precalculus

[proven marten]

Not your time

[mighty onyx]

no it's a homework I got a quiz friday my test is next monday or wednesday

[proven marten]

please wait a little

[old nimbus]

sorry

[viscid thistle]

Try evaluating it using a calculator

[viscid thistle]

Wtf

[mighty onyx]

all day man

[old nimbus]

is it my turn now?

or you still going?

[proven marten]

not really

@mighty onyx Use log laws

[old nimbus]

oh ok

[viscid thistle]

Prime Minister please wait

[proven marten]

Yes

I was about to say that

Btw

[viscid thistle]

How

Are you guys getting 1620

[proven marten]

Whats wrong with u dont get sick next time

^

[viscid thistle]

Like wtf

[old nimbus]

???

[proven marten]

notice they have a negative index

Mr Boriš

[viscid thistle]

Like wtf
?

[proven marten]

1/b^-4 = 1/1/b^4

@mighty onyx

Im going to ask u to simplify

Mr Boris

[old nimbus]

I did it the way Albot did it and go 1620 if that helps?

[proven marten]

Would it be fine if we pung you when we finished

[old nimbus]

*got

[viscid thistle]

1620=2^2x3^4x5

[proven marten]

ok I use pen and paper now

[mighty onyx]

I tried expanding the original problem and I was told abunch of different things

[proven marten]

its

c^-1 right

[viscid thistle]

No, c^(-1)

[mighty onyx]

yeah

[viscid thistle]

Lol

What is going on

[old nimbus]

lol

[mighty onyx]

$ \ln\left(\frac{a^2}{b^{-4}c^{-1}}\right) $ expanded is $\ln(a)^2 - \ln(b)^-4 +\ln(c)^-1$ right?

[viscid thistle]

$\ln(\frac{a²}{b^{-4}c^{-1}})=\ln(a²b⁴c)=2\underbrace{\ln(a)}{2}+4\underbrace{\ln(b)}{3}+\underbrace{\ln(c)}_{5}$

[obsidian monolith]

Albot1288:

[viscid thistle]

Idfk what you guys did

[proven marten]

hm

why was there still ln

in your final answer

[mighty onyx]

So I should end up with this?

[viscid thistle]

why was there still ln
@proven marten the next step is to sub for the values lol

What are you doing @mighty onyx ?

You have the values of log(a) , log(b) and log(c)

[proven marten]

^

@proven marten the next step is to sub for the values lol
@viscid thistle why not just keep it in the calc...

[viscid thistle]

?

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

I understand what I did wrong, I was plugging in ln(a) into a and then getting ln(1620) or 7.39

[viscid thistle]

Wtf

You said a=2

I got scammed

Lmao

I was like wtf

:(

[mighty onyx]

well in my defense I posted the problems

[viscid thistle]

True

Its late for me

[proven marten]

scammer gets scammed

[mighty onyx]

so it would be $2(2)+4(3)+5$ which is 21

[obsidian monolith]

Albot1288:

[viscid thistle]

Yes im sorry

[mighty onyx]

@old nimbus your turn

[old nimbus]

cheers

[viscid thistle]

Yes, correct.

Go on mr prime Minister

[old nimbus]

how do you integrate 3 - 9/ (x+1)

[proven marten]

Thats not pre calc

[old nimbus]

does the 9/(3+1) become 9ln|x+1|

what is it then?

[viscid thistle]

Yes

[proven marten]

Calculust

[old nimbus]

and the 3 becomes 3x

[viscid thistle]

Yes

[old nimbus]

so 3x - 9ln|x+1|

[viscid thistle]

Correct

[old nimbus]

cheers

[viscid thistle]

Np

[proven marten]

🍻

[old nimbus]

is integration not precalc?

[proven marten]

Do me a favour and go to the calculus channel and check the description of it

[old nimbus]

well I went there and they said go to precalc for your problems

[proven marten]

Who

Dont say

polynomial

wait no

polynomial would say algebra

[old nimbus]

???

[proven marten]

polynomial is a user that tells people to go to algebra when they post calc questions

[viscid thistle]

polynomial would say algebra
Poly left the server lol

[proven marten]

oh

what happened

[old nimbus]

lol

[viscid thistle]

Idk but he left that's all i know

Thank god

[lime bolt]

no he was banned

[low night]

Yo can i get help on my hw tomorrow at 2:00pm

I wanna do it with someone to ensure im doing everything correctly

[viscid thistle]

@lime bolt really?

[proven marten]

First of all, why at 2 pm

[low night]

cuz thats the time im free

[lime bolt]

yea, he wasnt even warned

straight to ban

[proven marten]

Second of all, 2pm where?

[low night]

est

[proven marten]

Im going to

yes

I thoughty ou were american

[low night]

i am

dude what school is this

bruh wtyf

is this gmu

this is gmu math discord right?

[lime bolt]

?

[proven marten]

No

This is an international math discord server

[old nimbus]

yes

[proven marten]

As you can see

[lime bolt]

hmm i just searched up and joined a server called "gmu stem"

[proven marten]

We even have the Prime Minister

hmm i just searched up and joined a server called "gmu stem"
@lime bolt Invite pls

[old nimbus]

yeah lol

[lime bolt]

lol it doesnt look great but if u want

ill delete the link now

[mighty onyx]

Solve thelogarithmic equation for x. $\log_9x=32$.

[obsidian monolith]

Albot1288:

[mighty onyx]

it is not as simply as solve for x as I thought

[proven marten]

9^32 = x?

[mighty onyx]

That's what I put

[proven marten]

and

use the TI 84 Plus

that costs a lot of money

[mighty onyx]

I know but It doesn't like viarables

Solve the exponential equation for the exact value of x. $\ln(\ln x) = 3$

[obsidian monolith]

Albot1288:

[mighty onyx]

I got $e^{e^3}$

[proven marten]

looks about right Kakarot

[mighty onyx]

but it says it's wrong rip

unless

nope idk what he put

[obsidian monolith]

Albot1288:

[viscid thistle]

Correct btw @mighty onyx

[mighty onyx]

Yay but it says it's wrong I've been doing math since 10am so I'ma take a break

[fervent lagoon]

i don't understand how I could possibly getting this wrong

i got the correct angle but it says my angle is wrongg

i just did beetle1's total run minus beetle2's first run

[noble urchin]

,help

[obsidian monolith]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[noble urchin]

,list

[uncut mulch]

show work

[delicate rivet]

How do you do question a)? -the test is over dw

[tardy ridge]

The surface area of your cubiod is 2x^2+4xh

and 2x^2+4xh=150. Find h in terms of x.

