precalculus

[wide lynx]

the activity is pepega

[burnt grail]

Can anyone tell me how to do this

[fleet yew]

@burnt grail just let x^4 equal that number

[pale bison]

@wide lynx ??????

[wide lynx]

i dont understand the activity

so i came here

?????

[pale bison]

what were the "three ways" that you were shown to?

how am i suppose to know what technique is required of you to solve it

[wide lynx]

it's estimating instantaneous roc with average roc

ill just watch videos

[pale bison]

bye

[burnt grail]

@fleet yew how do I get 4 values for z doing that though?

[fleet yew]

@burnt grail those values of z are the values of x

Just let z^4 equal that number

@burnt grail you are familar with roots of unity?

[burnt grail]

No I’m clueless

[fleet yew]

@burnt grail just as an example: what are the 4th roots of 16?

ping me

[burnt grail]

2 @fleet yew ?

I’m stupid sorry

[fleet yew]

there's four of them aren't there?

think about it

let's say that z^4 = 16

here's a little trick: 2 + 2 = 4.

it is usually easier to take the square root twice than it is to take the 4th root

so try letting a=z^2

[burnt grail]

So it gets confusing when I have square roots and i

[fleet yew]

hold on a sec

so now you have

a^2=16

solve that

[burnt grail]

A = 4

[fleet yew]

and?

you take the square root, there should be two answers

a = 4, and a = -4

[burnt grail]

Right so +-

4

[fleet yew]

so then we substitute back in for z

so we have two equations

z^2 = 4, and z^2=-4

does this all make sense so far

[burnt grail]

Yes it does

[fleet yew]

solve those two equations

[burnt grail]

+- 2

[fleet yew]

yes, that's the first equation

now solve the second one. z^2 = -4

[burnt grail]

Would it just be -2

[fleet yew]

?

what is (-2)^2

[burnt grail]

How would you do the second one and get a negative value

[fleet yew]

lol i don't think you should be doing this problem no offense

square roots of negatives are imaginary numbers

[viscid thistle]

the square root of -4 is 2i, no? or am i imagining things

[burnt grail]

Oh well I mean I thought you were asking for a literal value

[viscid thistle]

gotem

[fleet yew]

don't spoil it

[burnt grail]

So I was confused

[fleet yew]

all i'm asking you to do is solve z^2 = -4

there are only two answers

[burnt grail]

Well it’s obvi -2i and 2i

But how am I gonna do this with a square root

[fleet yew]

yes ok we got it

so now we know that there are 4 solutions to z^4=16

2, -2, 2i, -2i

[burnt grail]

Btw when we do get there, I have to put it in exponential form

[fleet yew]

yeah ik we're getting there lol

now do you notice anything about those numbers

squid don't spoil it

[viscid thistle]

did y'all get my corny joke? and so sorry for interrupting it was just irresistible

[burnt grail]

Z0 z1 z2 z3

[viscid thistle]

the square root of -4 is 2i, no? or am i imagining things

heck it's too corny

[fleet yew]

@burnt grail what do you notice in common about these 4 numbers: 2, -2, 2i, -2i

here's a hint: do you know how to take the absolute value of a complex number?

[burnt grail]

I figured out my problem thank you though 🙏🏽

I Fucked up my cos

[fleet yew]

alright gl

[reef jasper]

how do i set up the law of sines with this

this is my pic so far

[wide ocean]

polynomial

question

does anyone know?

not sure

[oblique thicket]

@wide ocean construct an equation like this: $\frac{(1+x)(2+x)(4+x)}{1 \cdot 2 \cdot 4} = 9 $ After quick simplifications, we can get $x^3+7x^2+14x+8=72$ which leads us to $x^3+7x^2+14x=64$ Which is already easily solvable.

[obsidian monolith]

Dmytr:

[wide ocean]

@oblique thicket thank you so much!

[oblique thicket]

you are welcome

[wide ocean]

the answerkey says x=2 (increase dimension by 2)

[oblique thicket]

yep

[wide ocean]

and i'll see if it's solvable with the equation above, thanks I hope it works

[oblique thicket]

It is solvable and you'd get exactly 2

[wide ocean]

thank you! bless

[oblique thicket]

Thanks

[viscid thistle]

may I get help with this

these where my answers and im not too sure whats wrong

[alpine basin]

How do I use synthetic division on this problem with its dividing by an x^2 +4?

[gilded stirrup]

You can’t use synthetic

You have to use long division

Then write coefficients

And put an 0 for the x term of x^2+4

[alpine basin]

put a zero where?

[willow bear]

5x^4 - 3x^3 + 2x^2 + 0x - 1

[gilded stirrup]

Cause it’s x^2+0x+4

Yeah + that

[alpine basin]

How would I divide that once I have it set up? Im having a bit of trouble figuring it out

[viscid thistle]

https://youtu.be/SbUiZx5a0Ok

I personally like this one

There are loads of videos tho

@alpine basin

[uncut mulch]

similar process to long division with numbers

instead you have powers of x instead of units,tens,hundreds,etc...

[alpine basin]

@viscid thistle How do I know when I can use synthetic division? Would this problem have to be normal long division as well?

[viscid thistle]

The divisor has to be x - a

For u to use synthetic

[alpine basin]

okay thanks

[viscid thistle]

Just always use long

[velvet garden]

what is synthetic division

the one that uses Ruffini's rule?

[viscid thistle]

Yes

Faster but so mnemonic

[velvet garden]

I see that as an absolute win

[viscid thistle]

In the short term

[alpine basin]

I got 4 for this problem, but the answer is 6, can someone show me how I'd solve this?

[uncut mulch]

wdym by

the answer is 6

[alpine basin]

oh my bad

it wants me to find the remainder using the remainder theorem

[uncut mulch]

if a polynomial of x: P(x), is divided by (x-a),
the remainder will be P(a)

but the answer won't be 6

here, if $P(x) = x^4 - 9x^2 + 14$ is divided by $(x-2)$, the remainder will be $P(2)$

[obsidian monolith]

ramonov:

[fervent girder]

Does this solution appear wrong? If -4 is a zero, should it not be listed in the final zeros? I am confused by the chart as well

[pale bison]

this is some thonkery

is not for the polynomial x^3 - 19x - 30

[fervent girder]

...i've been stuck on this for 2hrs... okay, so the solution is wrong for that chart?

