help-13

[sweet ridge]

But

Imma trying to conclude my understanding in these drawing...
Anything wrong?
Its about set by the way

[kindred storm]

You didn't label A and B in the second one.

If they're the smaller two, then it's the combination of the circles, not including anything outside the circles.

[sweet ridge]

Ogey

Anything wrong again?

[kindred storm]

Yes. You've labelled the union as some elements outside of A and outside of B.

[mild sundial]

union set is surely wrong

[sweet ridge]

Ogey (????)

[mild sundial]

It needs to be like this

[kindred storm]

The union doesn't include anything outside of A or B, so you probably shouldn't label the section that includes things outside A or B as the union.

[kindred storm]

No, that's also wrong.

[mild sundial]

also name what you're trying to show- the belongs to

[mild sundial]

it's correct

[kindred storm]

You're labelling lots of elements that aren't inside A or inside B as being in the union.

[mild sundial]

when did i label

[kindred storm]

You have a square.

You labelled the square as the union of A and B.

[mild sundial]

that's a way of showing it

[kindred storm]

The square contains lots of elements not in A and not in B.

[mild sundial]

yes

[kindred storm]

That's why it's wrong.

[mild sundial]

a n b sets belong to a universal set wym

I'm correct.

[kindred storm]

You're labelling the universe as the union, when it's not.

[mild sundial]

i ain't labelling it as a union

I clearly wrote

a U b

[sweet ridge]

Ill sit for a while trying to understand what is not my native language

[kindred storm]

You said A U B underneath a square, meaning that the square is A U B.

The square isn't A U B.

[mild sundial]

bruh

we r clearly speaking about set a and b

so a U b makes sense

[kindred storm]

I have no idea what you're talking about.

A U B doesn't mean something that also includes A U B. It means something limited to exactly A U B.

[crystal raptor]

You can't just draw two unshaded overlapping circles and then write A U B near it. You need to shade or properly indicate which part is A U B

[kindred storm]

You didn't limit it in any way to exactly A U B.

[mild sundial]

yes

[kindred storm]

You didn't shade A U B.

You didn't indicate in any way what A U B is.

[mild sundial]

bruh.

okay now I get where u going

[kindred storm]

You merely labelled a rectangle as A U B, when that rectangle includes more than A U B.

[mild sundial]

fine but I was just showing him u need to intersect the sets

[sweet ridge]

Okay...
So like...
U make a first circle...
Then it filled with 2 circle only "a and b"
So the first circle called union?

[kindred storm]

The union is the combination of all elements in A and in B.

If it's in A, it's in the union. If it's in B, it's in the union.

[sweet ridge]

So we cant draw it with circle???

[kindred storm]

The shaded area is A U B.

[sweet ridge]

Okay....

Even its seperated, it will be called union if it's told to be it?

Ummm

[kindred storm]

If they're separated, there's just no elements in the overlapping part.

They should be drawn overlapping to be more general.

Overlapping doesn't mean the overlapping part contains any elements.

But if you draw them apart, you're not handling situations where they both contain the same element.

[sweet ridge]

So overlapping them to make it look more "united"?

[kindred storm]

No, it's to work for all sets.

[earnest radish]

seeing a face

[kindred storm]

The sets that don't contain any common elements will merely have nothing in the overlapping part.

So, it can handle all situations.

[fervent jackal]

if it doesnt overlap then A & B is the empty set

[kindred storm]

But if you don't overlap them, you're not covering the situation where they contain any common elements.

[fervent jackal]

we make it overlap to make it more general

because most cases two events have an intersection

[kindred storm]

We're talking about unions.

[fervent jackal]

Ik

[kindred storm]

That's why unions are drawn like here: https://en.wikipedia.org/wiki/Union_(set_theory).

If you're trying to handle all possible sets.

If you're drawing it for two particular sets, you can do otherwise.

[sweet ridge]

Learning math

Ogey...

Im starting to understand

Btw thank you very much ! @kindred storm @fervent jackal @mild sundial

I hope we meet again!

[kindred storm]

No problem.

[sweet ridge]

.close

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[ionic urchin]

extremly lost

can someone help

[gentle flower]

looks like a trig substitution problem

[ionic urchin]

wait heres a better picture

[ionic urchin]

but yea...idk how to do that

[tropic oxide]

u := arccos(x) suggests itself

[ionic urchin]

...what that mean cuzzo

[tropic oxide]

the substitution u := arccos(x)

[ionic urchin]

yea ion think im that ahead bro

[cedar kiln]

@ionic urchin Has your question been resolved?

[raven shard]

(do you know what a u-substitution is?)

[ionic urchin]

i missed that class

prolly should learn that or can yall jus help

ill learn after too

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[frail cosmos]

Stuck in the first one I'll translate it

A, B, C and D are points on the circle
AB and CD meet together at point F.
<DAC=<DBC
<ACD=<BCD
Prove that DC is a diameter (already did)

Prove that AB is vertical to CD

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@frail cosmos Has your question been resolved?

