help-28

[empty sapphire]

see here

https://byjus.com/maths/difference-between-linear-and-nonlinear-equations/

[sinful canyon]

Sorry but like for example 9w-2=9w-2 is basically what it’s asking for i don’t know why my teacher is calling it that or my math book

[gritty rose]

If you're working on a problem, just post it here with a picture or screenshot

[empty sapphire]

so you're wondering why it has infinitely many solutions?

Can you clarify where you're confused here?

[sinful canyon]

I’m mostly confused on how I determine if it does or doesn’t have a solution

I’m sorry if it doesn’t make since I’m learning two years of math in a year and the class is too fast for me and I get confused easily

[empty sapphire]

no worries, just need clarification 🙂

Ok so there's only really 3 things that can happen here

bare with me one moment I'm typing up examples to help solidfy this

[sinful canyon]

Alrighty

[empty sapphire]

1. There is one solution. This is when we can solve for and isolate x.
   For example: 2x - 3 = 3x - 4. Solving this we see that x = 1 and that is the only solution here. Graphically, we see this in this photo. They intersect at x = 1 and the point in where they interect is (1, -1).

does that make sense so far?

[sinful canyon]

Yes

[empty sapphire]

ok good

so that should make sense

now let's take your example above

2. we have 9w-2 = 9w-2. If we try to solve for w, we get 0w = 0 or 0 = 0. What this tells us is that no matter what value we put in for w, the equation will always result in 0 = 0. For example say if w = 1, then:
   9(1) - 2 = 9(1)-2 ---> 7 = 7 ----> 0 = 0.

So, what this means is that no matter what we put in for w any value of w works. This should make logical sense, right? I mean we're saying that 9w-2 is equal to itself, of course whatever we put in for w the left hand side and the right hand side wil agree: they're the same line! Graphically it looks like this. Notice the two separate colors to indicate the lines are sitting on top of one another.

This is why the answer is "infinitely many solutions"

ok let me know if you get stuck here before I move onto the last acase

[sinful canyon]

So basically it’s infinitely many solutions because for example 9x4-2 = 9x4-2
And because of that w can be basically any number and it would be the exact same every time

[empty sapphire]

yeah exactly

[sinful canyon]

Alright

[empty sapphire]

it's literally the same line

ok

so for the last case

3. No solutions.

This occurs when we get something algebraically illogical.

For example, consider 3x -2 = 3x - 3. If we try to solve the equation, we get 1 = 0. This is obiviously not correct and so this is an erroneous answer. What that tells us is that there are no solutions for this.

Which, should make sense, right? Linearly, these lines are parallel and parallel lines never cross. So there will be no point of intersection ever.

Graphically it looks like this:

As you can see, there is no point and time where the parallel lines will cross, so we no x value in which they will meet.

So there are no solutions

[sinful canyon]

So if there parallel there’s no solution but if they aren’t that means there is basically

[empty sapphire]

yep!

it was one of Euclid's most important points when he wrote "Elements" that caused some controversy

even if the angle between two lines is .0000001 degree off, eventually the lines will intersect

so with linear equations, either they have 1 solution, no solutions, or infinitely many solutions

[sinful canyon]

Yup

[empty sapphire]

ok welp, glad that made sense

[sinful canyon]

Thanks

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[ornate wadi]

need help with this question

[ornate wadi]

<@&286206848099549185>

[next sail]

don't ping helpers unless no one's helped within 15 mins

[ornate wadi]

Hey!

[next sail]

what have you tried

[ornate wadi]

i had it opened in another help channel and no one responded in 15 minutes so i closed it and opened this one and pinged

sorry for the confusion

[next sail]

ah

not ideal

anyway this is a fun question

what have you tried

[ornate wadi]

this is what i have gotten so far

but im not sure if im going in the right direction or not

[next sail]

how did you get that

[ornate wadi]

i plugged in -2

[next sail]

why -2

[ornate wadi]

because the limit is x---> -2

[next sail]

ok...

[ornate wadi]

yeah im not too sure where to start

[next sail]

but it's x -> -2

not n

[ornate wadi]

ah

you are right

[next sail]

so you shouldn't necessarily make n -2

[ornate wadi]

should i plug it in for x instead

damn im dumb

my bad

[next sail]

it's fine

[ornate wadi]

didnt even realize i did that

[next sail]

anyway yeah trying x = -2 is a good sorta test

[ornate wadi]

so

i get

2n+15

-5

if i plug -2 in for x

[next sail]

that's not right

[ornate wadi]

oh shoot

hold on

[next sail]

the bottom is wrong

[ornate wadi]

is it undefined

[next sail]

?

[ornate wadi]

if i plug x = -2

is it undefined

[next sail]

is what undefined

the bottom is x^2 + x - 2

so if you put x = -2

(-2)^2 + (-2) -2
= 4 - 2 - 2
= 0

and the top is like

[ornate wadi]

i got

12-n+3 / 0

[next sail]

not quite

the top is 3x^2 + nx + n + 3

so that's like

3(-2)^2 + n(-2) + n + 3

[ornate wadi]

yea

[next sail]

oh wait yeah you're right

ok ok

[ornate wadi]

hehe

[next sail]

but you should combine the 12 and the 3 into 15

[ornate wadi]

15-n/0

[next sail]

but yeah basically the problem is that you're sorta dividing by 0

[ornate wadi]

yeah

[next sail]

so you don't get anything finite

[ornate wadi]

mhm

[next sail]

so why are you dividing by 0

is there a way to get around that sorta

[ornate wadi]

im not sure

[next sail]

why is x^2 + x - 2 = 0?

urgh i'm not saying it right

lemme try and explain it...