[pale bison]

are there more polynomials?

[fervent girder]

that is the complete solution

[pale bison]

you should only focus on how -2 is a zero leads to the factorization

can you take a pic of question 21?

[pale bison]

so they mixed 20 and 21 together for some reason

no need to worry

[maiden pebble]

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

Does sin^2(x) seem to hump the x-axis because it is squared, so it never passes through an axis

Like double roots (I think thats what they are called)

[hexed ermine]

Right because its squared, any negative part gets squared so it's always positive

[maiden pebble]

OOOH is that why double zeros dont pass axis?

[hexed ermine]

Yep

[maiden pebble]

🤯

[hexed ermine]

If you have (x-a)^2 theres a double root at a, and you can't get a negative number outta it

Because x is in the reals

[maiden pebble]

https://i.imgur.com/1e8sRBx.png

could I write this as sin*sin(2x)

x being theta

[pale bison]

$\sin\cdot\sin(2x)$...?

[obsidian monolith]

Publius:

[pale bison]

the answer is no^

[maiden pebble]

ok

thx

https://i.imgur.com/khEIt7t.png

I have simplified the bottom to cos^2(x) but dont know what to do on the top

[willow bear]

double angle identity sounds like it might help

[maiden pebble]

how?

I turned it into 1-cos^2(2x)

wait ur a genius

Do you mean (sin^2(2x))/cos^2x --> tan(2x)

[willow bear]

i,,, no?

$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

[obsidian monolith]

Ann:

[willow bear]

this is the identity i meant

and here, $\sin^2(2\theta)$ would become $4\sin^2(\theta)\cos^2(\theta)$ upon applying it

[obsidian monolith]

Ann:

[maiden pebble]

Oh, so I would square the 2sin(x)cos(x)

thanks

[lofty prism]

f(x) = 1/2 x
prove that f(8) approx equal to 4

i do it like this:
f(4) = 1/2 * 2 = 1
it doesnt give me 4
please any help???

[hexed ermine]

What?

if you plug in 8 you get half of 8 which is 4

[lofty prism]

yeah i tried this but it doesnt give me approx

it gives equal

[viscid thistle]

Your questions are weird...

^

[lofty prism]

what?? but nevermind i found the answer
f(8) ~ 8*1/2 ~ 4
even if you didnt help me i appreciate you guys wanting to so thanks!

[heady jewel]

...

[viscid thistle]

...

Uh i hope you are kidding

[lofty prism]

uhhh what

[peak badge]

Looks like there's some miscommunication over here.

If you have cleared your doubts well and good. Otherwise, can you please repeat your question?

[viscid thistle]

if you plug in 8 you get half of 8 which is 4

what?? but nevermind i found the answer
f(8) ~ 8*1/2 ~ 4
even if you didnt help me i appreciate you guys wanting to so thanks!

...

[fossil crown]

From

#❓how-to-get-help

Rule 5

While asking questions, make sure you mention all relevant details, including the context, what you have tried and what you're stuck at. Do not expect others to simply solve your questions for you.

[lofty prism]

f(x) = x * (1/2)
show that f(8) ~ 4

[peak badge]

Ah, okay.

[viscid thistle]

f(8)equals4

[uncut mulch]

no like seriously, what class is this supposed to be?
why are you asked to show that approximations are equal
and equal stuff are approximations?

[peak badge]

Basically how to use functions in math, I guess?

[uncut mulch]

and/or are you grossly misrepresenting the problems

[lofty prism]

i mean the question was f(8) = 4 but its better if you write it with ~

[viscid thistle]

....

No.

[lofty prism]

but it's quite the same but approx is better

No.
@viscid thistle for what?

but i will put an = if you want to

however approx generalizes it more...

[willow bear]

hold on what

[uncut mulch]

the use of approximation sort of implies non-equality

[viscid thistle]

Hell no.

[willow bear]

can you show the question exactly as it was stated @lofty prism

screenshot maybe

[lofty prism]

Let f be a function defined in IR such that f(x) = x/2
1- Show that f(8) = 4
thats exactly what it says i cant photograph it

[viscid thistle]

....

[willow bear]

why can't you take a screenshot

[lofty prism]

the use of approximation sort of implies non-equality
@uncut mulch i dont think so

[viscid thistle]

.....

[lofty prism]

why can't you take a screenshot
@willow bear it's not on my computer

but its okay i got the answer tyanks for your help

[viscid thistle]

No it's not okay, you think that using ≈ is "better" or whatever while = is the appropiate use

[uncut mulch]

and/or are you grossly misrepresenting the problems
seems that's a yes

[lofty prism]

uh thats an arithmetics fact though

[uncut mulch]

if the question use an $=$, you should transcribe/represent it as such \
if the question use an $\approx$, you should transcribe/represent it as such

[obsidian monolith]

ramonov:

[lofty prism]

guys just to test your knowledge, put the correct sign here :
1+1 # 2

[uncut mulch]

we're not the ones that need testing

[peak badge]

That's not the point, sucette

[lofty prism]

could you answer.plzase

[viscid thistle]

7

[lofty prism]

answer:
both = and ~ are correct but ~ is more correct

so yeah lol

[viscid thistle]

??????????

.......

[peak badge]

[uncut mulch]

...

[lofty prism]

7
@viscid thistle huh

[willow bear]

...

[peak badge]

Let's not jump the gun bois

[lofty prism]

guys if theres anything wrong with what i say feel free to correct me but i dont think i told shit here

[peak badge]

At this rate,we need to VC to understand this

[viscid thistle]

Can you define what ~ is for you?

