct

[neon frigate]

it is?

[vapid shuttle]

yes

you can double check yourself

try some of the points, like x=4

2(4)+9=17

checks out

[neon frigate]

it is correct

oh snap

i got one with a minus siogn with it

i got negative one

-1

[vapid shuttle]

Show it?

[neon frigate]

is this one different too

or same

[vapid shuttle]

Same

don't let the minus sign get you mixed up

it just got a headstart in the opposite direction

nothing new here

[neon frigate]

what does that mean

[vapid shuttle]

nothing really, just trying to make it make sense to you

[neon frigate]

oh ok

[vapid shuttle]

In general what you are supposed to do with these

is find the slope

which is what we have been doing first

when x goes up 1, y goes up how many?

that relationship, is the slope

and then you need to find the y-intercept. This is what I have been referring to as the "headstart"

it is when x is 0, how big does y start as?

[neon frigate]

wait

y goes up by 6?

[vapid shuttle]

not 6

[neon frigate]

how mcuh

i dont get it

cus one of them is

[vapid shuttle]

what'd you try?

[neon frigate]

1,-1

[vapid shuttle]

no it is -1 to 8

[neon frigate]

2,8

[vapid shuttle]

yes, y goes from -1 to 8

when x goes from 1 to 2

[neon frigate]

then its 9

[vapid shuttle]

yes 9

[neon frigate]

oh ok

[vapid shuttle]

so our initial guess is what?

[neon frigate]

in y or x?

[vapid shuttle]

y=...

[neon frigate]

y=9

[vapid shuttle]

no

everytime x goes up 1

y goes up 9

y is related to x

[neon frigate]

yeah

[vapid shuttle]

by a factor of 9

y is not just always 9

[vapid shuttle]

so fix this

[neon frigate]

y=9x+1

[vapid shuttle]

okay that's our first idea

[neon frigate]

because of the headstart

[vapid shuttle]

not quite

slow down a little

"9" is the slope

that tells our equation will be y=9x+b

[neon frigate]

yeah

[vapid shuttle]

now we need to find out what b is

so plug in, x=1

at x=1, y should = -1

-1=9(1)+b

so what is b?

[neon frigate]

8

[vapid shuttle]

no

9+8 is not -1

[neon frigate]

-10

[vapid shuttle]

yes

[neon frigate]

so its

[vapid shuttle]

so now we've found b to put back into this

[neon frigate]

y=9x-10

[vapid shuttle]

yes

[neon frigate]

ohhh

okk

thank you so mcuh

this is way harder right?

[vapid shuttle]

no, not way harder

we know how to find the slope and y intercept of A

that's what we have been doing

right?

[neon frigate]

yes

[vapid shuttle]

do you know how to tell what the slope and y intercept of B are by looking at it?

[neon frigate]

the slove is

5/2

[vapid shuttle]

yes

[neon frigate]

yeah

[vapid shuttle]

and the y-intercept?

[neon frigate]

all i have to get is the slope of function a

[lone heart]

Channel closed

The owner is missing!

[neon frigate]

-4

[vapid shuttle]

yes

[vapid shuttle]

so yes, all you have to do is find the slope and y-intercept of A as we have been doing with the rest

and then compare them

[neon frigate]

yeah

how do i find them

[vapid shuttle]

you know how!

we just did it like 10 times

[lone heart]

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[alpine sable]

did i do reverse right

[limpid turret]

https://tenor.com/view/jake-peralta-squinting-read-trying-to-read-brooklyn-nine-nine-gif-13387080

[alpine sable]

is my handwriting that bad or is it smol?

or both

[paper mango]

I presume we can skip the letters right

[alpine sable]

give me a moment

[paper mango]

it's readable :D

[limpid turret]

Prove $A\cup B\subseteq C$ given $A\subseteq B \subseteq C$?

[ocean seal]

SWR

[paper mango]

both ways <=>

[alpine sable]

is this better?

11th just the reverse

[limpid turret]

what is this?

[paper mango]

yes, but did we get the task right

[limpid turret]

$x\in C \vee x\in C$ (assumption)?

[ocean seal]

SWR

[alpine sable]

yes

[paper mango]

does that mean A subset B AND B subset C

[alpine sable]

no

[paper mango]

so OR?

[alpine sable]

oh which one

yours or swr

[limpid turret]

here

[alpine sable]

and

[paper mango]

ah k

[alpine sable]

sorry it's just for me so i started in my language

lmao

[limpid turret]

I would clarify your assumptions here

I get x in B means x in C

but x in A -> x in C is not one of your assumptions

[paper mango]

the reverse direction is correct until here, the line after seems a bit unclear

why you jump to:

[alpine sable]

oh i mean i thought because of transitivity of relation

[limpid turret]

That is correct. But you need to state that. It is not an assumption.