[shadow ember]

quick question

not a math question

but a question about math

(if that makes sense)

can I do calc without a graphing calculator? Our teacher said that they aren't permitted and we will be taught ways to do it without a graphing calculator

Will the class be a pain in the ass or will I not be learning real calculus?

[delicate rivet]

and 2x^2+4xh=150. Find h in terms of x.
@tardy ridge ohhhhh ok ty

[viscid thistle]

@shadow ember I wouldn’t say you “need” anything more than online graphing tools to help build intuition and visualize some stuff

And even then you don’t “need” it

[shadow ember]

I can't visualize stuff 😦

[viscid thistle]

You can do everything without a graphing calculator if that’s what the syllabus says

It’s certainly possible to do calc like that

[regal sleet]

How do I evaluate this and find the equation for the line since cot pi is undefined?
y-3=cot(pi)(x+1)

[uncut mulch]

is this part of a bigger question?

[viscid thistle]

@short sorrel im here now

Hello everyone this is my first time here

i need help on proving this pascal's rule

i need to use the nCr formula to do it

[limpid moss]

could you show the step you got stuck on

[viscid thistle]

ok.

https://gyazo.com/6c6638b69f6c581031a7804f273ac9eb

I essentially

plugged em in

to

https://gyazo.com/677e07971ae9d1798aa70ee7a475c5f1

(@limpid moss)

[limpid moss]

yes theres no problem with the method youre using

[viscid thistle]

im stuck on how to further simplify

ik i need common denominators

[limpid moss]

consider the relation $(x+1)!=(x+1)x!$

[obsidian monolith]

powerblo:

[limpid moss]

it should help yield common denominators

[viscid thistle]

wwait whaaat

is that only true for some number + 1

[limpid moss]

its not anything special, it just comes from how you write the factorial

[viscid thistle]

or can it be (x+y)! = (x+y)x!

[limpid moss]

no, if that confuses you try writing down factorials in their product form for yourself

[viscid thistle]

alright thanks so much

i'll try simplifying at let you know how it goes

really appreciate the help 😄

wait a second if its

(x-1)!

then is it the same as

(x-1)x!

[limpid moss]

not quite, as i said try $(x+y)!=(x+y)x!$, its best to see for yourself why it doesnt work for $y\ne1$

[obsidian monolith]

powerblo:

[viscid thistle]

hmmm

ok

lemme write it out

ok

i see

now in my case i only have -1's

so do I have to like factor a -1 out

i dont think i did it right lol

[limpid moss]

you sure you writing the factorial out properly?

[viscid thistle]

https://gyazo.com/cd24b2c5c165e9aa0b3d4b63d59e5493

there was only one (r-1)! in the denominator

[limpid moss]

well the equivalent of $(x+1)!=(x+1)x!$

[obsidian monolith]

powerblo:

[limpid moss]

would be $x!=x(x-1)!$

[obsidian monolith]

powerblo:

[limpid moss]

again, you should really try this out for yourself if you want to get used to shuffling factorials around in combinatorics

[viscid thistle]

im so confused

wiat

wait*

is what i did up there correct

[limpid moss]

you tried to use $(r-1)!=-1(r+1)r!$, right?

[obsidian monolith]

powerblo:

[viscid thistle]

yea

[limpid moss]

yeah thats not correct

[viscid thistle]

thonk

[limpid moss]

before tackling the problem above

try to prove the statement $(x+1)!=(x+1)x!$

[obsidian monolith]

powerblo:

[limpid moss]

and see if you can do the same thing for $(x+y)!$

[obsidian monolith]

powerblo:

[limpid moss]

i.e. as $(x+y)!=Ax!$

[obsidian monolith]

powerblo:

[viscid thistle]

ok

ok i got this

https://gyazo.com/64dc35ce39fd08eb70133fb969a5dee2

[limpid moss]

ayy youre on the right track

you can do that one more time for another factor in the denominators, and youll have a common denominator

[viscid thistle]

ohh

wait

(n - r - 1)!

i have to break that down probably

wait

is that the same as

(n-r)! - 1!

[limpid moss]

you sure you did the thought process behind $x!=x(x-1)!$?

[obsidian monolith]

powerblo:

[regal sleet]

@uncut mulch yeah it’s finding an equation for a tangent line to a point in a parametric equation

[viscid thistle]

im not sure how to use that here

[limpid moss]

well its not much different

[limpid moss]

but you should be able to at least tell if (n-r-1)! equals (n-r)! -1!

[pallid tapir]

does anyone know what the function for that is

[limpid moss]

looks like a parabola

[viscid thistle]

wait does it

or was it jsut a coincidenc

[pallid tapir]

do you know what the f(x) is?

[limpid moss]

well it doesnt to start off

[viscid thistle]

ok

[limpid moss]

do you know what the f(x) is?
just telling you the answer kinda defeats the purpose

[pallid tapir]

ye lol

[limpid moss]

but you can see it goes through 3 points

[pallid tapir]

yes

[viscid thistle]

0 and 4 are solutions to your quadratic

[limpid moss]

a parabola that goes through 3 points is uniquely decided

[viscid thistle]

and the first term is negative since its going down

so u can solve for it then

but anyways back to my mess

aaa im stuck man i have no idea how to figure this out

yep im lost

only the second week of school rip

@limpid moss do you think I should try the other side

or is it the right step by simplifying (n-r-1)!

[limpid moss]

well the question is if you can express (n-r-1)! in the form of (n-r)!

if you can, by all means give it a try

[viscid thistle]

i dont have that much time lol

this is the challenge question in a 30 question packet and its due in 1 hour

[limpid moss]

hmm

well a hint is to use the same $x!=x(x-1)!$ equation

[obsidian monolith]

powerblo:

[limpid moss]

you found that theres r! and (r-1)! in each of the denominators, so you could factor r out and make a common denominator

you can do the same thing for (n-r)! and (n-r-1)!