[rugged linden]

~ implies that this is just about the answer, it's not actually the answer

[uncut mulch]

$1 + 1 \approx 2$ has implications that $1 + 1$ could be a value that isn't $2$

[rugged linden]

it's an approximation

[obsidian monolith]

ramonov:

[rugged linden]

yeah

it doesn't generalize anything, you're just saying something that is untrue

it's like saying 5 is approximately 5

it's untrue

[lofty prism]

basically = is used for values that are perfectly equal such as
1=1
3+3=6
2^2=4
etc
~ is used to compare two values which dont need to be the same or perfectly the same
1 ~ 1
4 ~ 4.5
3 ~ 10
3+4 ~ 11
idk if im clear lol

[peak badge]

Much better

[rugged linden]

Yeah

[lofty prism]

edited please check l1stest update

[rugged linden]

in your earlier example where f(x) := x/2 and f(8) you said was apprximately 4

[viscid thistle]

Bruh

[rugged linden]

oh wait you're using it in the inclusive sense? That's the misunderstanding i suppose

[peak badge]

At this rate I think it's more of a language barrier/definition issue...

[rugged linden]

The ~ sign doesn't include the = in it's definition, if something is equal it is =, if it's around the value then it's ~, they're used for seperate things

[obsidian monolith]

whzup:

[lofty prism]

what sources are you using to state this definition . which arithmetics sources .

[peak badge]

Okay, you are a high schooler right?

Lemme try this out

[rugged linden]

I mean there isn't really a standard convention which convenes to determine what ~ and = means

[lofty prism]

I don't think $\approx$ defines an equivalence relation, does it?
@opal jacinth in a sense it does

[obsidian monolith]

💣♧ 𝐬𝐔ς乇ţ𝐭𝐄 ✊🍧:

[rugged linden]

it's just notation, we're just telling you what is normally done

[uncut mulch]

with no specified accuracy.

anything can be considered approximately equal to anything

[lofty prism]

yeah so you are basically saying we cant say 1~1

[rugged linden]

big mind

[lofty prism]

hats the dertivative of x^2

whats*

[fluid shore]

nothing

formulas don't have derivatives

[peak badge]

= is used to define when two particular quantities are exactly the same.

~ is used to define when two particular quantities are approximately the same.

that is, there is some chance that it need not exactly be the same but the "measure" of two quantities are close to each other.

@lofty prism

[fluid shore]

the real valued function of a real variable f(x) = x^2 has a derivative f'(x) = 2x

[lofty prism]

it's 2x

just saying

= is used to define when two particular quantities are exactly the same.

~ is used to define when two particular quantities are approximately the same.

that is, there is some chance that it need not exactly be the same but the "measure" of two quantities are close to each other.

@lofty prism
@peak badge thats why i hate high school programs lol, they really mess students minds up

[peak badge]

it's 2x
@lofty prism

Your statement is true if you consider that we are taking the derivative w.r.t X.

However, if we took the derivative w.r.t say, Y - then the derivative will be 0.

[lofty prism]

|not| (means absoluetly not lol because || is absolute value sign)

@lofty prism

Your statement is true if you consider that we are taking the derivative w.r.t X.

However, if we took the derivative w.r.t say, Y - then the derivative will be 0.
@peak badge depends if x is dependent on y

[uncut mulch]

However, if we took the derivative w.r.t say, Y - then the derivative will be 0.
depends if x is dependent on y

[peak badge]

Yes

[fluid shore]

dafuq

sucette, you just changed what you said to what ramonov said

-.-

[peak badge]

Just leave it, Abhi

[lofty prism]

no because 1+1 isnt superior to 1

2*

[fluid shore]

??

1+1 >= 2 is a true statement

1+1 = 2 is a true statement

[peak badge]

@lofty prism so you hate High school and I presume you don't get the math as well?

[lofty prism]

@lofty prism so you hate High school and I presume you don't get the math as well?
@peak badge ?? i m not in high school though

[peak badge]

So, what's your educational background as of now? That way, I can better explain things to you

[lofty prism]

this is personal information i would not like to leak, but you must know i am not a high school student.

[peak badge]

Sure, your wish.

I am pretty sure you aren't in high school and by your responses till now, I can probably say that English isn't your first language as well.

[viscid thistle]

What is wrong with that?

[lofty prism]

duh

[peak badge]

I think all of us have been pretty clear in your responses to your queries but there still exists to be miscommunication between us, maybe that's one of the reasons.

[lofty prism]

what is queries

[peak badge]

What is wrong with that?

Some people understand stuff better in their native language .

Questions

[fluid shore]

yes, i do math better in Hindi

I prefer math being written using Hindi instead

[lofty prism]

youre hindi?

[viscid thistle]

That's the opposite for me, I prefer English for maths

[peak badge]

I am just saying. I have met people who don't grasp English as well and prefer things to be in their native language.

[lofty prism]

what is grasp

[peak badge]

Ability to understand

[lofty prism]

ok

i dont grasp you

[peak badge]

That's fine. Usually nobody in most of my circles do

[lofty prism]

circles?? are you challenging me in eulidcian geometry.

(jk)

[peak badge]

...

[lofty prism]

i was just kidding lol

dont take it too harsh

[odd abyss]

does someone know how to write the partial fraction decomposition of a rational expression

I need help understanding

[jagged glade]

Hello @odd abyss , which part of the partial fraction do you need help with?

https://www.mathsisfun.com/algebra/partial-fractions.html
There are some basics for partial fractions in the above link, feel free to take look~😁

[odd abyss]

yes hello

[jagged glade]

Yes?

[odd abyss]

let me take a look at this article

[jagged glade]

Sure! Take your time

[odd abyss]

so my first question is how to factor the bottom

after foiling

[jagged glade]

Well, since x²+2 has complex roots, we don't have to factor it.

For partial fractions

[rugged linden]

Isn't what they're going to have to do polynomial long division?

with one of the (x^2 +2)

hm

[jagged glade]

But for the (x²+2)²,we have to separate into (x²+2) and (x²+2)²

[rugged linden]

I think u made a typo with the (x^2 + 2)^2

[odd abyss]

so im not foiling

[rugged linden]

but from there you can't do partial fractions because the degree of the numerator is larger than the denominator

so they're gonna have to do polynomial long division with one of the x^2 + 2 terms

[jagged glade]

So. It would sperate into
$\frac{Ax+B}{(x^2+2)}+\frac{Cx+D}{(x^2+2)^2}$

[obsidian monolith]

Biscuit:

[odd abyss]

monkas

[rugged linden]

oh wait sorry yeah you didn't make a typo with the (x^2 + 2)^2 thing

i forgot about that rule

[jagged glade]

Sorry, I was focusing on the LaTeX, can't really read what you guys were typing just now :P

[rugged linden]

bahah its okay

[odd abyss]

ok

so

[jagged glade]

Yea?

