help-23

[astral glacier]

(also you can combine 3 and 5 without consequence, just throwing that in here)

[desert pasture]

$\Z_2 \times \Z_2 \times \Z_2 \times \Z_3 \times \Z_5$ and permutations of the same

[astral glacier]

Nope

[flat frigate]

wai

[astral glacier]

Firstly permutations would fall into the same isomorphism class

[desert pasture]

and secondly?

[astral glacier]

So this is one of them

Secondly, think about how you can combine these to keep three elements of order 2

[desert pasture]

I could combine Z_3 and Z_5

[astral glacier]

That's one

You can combine Z2 and Z3 too

[desert pasture]

or combine 3 and 2

yea

[astral glacier]

Or 2 and 5

Or 2 and 15

[desert pasture]

2 and 15 feels off

[astral glacier]

As long as you only combine one 2

[desert pasture]

oh, nvm

[astral glacier]

2 and 3 and 5

[desert pasture]

yea , makes sense

[astral glacier]

Now see what classes you get

Also hold on why do we know G is abelian

[desert pasture]

I think we assume that here

this was in a quiz today

( that some other batch wrote)

[astral glacier]

That's weird

And sad

[desert pasture]

The previous question was state the fundamental theorm

[astral glacier]

At least mention it in the question smh

[peak estuary]

list all elements of order 2 that you found

[desert pasture]

for each isomorphism class?

[peak estuary]

you only found one class

Z_2^3 times Z_3 times Z_5

[astral glacier]

We talked about other ones but yea, he only listed ons

[peak estuary]

I mean this doesnt do shit

same class

[desert pasture]

fair

[desert pasture]

(4,0,0)

[peak estuary]

thats not an element of that group

well ok I should have put () I guess

(Z_2)^3

not Z_8

[desert pasture]

(1,0,0,0,0)(1,1,0,0,0),(1,1,1,0,0)

[peak estuary]

what about (0,1,0,0,0) ?

[desert pasture]

that's equivalent to (1,0,0,0,0)

[peak estuary]

no

its a different element

[desert pasture]

okay

so what am I doing wrong

[peak estuary]

you found the wrong class

try again

two other classes to check

[desert pasture]

$Z_2 \times Z_2 \times Z_{10} \times Z_3$

[flat frigate]

wai

[peak estuary]

no

thats the same thing

do not combine different primes

[desert pasture]

combine the same prime?

[peak estuary]

the other options are Z_8 times Z_3 times Z_5 or Z_4 times Z_2 times Z_3 times Z_5

the only question is how you distribute the 2's

2*2*2 or 4*2 or 8

[desert pasture]

In Z_4 \times Z_2 \times Z_3 \times Z_5 consider (2,0,0,0),(0,1,0,0),(2,1,0,0)

[peak estuary]

check for typos

[desert pasture]

I can't seem to find anything

[peak estuary]

4=0 in Z_4

did you really mean that?

[desert pasture]

I thought we were in Z_8 for some reason

my ba

*bad

thanks

[peak estuary]

good

and for fun do the last class as well

[desert pasture]

Will do after dinner

Thanks

.close

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[worn geyser]

hi, i need help with letter (a).

[worn geyser]

the y = 6 confuses mee

thanks in advance

[frigid spruce]

graph of y = 6 is basically a straight line

[worn geyser]

yes i know

[frigid spruce]

have you tried plotting it on a graph paper yet?

along with the graph of f

[frigid spruce]

yep

[worn geyser]

so this is the area i need to find?

okai

ohhh okai

well then i need to find the integral of y=6 from 0 to 2 and then the integral of f from 0 to 2 as well

and take the integral of y minus the integral of f

right?

or noo

i am not sure how to solve for (a)

[frigid spruce]

just use 1/2 lb

[worn geyser]

but i can actually do this

i think

[worn geyser]

what does lb mean?

[frigid spruce]

yeah, you can do that too

[worn geyser]

but what do you mean by this?

[worn geyser]

okie

[frigid spruce]

the basic formula to find the area of a triangle

half times length times breadth

1/2 * 2 * 6

ur gonna get the same answer

[worn geyser]

yayy

i got 6 as well

[worn geyser]

ohh

[frigid spruce]

yeah, both the methods work

[worn geyser]

but for (b), they want me to find the volume of this?

when the figure is rotated around the x-axis

[frigid spruce]

that would be a cone

[worn geyser]

yess

[frigid spruce]

did they specify that you’re supposed to solve the questions using calculus

[worn geyser]

i am not sure, but i am supposed to find the volume using integrals

[frigid spruce]

oh alright, i see

brb

[worn geyser]

okie

[worn geyser]

i did this

but its wrong

[worn geyser]

ohh they want this one it seems

then i need to find the volume of y

and then of f

and then take volume of y minus volume of f

i think thats it

wait let me try

[frigid spruce]

nah

that's not how you do it

you need find the volume of the black cone

[worn geyser]

hmm but i got the correct answer though

[quick minnow]

What is the name of this topic?

[worn geyser]

the first one is incorrect

[worn geyser]

uuh yes exactly

this is what i found the volume of

[worn geyser]

i am doing integration

[frigid spruce]

applications of calculus ig

[frigid spruce]

oh okay, let me check your working

[worn geyser]

okie

thanks

[frigid spruce]

why squared?

[worn geyser]

because we want to find the volume

i am not sure why the formula is like this but thats what we learned

[worn geyser]

they explain it here i believe

but i will take a closer look i to it later

@frigid spruce hope its ok i pinged youu

[frigid spruce]

yea it's okay

https://www.youtube.com/watch?v=drpxZ1aztWE

could you watch this video

your working isn't right. you're missing something deeper and this video explains it really well

it's pretty much the same thing, except in our case h = 2 and r = 6

[worn geyser]

sure

[frigid spruce]

similarly you can solve its second part too

[safe radish]

@worn geyser Has your question been resolved?