[viscid thistle]

wait whaaaat

omg im so

ok wait

https://gyazo.com/87df37067c8e98b2c2a2cf973f6b8c09

this is what i have rn

i dont have r! in denominator

[limpid moss]

have you tried x = n-r

[viscid thistle]

yeah i just ran in circles

idk it was a challenge after all, i'll just wait for the acitivty in class tomorrow when we do it together

if u solved this problem u get no hw for a week

i tried

lol

gn its late for me

[limpid moss]

aight gn fam

[novel cargo]

mornin frens

[novel cargo]

checking 4cos(x) = 1 + 2cos(x) on wolfram, under alternate forms I get:

what is this?

where and when do I learn this?

just curious

[willow bear]

ah yes complex exponentials

guess you'll learn that when you get to complex numbers

[novel cargo]

thansk

[novel cargo]

sin(x)^3 = sin(x)

sin(x)^2 = 1 | divide by sin(x)

sin(x)^2 - 1 = 0 | subtract 1

(sin(x) + 1) * (sin(x) - 1) = 0 | factor

[past meadow]

forgetting about the case where sin(x)=0 when you divide by sin(x)

btw

[novel cargo]

hahah

thanks

this is what I learned today the hard way

just wanted to share here

[past meadow]

oh i thought you were leading up to "why is there the extra solution of x=npi" lmao

[novel cargo]

yeah,

I wanted to ask to see someone solve it

you already saw it on first glance

[past meadow]

sorry i ruined the fun lol

[novel cargo]

btw, here if you divide by zero, you will miss some solutions

np

2 cos(x)^2 + 3 cos(x) = 0

I got: x = pi/2 + 2npi, x = -pi/2 + 2npi

checking on wolfram I get:

so, I have the first solution, but I removed the second and third because range cos is [-1, 1]

what are these? curious

complex nums?

also, the domain of inverse cos is [-1, 1]

[near monolith]

Yes, those are complex numbers

[novel cargo]

is there a different definition for inverse trig functions then?

[willow bear]

well it basically boils down to the complex logarithm

with some extra work

[near monolith]

cos(x) when x is a real number will be in the [-1,1] interval, but if you define it for complex numbers things get complicated.

cos(ix) = cosh(x) and sin(ix) = i sinh(x)

[novel cargo]

how much complex numbers do I need to know, precalc?

just the basics?

[near monolith]

In pre-calc you only use real numbers and can ignore those solutions

[novel cargo]

learning math is an emotional roller coaster

3 csc(5x)^2 = -4

does this have real solutions?

[willow bear]

no

[novel cargo]

thanks

2 tan(x)^2 - 3 tan(x) = -1

[daring yarrow]

do you need help solving?

[novel cargo]

I got: x = -pi/4 + npi, x = pi/4 + npi

but wolfram gives somewhat different

[daring yarrow]

how did you solve?

[novel cargo]

tan(x) ( 2 tan(x) - 3) = -1 | factored

Oh my

I set tan(x) = -1 and solved

as if there was a zero

[willow bear]

[novel cargo]

[daring yarrow]

[willow bear]

that's like solving the equation 2u^2 - 3u = -1 by writing u(2u-3) = -1

[novel cargo]

please console me by saying that you guys also make such mistakes

[willow bear]

not of this caliber no

[daring yarrow]

i'll console you by saying i despise trig equalities

[novel cargo]

hahah

[novel cargo]

sec(x)^2 + 6tan(x) + 4 = 0

I got: x = arctan(-5)+npi, x = -pi/4 + npi

but apprently it's wrong (or I don't understand)

from wolfram

[novel cargo]

<@&286206848099549185>

[blissful ridge]

Show your working

[novel cargo]

OK

tan(x)^2 + 1 + 6tan(x) + 4 = 0 | pythagorean

tan(x)^2 + 6tan(x) + 5 = 0 | combine like terms

(tan(x) + 5) (tan(x) + 1) = 0 | factor

tan x = -5, tan x = -1

[blissful ridge]

I'd like to see the original question and your input in WA

[novel cargo]

OK

#46

[blissful ridge]

Your input in WA

[blissful ridge]

WA is doing some high level shit

Your answer is probably correct

[novel cargo]

thanks

I wish I knew how to change WA axes to radian

[mighty onyx]

lemme just test something

$e^(e^3)$

[obsidian monolith]

Albot1288:

[mighty onyx]

that's what I thought

[blazing parrot]

need curly braces

[viscid thistle]

{}

[obsidian monolith]

Albot1288:

[willow bear]

$e^{e^3}$

[obsidian monolith]

Albot1288:
Compile Error! Click the reaction for details. (You may edit your message)

Ann:

[mighty onyx]

I know that it's wrong but my teacher put in $e^(e^3)$

[obsidian monolith]

Albot1288:

[mighty onyx]

like why?!?!

well just wanted to rant and test my latex

[novel cargo]

your latex needs improving but rant +1

[mighty onyx]

lol

what's worse is he says to put in the exact number but when I do he put it in wrong

I wanna see if I'm putting it in wrong

$-(1/8)$

[obsidian monolith]

Albot1288:

[novel cargo]

is this fraction?

[viscid thistle]

does your homework have latex?

[mighty onyx]

sometimes

[viscid thistle]

\frac{}{}

[willow bear]

$-\frac{1}{8}$ if you want fractions

[obsidian monolith]

Ann:

[viscid thistle]

or even -\frac18

[mighty onyx]

well I don't think he put it in latex because he forgets his parenthesis sometimes in the answer

[viscid thistle]

that's messed up

[mighty onyx]

ye

[novel cargo]

is this a plain text compare?

[mighty onyx]

they way he does it I think is if you wrote it on paper and then typed it out exactly

but people write differently

like $8 * 10^{32}$ is the same as $8e^{32}$

[obsidian monolith]

Albot1288:

[novel cargo]

how's this the same?

