help-23

[scenic swallow]

thanks

[calm bridge]

do you need help with anything else?

[scenic swallow]

i just started this exercise

so idk now

if i have i will chat here or to you

if its okay

[calm bridge]

ok, np

[safe radish]

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[lean otter]

How am i supposed to solve this?

[lean otter]

My book said some stuff about. put your right hand in ??????????????????????????

anyways, my book didn't teach this

[devout shale]

the right hand rule

[quasi bison]

arrange your thumb, index and middle fingers like this, vaguely in the shape of an orthogonal triad

thumb = a
index = b
middle = a × b

in your case specifically, thumb = v, index = B and thus middle = F

(if q > 0)

[stoic dune]

Proposition: Ann has at least one hand

[quasi bison]

proposition: Ann has at least one right hand

[stoic dune]

Even better!

Or maybe the picture was flipped

[lean otter]

WHAT

[quasi bison]

what?

[lean otter]

I do not understand any of that

Is there a math way to do this?

[quasi bison]

this is the right hand rule

illustrated with my own flesh and blood right hand

[devout shale]

are we sure you didn't use a cadaver

[lean otter]

There is no right hand in my brain

so i can't do it

[quasi bison]

do you have a right hand or are you an amputee

[lean otter]

???

[quasi bison]

you can't really draw this.

[lean otter]

or is ti like this

[quasi bison]

this looks vaguely phallic.

anyway like

in these drawings

treat the x axis as going left to right

y as bottom to top

and z as directly towards you

[lean otter]

😭

So

the arrow

correct?

[quasi bison]

yes

for q>0

do them all for q>0

then the ones for q<0 are just opposites of those

[devout shale]

these drawings are from a fever dream

[quasi bison]

this is REALLY phallic.

[lean otter]

NO THEYRE NOT

You're mind is just in the wrong plae

It's literally a hand

[quasi bison]

a very artistic rendition thereof.

[lean otter]

the

B vector is pointing going towards positive z when q > 0 right

I hate this

[solar hazel]

i'm dead

[lean otter]

https://tenor.com/view/ตายแพร๊บ-cpr-revive-cat-gif-15837553

wait

cross products

aren't like

$$ \vec{a} \cross \vec{b} \neq \vec{b} \cross \vec{a}$$

[flat frigate]

One person

[lean otter]

so

(1) and (2) aren't the same?

So

$$ V_z > 0$$
so that means
$$ B_x > 0$$ as well

so

?

[flat frigate]

One person

[lean otter]

<@&286206848099549185>

$$\vec{a} \cross \vec{b} = (a_x \vec{i} + a_y \vec{j}) (b_x \vec{i} + b_y \vec{j})$$

[flat frigate]

One person

[unkempt rain]

One Person

[lean otter]

$$\vec{i} \vec{j} = \vec{k}$$
$$\vec{j}\vec{i} = -\vec{k}$$

[flat frigate]

One person

[unkempt rain]

I think I can help

[lean otter]

Please do

Since $$ v_z \vec{k} > 0$$
$$ b_x \vec{i} > 0 $$ as well?

graph (1)

[flat frigate]

One person

[lean otter]

.close

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[lean otter]

Does anyone have a better explanation for this problem?

[lean otter]

Everything that I have found just says

" Use the answer to solve the problem "

Do i just like

create a list

and see if any of the vectors I have will put me on teh target square

and if there are none

I extend the list by 1

like

The vectors in the red being the list of positions i can reach with just two vectors added together

I would then check if its equal and if its not add another variable to one side

then try to solve for that variable using the vectors I have already

[minor goblet]

if you didnt have the "cant step on" condition itd be easy no ?

[lean otter]

and if none can be used I create another list of vectors that use three vectors added together instead of just two

[lean otter]

maybe

[minor goblet]

you could just take 2 displacements and make them your basis vectors then find out the coordinates of the target vector in terms of your basis

sorry 3 displacements*

now you can just repeat that until you find the combination that works

(maximum 4 repeats)

thatd be a technical way to solve it

im pretty sure if you solve it like that by ignoring the "cant step on" condition you can just rearrange the terms to avoid the point

[lean otter]

okay

thank you

.close

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[slate gale]

Hello! This is my first time asking for help here. I have a math degree from 10 years ago so my calculus skills are very rusty. I have attached my work and wrote my entire thought process so far to this question. Mostly, I want to make sure the next direction(s) I am trying to take make sense? If not, what should I consider instead? Thank you!!

[safe radish]

@slate gale Has your question been resolved?

[frozen veldt]

I haven't looked terribly closely at the particular function you have but in general a good idea is to solve for when the gradient function equals zero

and the gradient function is a function that consists of the partial derivatives

so essentially you want both of the partial derivatives to equal zero and that should give you a system of equations

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[stone crypt]

Why is it that when we solve for (x^2 -1 ) and get -1 the limit is approaching -1 from the positive side

[final halo]

x^2 - 1 is always bigger than or equal to -1

so we must be approaching from the positive side

[stone crypt]

but aren't we finding the limit as x approaches 0 which technically will be less than -1

because say we use -0.0001^2 - 1 with 0.0001 being x

[final halo]

x^2 is always positive or zero, x^2 - 1 will never be less than -1

[stone crypt]

ohh okay

that makes sense

thank you

.close

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[flat frigate]

qingfengmingyue

[broken yew]

question?