[ornate wadi]

okay

[next sail]

ok so yeah so the hint says also

given that the bottom is 0, what does the top have to be for there to maybe be a finite limit

[ornate wadi]

would n be 15 then

[next sail]

right

because then you would get 0/0

[ornate wadi]

yeaaa

[next sail]

so the limit could be finite

[ornate wadi]

i see

[next sail]

whereas if you have anything else / 0 then that's always gonna be infinity or something

[ornate wadi]

yeahh

[next sail]

so yeah what do you think now

[ornate wadi]

do i plug 15 in for n

[next sail]

yes

[ornate wadi]

is the answer -1

[next sail]

uhhhh

lemme think

[ornate wadi]

mk

[next sail]

that's not what i got

[ornate wadi]

whatd you get?

[next sail]

i don't want to just give answers

[ornate wadi]

okok

[next sail]

ok one sec

ok so one thing is sorta like

if n = 15, it should be like 3x^2 + 15x + 18 on top right

so the top should divide by 3

wait no

ok i'm dumb it is -1

i got -1 on the bottom instead of -3

you're right

[ornate wadi]

ahh okok

so is that the answer for a or b

im confused on that

[next sail]

?

the answer for a is the n you got
the answer for b is the limit

[ornate wadi]

ohhok

i gotcha

thank you

also

i got another question

if you dont mind

i appreciate your help a lot

[next sail]

uhhh

[next sail]

i'm sorry i'm going to be dragged away any moment now by someone, gotta go somewhere

[ornate wadi]

oh okay

no worries

thank you !!

[next sail]

open another channel for this new question and probs someone else will help

[ornate wadi]

okok

ty ty

.close

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[torn jolt]

can someone show me how to do the work here, i get stuck at common denominator part after putting 2/(x+h)+3 - 2/x+3 / h

[spiral vigil]

let me write it in tex so we can see

$\frac{\frac{2}{x+h+3} - \frac{2}{x+3} }{h}$

[glossy valve]

hayley!

[spiral vigil]

right?

[torn jolt]

looks perfect

thats the part im stuck at

idk how to do common denom from there

cause +3 messes it up :/

[spiral vigil]

easiest way to get a common denominator is to just multiply both of them together

[rapid rain]

And so yeah the common denominator would be $(x+h+3)(x+3)$

[glossy valve]

rafilou2003

[torn jolt]

alr let me foil that

[rapid rain]

Nonono no need to foil

[torn jolt]

why not

[rapid rain]

Keeping the denominator factored here will reduce the amount of useless calculations

[torn jolt]

what would the equation look like then, im kinda lost

if (x+h+3)(x+3) is common denom, how would u write that in the entire equation

[rapid rain]

So you would write $\frac{\frac{2}{x+h+3} - \frac{2}{x+3} }{h} = \frac{\frac{???}{(x+h+3)(x+3)}}{h}$

[glossy valve]

rafilou2003

[torn jolt]

alr let me write that down rq

question marks just 2 or??

[rapid rain]

It's up to you to find what goes in the question marks

[torn jolt]

2-2 = 0

so is it 0

[rapid rain]

No unfortunately

[torn jolt]

so is it 2

[rapid rain]

Neither

[torn jolt]

or 4

2)(2)

4

[rapid rain]

Do you know how to add fractions together ?

$\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$

[torn jolt]

brain farting

[rapid rain]

Ok I'm gonna write the whole formula but it's weird if you've never seen it

[torn jolt]

so lost

[glossy valve]

rafilou2003

[torn jolt]

ooooh

[rapid rain]

It's like the cross product

[torn jolt]

2x+6+2x+2h+6

[rapid rain]

Except there's a minus

So same formula but with - the other fraction

[torn jolt]

so u would do 2x+6 - 2x+2h-6

allowing u to cancel 2x and 6

and u keep h

[rapid rain]

Uh not quite, plus don’t forget parentheses

[torn jolt]

hmmmmmmmmm

[rapid rain]

(2x+6) - (2x+2h+6)

[torn jolt]

2(x+3) - 2(x+h+3)

yeah

[rapid rain]

Yes that's better

[torn jolt]

2x+6 -2x-2h-6

[rapid rain]

+6*

Yes and after simplifying ?

[torn jolt]

o

cross out x's

and 6's

left with -2h?

[rapid rain]

Yes!

So we're left with $\frac{\frac{2}{x+h+3} - \frac{2}{x+3} }{h} = \frac{\frac{-2h}{(x+h+3)(x+3)}}{h}$

[glossy valve]

rafilou2003

[torn jolt]

correct

then what

[rapid rain]

Well you can now simplify some more

[torn jolt]

cross out all the h's?

or

multiplying the polynomials actually

[rapid rain]

Uh which hs

Nono the first denominator keep it like that

[torn jolt]

am i able to cross out every h?

[rapid rain]

Which h cancel out ?

[torn jolt]

all of 'em?