[frail cosmos]

<@&286206848099549185>

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[frigid hemlock]

,

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[jovial glacier]

I have 8 coins, and I will pick 2 of them and add their values up. What can I set the denominations of the coins to be so I can get the most consecutive values starting from 0? (With one of coins being 1, and another coin being 0)

[jovial glacier]

I know the maximum I could theoretically get would be 36 or (8 * (8 + 1)) / 2. I solved it through a brute force search and got 0,1,2,5,8,11,12,13; which can cover every value from 0 to 26. How can I solve it without checking all 2 billion combinations?

0,1,2,3,4,5,6,7 can do 0-14
0,1,2,5,7,10,11,12 can do 0-24, which was the best I could come up with by guessing

[frail thistle]

i'm a little confused by your description of the problem, because assuming you can't pick up the same coin multiple times 0, 1, 2, 5, ... , 12 cant do 4

[cedar kiln]

@jovial glacier Has your question been resolved?

[jovial glacier]

You can pick up the same coin both times*

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[latent hazel]

Hi, is there any way to obtain 3x3 matrix that have 3 eigenvalues (1, 2, 3) and has 7 non zero elements?

[floral terrace]

!occupied

[cedar kiln]

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).

[sage forge]

Yes, there is!

Hint: You can choose A_12=A_13 = 0 and A_11 = 1

[latent hazel]

I tried something simmilar but I am always stuck with a case that determinant is -1

basically I do not know what to put in other places

[sage forge]

Ok, do you know what the advantage of this choice is?

[latent hazel]

no I do not

[sage forge]

If we want to calculate the Ev we need det(A-x*1)

Since there is only one non 0 entry in first row we can apply Laplace

Ok?

[latent hazel]

ok we have something like a polynom that needs to be zero

in order to get zero as determinant and have eigenspaces when working out kernell

but well I do not know what you mean by Laplace

[sage forge]

Laplace method to calculate the determinant?

If there is only one non 0 entry in a row the determinant can be easily calculated

[sage forge]

Never heard of that?

[latent hazel]

no never

I only heard about Sarrus

[sage forge]

Ok, no problem. I don't have a lot of time explaining the method to you though

[latent hazel]

Oh I see I believe Laplac is when you choose a line or column and you could calculate detminant as sum of sub determinants

am I right>

?

[sage forge]

Exactly

[latent hazel]

we just call it differently

[sage forge]

And here we go over the first row

Great that you have it

Can you apply it?

[latent hazel]

yes

now I have something like this [(1, 0, 0),(1, 2, 1),(1, 1, 3)]

[sage forge]

Doesn't work

[sage forge]

Ok, since I am a bit in a hurry: If you applied Laplace for Det(A-x1) for the given entries you would get (1-x)• some polynomial

[latent hazel]

ok thanks you helped me a lot

I will try rest on my own

[sage forge]

Ok, sorry that I couldn't explain in more detail. Good luck

Now there are only 4 entries to worry about

[latent hazel]

ok I belive I will manage

[latent hazel]

do not worry you were helpfull

[sage forge]

You will for sure 🙂

[latent hazel]

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[glossy sigil]

Where am I wrong

[cedar kiln]

@glossy sigil Has your question been resolved?

[glossy sigil]

<@&286206848099549185>

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@glossy sigil Has your question been resolved?

[rancid pollen]

Have you subbed in the initial conditions

[glossy sigil]

Yup

I did. that first

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@glossy sigil Has your question been resolved?

[bronze crag]

@glossy sigil Where in your equations do you use 1.1million?

[glossy sigil]

e^c=A

=1.1million

[bronze crag]

so then what did you get for c?

Normally you would expand "e^(.05t + c)" to "e^c * e^.05t"

then you can just use a different constant: y= Ce^.05t + 1million

that way solving for C is easier when you plug in your initial population: 1.1million = Ce^(.05*0)+1million

(since e^0=1)

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[versed fulcrum]

Hello

[versed fulcrum]

Does anyone know the second symbol name?

[mighty drift]

lambda, sigma, rho

[versed fulcrum]

Lowercase sigma?

[mighty drift]

lowercase yeah
Uppercase is the summation symbol

[versed fulcrum]

Oh I see

Thanks

[mighty drift]

it's also very frequently used to represent a permutation

or in probabilities for the variance or standard deviation (std I think, not sure though)

[dusty hazel]

variance is sigma^2

[mighty drift]

yeah that's what I thought

[versed fulcrum]

😮

.close

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[placid plank]

how does this work?

[placid plank]

is it just divided by log?

[tropic oxide]

no

"dividing by log" is an error so serious it's a real DROP EVERYTHING, THIS IS AN EMERGENCY situation

log(x) isn't the product of x with something called "log"

log is a function and not a number!