Ok firstly, I'm going to be a bit picky on notation

So its about how you use that mod n

The way you've written it is uh... a bit like how a CS student would

,,2+3\equiv 0\pmod 5

oops

[broken yew]

now the problem is

you cant really use this notation I suppose?

because this notation kindof assumes youve proven its a group and whatnot

im just tryna clear that up in my mind but yeah

[broken yew]

In any case, I don't like this notatoin.

For 10a, I probably would've gone contradiction/contrapositive approach

Suppose x * y is outside Z_n

Then show its not the least non-negative remainder

Oh wait shit i just realised

[broken yew]

Ok yeah, that notation is 'ok' given how they defined the operation in the question 😒 (ew)

but thats not how modulo arithmetic is usually introduced in group theory

😒 😒 😒

Which parts of group theory have you done up to this point. Ill have a look at b

i still don't feel like you've proven commutativity/assoc properties properly

this is the approach and what I would rely on to do it

thats alrightt hen

xy = yx = qn + r

wouldve been commutativity

justifying its the same r

associativity is slightly more pain.

=================
moving onto b

Am finding it hard to read - you happy about the identity element you found?

its just 1 I hope

yh I think you should rewrite it making that clear

Your proof should just prove 1 is the identity

oh nvm u put the associativity proof first no wonder i was confused

[broken yew]

but yeah still using this

x . 1 = 1 . x = x = 0n + x

associativity comes from part 10a, no need to restate it.

sure.

[flat frigate]

qingfengmingyue

[broken yew]

So bezout's won't make much sense til you figure what variable needs to be what

ax + by = 1

I would try writing out the final result you're after and maybe it'll be clear.

again, use this

what must x^-1 satisfy

[broken yew]

still related to this

x' * x = 1 yes

using that operation *

but how to express in the standard operations in Z

[broken yew]

You need this

what is x' * x

sorry

what is x * y

how did we define it

the result is r

where q is taken as big as possible

keeping r the smallest

ok?

r is the least non-neg rem

so now write that for x, x'

(im using x' in plaintext as easier, but obvs dont in handwriting)

xy = qn + r -> x * y = r

its not necessarily 0n

for example integers mod 5

2 and 3 are inverses

2 . 3 = 1 . 5 + 1 -> 2 * 3 = 1

yeh

x' . x = k . n + 1 for some k in Z
-> x' * x = 1

now do some rearranging

lets write x' as y for now so tis easier

xy = kn + 1

rearrange it so it looks like bezout (wont be same letters)

right

xy + (-k)n = 1

We've done a backwards proof basically

Now you need to start with bezout, and work forwards

to show what x^-1 must be.

but you have all the steps

yes

if its associative in some superset

it must also be in a subset

you can see this with quantifiers

(forall a, b, c in X) something

implies

(forall a, b, c in subset of X) something

=====

ig u can justify this with a quick one-liner but yeah

basically

but not really

you've miswrote it

So we defined our operation in a pretty convoluted way

x * y is the least non-negative remainder when xy is divided by n

y * x is the least non-negative remainder when yx is divided by n

but xy = yx

therefore x * y = y * x

==========
You could also define:

x * y is the smallest possible non-negative r of the integer tuples (k, r) which solve xy = kn + r

I think this works???

but in any case, this definition is by no means nice

well you see in all these definitions and proofs I'm writing, I've avoided saying xy mod n

Its convoluted if we want to be precise with our words

this has right idea but isnt precise because it misses details about k, r

[broken yew]

(forgot r could be 0)

yeah

[broken yew]

these 2 lines were the most clear for commutativity

But you will need the algebra for associativity

oh im high maybe lemme reread

[broken yew]

This still works, because its impossible to get remainder 0

with the set 1, 2, 3, ..., p-1

under division by p

And that actually needs to be part of your proof in b

To show this operation is well defined on U_p

====================================
x * y is the least non-negative remainder when xy is divided by n
y * z is the least non-negative remainder when yz is divided by n

(x * y) * z is...
x * (y * z) is...

Yeah actually this is the best way, I just realised.

maybe

actually maybe not nvm - u defo need the algebra

=========================
x * y is the least non-negative a of the integer tuples (p, a) which solve xy = pn + a
y * z is the least non-negative b of the integer tuples (q, b) which solve yz = qn + b

(x * y) * z is the least non-negative c of the integer tuples (s, c) which solve az = sn + c
x * (y * z) is the least non-negative d of the integer tuples (t, d) which solve xb = tn + d

xy = pn + a -> xy - pn = a

then substitute that back in.

is the approach

bruh this is painful 😒

uhhhhhhhhh

So x(yz-qn)=xyz-xqn=tn+d

Thats half of it

what about the other substitution

xy = pn + a -> xy - pn = a
yz = qn + b -> yz - qn = b

you use both of these to show (x * y) * z = x * (y * z)

😵‍💫 i need 5-10 mins lmao

You get these equations after substituting. Can you rearrange them so that they're both xyz = ???

(xy - pn)z = sn + c
x(yz - qn) = tn + d

good good

a lot of letters

[broken yew]

but you can carefully read the formulations we had

to figure out how to justify c = d

without referring to mod anything

(s+xp)n+c

(t+qb)n+d

essentially u gotta show s+xp = t+qb

[broken yew]

to show that, u gotta show this

The way to think of it is --- they will be equal, because you are 'able to make them equal'

xp and qb will be some random numbers. You are able to pick s and t to be anything

to make c and d as small as possible

[broken yew]

^

maybe there was some simpler way to do all of this, but yeah, idk honestly. That definition sucks 🙄

not really.....