[lime bolt]

[mighty onyx]

because in scientific notation e can represent $10^{x}$

[willow bear]

8e32 does not refer to 8 * e^32

[obsidian monolith]

Albot1288:

[willow bear]

as in

8e32 is not 8 * 2.71828...^32

[novel cargo]

ahh, I read it differently

e like the number e

[lime bolt]

he means "scientific notation"

[novel cargo]

interpreting math answers is very complicated

can't be done by comparing plain text letter for letter

[mighty onyx]

exactly

[novel cargo]

lol, loved this

[mighty onyx]

I just now gotta figure out what he used for a product

[viscid thistle]

please tell me they didn't use x for multiplying

[mighty onyx]

oh god lemme check

[novel cargo]

xxx = x^2

[lime bolt]

[novel cargo]

don't forget curly brace

[mighty onyx]

It doesn't look like it

$10e^{-8x} = 3.$ I should be getting $-(1/8)*ln(3/10)$

[obsidian monolith]

Albot1288:

[willow bear]

$-\frac{1}{8} \ln(\frac{3}{10})$

[obsidian monolith]

Ann:

[mighty onyx]

but I should be getting that right

[willow bear]

is your homework system rejecting this answer?

[mighty onyx]

yeah

[willow bear]

can i see the answer box, exactly what it looks like, and exactly what you're putting into it

this is no longer an issue with the math but with the homework system so i want to deal with that

[mighty onyx]

I got another problem so yall can see the "methods"

[willow bear]

Do not use any space

is there a space right before the minus

[mighty onyx]

No

[willow bear]

or after the )

[mighty onyx]

No

[willow bear]

huh

report this as a bug?

[viscid thistle]

maybe they want an alternate answer

[willow bear]

they made no indication of what form they want the answer in

so this is either a bug in the system or a fuckup in the wording

[mighty onyx]

I used a calculator to find an alternative answer and got 0.150497

but it still say's it's wrong I tried every combo I could of those numbers I just don't know what he put as "right"

[viscid thistle]

hm

do you have unlimited attemps?

[novel cargo]

oh, nooo

[mighty onyx]

yeah

[viscid thistle]

try with this then

[willow bear]

have you tried writing the ln as log?

[novel cargo]

smart move

[willow bear]

like just replacing the letters ln with log is what i mean

[viscid thistle]

$\frac18\log(\frac{10}{3})$ try with this too

it is equivalent

to what you had

[obsidian monolith]

Al𝟛dium:

[mighty onyx]

what if he put the ln first?

[viscid thistle]

have you tried what i wrote

[novel cargo]

the possibilties are exponential

[mighty onyx]

yeah incorrect

[viscid thistle]

the natural way of writing it is with the log on the right so that it does not create ambiguity

whatever, there are some quite of possibilities and it's dumb to keep trying to change stuff. Your math is correct and that is all that matters

[lilac storm]

How does it turn up as (x+2) (x+2-3)? I know that (x+2) is the common factor which we out, but I don’t understand how it’s possible for the -3 to just join the second (x+2)

[novel cargo]

see it as two terms

first term = (x+1)^2

second term = -3(x+2)

now, you factor out (x+2)

[lilac storm]

The first bracket is (x+2)^2

[novel cargo]

sorry, I mistyped

that's what I meant

[lilac storm]

Yeah np I appreciate it

Continue pls

[novel cargo]

when you factor out (x+2)

what remains is (x+2) - 3

you follow?

[lilac storm]

Yes

[novel cargo]

that's it

now you have: (x+2) * ((x+2) - 3)

remove the inner parentheses

they're redundant

[lilac storm]

So if something’s like (x+-x) -/+ x it’s okay for it to be (x+/-x+/-x)

[novel cargo]

that's ambiguous

I can't say

maybe Ann knows

[viscid thistle]

$\overbrace{(x+2)^2}^{a^2}-3\overbrace{(x+2)}^{a}$ sub in a:=x+2, $\ a^2-3a$ factoring out a $\underbrace{a}{x+2}(\underbrace{a}{x+2}-3)$ sub back $\ (x+2)((x+2)-3)\implies (x+2)(x-1)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

a clarification

[novel cargo]

wow, gj

[viscid thistle]

the whole point of the sub is to make clearer the factorising

[lilac storm]

Okay I understand now

I forgot that factoring out something from an equation multiplies it by everything left in it

So it’s basically (x+2) ((x+2)-3)

Tsm

[novel cargo]

is it OK to not memorize the trig sum and diff formulas?

just being aware of them and recognize if some are possible and look them up?

[viscid thistle]

Angle sum formulas are necessary

Not okay at all

difference formulas are just sum formulas sin(a-b) = sin(a+(-b))

@novel cargo you need to memorize all the principal formulae

like identities and values of standard angles

Those formulas are actually pretty complicated to prove

So you should memorize them

[novel cargo]

oh man, no wiggle room it seems

put yourself together az

[viscid thistle]

Just 2 formulas, it's very doable

I think memorizing as a very helpful tool, if you can just take some time out of the day and memorize important formulae, identities, rules and values, you can save a lot of time and effort instead of looking those up again and again

[novel cargo]

how do you reduce it to two

?

[viscid thistle]

Also I don't think it's difficult to memorize them, as when you will solve questions while looking it up a few times it gets automatically drilled inside the head at least for me

[novel cargo]

yeah

I have the sin and cos sum and diff almost down

tan a little bit needs more

[viscid thistle]

if you have sum, you have diff

difference formulas are just sum formulas sin(a-b) = sin(a+(-b))

tan(a+b)=sin(a+b)/cos(a+b)

[novel cargo]

you sure that works with tan?

[viscid thistle]

Yes

You just need a little trick

@novel cargo btw I don't put a conscious effort in memorizing formulas and stuff, what I do is first understand the concept, glance at the derivation, then try to derive it myself. While going through the process the formula is already embedded in my head if you get what I mean. It works for me maybe it will work for u too

Let me make texit

[novel cargo]

thanks

[viscid thistle]

Bro I was making big text

[novel cargo]

you continue

pretend nothin happened

[misty quest]

so many people in precalculus

[viscid thistle]

x^3 - 3x^2 + 4 factoring hint?

x = - 1

kthx

[viscid thistle]

Align was bugged so this is the best I could do

@novel cargo

[novel cargo]

@viscid thistle CHAD

[viscid thistle]

There's another mistake

Shit

[novel cargo]

sin(a)cos(a)?