[magic solar]

10^LHS=10^RHS

[tropic oxide]

and here specifically,

log(x) = 0 => x = 1

[tropic oxide]

or that, assuming the log is implied as log_10

[magic solar]

Unless log refers to log base e

[placid plank]

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[tropic oxide]

doorslam!

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[maiden cloak]

I don't understand this. why is W a subspace of Rn?

[tropic oxide]

which step confuses you?

[maiden cloak]

W is a set containing vectors of an equal amount to the dimension of R. each a is a coefficient of each vector. if if there are some a(s) that let you set all the vectors to 0, its a subspace?

whats the "for all i, 1 <= i <= m" part for?

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[maiden cloak]

<@&286206848099549185>

[tropic oxide]

i think you misunderstand...

ok so like

you have here essentially a system of linear equations

a homogeneous one at that

the statement is that its solution set is a subspace of R^n

[maiden cloak]

ok

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[serene basin]

why is the hessian the jacobian of the gradient, rather than the jacobian of the jacobian? i know that the jacobian is the generalization of the derivative for functions R^n to R^m, but for multivariate functions R^n to R^1 for some reason we use the gradient instead (the transpose of the jacobian). i am confused generally about why this transpose is necessary.

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[inner solar]

let rotation be defined as this

[inner solar]

how can i write this rotation in a form of matrix?

(e_1,...,e_n) is standart basis

[upper abyss]

I'm not clear on the rotation that's happening here

Oh wait, where we let e2 be fixed?

We find the matrix for any transformation the same way:

• Plug the basis members into it
• The results we get form the columns of the matrix

[inner solar]

what do you mean?

can it be like this? @upper abyss

in each column i wrote mapping of that basis

yes it is correct

.close

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[dapper imp]

how do i solve this to the x1+x2+x3-b=0 form? (the topic is vectors)

is it 3x1-x2+x3-5=0 ?

[dapper imp]

arigatou

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[wraith dagger]