[rapid rain]

First it might help to write :

[rapid rain]

$\frac{\frac{2}{x+h+3} - \frac{2}{x+3} }{h} = \frac{\frac{-2h}{(x+h+3)(x+3)}}{h} = \frac{-2h}{h(x+h+3)(x+3)}$

[glossy valve]

rafilou2003

[rapid rain]

This is not correct without taking the limit

[torn jolt]

how does that work???

bottom h just dissapears

i assumed u factored out other h, but its still there

[rapid rain]

Well, we divide -2h by something, then we divide by another thing, so it's the same as dividing by their product

(a/b)/c = a/(bc)

So yes top h and bottom h cancel out

[torn jolt]

but its still there

-2h

only that h was removed

it wasnt canceled with anything else

[rapid rain]

What not at all ! Look at the new denominator

The h is in front of the denominator

[torn jolt]

it was multiplied

so get rid of the bottom

but wouldnt u also multiply the -2h

making it -2h^2

[rapid rain]

Because of this formula

[torn jolt]

-2h is unaffected from that?

so like that

[rapid rain]

The formula : $\frac{\Big(\frac{A}{B}\Big)}{C} = \frac{A}{BC}$

[torn jolt]

never heard of that formuka

imma take note of it

[rapid rain]

That's false

You can't put a parenthesis lile that on the bottom

[torn jolt]

but multiplying bottom by h would make it dissapear

[glossy valve]

rafilou2003

[torn jolt]

ohhh

so like this

[rapid rain]

Yes

[torn jolt]

and so from there, do we cross out numerator and denom H, or do we simplify entire denom

cross out num and denom h, then factor in h = 0?

[rapid rain]

Yes!

And what do you get?

[torn jolt]

so rn im at -2 / (x+3) (x+3)

and i can foil that

then plug in c after that

[rapid rain]

Well no need to foil

[torn jolt]

true, but i alr did :/

[rapid rain]

$\frac{-2}{(x+3)²}$

[glossy valve]

rafilou2003

[torn jolt]

i got -2 / 36

= -1/18

[rapid rain]

Yep

[torn jolt]

perfect

thank you so much

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[desert ingot]

f(x,y)=
x
1
​
+
4+x
​
−
1+x
​can anyone sketch this

[void nova]

Am I the only one seeing this?

[onyx glen]

no you're not

[desert steppe]

No

[onyx glen]

@desert ingot got a picture?

we can't read this

[void nova]

I believe they wanted to make it look like a fraction but I don't get how terms are put together

[desert ingot]

This

[void nova]

Why didn't you write any square roots in your message? 😅 And there was no y, either

[desert ingot]

Copy pasted badly

[void nova]

Ok dw, remember that you should always check what you write before sending

Or to be sure, just send the pic so that is easier both for you and for everyone reading

Anyway, I believe this function is very hard to be sketched by hand

[onyx glen]

it's a function of two variables, for a start!

[void nova]

That's the main reason for which it's very difficult yeah

[desert ingot]

Even a rough shape helps

[void nova]

@desert ingot Are you sure you have to graph it? Or is that something for your curiosity?

[desert ingot]

Sketch a domain it says

[void nova]

Ahnn the domain only okok that's way easier lol

Let's start by watching the function: are there any denominators? Are there any logarithms? Are there any arcsin, arccos, tan...? Are there any roots with even index?

[desert ingot]

Yeah

[void nova]

So what are the conditions needed for those terms to exist?

For example, take $\frac{1}{x}$. What are the values of x you can put into that?

[glossy valve]

Alberto Z.

[desert ingot]

Eh it is what it is i can give gou range and domain tho

[void nova]

?

Can you tell me the domain of $\frac{1}{x}$?

[glossy valve]

Alberto Z.

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[signal zenith]

Hiya! I need some help with finding the length of D-F and D-E,

I also am not sure on how to find the areas of ADFG and DEGF

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[normal thicket]

Is there any software to see the locus sphere?

[normal thicket]

I wanted to see these questions geometrically

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[normal thicket]

<@&286206848099549185>

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@normal thicket Has your question been resolved?

[normal thicket]

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[frozen plover]

Let A be a subset of R so for all x reals it exists a inside A so that x<a is it true ? Sorry if it's not very precise

[fast peak]

is what true. that such an A exists? that any A satisfies this?

[jolly terrace]

sounds like the supremum

[frozen plover]

Ye it's the negative of supremum

[vast fossil]

the set A has no supremum though

[hidden harbor]

As you've stated the problem, that's not a question of if its true or not.
You're saying: We got a subset A, and it has this property

[vast fossil]

Yeah you can't question a definition

[frozen plover]

Ye it's what I thought

Is there a def of true or false in this assertion

[hidden harbor]

can you please check if that's exactly what you wanna say and rephrase if it isnt?

[frozen plover]

Yes

[hidden harbor]

Then that's not a statement, thus it cannot have a true/false value.
It's an assertion, so it's like that because you have defined it like that

[frozen plover]

Ok

[hidden harbor]

a statement with T/F values would be the questions Denascite has posed, for example.
Is there a subset A that satisfies that assertion?

[frozen plover]

I see thanks

👍

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[wispy sand]

Generating pattern
I need to find the general term for this sequence
"3, 9, 27, 81,"
I guessed that it's 3^n
But how to get 3^n as an answer?

[fast peak]

well what exactly do you mean with "get"

you can notice that the common ratio is 3 and from there get 3^n

but there is no proof that the "only" pattern this follows is 3^n, because that is false

[wispy sand]

I wanna know the exact solution on how i got 3^n

The teacher explained on how to get the general term if the sequence is shown, but i kinda didn't got to follow

[hollow wharf]

Are you asking how you come up with 3^n, or how to show 3^n is a general expression of the sequence?