[viscid thistle]

Another one

Fuck

[obsidian monolith]

HoboSas:

[novel cargo]

yo, don't correct them

let me find and learn

[viscid thistle]

Too late I guess

[novel cargo]

np

perfect

[lilac storm]

I don’t understand the question

What are linear factors?

[novel cargo]

probably means degree one

[lilac storm]

Which is...?

[novel cargo]

x^1 is degree one, linear

x^2 is degree two, not linear

[lilac storm]

Ummm...

So would that be something like x(x+/x)?

[novel cargo]

yeah, I'd say

[lilac storm]

Okay cool

[novel cargo]

but I may be wrong tho

I don't know the context of your classes

[lilac storm]

Thanks for the help nonetheless

It’s just facotrising and expansion

How do I stop myself from solving (7)^2 as 14...

I always seem to do it

[novel cargo]

rewrite it as 7*7 until you sure you can do it mentally

[lilac storm]

Yeah that’ll help

[novel cargo]

I do it with other things myself

[lilac storm]

Brains are weird smh

[blissful kayak]

^

Sometimes they do work, other times they sorta work, and other times they don't work

It do be like that... ;/

[viscid thistle]

what would be the most intuitive way to solve
$(1.23 * 10^-4) + sqrt(4.5 * 10^3)$

[obsidian monolith]

Deleted User:

[viscid thistle]

I know I can just use my calculator

but I want to learn to do it by hand

I've already tried squaring all terms to get rid of the square root

but that just leaves a mess

[viscid thistle]

You can't square all terms because you want to, this is not an eqn, it's an expression

@viscid thistle you there?

$(1.23\cdot 10^{-4})+\sqrt{4.5\cdot 10^3}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

And it'll be useful to express such numbers as this: $2\cdot 10^{-2}=0.02$ on decimal form

[obsidian monolith]

Al𝟛dium:

[spark veldt]

Someone tell me what I’m supposed to do here

Please

[blissful kayak]

Oh I think you're supposed to eliminate the parameter

For |a|=|b|, the graph is a circle. If |a|=/=|b|, the graph is an ellipse

And on the bottom left quadrant (Bottom left for us, bottom right for the paper), we solve for sine and cosine

And leverage the Pythagorean Identity

Which is guided to you on the top

Here's my work

For the first one

[spark veldt]

Thanks for doing it out

[blissful kayak]

Of course!

You need help for the last one?

[spark veldt]

I think I can handle it

[blissful kayak]

Great!

[old nimbus]

you guys dont help with integration here aye?

[blissful kayak]

Not yet

I'm gonna get back into CalculusLand tomorrow!

I'll ask for help when I need it

So stay tuned :3

(I left off last time at U-substitution)

[copper vigil]

is there any theorem about limits of rational functions and their limits at plus or minus infinity
like if you have some polynomials P(x)/Q(x) and the limit is inf/inf or inf/-inf etc

[harsh smelt]

@copper vigil their limit equals to ratio of leading coefficients if polynomials are of the same degree

otherwise if P(x) has degree > Q(x) limit is infinity (+-) and 0 if Q(x) has bigger deg

[copper vigil]

so how do i know if it's plus or minus inf

if P(x) has a higher degree than Q(x)

so inf/inf and -inf/-inf would be a +inf limit

and -inf/inf and inf/-inf would be a -inf limit

yeah that makes sense thank you

just needed to refresh myself on that

so basically the other terms don't matter right? it's the same as finding the limits of just the highest degree terms on the top and bottom of the fraction

[harsh smelt]

for limits at infinity they do not matter

since they become neglectible

[azure hedge]

yo how do u find a quadratic function if you are given the zeros and the vertex point. Lets say the zeros are x=-1,3 and vertex point is (1,2)

[fleet yew]

well every quadratic function can be written in the form y=a(x-x1)(x-x2)

where x1, x2 are the zeros

and a is the leading coefficient that scales up the function by a certain amount

@azure hedge

[azure hedge]

ok

so in my case x1 is 1 and x2 can be -3

[fleet yew]

look at the equation again

minus signs inside the parantheses because the zeros are subtracted from x

so x1 and x2 are your zeros

[azure hedge]

ok

how do u find a

[fleet yew]

you know everything but a

so all that's left to do is plug in

x = 1, y = 2, and x1 and x2 are the zeros

[azure hedge]

ooooooooo

i get it

ty my g

[ornate wolf]

any idea how i should start?

[fleet yew]

@ornate wolf let z = a + bi

[ornate wolf]

@ornate wolf let z = a + bi
@fleet yew should i sub it in?

[fleet yew]

Yeah

[ornate wolf]

what should i do afterwards? i tried that actually but got stuck

do i try to find value of a and b?

[blissful kayak]

No

You find its modulus given the structure of z

Because if it wants you to find $z\times\bar{z}$, it gives you a purely real number. And that number is its modulus

[obsidian monolith]

Coleculus:

[ornate wolf]

yeah

z x z- = |z|^2

[blissful kayak]

Mhm!

[ornate wolf]

i think my issue is, i don't know the steps to get there

[blissful kayak]

So you can just square the expression even

[ornate wolf]

I sub in the a+bi, I get

6a + 6bi + 2/(a+bi)

[blissful kayak]

6a+6bi

[ornate wolf]

yup

[blissful kayak]

Okay, so you can put that in parenthesis. Then next to it (In separate parenthesis), change all the bis to -bi

And then you can just expand it out like a difference of squares

(Am I making sense?)

[ornate wolf]

sorry so I have to do the following:

6a + 6bi, the denominiator is 1

so I first have to make it a+bi

so i multiply the numerator as well

then i add them with 2/a+bi

then mulitply it by the conjuage

[silent mulch]

Can someone help im stuck

I wannq say it's -3 but im not sure and it's not multiple choice

[ornate wolf]

(Am I making sense?)
@blissful kayak i'm kinda confused now actually, do i add them together, then multiply by its conjugate?

[novel cargo]

average rate of change is max(f(x)) - min(f(x)) divided by delta x

[fleet yew]

@ornate wolf you figure it out

?

[ornate wolf]

no

[fleet yew]

Ok i think this will help

$6(a+bi) + \frac{2}{a+bi} = Real$

[obsidian monolith]

AMD:

[fleet yew]

That "Real" is whats important

It means you can just get rid of things on the left hand side

[ornate wolf]

so it's just 6a + 2/a?