[worn geyser]

okie

i will take a look on it in a bit

but i struggle with letter (c) now

@frigid spruce

<@&286206848099549185>

[slow zodiac]

Formula for surface area of a circle?

(Its perimeter)

[frigid spruce]

it's the same thing but instead of integrating the volume of the small disks, you need to integrate the the surface area of the disks (excluding the surface area of the 2 flat surfaces)

ie, its perimeter

[worn geyser]

oh really

okai

[frigid spruce]

yep

[worn geyser]

but why

not the small area

like i did for the volume

i took the volume of y minus the volume of f

wouldnt i do the same for the area?

area of y minus the area of f

[frigid spruce]

does this help?

[frigid spruce]

huh

did you not watch the video? 😭

[worn geyser]

noo not yet

[frigid spruce]

damn ok

[worn geyser]

bahah i will do it now

[frigid spruce]

so you're supposed to do it the way the guy did in the video 😭

[worn geyser]

but it worked this way

[worn geyser]

oh

[frigid spruce]

i have no idea how that worked

it isn't supposed to be done like that for part b and c

[frigid spruce]

yeah, pls

[worn geyser]

i also checked the conclusion, and they did the same

[frigid spruce]

hm that's surprising

[worn geyser]

see

its actually the same

i just did it separately

[frigid spruce]

that's cool

[worn geyser]

riight

[frigid spruce]

i did not know it could be solved that way

[worn geyser]

and it makes sense

thats why i did it like that

but for letter (c) i thought i could do the same

but no

they did like this for c

[slow zodiac]

They wanted surface area

[slow zodiac]

This here is volume

[worn geyser]

i know, but my question is why they didnt take Ay - Af

like they did for the volume

[safe radish]

@worn geyser Has your question been resolved?

[safe radish]

@worn geyser Has your question been resolved?

[worn geyser]

<@&286206848099549185>

[merry owl]

that is because the surface area when taking Ay-Af remains jagged even if you take the limit; in surface area, jaggedness does not dissapear in the limit like it does for volume

so you have to take the slant height , sqrt(1-f(x)^2)

[safe radish]

@worn geyser Has your question been resolved?

[safe radish]

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[wide trellis]

I need help with science if anyone got me

[wide trellis]

Science hw

[safe radish]

@wide trellis Has your question been resolved?

[languid ermine]

also, you may have better luck in one of our science servers! see #old-network

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[frozen quarry]

Can anyone help me?

[frozen quarry]

Sketch a graph in a coordinate system that clearly shows a function, and another graph that clearly does not show a function.

[warm warren]

what have you tried?

[frozen quarry]

So for example I can do…

y = 3x^2 + 6x + 2

And that is a function?

When a function is like a line?

[molten acorn]

A function is linear when it is in the form ax+b

[molten acorn]

And quadratic ( a parabola) when it's in the form ax^2 + bx + c, this is an example

[molten acorn]

Also, the part of the question that asks "clearly does not show a function"... It's quite ambiguous since functions can really be almost anything (this is imo, I would be rlly happy if you could clarify this)

[frozen quarry]

so without ax^2 + b?

if its cubed then its like parabola one?

[molten acorn]

If it's cubed, it's shaped like an s

[molten acorn]

Linear means it looks like a line

[frozen quarry]

what?

no...

[molten acorn]

Wdym?

You can check on desmos

[frozen quarry]

what is y = x^2 = 1 then?

[molten acorn]

That does not seem to be a function to me

[timid escarp]

why are there two equal signs here?

[molten acorn]

That's more of an equation

[frozen quarry]

y = x^2 -1

i mean this

[molten acorn]

Ooh

It's parabolic

[frozen quarry]

okay what needs to be a function?

What is considered as linear and parabolic?

[molten acorn]

Linear if its ax+b
Parabolic if it's ax^2 + bx + c

And a =/= 0

Well, if the function has the term x^2 then it's parabolic

[frozen quarry]

so it an be y = x^2 + 1

[molten acorn]

Yes

It can be anything with the term x^2 (but no x^3 or any higher powers)

[frozen quarry]

so like

parabolic = contains X^2?

can it also be y = x^2?

@molten acorn

[timid escarp]

as long as x^2 is the highest power, that works.

[frozen quarry]

fiine

and linear ANYTHING if it doesn't have x^2?

[timid escarp]

if there's only x and no other higher power around, then it's linear.

[frozen quarry]

fine.

okay.

Sketch a graph in a coordinate system that clearly shows a function, and another graph that clearly does not show a function

[molten acorn]

Mhmm

[frozen quarry]

what is the question?

[molten acorn]

Wdym?

[frozen quarry]

i mean what do you want?

[molten acorn]

What is the definition of 'function' ?

In your book

[frozen quarry]

ill just send you whole page

[molten acorn]

Okay

[frozen quarry]

9.1 Functions

In mathematics we often encounter equations that contain numbers and variables, which are usually denoted by x and y. The simplest examples are equations of straight lines and parabolas. Now we want to define the relationship between these variables in such a way that it describes a function.

A function describes a relationship between two variables (e.g., x and y) such that for every single value of x that we put into the function, we obtain exactly one value of y from the function. The variable y depends on the value of x. Then we say that y is a function of x. The value that y takes is called the function value.

An example of this is the function
y = x^2 - 1.
If we choose x = 0, then the value of y is -1.
If we choose x = 2, then the value of y is 3, and so on.

