help-0

[wintry sluice]

yes

[keen glade]

looking at the options P(B)+P(B|A)=1 can't be right cause it would be 0.8 not 1. right?

[wintry sluice]

how

did you find p(B|A)?

[keen glade]

P(b|A)= P(A|B) / P(A)

[wintry sluice]

yes

[keen glade]

which gives me 0.3 then if i add P(B) which is 1-0.3=0.7 gives me: 0.3+0.7=1 so it's actually correct?

[wintry sluice]

wait also

this is wrong?

wait no

nvm

sorry

its a intersection b

by p(A)

p(B|A) = p(A∩B)/p(A)

[keen glade]

so should i consider P(B)+P(B|A)=1 right?

[wintry sluice]

why is that so

[keen glade]

P(B) is equal to 0.7, P(B|A) is equal to 0.3 which gives me = 1

[wintry sluice]

how is p(B|A) equal to 0.3

shouldnt it be 0.1?

[keen glade]

p(A|B) is equal to 0.1

[wintry sluice]

did you find p(A∩B)?

yes

[keen glade]

no

[wintry sluice]

first find p(A∩B)

and then solve p(B|A)

[keen glade]

how do i do that?

[wintry sluice]

from the given things

you know that p(A) is 0.3

and p(B)

[keen glade]

P(A) x P(B)?

[wintry sluice]

is 0.7

and p(A|B)

is 0.1

so p(A|B) = p(A∩B)/p(B)

you know this formula right

?

[keen glade]

ohh i thought that the formula for P(A|B) was P(A)+p(B) / P(B

[wintry sluice]

bruh 💀

[keen glade]

im sorry :(((((((

[wintry sluice]

nah nah

its ok

try finding the answer now

[keen glade]

P(B|A) is 0.7?

[wintry sluice]

no :(

what did you do

[keen glade]

P(B|A) = P(A) x P(B) / P(A)

[wintry sluice]

NO

[wintry sluice]

this is the formula

why are you multiplying p(a) and p(b)

[leaden kraken]

You seem good at stats and probabilities, could u pls help me w a question ab permutation and combination. My question is on help forum section, but no one answered it yet. Pls help me

[wintry sluice]

just a minute

[keen glade]

but how i do p(a ∩ b)?

do i have to divide p(a) and p(b)?

[wintry sluice]

ok so im gonna start from the beginning

these are given

p(A) = 0.3

p(B) = 0.7

p(A|B) = 0.1

right?

[keen glade]

yes

[wintry sluice]

so

what is the formula

of p(A|B)

ill say it

p(A|B) = p(a ∩ b)/p(B)

so

to find (a ∩ b)

you must substitute the values

here

get it?

[keen glade]

yes till here yes

[wintry sluice]

so find a ∩ b)

[keen glade]

0.21?

[wintry sluice]

you mind if i write it and send it?

[keen glade]

no not at all

[keen glade]

ohhh okay

so you have to plug in the info that you have

[wintry sluice]

yess

[wintry sluice]

......

[keen glade]

0.07/ 0.3= 0.23

[wintry sluice]

??

[keen glade]

now to find P(B!A)

i'll take 0.07 and divide it by P(A)?

[wintry sluice]

yes

[keen glade]

so 0.07 divided by 0.3 which gives me 0.23

[wintry sluice]

yes

[keen glade]

okay so given that P(B)+P(B|A) = 1 is not right

given these: P(B)= 3 P(B|A) is also not correct

wait no it could be correct since P(B) = 0.7 and 3 P(B|A)= 0.69

[wintry sluice]

yes

approximation

[keen glade]

okay so that's the right option

[wintry sluice]

WW

nice

is it correct?

[keen glade]

i can't check it

[wintry sluice]

oh

[keen glade]

it's homework

[wintry sluice]

oh

alr

which grade are you in

[keen glade]

7th

[wintry sluice]

oh

wtf

[keen glade]

thank youuuuu

thanks for being patient!!!

.close

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[wintry sluice]

anytime

[lone heart]

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[fluid zealot]

Can someone pls check if I did the first two parts correctly? Also how would I do the last part?

<@&286206848099549185>

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@fluid zealot Has your question been resolved?

[fluid zealot]

Okay I think I fixed the second part but still don't get the show that part

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@fluid zealot Has your question been resolved?

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@fluid zealot Has your question been resolved?

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[violet ore]

How do I show $\beta'$ is a basis for $V$?

[ocean seal]

o.O

[past scroll]

well you have that $\beta'$ has the same number of elements as $\beta$ so you need to show either that the span of $\beta'$ is $V$ or that $\beta'$ is linearly independent. linear independence is probably easier here

[ocean seal]

Awesam

[violet ore]

So to approach showing linear independence by finding a unique solution for the linear combination that equals the zero vector is infact the zero vector? i.e. linear independence implies Ax=0 iff x = 0

[fierce dune]

Someone help me I’m too tired to do this 😭

[past scroll]

yes you should show that if a linear combination of the $x_n'$s are 0 then all the coefficients are 0

[ocean seal]

Awesam

[violet ore]

So we know that each $\beta'j$ is the linear combination $\sum^n{i = 1}{Q_{ij}x_i}$

[ocean seal]

o.O

[past scroll]

yes that's the definition

[violet ore]

so should I first approach by finding Q?

no, maybe not

So suppose $a_j \in \mathbb{F} \forall 1 \le j \le n$

$$a_1\beta'_1 + a_2\beta'_2 + \cdots + a_n\beta'_n = 0$$

[ocean seal]

o.O

[past scroll]

no Q is arbitrary

technically it's $x'_j$ not $\beta'_j$

[ocean seal]