[fleet yew]

Because if you add, subtract, multiply, or divide a real number by a real number, the resulting number is also real

No

First mulitply the top and bottom of the fraction by the conjugate

[ornate wolf]

First mulitply the top and bottom of the fraction by the conjugate
@fleet yew yup

[blissful kayak]

Oh okay I guess I was completely wrong LMAO

[ornate wolf]

@fleet yew yup
@ornate wolf 2/(a+bi) * (a-bi)/(a-bi)

yeah?

[fleet yew]

Ok then what do you get

[ornate wolf]

Ok then what do you get
@fleet yew 2a-2bi / (a^2+b^2)

[blissful kayak]

Yes

[fleet yew]

Now bring everything on the left hand side into that fraction

[ornate wolf]

sorry I don't understand

[fleet yew]

X + Y/Z = (XZ + Y)/Z

[blissful kayak]

^

[ornate wolf]

ah

[blissful kayak]

Shouldn't there be a name for that process?

I feel like there is but I might be wrong lol

[ornate wolf]

okay I've got:

(6a^3 + 6ab^2 + 6bia^2 + 6b^3i + 2) / a^2 + b^2

[fleet yew]

it's called bringing it to a common denominator

X = X/1

ok so now here comes the part where the "Real" matters

$\frac{6a^3 + 6ab^2 + 6bia^2 + 6b^3i + 2}{a^2 + b^2} = Real$

[obsidian monolith]

AMD:

[fleet yew]

just any arbitrary real

whose value doesn't really matter

this is basically just comparing real and imaginary parts

because we know that the imaginary part of the right hand side is equal to 0

so you can just multiply both sides by a^2+b^2

because a and b are real

[ornate wolf]

does that leave us with only 6a^3 + 6ab^2 + 2?

[fleet yew]

no

it leaves you with the numerator of the fraction

[ornate wolf]

if imaginary is 0, doesn't that means bi = 0?

[fleet yew]

you were specifically given that Im(z) does not equal 0

so we must remove that from the set of solutions

[ornate wolf]

but in this equation, the RHS is real, which means the Im must be 0?

sorry man.. i'm so lost

[fleet yew]

if you really want, just call the right hand side x + 0i

or just x

there's one thing you need to know

the central rule to these sorts of problems

$a + bi = c + di$ if and only if $a=c$ and $b=d$

[obsidian monolith]

AMD:

[fleet yew]

just for any arbitrary complex numbers

[ornate wolf]

okay

[fleet yew]

ok i can tell you're a bit lost huh

let's go back up for a sec

[ornate wolf]

yeah.. sorry man, im really terrible at math and i've spent hours sitting down but my brain is just............

[fleet yew]

$\frac{6a^3 + 6ab^2 + 6bia^2 + 6b^3i + 2}{a^2 + b^2} = Real$

[obsidian monolith]

AMD:

[fleet yew]

this all makes sense right?

[ornate wolf]

yup

[fleet yew]

and you would agree that a^2+b^2 is a real number

because a and b are real

[ornate wolf]

yes

[fleet yew]

and a real number times a real number is a real number

[ornate wolf]

yup

[fleet yew]

$6a^3 + 6ab^2 + 6bia^2 + 6b^3i + 2 = Real$

[obsidian monolith]

AMD:

[fleet yew]

so this is also an equally valid statement

[ornate wolf]

okay i have a confusion here, we have 6bia^2 +6b^3i

in the equation

doesn't it make those numbers imaginary?

if 6bia^2 +6b^3i is imaginary, and the other parts of the equation are real, how can real + imaginary = real?

sorry for the dumb question

[fleet yew]

it cancels out nicely

also i think you made an algebraic error

what you should have is

$6a^3 + 6ab^2 + 6bia^2 + 6b^3i + 2a - 2bi = Real$

[obsidian monolith]

AMD:

[ornate wolf]

it cancels out nicely
@fleet yew the imaginary part cancels each other?

[fleet yew]

just be patient i'll show you

so what we have now is that equation that i just wrote above

[ornate wolf]

yes you're right i missed out 2a-2bi

[fleet yew]

now you should know that: Real/2 = Real

[ornate wolf]

yup

[fleet yew]

so yeah you can divide through by 2 to get

$3a^3 + 3ab^2 + 3bia^2 + 3b^3i + a - bi = Real$

[obsidian monolith]

AMD:

[ornate wolf]

yup

[fleet yew]

so this is a real number

so that means that the sum of the imaginary parts must be 0

[ornate wolf]

correct

[fleet yew]

do that and try to solve

it comes out real nice

$3a^2bi + 3b^3i - bi = 0$

[obsidian monolith]

AMD:

[ornate wolf]

do I write the above as:

$i(3ba^2 + 3b^3 -b)$

[fleet yew]

yeah

[obsidian monolith]

wiz:

[fleet yew]

factor out the b as well

3a^2 + 3b^2 - 1 = 0

i'm just gonna give you the answer bc i gtg

but you solve that and get that a^2 + b^2 = 1/3

and (a+bi)(a-bi) = a^2 + b^2

which equals z times conjugate(z)

so the answer is 1/3

[ornate wolf]

Ok I'll try to digest it

Thanks so much man for sticking throughout, really appreciate it

[blissful laurel]

Can I get some help with this one

[novel cargo]

more #calculus it looks

[blissful laurel]

?

you mean its supposed to go in that channel?

[novel cargo]

this precalculus

[blissful laurel]

oh okay thanks

[novel cargo]

np

[ornate wolf]

so the answer is 1/3
@fleet yew yup it makes so much sense, thank you so much!

However when going through all the steps, I don't know how to have the intuition to derive those steps? I wouldn't have thought of doing it like that.,. any tips?

[fleet yew]

@ornate wolf the process i showed you is called "comparing real and imaginary parts"

remember what i said earlier

if a+bi = c+di then a=c and b=d

[ornate wolf]

if a+bi = c+di then a=c and b=d
@fleet yew yeah... but those things don't even pop up in my mind

any hints on how to start?

Do I square the terms so i can get e.g a^2 + b^2 for each mod sign?