[odd abyss]

how are you allowed to make (x^2 + 2)^2 into that

[rugged linden]

well

you know how when you have normal fractions you mutiply the diagonals and multiply the denominators when you want to brute force normal fraction addition, we're doing the opposite here

[jagged glade]

We would just let $\frac{x^3+x^2+8}{(x^2+2)^2}=
\frac{Ax+B}{(x^2+2)}+\frac{Cx+D}{(x^2+2)^2}$

[odd abyss]

regarding @rugged linden I actually dont know when you have to that

[uncut mulch]

missing ^2 on the lhs

[odd abyss]

and @jagged glade I will just go along with it

[rugged linden]

you don't have to but you sometimes do it

aaa its hard to explain over text

[jagged glade]

$\frac{x^3+x^2+8}{(x^2+2)^2}=
\frac{(Ax+B)(x^2+2)}{(x^2+2)^2}+\frac{Cx+D}{(x^2+2)^2}$

[odd abyss]

when you add fractions you find the gcf of the denom

is that what youre saying

[rugged linden]

Yeah kinda

[odd abyss]

ok biscuit

[jagged glade]

And we now compare the coefficients of the x terms in the numerator

[odd abyss]

a and c

[rugged linden]

Okay wait im just making this more confusing bahahah just listen to biscuit

[jagged glade]

$x^3+x^2+8=
(Ax+B)(x^2+2)+Cx+D$

[obsidian monolith]

Biscuit:

[jagged glade]

And by comparing coefficients we can find A,B,C and D

[odd abyss]

why do we need c and d

[jagged glade]

Well, it's a cubic polynomial. We shouldn't miss out the x^1 term and x^0 if this is what you mean :)

And the partial fraction guides us into this equation.

[odd abyss]

ok

[jagged glade]

$x^3+x^2+8=
(Ax^3+Bx^2+2Ax+2B)+Cx+D$

[obsidian monolith]

Biscuit:

[jagged glade]

$x^3+x^2+8=
Ax^3+Bx^2+(2A+C)x+(2B+D)$

[obsidian monolith]

Biscuit:

[jagged glade]

Can you find the values of A , B , C and D?

[odd abyss]

let me write this

[obsidian monolith]

Biscuit:

Biscuit:

[jagged glade]

Sorry, just now got typo

ty ramonov

[odd abyss]

the texit disappeared

earlier

[jagged glade]

Yes, because I edit my text because of the typo

[odd abyss]

the ones way earlier

so you want me to find a b c d

[jagged glade]

Yep

[odd abyss]

which one is right

the 1st or 2nd texit

[jagged glade]

$x^3+x^2+8=
Ax^3+Bx^2+(2A+C)x+(2B+D)$

This

[obsidian monolith]

Biscuit:

[odd abyss]

so solve that

[jagged glade]

Comparing coefficients yea

[odd abyss]

so im not solving

[jagged glade]

Okay. So do you get the gist of doing partial fractions?

Well, it's solving for A,B,C and D, but not x.

[odd abyss]

so I just add them together

I get the "gist" of doing them

I know why

not the how

[jagged glade]

I see.

[rugged linden]

maybe try a simpler example first?

[odd abyss]

ill try one and come back

[rugged linden]

good luck :))

[odd abyss]

cause I dont know how to solve a b c d

[jagged glade]

Oh, I can show.

Like for LHS, the coefficient of x³ is 1.

And for RHS, the coefficient of x³ is A

So by comparing coefficient, we have A=1

[odd abyss]

is this question an improper or proper rational expression

firstly

the denom is bigger right

[jagged glade]

Should be proper, since the numerator is of degree 3 and the denominator is of degree 4

[odd abyss]

so in order to solve were making the bottom smaller

[jagged glade]

But this since example is has a square term, we have to break it into (x²+2) and (x²+2)²

[odd abyss]

ok

[odd abyss]

ah

so we end up with the a b c d thing

[jagged glade]

Yep

[odd abyss]

so we get

[ x^3 + x^2 + 8 / (x^2 + 2)^2 ] = [ (Ax + B)(x^2 + 2) /(x^2 + 2)^2 ] + [ (Cx + D) / (x^2 + 2) ^2 ]

[jagged glade]

Missing a C, but it's correct.

[odd abyss]

where does the c go

[jagged glade]

Cx+D

[odd abyss]

yea

ok

so now you multiply the A and B binomial with x squared + 2

[jagged glade]

Yes

[odd abyss]

ok now what

[jagged glade]

Now we compare the coefficients of the numerators, since the denominators are the same

[odd abyss]

yep

so how do we compare

[jagged glade]

Lemme repost the result

$x^3+x^2+8=
Ax^3+Bx^2+(2A+C)x+(2B+D)$

[obsidian monolith]

Biscuit:

[jagged glade]

So, first, we can compare the coefficient of x³ term.

For LHS, the coefficient of x³ term is 1.

For RHS, the coefficient of x³ term is A.

So, we get A=1

[desert pendant]

Guys, if we have a function inside of a summation, can you take the function out

like i have the sum of the imaginary part of something

so can i take the imaginary part out

[fluid shore]

how about showing the specific question

[odd abyss]

I posted it

earlier

whats LHS and RHS

left

and right

ok

[jagged glade]

LHS is left hand side

RHS is right hand side

[odd abyss]

quick google

[fluid shore]

i was talking to kiesh

[desert pendant]

nvm

[odd abyss]

so A=1

B = 1

right

[jagged glade]

Right

[odd abyss]

ok

do I need c and d

[jagged glade]

Yes.

[odd abyss]

c and d = 8?

[jagged glade]

2A+C=0

2B+D=8

[odd abyss]

why

[jagged glade]

For LHS, the coefficient of x term is 0.

[odd abyss]

ohhh

[jagged glade]

For RHS, the coefficient of x term is 2A+C.

[odd abyss]

since its not therwe

[jagged glade]

Yea, therefore it's 0

[odd abyss]

so I rewrite as x^3 + x^2 + 8 = Ax^3 + Bx^2 + 8

[jagged glade]

Yea, you can do that, but not necessary.