Awesam

[violet ore]

Okay

How do I show $x'j = 0$? I can see that its true if $Q{ij} = 0 \forall i, j$ as that is trivial

[ocean seal]

o.O

[past scroll]

no you don't show that the $x'_j$s are 0 you show that the $a_j$s are 0

[ocean seal]

Awesam

[violet ore]

oh, right,

[past scroll]

you should write out each $x'_j$ based on the definition provided and you need to use that $Q$ is invertible and the $x_i$s are linearly independent (because they're part of a basis)

[ocean seal]

Awesam

[violet ore]

Is it true that

$\left(\sum^n_{i = 1}{Q_{ij}^{-1}x_i}\right)x'j$ = \left(\sum^n{i = 1}{Q_{ij}^{-1}x_i}\right)\left(\sum^n_{i = 1}{Q_{ij}x_i}\right)$

[ocean seal]

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

[violet ore]

Then I can rearrang the sum on the right side and reduce the inner sum to x_i

[past scroll]

there did you get the inverses from? that's not necessary. but yes what you wrote is true

maybe that works though

it's not the way I did it but it could still work I suppose

[lone heart]

@violet ore Has your question been resolved?

[lone heart]

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[crisp turtle]

ive been stuck on this for a while

S is a random point outside of the ellipse
the values given are OA, OB, OS, and SOA
SP is perpendicular to PT, the tangent to the ellipse at P
i need to find SP'A

[crisp turtle]

i know how to find SP'A (denoted by x) given POA (denoted by p)
tan(p) = a^2 / b^2 tan(x)

and OP (denoted by r) can be found with a, b, and p

and i also found that:
x - p = OPP' and x - d = OSP'
so
r sin(x - p) = D sin(x - d)
because they are both equal to OH

but i.. dont know how else to go forward

[lone heart]

@crisp turtle Has your question been resolved?

[charred flint]

the nongeometric way is using the equation for the tangent line of an ellipse, but taking the negative reciprocal of the slope to make it the perpendicular line, then solving for the normal line equation that has S on it. then the answer is just the x-intercept of that line

[crisp turtle]

hm.. so

[charred flint]

oh the coordinates of S aren't even given, nevermind then

[crisp turtle]

well

its Dcos(d), Dsin(d)

-dx/dy for an ellipse is a² y / (b² x)

so i have to solve for a point where

the slope from (x, y) to (Dcosd, Dsind) = a² y / (b² x)

thats.. simpler than i imagined

i guess im not solving this geometrically then

i was tryna go at this geometrically and

failing

repeatedly

wait but this is only one equation for two variables

[charred flint]

uh is your ellipse derivative 2-variable, there's a 1-variable version

[crisp turtle]

oh?

thanks

[void sable]

oh nah im doing that in hs

[lone heart]

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[glacial yoke]

How does the coefficient of 3 make 3/2 one half?

[frozen tiger]

(3/2)/3=1/2

ab=c
b=c/a

3cos(4theta)=3/2
cos(4theta)=(3/2)/3

[glacial yoke]

Will it always be the numerator thats being divided?

[frozen tiger]

ye

[raw stream]

4 (theta) = 60 as cos 4 theta = 1/2
(theta) = 15

.close if u done

[glacial yoke]

.close

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[alpine sable]

do you solve the integral of 1/(x-1)^2 as 1/(x-1)^2 or 1/(1-x)^2?

1/(x-1)^2 and 1/(1-x)^2 are equal, but their integrals are not, which one do you choose?

[gray isle]

how are you getting different results

[alpine sable]

with 1/(x-1)^2, t=x-1, dt=1, you just apply regular substitution, multiply it with dt and get the expected result
with 1/(1-x)^2, t=1-x, dt=-1, you have to multiply with dt which here is -1 so you get a diff. result

[gray isle]

what results are you getting

[alpine sable]

1/(x-1)^2 -> -1/(x-1) + C
1/(1-x)^2 -> 1/(1-x) + C

oh

nvm im stupid

.solved

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[livid tundra]

What should I do using cramers rule

the 3x3 matrix thing it is if im not wrong

[pallid scarab]

Yes, write the systems of equations as a vector equation AX = B

[lone heart]

@livid tundra Has your question been resolved?

[polar slate]

write out matrix A then find det(A)

write out matrix A1, A2, A3 (with the b (or constants vector) replacing the column indicated by 1, 2, 3), then find the det(A1), det(A2) and det(A3)

[pallid scarab]

As in let X = (x,y,z), etc...

[polar slate]

x1 = det(A1)/det(A)

x2 = det(A2)/det(A)

x3 = det(A3)/det(A)

[pallid scarab]

That's a bit like handing out the solution

[polar slate]

its just the formula..

[pallid scarab]

Sure but that's basically explaining the whole process

[polar slate]

is that not allowed? i thought its just blatantly handing out solutions like numbers etc

im just describing what they have to do in order to get to the correct solution

i dont see anything wrong with that personally

[pallid scarab]

Giving the whole directions is just like chewing the work for them

So there's no incentive to figure some things out for themselves

[mossy swift]

lol

[polar slate]

either way, my intention is to fast-forward their learning instead of having them to figure out a formula for a long time. i'd argue that in their position, i would appreciate that if someone were to either summarize "textbook" information or formula so that they dont waste time

it's cramer's rules anyway. its literally just following formulas. im not guiding them through how to find dets of each matrices. im literally just giving them the formula on how to find 'x'

i aint trying to be the riddler when it comes to helping someone learning. i want to direct and concise as possible

[lone heart]

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[tawdry sentinel]

Point of clarification...