An example of an equation that is not a function is
x = y^2 - 1.
If we choose x = 0, this gives two values of y: y = 1 or y = -1.
Since a single x-value corresponds to two different y-values, this is not a function.

A function is usually given a name, and f is the most common, though other names such as g and h are also used.
An example of a function is the straight line
f(x) = x + 1.
We say “f of x equals x plus 1.”
The name of the function is f, and x is the variable that takes on the values we evaluate in the function.
Thus we can write
f(2) = 2 + 1 = 3.

So far we have written functions in the form y = x + 1.
This tells us that y = f(x), and individual points on the graph of the function have coordinates
(x, f(x)) = (x, y).

⸻

Example 9.1

Which of these equations represent a function?

[molten acorn]

I see

[rare briar]

Right, so from this text, can you see what makes an equation a function and what makes an equation not a function?

[frozen quarry]

so a function must be a parabola and linear

and it cant be circled?

[molten acorn]

No that's not how it works

From the definition of your book, an equation is not a function if :
There can be 2 y values for a certain x value

[frozen quarry]

and is that true?

[molten acorn]

Pretty much so

But as you go to higher levels of math this definition may change a little bit

[frozen quarry]

Sketch a graph in a coordinate system that clearly shows a function, and another graph that clearly does not show a function.

shows a function

so i can do

y = x^2 + 5x + 9

[molten acorn]

Yes

[frozen quarry]

okay how do i sketch it?

the normal way?

like first finding d and then p q etc

[molten acorn]

Just do whatever way you are more comfortable with

[velvet nebula]

You can always pick a simpler function

[frozen quarry]

why cant this be a function?

[velvet nebula]

What would be the value at 0

[frozen quarry]

??

-4

@velvet nebula

[rare briar]

At x=0 I think they mean

[velvet nebula]

Yeah at x=0

[frozen quarry]

I’m too lazy to find it now

I have to calculate it

[velvet nebula]

You see two different values of y

Right?

But a function needs to have one output.

[frozen quarry]

f(x)?

So if we know x = 2

f(2)

y = x^2 + 5x + 9

Is that how functions work

It’s basically putting numbers in x?

[velvet nebula]

Yeah it's a thing you can plug a value in to get a output

You got the main idea

[frozen quarry]

@velvet nebula do you know Geogebra?

How to use it

<@&286206848099549185> ?

.close

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[somber cape]

having trouble showing if a < b < c < d then c - b < d - a

[somber cape]

feel like I have to do some translation hmm

by translation I mean using the fact c < d and that b < c and a < b

let me try that

[glacial cairn]

You could just say c < d and -b < -a

[somber cape]

o that works too

[somber cape]

I was able to figure this one out also

thanks for the help

.solved

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[frozen quarry]

hey! so i need to learn on how to see if a equation is a function. Chatgpt says its either:

1. Algebraically – check how many solutions y has for a given x
2. Vertical Line Test (Graphical)

How do I do the 2nd one?

[night magnet]

graph the the function in the xy plane. if a vertical line (perpendicular to x axis, or parallel to y axis) only passes the function graph once, it's a function with respect to x

[brave wolf]

take the graph, if some vertical line intersects it at more than 1 point, then its not a function

The left one is a function, the right one isnt

another example, the left one isnt a function, the right one is

and one more

you get the point ig

[quasi bison]

!noai btw

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[frozen quarry]

but in this case its correct

[quasi bison]

this was a general policy reminder

regardless if it is sometimes correct or not

if you dont know what is bullshit and what isnt -> you cant tell if AI has said bullshit
if you DO know what is bullshit and what isnt -> you dont need AI to parrot it to you

[frozen quarry]

okay, the blue line is y and pink is x?

[elder moss]

not really

the vertical line represents a line parralel to y axis, as generally a function is expressed as y = f(x). the red lines/curve represents the equation

[left gyro]

the blue vertical lines are not functions, because they have the equation x = number instead of the usual function equation that says y = f(x)

what you can think of is if you have a function y = f(x)
then that already shows theres one y for a given x
so at any x, you can only have one y
so if you drew a vertical line x = number, it can only cross the function once at the one y-value it has
so the vertical line crosses the graph once and so it passes the vertical line test

another way you can think of it is that:
if a vertical line crosses the graph twice or more,
that would mean theres at least two points with the same x-coordinate but different y-coordinates
call them (x1, y1) and (x1, y2)
then if you had y = f(x), that would mean f(x1) = y1 and f(x1) = y2
that would mean f(x1) is equal to two different numbers which cant happen
so the graph cant be a function

[frozen quarry]

poop

i understand a little bit

so by your definition

[left gyro]

doing good so far, try sticking to a definition closely to see what happens

[frozen quarry]

function =

x = number, becasuse it is a vertical line

[left gyro]

thats not what I said

maybe you can elaborate

[frozen quarry]

okay can you give me an equation that isn't a functiin

ill put it in geogebra

[open wedge]

$y = x^2$

[flat frigate]

1 divided by 0 equals Infinity

[frozen quarry]

no, like full one

[open wedge]

$f(x) = x^2$ you mean?

[flat frigate]

1 divided by 0 equals Infinity

[frozen quarry]

x = y, why is it exactly there?

[open wedge]

?

[elder moss]

can you elaborate

[frozen quarry]

how is x = y there?

why exactly 0

[elder moss]

put x=0 what do you get as y

x=y=0

[frozen quarry]

why 0 exactly there?

why isnt it for example 5?