[odd abyss]

so further would be x^3 + x^2 + 8 = x^3 + x^2 + 8

[jagged glade]

There is a PDF for comparing coefficients, you can have a look

[odd abyss]

On completion of this worksheet you should be able to evaluate
constants given two identical polynomials.

[jagged glade]

Correct

Your descriptions are always decent and accurate, I like that

[odd abyss]

👉 👈

[jagged glade]

[odd abyss]

so we already got the values of a b c d

[jagged glade]

Yes, then we can rewrite our fraction into the partial fraction form.

[odd abyss]

how

this document is 20 years old btw

pretty outdated method

[jagged glade]

https://cdn.discordapp.com/attachments/488104216678760469/726451856464805888/681190496273432663.png

[odd abyss]

these are the partial fractions?

[jagged glade]

Yea, this is the partial fraction decomposition.

To be honest, if this is your first time doing partial fraction decomposition, this is not a good example since it looks like we didn't do anything much.

And it's actually a hard example.

[odd abyss]

it looks pretty hard

but one more thing

https://cdn.discordapp.com/attachments/363224154469826562/726446579648823386/681190496273432663.png

[jagged glade]

Yes?

[odd abyss]

how do we go from there

to https://cdn.discordapp.com/attachments/488104216678760469/726451856464805888/681190496273432663.png

just the top part

[jagged glade]

https://cdn.discordapp.com/attachments/363224154469826562/726439680148504586/681190496273432663.png

[odd abyss]

ok

[jagged glade]

Substitute the A,B,C and D that we have found into that

[odd abyss]

so is this https://cdn.discordapp.com/attachments/363224154469826562/726439680148504586/681190496273432663.png

just something you do when you find all the letters

[jagged glade]

Yep

[odd abyss]

A1x + B + A2x + B2

[jagged glade]

Works too

Typo B1

$\frac{(A_1x+B_1)}{(x^2+2)}+\frac{A_2x+B_2}{(x^2+2)^2}$

[obsidian monolith]

Biscuit:

[jagged glade]

Do you mean this?

[odd abyss]

yea

that format

a b c d

ax b cx d

[jagged glade]

It works too, it's really up to you. Whichever you are comfortable with

Or this too

$\frac{(A_1x+A_2)}{(x^2+2)}+\frac{B_1x+B_2}{(x^2+2)^2}$

[obsidian monolith]

Biscuit:

[odd abyss]

yea I dont think im gonna be able to test out of this

this is the next question\

[jagged glade]

This doesn't seem too hard comparing to the previous one

[odd abyss]

I mean if all I have to do is multiply the 2 together then it should be easy

thank you for all the hard work

[jagged glade]

You're welcome!

[urban harness]

hi do u guys know what these types of problems mean???

i literally have no idea

like y= max{x, 6-x} what does that mean?

[uncut mulch]

at a certain x, y will be the max value of x or 6 - x

eg, at x=0, x and 6-x will have the values 0 and 6 respectively

[uncut mulch]

6 is the larger of the two and thus y=6 when x = 0
the graph will effectively be piecewise

[wide sequoia]

is the range for the function (-inf,0] U (3,3)

[wide sequoia]

Hello?

[stuck lark]

you want to denote that 3 is a point in the range. () denotes open intervals and (3,3) doesn't denote a valid interval. just write the single point in a set. (-infty,0]U{3}

[wide ocean]

I squared and zeroed the equation to use Quadratic formula

But all my answers no real solutions

I'm confused?

Thanks if you know how to answer this..

When -5 is squared, is it +25 or -25?

[viscid thistle]

25

(-5)^2 is positive 25

[wide ocean]

@viscid thistle thank you

That solved my question

[sour eagle]

You should also look at the condition of 2x + 5 and x + 3 being positive ( since square roots cannot be negative nor take negative values when solving over real numbers). So looking at a), x = -2.75 is not a solution.

[wide ocean]

Now the quadratic works

@sour eagle Thanks!

[elfin crescent]

are all sqrt functions neither even nor odd functions?

[uncut mulch]

no

something like:
f(x) = sqrt(x^2) (which is also |x|)
is even

[fluid shore]

Hmmm, the function:

$$\sqrt{|x|}: \bR \to [0,+\infty)$$

is an even function.

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

oh nice, rekt

@elfin crescent

[stuck lark]

\sqrt{|x|}: \bR \to [0,+\infty)

[fluid shore]

?

what

[elfin crescent]

o ok

[willow bear]

$x \mapsto \sqrt{|x|}: \bR \to [0, +\infty)$

[obsidian monolith]

Ann:

[willow bear]

@fluid shore

[stuck lark]

[fluid shore]

....................................................

needless, the meaning was clear

i mean, unless you need it clarified

[elfin crescent]

so that function will be an even function then right?

[fluid shore]

Yea that's an even function

[elfin crescent]

k

[uncut mulch]

,w graph sqrt(x^2)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[uncut mulch]

,w graph sqrt(|x|)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[fluid shore]

here, let me clarify what function i was referring to

[stuck lark]

$\fxn{f}{\bR}{[0,\infty)}{x}{\sqrt{|x|}}$

[obsidian monolith]

RokettoJanpu:

[fluid shore]

$f = {(x,y) \in \bR^2: y = \sqrt{|x|} }$

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

Or more specifically, for my case:

$f = { (x,y) \in \bR \times [0,+\infty): y = \sqrt{|x|} }$

[obsidian monolith]

Abhijeet Vats:

[viscid thistle]

can I get some help with this one? thamks

btw i'm not supposed to know the answer

the question is: solve this

without the tan thing

[uncut mulch]

did you copy down the question properly?

[viscid thistle]

yes

i've looked over it several times today and yesterday

it baffles me cause it's the only weird one out of all of them

[uncut mulch]

it (the series) doesn't converge

[viscid thistle]

how do i prove that

wait why wouldnt it converge?

[uncut mulch]

,w lim n to inf of lim k to n of (4n^2)/(4n^2 + (2k-1)^2)

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[uncut mulch]

summand doesn't approach 0

[viscid thistle]

,w lim n to inf of lim k to n of 1/(2n + (2k-1))

[obsidian monolith]

Display results online and refine query

Requested by Ichikaru#2677

[viscid thistle]

w h a t

my book says that the second one tends to ln(sqrt2) what the fuck

[fluid shore]

That left-hand side looks like it could be written as a Riemann integral.