[tawdry sentinel]

The limit for a sequence exists if the limit for f(x) = a_n exists?

And those limits are hence equal?

[alpine sable]

yes

[tawdry sentinel]

Thanks!

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[paper mango]

IF it exists then they are equal

but if you refer to, let's create a function f:R->R which mimics the behavior of a_n

[tawdry sentinel]

.reopen

[lone heart]

✅

[paper mango]

maybe I have misunderstood your f(x) = a_n though

[tawdry sentinel]

So, in this case, I'm more concerned with the sequence {a_n}

{a_n} isn't a function.. but it looks and behaves like one.. at least that's my understanding

[paper mango]

ys if you create a function that mimics its behavior then this argument goes only one-way

[paper mango]

yeah a sequence can be regarded as a function

a: N->R

which maps the natural numbers onto the real

and we consider the input for the function

as the index of the element within the sequence

extending that function onto R->R doesn't mean these two functions behave the same regarding limits

[tawdry sentinel]

Difference being, (and I'm going to butcher this... there are too many different sets of numbers) x can occupy any value in the entire catalog of real numbers.. -infinity to infinity, and every irrational value in between

n only ever occupies the natural numbers from 0 to infinity.. no fractions, no irrationals, no decimals

[paper mango]

yes, and that extension to real (or also inf./-inf.) leads to potentially different limits (and their existence)

e.g. take some periodic cos curve and let a sequence depict its local maxima

then the sequence is just constant

{a_n} = {cos(n*2pi)}

if we extend it to a real function

f(x) = cos(x*2pi)

then its limit for x->inf isn't defined

however the limit for the sequence is defined for n->inf

[tawdry sentinel]

because a_n just oscillates... ?

[paper mango]

ys

well f(x) oscillates

{a_n} is just (1, 1, 1, 1, ...)

[tawdry sentinel]

Aha.. ok

[paper mango]

using above extensions can be useful for analysing sequences, but I'd be careful using it for implications

"if f(x) does that then {a_n} does that" may not be true

[tawdry sentinel]

So.. you had sin up there initially and changed it to cos... and I'm curious, because the coursework I'm going through always uses cos as an example.. never sin...

but wouldn't sin suffer the same issue?

[paper mango]

doesn't matter, just made it faster for me to write

we can also take

[tawdry sentinel]

ok

[paper mango]

{a_n} = {sin(pi/2+n*2pi)}

which is just (1, 1, 1, 1, ...) as well

[tawdry sentinel]

that's what I was thinking, but it's just odd that everyone is using cos in this topic

you, my textbook, my instructor, etc.

[paper mango]

sin and cos are the same functions, just shifted by pi/2, so depending on what you want to show with an example, it's slightly quicker to do it with cos :P

and sometimes with sin

[tawdry sentinel]

gotcha

thanks for the input!

[paper mango]

np

[tawdry sentinel]

.close

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[lone heart]

@gritty pond Has your question been resolved?

[gritty pond]

<@&286206848099549185>

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@gritty pond Has your question been resolved?

[gritty pond]

<@&286206848099549185>

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@gritty pond Has your question been resolved?

[gritty pond]

.close

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[tawdry sentinel]

https://tenor.com/view/donkey-boulder-shrek-donkey-rock-gif-27505603

What's your question?

[small lance]

!status

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@lone carbon Has your question been resolved?

[lone heart]

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

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[bronze thicket]

i need help whit this

let $(Xn)$ be defined in the form $x_1=a$ and $X{n+1}=\sqrt{X_n+b}$, where $a^2<a+b$ , a and b are positive. Calculate limit $(X_n)$

[ocean seal]

leycors

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@bronze thicket Has your question been resolved?

[spiral coral]

Hi i Will like to be better at division

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[dapper smelt]

I don't know how to do this at all

It would be appreciated if someone can explain how to do it

[rose void]

one method to approach: set length PR as "1" and thus the length of QR would be .6 and you can solve for the missing side length, and then you can use sin=opp/hyp

[dapper smelt]

So what are are tryna find here?

When it says the Value of sine qpr

What does it mean by that

[rose void]

well sin is just opposite/hypotenuse

we can solve for hypotenuse then sin

[dapper smelt]

Yeah

So is it

Asking for the angle?

I'm confused

How do we solve for the hypotenuse

[rose void]

well we know length QR is 60% of length PR

we can set length PR to 1

[dapper smelt]

Yeah

[rose void]

then we can use the theorem to solve for the missing side length

[dapper smelt]

Yeah

And then

That's it?

[rose void]

pretty much

[dapper smelt]

Alright then

Thanks!

How do I close the channel

[rose void]

".close"

[dapper smelt]

.close

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[delicate furnace]

Coudo someone help with this fraction?

[lone heart]

Please don't occupy multiple help channels.

[vapid drift]

you need to simplify this?

[delicate furnace]

Alright, how do I start?

[vapid drift]

factor out 2ab from the numerator

[delicate furnace]

That leaves us with...

ab-2a(square)?

[vapid drift]

that leaves us 2a-2b

exatly 2ab(2a-2b) in the numerator

[delicate furnace]

One moment lemme try to write that down

Alright, but I don't understand how the numerator got turned into 2a-2b

How did you factor it out?