sofficino9188

LaTeX source sent via direct message.
```Compilation error:```! LaTeX Error: Unicode character ∈ (U+2208)
               not set up for use with LaTeX.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.65 so $c+(1/n)∈
                   A$, which is absurd because $c$ is the largest of all the...

You may provide a definition with```

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[merry nebula]

.close

.close

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[inner solar]

Prove that if $A$ is a regular matrix, then $A^T$ is also a regular matrix, and its inverse matrix $(A^T)^{-1}$ is equal to the transpose of the inverse of $A$, i.e., $(A^{-1})^T$, where $A^{-1}$ is the inverse matrix of $A$.

[wraith dagger]

Slowaq

[inner solar]

this isnt sufficient is it?

i did not show that A^T is regular matrix

how should i do that?

.close

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[zealous niche]

How do you solve this?

[zealous niche]

mhm?

[silver fable]

That's just the definition of e

[zealous niche]

ye what about it

like

how do you know it's equal to e

@silver fable

[silver fable]

It's just the limit definition of e

In general :

[zealous niche]

how do i prove it

[silver fable]

idk how to prove it, sorry

https://youtu.be/Q2HiFA5dnY0?si=E1mLl3WU-paEypSu

[silver fable]

i don't think u have to prove it
by definition, e is equal to that limit

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[primal stag]

Can somebody help me find the time complexity of this Dijkstra algorithm implementation in python?

def dijkstra(graph, start):
    previous = {v: None for v in graph.adjacency_list.keys()}
    visited = {v: False for v in graph.adjacency_list.keys()}
    distances = {v: float("inf") for v in graph.adjacency_list.keys()}

    distances[start] = 0
    pq = PriorityQueue()

    pq.add_task(0, start)

    while pq:
        removed_distance, removed = pq.pop_task()
        visited[removed] = True

        for edge in graph.adjacency_list[removed]:
            if visited[edge.node]:
                continue

            new_distance = removed_distance + edge.distance

            if new_distance < distances[edge.node]:
                distances[edge.node] = new_distance
                previous[edge.node] = removed
                pq.add_task(new_distance, edge.node)

    previous_str = {str(k): str(v) for k, v in previous.items()}
    visited_str = {str(k): v for k, v in visited.items()}
    distances_str = {str(k): v for k, v in distances.items()}


    print("Previous:", previous_str)
    print("Visited:", visited_str)
    print("Distances:", distances_str)

    return

[cedar kiln]

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[dire geode]

#old-network for CS server

Or find a python server

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

(-i)^2 = -1

but

(-1)^2 = 1

how is this broken down please? because idk if I am right here, I deduced this for the 2nd equality: (-i)^2 = (-1^2) * (I^2) = 1 * -1 = -1 is this right?

[storm seal]

i is the sqrt of -1

So when it's squared the sqrt is cancelled out

(-1) squared is 1 because the negatives cancel out

[crimson sedge]

ahhhhhh thank you!

.close

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[cerulean sail]

Interesting question do you have any idea of how to start it off?

(did you try the hint they gave you?)

[restive leaf]

i tried replacing Q^T and Q from the hing with the given equality for Q

but I only got the transpose part to I + vv^T alpha

so it's still not the same as Q

[crimson delta]

alpha is a number

so you can just move it

[restive leaf]

i did Q^T = Q

[restive leaf]

yeah, i just realized that

i am trying to expand Q^TQ now

any ideas how to simplify this?

i got it

thanks

.close

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[tawny forge]

R = {(n,|n| + 2) : n is an element of Z} is a subset of Z x N

[tawny forge]

I understand this as a relation. What im having trouble with is understanding is how a function can go from Z -> N

Z being set of intergers and N set of natural numbers

[clever mango]

Do you mean overall how can there be a function f: Z -> N

[tawny forge]

Well, I know in the set it mentions n can be integers and when you plug in an integer to |n| it will be a natural number

its just odd that after the set they mention is a subset of ZxN

could you still write that if y = n + 2 instead?

[clever mango]

I mean this is not because through this relation every element from Z relates to some element from N

not odd*

[tawny forge]

true

but if n is an element of Z why not just write its a subset of ZxZ?

because regardless, if you enter an interger for |n|+ 2 youre going to get a natural number

[clever mango]

well yes, you could write as well f: Z -> Z and it would be ok because Z contains N

its just.. if we know that every value that the function can have is positive then we would as well denote it as f: Z->N

[tawny forge]

gotcha, that makes sense

This answers my question, thank you

.close

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[normal thorn]

I have done A but need B

[uneven iron]

Derivative equal to 0, study intervals between critical points

[normal thorn]

I cant get the derivative right though

You use product rule right?

[uneven iron]

Yes, but you don't need it

[normal thorn]

Dp/dt = -4.1sin(PIt/2.3)

[uneven iron]

wrong

Remember chain rule

[normal thorn]

Dp/dt = -4.1Pi/2.3 cos(Pit/2.3) ?

[cedar kiln]