[viscid thistle]

i'm trying to do that

[uncut mulch]

,w lim of n to inf of (sum from k=1 to n (1/(2n + (2k-1))

[obsidian monolith]

Display results online and refine query

Requested by ramonov#0921

[viscid thistle]

makes sense

[uncut mulch]

that one looks legit

[viscid thistle]

,w lim of n to inf of (sum from k=1 to n (4n/(4n^2 + (2k-1)^2))

[visual finch]

https://i.imgur.com/mbsuT1c.png

What form is this in

It's not standard

it looks like standard circle form

[sour eagle]

Just divide by 16 and you get the standard form

[visual finch]

ye

i got that

but im curious what this form is called

nvm

[alpine basin]

How would i solve this problem? im having issues with it

[viscid thistle]

What are the coordinates of the vertex?

In terms of a, b and c

[alpine basin]

You mean -3, 2?

[viscid thistle]

In terms of a, b and c
^

[alpine basin]

3 and 2?

[viscid thistle]

In terms of a, b and c
^

Never heard of -b/2a?

[alpine basin]

I got -2 = a(3 + 3)^2 + 2

and then arent I supposed to solve for a?

[viscid thistle]

How did you get that

[alpine basin]

i just plugged in h,k and x,y

[alpine basin]

Im supposed to write a formula for this graph, I got (x+3)(x+2)(x-1)^3, but it's supposed to be multipled by a 1/6 as well, how do I find that answer based on this graph?

[viscid thistle]

Probably with the point (0,-1)

[alpine basin]

What would that have to do with 1/6?

[viscid thistle]

Plug x=0

Try

[alpine basin]

Wait

Solve for y by plugging in x = 0?

[viscid thistle]

You want y=-1

[alpine basin]

i get -1=6

-1=-6***

[viscid thistle]

You want to find a coefficient

[alpine basin]

yes

[viscid thistle]

So it should be - 1=6c

[alpine basin]

oh i see solve for c

[viscid thistle]

Yes

c(x+3)(x+2)...

[alpine basin]

yeah makes sense

thanks

[viscid thistle]

Np

[kindred whale]

Not sure if I’m doing this right anyone help?

[craggy dune]

,rotate

[craggy dune]

although you have to input the value

b is wrong

[kindred whale]

im inputting the functions in and mulitplying then multiply by -2?

[craggy dune]

f(2) is the function f evaluated at 2

that means if f(x) = x

then f(2) = 2

if f(x) = x^2 + 2x + 2 then f(2) = 2^2 + 2 * 2 + 2 = 10

[kindred whale]

how do i determine then

sorry im lost

[craggy dune]

no worries

what does f(x) * g(x) look like

?

ignore the -2 for now

[kindred whale]

am i taking 2x+6 times 1/x

[craggy dune]

yes

[kindred whale]

cuz if so then 2x+6/x

[craggy dune]

are you sure about that?

you can multiply your result with x that should give you f

[kindred whale]

im not sure thats why im lost

[craggy dune]

(2x + 6)* 1/x

= (2x + 6)/x

what comes next?

[kindred whale]

*-2

[craggy dune]

no

okay you should have (fg)(x) = 2 + 6/x

that makes sense no?

if you divide 2x + 6 by x

because 2x/x = 2 and 6/x = 6/x

[kindred whale]

yes

[craggy dune]

so now you want to know whats (fg)(-2)

so you evaluate the function at x = -2

can you give me the result?

[kindred whale]

think so give me second

2(-2)+6/-2 plugging it in

-7

[craggy dune]

no

you have (fg)(x) = 2 + 6/x

the idea is right but you used the wrong function

[alpine basin]

How do I find all complex solutions of this problem?

[craggy dune]

you can sub x^2 with y

then just factorize

[alpine basin]

what you mean

[craggy dune]

solve the equation y^2 - 7y -144 = 0

after you're done replace the y with x^2

[alpine basin]

Im confused

I sub the x^4 as well?

[sour eagle]

x^4 = (x^2)^2 = y^2

[craggy dune]

its not really needed but it will make the problem look like something you should have solved many times before

[alpine basin]

im not really too sure what you mean

subbing in y gets me all complex solutions?

[craggy dune]

no factorization

you know something like x^2 + 2x + 1 = (x + 1)^2

have you seen something like that before?

[alpine basin]

yea

[craggy dune]

thats the problem type you're dealing with right now

[alpine basin]

arent i supposed to do it without factoring?

[craggy dune]

no

its basically a more advanced factoring problem

with subbing you get the first step

then you replace the y with x^2 and with that comes the slightly more difficult step

[alpine basin]

can you show what that would look like

I dont really know what to do

[craggy dune]

well factor (y^2 - 7y - 144)

that means you need to find a and b so that (y + a)(y - b) = (y^2 - 7y - 144)

with a - b = - 7 and a * - b = 144

once you have that you turn that into (x^2 + a)(x^2 - b)

then (x^2 + a) = (x + ci)(x - ci)

and (x^2 - b ) = (x + d)(x - d)

there are some pretty strong hints in there

especially as you can assume that the factors are nice

[alpine basin]

what is ci

[craggy dune]

c * i

[alpine basin]

I got (x^2-16) = (x+4)(x-4)

[craggy dune]

good

[alpine basin]

but im not sure howd Id factor (x^2+7)

[craggy dune]

x^2 + 7 ?

[alpine basin]

oh

its 9

[craggy dune]

okay good

(x^2 + 4) = (x +2i)(x-2i)

please don't just believe me

verify

take the right side term apart and see if it equals x^2 + 4

[alpine basin]

yeah seems to work when distributing

so it'd be (x+3i) (x-3i)

?

[craggy dune]

yes

so you have a total of 4 solutions

which works because of polynomial with deg 4

[alpine basin]

oh thats a good way of checking

Thanks

Can I ask you about another problem on my practice exam?

Im supposed to find an equation for this, I got part of the answer which is (x-3)/(x+3)(x+4) but I couldn't get the -8 thats supposed to be multiplied on the (x-3)

[uncut mulch]

you should also account for the scaling factor

$y = \frac{a(x-3)}{(x+3)(x-4)}$

[alpine basin]

The scaling factor?