[exotic canopy]

$2a^2 2b - 4ab^2 = 2aa\cdot2b - 4 a b b = (2ab)(2a)-(2ab)(2b)$

[ocean seal]

artemetra

[vapid drift]

you can just multiply 2ab by (2a-2b) and see its true

[delicate furnace]

I know its true, im just to dumb/unknowing to know how to factor the big one wodn to that

and one other thing, how do I stop occupieng a channel, because i have two at the moment

[vapid drift]

i know that in 2a^2* 2b and in 4ab^2=2* 2ab^2 there is number "2" number "a" and number "b" so i can factor them out

[vapid drift]

type ".close" in the channel to close it

[delicate furnace]

how do i factor them out, im not used to hearing this stuff in english

do i just subtract them from each other?

[vapid drift]

so maybe look at easier example
if you want factor out for example 2a+4ab you get 2a(1+2b) because 2a/2a = 1 and 4ab/2a=2b

so you divide

both monomials by the thing you wanna factor out

in this case those monomials are 2a and 4ab and the thing you wanna factor out is 2a

in your example your monomials are 2a^2*2b and 4ab^2

you factor out 2ab

so you divide those monomials by 2ab

[delicate furnace]

so if I wanna factor out "a" for example, i divide both monomials by "a"?

[vapid drift]

yes

[delicate furnace]

alright im startring to get it

[vapid drift]

and you write a(monomial1/a + or - monomial2/a)

• or - depending if you add or substract originally those monomials

and you can do it with as much monomials as you there are

in this case 2

but can be 3,4,10 or even infinite monomials

the formula is the same

[delicate furnace]

but the monomials are the two sides of the question in this case

[vapid drift]

dont understand what you mean by that?

[delicate furnace]

that 2a^2b is one and -4ab^ is another

[vapid drift]

yes

[delicate furnace]

the two sides of the question

alright

so we factor out 2ab

[vapid drift]

yes

and we get a product in the numerator 2ab(2a-2b) and 2ab in denominator

they cancel out

and you get 1*(2a-2b)

and that is just equal to 2a-2b

[delicate furnace]

with my head that would be a^b-2a...

something

i dont understand how you divide those big numbers by 2ab

2a^2b divided by 2ab

a+a*a+b+b divided by a+a+b

[vapid drift]

everuthing canel out

2/2=1

a^2/a=a

[delicate furnace]

ooooohhhhh

[vapid drift]

and b/b=1

[delicate furnace]

now I understand

we divide the top with the bottom, i dont know what i was thinking

[vapid drift]

it is confusing if you look it as addition

[delicate furnace]

but then we have the -4ab^ left

[delicate furnace]

oh alright

[vapid drift]

ok im gonna write the full sol

to show you how its done

cuz your still confused

[delicate furnace]

thank you man

im understanding more once i got my mind straight

[vapid drift]

ill actually show you two ways to solve it

one with factoring

one without

[delicate furnace]

okay

[vapid drift]

tho you'll need factoring is harder questions

[delicate furnace]

for harder questions?

[slender granite]

fwiw this is wrong

[delicate furnace]

oh yeah i relized, but thanks for pointing out

[vapid drift]

this is how to solve factoring

this is without

[delicate furnace]

but arent we left with 2^b?

[vapid drift]

why?

[delicate furnace]

arent those the last numbers who are not crossed over?

2 ^ and b

[vapid drift]

its not 2^b

its 2*b

[slender granite]

2(a^2)2b/(2ab) is just 2*a*a*2*b/(2*a*b), and there's no point in looking at it as if it was addition. So literally just remove one factor of 2, one factor of a and one factor of b from the top which gives a*2*b=2ab.

The whole thing:
2*a*a*2*b-4*a*b*b/(2*a*b)=a*2-2*b=2(a-b)

[vapid drift]

oh yeah you can factor out 2 as well at the end mb

[slender granite]

note: * is the symbol for multiplication and ^ is the symbol for exponentiation

[vapid drift]

all that was said here before

[slender granite]

He was confusing it with addition and stuff

[vapid drift]

yeah i told him

[slender granite]

I wrote it like this to make it clear that all you literally have to do is think of it as if it was expanded and just cross them over

[delicate furnace]

but wouldnt that levae us with 2b?

and not 2a -2b

[slender granite]

Why do u think that

[vapid drift]

i didnt cross out 2a

so why wouldnt be there?

[delicate furnace]

oh i looked at it wrong

the umbers left are: a,2 ^, b

right?

[slender granite]

the numbers left are, as thesnake wrote in the image, 2*a-2*b

..

[vapid drift]

here left numbers are a and 2

and here are 2 and b

4/2 is 2

[delicate furnace]

did the ^ turn into a * when it didnt have a number to hang onto?

[slender granite]

the ^2 turned into ^1

[vapid drift]

its just lazy way to show that a^2/a=a

[slender granite]

AdiDaddi think about what happens when you expand these.
(4ab^2)/(2ab) = (2*2*a*b*b)/(2*a*b)

So if you just remove one 2, one a and one b from 2*2*a*b*b, you get your result

Doing that gives u 2*b.

[delicate furnace]

4A2B

[slender granite]

?

[delicate furnace]

2x2=4 4xa=4a

and so on

hoiw should i think?

[vapid drift]

remove one "2" one "a" and one "b" from set {2,2,a,b,b}

what do you get?

