help-39

[rapid wagon]

But if I substitue these I still have 2x mixed with the u variables

[rare burrow]

oh this looks dirty

yeah what if you integrate by parts twice

and then it should work out I think?

wait let me look at something

[rapid wagon]

What do you mean by that?

Also can someone remind what is integration by parts? is it the integration method using multiplication rule of derivatives?

[rare burrow]

like, maybe pick the $u$ that hylke suggested and $dv=\frac{1}{1+x^2}$

[jolly parrot]

00100000

[rare burrow]

it's basically $\int u~dv=uv-\int v~du$

[jolly parrot]

00100000

[rapid wagon]

Ah yes

[toxic lichen]

\dd{v}

looks better

[rapid wagon]

How you know to use this one and not substitution method or something else? experience? or I can tell by something specific here

[rare burrow]

oh, thank you

$\dd{v}$

[jolly parrot]

00100000

[rare burrow]

oh that's hot

$\ddv$

[jolly parrot]

00100000
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rare burrow]

damn you do need the {}?!?

oh wait no

[toxic lichen]

yes you do

[rare burrow]

$\dd v$

[jolly parrot]

00100000

[rare burrow]

there we go!

[toxic lichen]

the thing about LaTeX commands is that they look \like{this} generally, and the name of the command is a sequence of letters without spacing between them

[rare burrow]

lmao allow me to use dirty syntax on mathcord

[toxic lichen]

so if some other letter brushes up against that

you can get unrecognized commansd

commands*

e.g. \sinx fails

[rare burrow]

ye ye ik ik. it's like the whole $\frac 12$ thing

[jolly parrot]

00100000

[toxic lichen]

but a number, bracket or whatever else breaks the sequence

[rapid wagon]

,rccw

Integration by parts again?

,rccw

[rapid wagon]

I'm stuck🫡

[hazy pilot]

This is good

notice how you have - of the original integral

[rapid wagon]

Omggg

Nvm

[hazy pilot]

owh

[dapper spruce]

IBP is incorrect

v' = 1/(1+x²) , so , v=arctan(x)

[rapid wagon]

oh

,rccw

[jolly parrot]

Couldn't find an attached image in the last 10 messages.

Apr 4, 2025, 4:13 PM

[rapid wagon]

Back to stuck idk

,rccw

[rapid wagon]

Oh wait i have arctanx

I want arctan^2x so i can get I(x)

How?

Oh wait if i do integration by parts again

Nvm

[rare burrow]

v is wrong here

[dapper spruce]

you can use IBP on I(x)

[rare burrow]

oh wait you fixed it, my bad

[dapper spruce]

I(x) = int of arctan²(x)dx

put , u=arctan²(x) , du = 2arctan(x)/(1+x²)dx , dv =dx , v =x

[dapper spruce]

and compare it with this

[hazy pilot]

it's imaginary I think

[rapid wagon]

How u=arctan²(x) when my expression only has arctan(x)

[rapid wagon]

What is

[dapper spruce]

i mean this .

use IBP on the first integral

[rapid wagon]

I'm a bit lost ngl

[rapid wagon]

Didn't I already do this

[dapper spruce]

you did this with the second integral

$$\int \frac{\ln(1+x^2)}{1+x^2}dx$$

[jolly parrot]

ΔCΣ♠

[dapper spruce]

this is the second integral

,, \int \arctan^2(x)dx

[jolly parrot]

ΔCΣ♠

[dapper spruce]

and this is the first integral

[rapid wagon]

Ohh

Ohhhh

[dapper spruce]

use IBP on the first

[rapid wagon]

Ok

[dapper spruce]

and keep this

to compare them

[rapid wagon]

dv= dx ?

[dapper spruce]

yep

[rapid wagon]

U can do that?

[dapper spruce]

why you dont do it

its not hard to you

[rapid wagon]

No i just thought u need a function dx is not function

[dapper spruce]

the function is 1

1*dx = dx

arctan²(x) = 1*arctan²(x)

[rapid wagon]

Oh

Sorry im new to integrals

[dapper spruce]

No problem

[rapid wagon]

Is this ok?

[dapper spruce]

true

[rapid wagon]

Tysm!!

[dapper spruce]

No problem

[rapid wagon]

.close

[dapper spruce]

if you have more integrals you can send them

[pearl pond]

Channel closed

Closed by @rapid wagon

Use .reopen if this was a mistake.

[rapid wagon]

Ok!

[pearl pond]

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[desert solar]

$\rho=4, \theta=\frac{\pi}{3}$

[jolly parrot]

Task Bot

[desert solar]

corresponds to $z = 2+2 \sqrt {3}i$?

[frozen lantern]

ok

....................what.

ah rho for r

okay

[jolly parrot]

Task Bot

[frozen lantern]

,w 4e^{i pi/3}

[jolly parrot]

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Requested by percykel

[frozen lantern]

,calc 2 sqrt(3)

[jolly parrot]

Result:

3.4641016151378

[frozen lantern]

seems okay.

[desert solar]

Because the solution says $2\sqrt{3}+2i$

[jolly parrot]

Task Bot

[desert solar]

😭

[frozen lantern]

misprint maybe

[desert solar]

Okay!

[frozen lantern]

or maybe you read the argument wrong

[desert solar]

Write in algebraic form the complex numbers they have as a form and
as the main topic the pairs of numbers indicated below

It says this

[frozen lantern]

hmm

[pale frost]

book is wrong

[desert solar]

Thanks

.close

[pearl pond]

Channel closed

Closed by @desert solar

Use .reopen if this was a mistake.

[pearl pond]

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[dapper spruce]

.open

[dapper spruce]

i want to see if my steps are true

[hazy pilot]

I don't see any mistakes

[toxic lichen]

where are you typesetting this

the exponents are very very indistinct rn

like i legit thought they were all asterisks

[dapper spruce]

$(a+ib)^n , (\sqrt{a^2+b^2})^n$

[jolly parrot]

ΔCΣ♠

[toxic lichen]

this part

[dapper spruce]

?

[toxic lichen]

relies on an unspoken assumption that a>0

like what you've essentially recreated is de moivre's formula except arg(a+ib) ends up off by half a turn if a < 0.

no, $\sqrt{a^2} = |a|$ not just $a$.

[jolly parrot]

Ann

[toxic lichen]

\a is not a command.

[dapper spruce]

so are there problem too ?