[broken yew]

smallest possible while taking into account s and t are integers

ngl im half lost in proof as well at this stage sigh

But basically yeah you do have to justify c = d

and the original question did ask you to do it using associativity property in Z

😢

The other thing I'm gonna say at this point is that this proof is not terribly important.

In terms of revising/your course

like dw too much about it

The important part was bezout

which we did

modular arithmetic is usually defined through group quotients

which is much nicer to use

i think u need to do that as well technically

luckily this ones quick

[safe radish]

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[lean otter]

Show that if $G = (V,E)$ is a simple graph with $n$ vertices and more than $\ds \binom{n-1}2$ edges than $G$ is connected

[lean otter]

Been grinding my gears for this one

Really can't figure out what the hell I'm supposed to do

[solar hazel]

pigeonhole principle?

[lean otter]

How exactly?

[minor goblet]

not sure about the (n-1) instead of (n) but if it was n you could just say that you have n vertices and you need to choose 2 per edge

so nC2 edges in a graph will obviously have one edge at least per choice of two vertices

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[vale elbow]

tan(-5π/6) = tan(-pi + pi/6) = tan(-180+30) = tan(-150) =tan(210) = tan(270-210) = tan(60) = sqrt 3 I am trying to find the exact value of this and dont know where i went wrong please help
p.s I checked and the answer is meant to be 1/sqrt3

[safe radish]

@vale elbow Has your question been resolved?

[vale elbow]

<@&286206848099549185>

[sly hollow]

Try expanding tan(15-90) in terms of $\frac{sin}{cos}$ and then expand sin and cos expressions using known indentities.

[flat frigate]

NZzska

[sly hollow]

at least that's what I'd do

[vale elbow]

i see

thanks @sly hollow

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[upbeat ridge]

Can someone help me with this?

How can I answer in exact value when sin(π/8) doesn't give an exact value?

[lean otter]

there's a certain double angle identity to be used here

[upbeat ridge]

(cos(x))^2?

[lean otter]

nope

but it has something to do with cos

[upbeat ridge]

I'm confused 😢

[lean otter]

i suppose you haven't been taught the identity but you're supposed to use cos2x=1-2sin^2x

x here is pi/8

[upbeat ridge]

Ohhh

I forgot that existed

[lean otter]

so all of this turns into cos(2*pi/8) or cos pi/4 which is 1/sqrt(2)

[upbeat ridge]

Then it would become cos(π/4)!

[lean otter]

that's right

[upbeat ridge]

yaaaay

<@&286206848099549185>

What was the command for trig identities?

!rocket

!help

[safe radish]

Please read #❓how-to-get-help

[safe radish]

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[jovial scroll]

here, i think the most valuable knowledge to have is where 5pi/3 is on the unit circle. In other words, Q1, Q2, Q3 or Q4. I cannot visualize it however, easily, where 5pi/ is on the unit circle. I can only think if it was 6*pi/3, it wouldve been 3pi, which is out of the unit circle lol. Any advice

[quasi bison]

6pi/3 is 2pi not 3pi

and 3pi isnt out of the unit circle it is simply at the same point as 1pi

[jovial scroll]

oh yeah lol

my bad too quick to say that

but still

is there any good way to visualize where 5pi/3 is on the unit circle

besides memorization

like i know the values up until pi over 2 in my head, just where to place the rest

[quasi bison]

pi/3 is one sixth of the way around the unit circle

so 5pi/3 is five-sixths

it'll land in the fourth quadrant

[jovial scroll]

nice, i like that

and every value pi/x , we can see the progress as 1/2x

so pi/3 = 1/6 progress

andthus pi*5/3 = 5/6 progress

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[sour patrol]

This answer of mine does not feel right. I feel like it doesn't take into consideration if one of the 10s is a spade. Thoughts?

[sour patrol]

<@&286206848099549185>

[dapper venture]

what's your thought process?

[sour patrol]

Sorry, I was missing this part

there are 6 cards that are spades. there are 4 cards that are 10s. there are 21 remaining cards. i need 2 spades, 1 ten, and 2 other cards to make up 5 cards in a hand

@dapper venture

[dapper venture]

yeah you didn't considerate the case when you have a spade 10

try breaking it to 2 cases

case 1 with spade 10 and case 2 without spade 10

[sour patrol]

In that case... C(3,1) C(1,1) C(22,3) plus C(3,2) C(6,1) C(21,2) seems to be right maybe?

[dapper venture]

3?

[sour patrol]

explanation: the first is the case where i need 1 ten that isn't a spade, a 10 of spades, and 3 other cards. the second is where i get 2 tens (not spades), a spade, and 2 other cards

[sour patrol]

which 3 are you questioning?

[dapper venture]

ok

nvm that

[sour patrol]

in the second part, shound the C(6,1) actaully be C(5,1) because I dont want one of the spades considered? (the ten of spades)

[dapper venture]

should be C(5,1) instead of C(6,1)

[sour patrol]

are we referring to the same thing?

[dapper venture]

yeah

it look fine

[sour patrol]

haha. ok.

thanks for the help @dapper venture do you see any other flaws in this?

[dapper venture]

no

[sour patrol]

cool. thanks!

.close

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[tall crow]

I'm working with matrixes yeah, and I've been told to add multiple matrixes/subtract matrixes which do not have the same dimensions

[tall crow]

Do I just say "Not possible?"