[delicate furnace]

2ab

[slender granite]

😄

[vapid drift]

no

[slender granite]

2b

[vapid drift]

just {2,b}

[delicate furnace]

oh sorry

[vapid drift]

but yeah

[delicate furnace]

i agree

[vapid drift]

that the idea

[slender granite]

it's important to realize that the notation which might be confusing you is really just 2*2*a*b*b

[vapid drift]

when i "cross out numbers" they turned into *1 so its like they "vanish"

[vapid drift]

they vanish how we did here

[delicate furnace]

what do you mean

[delicate furnace]

oh ok

[slender granite]

adidaddi do you agree that (a*b)/b=a?

[delicate furnace]

hm

where did those nu\mbers come from

[vapid drift]

its an example

[slender granite]

These are just arbitrary variables to be clear

[delicate furnace]

oh alright

[slender granite]

Let the variables a and b be absolutely any numbers, but b cannot be zero

[vapid drift]

they can be any numbers

[slender granite]

(but b can't be zero)

[vapid drift]

yes yes

you cant divide by 0

[delicate furnace]

(2*3)/3=2

[slender granite]

So AdiDaddi do you agree that the equation (a*b)/b=a is true for any numbers a,b where b is not 0?

[vapid drift]

(2*3)/3=2

[delicate furnace]

yeah i would think so

[slender granite]

if you don't fully agree here, you cannot possibly agree with anything else regarding your question btw

This is the only thing you need to understand to understand your question fully, too

[delicate furnace]

i agree that 2*3= 6 6/2=3

meant 6/3=2

[vapid drift]

its important that you know you can do this: (a* b)/b=b/b* a=1*a=a

for any numbers a,b and b=/=0

[delicate furnace]

alright

[slender granite]

Yep that's a good way to explain it. If you understand factorization, you will also understand that (a*b)/b=a for any numbers a,b, where b isn't 0.

And so

with this knowledge

Let's apply this to your example.

[delicate furnace]

yup

[vapid drift]

also, you often write this by "crossing out" the numbers

[slender granite]

So we know that (x*y)/y=x. If you let x=4ab and y=b you get (x*y)/y=x <-> (4ab * b) / b = 4ab. So you just took 4ab^2 and divided that by b and with the knowledge you just learned, you know for sure that the result of that must be 4ab.

[delicate furnace]

oh ok

[slender granite]

Now you can continue doing this for all the other numbers.

For example, let's do the full (4ab^2)/(2ab)

[delicate furnace]

i understand

[slender granite]

4ab^2=2*2*a*b*b. If you divide this by 2*a*b, all you're doing is removing one 2, one a, and one b, from 2*2*a*b*b which gives you 2*b, as we saw earlier, and this can be explained using the fact that (x*y)/y=x because e.g:

4ab^2 = 4ab * b.
So (4ab * b) / b = 4ab according to the equation (x*y)/y=x.
And: 4ab = 4b * a. So (4b * a) / (a) = 4b.
And: 4b = 2 * 2b. So (2 * 2b) / (2) = 2b.

And these last three lines were really just me doing (4ab^2)/(2ab). You can also do this in one step by realizing that 4ab=2ab * 2b, so (2ab * 2b) / (2ab) = 2b.

[delicate furnace]

i dont really understand, but I dont have anymore time unfortuantely. Thank you so much for helping out thoug. I atleast got the idea.

@vapid drift thank you so much to

[slender granite]

Wait first im gonna tell u one quick thing

[delicate furnace]

alright alright

[slender granite]

Do you agree that 4ab^2 = 2*2*a*b*b?

or at least 4*a*b*b?

[delicate furnace]

yep

[slender granite]

And do you agree that a*b = b*a (in other words order doesn't matter in multiplication)?

[delicate furnace]

ÿep

[slender granite]

ok. So to factorize something, this is all you need.

E.g. if you want to bring out the factor 2ab from (2a^2)*(2b), all you need to do is expand (2a^2)*(2b) in your head, which gives you 2*a*a*2*b, and then reorder it so that 2*a*b comes first, so: 2*a*b * 2*a. Now, with the rule (x*y)/y=x, you can see that (2*a*b * 2*a) / (2*a*b) = 2*a.

[delicate furnace]

okay

i understand aliiitle

[slender granite]

I think you'd understand perfectly if I wrote this on paper but I can't do that atm unfortunately

Hope u have a good day though

[delicate furnace]

you to man, thank you so much

[slender granite]

np, TheSnake did a really good job of helping you, I just wanted to add a few things that imo make it super easy to understand but it can be a bit confusing when it's written like this on discord

[lone heart]

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[split flume]

does this look correct?

[neon bronze]

No, you’ve calculated B for A and A for B

C is correct

[split flume]

oh

so I just swap em?

[neon bronze]

Yes

[split flume]

alr

thank you for pointing that out

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[reef flame]

the way I'm trying to prove this

is by splitting the graph into two

so I have one node of degree two connected to a cloud of degree four nodes

how can I prove that the amount of edges going from the vertex of degree two all through the cloud of degree four nodes has an odd edge count?

[lone heart]

@reef flame Has your question been resolved?

[reef flame]

<@&286206848099549185>

these are unicursal so we have can have multiple edges from one node connecting to the next

like this

which clearly has an odd amount of edges I'm just unable to prove it

[charred flint]

not sure how to get your graph theory approach to work, just gonna throw in some ideas

one big thing about these curves is you can 2-color them, so maybe there's some parity stuff there

another way to look at it as how a path that hits an intersection is going to come back to the point later

if you can prove it has to pass through an even number of other intersections before traveling back to the blue point, you'd be done

your graph theory approach might be impossible if the proof requires something about it being a planar curve

@reef flame

[lone heart]

@reef flame Has your question been resolved?