[willow bear]

z + conj(z) = 2Re(z)

[ornate wolf]

z + conj(z) = 2Re(z)
@willow bear for Z - conj(z) it is = 2Im(z) too right?

[willow bear]

not quite

it's 2i Im(z)

[ornate wolf]

it's 2i Im(z)
@willow bear ah, youre right

any other hints? my brain can't think of anything

[willow bear]

$2 |{\Re(z)}| + 2 |{\Im(z)}| = 2\Re(z) + 8 + 2i \Im(z)$

[obsidian monolith]

Ann:

[willow bear]

clearly Im(z) must be 0 from this

[ornate wolf]

Ann:
@obsidian monolith why is it not :

$ |2 {\Re(z)}| + |2{\Im(z)}| = 2\Re(z) + 8 + 2i \Im(z)$

[obsidian monolith]

wiz:

[ornate wolf]

or it doesn't matter?

[willow bear]

it doesn't matter

since the absolute value function is multiplicative and |2| = 2

[viscid thistle]

can a limit not exist even if both of the one sided limits r equal

[copper vigil]

if the function is not continuous

[willow bear]

no @copper vigil

[copper vigil]

wait nvm you're right the limit would exist

[willow bear]

@viscid thistle if lim[x->c+] f(x) and lim[x->c-] f(x) both exist and are equal then the two-sided limit also exists and is equal to them both

[viscid thistle]

ok ty

[ornate wolf]

it doesn't matter
@willow bear thanks, then in this case do i just write 8?

[willow bear]

wym "write 8"

[ornate wolf]

wym "write 8"
@willow bear wait i mean z= 4?

since im must be 0?

[willow bear]

you have 2|z| = 2z + 8

with z now real

[ornate wolf]

um, how do i proceed to express it in terms of z?..sorry i'm still confused

[willow bear]

wym "express it in terms of z

z is now confirmed to be a real number and i am asking you to solve the equation 2|z| = 2z+8

[ornate wolf]

z is now confirmed to be a real number and i am asking you to solve the equation 2|z| = 2z+8
@willow bear can i simply divide 2 on LHS and RHS?

[willow bear]

yes

[ornate wolf]

okay so then I have

|z| = z + 4

|z| = a+4

z = (a+4)^2 ?

[willow bear]

wat

where did a come from

also |z| is not sqrt(z)

bruh you're overthinking it

[ornate wolf]

where did a come from
@willow bear z = a +ib?

and ib is now 0

so becomes a +4

[willow bear]

z is real

are you telling me you can't solve the equation |x| = x+4?

[ornate wolf]

don't we need to sub in the a+bi?

sorry

[willow bear]

bruh

you're overthinking it

Im(z) is 0

numbers with zero imaginary part are real

z is real

z. is. REAL

[ornate wolf]

thanks man

coming back to it now

it makes sense

[lime bolt]

[viscid thistle]

@viscid thistle you there?
Hi went to sleep. Thanks letting me know about the squaring thing. So you just recommend turning them into decimals w/o scientific notation and just solving that way?

[viscid thistle]

in this specific case, yes

you can operate numbers easily

[novel cargo]

oh man

precalc is more work than calc change my mind

nvm

[lime bolt]

bruh

[cedar cipher]

Probable yes

[narrow marsh]

Just want to be pointed in the right direction to approach this Q:

There are two jars with a, b balls respectively, each has one golden ball, rest are black. Balls are taken without replacement from jar 1. Each time the golden ball in jar 1 is taken, the balls are replaced in jar 1 and a ball is taken without replacement from jar 2. Calculate the chances of getting the golden ball in jar 2 if you have k tries in total of taking balls from 1.

(Edited the right Q in, mb)

Not from textbook - is from game situation. Wondering if a general formula dependent on k could even easily be found. If not, even finding the probabilities for specific k doesn't seem simple.

[echo wagon]

Am I being dumb? How does the number of balls you took from the first jar affect your chances of getting the golden ball in jar 2?

[narrow marsh]

When you take the golden from jar 1, you take a ball from jar 2

[echo wagon]

Yes, but there will always be b balls in jar 2

[narrow marsh]

Without replacement from 2

[echo wagon]

So the probability is 1/b

[viscid thistle]

Lol, I think it’s invariant as written

[narrow marsh]

eh?

[echo wagon]

I might be misinterpreting the question

[viscid thistle]

There might be something missing/written incorrectly

Ah wait I might see

You may be able to draw more than once from jar 2

This chance goes up as k increases

Is this right shuri?

[narrow marsh]

Yeah, I edited it to try to make it clearer

[echo wagon]

Oh, like you draw balls from jar 1, then when you get the golden ball you put them all back and draw one ball from jar 2, then you draw balls from jar 1 again, and so on

[narrow marsh]

yh.

[echo wagon]

That does seem hard

[narrow marsh]

In my specific example a=9, b=10, k=7 but that doesn't look easy to me (and there are a few other k's I'm interested in)

[echo wagon]

Well, I can help you with a specific case. If k = 1, it is 1/(ab)

[viscid thistle]

I will see if I can come up with anything but no promises

[echo wagon]

Lol

Same, but I'm very bad at probability

[narrow marsh]

Sounds like monte carlo sim is quicker way to estimate 🤔

[viscid thistle]

True, but unsatisfactory

[narrow marsh]

Mentions on anything on this r 👌

[viscid thistle]

Seems like some binomial but without replacement shenanigans might work

[narrow marsh]

Best approach I can think of atm is to split it into 2 problems... 1st jar and 2nd jar... hm

[viscid thistle]

Well yes the main problem is the first jar

If you can find the expected number of golden balls obtained from the first jar, then it’s easy from there

[narrow marsh]

ah expected is good enough, didn't realise

Isn't that then just 2k/a

[viscid thistle]

Uh no I don’t think it could be that simple

How’d you get that?