[burnt notch]

Yes

[tall crow]

Thank you

How do I find the matrix of cofactors and adjoint of a matrx? @burnt notch

[burnt notch]

I haven't seen those for years, I don't remember how to do that actually

[tall crow]

Oh

[sage shore]

it's quite a procedure haha

check out this youtube video, about 7 mins only

https://www.youtube.com/watch?v=YvjkPF6C_LI

[burnt notch]

Yes, I know but I don't remember the steps lol

[sage shore]

This video should be enough to illustrate the steps required!

[tall crow]

This says finding the inverse is it the same thing with cofactors and adjoint?

[sage shore]

Yes, to compute the inverse, you need the adjoint

And to compute the adjoint, you need the cofactors

[tall crow]

Interesting

[sage shore]

The three are related, inverse adjoint cofactor

[tall crow]

Can I leave this open till i'm done?

[sage shore]

7 mins only

[sage shore]

Sure!

$$A^{-1}=\frac{Adj(A)}{det(A)} \hspace{2cm} det(A)=|A| \ne 0$$
$$With: , Adj(A)=[ , Cof(A) ,]^{T}$$

[flat frigate]

LUNA

[tall crow]

Quick question, if I'm multiplying 2 matrices where the columns of matrix A is not the same as the rows of Matrix B can I still multiply?

[sage shore]

No D:

[tall crow]

Ah, okay I was scared

So what do I write? Not possible again?

[sage shore]

Yeees

$$A(m , \times , n) \hspace{2cm} B(p , \times , s)$$
$$You , can , multiply , A , by , B , only , when , n=p$$
$$And , the , resulting , matrix , will , be , (m , \times , s)$$

[flat frigate]

LUNA

[tall crow]

So

I have Matrix A(1x1) and Matrix B(1x3)

Does that mean I can only multiply A x B

but not B x A

[sage shore]

Exactly!

Matrices multiplication is not commutative

it's not like real numbers, 2 x 3 or 3 x 2 same thing

[tall crow]

Okay okay nice, just like cross product of vectors?

[burnt notch]

Yes

But cross product is a bit nicer because it is almost commutative, it's only off by a sign when you swap the two vectors (we say it is anticommutative)

[tall crow]

Interesting

I might've run into a bit of an issue

[sage shore]

What

[tall crow]

So I'm multiplying thee two matrices

[mental needle]

the length of rows of first matrix and the length of columns of second matrix must be equal

[tall crow]

I understand it's row multiplies column

So when I do 3x1

Do I do 3x2

or 5x2

[worthy hemlock]

Across the row, down the column

[sage shore]

I know lol, i was telling him that matrices multiplication is not commutative like real numbers multiplication

[worthy hemlock]

So this

[sage shore]

These were not dimensions, but real numbers

:p

[mental needle]

was supporting your argument

with a clear image

[tall crow]

Okay so each number takes another

What is the case when one number in a row takes the entire column?

When it's a 3x1 matrix and 1x3?

[burnt notch]

Yes that's an example

[tall crow]

What?

It would be 3x3

[burnt notch]

No, a 3x3 matrix

[tall crow]

Do i keep going forward?

Like my 5x2 will be 10

[burnt notch]

If it was 1x3 times 3x1 then yes it would result in a scalar (a.k.a. a 1x1 matrix)

[mental needle]

[1]
[2] x [1 2 3]
[3]

Answer would be a 3x3

[burnt notch]

No

[sage shore]

No

:p

[worthy hemlock]

No, 3 x 3

[tall crow]

That's if it's 1x3 and 3x1

[tall crow]

So what do I do from here 😭

After I complete 3x1 and 5x2

[worthy hemlock]

This

Continue the process

[tall crow]

Do I go to the next row?

Second row and second column

[mental needle]

yea 3x3

i was mixing things p a bit

[worthy hemlock]

Yes

Well, depends

[burnt notch]

Your result needs to be a 3 x 4 amtrix in the end (just to check the answer in case you need)

[tall crow]

I know iot will be 3x4 I just don't know how to arrange it

I've multiplied all numbers in the first row by the numbers in the first column

[burnt notch]

Continue the process you began

[tall crow]

Do I go to the second row and second column?

[burnt notch]

No, you first have to multiply the first row with all the columns. Having done that, you can go on with the second row

[tall crow]

oh wow

So I start from 3 again

3x5

and 5x1

Wait no

[tall crow]

I always have to add?

[worthy hemlock]

The row times columns would be the element in the new matrix. Because you know it's a 3 x 4, the first row by the second column would result in element (1, 2) of the new matrix

[worthy hemlock]

That's how matrix multiplication works

[tall crow]

So the next step is 3x4 and 5x3

Then 3 x 5 and 5 x 6

[worthy hemlock]

Do you notice how the multiplication of the first row in the first matrix, times the first column in the second matrix results in the element (1, 1) in the new matrix?

[worthy hemlock]

Same logic for the rest, if you look at that image, first row by the second column is element (1, 2) in the new matrix, etc

[tall crow]

I think I understand this and the one you posted dldh06

So when I finish one roqw by column

I add it up

and the do the column with the same row

[worthy hemlock]

https://youtube.com/watch?v=vzt9c7iWPxs
If it helps watch this

[mental needle]

when you have free time, you can also explore something a little advanced but more fun
Strassen Algorithm (You won't use it for now)

[tall crow]

Okaky

,rotate

[tall crow]

@mental needle @worthy hemlock

so like this?

[worthy hemlock]

Yep

Looks good

[tall crow]

I have Symbolab premium but for some reason I can't solve Matrix questions with it

I can only use the examples

[worthy hemlock]

I like using this site, www.matrixcalc.org

[tall crow]

,rotate

[tall crow]

Do i have to add?