[reef flame]

frick

thank you for the help

I will try these things

very tricky problem...

.close

[lone heart]

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[alpine sable]

could someone help me with this real quick i just dont know how to set it up

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

thx

[amber patio]

Hello I need help

[lone heart]

@visual finch Has your question been resolved?

[lone heart]

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[slow spear]

need help pls

[jagged cobalt]

what subsitution do you think is sensible

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[forest raft]

In a population, individuals' Body Mass Index (BMI) is normally distributed with a mean value of 23 𝑘𝑔/𝑚^2 and a spread of 3.8 𝑘𝑔/𝑚^2
a) Determine a prescription for the density function for the distribution of the individuals' BMI in the population.

[lone heart]

Please don't occupy multiple help channels.

[forest raft]

b) Determine the probability that an individual's BMI is between 22 𝑘𝑔/𝑚^2 and 26 𝑘𝑔/𝑚^2

c) Determine the normal outcomes for the individuals' BMI.
I only need help for c

plls :((

[reef kite]

What have you tried so far?

[forest raft]

I did a and b but i didnt understood the question

c

@reef kite

<@&286206848099549185> pls help me :(( <33

<@&286206848099549185>

pls

<@&268886789983436800>

[real gazelle]

saying stuff like this is unacceptable here

Can you DM a screenshot of what he DMed you to @fierce herald if it was inappropriate

I muted him, for the time being

[forest raft]

then he blocked me

@real gazelle

In a population, individuals' Body Mass Index (BMI) is normally distributed with a mean value of 23 𝑘𝑔/𝑚^2 and a spread of 3.8 𝑘𝑔/𝑚^2
a) Determine a prescription for the density function for the distribution of the individuals' BMI in the population.
b) Determine the probability that an individual's BMI is between 22 𝑘𝑔/𝑚^2 and 26 𝑘𝑔/𝑚^2
c) Determine the normal outcomes for the individuals' BMI.
I only need help for c
plls :((

<@&286206848099549185>

[lone heart]

@forest raft Has your question been resolved?

[lone heart]

@forest raft Has your question been resolved?

[onyx canopy]

Hi

[lone heart]

@forest raft Has your question been resolved?

[forest raft]

hi

[brisk wagon]

help

[forest raft]

<@&286206848099549185>

hii

[brisk wagon]

HI

HELP PLEASE

how do i get help with math rn i been struggling for an hour with frequencies

[forest raft]

broooooooooooooo

go to another channel

[brisk wagon]

oh sorry it just sent me here

[fiery dust]

you thought someone was finally helping you lmao

i would have if i knew the what o do

[glossy finch]

T_T

[open wasp]

@forest raft Whats your question sweetheart? ❤️

[forest raft]

OMGGGGGGGGGGGGGG

THATS SO CUTEEE

❤️

AWWWWW

[open wasp]

your question??

okay nvm ig

[twilit loom]

:/

[mossy totem]

Can you translate

[echo wolf]

Can some1 plsssss help me out w/ this one

[ebon sparrow]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[ebon sparrow]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[lone heart]

@forest raft Has your question been resolved?

[lone heart]

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

How do i do these

[ripe needle]

Well i did it till this

But like idk what to do next

[torpid dirge]

for minutes to hours, you "carry over" one for each 60 minutes

[ripe needle]

Alr

Its 17 mins more then 60 tho

[torpid dirge]

77min = 1h 17 min

[ripe needle]

Oooooo

[torpid dirge]

try to apply it here

[ripe needle]

So the ans would be 17:17?

[torpid dirge]

yea

[ripe needle]

Okii

Tyy

[torpid dirge]

np

[ripe needle]

Is this right?

[torpid dirge]

<@&268886789983436800>

[torpid dirge]

i think so

[ripe needle]

Go learn math 💀

[ripe needle]

Alrr tysm!!!

[torpid dirge]

probably hacked

[ripe needle]

I guess

[torpid dirge]

math enthusiast

[ripe needle]

Idek him

I js wanted to say smth 😭🤡

[ripe needle]

Nah bro i hate math 😭😭

[torpid dirge]

you want others to feel your pain

[ripe needle]

Frr 😭😭

[torpid dirge]

evil

[ripe needle]

If i didn't have a exam tmr i would be chillin

😭😭

[torpid dirge]

exams bad

[ripe needle]

Fr

Do yk ho to take out percentage loss?

How*

[torpid dirge]

send the question

[torpid dirge]

if im in the mood and know how to do i might help

[ripe needle]

Also tysm for helping since morning 😭😭

[torpid dirge]

loss of 5% so he sold it at 95% the original price

[torpid dirge]

[ripe needle]

How would i like solve that

Ratio?

[torpid dirge]

an example would be, if im selling a car i bought for 1000 at a 20% loss, 80% original price, it would be 1000*80% = 1000*(80/100)=800

[ripe needle]

Alrrr

?

[torpid dirge]

seems good

[ripe needle]

Tysm!!

[torpid dirge]

np

[ripe needle]

I got a q

Since i kinda need more help

With other questions

Do i js keep this thing open n keep asking or make a new one?

[torpid dirge]

sure ill help here

[ripe needle]

Okii tysm bro i owe u my life 😭

[torpid dirge]

if im gone just create another lol

[ripe needle]

Lol alr

[torpid dirge]

are you last min cramming

[ripe needle]

I think I did this right

[ripe needle]

Yes-

I didn't study b4 cuz i couldn't and its my mock

[torpid dirge]

this is wrong

[ripe needle]

Oo-

[torpid dirge]

what is the coordinates of A?