[toxic lichen]

i just told you exactly where i saw a problem.

if you want, i can repeat myself n times.

i'd rather not do that though.

[dapper spruce]

ok dont be mad

[toxic lichen]

i am not mad.

[dapper spruce]

we can try for some values

[toxic lichen]

@midnight haven why are you shushing me?

[toxic lichen]

try a=-1, b=0. your y ends up as 0 and you get (-1)^n = 1.

[dapper spruce]

i see , so "a" should be more than 0

[toxic lichen]

this is what i said all along.

it would have been much better if you heard and understood me the first time i said it.

[dapper spruce]

hmm , that doesnt matter , have a nice day and dont be mad

.close

[pearl pond]

Channel closed

Closed by @dapper spruce

Use .reopen if this was a mistake.

[toxic lichen]

the more you say "don't be mad" the more likely it is that the other person will actually get mad because of that

[pearl pond]

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[desert solar]

Factor $z^4-2z^3+12z^2-14z+35=0$

[jolly parrot]

Task Bot

[robust oar]

!status

[pearl pond]

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

[desert solar]

$z(z^3-2z^2+12z-14+35/z)=0$

[jolly parrot]

Task Bot

[toxic lichen]

this is a fake factorization and is completely in the wrong direction

[desert solar]

Wdym

[toxic lichen]

i mean that this is useless.

because you introduce a 1/z

you cannot have that

[desert solar]

I can try to divide by Z-1

[toxic lichen]

do it, see if it works.

try other rational roots.

[desert solar]

$(z^4-2z^3+12z^2-14z+35)/(z-1)$

[jolly parrot]

Task Bot

[desert solar]

,w (z^4-2z^3+12z^2-14z+35)/(z-1)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sofficino9188

[desert solar]

$(1-2i)(1+2i)(-i\sqrt{7})(i\sqrt{7})=0$

[jolly parrot]

Task Bot

[desert solar]

,w graph z^4-2z^3+12z^2-14z+35

Here you have to use Wolfram

How to read these graphics

😭

,w graph z^4-2z^3+12z^2-14z+35=0

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sofficino9188

[desert solar]

.close

[pearl pond]

Channel closed

Closed by @desert solar

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[pearl pond]

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[spice nest]

For quetsions like the following:

Where y = equation find the points on the parabola or equation which are closet to point (a,b)

when do I need to do (x+a) or (x-a) added to (y+b) or (y-b)?

[rough stream]

There's many ways such a question could be handled, but you seem to have a method in mind. Does the method have a name? Or do you have an example of that method being used?

[spice nest]

This is how it is done in answer keys and I was just curious on how I should be handeling it or if their are other alternatives, etc...

[rough stream]

Did you have a question about this solution?

[spice nest]

Yes I do

Where they do (y-3) (where y is replaced with 6 - x^2)

when do I know

I need to do y-3 as opposed to y+3

[rough stream]

This line is the distance between (x,y) and (0,3)

It could have been written:
d² = (x - 0)² + (y - 3)²

You'd never write + 3

Note they are just finding the distance between (x,y) and (0,3), then minimizing that distance by taking a derivative in terms of x

[spice nest]

what if it was (x,y) and (-1,-3)

would I be doing

d^2 = (x+1)^2 + (y+3)^2

[spice nest]

Yes I understand that part clearly

Alright then thanks a lot!

.close

[pearl pond]

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[pearl pond]

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[heady vault]

I'm so confused by this question on definition integrals. It's for finding the rectangle approximation problems, like you would find in 5.1 [https://openstax.org/books/calculus-volume-1/pages/5-1-approximating-areas]

But what the fuck man, I come up with insaaaane numbers and it wants some equation form?
https://i.imgur.com/ICZJMEs.png - this isn't a homework question just practice.

I'm looking at my notes and the textbook and I don't see /anything/ remotely similar on generating an expression for A=

[heady vault]

is it just asking for like integral lower bound 1, upper bound 6 f(x)dx?

[rancid depot]

it is asking for the sum version

[full relic]

Yes

[heady vault]

isn't that just like

f(xi)dx

[full relic]

It’s riemann integration

[sharp vigil]

it gives you an explicit formula for both $x_i$ and $\Delta x$, which you can simply plug into the given definition

[full relic]

Yeah

[jolly parrot]

cloud

[full relic]

It wants the limit

So: the integral

[heady vault]

oh god

I spent an hour overcomplicating this

lmao

[full relic]

Yes

Sometimes you gotta not overthink

[heady vault]

why would it ask me for something that has the answer higher up though

[sharp vigil]

note that it does want you to calculate this as the limit of the sum, not using the fundamental theorem of calculus

[heady vault]

interesting

[sharp vigil]

so you'll have to plug in f, x_i, Delta x, and find the limit

[full relic]

But it’s still just an integral

[heady vault]

so it's 1+(3/4(8x/x^2+1))

because n is just 4 for a rectangle no?

[full relic]

What do you mean n is 4

N is how many rectangles you use

[heady vault]

woops

so 3/n

What the hell do you use for n?

[sharp vigil]

you compute the sum as a function of n and then take the limit as n approaches infinity

that's what the definition says

[heady vault]

This is actually needlessly complicated, if b-a/n is just 5/n
xi is just 1+ (1 through 6) times 5/n

Then you have to plug in n, n(n+1)/2, up to 6....

[sharp vigil]

i is an index variable which varies between 1 and n (in the sum)

[heady vault]

right but it's between 1 and 6 because of the interval

[sharp vigil]

i is the index of a particular rectangle, so its range (1 to n) depends on the number of rectangles (n), not the length of the interval

broad overview of your steps:

1. calculate the expression in the sum as a function of i and n
2. sum that expression with i ranging from 1 to n
3. take the limit as n approaches infinity

[heady vault]

yeah I'm completely lost none of this is making any sense

the textbook doesn't match what it's asking

my notes are about an example of x+x^2+x^3 which just translates into pure numbers not an expression

it's not (5/n)(1+i/(5/n)) so I don't see what I'm doing wrong

delta is is 5/n, a is 1

[sharp vigil]

https://youtube.com/watch?v=owqpnn4bmKs

[sharp vigil]

that nearly correct for $\Delta x \cdot x_i$, but what they are looking for is not that but $f(x_i) \cdot \Delta x$

[jolly parrot]

cloud

[sharp vigil]

looking at the question more closely i don't think you actually have to evaluate the sum/limit, but you do need to write it out correctly

so here is an example just focusing on writing the sum itself: https://youtube.com/watch?v=eidemLqiPyQ

[heady vault]

yeah it's just not computing in my head how to go through it

I'll get it but this example was not explained at all in class

[sharp vigil]

you have all of the information you need in the problem image, it's just a matter of applying it

[heady vault]

yeah that first video helped a lot I think, if this is the right way

but wtf how does a fraction change it because he's taking merely addition

like it's i^2 + i + 1 at the end, whereas mine is i/i^2+i+1

i's cancel out?