[worthy hemlock]

No

[tall crow]

Why don't I have to add for this scenario

[worthy hemlock]

Because it's one element times the other

It's just 2 * 3

There's no more elements in the column to sum

https://matrixcalc.org/
That site btw is good to use

It shows the work

[tall crow]

Interesting

thanks

So it's always

One row multiplies an entire column

Then you add

Then you repeat it w the remaining columns

[worthy hemlock]

Yeah

[tall crow]

@sage shore

[sage shore]

wassup

[tall crow]

The vdieo you gave me I've completed it and got the inverse

How do I know which ones are the co factors or adjoint

Should I send a pic?

[sage shore]

The cofactor matrix is the first matrix you compute

The one which resulted from the 9 little determinants with signs next to them

That's the cofactor matrix

[sage shore]

The Adjoint matrix is its transpose

Keep diagonal of cofactor matrix, and swap others

[tall crow]

okay

That's the adjoint

?

[sage shore]

Yes

The adjoint is the transpose of the cofactor

[tall crow]

So the matrix I have with the -48 and +6 and +12

etc

Those are the cofactors?

[sage shore]

Yes, that is the cofactor matrix

Transpose it for the adjoint

[sage shore]

Noice

But why don't you name them, to be clear in your answer sheets

$$A^{-1}=\frac{Adj(A)}{det(A)} \hspace{2cm} det(A)=|A| \ne 0$$
$$With: , Adj(A)=[ , Cof(A) ,]^{T}$$

[flat frigate]

LUNA

[sage shore]

You can call the first Cof(A)=...

And the second Adj(A)=...

[tall crow]

Okay

I don't klnow why this is under Business Administration

I have to simplify the fractions yeah? @sage shore

[sage shore]

As you wish, or as your teacher does..

Some would divide all the entries by 60 and simplify

[tall crow]

All of them I suppose

[sage shore]

Some (including me) would leave it as (1/60) Adj(A)

[tall crow]

Well this has been fun

Somehow finding the inverse is easier than multiplication I think

How do I know which numbers to switch when I'm transposing the Cof(R)?

[sage shore]

No way!

[sage shore]

The guy has shown you in the video

[tall crow]

No I mean in other examples

Is it the same proces?

[sage shore]

Yeah, diagonal stays the same

And the "mirrors" numbers get swapped

kind of mirror the numbers which are not in the diagonal

Also, transposing a matrix is making its rows its columns, and its columns its rows

Yeah, this is more common

[tall crow]

I see

When I was finding the cofactors and I found those mini determinants, why didn't I put the big number outside

Like when you're normally finding the detemrinant there's a number outside no?

[sage shore]

Because those are not exactly determinants

More like the minors

They are called minors

[tall crow]

oh I see

So when finding minors for the co-factors I do not have to put the number outside

[sage shore]

And when you add the alternating sign, they become called cofactors

[sage shore]

Yes, no need!

[tall crow]

Ah yes the signs

s it always + - +, - + - and + - +

[sage shore]

yes, alternating

[tall crow]

What happens if the determinant of a matrix is 0?

[sage shore]

Then the matrix does not have an inverse

[tall crow]

I'm not trying to do the inverse right now

I'm trying to do Cramer's Rule

To find the values

of 3 variable equations

[sage shore]

Then Cramer's rule cannot be applied

[tall crow]

Interesting

[sage shore]

Because either there is no solution

Or infinity of solutions

Not one unique solution that you can find using Cramer's rule

[tall crow]

So did I make a mistake or not?

Look at D_x1

[sage shore]

Waiiit, nooo

Cramer's rule is not applicable when D=0

Not Dx1 and Dx2...

Dx1, Dx2... can be 0, it's okay

Here, D=-12, so it's okay to proceed

[tall crow]

Oh i See

[sage shore]

You did not make a mistake, it is 0

So, x1=0

[tall crow]

Alrighty

Thank you so much

.close

[safe radish]

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[pine dove]

I don't understand how the answer is p = 0.002. Also, I'm unsure what table to use

[safe radish]

@pine dove Has your question been resolved?

[pine dove]

<@&286206848099549185>

<@&286206848099549185>

[pine dove]

.close

[safe radish]

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[hoary breach]

how do i even start this im so lost

mb 😭

[uneven epoch]

it's fine🥰

[uneven epoch]

im stuck and i think i did it wrong lol

[pseudo scroll]

What happened between lines 2 and 3

[uneven epoch]

idk i tried changing the denominators to add them up

but they didnt work

[pseudo scroll]

[uneven epoch]

i just noticed a mistake💀

sry

lol

lemme try again

ok nvm i still cant get it @pseudo scroll

[pseudo scroll]

hmm this is quite the problem

,w cot(x) + 6sin(x) - 2cos(x) = 3

[flat frigate]

Results provided by WolframAlpha

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

Requested by neonperseus

[pseudo scroll]

Apparently it has decent solutions

[uneven epoch]

how😭

[calm bridge]

how is cotx = cosecx - 1

[uneven epoch]

ik i just said i made a mistake

[calm bridge]

oh ok

[uneven epoch]

but even after that i still cant solve it🤡

[calm bridge]

lemme try

yeah well, just just cot to cos/sin

and multiply whole equation by sinx

and then you can easily factorise the equation

[pseudo scroll]

Ofcourse you can

[uneven epoch]

bruh i just realised my teacger gsve thr working

😭

im so stupid omg

[calm bridge]

[uneven epoch]

thx for ur help guys

im so sry

[calm bridge]

no problem!