[wispy sand]

How i come up with 3^n

Bro?

I can just put it as an answer with no solution

But idk if the teacher will allow it....

[wary stag]

what is the full question?

[wispy sand]

Letter C

So i got 3^n randomly

[wary stag]

im quite sure its just Tn = 3^n

theres no working

[wispy sand]

I wanna know the solution on how i got the answer 3^n

[wary stag]

what do u mean how you got it?

do you mean why is it 3^n?

[wispy sand]

There's solutions on how you get the general term

[wary stag]

something like that u mean?

[wispy sand]

Nth also means general term?

Nth term*

[wary stag]

yes

[wispy sand]

Then yes

Somwthing like that

[wary stag]

$Tn = 3 * 3^{n-1}$

[wispy sand]

Does T means a?

[wary stag]

yep

[glossy valve]

MrZhongZuChongTu

[wary stag]

its basically this

then u simplify it to

3^n

[wispy sand]

Ohh cuz 3⁰ is 1

[wary stag]

u understand how this = 3^n?

[wispy sand]

Yeah

Is that all?

I was waiting for the solution on how we got 3^n

But it's late rn

I needa sleep

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[wispy sand]

Thanks anyways

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[rigid belfry]

Do we use this formula for this question?

[simple ridge]

what is the question

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[proud spear]

Just a quick question :
How can one rigirously define irrational numbers ?

[jolly terrace]

Yes

[vast fossil]

This is not a yes/no question

[jolly terrace]

Did not see the " how" 😂

You use the definition for rational, which is any number that can be written as a ratio of integers

So anything that doesn't fit that can be defined as irrational

[vast fossil]

No, that can't be the definition

[thick hedge]

an irrational number is one that cannot be expressed as $\frac{a}{b}$ where a and b are rational

[glossy valve]

physicsrocks

[vast fossil]

Then the imaginary unit would be irrational

Forcing every complex number to be real

[thick hedge]

wait, what?

how

[vast fossil]

Irrational numbers are defined to be real numbers that are not rational

Rather than just "not rational"

[proud spear]

Ok, but this has only shifted the problem to: How can be real numbers rigorously defined?

[jolly terrace]

Are we talking in the context of real analysis or complex?

[vast fossil]

No, it shifted the question differently We need to define real numbers

[proud spear]

Whoops, I mistyped

[vast fossil]

One way of constructing the reals involves Cauchy sequences

[jolly terrace]

Eew

[vast fossil]

That's the only consturction of the reals that I know so

[proud spear]

That does not seem like something a 7th grader would know

[vast fossil]

[thick hedge]

I'd probably define it as a number that can't be expressed in the from of a+ib ; except when b=0. where i =√-1

[proud spear]

looks very straight-forward

thanks

[thick hedge]

or as math stack exchnage puts it, any number that can find a place on the number line

[jolly terrace]

the ratio thing is probably the intuitive way of thinking about it

if you can't write it as a ratio of integers, it's irrational

[proud spear]

I think I'll go with this approach,
thanks again everyone

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[royal halo]

hello

[royal halo]

I do not know how to solve these to

I could not start off

How should I start off with number 13

[plain fox]

the question tells us that Q is the midpoint of line PR

this means that
PQ = QR
because mid point is at the middle of the line and both the lines must be equal

[plain fox]

so just consider
PQ = QR
and solve for x

[royal halo]

should x be 6>

?

[plain fox]

yes

[royal halo]

Thankyou!

Am I allowed to ask you one more questions which is not number 14?

[plain fox]

ask

[plain fox]

if i know ill answer otherwise wait for another helper

[royal halo]

this one

[plain fox]

this is the diagram

a rough sketch , what do u think should be the answer ?

[royal halo]

should I do

the square root of y2-y1 squared plus x2-x1 squared?

[plain fox]

yeah but that gives distance between r and s

and what do u know about mid point (the last question)?

[royal halo]

im not quite sure what you mean

[royal halo]

would that mean that it is also the difference between s and t since the midpoint is s?

[plain fox]

yeah kinda of

consider t as (x,y) and then use the distance formula between s and t

and then consider both equal cuz midpoint

[royal halo]

8.2?

[plain fox]

idk i didnt solve

also u should get 2 answers , 1 for x and 1 for y

cuz there are 2 values in determining the coordinates of a point

[royal halo]

will it be 7/2, 1/2

[plain fox]

it seems that
x should be 9
y should be -4

this can be wrong tho

can u send ur working to figure out ur mistake

[royal halo]

oops will the answer be

(-7/2, 3/2)

[plain fox]

use the distance formula

wait are u allowed to do it that way?

thats the "not so correct and not so wrong " way

[royal halo]

I thought I should have used the midpoint formula

[plain fox]

yeah that

x2 + x1 / 2

[royal halo]

yes yes

that one

wdym by not so correct and not so wrong?

[plain fox]

but use it between r and t

[plain fox]

dont mind it , im dumb

[royal halo]

how should I use it

between r and t if I am supposed to find t

[plain fox]

just use it

t is (x,y) btw

[royal halo]

so the way I did and the answer was correct?

[plain fox]

no

[royal halo]

is it possible for me to see your solution method?

[plain fox]

idk

!nosols

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[royal halo]

i see

[plain fox]

ill explain it to u

[royal halo]

ok

[plain fox]

what do u think the midpoint formula gives?