[elder moss]

eqn is y=x so the value of y will be equal to x , if you put x= 5 => y =5

[open wedge]

so if you put x = 0, then y = 0

[frozen quarry]

but theres no 0..

so why do they assume im gonna put 0

[open wedge]

elaborate please?

oh i got 1 equation for you

$x = 1$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

this one should do it

[elder moss]

the eqn relates y with x, if you substitute a value of y in any equation you get corresponding y

[frozen quarry]

elaborate that this isnt a function @left gyro

[elder moss]

eg define a function f(x) = x
which means it returns the x value we substitue

now if we plot the graph of f(x) as y=f(x) { assigning f(x) value to y coordinate}
we get the corresponding value for y ( y=x)

[left gyro]

this is a function nenni

pleas dont twist others' words

that function is y = x^2

y = f(x) = x^2

[frozen quarry]

he literally said it isnt

[open wedge]

can you try the equation $x = 1$?

[flat frigate]

1 divided by 0 equals Infinity

[left gyro]

let me look around and see what youve misread this time

[open wedge]

i suggested y = x^2 but i used the wrong method for that equation

[elder moss]

.

[open wedge]

i think she used my words to twist others

[left gyro]

okay this is straight up a confusing way to say it

nenni hear the guy out, this is supposed to be a function

you couldve gone with like ±√x for a graph that isnt a function, or like x = y^2

this would be like the y = x^2 but sideways

if you draw a vertical line through it, it will cross x = y^2 twice, once above and once below

this is a way to see that when you square root something, you get two square roots, so you have to "pick one"

to find 4 = y^2, you end up with y = 2 or y = -2

[frozen quarry]

how to know how to wrtie y = x^2 in graph?

[elder moss]

wdym

[open wedge]

i think she means how to graph $y = x^2$

[flat frigate]

1 divided by 0 equals Infinity

[elder moss]

hmm.. it's a general eqn of parabola

$y = 4ax^2$

[flat frigate]

heisenberg

[frozen quarry]

what is like the difference between

y = x^2
x = y^2

[open wedge]

$y = x^2$ is a function while $x = y^2$ is not a function. easily proven using the vertical line method

[flat frigate]

1 divided by 0 equals Infinity

[frozen quarry]

and where is the "vertical line"

[open wedge]

try the vertical line $x = 1$

[flat frigate]

1 divided by 0 equals Infinity

[left gyro]

in the sense that if you drew one, itd cross the curve twice

the vertical line test says "not a function" if at least one vertical line crosses the graph twice

[frozen quarry]

YOU LITERALLY SAID that if it crossed twice, its a function but one time its not

[open wedge]

at least twice

[open wedge]

please don't twist other person's words

[left gyro]

can you name the one time

[frozen quarry]

so every parabole that is side ways is not function?

[left gyro]

sure

for now you can say that its not a function

[elder moss]

yes if you take y =f(x)

let's make it simpler

do you know what f(x) is?

[frozen quarry]

But if we draw verticals lines, it will only cut one time the parabola

@left gyro

[left gyro]

which one

y = x^2 is a function
x = y^2 is not a function

if you look at just the blue curve,
any vertical line you draw will only go through the blue curve once
and so the blue curve is a function

if you look at just the red curve,
drawing a vertical line through it will go through the red curve twice
and so the red curve is not a function

[frozen quarry]

This is x^2

So is it y = x^2?

if I have x^2

Only

Is it always y = x^2?

@left gyro

[left gyro]

x^2 and y = x^2 commonly mean the same thing

this is because when we graph functions, we commonly treat them as y = f(x) = x^2

so the "y = f(x) =" part is usually left out

[open wedge]

can you name an equation where using the vertical line method, there exists a vertical line that intersects at infinitely many points?

[left gyro]

if I want to "graph x^2 and x^3"
that means to graph two different curves: y = x^2 and y = x^3

[frozen quarry]

But what is the difference here in questions?

Why can’t y^2 be a upward parabola

[frozen quarry]

Not now

[left gyro]

youre saying to swap x and y and expect the same results?

a question like that sounds like saying "why cant x = y^2 be an upward parabola like y = x^2?"

as a reminder x is short for x-coordinate and y is short for y-coordinate
in most cases, if you swap the coordinates of your points, the point isnt going to be in the same place

swapping the coordinates flips the graph to look sideways

you can compare y = x^3 and x = y^3 for another example of this

[frozen quarry]

So function of y is always sideways?

[frozen quarry]

Isn’t that basically when delta < 0?

[open wedge]

can you elaborate please?

[left gyro]

if you take the function of x version as upright, the function of y version will both be sideways and mirrored

[burnt notch]

Do you know how the Cartesian plane work?
Because from this it seems not at all ...

[left gyro]

this is part of a general pattern of where the point (x, y) ends up when you change it to (y, x)

it will be rotated then mirrored

[frozen quarry]

F(x) = upward
f (y)) = sideways

[burnt notch]

Huh? Parabolas are only one among the infinite functions. Don't focus too much on the shape

[left gyro]

y = f(x)
x = f(y)

[frozen quarry]

Function of y is is f(x)

[burnt notch]

No

[left gyro]

thats not correct

[burnt notch]

Function of y is f(y)

[left gyro]

the wording of y = f(x) is "y is a function of x"

be careful with where youre getting these words from

x = f(y) would be "x is a function of y" instead

[frozen quarry]

So function of y is sideways

[burnt notch]

Not necessarily ....