[uncut mulch]

yes. also there's another mistake

should be (x-4) in the denominator

[obsidian monolith]

ramonov:

[alpine basin]

Oh right

[uncut mulch]

scaling factor in this case a

[alpine basin]

so do i solve for a?

[uncut mulch]

yes

[alpine basin]

Also, could there be a time where a is on the denominator

?

[uncut mulch]

you could start with a on the denominator

you'd end up with a different value for the variable

but the end result would be the same

[alpine basin]

hm

[uncut mulch]

we usually put it on the numerator in these situations for convenience

[alpine basin]

so either way just have to solve for a? And then I would plug in, in this case, -2 for y and 3 for the x? Then solve for a?

[uncut mulch]

no

[alpine basin]

o

[uncut mulch]

(3,-2) is not a point that lies on the curve

[alpine basin]

oh do you just pick any point

so I could do (3, 0)?

I was looking at the intercepts

[uncut mulch]

(3,0) doesn't really help though

since that's how you got the factor (x-3)

consider using the y-intercept

[marble osprey]

ok guys

i got a question for yo

[alpine basin]

so (0, -2)?

[marble osprey]

would you like to hear it

[uncut mulch]

yeh. plug (0,-2)

[alpine basin]

@uncut mulch Would that always be the case where I should use the y intercept for the point to plugin?

[uncut mulch]

no

[alpine basin]

hmm then how would I know which one to choose

[uncut mulch]

use whatever points you can easily identify and/or are given

eg if they explicitly label a point, you could use that

[alpine basin]

Hmm

I thought technically the x intercept and y intercept both label a point

I guess what im tryna ask is how I can tell which one is more useful to plug in

[uncut mulch]

well in this case, the x-intercepts and asymptotes determined the linear and reciprocal factors

you'd need to use other info to determine other properties

the curve won't always have a nice y-intercept

[alpine basin]

hmm okay

Thanks

[alpine basin]

@uncut mulch could you help me with this one as well? if you dont mind xD

[uncut mulch]

perform polynomial long division

[alpine basin]

Ok lemme do that

@uncut mulch I got 2x-8 remainder 41

[uncut mulch]

show your work

[alpine basin]

uh lemme take a picture one sec

Sorry if my hand writing's bad

[uncut mulch]

$-6 - (-10) \neq -16$

[obsidian monolith]

ramonov:

[alpine basin]

oh ur right

like this?

[uncut mulch]

check your signs again

[alpine basin]

oh +10

yeah I keep messin that up mb

[uncut mulch]

1- (-10) = 11

[alpine basin]

yeah

[uncut mulch]

$\polylongdiv{4x^2-6x+1}{2x-5}$

[alpine basin]

The answer says its just 2x+2, when finding the equation of the oblique asymptote, does the remainder not matter?

[uncut mulch]

$f(x) = 2x+2 + \frac{11}{2x-5}$

[obsidian monolith]

ramonov:

[alpine basin]

Yeah the answer key just has this

which is why i was curious

[uncut mulch]

for extreme values , 11/(2x-5) approaches 0

and f(x) will behave like 2x+5

[alpine basin]

Ah

Thanks for the help

[odd helm]

how would i do this problem

[viscid thistle]

could anyone explain the max thing to me. or how to solve these problems. I'd appreciate any help

[rapid lance]

@odd helm do you know how to add vectors?

it would help to graph the vectors to get a sense if you feel confused

[viscid thistle]

{5a+2b : a,b ∈ Z}
{6a+2b : a,b ∈ Z}
how do I do those

[rapid lance]

what are they asking for?

[viscid thistle]

set

it's an infinite one obviously

but how do I do the '...' for it

[rapid lance]

i think its the same as any other set

[viscid thistle]

this is the answer apparently

{5a+2b : a,b ∈ Z} = {...,−2,−1,0,1,2,3,...} = Z

wouldn't it be the same for the second one?

[rapid lance]

can you get odd multiples of 5 in that one?

or in that case odd numbers?

[viscid thistle]

guess not

[viscid thistle]

{3,6,11,18,27,38,...}
how do I express it in set notation?
guess x ∈ N, but what else?
I see it's +3,+5,+7,... but not sure how to express it algebraically, since in 3 x would have to be 0, and to acquire the result of 3 you'd have to add 3, which breaks the pattern?

[harsh smelt]

6-3=3
11-6=5
18-11=7
27-18=9

[viscid thistle]

I see it's +3,+5,+7,...

[harsh smelt]

reasonable assumption may be that it is sequence given by rule $\
a_{k+1} = a_k +2k+1$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

and $a_1 = 3$

[viscid thistle]

how have you come to this assumption

[harsh smelt]

i just looked at first terms

[obsidian monolith]

Commander Vimes:

[harsh smelt]

i mean look at first terms, to be more precise at their differences

it is just odd numbers

solving this recurrence will give u exact form of set elements

[viscid thistle]

thank you

[viscid thistle]

{(x, y) : x ∈ [1,2], y ∈ [1,2]}
what am I supposed to do

[fluid shore]

?

[viscid thistle]

it says 'sketch the following sets of points in the x-y plane'

[fluid shore]

What is the question?

[viscid thistle]

but wouldn't that just be a straight line

[pale bison]

what have you tried?

[fluid shore]

You're just sketching $[1,2] \times [1,2]$, where $\times$ is the cartesian product

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

So, it'll just be the set of ordered pairs such that $1 \leq x \leq 2$ and $1 \leq y \leq 2$. That's going to be a rectangle.

[viscid thistle]

ok but wouldn't it just be a line starting at [1,1] and ending at [2,2] ?

[obsidian monolith]

Abhijeet Vats:

[fluid shore]

??

ok but wouldn't it just be a line starting at [1,1] and ending at [2,2] ?
What do you mean by "a line starting at [1,1] and ending at [2,2]"?