[ripe needle]

0,-2

[torpid dirge]

yes, and C

[ripe needle]

4,4

[torpid dirge]

yes

put them in the formula you wrote

idk did you use the coordinates of B or something

[ripe needle]

No

I just took 2 random point between em-

[torpid dirge]

o lol

well use the coordinates from points A and C

[ripe needle]

?

Iike this?

[ripe needle]

Ohkk

[torpid dirge]

yep u got it 👍

[ripe needle]

Yess

N if ABCD is a square

D's coordinates would be 5,0?

[torpid dirge]

5,-1

[ripe needle]

Oo

[torpid dirge]

5 in one direction and 1 in the other

[ripe needle]

How would i do this-

[ripe needle]

Mhm

[torpid dirge]

beetle strong fr

[ripe needle]

Ikrrr

[torpid dirge]

convert tonnes to kg

[ripe needle]

Uh, idl how to

Idk*

[torpid dirge]

yea...

same, i googled

[ripe needle]

We haven't done that

[ripe needle]

Lol

[torpid dirge]

i think its 1000, but us tones whatever that is is 907 ish...

[ripe needle]

Alrr

But bro what if this comes in the exam n we haven't done that-

Only conversations we have done

[torpid dirge]

idk qwq

just guess

[ripe needle]

Imma be ded-

[ripe needle]

Me for half my exam fr

[torpid dirge]

its probably some multiple of 3 in power

[ripe needle]

Ohkk

[torpid dirge]

you'll be fine

[ripe needle]

Hopefully 😭

This alr?

[torpid dirge]

allahu akbar

[ripe needle]

Muslim?

[torpid dirge]

to this

[ripe needle]

Oo

[torpid dirge]

pray to Allah

[ripe needle]

I am bro 😭

[torpid dirge]

180cm

not m

[ripe needle]

Ohk

[torpid dirge]

oh yea, whats the first letter of your disc name

is that an f?

[ripe needle]

S

[torpid dirge]

shay?

[ripe needle]

Mhm

Its a nickname

[torpid dirge]

i can't cursive

[ripe needle]

Lol

I write in cursive usually

[torpid dirge]

same

[ripe needle]

Ww

[torpid dirge]

very unreadable

[ripe needle]

Lol

[torpid dirge]

ww

[ripe needle]

What do i do next

I fr gotta rabi zidni illma my exam tmr 😭💀

[torpid dirge]

u what

[torpid dirge]

i haven't seen that in awhile

[ripe needle]

Uhh nvm- sorry

[ripe needle]

Lol

[torpid dirge]

the question

whats the other length of the wood???

[ripe needle]

Wym

He has 2 meters

He needs 2 peices of 45cm n 2 pieces of 60cm

Does he have enough

[torpid dirge]

idk like

it could be 2m by 2m

or 2m by 10cm

bad question

[ripe needle]

Ig u can make ur own scale

[ripe needle]

This is a "easy" question

😭

[left crane]

could you help me with a math problem ?

[ripe needle]

Thats like spanish

[torpid dirge]

na

[torpid dirge]

its just.. badly worded

badly made

[ripe needle]

Nah bro its plane hard 😭

[torpid dirge]

ill probably get going

[ripe needle]

Ohk

Tysm for helping so much today!!

[torpid dirge]

np

[ripe needle]

.close

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[short delta]

First 2 are right

[lone heart]

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

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

[crisp widget]

can i get help please

[lone heart]

@crisp widget Has your question been resolved?

[lone heart]

@crisp widget Has your question been resolved?

[charred flint]

I'm not sure how the diamond is supposed to work, do you have another example? I'm mostly not sure what the connection lines are supposed to mean

[lone heart]

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[tidal quartz]

try to factorize all four polynomials

some cancellation should happen

i.e. write x^2+7x+10 as (x+5)(x+2)

[tidal quartz]

just simplify then write the answer 🙂

[torpid roost]

wdym?

[ebon sparrow]

,rccw

[ebon sparrow]

@torpid roost here where it comes the fun part!

cancelling

[torpid roost]

what can i cancel out for each of them?

[ebon sparrow]

ye

you have like two

[torpid roost]

do they need to both be numerators/denomerators?

[ebon sparrow]

yep

[torpid roost]

so then i only can cancel out x+5 and x+4?

[ebon sparrow]

x+5 and x+2 only

[torpid roost]

why x+2?

[ebon sparrow]

also i see another one

actually nah it should be like this

[torpid roost]

so to cancel them out they cant both be denominators or both be numerators they have to be different?

[ebon sparrow]

if numerator and denominator are same then yes

but you can't with different

[torpid roost]

so now its (x+5)^2 divided by (x+4)^2

[ebon sparrow]

isn't x-5 on the second one tho?

[torpid roost]

its a plus

[ebon sparrow]

oh

it looks like minus but yea your work is correct

[torpid roost]

👏 YAY

I have aother one that is the same idea but looks kinda different

[ebon sparrow]

go for it

[torpid roost]

how do i begin?

[ebon sparrow]

same thing

apply the fraction rule and then cancelling

@torpid roost

[torpid roost]

how do i factor tho with the big exponents?

[ebon sparrow]

you don't need to

jusst apply the exponents of law

[torpid roost]

the what?

i dont know what that means

[ebon sparrow]

,tex .exp rules

[ocean seal]

アキラ (>_<)

[ebon sparrow]

the quotient rule one

before you do that apply the fraction rule first then this

[torpid roost]

whats the fraction rule?