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@normal thorn Has your question been resolved?

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[neat patio]

How do you find the inverse derivative of this function at a?

[dire geode]

Use inverse function theorem

[neat patio]

yeah I know that

but you need the value of f^-1(2) in order to use inverse function theorem

so I know I could set

2 = x^3 + 3sinx + 2cosx

but from here idk how to solve this

[dire geode]

plug in easy values for x

[neat patio]

oh like 0 or something

wait

no fucking way

I accidentally found the answer

[cedar kiln]

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[kindred bridge]

https://tenor.com/view/yes-milk-gif-20891953

[cedar kiln]

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[proper rivet]

Hi, I am optimising the above with Lagrange but didn't get n in the system of equations. How do I solve then?

[cedar kiln]

@proper rivet Has your question been resolved?

[proper rivet]

<@&286206848099549185>

[dire geode]

You're missing equations in the gradient

you only took partial with respect to m. you also need to for n

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[crisp sand]

https://i.imgur.com/1XEiTdo.png

[crisp sand]

ignore the pictures of work but i am pretty confused with how to approach this problem.

not sure if we need to use 2d forces in this or projectile motion kinematics

[cedar kiln]

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[crimson sedge]

Help

Oh srry

[stiff totem]

the second vector having zero as its third component lets you rule out a bunch of the options

like, the first vector [1,1,1] would require having -1 copies of u, and the first two components obviously can't be reached using v

also one of the options is in a very similar form to your given vectors

yeah

[minor ocean]

kk ty, one more thing tho

[stiff totem]

this might depend on your textbook

but zero vector should be parallel to everything

if you define parallel vectors to be vectors with the same span

then they aren't

but that's rare

last one as well

well

actually, if "same direction" means parallel

then yes

if you have an alternative definition that may be wrong

that's why I'm asking

"direction" isn't really a standard term -- it could be different depending on the course

[cedar kiln]

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[azure lagoon]

can anyone help me with this

[cedar kiln]

@azure lagoon Has your question been resolved?

[stiff totem]

complete the square on the denominator

the result should be an arctan integral or something similar

[azure lagoon]

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[crisp sand]

https://i.imgur.com/1XEiTdo.png

[cedar kiln]

Please don't occupy multiple help channels.

[dire geode]

close one of your channels

[cedar kiln]

@crisp sand Has your question been resolved?

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[spice light]

how would you parametrize a cylinder x^2 + y^2 = 9 in cylindrical unit vectors?

[cedar kiln]

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@spice light Has your question been resolved?

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@spice light Has your question been resolved?

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[flint python]

Help with this question please

I got u=1-x^2+y^2

And v=-2xy

Is that correct

For part A

<@&286206848099549185>

[valid tendon]

yes

[cedar kiln]

@flint python Has your question been resolved?

[flint python]

So, the locus of Q when P moves along the line y=4x 16x^2–8x^2=8x^2

For part b

@valid tendon

<@&286206848099549185>

[tropic oxide]

this looks like it might have been GPT'd.

[flint python]

It was

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[mystic wren]

so i must select which of steps are invalid in this attempt at showing that the equation has no real number solutions, but they all look right to me... 😭 any help appreciated

[mystic wren]

im pretty confident that 1 and 2 are correct

[vital wing]

check step 3 closely

after step 2, we have
(x³ - x² + 1)(x² + x + 2) = 0

what can we conclude about it?

[mystic wren]

both things in the parenthesis must equal 0? but i thought thats what they said uhhhhh

[raw gulch]

think about the solvability of a third degree equation

[vital wing]

what is the product of 0 and 2?

[mystic wren]

0

[vital wing]

were both factors 0?

[mystic wren]

alright

[vital wing]

I'm sure that's not how you approach the question

[mystic wren]

:0

[mystic wren]

wdym?

[vital wing]

a × b = c

a and b are factors

were both factors here 0?

[raw gulch]

think about the solvability of a third degree equation

[mystic wren]

wait where did the 0 and 2 come from?

[vital wing]

example

[raw gulch]

a third degree polynomial always has at least one real solution

[mystic wren]

i forgor abt that

[vital wing]

-.-
you are not helping

yes, you're right
but that doesn't help if the asker cannot use it because step 4 ruins it

[raw gulch]

so that helps you to think about the given equaiton

[mystic wren]

OH i seee what u mean

[vital wing]

anyway, if 0 × 2 = 0, do you need both factors to equal 0?

[mystic wren]

as in, both factors do not need to be 0 in order to equal 0

yes

[vital wing]

that's it!

[mystic wren]

okk

[vital wing]

step 3 is the offender :D

[mystic wren]

what about 4 5 6?

it is a select all :0

[vital wing]

it only proved that the second factor can't equal to 0
but per this, the first factor can be 0