[sharp vigil]

well the placement of i is based on where x was

for this problem they only ask for "an expression" so i think you can just write out the sum without evaluating it

[heady vault]

Got the submission wrong so working through new one

This shit is fucking aids tiny annoying steps

https://i.imgur.com/htkdXPz.png

that's wrong apparently

even following the videos steps it's not right lol

[sharp vigil]

what function and region are you trying to compute for?

[heady vault]

https://i.imgur.com/1hU6Ls7.png

So I'm starting with 6(2+(3/n)i) on the top, then on the bottom I'm getting (9/n^2)i^2 + (12/n)i + 4

[fickle schooner]

Missing the +1

On the denominator

[heady vault]

so +5?

[fickle schooner]

Yh

[heady vault]

https://i.imgur.com/vJR9QOU.png

I don't get this

at all

let me try cnacelling out the (/n)i

nope

what in the fuck is this question even asking me man

[fickle schooner]

You missed something

Delta x

[sharp vigil]

webwork generally doesn't require you to simplify your answers

[heady vault]

oh

I forgot to multiply by 3/n

that was the most aids question I've done all year

https://i.imgur.com/vSdY3Fw.png

so you take x subscript i to substitute for X, map it out algebraically then multiply that by delta x

[pearl pond]

@heady vault Has your question been resolved?

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[frosty grail]

I am learning how to find the derivative of exponential functions
For this example,
y = 2^x + 3^x

[frosty grail]

The final answer in the answer key is (2^x)(ln2) + (3^x)(ln3)
Is this really how far you can go with it?

[open rivet]

yup

in general: $\dv{x}a^x=\ln(a)\cdot a^x$

[jolly parrot]

Bonk

[frosty grail]

Ok, thanks so much!

I have another question

So for f(x) = 3e^2x - 2^3x,
I got 6e^2x - (2^3x)(ln2)(3)

Can I multiply the 3 and the 2^3x?
So I would get 6e^2x - (6^3x)(ln2)

?

[open rivet]

no

[frosty grail]

Why not?

[open rivet]

remember what exponents are

2^3=2 * 2 * 2

[frosty grail]

I thought that would matter when you are collecting light terms through addition or subtraction.

Oh

[open rivet]

if i do 3 * 2^3 i get 3 * 2 * 2 * 2 which is not equal to 6 * 6 * 6 which is 6^3

[frosty grail]

I see, that makes sense

So it changes the exponents

It would lose its original meaning

Ok thank you

[open rivet]

if you had 3^3 * 2^3, then you can make it 6^3

[frosty grail]

So if the exponents are the same, then you can multiply them?

[open rivet]

yes

and you can see this once again when you expand them

$3^3=3\cdot 3\cdot 3\2^3=2\cdot 2\cdot 2\3^3\cdot 2^3=3\cdot 3\cdot 3\cdot 2\cdot 2\cdot 2=(3\cdot 2)\cdot (3\cdot 2)\cdot(3\cdot 2)\=6\cdot6\cdot6=6^3$

[jolly parrot]

Bonk

[frosty grail]

I see!

But if we had something like 3^2 and 2^3, then it wouldn't work, right?

3 * 3 * 2 * 2 * 2 = 72

And then if you somehow multiply them together, I'm not even sure how that would work out cuz you can't do it

Ok thank you so much, I get it now

[edgy wren]

you could rewrite it as 3^2 * 2^2 2 and then
(32)^2 (2)

[open rivet]

indeed, but what you can do is something like 3^2 = 3^3/3 => 3^2 * 2^3=(6^3)/3

[frosty grail]

Oh wait that's right

OHH you guys are right

So you're saying I would have 2^2 * 3^2 = 6^2
then just have the other 2 multiplied alone?

So like 6^2 * 2

[frosty grail]

Wait I am having a bit of trouble reading that math, is it okay if you can do it on that math drawing thing

[edgy wren]

$3^2=\frac{3^3}{3}, \frac{(3\cdot2)^3}{3}$

[jolly parrot]

yoboiqimmah

[open rivet]

\begin{align*}&3^2=\frac{3^3}3\&3^2\cdot 2^3=\frac{3^3}3\cdot2^3=\frac{6^3}3\end{align*}

[jolly parrot]

Bonk

[open rivet]

clearer?

[frosty grail]

Yes, that's much better! But I don't understand what the second line is trying to do. I don't understand what 6^3/3 is supposed to mean or do

[open rivet]

idk either

just a different way of writing it

doesnt necessarily need to have a use

what does have a use is being able to recognise that you can rewrite it

[frosty grail]

Mmm

So what is 6^3/3 supposed to represent

I get the first line, with the rewriting of 3^2

[open rivet]

since perhaps you get an exercise where having it has 6^3/3 is more useful than 3^2*2^3

[open rivet]

,calc 6^3/3

[jolly parrot]

Result:

72

[frosty grail]

Oh, I see

3^2 * 2^3 isn't specific to my original question, you were showing how that expression can be rewritten

That makes sense now, quite new too

[open rivet]

yeah

and being able to rewrite equations is a very useful skill to have

[frosty grail]

That's really true
I can't tell you how many times I am doing math problems and I look at the way someone solved it and I'm like, "that's possible to do??"

It feels like illegal math but somehow you can do it

Very cool

[fossil jewel]

lol youll have this feeling for the rest of your mathematical career

stronger year by year if anything haha

[frosty grail]

LMAO

[open rivet]

true

[frosty grail]

Do you guys happen to know if it is possible to save a PDF of my convos here so I can refer back to them

Is that possible?

Aside from taking screenshots

[open rivet]

i dont think so

but you can keep the message link

#help-39 message like so

[frosty grail]

Oh message link?