[viscid thistle]

oh nevermind

[fluid shore]

You're confusing interval notation with ordered pair notation

[viscid thistle]

yeah

forgot that americans use [] instead of <>

[fluid shore]

wot

[viscid thistle]

<1,5> is 1<= x <= 5 here

as in interval notation

[sour eagle]

[fluid shore]

I've never seen that notation for intervals

[viscid thistle]

well we use it

for 1<x<5 it's (1,5)

[fluid shore]

My dude, I'm not even American lol

[ , ] is typically what's used for closed intervals

it can also be used for other things but it's clear from context what it's supposed to mean

[viscid thistle]

what about this lol

{(x, y) : x, y ∈ R, y ≥ x^2 - 1}

all I can get from it is that it starts at [0,-1]

[viscid thistle]

or this one
{(x, y) : x, y ∈ R, x^2 + y^2 ≤ 1}

[pale bison]

did you solved the previous one?

[viscid thistle]

no

[pale bison]

ok

{(x, y) : x, y ∈ R, y ≥ x^2 - 1}
this means, x and y are real numbers, and you're looking at the x y plane

(x, y) : x, y ∈ R^

[viscid thistle]

yes ik

[pale bison]

and you're picking points x and y such that y >= x^2 - 1

i.e. points from the curve x^2 - 1 and points above it

[past meadow]

n=choose an unoccupied channel

[viscid thistle]

ok

i'll draw it publius

give me a sec

[sour eagle]

@viscid thistle the solution has already been suggested in #prealg-and-algebra ?

Why would you ask again?

[willow bear]

what is this

is this meant to be {y ≥ x^2 - 1}

[viscid thistle]

yes

[willow bear]

the boundary is a parabola. yours looks off at the bottom

why are the points (±1, -1) in there

[viscid thistle]

dunno

[harsh smelt]

what is the task. the one Ann mentioned?

{y ≥ x^2 - 1}

[viscid thistle]

is everything okay except of those 2 points?

yea

[willow bear]

i mean

your parabola looks weird

like

the boundary of your region should pass through the points (±1, 0)

[harsh smelt]

why bottom of parabola is "cut off"

[pale bison]

it's a parabola after an uppercut

[viscid thistle]

That's a vase

[harsh smelt]

can u first draw just y = x^2-1?

[willow bear]

vimes thats what im trying to point out

[viscid thistle]

oh

I understand now

btw

it passes through (±1,0)

I meant for it to pass through them at least

[willow bear]

well

correct your sketch

draw the parabola y = x^2 - 1 properly

then shade in the region north of it

[pale bison]

still a bit pointy but it's acceptable

[viscid thistle]

wdym

[pale bison]

,w graph y = x^2 - 1

[obsidian monolith]

Display results online and refine query

Requested by Publius#2314

[viscid thistle]

diff scale

[pale bison]

i mean ok the scale is a bit off, but it is not pointy at the bottom

well anyway

{(x, y) : x, y ∈ R, x^2 + y^2 ≤ 1}

now that you've completed the previous problem, can you do this one on your own?

[viscid thistle]

I'll try

[harsh smelt]

it is better to first draw a boundary

and then to see the behaviour of remaining parts

[viscid thistle]

I drew the points

[pale bison]

i mean, does the equation x^2 + y^2 ≤ 1 remind you of anything

you should not apply the "plot point and connect" technique here

[harsh smelt]

you should not apply the "plot point and connect" technique here
would better to use technique "see relation and show it"

haha

[willow bear]

bruh

southern halfplane: am i a joke to you

[viscid thistle]

wouldn't it just be a circle then

[harsh smelt]

yes it would

[viscid thistle]

what's the difference between it and x²+y² =1

[pale bison]

none read it wrong

[willow bear]

no it'd be a disk

not a circle

x^2 + y^2 = 1 is a curve

[harsh smelt]

what u have plotted firstly is closed region 0 <= y <= sqrt(1-x^2)

[willow bear]

x^2 + y^2 <= 1 is a region, specifically the region enclosed by said curve

a disk in this case

[viscid thistle]

@viscid thistle the solution has already been suggested in #prealg-and-algebra ?
@sour eagle where

[sour eagle]

#prealg-and-algebra message

[harsh smelt]

@viscid thistle u can remember (or recall if u heard this before) that x^2+y^2 = a (a >= 0) always forms a circle

[viscid thistle]

i know the channel, but where is the solution?@tilde

[willow bear]

@harsh smelt what if a=-1 lol

[harsh smelt]

fixed

[sour eagle]

Dude, HoboSas literally said what you should do. It's not the whole solution, but from there it's easy.

[harsh smelt]

but if a =-1 it would some region on complex plany i guess

[viscid thistle]

I still don't understand why it's a disk and not a circle

[pale bison]

@sour eagle hobosas was wrong, this is not a "prove this identity" question
wops i read it wrong

[willow bear]

x^2 + y^2 ≤ 1 includes the points inside the circle.

[harsh smelt]

circle is boundary

[viscid thistle]

ah

[willow bear]

the difference between a circle and a disk is the same as between a sphere and a ball

[viscid thistle]

oh I get it

thank you all

[sour eagle]

Oop nvm

[viscid thistle]

if x>1 then it's an entire second quadrant?

{(x, y) : x, y ∈ R, x > 1}

[pale bison]

no

[harsh smelt]

is second quadrant > 1

???

[pale bison]

that can't be more wrong, points in second quadrant have negative values of x

[viscid thistle]

I mean the 1st quadrant

except of the [0,1] part

[smoky pagoda]

if x>1, it's the right half of the disk

not the first quadrant

[viscid thistle]

for
{(x, y) : x, y ∈ R, x > 1}

[smoky pagoda]

yeah

[viscid thistle]

oh

[smoky pagoda]

because y is free to be anything

[viscid thistle]

so 1st and 4th except of the [0,1],yes?

[smoky pagoda]

what's wrong with the [0,1] interval?

x can belong to the (0,1] interval in this case

[viscid thistle]

cause x > 1

[smoky pagoda]

oh wait nvm

im retarded

[viscid thistle]

so x can't be below and including 1

[smoky pagoda]

no fi x>1

then it just inst part of the disk

since the disk has radius 1

[viscid thistle]

it's a diff question

[smoky pagoda]

ah

[viscid thistle]

no disk here

[smoky pagoda]

what's the q

[viscid thistle]

{(x, y) : x, y ∈ R, x > 1}