please find the roots to make sure that 6 is also wrong ;)

[raw gulch]

yes ok, just only wanted to realize the fact ab cube polynomial 🙂

[mystic wren]

oo okay

[vital wing]

we must resolve the actual question before proving that the final step is incorrect
this time, the offenders are step 3 and 6

[mystic wren]

ok so 6 is wrong simply cuz theres a root

thank you for your help

[vital wing]

mhm

[mystic wren]

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[mystic wren]

how does the claim get to the last part?

[mystic wren]

like why did the (n+1) turn into just n for the sequence 0,1,...,n

[cerulean sail]

a0 = 0 and an = n

So you have (0 + n), which is n

If that's what you mean?

[mystic wren]

:0

ohhhhh

i thought the (n+1) turned into n, turns out its the a0+an that did

thanks!

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[cedar kiln]

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[cerulean sail]

A pleasure

[mystic wren]

[cedar kiln]

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[uncut roost]

I NEED HELP!

[crimson sedge]

Use the second equation

try making the bases equal

so like, 9^(something) = 9^(something)

do you know how to turn the 3^(10x) into 9^(something)

[uncut roost]

here is the answer

[crimson sedge]

yea thats fine

[uncut roost]

what is happeneing in the answer

[crimson sedge]

answer my question though

[uncut roost]

no

wdym

[crimson sedge]

they made a bunch of subs

[crimson sedge]

okay so like

do you know about the exponent identity $(a^b)^c = a^{b\cd c}$

[uncut roost]

no but i know now

[crimson sedge]

okay u should remember that, it is important

so like

[uncut roost]

okay ill write that down

[crimson sedge]

3^(10x)

think of 10x as the product b*c in what i sent

[uncut roost]

ohhhh

[crimson sedge]

i want you to seperate it into two

so that you can get something that gives u 9

so like

3^(something) = 9

what is something

[uncut roost]

10

2

[crimson sedge]

okay good

[uncut roost]

why do we need 9

[crimson sedge]

so 3^2 = 9

i will explain in a bit

so

[crimson sedge]

we want our b = 2 here because 3^2

[uncut roost]

oh for the other side?

[crimson sedge]

so what should our c be?

like

[uncut roost]

5?

[crimson sedge]

almost!

you forgot the x

so 5x

[
3^{10x} = \p{3^2}^{5x}
]

[crimson sedge]

agreed?

[uncut roost]

yes

i agree

[crimson sedge]

okay so we get

9^(5x)

NOW the reason we did this is because

there is another exponent identity

[uncut roost]

wait

oh

nvm

[crimson sedge]

[
a^x = a^y \Iff x = y
]

[uncut roost]

ohhh

[crimson sedge]

if two numbers have the base, and are equal, their exponents are equal

note this down too

[uncut roost]

cancel the basses ees

bases

[west dirge]

science?

[crimson sedge]

yes

[west dirge]

teach me

[crimson sedge]

!occupied

[cedar kiln]

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).

[west dirge]

im form 1

[crimson sedge]

so taffy like

[uncut roost]

yes

now x = y

[crimson sedge]

[
9^{y-11} = 9^{5x}
]

[crimson sedge]

how does this simplify

[uncut roost]

what about the other equation

y-11 =5x

[crimson sedge]

wdym?

[crimson sedge]

yes!

[crimson sedge]

so keep this in mind alright?

[uncut roost]

this is the other equatuoin

[crimson sedge]

yes

[uncut roost]

equation*

[crimson sedge]

i was about to get to that

so like

[uncut roost]

okay

[crimson sedge]

another important identity relating to logarithms, (this is their definition): [
\m{\log_a}{b} = c\Iff b = a^c
]

[crimson sedge]

try doing this here. what do u get

[uncut roost]

y = x+1)^2

[crimson sedge]

yeah exactly

[uncut roost]

yay

[crimson sedge]

but remember

[crimson sedge]

we have this as well

[crimson sedge]

so how about you plug in the y here

into y - 11 = 5x

what do uget

[uncut roost]

what

wdym

[crimson sedge]

you know that y = (x+1)^2

and you have a y in y - 11 = 5x

which you can replace with (x+1)^2

still doesnt make sense?

[uncut roost]

why are we replacing it

[novel seal]

i had logarithm for exam today, it was nice

[crimson sedge]

because the issue is that to solve an equation we need to have one variable in it

[crimson sedge]

!redir

[cedar kiln]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[crimson sedge]

if i give you an equation with 2 or more variables

you cant "solve" for a variable and get a number

but if you have 1 variable, you can solve for that variable and get a number

[uncut roost]

you combine them?

[crimson sedge]

y - 11 = 5x has two variables, so we cant get a value for x

[uncut roost]

yes

[crimson sedge]

(x+1)^2 - 11 = 5x has one variable, so we can get a value for x

does it make sense?

[crimson sedge]

yes

[uncut roost]

yes

[crimson sedge]

ok great

so now

expand (x+1)^2

[uncut roost]

but it has 2

[crimson sedge]

yeah?

whats the issue?

[uncut roost]

you said it had 1

[crimson sedge]

one variable, which is only x

im not talking about the degree of the variables

it only has one letter-type

if you want me to say it that way

[uncut roost]

wait what are the 2 equations

[crimson sedge]

y - 11 = 5x
y = (x-1)^2

[uncut roost]

okay

see they both have x and y

[crimson sedge]

okay but thats what i am saying we can get rid of the y by substituting the second equation in the first

we can reduce the first equation from having two variables to only having one

[uncut roost]

5x+11 = (x-1)^2

?

[crimson sedge]

okay first of all, the 11 magically got an x and the 5's x vanished

and also, why is 11 negative?

[uncut roost]