Ohhhhh

[open rivet]

this works as long as the messages dont get deleted

[frosty grail]

Do these channels get deleted or do they keep getting recycled

I saw some ones got wiped

[open rivet]

they get recycled

very rarely they get deleted

[frosty grail]

Alright good to know, thanks so much for your help

[open rivet]

!done

[pearl pond]

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[storm verge]

hello

[storm verge]

Find all odd positive integers
n > 1 such that if a and b are relatively prime divisors of n, then a + b − 1 divides n.

.close

[pearl pond]

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[umbral lodge]

How can a slope be equal to 0 in a linear equation graphed if Ax + By = C, so By = C - Ax, so, if A, B, C are constants, C - Ax_1 must be different of C - Ax_2 ?

[umbral lodge]

If x_1 isn't equal x_2

[thick raven]

What is the slope of the line graphed by Ax + By = C?

[umbral lodge]

I probably used the wrong terms, but if you have a graph like this one

It is a linear equation correct

[thick raven]

Correct

[umbral lodge]

(x, y)

And the slope is 0, so ( 4-4 ) / ( -4 - 5 ) = 0

[thick raven]

Good

What I mean is

In the general case Ax + By = C, what is the slope?

In terms of A, B, and C

[umbral lodge]

The

[thick raven]

For that line it's 0, but in general what its it?

Yes it's always a straight line

[umbral lodge]

idk in terms of A B C

but in terms of x and y is

[placid geyser]

In this case, for the slope to be 0, if x1 ≠ x2,
B(y1-y2) = (C-Ax1) - (C-Ax2)
B(y1-y2) = A(x2-x1)
As slope = (y1-y2)/(x1-x2),
slope = -A/B
A can be 0
Hence, if A is 0, slope is 0

[thick raven]

Unless A=B=0 in which case it's either the entire 2d plane or none of it

[umbral lodge]

( y_2 - y_1 ) / ( x_2 - y_2 )

[thick raven]

Yes

But in terms of A, B, and C?

[placid geyser]

??

[thick raven]

I mean yeah that's basically the answer, much better to let him work it out though than just telling him

But yes it's -A/B

[placid geyser]

I was asking whether he understood that process

[placid geyser]

I thought samuel asked this thing so I replied with the answer

Then I realised it and deleted it lol

[umbral lodge]

damn

[umbral lodge]

I don't quite understand this one

[placid geyser]

Which step?

[umbral lodge]

Primarily the first one, I knew slope was y2 - y1 / x2 - x1

How does taht work with A B C

[placid geyser]

So, you see,
By1 = C - Ax1 and By2 = C - Ax2

Correct?

[umbral lodge]

Yes

[placid geyser]

$slope=\frac{y_1-y_2}{x_1-x_2}$

$slope=\frac{B(y_1-y_2)}{B(x_1-x_2)}$

[umbral lodge]

don't worry

[jolly parrot]

Suika

[umbral lodge]

I got it

nice

[jolly parrot]

Suika

[umbral lodge]

ok

[placid geyser]

$slope=\frac{By_1-By_2}{B(x_1-x_2)}$

[jolly parrot]

Suika

[placid geyser]

$slope=\frac{(C-Ax_1)-(C-Ax_2)}{B(x_1-x_2)}$

[jolly parrot]

Suika

[placid geyser]

$slope=\frac{(C-C-Ax_1 + Ax_2)}{B(x_1-x_2)}$

[jolly parrot]

Suika

[placid geyser]

$slope=\frac{(Ax_2 - Ax_1)}{B(x_1-x_2)}$

[jolly parrot]

Suika

[placid geyser]

$slope=\frac{-A(x_1 - Ax_2)}{B(x_1-x_2)}$

[jolly parrot]

Suika

[placid geyser]

$slope=-\frac{A}{B}$

[jolly parrot]

Suika

[placid geyser]

So, if A = 0, it's possible that the slope is 0

[umbral lodge]

Alright

That makes a lot of sense

[umbral lodge]

Wait

Wouldn't it be $slope=\frac{A(x_2 - x_1)}{B(x_1-x_2)}$

[jolly parrot]

samuel

[umbral lodge]

So here you use $slope=\frac{B(y_2-y_1)}{B(x_2-x_1)}$

[jolly parrot]

samuel

[umbral lodge]

Does that make sense?

[placid geyser]

Yes

[umbral lodge]

Awesome

[placid geyser]

It is that

I just switched the x1 and x2 and put a minus there

[umbral lodge]

I understand nwo

So I could simply say that C - Ax_1 is equal to C - Ax_2 if A is 0

That was pretty simple

lol

thanks a lot Suika and depression

.close

[pearl pond]

Channel closed

Closed by @umbral lodge

Use .reopen if this was a mistake.

[placid geyser]

Sounds weird

Thanking depression

[umbral lodge]

indeed, but it is their name

[placid geyser]

Yea lol

[pearl pond]

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[cosmic saffron]

Assume there is an interger number of a very large digits (it is unknown how large the number is).
We will call this X.

Then there is also Y, which are an interger number from number 1 to 99.

Assuming X and Y always positive {X > 0; Y > 0}
F = X - Floor(X/Y) * Y

F will only result in interger number of 0 to 10.

is that statement true?

[thick raven]

Try it with Y=99000000000000000 and X = 99

[cosmic saffron]

Y is 1 to 99

oh ... wait

I put wrong formula

sorry

[thick raven]

Yeah sorry I meant the other way round

[cosmic saffron]

there fixed

it should be
F = X - Floor(X/Y) * Y

i only changed that ^

[placid geyser]

I think it's 0 to 99 instead

Not 0 to 10

[odd surge]

yeah

[cosmic saffron]

how do you achieve 99?

[odd surge]

wait hold up

[cosmic saffron]

no, that is 9

[odd surge]

if Y is 1 to 99 then F is 0 to 98

you can't get 99

[cosmic saffron]

how to achieve 98 then

I don't think that possible either

[odd surge]

X = 197, Y = 99

then X/Y is less than 2 so floor(X/Y) = 1

F = 197 - 1 * 99 = 98

[cosmic saffron]

oh... you are right

Ty all ❤️

[pearl pond]

@cosmic saffron Has your question been resolved?

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[pearl pond]

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[sharp smelt]

\bf{Prove that every convergent sequence is bounded}
\
It follows by definition, If a sequence converges to $L$, that $\forall \varpesilon >0, \exists N \in \N$, such that if $n \geq N$, then $\abs{a_n-L}< \varepsilon$.
\
If a sequence is bounded, by say, $M$, then $\abs{a_n} \leq M; \forall n \in \N$
\
\t proof: By the reverse triangle inequality, it follows that $\abs{a_n} -\abs{L} < \abs{ \varpesilon}$
\
OR
\
$\abs{a_n} < L + \varepsilon$

[sharp smelt]

I'm stuck beyond this

[jolly parrot]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rough stream]

Your start is great.

an < L + e. So for all n > N, an is bounded.