Whatever that means

[left gyro]

x = f(y) is sideways, assuming y = f(x) is upright

I think you should see more reasons than just repeating what exactly the magic words are

[frozen quarry]

But you said if you said if something is sideways it isn’t a function

[left gyro]

yes, and thats because x = f(y) is not a function of x

its a function of y instead

I think youve got to recognize here:

you were already on relatively thin ice in the sense that youre just learning the bare basics on what can be considered a function

the moment we change something thats not in the school curriculum, everything becomes more complicated

[left gyro]

sideways things are a function in some directions but not others

this sort of mess is very hard to get right

and it leads to active vs. alias which imo is one of the most tedious problems you can come across

[frozen quarry]

Ok that’s weird

[left gyro]

because of that, dont consider x = f(y) for now

stay at y = f(x)

[elder moss]

ig you need to work on basics of functions adn graphs

[left gyro]

upright things are also a function in some directions but not others, theyre just rotated versions of the sideways things

sideways or upright in this sense dont really mean anything other than to rotate the directions

you should ignore all this and just keep it simple

y = f(x), x is the input and y is the output

if you take a function y = f(x) and you rotate it sideways, it might not be a function of x anymore

like as you saw in y = x^2 turning into x = y^2

there are two ways you can see that x = y^2 is not a function

algebra: I put in an input like 4
I get 4 = y^2
this means y is either 2 or -2
functions shouldnt need me to choose a value
so this is not a function

graphing: I graph x = y^2 on the graphing calculator and see a < shape
if I drew or graphed a vertical line through it, it would cross the shape twice
so this is not a function

[frozen quarry]

,w x=3

[flat frigate]

Results provided by WolframAlpha

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

Requested by nenniegs

[frozen quarry]

Why isn’t that a function

If y is 3

And x is 3

[charred jetty]

You're looking at it through the view of y, I would say.

[left gyro]

algebra:
I put in an input like 3
3 = 3
this doesnt tell me a y-value
so this is not a function

graphing: if you drew a vertical line through x = 3,
It would completely overlap x = 3, so it doesnt "cross" it just once, it crosses everywhere
so this is not a function

[burnt notch]

Where do you see that y = 3??

Too advanced now I'd say

Let's keep it simple for nenni

[open wedge]

i agree

[frozen quarry]

I probably don’t need even learn it

University level

Guys I’m not undergraduate

x(x-3) = y

x^2 - 3x = y

How to know that it’s not a function

Because y will be Square rooted?

[burnt notch]

y = x(x - 3) IS a function!!!

Who told you it's not??

[left gyro]

this already solves for y

do you know what "solving for y" means

[frozen quarry]

Ik it

[left gyro]

as before, if you can solve for y, you have a function

[frozen quarry]

Okay tell me how you did spot it

[left gyro]

you can see that y was solved for

since you know what solving for y means, you know what to look for

(you look for y = on one side and an expression that doesnt have y on the other)

[frozen quarry]

x^3 = y^2

Is that a function

[left gyro]

thats not y =, so you technically need to solve it

what happens if you solve it?

[frozen quarry]

I get 2 values?

Of y

Cause square root

@left gyro

[left gyro]

yea, +y and -y

so its not a function

youd get y = ±√(x^3)

[frozen quarry]

But where’s the “linear” method?

[left gyro]

its called the vertical line test, call it by its name

say that or say that you mean something else by the linear method

[frozen quarry]

Okay vertical line test

Where is it

How to use that for that

[left gyro]

first thing first

graph x^3 = y^2 by typing it into the graphing calculator

go screenshot what you see

[frozen quarry]

,w x^3 =y^2

[flat frigate]

Results provided by WolframAlpha

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

Requested by nenniegs

[left gyro]

there we go thats a graph

you can see its calling it an "implicit plot"

bit of a hint there that x^3 = y^2 is also called an implicit too, but thats not too related right now

[frozen quarry]

Why isn’t it a function ? If I draw a line it’s going though one point

[left gyro]

you need to be drawing vertical lines

and if at least one of those vertical lines goes through more than one point, its not a function

looking at this picture,

[frozen quarry]

Okay but what is the reasoning that if a line crosses two points, it’s not a function?

[left gyro]

you dont seem to organize your questions very well

anyways

lets say you draw this vertical line and it crosses two points

lets call the two points (x1, y1) and (x1, y2)

one point is vertically above the other one

so they have the same x-coordinate

but their y-coordinates y1 ≠ y2

now as before, a function would be viewed as y = f(x)

so the y-coordinates would be our f(x1)

(x1, y1) would mean y1 = f(x1)
(x1, y2) would mean y2 = f(x1)

or y1 = f(x1) = y2

now you can see here y1 and y2 cant be the same

because we said it was two points instead of just one

[elder moss]

you can start from 5:45
https://youtu.be/kvGsIo1TmsM?si=BlnSNaiQ96iysoYu

[left gyro]

and so we know whatever this graph is, it cant be a function because it just tried to give two outputs (y1 and y2) for an input x1

[left gyro]

if you go to 5:45 here, itll give an example of my reasoning
with the not-function being x^2 + y^2 = 4
and with the two vertical points being (1, √3) and (1, -√3)

[elder moss]

this might clear some doubts
https://youtu.be/wbBY2tTqXDA?si=mpYigrrrjaSRvBhG

[frozen quarry]

Okay

Anyway

y = 0

Is that a function

@left gyro

[left gyro]

yes

[frozen quarry]

Why

[left gyro]

theres no place to put an input, but the output is always 0

I put in x = 0, it gives out y = 0

I put in x = 1, it gives out y = 0

[frozen quarry]

How do you know if 0 is x?