@uneven epoch if you don't need any further help, type .close

[uneven epoch]

.close

[safe radish]

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[safe radish]

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[limber pine]

back. i need some help for question 2

[tepid walrus]

!status

[safe radish]

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

[limber pine]

2

[tepid walrus]

What have you done?

[limber pine]

ill send

i know something definitely went wrong here though

[tepid walrus]

Wait but, you sent #3 no?

[limber pine]

yeah

its the same question

i was just doing other questions before it

hence 25

[tepid walrus]

Just to be sure, are you trying to do 12.2 or 12.3?

In the first image you sent

[limber pine]

12.3

[tepid walrus]

I do see a small mistake in your answer here

[limber pine]

its not my answer

its a step but

[tepid walrus]

when you brought the -2 inside the log to the exponent you forgot the -

[limber pine]

i felt something was wrong so i didnt continue

[limber pine]

oh

lemme look

[tepid walrus]

but your general idea is right

you're slowly isolating y

[limber pine]

so is the quadratic part supposed to happen?

[tepid walrus]

depends on the way you're doing it, but in the end, it will give the same answer

I recommend you don't factor the polynomial though

And also, you replaced y with x, but didn't replace x with y, so you have x'es on both sides which is not what you're supposed to do

[limber pine]

oh my bad

bad*

so (y+2)(y+2)

wait

factoring it was a waste of time

i just got what i had before

[limber pine]

a step 3, where its x-2 = log-2(y-1)^2

can i just distribute the -2?

i thought i couldnt, which is why i factored it

but when i factored it, i ended up getting what i wouldve gotten if i just distributed the 2

[tepid walrus]

Why is there a - inside the log?

[limber pine]

as in the (x-1)?

[tepid walrus]

no, as in -2()

[limber pine]

i dont know

thats just what the question said

[tepid walrus]

$y = -2\log(2(x-1)) + 2$

[flat frigate]

imTypθ

[tepid walrus]

There is no -2 inside the log?

[limber pine]

oh that

wait a minute...

the whole thing is wrong now

because -2log2(x-1) = log2(x-1)^-2

negative exponent

[tepid walrus]

yes

[limber pine]

so what now?

[tepid walrus]

$x -2 = \log((2(y-1))^{-2})$

You have this right?

[limber pine]

i guess i should apply the negative exponent property

[flat frigate]

imTypθ

[tepid walrus]

Yeah, you can distribute the exponent

[limber pine]

1/2(y-1)^2 right?

on the right hand side

[tepid walrus]

sure. Be careful because the 2 is inside the exponent too

[limber pine]

oh

so if theres an exponent and log like: clog_a(b)^n, then the exponent would apply to the entire log including c right?

just to clarify

so the right hand side would be 1/(4(y-1))^2 right?

[tepid walrus]

looks right

[limber pine]

ok

then distribute the 4?

?

[tepid walrus]

Just do your usual algebra to isolate y.

[limber pine]

ok

i dont think i can do it

ill just not do this question and go to another

[tepid walrus]

what? Why? I think you can

[limber pine]

after the step i mentioned earlier, i did:

10^x-2 = (4y-4)^2
but since i didnt know how id get y on top, i had to bring it to the left hand side

(4y-4)^2) 10^x-2 = 1

wait

ill try to solve this further

wait, i cant

and this is looking nothing like the answer on the sheet

so i cant do it

[tepid walrus]

You can...

I know exactly why you have a different answer, I'm just waiting for you to get there

What step are you currently on?

[limber pine]

step 7 ^

[tepid walrus]

$10^{x-2} = \frac{1}{[2(y-1)]^2}$

[flat frigate]

imTypθ

[tepid walrus]

That's where I left of. What did you do next?

[limber pine]

10^x-2 = 1/(4(y-1))^2
(4y-4)^2) 10^x-2 = 1

[tepid walrus]

Sure, that's one way of doing it.

What do you want to do next?

[limber pine]

i dont know what to do next

the only thing i can think of is divide both sides by 10^x-2

[tepid walrus]

well that's a good start

[limber pine]

but how am i going to get y by itself?

[tepid walrus]

$4(y-1)^2 = \frac{1}{10^{x-2}}$

[flat frigate]

imTypθ

[tepid walrus]

Do you not see the way to isolate y?

Also, you can simplify the RHS so that it looks more like your answer key

[limber pine]

(y-1)^2/4?

on the lhs

[tepid walrus]

what you do to one side you do to the other

But yes, dividing by 4 is the way to go

[tepid walrus]

However, I'd consider doing this first

[limber pine]

is it 10^-(x-2)?

[tepid walrus]

yep!

good job finding that one

[limber pine]

(y-1)^2 = 10^-(x-2)/4

[tepid walrus]

$(y-1)^2 = \frac{10^{-(x-2)}}{4}$

[flat frigate]

imTypθ

[tepid walrus]

You're almost there

[limber pine]

ill try something

ok nevermind

the only way im thinking of to get rid of the square is to do a sqroot

but the answer has no sqroot

[tepid walrus]

It has one, but it's kind of hidden. How else can you represent the square root?

[limber pine]

^1/2

[tepid walrus]

There you go!

[limber pine]

y-1 = (10^-(x-2)/4)^1/2

y = (10^-(x-2)/4)^1/2 + 1

but thats still nowhere close to the answer unfortunately

[tepid walrus]

it is, you just don't see it :) What can you do when you have a fraction raised to an exponent?