[royal halo]

it gives the distance of the point and the midpoint

[plain fox]

no

it gives the MIDPOINT between TWO points

@royal halo u here?

[royal halo]

yes

trying to figure out

[plain fox]

ok and what is the midpoint between the line RT?

answer my questions and u will understand and figure out the answer

[royal halo]

s

[plain fox]

so if u used the midpoint formula between R and T , what would u get?

[royal halo]

(-9+x/2, 4+y/20)

[plain fox]

no

not the formula , what point do u think u would get when using midpoint formula between R and T?

[royal halo]

s

[plain fox]

yes correct

[plain fox]

so when u use mid point formula u get , this

but u also know that using mid point formula u get coordinate of S

so just take them equal

[plain fox]

(also its 4+y/2 and not 20)

[royal halo]

mhm

so basically I can

-9+x/2= 4+y/2

[plain fox]

no

[royal halo]

this is hard

[plain fox]

coordinates of s are (2 , -1)

using the mid point formula the coordinate u get are (-9+x/ 2) , (4+y / 2)

so take the respective coordinate equal
-9+x/2 = 2
4+y/2 = -1

[royal halo]

oh I get it

[plain fox]

its not that hard if i could teach u using solution , but this server dont allow

[royal halo]

can I friends request?

[plain fox]

no , but u can dm

i wont dm answer tho

this server wants ppl to properly understand questions so they can do the rest by them selves

and not for ppl who are higher in level to give out answers

[royal halo]

i see

which grade are you in?

[plain fox]

its different across countries

and educational systems

btw if ur question solved , close this channel with ".close"

[royal halo]

is it

(13,-6)

I have a doubt

[plain fox]

thats the answer im also getting

[royal halo]

ok

thankyou

you were very helpful

😉

[plain fox]

ty ♥️

[royal halo]

see you

.close

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[low bloom]

in the domain of f, can x be 1?

[low bloom]

after simplification f(x,y) is $sin^{-1}(y) - sin^{-1}(x)$

[glossy valve]

Swas.py

[low bloom]

so domain should be $x, y \in [-1, 1]$

[glossy valve]

Swas.py

[low bloom]

or should the boundries be not included

[vast fossil]

You may include them

[low bloom]

but wouldn't the integral be undefined at x = 1

[vast fossil]

Hm?

,w integrate 1/sqrt(1 - theta^2) from theta = -1 to theta = 1

[glossy valve]

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[somber drum]

am i doing this right?

[somber drum]

<@&286206848099549185>

[full forum]

@somber drum Has your question been resolved?

[rocky vale]

It looks like you've got the right general idea

but something is wrong

2.324 and 3.550

Where did those numbers come from?

@somber drum

in the second-to-last line

[somber drum]

i cube rooted 0.6 and 1.4

[rocky vale]

,calc 0.6^(1/3)

[glossy valve]

Result:

0.84343266530175

[rocky vale]

,calc 1.4^(1/3)

[glossy valve]

Result:

1.1186889420814

[somber drum]

huh i guess i didnt put it in correctly

[rocky vale]

yeah I was looking at this for a minute trying to figure out what went wrong lol

cuz you should have ended up with something negative on the left side

at the very end, after subtracting 1

[somber drum]

ok so my answer should be delta= 0.119 if i round it to 3 decimals right?

[rocky vale]

Yeah, that should work

Are you choosing that one instead of 0.15 for any particular reason?

[somber drum]

0.15? where did that come from?

[rocky vale]

what's the full inequality you end up with?

with all three expressions

[somber drum]

-1.843<x-1<0.119 right?

[rocky vale]

on the left, 0.843 - 1 = -0.157

not -1.843

[somber drum]

ah, im dumb

acutally its both right? since im looking for the values above and below 1

[rocky vale]

what you want to do is choose a delta that guarantees that
|x-1| < delta implies -0.157 < x-1 < 0.119

there isn't just one right answer

if you choose delta as 0.157, then that doesn't work

because |x-1| < 0.157 doesn't guarantee that x-1 < 0.119

because for example x-1 could be 0.156

ik it's kinda confusing with these decimals like this

[somber drum]

so why does 0.119 work?

[rocky vale]

because of the two, -0.157 and 0.119

0.119 has the smaller absolute value

you need a delta that is smaller than both absolute values

so even something like delta = 0.1 would be okay

[somber drum]

ah i see

[rocky vale]

in fact, I probably wouldn't round up to 0.119 for that reason, you want delta to be smaller than the numbers you get

[somber drum]

i pick which ever one is the smallest absolute value so that it for sure wont be bigger than |x-1| right?

[rocky vale]

yeah, and I'd also round down a bit for good measure

like, think about this

if you choose delta as 0.1

|x - 1| < 0.1 guarantees that -0.157 < x-1 < 0.119

right? because if x-1 is between -0.1 and 0.1

then it's definitely between -0.157 and 0.119

[somber drum]

yea i understand that now

thanks!

[rocky vale]

yeah, that's all you need, is any delta that guarantees that it falls somewhere in the right interval

no problem 👍

[somber drum]

also before you leave, i went ahead a bit and have another question

[rocky vale]

sure

[somber drum]

if you dont mind that is

on this one i got to part c but i dont really know what to do

[rocky vale]

can you show your work for part b?