[left gyro]

nenni what I am giving here are called "examples"

[elder moss]

do you know independent variable?the x in f(x)

[left gyro]

I use them whenever I need to mention some values of x that dont have anything special to them, they can be anything but I like choosing simpler values

I can also put in other values of x if you want

but all of those values would give out y = 0 as an output

you can see here the function isnt doing much but its still doing its job of outputting a number

[left gyro]

its not trying to output two numbers at the same time (√3 and -√3)

so its a function, a very boring one

[frozen quarry]

Okay anyway

That’s a good made up

y-2 = x/y

[frozen quarry]

@left gyro

How to know if it’s a function

[left gyro]

hmmm I wonder

its almost like I told you exactly what you can do

lets go through both methods

first, the graphing one, its faster and easier

remind me what I said about how you use the graphing one?

[frozen quarry]

Hmmm

[left gyro]

whats step 1?

[frozen quarry]

Simplify the equation

[left gyro]

thats not the graphing way unfortunately

lets try again

[left gyro]

remember?

the graphing way is called the "vertical line test"

now whats the first step of the vertical line test?

[frozen quarry]

To draw the thing?

The parabole

[left gyro]

to graph it, theres no way you can draw this by hand

what does the computer say y - 2 = x/y looks like

[frozen quarry]

So I need to calculate it?

[frozen quarry]

Without the computer

I can’t use computer

[left gyro]

nenni I want to make sure you know both methods

[elder moss]

A relation

[left gyro]

just go do the easier one first and get it out of the way, answer what Im asking

then we go over to the harder way that youre expected to do for the problems

what was step 1 of the graphing way (the vertical line test) that I said to you?

@frozen quarry hello?

[frozen quarry]

Hold on

[charred jetty]

What is your main question?

Are you trying to understand vertical line test or just functions, in general?

[frozen quarry]

Honestly it’s so difficult to understand yall through text…

And it takes a lot of time reading this

And typing out

[elder moss]

Hmm.. better watch a lecture or ask f2f

[charred jetty]

😔

[elder moss]

Face to face

🥀

[left gyro]

honestly you shouldve said you preferred screensharing hours ago or we wouldnt be in this situation

I need to go to sleep

[elder moss]

Gn

[frozen quarry]

Gn

[elder moss]

Lit this been going on for a while

[charred jetty]

Diabetical dreams

[safe radish]

@frozen quarry Has your question been resolved?

[safe radish]

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[keen moon]

I'm suppose to do integration by parts I'm pretty sure but I forgot what we look for when picking for u or v

[ornate wyvern]

ilate

here

[keen moon]

you pick this for u right

[ornate wyvern]

sure

[keen moon]

ok thx

.close

[safe radish]

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[safe radish]

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[frozen quarry]

How to write easy equations like x = y + 3

[frozen quarry]

on graph

[elder moss]

you are back

[quasi bison]

https://www.youtube.com/watch?v=yaOf7ioCgFQ&list=PL0o_zxa4K1BVzbQjmhQxe5eQwuFQx2oV_&index=2

(playlist)

[safe radish]

@frozen quarry Has your question been resolved?

[frozen quarry]

Do is something like this straight line?

[mighty mango]

rewrite the equation by making y "alone"

then plug in random valeus for x, get the values for y, then plot the points like (x , y) then connect them

if you dont know how to do that/didnt understand

[mighty mango]

refer to this

[frozen quarry]

So like

If I make x 1

Then it’s 1 -3= y?

[elder moss]

yes

[safe radish]

@frozen quarry Has your question been resolved?

[frozen quarry]

But for x… we actually decide where it’s placed

The points for parabole

The equation doesn’t do anything

[frozen quarry]

…

[elder moss]

Wdym

[burnt notch]

You're asking totally nonsensical questions, fyi

I have the hunch you don't even know what you want to ask 😅

For example, what exactly do you want/need to do with the function x = y + 3

Also, if you have an original task written by your teacher, send it please. So that we understand better from the whole context

[safe radish]

@frozen quarry Has your question been resolved?

[magic junco]

@frozen quarry Have you watched it yet??

[mighty mango]

NENNI

WATCH THE DAMN VIDEO

[safe radish]

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[lean otter]

\int_{-20}^{-10}\left(\frac{x^{2}-x}{x^{3}-3x+1}\right)^{2},dx
;+;
\int_{,1/21}^{,1/11}\left(\frac{x^{2}-x}{x^{3}-3x+1}\right)^{2},dx
;+;
\int_{,21/10}^{,11/10}\left(\frac{x^{2}-x}{x^{3}-3x+1}\right)^{2},dx

[junior raven]

what have you done so far

[lean otter]

Reduced them to these two integrals

And it's I1-I2

<@&286206848099549185>

Please help

[merry owl]

!show

[safe radish]

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

[lean otter]

This is the closest I reduced

I'm using laptop that's why I can't show

[empty gyro]

I think u-sub will make this easy

[lean otter]

Nope

Tried that

Not helping

[empty gyro]

ooops wait nvm

I just put this in a calculator. You can't solve this integral without finding the roots to the cubic equation in the denominator, which makes this a rather tedious problem

Are you being asked to find the exact solution, or just a numerical approximation?

[lean otter]

<@&286206848099549185>

Help please

[low echo]

You need value of I?

[low echo]

The sum of those 3

@lean otter

[low echo]

I'm asking if he just needs to find the sum of those 3

That's it

?

[quasi timber]

They’ve waited.

as SWR as stated previously, you will need to find the roots of the cubic equation in the denominator.

@lean otter

[lean otter]

Nahh

That's wayy too complicated

There's supposed to be an easy way out

[low echo]

My calculator is thinking for a long time

[fierce ravine]

= 0

[low echo]

It's been several seconds

And the answer is...

Math error

[quasi timber]

[low echo]

It gave the answer of the 1st 2 easily

It's thinking at the 2nd

yup math error

alright bye

[quasi timber]

also are you sure the integral even converges @lean otter.