[alpine sable]

how would i state that? like notation wise

[limpid turret]

Add a new line

[paper mango]

you actually don't need to, because you already had:

which directly becomes xeC

due to the assumption:

[alpine sable]

okay wait wait

so what do i do?

[limpid turret]

I'm talking about forward proof, not backwards

[paper mango]

but he asked did i do reverse right

[alpine sable]

oh but the forward proof is directly taken from my class

[paper mango]

maybe the forward proof has flaws too

[alpine sable]

i am surprised it's wrong

because it's not mine lmao

[limpid turret]

This was the only flaw I found in the forward proof.

It's not necessarily a flaw, but it should be more explicit about how it gets x in C

[alpine sable]

i thought that too but my prof just wrote (by hypothesis)

[paper mango]

k, we'll do backwards first and then expand the forward

you start the backward proof with:

where you show that xeA -> xeB

which is right

then you also need to show

xeB -> xeC

[alpine sable]

did i not do that?

i thought i did that

[paper mango]

well yes, it's just a bit unclear where you start that subproof

you could separate it e.g. by a line and then ```

xeB
xeB OR xeA
xe(AUB)
xeC

which is what you did, just with a jump in between

[limpid turret]

Side note: There's a more trivial proof of the forwards if you have already proven $A\cup B=B\leftrightarrow A\subseteq B$

[ocean seal]

SWR

[alpine sable]

okay wait let's slow down a bit fellas

[paper mango]

xeA -> xeB is what you have correct for the backwards proof

you only need to separate it a bit from the xeB -> xeC

[alpine sable]

So literally put a vertical line between a subset c and y element of b?

or horizontal on this pic

but vertical cuz my tablet is weird

[paper mango]

like always start with the assumption to clearly indicate it's the starting point

because you don't start from xeB in any line

[alpine sable]

so like this?

[paper mango]

yes that's way clearer

[alpine sable]

okay

[paper mango]

you can use x as well for both btw

if you want to impress them you can also use implication arrows :3

[alpine sable]

@limpid turretnow what were you saying about forward, if you wouldn't mind repeating, please?

[limpid turret]

I was basically saying only this and nothing else.

[paper mango]

assume yeB
==> yeB OR yeA
==> ye(AUB)
==> yeC   [proven]

[alpine sable]

okay but i meant this

[alpine sable]

and say what?

[paper mango]

I presume he meant the step xeA to xeC

where it'd be first xeA => xeB and then xeB => xeC

here

but it's fine

set theory proofs tend to be very elaborate when it comes to the basics, not to skip any steps

once you have the basic proofs done, you can use them for other proofs, simplifying them

[alpine sable]

okay but could you explain why my prof wrote xec V xec(by hypothesis) from what i get it is because AUB is a subset of c and since X e AUB X e c

[limpid turret]

Say that x in C due to the assumptions of A subset B, B subset C, and transitive property of subsets

[alpine sable]

i think that's clear, no?

did i get the logic right, does it hold?

[limpid turret]

yeah this reasoning is incorrect.

[alpine sable]

okay, can you explain why?

[paper mango]

since it's no hypothesis in the first proof

[alpine sable]

because a subset is a relation between two sets where A subset B if all elements of A are in B. and we know that x is an element of the union between A and B and that the union is a subset of C which means that all elements in A U B must be in C

[paper mango]

you have xeA and xeB

A subset B and B subset C does of course imply A subset C

but it's not directly written out as assumption

meaning the step xeA -> xeC

misses the substep

xeA -> xeB (by assumption) -> xeC (by assumption)

[alpine sable]

okay but he said my reasoning is incorrect

not the proof

[alpine sable]

and this was my reasoning

[paper mango]

hm thought he referred to the brackets

[paper mango]

why did you state that we know that the union is a subset of C

before it's proven we don't know that

after it's proven then yes, all elements in xeA and xeB are in xeC

[alpine sable]

i assumed we assume the part we have and prove the part we don't

and vice versa

[paper mango]

yes, for forwards we don't yet have that AUB is a subset of C

we know A is a subset of B

B is a subset of C

[alpine sable]

so if it says prove Prove: A ⊆ C and B ⊆ C if and only if A ∪ B ⊆ C

i assumed the part after iff holds for forward

[paper mango]

before

A ⊆ C and B ⊆ C holds, A ∪ B ⊆ C to be proven for forward

A ∪ B ⊆ C holds, A ⊆ C and B ⊆ C to be proven for backwards

[alpine sable]

AH god i am dumb

[paper mango]

nw :D

[alpine sable]

also wait wait

isn't disjunction an inclusive or

[paper mango]

yes

[alpine sable]

because if it is then x e c or x e c holds by assumption

[paper mango]

ofc, but you usually write in brackets which rule you use

[alpine sable]

as long as x is in either than x is in c

and we obviously see that x is in either a or b

[paper mango]

like ```
Assumption
Step 1 (due to Rule X)
Step 2 (due to Rule Y)
Step 3 (due to Rule Z)
...