[somber drum]

i know there isnt work but i kinda thought about it and if the error tolerance is 5, then the absolute value of the radius +5 minus the radius will give the exact error tolerance

[rocky vale]

sorry may need just a min

[somber drum]

take your time

wait would my f(x) be pi * r^2?

because thats the formula for a circle which would get me my limit in b

[rocky vale]

So, I get something that looks different for part b, it might be equivalent but

what we want to do is make this look like the definition of the limit

so we can label everything in part c

[somber drum]

yea i got that

[rocky vale]

obviously the e/d def of the limit is what you guys are studying right now, is it fairly new?

[somber drum]

yup just learned about a couple hours ago

[rocky vale]

ok, yeah this is one of those things that throws everybody at first

those absolute values in the definition, think of those as distances

when you read |x-a|, think "the distance between x and a"

[somber drum]

yea thats what my prof said as well

[rocky vale]

awesome, so the first thing they told us in part b is that we have an error tolerance of 5 cm

which means

the distance between the area we want and the area we get must be less than 5

in other words

|A - 623| < 5

[somber drum]

the distance from the area to 623 must be less than 5

ah ok i got it

[rocky vale]

so in this case A is like your f(x)

so yes, what you said before is true, f(x) = pi*r^2

and we have |pi*x^2 - 623| < 5

so, if f(x) is pi*x^2

can you tell me what 623 and 5 are?

for part c

[somber drum]

623 is the limit and 5 is what x is approaching?

[rocky vale]

623 is the limit L, yes

look at your limit definition

|f(x) - L| < what?

[somber drum]

epsilon?

[rocky vale]

yep, epsilon is always your "tolerance"

when you say, I need this function to be at least this close to the limit

you're saying, I need it to be within epsilon of the limit

in this case, we're allowed to be up to 5 cm off

[somber drum]

ok so if epsilon is 5, how do i find my delta?

[rocky vale]

like you did in the previous problem

for part a, you found that the exact radius necessary to make a perfect disk was sqrt(623/pi)

just like how epsilon is your tolerance for how far you're willing to let the Area of the disk vary, delta is the corresponding tolerance for how much you can allow the radius to vary

[somber drum]

ah ok give me a min to think about this

[rocky vale]

okay

think about what a should be before you worry too much about delta

[somber drum]

a should be the radius?

[rocky vale]

a is the ideal radius

what you calculated in part (a) of this question

x is the actual radius, so

|x - a| < delta means "the distance between the actual radius and the ideal radius better be less than..."

some number, delta

[somber drum]

it should be less then the absolute value of the sqrt(623-5/pi)?

[rocky vale]

that's not quite what I get

so, a is sqrt(623/pi)

do we agree?

[somber drum]

yea

[rocky vale]

so, starting from the |f(x) - L| part, we had

|pi*x^2 - 623| < 5

-5 < pi*x^2 - 623 < 5

and then, did you solve this inequality for x?

[somber drum]

ah wait i was like conceptually doing this in my head

let me do that

14.029?

[rocky vale]

that's roughly right, but they want exact values

and wait, what's the whole inequality after you solve for x?

[somber drum]

so sqrt(618/pi)

[rocky vale]

on the left, yes, and on the right it should be sqrt(628/pi)

[somber drum]

yea i got that too

[rocky vale]

sqrt(618/pi) < x < sqrt(628/pi)

awesome, but now we want x - a

so we can get |x - a| < delta

[somber drum]

so |x-sqrt(623/pi)|<sqrt(618/pi)

[rocky vale]

so close

you subtracted a in the middle, you need to subtract a on the sides as well

so $$\sqrt{\frac{618}{\pi}} < x < \sqrt{\frac{628}{\pi}}$$

becomes

$$\sqrt{\frac{618}{\pi}} - \sqrt{\frac{623}{\pi}} < x - \sqrt{\frac{623}{\pi}} < \sqrt{\frac{628}{\pi}} - \sqrt{\frac{623}{\pi}}$$

[glossy valve]

tatpoj

[somber drum]

ok i think i understand

and then dela should be which ever one is the smaller absolute value

[rocky vale]

yes

which is kind of annoying with an expression like this but

you can use a calculator if you're allowed

or

notice that the square root function increases slower and slower as it goes on

[somber drum]

soo the right one

[rocky vale]

yep

[somber drum]

god problem is trying to teach the entire lesson

anyways thanks a lot, i really appreciate it

[rocky vale]

no problem 👍

at the end you should have something like

$\left| x - \sqrt{\frac{623}{\pi}} \right| < \sqrt{\frac{628}{\pi}} - \sqrt{\frac{623}{\pi}}$

[glossy valve]

tatpoj

[rocky vale]

basically, if you're within that range of the ideal radius

you'll be within 5 of the ideal area, as required

[somber drum]

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[fervent anchor]

I understand the concept of a power set, but these questions are confusing to me

I need to do numbers 16, 18. Could somebody give me some sort of direction so I have a path to think about this?

i think 16 is 3 but im not sure

no, 4 i left out the empty set

so it would be { {}, {m}, {n}, {mn} } for 16 which has cardinality 4

[fast peak]

why do you think it is a constant number

[spiral vigil]

your answers to almost all of these should be expressions involving m and n

[fast peak]

you misunderstood what |A|=m means

it means that A has m elements

[fervent anchor]

a

i see

cardinality of A is A^m

right?

[fast peak]

the cardinality of A is m

A^m is again a set

its something else

[fervent anchor]

right

im dumb

[spiral vigil]

somewhere in your notes should be something about how power set interacts with cardinality

(size)

[fervent anchor]

this?