[lean otter]

Yup man

Our teacher gave it

[quasi timber]

Well i’d actually suggest taking another look at the last integral.

[lean otter]

It's a genuine question

[quasi timber]

I think you HAVE to find the zeros of 𝑥³-3𝑥+1

[lean otter]

.

Are there any real zeroes ?

Idk

[quasi timber]

Yes.

[ivory flame]

Try simplifying using partial fractions

[lean otter]

Nothing is useful man 😭😭

[ivory flame]

Ur lower bounds of the definite integral are greater than the upper too, which might be a problem

Also you should look at each integral separately as opposed to combining them as they are all different definite integrals

[lean otter]

I've tried everything man
Js give me the answer atp 😭😭

[ivory flame]

Calculate each integral separately

[brisk shard]

put x = 1/(1-t) in second and x = 1-1/t in third

should make the limits identical as the first integral

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[quiet plume]

!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 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

[kind beacon]

Im trying to figur eout how to solve this

i am thinking of proving it is True first using hilbert

I will prove this first
⊢(A→(B→C))→(B→(A→C))

cuz hilbert doesnt allow assumptions

then i use deduction theorem

[safe radish]

@kind beacon Has your question been resolved?

[kind beacon]

Did I do this right?

[safe radish]

@kind beacon Has your question been resolved?

[safe radish]

@kind beacon Has your question been resolved?

[split kayak]

yeah, id swear thats correct

completely different approach, but you could argue that in boolean algebra, addition is associative and commutative.

[safe radish]

@kind beacon Has your question been resolved?

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[proven wharf]

why is this wrong?

[mossy lotus]

what happened to (3x-1) in the numerator

and the x in denominator

[visual kraken]

It looks like you used v^2 instead of v' when you were filling it out

[proven wharf]

thankz

[visual kraken]

Otherwise I think your stuff looks good, just gotta fix that

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[clear garnet]

I need help this dont look correct

[clear garnet]

i did part c and it dont seem rigth

[keen talon]

yikes physics

[clear garnet]

ikr

so no help?

[mossy lotus]

well, your problem says the disc is parallel to the surface of the table

In your diagram, it doesnt look like that at

[clear garnet]

where u see this at

[mossy lotus]

they said the axle is vertical and disc is horizontal

you flipped it

[clear garnet]

i dont think that changes my maths tho

but u right about that

im confused tho

[mossy lotus]

your alpha went inexplicably missing

[clear garnet]

oop

that helps i think

[mossy lotus]

also, why do you have two object's torques in same equation? The torque on one object should be balanced by the moments of all the forces on it. So I*alpha = T*r or smthn.
or maybe I am being a big dumdum, afterall it has been years since I had to solve this kind of a problem before

[clear garnet]

dont i have to calculate both of the disks inertia to find the angular velocity of the equation

[gaunt socket]

is this rotational kinematics lmao

[mossy lotus]

also, considering that the two objects have different radii, you would have different angular accelerations for the disc and the pulley

[gaunt socket]

we were doing this shit in class today

[mossy lotus]

only the R*alpha products would be equal for the two

[clear garnet]

im confuzed

[mossy lotus]

well, the rope connects the pulley and the disk

so whatever rotation what takes place is imparted by the movement of the rope

so the tangential accelerations at the points of contact would have to match the acceleration experienced by the rope

so r_disc * alpha_disc which is the tangential acceleration is same as the acceleration a experienced by the mass mu and so on.

otherwise there would be slipping between the rope and pulley as well as rope and disc

[clear garnet]

im confuzed

me no comprehendo

[safe radish]

@clear garnet Has your question been resolved?

[safe radish]

@clear garnet Has your question been resolved?

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[tardy lark]

https://cdn.discordapp.com/attachments/1077788560805089433/1439055254904635492/image.png?ex=69192025&is=6917cea5&hm=a631693582fa00e168aca7d73efce52f6ef7b8d17a9df9d7e85ab5957a976678&

[tardy lark]

Why does this system of equation have no solution

4 unknowns, 5 equations

I used 4 equations for the system. But it doesn’t work

A,B,C and D are the unknowns btw

[safe radish]

@tardy lark Has your question been resolved?

[median vigil]

what makes you believe there are no solutions?

[tardy lark]

I used an online matrix solver

I also used real values instead of the constants in an excel spreadsheet and it gave me an error

[median vigil]

hm yeah now that i look at it, it probably is statically indeterminate

[tardy lark]

How can we solve this then?

There must be a solution

I’ve been thinking about, A=D and B=C since it’s symmetrical

However, what if c or b was slightly shifted to the right? There would be no solution?

And this example seems so simple, how would we solve complex beam supports

[median vigil]

well the issue is that you have too many supports

you have too many places where the moments have to sum to 0. if you had just 2 then you would be ok

actually now that i'm entering in the equations to my own calculator (using dummy numbers), i do get a solution?

[tardy lark]

Hmm

When i used real values it didn’t

Is there a tool that can solve the matrix ?

[median vigil]

it should work regardless as long as you set up your equations correctly

[tardy lark]

To reduced echelon form or smth

[median vigil]

any programming language with a linear algebra library

also most graphing calculators

[tardy lark]

I meant without replacing the variables with anything

Wait let me show you my excel

with =MMULT(MINVERSE(R8:U11),W8:W11)

[median vigil]

well i wouldn't expect it to be invertible

you have too many supports, so there will not be a unique solution for the forces

there are infinitely many different solutions

[tardy lark]

i see

what if i say A=D and B=C

from a simulation^^

[median vigil]

if you do that, then you can enter those as two of the equations

[tardy lark]

alright let me try

[median vigil]

i would still not necessarily expect unique solutions, though

[tardy lark]

Slightly better, but here we can see the issue

[safe radish]

@tardy lark Has your question been resolved?