[sharp smelt]

thought tbf, the contrapositive may be easier to prove

[sharp smelt]

yea

but what about n<N

[rough stream]

There's finitely many. So, there's a maximum element.

There might be a better way to say that haha

[sharp smelt]

I'll have to show that I think

sounds like a modified version of WOP

A set with finitely many elements has a least and greatest element

No idea how I'd prove that though

[toxic lichen]

this is much simpler

[sharp smelt]

hmm?

[sharp smelt]

Does this follow from some sort of axiom?

[lilac jackal]

you can try proving it by induction later. for now just take it for granted

[sharp smelt]

Cool, thanks!

[sharp smelt]

Wait, I'm sort of confused here, epsilon can blow up to infty right

so how is it bounded

[toxic lichen]

epsilon doesnt blow up to shit

you can fix ε = 0.42069

then you get that a tail of your sequence is bounded by |L| + 0.42069

and before the tail you have only a finite number of terms

[sharp smelt]

Yeah, but the defn says for a;ll epsilon>0

[sharp vigil]

if you know it converges already, then you can pick any epsilon you want and be guaranteed an N for it

[toxic lichen]

for all positive epsilon, and in particular for epsilon=0.42069

[sharp smelt]

Ah, got it

Thanks

I would like to do another proof now( of the algebric limit theorm) now

More specifically for the product

We wish to show that $\abs{a_nb_n - ab} < \varepsilon$

[jolly parrot]

What a wonderful world !

[toxic lichen]

you want to do it w/ epsilonics directly?

[sharp smelt]

yea

[toxic lichen]

cause there's an easier and less crunchy way to do it

[sharp smelt]

if possible

sure

[rough stream]

"epsilonics" I like that

[toxic lichen]

thats like, more structured by being broken down into a bunch of lemmas

would you like to hear me out on that

[sharp smelt]

yea

[toxic lichen]

Definition: A sequence (x_n) is said to be null if lim x_n = 0.

Lemma 1: If a sequence (a_n) converges to a finite limit L, then the sequence (a_n - L) is null.

Lemma 2: If (x_n) is a null sequence and L is a real number, then lim (x_n + L) = L.

Lemma 3: The sum of two null sequences is another null sequence. [That is, if (x_n) and (y_n) are null sequences, then (x_n + y_n) is null too.]

Lemma 4: The product of a null sequence and a bounded sequence is a null sequence. [That is, if (x_n) is a null sequence and (b_n) is a bounded sequence (which actually need not converge!), then (b_n x_n) is null.]

[sharp smelt]

Hmm

I'll have to prove all these lemma's first though, I suppose?

[toxic lichen]

yes

but all of these lemmas have very easy proofs that you should not overthink.

ok there we go all formatting and typos fixed

once you have all of those lemmas, use your "convergent implies bounded" result to get the product thing:

let $\lim a_n = L$ and $\lim b_n = M$. consider the sequences $a'_n := a_n - L$ and $b'_n := b_n - M$, both of which are null by lemma 1. then write: $$a_nb_n = (L+a'_n)(M+b'_n) = LM + Lb'_n + Ma'_n + a'_nb'_n$$
each term other than the first is a null sequence by lemma 4 and the ``convergent implies bounded'' theorem. by lemma 3, the sequence $(Lb'_n + Ma'_n + a'_nb'_n)$ is null, and so by lemma 2 we get $a_n b_n \to LM$, as desired.

[jolly parrot]

Ann

[toxic lichen]

(note that this proof is quite terse so do not try to skip over lines you don't understand)

[sharp smelt]

No, it all makes sense to me

[toxic lichen]

glad it does.

the spirit of what im trying to convey here is that the idea of breaking a big problem down into manageable chunks pays rather big dividends.

at no point did i have to do any heavy epsilon lifting

[sharp smelt]

Yea

got it

[toxic lichen]

(i do acknowledge that a breakdown may not always be obvious, which is why i am sharing mine here)

[lilac jackal]

another classic divide n conquer is "all polynomials are continuous"

[sharp smelt]

thanks

[toxic lichen]

there's another one in probability theory: E[XY] = E[X]E[Y] for X, Y indep

[inland ivy]

is that not how independent events are defined?

[toxic lichen]

RVs not events

[inland ivy]

right

but my question still stands

[toxic lichen]

no it's not

[lilac jackal]

its a theorem not a def

[toxic lichen]

the definition of two RVs X and Y being independent is that for any borel-measurable sets A, B ⊆ R the events {X ∈ A} and {Y ∈ B} are independent-as-events

or to put this in a less formal way

to say X and Y are independent is to say that any event having to do with just X is independent of any event having to do with just Y

[inland ivy]

ahh okay that makes sense

[sharp smelt]

I want to prove something quite simple now

if $a_n \geq 0 \forall n \in \N$, then $a \geq 0$, where $lim(a_n)=a$

[lilac jackal]

u didnt define a

[sharp smelt]

oops

[jolly parrot]

What a wonderful world !

[sharp smelt]

Contradiction

actually, I'm not too sure

[lilac jackal]

a drawing may help

[sharp smelt]

hmm

We have $\abs{a_n} -\abs{a}< \varepsilon$

[jolly parrot]

What a wonderful world !

[sharp smelt]

I let $\varepsilon = \abs{a}$

[jolly parrot]

What a wonderful world !

[lilac jackal]

bad use of reverse triangle but ok choice of e

[sharp smelt]

yea, my bad

$\abs{a_n - a} < \varepsilon$
This means $a_n<0$, which isn't possible

[jolly parrot]

What a wonderful world !

[lilac jackal]

exactly how do u get a_n<0

[sharp smelt]

one minute

made a mistake

okay, so $\varepsilon = \abs{a}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so $\abs{a_n - a} < \abs{a}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

if $a_n>0$, then $\abs{a_n-a} > \abs{a}$

[jolly parrot]

What a wonderful world !