sorry wait

okay now

[crimson sedge]

okay yes

thats correct

do you get what happened?

[uncut roost]

they are both y

so we make them eqaul

equal

[crimson sedge]

yes

do u see how we reduced our two variables to only 1? thats what you always want to do

if you have two or more equations and you are trying to solve for a variable, try to substitute one of the equations into the other, by solving for one of the variables, thereby reducing the amount of variables

okay great

[crimson sedge]

do u know how to solve this?

[uncut roost]

yes

[crimson sedge]

ok do it

[uncut roost]

x^2 - 5x -12

[crimson sedge]

also

hold up

[crimson sedge]

i just noticed

[uncut roost]

okay

[crimson sedge]

but why did you turn the (x+1)^2 to (x-1)^2 lol

[uncut roost]

i slept for 2 hours

[crimson sedge]

fair enough

[novel seal]

because you put it that way

[novel seal]

here

accidentally

[crimson sedge]

oh okay sorry for that

so expand 5x + 11 = (x+1)^2

[uncut roost]

i dont understand

that is not what the answer says

[crimson sedge]

it is

exactly it

[uncut roost]

where

[crimson sedge]

they just, for some reason, retained the 2

divide both sides of 2(x+1)^2 -22 = 10x by 2 you get (x+1)^2 - 11 = 5x

which is the same as we have

[uncut roost]

oh i get it

[crimson sedge]

the 2 is totally irrelevent

[uncut roost]

they changed the other side

[crimson sedge]

i dont know why they retained it but oh well

[uncut roost]

okay i get it now

[crimson sedge]

ok

[uncut roost]

wait

[crimson sedge]

solve the quadratic

[uncut roost]

do you know the log rules

[uncut roost]

there is no time

[crimson sedge]

yes, what about them?

[uncut roost]

i have to get to school

[crimson sedge]

understandable

[uncut roost]

can you tellme them

thank you so so much qylo

[crimson sedge]

here

[uncut roost]

THANK YOU

i have to go now

[crimson sedge]

also the exponents' laws for good measure, because u said earlier u didnt know all of them:

[crimson sedge]

oki bye

.close

[cedar kiln]

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[gray oasis]

I'm taking Calculus. For A. we are to get the domain of the functions.

For C. I don't know what's the way to solve that because our teacher didn't teach it.

[crimson sedge]

is the top thingy irrelevent?

okay

is 'a' specified to you?

h(a^2+1) could fall into either of your function's cases

so we need to know if anything about a is specified

[gray oasis]

Nope. There's nothing more there.

[crimson sedge]

okay so i assume they want u to generalise it for both cases

alright i guess

lets start with $h(x) = \s{x-5}$

[crimson sedge]

what is h(a^2 +1) for this?

[gray oasis]

Sorry what?

Where?

Do I press the texit?

[crimson sedge]

uhhhh

what

im asking you a question

[gray oasis]

There's a bot that appeared

[crimson sedge]

and i want ur answer to it lmao

[gray oasis]

Or is that irrelevant?

[crimson sedge]

yes i compiled that

it shows u the math equations

[gray oasis]

Oh okay sorry I'm new here

It's loading hold on hehe 😅

[crimson sedge]

"lets start with h(x) = sqrt(x-5)"

thats what i said

[gray oasis]

h(a^2 +1) = sqrt (a^2 +1) - 5

is that right?

just substitute it?

[crimson sedge]

yeah but u r missing parentheses

sqrt((a^2 +1) - 5)

[gray oasis]

oh right sorry

how do i proceed from that?

[crimson sedge]

also im highly sus of the way u wrote this

sqrt(x-5) if x <= 5?

but thats only defined for x = 5

anything lesser than that and it is imaginary

are u sure u wrote that correctly

[hidden mural]

she is not sure

[gray oasis]

I just copied it from the sheet

[hidden mural]

neither am cuz idk the question

[gray oasis]

wait lemme check

[valid tendon]

theres no wae its complex

[gray oasis]

that was our quiz earlier. I just sneaked to copy it to answer it for later because I didnt do well on the actual quiz

[crimson sedge]

i think u probably copied it wrong because the domain of h just doesnt include values lesser than 5

so

[gray oasis]

h(x) = { sqrt X - 5. if X <= 5
{ (X + 4)^2. if X > 5

find h(a^2 + 1)

i copied it right

i asked from my classmates

[crimson sedge]

was it supposed to be $\s x - 5$

[crimson sedge]

ask them if it is sqrt(x) - 5 or sqrt(x-5)

[gray oasis]

sqrt(x-5)

[crimson sedge]

okay thats just weird then

[gray oasis]

how come?

[crimson sedge]

try putting in x = 4 what do u get

[crimson sedge]

in this

[gray oasis]

oh right it will be an error

[crimson sedge]

yes

this is ONLY defined for x = 5

so saying x <= 5 is just wrong

[gray oasis]

He didnt teach us how to evaluate that

[crimson sedge]

so i really dont know whats happening in there

[gray oasis]

I thought were just going to substitute it in 1?

[crimson sedge]

yes but the definition of the function itself is ambigious...

but whatever, i digress

[gray oasis]

what i did on my paper was to just substitue (a^2 + 1) in both equation

[crimson sedge]

yeah that should be it

its just where x is defined seems wrong, which is what i was commenting on

anyways we can work on the A's which are much less ambigious if you want?

[gray oasis]

yes please

[crimson sedge]

okay so $\ds \s{x^2-1} + \s{1-x}$

[crimson sedge]

lets consider each part of this

first of all the sqrt(x^2 -1)

do you know the domain of this?

[gray oasis]

no sorry

r we going to use the number line or the interval table there?

or they are not necessary?

[crimson sedge]

i mean we can do it however u wanna

ok i can work u thru it

so like the square root function can never have whatever is inside it be negative yes?