[limber pine]

you multiply it by that power?

so if the fraction is raised to 2, you multiply it by itself

but this is a half exponent so

wait a minute

nevermind. i was thinking to use the exponent property of (a^m)^n = a^m+n

m in this case being -(x-2)

[tepid walrus]

well, you will have to use that, but you confused two properties

$(a^b)^c = a^{b\cdot c}$

[flat frigate]

imTypθ

[tepid walrus]

But what I was talking to you about was this property

[limber pine]

oh my bad

[tepid walrus]

$\left(\frac{a}{b}\right)^c = \frac{a^c}{b^c}$

[flat frigate]

imTypθ

[limber pine]

thatd make it 10^-(x-2)^1/2 / 4^1/2)

[tepid walrus]

right. And i am sure you can evaluate the denominator

[limber pine]

yeah. itd be 2

[tepid walrus]

correct. As for the numerator, you can use this property

[limber pine]

alright

[tepid walrus]

So, does the answer look more like your answer key now?

[limber pine]

kinda, but

the exponent for the numerator,

since its raised to the half power,

-(x-2)^1/2 = -(-0.5x-1)?

[tepid walrus]

yeah. Multiplying by 1/2 = dividing by 2

[limber pine]

yeah

[tepid walrus]

however, your answer key doesn't do it, it just leaves the /2 there

[limber pine]

Wait a minute

I think I know how to get the answer now

Instead of doing (0.5x + 1) [since there’s a negative outside the bracket]

i can just do 1/2x + 1

[tepid walrus]

careful distributing the negative

[limber pine]

ok

so to get rid of the fraction

i can do x + 1(2) maybe?

[tepid walrus]

i'm not sure what you're trying to do

$y = \frac{10^{-\frac{1}{2}(x-2)}}{2} + 1$

[flat frigate]

imTypθ

[tepid walrus]

I think that's what you have?

[limber pine]

i actually didnt have it like that. i shouldve wrote it down

i think the exponent should be (-1/2x + 1) then?

[tepid walrus]

Sure, but I don't why you don't just leave it at that?

You can distribute the 1/2 in if you really want to, but it's not mandatory

[limber pine]

oh

well, its kind of because of this answer, but i think it probably has the same value as this

[tepid walrus]

it does have the same value. Just distribute the negative and you will see

[limber pine]

but when i distributed the half, i got (-1/2x + 1)

[limber pine]

i think ill just stick with this answer, to be honest

[tepid walrus]

I said distribute the negative, not the half. See how the answer key keeps the /2 outside the brackets?

[limber pine]

oh

[tepid walrus]

sorry if that wasn't clear

[limber pine]

ohh i think i get it now

[tepid walrus]

:D

[limber pine]

since its raised to the half, you can just write it as 1/2(10^-(x-2)

then when you distribute the negative

itd be (-x+2) in the exponent

so the exponent, if rearranged, would be (2-x)

this was one heck of an inverse function...

thanks for the help

[tepid walrus]

no worries! Those will become easier with practice :)

[limber pine]

yeah. ill keep doing them

.close

[safe radish]

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[patent musk]

is this correct here?

[marsh walrus]

looks good to me

any part youre worried about specifically or confused about?

answers all look correct

[safe radish]

@patent musk Has your question been resolved?

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[blissful phoenix]

Каква е вероятността Б да се случи, в предположение, че А не се е случило и Б може да се случи само ако А не се е случило?

English:
We have two events A and B. B can only happen if A did not.
P(A) = 0.2
P(B) = 0.1
P(B | not A) = ?

[tall crow]

брат, това е английски сървър, имаш ли екранната снимка на въпроса?

[blissful phoenix]

Знам, но знам, че може да има и човек, който да ми помогне. Въпросът не е от някаква задача, а мой собствен.

[tall crow]

добре, проблемът с въпроса ви е, че трябва да посочите вероятностите of a и b да се случат или не, преди да можем да го изчислим вместо вас

[blissful phoenix]

Вероятността може да е всякаква, но нека кажем, че на А е 0.2, а на Б е 0.1
Просто бих искал да знам правилната формула

[tall crow]

ще оставя въпроса ви на английски, за да може някой да помогне

[blissful phoenix]

Няма нужда, аз говоря английски също

[tall crow]

He is asking what is the probability B will happen assuming that A did not happen and B can only happen if A did not happen?

<@&286206848099549185>

[tall crow]

о, не знаех

[safe radish]

@blissful phoenix Has your question been resolved?

[safe radish]

@blissful phoenix Has your question been resolved?

[dull mesa]

1?

едно

[blissful phoenix]

How did you calculate it? What's the formula?

[dull mesa]

Ако A не се е случило, тогава вероятността B да се случи може да се определи чрез намиране на условната вероятност за B при условие, че A не се е случило. Това може да бъде представено като P(B | не A). Тъй като B може да се случи само ако A не се е случило, можем да приемем, че P(B | не A) = 1, което означава, че ако A не се е случило, тогава B със сигурност ще се случи.

English version:
If A did not happen, then the probability of B happening can be determined by finding the conditional probability of B given that A did not happen. This can be represented as P(B | not A).

Since B can only happen if A did not happen, we can assume that P(B | not A) = 1, meaning that if A did not happen, then B is certain to happen.

[blissful phoenix]

That doesn't sound right. The probabilities of A and B I gave are 0.2 and 0.1.

[tall crow]

Does 1 not mean that it will always happen?