[lilac jackal]

thats not clear

[sharp smelt]

hmm

[lilac jackal]

https://cdn.discordapp.com/attachments/908078029778083872/1357970864712323242/796939110815629322.png?ex=67f2246c&is=67f0d2ec&hm=04af7eb607d339772172f4dda12565da9c485de1b2212b807995b20cc3e075ee&

to be formal, is this true for all n?

[sharp smelt]

no

for all n greater than a certain N in n

[lilac jackal]

good

now directly show this implies a_n<0

[sharp smelt]

so $\abs{a_n-a}<\abs{a}$
a_n^2+a^2-2aa_n<a^2 \implies $a_n(a_n-2a)<0$. So $a_n<2a$ and $a_n<0$

[jolly parrot]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lilac jackal]

that looks complicated and probably wrong

yeah i dont believe the last step

redo n think simpler

[sharp smelt]

I mean one way is $a_n-a<-a$

[jolly parrot]

What a wonderful world !

[sharp smelt]

that's one case

so a_n<0

[lilac jackal]

its not a case

its just true

[sharp smelt]

okie

got it

thanks

[lilac jackal]

np!

[sharp smelt]

.close

[pearl pond]

Channel closed

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[pearl pond]

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[midnight haven]

to find the other roots

what i did

is i found the conjugate

and multiplied it with the normal thingy

then divided the answer of multiplication with the original equation

is this correct?

[fickle schooner]

Yep

[midnight haven]

but what is he doing here

x- whatever

[compact ridge]

after this step you should get x^2 + 2x + 3

[midnight haven]

where did he get that from

[compact ridge]

yeah then you just divide the quartic by x^2 + 2x + 3

[fickle schooner]

Is that not what you did?

[midnight haven]

this is not my work

im confused on where x- (-1 +square root of 2) came from

where did x- come from

[compact ridge]

if $r$ is a root of a polynomial, then $x - r$ is a linear factor

[jolly parrot]

south

[compact ridge]

say if 3, 5 were the roots of a quadratic

the quadratic would be y = a * (x - 3)(x - 5)
a being some real number: this is the root form of a quadratic

[fickle schooner]

You can decompose a polynomial into a product of its linear factors

Yh

Wait, that's a tautology

[midnight haven]

this

why did we add this to the multiplication

[fickle schooner]

In south's example

If you had a quadratic $x^2 + 5x + 6$ and decided to factorize it: you get $(x + 3)(x + 2)$

[jolly parrot]

StrangeQuarkAL

[midnight haven]

right

[fickle schooner]

the roots of that quadratic are -3 and -2

$(x - (-3))(x - (-2))$

[jolly parrot]

StrangeQuarkAL

[fickle schooner]

See how I've decomposed the quadratic as a product of linear factors?

[midnight haven]

yea

[fickle schooner]

Same thing here

$P(x) = x^4 + ...$
It is a quartic so it has 4 roots

$P(x) = (x - a)(x - b)(x - c)(x - d)$

[jolly parrot]

StrangeQuarkAL

[fickle schooner]

where a, b, c and d are the roots

[midnight haven]

i dont really understand but since my exam is tomorrow should i just do x-(z) * x-(z conjugate) to any similar questions?

[fickle schooner]

Technically yes but it's important to understand this and it's a little nuanced than that

Lemme at least finish my explanation

[midnight haven]

alright sure

[fickle schooner]

Do you agree that the quartic (4th degree polynomial) has 4 (not necessarily distinct) roots?

[midnight haven]

yea

[fickle schooner]

Good

So my decomposition makes sense too?

P(x) = (x - a)...

[midnight haven]

oh

yea i see it

wait let me see another question

[fickle schooner]

Alright

[midnight haven]

yup makes sense

thanks

.close

[pearl pond]

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[fickle schooner]

Epic

[pearl pond]

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[lucid moth]

part (d)

None of the previous parts a b and c are important, except for this bit

[compact ridge]

the slope m = tan(theta)

[lucid moth]

Doesnt the fact that its a point of inflection tell us that the slope is just -1?

[lucid moth]

Hi south ❤️

[compact ridge]

nope

[lucid moth]

But the slope is changing from negative to positive no?

I mean the slope of the point of inflection

[spare lark]

You can have an inflexion point with an always positive slope (x³)

[lucid moth]

Yeah when its changing from concave up to concave down, right?

[spare lark]

Yes

[lucid moth]

We can see an example in this graph

[spare lark]

But its not linked to -1

Its link to its sign

[compact ridge]

hopefully you get the idea

and the slope can be positive too at a point of inflection

[compact ridge]

yeah the slope can be anything, including 0

[spare lark]

For example

[compact ridge]

(think of y = x^3 for the slope being 0)

[lucid moth]

Hmm ok I see your point

[spare lark]

You can make a sign chart of your second derivative and see where it change sign

Mb

[lucid moth]

Whats that? I havent heard of a sign chart before

[spare lark]

Didn't see you was at (d)

[compact ridge]

so for the tangent line

that's how tan theta appears

[spare lark]

Just as an example

[lucid moth]

Hmm I dont think ive covered that yet

[spare lark]

We do it way before second derivative and all this stuff kekw

For solving inequality mostly

[lucid moth]

Ok so to solve I just take f'(pi/8) then add 45 degrees? (because its the y axis instead of x axis, right?

[compact ridge]

nope

[lucid moth]

Ahh im self studying so I might have missed some stuff hehe

[spare lark]

Ic

[compact ridge]

that's dimensionally incorrect

f'(pi/8) is not an angle

a slope is not an angle

[lucid moth]

Ah sry I mean we put the answer into arctan

Then add 45 degrees

[compact ridge]

good so far

[lucid moth]

But I guess thats wrong too

[compact ridge]

this is wrong

just look at the diagram

also just to make sure we're on the same page, you want the angle between the tangnet line and the y-axis

[lucid moth]

Ah so we take our answer from the arctan away from 180 degrees?

[compact ridge]

nope

so arctan(-2) is negative right?

[lucid moth]

Yes

-63 degrees

[compact ridge]

okay what's the green angle then?

[lucid moth]

Is -2 the slope?

[compact ridge]

yes

[lucid moth]

I thought that arctan(-2) would give us this angle

[compact ridge]

nope, that purple angle is not negative

[lucid moth]

The angle between the x axis and the slope

[compact ridge]

arctan always returns an angle in either the 1st or the 4th quadrants

[lucid moth]

Ok good rule im adding that to my notes haha

[compact ridge]

yes pls, it's super important

intuitively you can see that that obtuse angle can't possibly be 63 deg

[lucid moth]

And we know whether its the 1st or 4th by its sign right?