[alpine sable]

yes but my point is what does it have to do with transitivity

if disjunction is an inclusive or and x is in either than the truth value is always one

[paper mango]

A subset B & B subset C ==> A subset C?

or the AUB subset C?

[alpine sable]

this

[paper mango]

since here is no disjunction

[alpine sable]

swr said i should explicitly note this but by definition of union XeA or xeB

and it's an inclusive or

so ofc x is in c since we assume which i wrote here that A subset c, B subset C

I'll circle it

[paper mango]

it's understandable ys

[paper mango]

the exclusive or is denoted differently, so you can leave as is

[alpine sable]

yea XOR

here is my assumption

[paper mango]

wat

I thought it's the other way around

[alpine sable]

oh sorry yea

circled the wrong proof

LMAO

[paper mango]

yes :D

[alpine sable]

ty @limpid turret @paper mango

.close

[lone heart]

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[wanton beacon]

Help, plz

Just the “What percent paid at most…” part

[paper mango]

boxplot properties are quarters per segment

you see the intersection of third and fourth segment is 29

=> 75%

[wanton beacon]

These are each 25%?

[paper mango]

If I don't misremember yes

[wanton beacon]

Yeah, that’s what it says here

[paper mango]

the middle line is the median

[wanton beacon]

But then how does that help me?

[paper mango]

with what

75% of people paid at most $29

at A

[wanton beacon]

How do yk?

[paper mango]

since the first three quarters of the upper boxplot are included

each segment is a quarter 1/4

25% + 25% + 25% = 75%

[wanton beacon]

Are these the segments you’re referring to? @paper mango

[paper mango]

ys

[wanton beacon]

Got it, tysm!

Also

Are you familiar w/ recursion? @paper mango

[paper mango]

sure

[wanton beacon]

U see #6 here?

[paper mango]

not fully

[wanton beacon]

Would 45,00 be miu sub 0 or 1?

[wanton beacon]

That’s all right

My question’s abt the 45,000

[paper mango]

k

[wanton beacon]

This is my q

[paper mango]

wdym by miu sub

[wanton beacon]

The thing at the top

[paper mango]

ah the greek mu :D

what constant does it represent in your case

whether it stabilizes?

[wanton beacon]

Wdym?

[paper mango]

I don't know what that constant should refer to

different countries use different notations

@wanton beacon you're meant to calculate the stabilized population

maybe you have some formula involving that constant, but you can also instead set up an equation which solves it

x*0.79+13000 = x

[wanton beacon]

Oh, I see

[paper mango]

which would be equivalent to the population staying the same

if you solve for x

you get the population for which this property holds

[lone heart]

@wanton beacon Has your question been resolved?

[lone heart]

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[gentle hamlet]

Can I get help on #29?

I already found the root -7/2 but the answer says C is -63/4 and I’m not sure why

[rancid jay]

if u use relation between roots and coefficients , u'll get c to be -63/4

[gentle hamlet]

Isn’t it

Multiply both roots and you get c/a

[rancid jay]

yes

it is

[gentle hamlet]

So I did 3/2 * -7/2 and I got -21/4

[rancid jay]

but that is equal to c/3

so c is -63/4

[gentle hamlet]

Is root multiplied to equal c/an only if coefficient is 1 for x^2?

[misty mauve]

ah sorry i misread something in the question forget my explaination

[gentle hamlet]

But I’m confused on this, if c/3 = -21/4 how do you get -63/4

[rancid jay]

product of roots equals c/coeff of x^2

[rancid jay]

uhm just multiply both sides by 3

[gentle hamlet]

Ohh

Wait sorry

[rancid jay]

i think it wud help to write it down on paper

[gentle hamlet]

So product of roots equals C is only if it is x^2 and if it has a coefficient bigger than 1, I’d have to change the quadratic to be x^2?