[tall crow]

I feel it is 0.8

[dull mesa]

Oops sorry I didn't use the probabilities

If the probability of A is 0.2 and the probability of B is 0.1, and it is assumed that B can only happen if A did not happen, then we need to calculate the probability of B given that A did not happen.

If A did not happen, this means that A's complement (not A) happened. The probability of not A happening is 1 - 0.2 = 0.8.

Therefore, the probability of B happening given that A did not happen is 0.1.

P(B | not A) = 0.1

I feel it's also 0.8

Since the probability of A not happening= three probably the B happening

[blissful phoenix]

Still doesn't sound right. You didn't say why this is the case:

Therefore, the probability of B happening given that A did not happen is 0.1.

[dull mesa]

I don't think B is supposed to be assigned a probability

Since the probability of a not happening is already B

I said it was 0.1 because you gave me 0.1

[blissful phoenix]

The probability of not A happening is 1 - 0.2 = 0.8.
You said this yourself. P(B) = 0.1

[dull mesa]

Sorry I didn't think much of it

$P(B)=0.8$

[flat frigate]

TheHermit

[blissful phoenix]

That's ok. I appreciate the effort but can we leave this to someone who knows more about probabilities.

[flat frigate]

Jangler

[daring lily]

With the given information, we know that $P(B \cap A) = 0$ has to hold (as B can't happen if A happens)

[flat frigate]

Jangler

[daring lily]

Use $P(B) = P(B \cap \bar{A}) + P(B \cap A)$ and you get an expression for your wanted probability

[flat frigate]

Jangler

[blissful phoenix]

So since $P(B \cap \bar{A}) = P(B)$ can we say $P(B | \bar{A}) = \frac{P(B)}{P(\bar{A})} \implies P(B | \bar{A}) = \frac{0.1}{1-0.2} = \frac{0.1}{0.8} = 0.125$

[flat frigate]

Velizar

[daring lily]

Exactly

[blissful phoenix]

Ok, nice. So I checked the wiki page on conditional probability and can see where your first formula came from and the one after it also makes sense but where did the third one($P(B) = P(B \cap \bar{A}) + P(B \cap A)$) come from?

[flat frigate]

Velizar

[daring lily]

$B = (B \cap A) \cup (B \cap \hat{A})$ \
$(B \cap A)$ and $(B \cap \hat{A})$ are disjoint, so the probability of their union is equal to the sum of the probability of both sets. (Third Kolmogorov axiom)

[flat frigate]

Jangler

[safe radish]

@blissful phoenix Has your question been resolved?

[blissful phoenix]

Makes sense but how do we know the equality $P(B) = P(B \cap \bar{A}) \cup P(B \cap A)$ is true? We know $P(B \cap A) = 0$ but how can $P(B) = P(B \cap \bar{A}) = P(B) * P(\bar{A}) = 0.1 * (1 - 0.2) = 0.1 * 0.8 = 0.08$ be true? It seems necessary but weird to assume $P(\bar{A}) = 1$
What mistake am I making here?

[flat frigate]

Velizar

[safe radish]

@blissful phoenix Has your question been resolved?

[safe radish]

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[torn vessel]

if both of the function's domain is different why is the graph same ?

[tidal imp]

The graph is not the same everywhere

However—

Actually no, come to think of it

[torn vessel]

where is not same ?

[calm bridge]

cscx is never zero

[tidal imp]

Well, if you know what csc means

[devout shale]

1/csc(x) is equivalent to sin(x)

[torn vessel]

yea but the domain of it is different than that of sinx

[tidal imp]

No, the domain is all reals

[calm bridge]

cscx has a different domain

not the function 1/cscx

[torn vessel]

sinx's domain is R
1/cscx's domain is R-npi

[devout shale]

nope, it is R

x's domain is R, 1/1/x's domain is also R

that is really what you are doing here

writing 1/(1/sin(x))

this is just sin(x)

it has denom 1

it is defined over all of R

[torn vessel]

its written in my book 😳

the domain of cscx is R-npi

[devout shale]

of csc(x)

yes

but not of 1/csc(x)

Just like this

the domain of 1/x

is R-0

the domain of 1/(1/x) ?

all of R

it is just x

[peak estuary]

no

[devout shale]

yes

[dull sequoia]

Shouldn’t that still be R\0

1/(1/0) is still undefined

[devout shale]

disagree

[peak estuary]

if you cant calculate certain parts of an expression then how should it be defined there

[dull sequoia]

Not in the domain

[peak estuary]

1/(1/x) is not equal to x

[devout shale]

y=x is undefined at x=0 if we write it as y=x^2/x ?

this is dumb semantics

[dull sequoia]

^^

[peak estuary]

it is undefined at x=0

[dull sequoia]

Yes

[torn vessel]

so both of the domains are different right ?

[devout shale]

oh well then forget I said anything brist0l

[peak estuary]

you can continuously extend 1/(1/x) to x

but its a different function

with different domain

[dull sequoia]

The graph looks the same because you’re missing points

And Desmos can’t show you the points

Cos there are infinitely many points close to the ones missing

[torn vessel]

what does it mean " missing points" ??

[dull sequoia]

Missing as in they aren’t being graphed

[torn vessel]

so i need to zoom in to find those ?

[dull sequoia]

You can’t zoom in to find them

It will always be “unseeably” small

[tidal imp]

You have to glide over it to see the hole

[torn vessel]

yea lol

[tidal imp]

That’s the one weakness of desmos tbh

[torn vessel]

ayo