[spiral vigil]

yeah

you did the other problems, right? or are you only doing 16 and 18?

[fervent anchor]

onlu doing 16 and 19

i18

i did other power series problems but none regarding cardinality yet

16 & 18 not 19

so the power set of a would be 2^m ?

[spiral vigil]

the cardinality of the power set of A would be 2^m yes

[fervent anchor]

yes tahts what i meant sorry

[spiral vigil]

[in fact, writing 2^A is another notation for the power set instead of P(A) ]

[fervent anchor]

So now we have to cross 2^m and 2^n

then take the cardinality of that

right?

if we dont know any of the elements though how would we take the cartesian product?

or would it be

|A| * |B|

so we get 2^(n+m)

@spiral vigil does this seem right?

[spiral vigil]

we don't know what the elements are but we know how many there are

very conveniently, we have this identity:

$| A \times B | = |A| \cdot |B|$

[glossy valve]

hayley!

[fervent anchor]

thats what I did here, right?

2^n * 2^m = 2^(n+m)

[fervent anchor]

i think what I did is the same as this at least

[spiral vigil]

yeah

it is

[fervent anchor]

nice

so for 18 ive made some progress

ive done |A| & |P(B)| = m * 2^n = 2m^n

so now im with |P(2m^n)|

[spiral vigil]

did you say $m \cdot 2^n = 2m^n$?

[glossy valve]

hayley!

[fervent anchor]

yes

[spiral vigil]

because that's not true

[fervent anchor]

oh

why wouldnt it be?

[spiral vigil]

better question: why would it be?

[fervent anchor]

well

2 * m is 2m

[spiral vigil]

$1 \cdot 2^7 \red\neq 2\cdot1^7$

[glossy valve]

hayley!

[fervent anchor]

true

hm

so i guess just put the m first

[spiral vigil]

PEMDAS

[fervent anchor]

m2^n

would it be that?

[spiral vigil]

just leave it as m * 2^n

[fervent anchor]

ok

[spiral vigil]

m2^n looks weird

[fervent anchor]

so now we refer back to this

so the cardinality would be

2^(m * 2^n)

[spiral vigil]

yeah

[fervent anchor]

sounds good

thank you!

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@fervent anchor Has your question been resolved?

[fervent anchor]

.close

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[fervent anchor]

.reopen

[full forum]

✅

[fervent anchor]

So we have this here, I know X is a circle, but for some reason my mind is blanking and I cant draw Y

is it just a line between -1 and 0?

like y=x between -1 and 0 on the y axis

[glacial pasture]

its the region bounded by y=0 and y=-1, just the space between those two lines

[fervent anchor]

tyvm

.close

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[queen cobalt]

Can someone teach me parent functions

[royal charm]

That's a very broad question and we have no idea what you're actually asking. Are you asking about y=x^2? y=1/x? y=sqrt(x)? Is it the transformations from parent functions? The more specific you can be, the better. Right now, no one can help you because we don't know what you want.

[queen cobalt]

Yea

Like the x^2 and square root

[void nova]

Try to be more specific.. What do you want to do with these functions? Which math do you already know? Do you know what a cartesian plane is? Do you know how to graph lines (such as y = 3, or y = -x +5 and so on)?

[queen cobalt]

I have a test tomorrow

And it's about parent functions

Beginner parent functions though

[void nova]

You have a test and you don't even barely know what to ask 😅

[queen cobalt]

How to identify all parent functions

[void nova]

This doesn't mean a lot

[queen cobalt]

A table for the coordinate on the parent functions

[void nova]

Mmh I'm not getting you

[queen cobalt]

Ok can you reach me function operations instead

Teach*1

Also I have a question

[void nova]

https://www.youtube.com/watch?v=2gVC9GhOPUo

https://www.youtube.com/watch?v=HBVevKX0_5w

[queen cobalt]

Ok

Thank you

[void nova]

Those two videos might help you, if I understood your questions correctly, but I'm not sure

[queen cobalt]

Ima watch the videos

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[fair mist]

Hello, I was wondering if someone could help me with this problem

I know what all of them are is just I don't know if they want range and domain in appropriation notation

And then also I don't know the number since the coordinate plane

No numbers

<@&286206848099549185>

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@fair mist Has your question been resolved?

[fair mist]

<@&286206848099549185>

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@fair mist Has your question been resolved?

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@fair mist Has your question been resolved?

[unborn quarry]

Hello, notice that they are a countable number of ordered pairs (points), so the domain is just a list of numbers. It's the list of first members of each ordered pairs, and the range is the list of second members

[unborn quarry]

I guess you should assume each square is 1 unit

[fair mist]

So what would be the answer

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[tawdry grove]

how do i solve these

[jolly terrace]

2x+1

you wrote the wrong sign

[tawdry grove]

oh riight

then what do i do to the infinities

[jolly terrace]

Anyway, you can factor out an x from each of those parenthesis

$(x-a) = x (1 - a/x)$ like this

[glossy valve]

TooManyCooks

[jolly terrace]

this is the trick you want to get used to if you're evaluating limits at infinity

[tawdry grove]

whattttt how do we do that, thats weird. x goes out in front then we make it 1 minus a, waaa

and i need to do that too all the top and bottom terms?