[ebon sparrow]

,, \frac{a}{b} \times \frac{c}{d}=\frac{a \times c}{b \times d}

[ocean seal]

アキラ (>_<)

[torpid roost]

do i flip it since its division?

[ebon sparrow]

yes

[ebon sparrow]

,rccw

[ebon sparrow]

seems good

[torpid roost]

do i multiply them now then simplify?

[ebon sparrow]

no, apply the quotient rule

[ebon sparrow]

look here

[torpid roost]

like z^13x^5-6z^5?

[ebon sparrow]

no like we know that z is 13 and the second is 6 so subtract the exponent

same thing for x

[torpid roost]

huh?

there is 6z^5 and 2z^6?

[ebon sparrow]

think it about this $\frac{z^{13}x^5}{6z^5}\times\frac{x^4}{2z^6x^3}$

[ocean seal]

アキラ (>_<)

[ebon sparrow]

what happens when you cancel

recall that when canceling its always gonna be subtracting

[torpid roost]

would i do x^5 -x^3 =x^2?

[ebon sparrow]

z^6 goes away and what do you think the numerator will be for z^13?

[ebon sparrow]

yeah you got it

its what im saying here

also don't forget that you have 6 times 2

so it becomes 12

[torpid roost]

are the 6 and 2 just not important to the zs when subtracting the exponents since its all multiplacation?

[ebon sparrow]

yea

[torpid roost]

is it z^13-z^5-z^6???

[ebon sparrow]

no

[torpid roost]

is it z^13 -z^6?

[ebon sparrow]

not quite

[torpid roost]

im confused again

[ebon sparrow]

do you want to restart

[torpid roost]

perhaps

[ebon sparrow]

ok how about in this way $\frac{z^{13} x^5 4}{6z^5 \cdot 2z^6 x^3}$

[ocean seal]

アキラ (>_<)

[ebon sparrow]

you can also cancel

[torpid roost]

i think the fact that there is 2 zs in the denominator confuses me

[ebon sparrow]

oh this

it still applies to z

[ebon sparrow]

so like i said here above

nah wait

when you subtract 13 by 5 you get 8

and there's another 6 which makes it 2 in the final

[torpid roost]

isnt that what i did here?

[ebon sparrow]

yea mb didn't see that

[torpid roost]

so does that mean there is a x^2 in the numerator?

[ebon sparrow]

yep

[torpid roost]

so is the numerator z^2 * x^2?

and the denominator 12?

[ebon sparrow]

close but x should be x^6

[torpid roost]

is that cuz there is x^4 and x^5?

[ebon sparrow]

and also 3

when you multiple those you get 9 and then subtract by 3

so the final will be 6

[ebon sparrow]

exactly

good job

[torpid roost]

can i simplify it any more?

[ebon sparrow]

nope

[torpid roost]

yay

[ebon sparrow]

thats your final answer

[torpid roost]

I wanna do one more before I close to make sure im still not confused

[ebon sparrow]

alright alright

we need to factor this time

just like you've posted your first problem

you can always check your in wolfram or something else to make sure you factor correct

and boom boom with cancelling

[torpid roost]

can i cancel out x+4 and x+5 from each of them since they are both in the denominator and numerator

?

@ebon sparrow

[ebon sparrow]

,rccw

[ebon sparrow]

you can cancel 3x-2, x+4 and x+5

[torpid roost]

oh i didnt see that there were 2 3x-2s

[ebon sparrow]

[torpid roost]

so is the numerator now x-3 and the denominator 3x-1?

[ebon sparrow]

yeye

[torpid roost]

yippepe

i think i understand good now :)

thank you

[ebon sparrow]

np

[torpid roost]

.close

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[last spire]

How to prove set A = B iff id_A = id_B? id_A is the identity function

[placid zinc]

You've actually got two proofs here: the forward direction, and the backward direction. Does that make sense to you?

[last spire]

Yes

[placid zinc]

Forward is pretty easy I think:
A and B are the same set. So naturally, any function that inputs both needs to output the same thing. If not, that function wouldn't be well defined

[last spire]

What about the reverse direction?

[placid zinc]

But maybe you want to specifically mention the fact that the function is the identity

[last spire]

Since A = B, ∀a∈A, a∈B, and so idA = idB?

[placid zinc]

The reverse is harder, and the proof will depend on the function being the identity

[placid zinc]

Yeah that's basically the forward. If it weren't true, then you'd have id(A) ≠ id(A)

For the reverse direction, it's helpful to notice that id(X) = X

Or wait, I guess I am comparing a function to a set, your teacher might want to avoid that

Well, let's start by saying id(A) = id(B)

Then for every a in A, there exists an (a,a) in id(B).

But that means...

[last spire]

a in B

Thanks!

.close

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[vale remnant]

I don't get how to solve part B

I got part A fine

[lone heart]

@vale remnant Has your question been resolved?

[lone heart]

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[low drift]

i tried converting to cylindrical and got $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} \int_{r}^{\sqrt{2r^2}}rdzdrd\theta$

[ocean seal]

Jaxx

[low drift]

i got $ \frac{4\pi}{3}$

latex broke so sad

i got 4pi/3

answer was 4pi/3 ( sqrt(2) - 1 )

[livid sage]

hint bounds of theta in this integral incorrect

[low drift]

nevermind im silly

its 0 to pi/2