[misty mauve]

sum of the roots should be -6/3 = -2. If one root is 3/2 then the other must be -2 - 3/2 = -7/2. Products of roots is c/3 which is equal to 3/2 * -7/2 = -21/4. c/3 = -21/4 so c = -63/4

[rancid jay]

in the question in arguement, product of roots wud be equal to C only when coefficient of x^2=1

[gentle hamlet]

Ok thx magnusop and nova i understand now

[lone heart]

@gentle hamlet Has your question been resolved?

[lone heart]

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[white charm]

I have a quick question for math if anyone could help

[dusky condor]

depends on the question but go ahead

[white charm]

for a quadratic equation that is in standard form, how would I solve for the value of b and solve for the value of the other root when given the equation 2x^2 + bx - 24 and the first root that is given is -8

[dusky condor]

i got no clue soembody else is going to have to help you

[white charm]

alr

[thorn monolith]

@white charm 2x² + bx -24 = 0? @dusky condor
Because the thing you typed is an expression not an equation

Oops accidentaly pinged you av king

[white charm]

its good

[thorn monolith]

I mean judice*

A root is a value which satisfies the equation

[hot bluff]

x=-8 and solve? not sure about this

[thorn monolith]

When you let x= -8, this equation should satisfy

[white charm]

??

[thorn monolith]

What of this isnt clear?

[white charm]

you didn't explain how i would solve for B and the other x value

[thorn monolith]

This finds value of b

Which then helps you find value of other root

[white charm]

So what method would I have to use? would I have to use the quadratic formula or just simply use algebra

??

[thorn monolith]

For which step

[thorn monolith]

This is just equation solving, you should be able to do it

[white charm]

But how tho idk how

[thorn monolith]

Well do you know root of the equation x-5 = 0

For example

[white charm]

oh yea positive 5

[thorn monolith]

So x=5 satisfies the equation right

We call that point, a root

Similiarly for equation 2x-2 = 0
x=1 is a root

Now, these equations gave as an example have degree of 1, meaning they can have 1 solution at most

Quadratics have a degree of 2, meaning at most they can gave 2 roots

[static stag]

is b equal to -13?

@white charm do you have the answer to this question?

[white charm]

no thats why im asking

[thorn monolith]

This questions already tells you that both exists, and even gives you one of the roots, putting that value on the equation should satisfy it, similiar to how typing x=5 on equation x-5= 0 gives you 5-5=0

[static stag]

okay

[white charm]

@static stag I had this question on a test

[static stag]

oh okay

[white charm]

could you maybe write the steps on auto draw and screen shot it??

that would be much helpful

[thorn monolith]

So let x= -8, since its said to be a solution/root, it should satisfy the equation

I wont give you the solution myself, you have to work it yourself

[white charm]

alr

[thorn monolith]

If this isnt clear, look at example
x² = 4
x=2 and x=-2 satisfy them, if you type them in place of x, you get 4=4 which is equivalent to saying 0=0

But if you type x = 3 to same equation you 9 = 4 which is absurd

In fact any other value different than x = 2 and x= -2 results with same absurdity

In other words, 2 and -2 are the only values that satisfy the equation
x²-4 = 0
So these two values are called roots of the equation

[white charm]

Ohhhh.... ok I understand now Thanks I appreciate your help

and I got the answer

thx

[thorn monolith]

What is the value you found for the other root?

[white charm]

sorry I was afk I believe the answer is 4?

[thorn monolith]

There are 2 values to find

You found b = 4?

[white charm]

yes

[thorn monolith]

What about other root

[white charm]

2

[thorn monolith]

Well both arent correct, you prob made a calc error

Do you know the following formulas:

If $x_1$ and $x_2$ are the roots of equation $ax^2 +bx + c = 0$ then

$x_1.x_2 = \frac{c}{a}$

$x_1 + x_2 = - \frac{b}{a}$

[ocean seal]

Cyrenux

[thorn monolith]

Using first formula gives you the other formula

Using the latter formula after you find other root,will give you value of b

[lone heart]

@white charm Has your question been resolved?

[raw stream]

vieta's relations

[lone heart]

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[unkempt compass]

im trying to solve for part b using the hand written part but im not getting the right answer. can someone explain what im doing wrong?

[unkempt compass]

ik there are 2 solutions but my single solution isnt even right

[tacit arch]

Are you in degrees or radians when calculating arcsin

[unkempt compass]

radian

[tacit arch]

Is your part a correct

[unkempt compass]

i feel its correct? according to desmo the maximum and minimum points for that equation seem right

or is it wrong?

hmmm i need to check the graph i was looking at another

😬

well i just rechecked what i thought was wrong and now im thinking everythings right but youre saying ca value is wrong

what is wrong with it?