[jolly terrace]

here's the deal

when you have x or powers of x going to infinity

it's hard to see how the fucntion will behave

so you want those infinities to "cancel" each other out

That's what this trick is supposed to help with

Actually you didn't even need to factor the expression you were given

Are you with me so far? Or should I try to reexplain it

[tawdry grove]

waht if i multiplied the orgiginal expressinon by like 1/x^2

[jolly terrace]

Exactly!!!

[unreal fiber]

For number 1 you could only focus on the dominant terms

[tawdry grove]

bu thats rigged

[jolly terrace]

but you gotta do it for both the numerator and denominator

[tawdry grove]

and bullying

[jolly terrace]

so it's just multiplying by 1

[tawdry grove]

elaborate pls

[unreal fiber]

$\frac{7 + 2x - 5x^2}{2x^2 - 7x - 4} \approx \frac{-5x^2}{2x^2}$ if x is very large

[glossy valve]

chaddypioneers

[unreal fiber]

because the x^2 terms will dominate over the x terms (i.e. they will become much much larger to the point where the other terms are negligible)

[unreal fiber]

now you just find the limit of $\frac{-5x^2}{2x^2}$ as $x$ approaches infinity

[glossy valve]

chaddypioneers

[tawdry grove]

but you just ignored everything thing else in the expression just becasue they do not have x^2

[unreal fiber]

yes when you take limits to infinity you can do this

because the x^2 terms are the most important (they grow much faster than the others)

and as x goes to infinity the others become irrelevant

[tawdry grove]

assume i do use this, what do i do after that, would i still multiply by the 1/x^2

[unreal fiber]

you would simplify that

[tawdry grove]

-5/2

[unreal fiber]

yep

[tawdry grove]

uh

what that means

[unreal fiber]

that's the limit

as x goes to infinity

[tawdry grove]

and then what bout - infinty

[unreal fiber]

this would still apply if x is a very large negative number (like -10000)

because the x^2 would make it positive

and the x^2 terms would still dominate

[jolly terrace]

Think of it this way

$\lim{x\to\infty}\frac{-5x^2 +2x +7}{2x^2 -7x -4} = \lim_{x\to\infty}\frac{x^2}{x^2}\left(\frac{-5 + \frac{2}{x} + \frac{7}{x^2}}{2 - \frac{7}{x} - \frac{4}{x^2}}\right)\ = \lim_{x\to\infty}\left(\frac{-5 + \frac{2}{x} + \frac{7}{x^2}}{2 - \frac{7}{x} - \frac{4}{x^2}}\right)$

[unreal fiber]

so as x goes to -infinity it would still work

[glossy valve]

TooManyCooks

[jolly terrace]

as x goes to infinity, those 1/x terms go to 0

the only thing that remains are the constant terms

[unreal fiber]

that also works too

[tawdry grove]

1/x terms?

[jolly terrace]

Anything divided by x

or powers of 1/x

become vanishingly small when x is very large

[tawdry grove]

oghhhhhhhhhhhh, you are righ t

so basically you have e-5/2 again

but like

[jolly terrace]

so the constant terms remain

[tawdry grove]

how to find - infinity valu e

[jolly terrace]

what do you mean

oh i see

[tawdry grove]

as x goes to infinity the value is -5/2

when x goes to negative infinity value is what

[jolly terrace]

You basically put a minus sign on all of your x

so if you have f(x)

evaluate f(-x)

then do the same thing

evaluate at infinity

[unreal fiber]

as x goes to -infinity the 1/x terms also vanish as they approach 0

[tawdry grove]

either way the 1/x terms vanish then

[unreal fiber]

take this but replace $\infty$ with $-\infty$

[glossy valve]

chaddypioneers

[tawdry grove]

for positive infinity we plugged in positive values which cancles out the 1/x thing to 0

for negative we plug in negative values?

[unreal fiber]

what is $\lim_{x \to -\infty}\frac{1}{x}$?

[glossy valve]

chaddypioneers

[jolly terrace]

yes, but the whole idea of1/x terms vanishing stays the same

[tawdry grove]

i still dont understand,

"when x goes to negative infinity value is what"

[jolly terrace]

Are you comfortable with the trick though?

turning stuff to 1/x

or powers of it

[tawdry grove]

do i know how to do it yes. do i understand we a re essentially just multiplying by 1, yes

[jolly terrace]

Right. There are three cases that really matter here. What you need to look out for is the order of the polynomials.

If the order of the numerator and denominator are equal (like in your example), the limit at infinity (plus or minus) approaches the ratio of the coefficient of the leading terms (in your case -5/2)

[full forum]

@tawdry grove Has your question been resolved?

[full forum]

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[golden saffron]

Hi

[simple ridge]

hi

[golden saffron]

...

Give u a easy question

[torn jolt]

Well on the LHS or RHS?

[golden saffron]

Just let it be right

[torn jolt]

8x8x8 = 1x8x8x8

[golden saffron]

...

(8×8×8)^0

=1

[torn jolt]

We are both correct

That's why you gotta be more specific

[simple ridge]

i can say 8x8x8/(8x8x8)=1

you can do it in infinitely many ways

[full forum]

@golden saffron Has your question been resolved?

[full forum]

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

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[sleek apex]

what

anti derivative?

apply power rule for integration correctly

[languid fable]

u mean dy/dx?

[golden saffron]

@full forum nope

[languid fable]

or integral

[sleek apex]

int(x^n dx)=x^(n+1)/(n+1)

we have x^-2

n=-2 in this case

n+1=-1