[compact ridge]

yes

[lucid moth]

Okokok, got it

So its 90 degrees takeaway arctan(-2), right?

[compact ridge]

nearly

cause arctan(-2) is negative, we need to negate it to make the angle positive again

[lucid moth]

Ahh yeye

[compact ridge]

so -1 * arctan(-2), that's actually arctan(2)

90 - arctan(2), there you go, yes

[lucid moth]

Okkkk

I understand now

Which is 90 - 63 = 26 degrees

27, whatever 😝

I get the method thats the most important hehe

Thank you everyone!

❤️

.close

[pearl pond]

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[naive dirge]

You guys

[naive dirge]

Better explain

It

Explain why continuous value means I have to encode it with numbers

Or face the wrath of a loyal servant of Mickey Mouse

Choose wisely

[toxic lichen]

this is quite the combative tone from you

[hazy pilot]

I'd suggest watching the entire video

maybe he explains why in a later part

[naive dirge]

I hate it, I should be informed of that

[autumn fossil]

well text isnt very "continous"

firstly, you must always change a whole letter. You cant change 1/2 or 1/10th of a letter

[primal blade]

Yeah letters are quantised 😂

[autumn fossil]

and you will probably want words like "large" and "big" to be close together, because they have similar meanings

im pretty sure that 3b1b makes this point somewhere in the vid

so just keep watching

[toxic lichen]

yes he does.

[pearl pond]

@naive dirge Has your question been resolved?

[pearl pond]

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[brazen swift]

https://i.imgur.com/ESpyL08.png

[brazen swift]

C.) The length of the median to side CD in Triangle ACD is 22.44 . find k

[languid goblet]

we would appreciate if its english

[brazen swift]

took me a bit to type

[dapper kraken]

could you translate it?

[pearl pond]

@brazen swift Has your question been resolved?

[brazen swift]

Sure, i just translated whatever i thought was necessary

but sure ill translate all of it

Points A , B , C , D are on the Parameter of a circle. Given:
AB || CD , <Bad = α , <CAD = 20 , AE = k
A. 1) Express using α the angles of triangle ACE
2) Express with k and α the length of sides CD & AB
B) Given: Surface of triangle ABE is 6.41147 times bigger than the Surface of triangle CDE. Given: α >20 , calculate α .
**C.) The length of the median to side CD in Triangle ACD is 22.44 . find k **

I'm stuck on C)

https://i.imgur.com/Ggf9g5D.png

https://i.imgur.com/rcJITyw.png

[brazen swift]

sketch with everything that's given^

[pearl pond]

@brazen swift Has your question been resolved?

[brazen swift]

<@&286206848099549185>

[dapper kraken]

oh wait sorry i forgot about this

[dapper kraken]

what did you get for B?

[dapper kraken]

ok i think the idea is with B and law of sines, you can find the ratio of AC and CD, and you know AC^2+(CD/2)^2-AC×CD×cos(110)=22.44^2, so you can combine them to get a value of AC and CD, then you can use law of cosines to find AD

(hopefully you can ise a calculator for this)

[brazen swift]

yeah i can

[pearl pond]

@brazen swift Has your question been resolved?

[pearl pond]

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[jolly geyser]

how did they also get 129?

[snow sail]

you can get one from the other by reflection

,calc 180-51

[jolly parrot]

Result:

129

[cobalt hinge]

Yeah, sin(180-x)=sin(x), just to let you know.

[jolly geyser]

is it that thing with the quadrants

cast diagrams

[pearl pond]

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[pallid kestrel]

Wait guys so in trigonometry 1 is sinx square plus cosx square but what is -1 is it (1.-1) so [(-1).(sin2 + cos2)] sorry for the simple question I'm going over trigonometry again

[autumn fossil]

yes, you could write -1 = - sin^2(x) - cos^2(x)

[pallid kestrel]

Oh it's that simple?

Thanks alot 😃

[autumn fossil]

Yes, 1 = sin^2(x) + cos^2(x) and if you multiply both sides by -1, it's still valid

if you multiplied it by 2, it would still be valid

2 = 2sin^2(x) + 2cos^2(x)

if you added -2 to both sides, it would again be valid

[pallid kestrel]

Oh yeah

[autumn fossil]

-1 = sin^2(x) + cos^2(x) - 2

[pallid kestrel]

-2?

Oh I get it

[autumn fossil]

Just arbitrary example

the point is that you can do almost anything to both sides and it will remain valid

[pallid kestrel]

Thanks alot for the detailed explanation

Have a great day

[autumn fossil]

you too :)

[pallid kestrel]

.close

[pearl pond]

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[pearl pond]

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[next forum]

hey guys, i ran this through a calculator, i multiplied each anser choice by the denominators conjugate on top and bottom and didnt get anything. do i multiply the tops conjugate by the top, and the bottom's conjugate by the bottom or is there smth im missing?

[next forum]

please ping me if you respond :3

[terse tartan]

Have you rationalised the denominator?

[hazy crypt]

i multiplied each anser choice by the denominators conjugate on top and bottom and didnt get anything.

do this to the original expression instead of each of the answer choices

[autumn fossil]

Can you show your work with the conjugation?

oh, yeah, you gotta do it to the original expression

[hazy crypt]

@next forum

[terse tartan]

It's (2sqrtx +3)×(sqrtx)/4x-9

[next forum]

oh then i didnt do that

[celest cedar]

Don't give him answer, but lead him

[next forum]

so my ork is pretty irrelevent

[next forum]

Delete this please

[next forum]

i can sho you but i kno its rong no

sorry about my keys messing uo

[celest cedar]

$(a + b) \cdot (a - b) = a^2 - b^2$

[jolly parrot]

SLANDER

[celest cedar]

When you divide by a number, multiply by itself.

[next forum]

i think i may actually have a better lead elsehere
thank you for hel;ing me out as much as you did though

have a good day guyss!

.close

[pearl pond]

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[cobalt hinge]

!nosols btw

[pearl pond]

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.

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[iron karma]

question is to find the maclaurin series - how on earth is this right? isn't f(0) undefined?

[west sapphire]

presumably they're filling in the gap at x=0 by setting f(0) = log(1) = 0

[iron karma]

i mean i guess but let's say they didn't do that, would the expansion exist?