[deft token]

nvm

[unkempt compass]

i think my first value is correct so far for x = 116.77 or 117days, gonna solve for 2nd

[deft token]

how do you knwo your solution isn't right?

[unkempt compass]

checking using desmo's. plugged my formula into it to draw the sin chart, then to test for y=14 i enter that for a second function

[unkempt compass]

i "knew" it wasnt right because my original drawn function was the wrong formula from somewhere else

so it was the wrong assumption

had multiple graphs going on

[deft token]

nah i get what you're saying

[unkempt compass]

for the other part just to conclude the problem

.close

[lone heart]

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[deft token]

how to approach b?

[tacit arch]

!show

[lone heart]

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

[deft token]

completed a, but unsure how to approach B

for n = 1, a1 = 3 >2

and compared a(n+1) to prove that an > 2 for all n

do i follow similar steps for b?

[tacit arch]

You can try showing the difference is nonnegative or the ratio is less than 1

[deft token]

okay

pretty messy but

@tacit arch

do i just sub in an for 2 as i've already proved an >2

[tacit arch]

You'd have to show the fraction is less than 1 still

[deft token]

so when i sub in, LHS simplifies down to 24/25

and RHS is 2

that doesn't seem right

since i need to prove LHS is less than or equal to

since an > 2
if you prove that an+1 is < 2 does that solve the problem?

[deft token]

i feel like this attempt is completely wrong

[tacit arch]

Substituting an = 2 directly was wrong

[deft token]

should i even sub in an into an+2 to begin with?

[tacit arch]

Also the starting equation looks wrong

[deft token]

yeah

[tacit arch]

Where is a_(n+1) / a_n

[deft token]

okay ill redo

[deft token]

now sub in an = 2?

[pseudo ice]

I wouldn't "sub in" a_n = 2

Pretend that was instead a quadratic, say (u + 3)(u - 2) >=0 - which values of u would form a solution to that?

[deft token]

when u = or greater than 2

@pseudo ice

[pseudo ice]

Yep, that's one part - there is another, but not very relevant for our purposes

But if you're strictly greater than 2, you're of course greater than or equal to 2

[deft token]

yeah

so through this an+1 < an is proven?

can't exactly sub in, so how would you write it? in like maths terms lol

[pseudo ice]

Well, you have that $a_{n + 1} \leq a_n$ if and only if you have $(a_n + 3)(a_n - 2) \geq 0$, which would be true if you had either $a_n \leq 3$ or $a_n \geq 2$

[ocean seal]

@pseudo ice

[deft token]

okay

[pseudo ice]

But you have the latter true by your part a and that an is always (strictly) greater than 2, which allows you to conclude that the original decreasing statement was true

[deft token]

fanx

while you're here could you help with part C as well lmao

to start, replace an with limit(n-> inf) an?

[deft token]

damn imagine if i could write like this @pseudo ice

[pseudo ice]

Well you can - you know the sequence must converge to some limit (you've shown it's bounded below and decreasing)

[deft token]

oh right

[pseudo ice]

One way you can find the limit is to call it something like l

You should hopefully be happy with the fact that any subsequence of a convergent sequence must converge to the same limit as the original sequence...(?)

[deft token]

since a1 is 3, and an is >2 would the limit just be 2?

as N approaches inf

an approaches 2

but again i need to show working is where im losing it

[pseudo ice]

Well i mean it probably is - but be careful of that reasoning, it's also true that a_n > 1, but impossible for the sequence to converge to 1

[deft token]

hmm

[pseudo ice]

One frequent way you can find the limit

Is that you can take limits of the sequence definition here and all

[deft token]

@pseudo ice

thoughts

[pseudo ice]

Yep, that’s it

[deft token]

churrrr

[pseudo ice]

Has to be the limit 2 because if it were -3, that would imply there were negative terms of the sequence, which would clearly be less than 2

[deft token]

.close

[lone heart]

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[alpine sable]

Can someone help me figure out where I went wrong for this question (number 3)

[candid flicker]

u know what the answer is?

[alpine sable]

It’s 18

But I keep getting 1/8

I used the same formula for a different question and it worked but this formula isn’t working and I think the other one would

[candid flicker]

look at the question carefully

[tacit arch]

A is each angle and S is the total angle

You need to match the information to the variable in the formula correctly

[alpine sable]

So did I use the wrong formula for that one too

[candid flicker]

when it says 'each angle' u divide by n and if 'sum of all' then dont

[alpine sable]

Okay so this one says the sum of EACH angle is this

and the other one says the sum of all of them