[gray oasis]

but it can be zero?

[crimson sedge]

yes

sqrt(0) = 0

so like

[gray oasis]

is any function still valid if it is zero?

0 can be a domain?

[crimson sedge]

what does that mean

yes that can be a part of a domain

why not

it depends on ur function

anyways

[crimson sedge]

do u agree with this

[crimson sedge]

,rccw

[crimson sedge]

yes

what about it?

[gray oasis]

one more question. im really confused about this.

0 can be a legitimate output? in finding domain of function?

[crimson sedge]

yes...

take the simplest function

f(x) = x

f(0) = 0

that is a valid input and output

there is nothing cursed about 0 that makes it not be able to be the input of any function

[gray oasis]

OOOHHH just the denominator that cannot be zero then?

of a fraction

[crimson sedge]

yeah

1/x, x can never be 0

f(x)/g(x): g(x) can never be 0
sqrt(h(x)): h(x) can never be a negative number

i hope this makes it clear

[gray oasis]

YESS

[gray oasis]

yes i agreee

[crimson sedge]

okk great

so

[
\tss{for}\s{x^2 -1}: \q x^2 - 1 \ge 0
]

[crimson sedge]

im just describing what we said in mathematical terms

what i wrote is fine yes?

[gray oasis]

greater than or equal to zero right?

[crimson sedge]

yeah

anything not negative

[gray oasis]

yeah okayyy

[crimson sedge]

so i want u

to solve that inequality

find the values of $x$ that satisfies the inequality $x^2 -1 \ge 0$

do yk how

[gray oasis]

uhh

no sorry

[crimson sedge]

ok this will make ur life extremely hard for the next few questions

just a warning

[gray oasis]

😭

[crimson sedge]

but i guess i will try and help

so like

,align
x^2-1&\ge 0 \
x^2 &\ge {???}

[crimson sedge]

what is ???

[gray oasis]

1

[crimson sedge]

great

[gray oasis]

positive

[crimson sedge]

xo $x^2 \ge 1$

[crimson sedge]

how do u get rid of the exponent?

[gray oasis]

uhhh

im not sure

factor?

[crimson sedge]

hint: 🫚

ok thats a bad joke

but just take the square root of both sides

[gray oasis]

i dont even get the joke 😭

was the hint a root?

[crimson sedge]

its because it has "root"

yeahh lmao 😭

[gray oasis]

how do i do that?

[crimson sedge]

what is the square root of x^2?

[gray oasis]

x?

[crimson sedge]

not quite!

its |x|

are u familiar with what that means?

[gray oasis]

absolute value?

[crimson sedge]

yeah!

[gray oasis]

so always positive?

[crimson sedge]

and what about 1? whats the square root of 1

[crimson sedge]

yes indeed

[gray oasis]

1

[crimson sedge]

okay

[gray oasis]

if finding for sqrt of aything that is squared, just cancel both right?

the sqrt and square?

[crimson sedge]

[
x^2 \ge 1 \Iff \abs x \ge 1
]

[crimson sedge]

[
\s{x^2} = \abs x
]

[crimson sedge]

so u should keep that in mind

[gray oasis]

okay

x and 1?

[crimson sedge]

hm?

[gray oasis]

oh wait

[crimson sedge]

we are gonna continue from here

im just like

trying to keep up my pace with u so u dont get lost

lmao

but are u fine with everything up until now?

[gray oasis]

is x^2 - 1 the same as (x-1)(x+1)?

[crimson sedge]

yes

it is

you could also solve it with what u r doing

[gray oasis]

its another method? 😭

[crimson sedge]

yea

we can uh

do that if u want

its all the same tho

[gray oasis]

can we try urs first?

[crimson sedge]

suree

so

[crimson sedge]

r u fine with this

[gray oasis]

with this

[crimson sedge]

ye do u understand everything

[gray oasis]

yes

[crimson sedge]

with it

sick

so

lets use our definition of an absolute value function

[
\abs x = \begin{dcases*}
x &if $x \ge 0$ \
-x &if $x < 0$
\end{dcases*}
]

[crimson sedge]

you know this right?

[gray oasis]

omg noo 😱

[crimson sedge]

what?

[gray oasis]

really noo

[crimson sedge]

okay so

its really simple

what its saying is that

if x is bigger or equal to 0 then the value of |x| is just x

like |3| = 3

|4342424| = 4342424

is this fine by u?

[gray oasis]

yepp

[crimson sedge]

yeah so the second part

is saying like

if x is lesser than 0 then its equal to -x

sooo

|-3| = -(-3) = 3
|-3424324| = -(-3424324) = 3424324

the negatives cancel so u just get the number

does this make sense

[gray oasis]

so just everything under absolute is positive no?

[crimson sedge]

yeah

exactly haha

[gray oasis]

ok ok hehe

[crimson sedge]

yeah so u get what this means now i guess?

[gray oasis]

yeah 😅

[crimson sedge]

okokok

so back to what we have

|x| >= 1

so

we can seperate it into two cases

[\begin{cases}
x\ge 1, \q x\ge 0 \
-x \ge 1, \q x < 0
\end{cases}
]

[crimson sedge]

this may seem confusing, but what i did is use this pretty much

ask if anything doesnt make sense lol

[gray oasis]

the second part still confusing hehe

[crimson sedge]

hmmmm

so like do u agree that a negative number multipled by a negative is positive always

-(-x) = x

[gray oasis]

yes

[crimson sedge]

yuh

so thats what its saying

its basically saying

if you have a number lesser than 0 (for example -3 )

then that number is equal to -x (so -(-3)) which is equal to 3

[gray oasis]

ahh omg i disregarded the ngative i the -x.

okok

[crimson sedge]

okk

so i hope this is fine now?

[gray oasis]

yes

[crimson sedge]

ok so back to this

for $x\ge 1, \q x \ge 0$, i want you to find the intersection of both intervals

[crimson sedge]

intersection means like

the part they both have in common

[gray oasis]

how?

to find it?

[crimson sedge]

i want more than 0 apples, but i also tell you i want more than 1 apple, so in conclusion i want more than __ apple