[west sapphire]

no, you have to have f and its derivatives defined at x=0 in order to define the maclaurin series

[dawn dirge]

It is undefined but we know that $$ \lim_{n \to \infty} \frac{\sin x}{x} = 1$$, and since $\log$ is continuous, the limit of $f$ as $x \to 0$ is $\log 1 = 0$

[jolly parrot]

elliot

[dawn dirge]

Sorry should say $x \to 0$ in the limit

[jolly parrot]

elliot

[iron karma]

right yeah, thats why f(0)=0 is a good like 'fix' to fix the discontinuity

but then when trying to find f' f'' etc given f(0)=0 let's assume (as this isn't in the question), the same thing ends up happening again and again

where the derivative DNE but you can find some sort of 'this would make sense as a fix' value

[west sapphire]

yep you're gonna have to do more work to show that the values given by the coefficients are consistent with the derivatives of f (with the gap at x=0 plugged in)

[dawn dirge]

It’s not even a ‘fix’ really, it’s just the limit as x approaches 0, but you can think about it that way. You will just have to figure out the derivative as normal and figure out what happens near 0 in the limit, similar to above

That’s when you can use e.g. L’Hôpital’s rule

[iron karma]

slightly confused as what you mean by it's not a 'fix', i think i meant that in the way that the expansion DNE unless we define f(0) as the limit as x->0 of f(x)

am i wrong to say that the expansion DNE without defining f(0) like that? or am i misunderstanding something else

[west sapphire]

correct, the expansion DNE if f(0) is not defined

clearly f(0) has to be 0 in order to make f continuous

you have to do more work to show that this also makes it differentiable (at least 4 times) in order to verify the formula

[pearl pond]

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[glass jungle]

I don’t know what to multiply the equation by?

[celest cedar]

What's the question?

[glass jungle]

Solve for X.

[celest cedar]

$a^2 - b^2 = (a+b)(a-b)$

[jolly parrot]

SLANDER

[glass jungle]

I need to multiply the equation by a common factor from the denominator but how here?

[celest cedar]

$a^2 + a = a (a + 1)$

[jolly parrot]

SLANDER

[untold zephyr]

Hint: $x^2$ and $1$ are both squares, so apply the above identity

[jolly parrot]

Oliver

[untold zephyr]

See if you can spot a common denominator then

[glass jungle]

The left side is (X+/-1) but I don’t know about the right side.

[untold zephyr]

How does the LHS denominator factorise using this?

[celest cedar]

Try to find common denominator, and it would be x + 1 because of 7/(x+1)

[glass jungle]

What about the other right side term?

[untold zephyr]

Can you factorise that denominator as well?

The 2 coefficiets share a common variable

[celest cedar]

.

[glass jungle]

Is it X (X+1)(x-1)?

[celest cedar]

$x^2 - 1^2 = ?$

[jolly parrot]

SLANDER

[celest cedar]

$x^2 + x = ?$

[jolly parrot]

SLANDER

[celest cedar]

Factorise or simplify these parts

[glass jungle]

X=1

[celest cedar]

simplify not solve

Or Factorise

[glass jungle]

(X-1)^2

[celest cedar]

Wrong

[glass jungle]

X(X+1)

[celest cedar]

Here is the identity

[celest cedar]

Right

[glass jungle]

And that’s for the second one?

[celest cedar]

4/(x² + x)

[glass jungle]

How is this wrong for the first one?

[celest cedar]

U have a² - b² not (a - b)²

$a^2 - b^2 \neq (a - b)^2$

[jolly parrot]

SLANDER

[celest cedar]

a = x
and
b = 1

[glass jungle]

So how would I find the right side?

[celest cedar]

What does right side mean?

[glass jungle]

Would I multiply the equation by x(x+1)(x-1)?

[celest cedar]

Owh no

By ur common denominator

x + 1

[glass jungle]

The entire thing?

[celest cedar]

Factorise all denominatores so u can see x + 1, then multiply by it

[celest cedar]

$x^2 - 1 = (x + 1) (x - 1)$

[jolly parrot]

SLANDER

[glass jungle]

Why can’t I add an extra X at front to rid the X squares?

[celest cedar]

U have x² - 1 not x²

When you multiply by a number, you multiply the whole operation.

[glass jungle]

I’m aware.

[celest cedar]

Hmm

[celest cedar]

You can unify the denominator

[glass jungle]

?

I have the answer key.

[celest cedar]

$\frac{4x + 1}{x^2 - 1} = \frac{7}{x+1} + \frac{4}{x(x+1)}$

[jolly parrot]

SLANDER

[celest cedar]

$\frac{4x + 1}{x^2 - 1} = \frac{7x}{x(x+1)} + \frac{4}{x(x+1)}$

[jolly parrot]

SLANDER

[glass jungle]

That’s wrong.

The answer key says I have to multiply by x(x-1)(x+1).

[celest cedar]

Here

[celest cedar]

Multiply it here

[glass jungle]

But how with the sqaures on the demo?

?

??

❌

[celest cedar]

$\frac{4x + 1}{(x+1)(x-1)} = \frac{7x}{x(x+1)} + \frac{4}{x(x+1)}$

[jolly parrot]

SLANDER

[celest cedar]

@glass jungle

[glass jungle]

There isn’t an extra X under the 7x

[glass jungle]

Yeah?

[celest cedar]

We multiplied by x and divided by itself

[celest cedar]

Look at here

[glass jungle]

Yeah I see it

[celest cedar]

Multiply x(x+1)(x-1) here and calcul

[glass jungle]

I’m stuck

[celest cedar]

Do u know how to calcul Delta?

[glass jungle]

We’ve never done delta. This is algebra 2.

[celest cedar]

Something like b² - 4ac

[glass jungle]

DISCRIMMINANT

[celest cedar]

Use it

[glass jungle]

THAT IS FOR FIDNING DISCRIMINANT

Sorry autocorrect messed up the sentence

64

[celest cedar]

Get solutions x1 and x2

[plain finch]

what's the fucking delta, it's called discriminant

[celest cedar]

Call it what you want

[glass jungle]

.close

[pearl pond]

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[pearl pond]

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[gleaming nebula]

Im supposed to determine 2 equations relating to this sine wave, I know how to find amplitude and vertical phase shift, but because the x intercepts are equitidistant from the origin, how do I know which way it shifted horizontally?

[gleaming nebula]

2 equations in the form y = asin[b(x-c)]+d where a > 0 and another equation where a < 0