[safe radish]

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[elder cypress]

I tried assuming f(x) to be a polynomial

Didn't work

[zealous ingot]

f''(x) = $x^2$ - f'(x) - f^2(x)

[flat frigate]

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

[zealous ingot]

you can assume P is some point ($x_0,y_0$)

[flat frigate]

donkey

[zealous ingot]

plug in x_0 to that DE

[elder cypress]

x² - y² = y''

[zealous ingot]

right

and y'' must be less or eq to 0

cuz point of maxima

[elder cypress]

Yes

[zealous ingot]

a doesn't have to be 0

[elder cypress]

Yeah I got confused between x and x_o

[zealous ingot]

fair fair

[elder cypress]

Now I just need to apply properties of hyperbolae, right?

[zealous ingot]

you can write the condition for tangent passing through point

discriminant will give you the answer

[elder cypress]

Thank you

.close

[safe radish]

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[safe radish]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[drowsy slate]

Guys i need help in understanding how to do real life application of parabola and quadratic function.

[drowsy slate]

can anyone [plss help me

guysss

[stoic saddle]

wires, projectile motion

[drowsy slate]

should i send the question screen shot?

[stoic saddle]

sure

[drowsy slate]

ok

pls help

pls help brooo

[hard crest]

okay, can you figure out the equation of this parabola? they've helpfully drawn the coördinate axes in

[safe radish]

@drowsy slate Has your question been resolved?

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Use .reopen if this was a mistake.

[drowsy slate]

.reopen

[safe radish]

✅ Original question: #help-23 message

[drowsy slate]

no. pls help me

[hard crest]

can you identify any points that the parabola goes through? also have you done Q1 yet?

[drowsy slate]

yes

[drowsy slate]

do we need to?

[hard crest]

actually, what have you done on this?

[drowsy slate]

umm Q1

barely

i tried solving it

but couldnt

should i ask chat gpt?

[manic radish]

don't ask jipidy

[hard crest]

Q1 isn't really a "solving" thing as much as it is a use your brain to think about how the world works thing

[manic radish]

What three relevant engineering challenges and safety concerns have you identified?

[safe radish]

@drowsy slate Has your question been resolved?

[drowsy slate]

ceiling height, tunnel width and turns

[drowsy slate]

bro stop being mean im in 9th grade ok idrk how to

[quasi bison]

she's a girl. do not call her "bro".

[drowsy slate]

oh srry

didnt know

[hard crest]

no i'm not trying to be it's just -- there isn't really a right/wrong answer. your answer seems fine

[drowsy slate]

oh ok

js help in Q2

[hard crest]

yeah so let's look at that curve. see how it has a highest point?

[drowsy slate]

yes

thats the vertex?

[hard crest]

we said the maximum height of the tunnel was what, 9.8m?

[drowsy slate]

yes

[hard crest]

yep!

[drowsy slate]

hmm

[hard crest]

and then hmmm what else do we know about the tunnel, we know its width right?

[drowsy slate]

yes 9.8

but at the axis of symmetry it will be - and + 4.9

[hard crest]

what points would that help us to find? it's okay if you ok perfect yes

[drowsy slate]

umm idrk

[hard crest]

no you got it

[drowsy slate]

oh

[hard crest]

what points have we identified that the parabola goes through?

the first was the vertex, we can write that as (0, 9.8)

[drowsy slate]

yes

ok

[hard crest]

what about the points where it's at its widest?

[drowsy slate]

9.8,0?

[hard crest]

remember what you said about the axis of symmetry

[drowsy slate]

yes

soo 4.9,0

[hard crest]

yep!

[drowsy slate]

YAY

[hard crest]

and (-4.9, 0) but that's going to be a bit annoying to use so we'll stick with (4.9, 0)

[drowsy slate]

ok

[hard crest]

now it wants us to write it in standard form, right?

do you know what that looks like?

[drowsy slate]

y= ax^2+bx+c ?

[hard crest]

yea

actually --- do you know what vertex form is?

[drowsy slate]

i thimk so

[hard crest]

that's going to be easier to use here since we already have the vertex

[drowsy slate]

y= a+(x-h)^2+k

i think so

[hard crest]

a times not plus, but yea

[drowsy slate]

oh ok

[hard crest]

so we know h and k

now we just need to find a

[drowsy slate]

how?

[hard crest]

well that's the vertex

For a parabola with vertex $(\blue h, \green k)$, its equation in vertex form is
$$y = \pink a(x - \blue h)^2 + \green k$$

[flat frigate]

bug

[hard crest]

what did we find for the vertex?

[drowsy slate]

ummm

confused...

srry

[hard crest]

that's okay

remember the highest point is the vertex

and we found that was at (0, 9.8)?

[drowsy slate]

yess

[hard crest]

okay great, so that's our h and k; let's plug those in and see where we get

[drowsy slate]

ok

can u do the math text thingy

[hard crest]

sure, but you could also just use ^2 for squared

it does look cool though :)

$y = a(x - 0)^2 + 9.8$

[flat frigate]

bug

[drowsy slate]

ok

[hard crest]

now we still have a bit of a problem, we don't know what a needs to be!

[drowsy slate]

so ax^2 + 9.8?

[hard crest]

yep

but we need to find a

[drowsy slate]

yes

[hard crest]

we can use that other point we found

[drowsy slate]

4.9?

[hard crest]

point

points have an x and a y

they're written (like, this)

just like we found that the vertex was at (0, 9.8)

[drowsy slate]

mhm