[narrow marsh]

On average, you get 1 golden every a/2 tries, right?

ah nvm, I agree it isn't that simple for expected

[viscid thistle]

the without replacement messes this up a lot

I can maybe think of an extremely tedious way to solve this

[blazing parrot]

Getting golden ball from jar 1:
tries probability
0 0
1 1/a
2 (1 - 1/a)*1/(a-1) = (a - 1)/a * 1/(a-1) = 1/a
3 (1 - 1/a)*1/(a-2) = (a - 1)/(a - 2) * 1/a
t + 1 (1 - P(t))/(a-t) where P(t) refers to probability of getting the ball in t tries

$$P(t + 1) = \frac{1 - P(t)}{a - t}$$

[obsidian monolith]

Not Chezstick:

[viscid thistle]

probability of getting the ball in t tries is not the question at hand though?

probabilities for getting 2,3 ..., k golden balls are also needed

[blazing parrot]

if you can calculate P(t) could you be able to sum over all possible tries required / ball within k trie?

[viscid thistle]

this was along the lines of my idea for an extremely tedious way to calculate it lol

[blazing parrot]

also once we solve for 1 golden ball couldnt we repeat for the rest since the state of jar A gets reset

[narrow marsh]

I'm a bit confused on that table. I thought it was 1/a, 2/a, 3/a, ...

[viscid thistle]

without replacement strikes again

[blazing parrot]

indeed

[viscid thistle]

a formula for the probability to obtain a specific amount q, with 1 ≤ q ≤ k, of balls, in k tries would work I think

[narrow marsh]

So the table, uh, what does it represent?

eg. 2 tries, probability 1/a

[viscid thistle]

multiplying each q by it's corresponding probability and summing them up would give the expected number of balls in k tries I think

breaking it into these pieces seems to make the problem a bit easier but I am still not sure how to do that

[blazing parrot]

So the table, uh, what does it represent?
the probability of the golden ball being picked with exactly that many tries

[narrow marsh]

nvm

I get it 👌

[viscid thistle]

I'm not sure about the table. Why does P(1)=P(2)?

[narrow marsh]

Chance of failing on 1st pick, and succeeding on 2nd = (a-1)/a * 1 / (a-1) = 1/a

that's the idea right?

[viscid thistle]

I was more looking for a conceptual answer

and not sure about that reasoning anyway

[narrow marsh]

I'm thinking hard right now... isn't it 1/a for all of them

[viscid thistle]

no that doesn't make sense either

[narrow marsh]

The chance of failing on 1st 2 picks, and succeeding on 3rd is 1/a, no?

[viscid thistle]

no?

[narrow marsh]

(a-1)/a * (a-2)/(a-1) * 1/(a-2)

no?

P(fail 1st) * P(fail 2nd) * P(succeed 3rd)

[blazing parrot]

P(fail 2nd) includes P(fail 1st) i believe

you can confirm that P(1) = P(2) otherwise if a = 2, P(1) + P(2) != 1

[viscid thistle]

The chance of failing on 1st 2 picks, and succeeding on 3rd is 1/a, no?
yea it is then, nvm

[narrow marsh]

I'm a bit confused here. If a = 3, you have 1/3 + 1/3 + (2/1)*(1/3) != 1

[blazing parrot]

hm

i may be stupid

[viscid thistle]

I mean, if I am betting against someone in this game of luck to find the golden ball, and I tell them that I get 2 tries to draw a ball and they get 1

why would they think that's fair

[narrow marsh]

The table --- in the game, you have to get the ball on exactly the 2nd try (or nth)

That's what the table represents at least

You can pick any n and your chances of winning are 1/a

[viscid thistle]

doesn't seem to be what the description of the table says but ok

[blazing parrot]

P(fail 2nd) includes P(fail 1st) i believe
dammit i think i assumed this and everything went downhill from there'

[viscid thistle]

"probability of getting the ball in t tries" sounds like, you get t tries to find a golden ball. You can get it on your 1st, 2nd, ..., or t'th try

[blazing parrot]

sorry i meant exactly t tries

[viscid thistle]

ok that clears things up

so it's "probability of getting the ball on the t'th try"

[narrow marsh]

Uh, not sure if it helps, but I think this recursion is right? Where E_k is expected number of golden balls with k tries

[viscid thistle]

I think this is too much of a leap

we have not even talked about the possibility of getting multiple balls within k tries yet

which, again, is the thing that makes this problem hard

[narrow marsh]

Think I made a mistake, big thonk again

$$E_k = \sum^{k-1}{i=1}\left(\frac{1}{a}\right)\left(E{i}+1\right), E_1 = \frac{1}{a}$$

[obsidian monolith]

Shuri2060:

[narrow marsh]

I think its this

Uh how to explain it. . .

I condition on getting the golden ball on exactly the ith try (probably 1/a)

[viscid thistle]

I might also remind that when a golden ball is obtained, the bag resets

which is another thing that makes this complicated

is that accounted for?

[narrow marsh]

Wish I had paper. Pretty sure yes.

[blazing parrot]

correct me if im wrong, but if we have a 1/a chance on the kth try, wouldn't we have 1/a chance on the ath try, when there is only 1 ball left?

[narrow marsh]

Not sure what you're referring to

For getting on exactly the kth try, it's 1/a probability for all k<=a

[blazing parrot]

since we are drawing without replacement, if we keep drawing unsuccessfully we will reach the state where the only ball left is the golden ball, which you will then have a 1 (!= 1/a) chance of drawing

[narrow marsh]

For getting it with k<=a tries, then yes. If you have a tries, you are guaranteed the ball

[blazing parrot]

oh wait i see my error

sorry

[narrow marsh]

it is k/a proba

np all good

[blazing parrot]

$$E_{k+1} = \frac{1}{a}\sum^{k}_{i=1} E_i + 1$$

[obsidian monolith]

Not Chezstick:

[blazing parrot]

rewrite of ur equation for me to think through it better

[narrow marsh]

i think its wrong

my recursion

[blazing parrot]

im not sure why you would + 1

[narrow marsh]

ill look at it again in a bit

[blazing parrot]

actually would such an equation be possible

if E_q is like 5.7 for some number q, then what does that mean?

minimum 5 balls and 70% chance for a 6th one?

but theres always the chance that you draw the golden ball every time

i dont think theres a way to represent (x_0 chance for 0 balls, x_1 chance for 1 ball, ...) within a single number

to represent that i think we need a function that takes in the number of balls, k and a and gives us the probability of that number of golden balls being the result

[narrow marsh]

I've just realised where I've gone wrong