precalculus

[solid warren]

yeah

usually polar coordinates are (r, theta) and spherical is (p, theta, phi)

their r has no relation to r from polar notation

[nova wedge]

and r in polar coordinates is the hypotenuse on the xy plane

[solid warren]

yes

[nova wedge]

Classes all being online sure is rough 🙂

[maiden igloo]

yep

[viscid thistle]

Hi!! Sorry for the disturbance but how would I find the sum or difference of log(25w^3/sqrta)

[tardy ridge]

$log(25w^3/sqrta)$

[obsidian monolith]

Fishraider:

[tardy ridge]

$log(25w^3)-1/2*(loga)$

you want this?

[obsidian monolith]

Fishraider:

[unique hill]

then log(25w³) can be further expanded

[tardy ridge]

good point

2log 5 + 3log w - 1/2 * log a

[sly ether]

don't know if it's the right channel but...how do i solve an equation of the form:
a^x+b^x=c
@sly ether can you guys help me here

trying here

[jagged glade]

Hello tacos, do you have the exact question?

[sly ether]

yeah

a = 0.5417
b = 0.736
c = 0.56

but it's also a general question on how i would solve this

[jagged glade]

I see... I have no clue right now but to use numerical methods.

[sly ether]

meaning testing every posibility?

[jagged glade]

Yea, or use computer to solve on the graph y=a^x+b^x-c for zeros

That's what I got

[west shadow]

Okay so I dropped my math classes of the semester because I can't do online

I still have the textbook

And I'm going over the questions

I'm hella confused

[sour hemlock]

You shouldn't drop math class.

Any reason why you had to?

[west shadow]

Yeah I cannot do online

And the school allowed it to be cleaned from my records

Because of the pandemic atm

[sour hemlock]

Why can't you do it online?

[west shadow]

I'm a foreign exchange student

Time zone is way off

[sour hemlock]

And what grade level is your math?

[west shadow]

This is college pre calc

[sour hemlock]

I see

So high school math?

[west shadow]

Yeah they don't allow me to bring my records from my high school

And I entered the school a week before it began and so I couldn't take the exam

[viscid thistle]

Hey, can i get some homework help?

hmm

i need help with #1 part B

[formal iris]

@viscid thistle

[fleet yew]

@viscid thistle p=2pi/b

[viscid thistle]

heyy

[formal iris]

seems like AMD has answered it

[viscid thistle]

sorry i dont quite follow

[formal iris]

Period = how long it takes for the function to do 1 full round

[viscid thistle]

i dont care aabout the answer

[fleet yew]

Lmao

[viscid thistle]

i want to know how to solve it

[formal iris]

Yes what part u dont get

[viscid thistle]

finding out how to get the period

[fleet yew]

Ok

Lemme write this in latex

[formal iris]

alright so you are given cos 4x

[viscid thistle]

yes

[formal iris]

Imma let Amd explain

mans certified

[viscid thistle]

okay

[fleet yew]

$m+a\sin(b(x-s))$

[obsidian monolith]

AMD:

[viscid thistle]

what?

[fleet yew]

That is what we call the general form of a sinusoidal wave

Because sin and cos are pretty much the same thing, just shifted

[viscid thistle]

yea

[fleet yew]

m is the midline, a is the amplitude, s is the shift, and b is the factor that determines the period

[viscid thistle]

ok

[fleet yew]

So lets look at your specific example

[viscid thistle]

that would be helpful

[fleet yew]

cos(4x) i think

[viscid thistle]

yes

[fleet yew]

As you can see, the midline is zero, the amplitude is 1, and there is no shift

Because cos(4x) = 0 + 1cos(4(x-0))

All there is to deal with is "b"

[viscid thistle]

yes ok

anything else @fleet yew

[fleet yew]

So b is 4

[viscid thistle]

yea

so i juts put 2pi/4

[fleet yew]

so you can think of this function cos4x as being 4 times "faster" than the parent function cosx

[viscid thistle]

yes

[fleet yew]

So its period will only be 1/4 as much

[viscid thistle]

yea

so the period is 1/4?

im lost

no 4

[fleet yew]

So you divide the period of cosx by 4

2pi/4 = ?

[viscid thistle]

pi/2?

[fleet yew]

Yes

[viscid thistle]

ohhhhh

ok

[formal iris]

Amd I have a question is it possible for you to help me as well?

[fleet yew]

Sure

[formal iris]

Imma wait till Disabled is done

[fleet yew]

@viscid thistle you good?

[viscid thistle]

uhh

i have a question about a cosine problem

but if parths problem is eaiser help him first

easier

[fleet yew]

K just ask it here

[formal iris]

go ahead i feel like mine will take a bit

[fleet yew]

@formal iris go to questions K

Or whatevers free

[viscid thistle]

help with C

[formal iris]

the aim of the that question is for u to combine the sine & cos then take the peroid

[viscid thistle]

ummm

loat

lost

so i find the periods of each?

[formal iris]

There is a way to combine the two into one cos or sin

Have u taken preclac yet?

[viscid thistle]

nope

today was my second class

[formal iris]

Jumping into challange questions is not efficent if you do know material

[fleet yew]

There is an easier way lol

[viscid thistle]

this is just my hw

[fleet yew]

But it involves testing values

First let x=0

[viscid thistle]

let me write this down

also

would you prefer to join mathematics voice?

[fleet yew]

I dont have mic lol

On my pc rn

[viscid thistle]

oh ok

ok im ready

[fleet yew]

Ok

first find out what value the function has when x is 0

[viscid thistle]

ok

[fleet yew]

What is it

[viscid thistle]

0?

is that wrong?

[fleet yew]

You sure about that?

[viscid thistle]

nope

[fleet yew]

sin(3x)+cos(4x)

Plug in 0

[viscid thistle]

sin(0)+cos(0)

ohhh

its 1

@fleet yew

[fleet yew]

Yep

Now find another point when it equals 1

[viscid thistle]

ok

isn't it at (2pi,0)?

[fleet yew]

(2pi, 1)

[viscid thistle]

1

ok

[fleet yew]

So heres the thing

[viscid thistle]

yea

[fleet yew]

Sinusoidal functions can have the same value twice in a period

So you dont know if that is actually the period or not

[viscid thistle]

oh

ok

so how do you tell if its the period?

[fleet yew]

If 2 pi is the period, then h(0)=h(2pi), and h(a)=h(2pi+a)

So lets use a test value

To make things simple lets use pi/2

[viscid thistle]

4

oh ok

[fleet yew]

Its in radians lol

So integers will make it ugly to look at

[viscid thistle]

yea lmao

[fleet yew]

So let a=pi/2

[viscid thistle]

right

[fleet yew]

Does h(0+pi/2)=h(2pi+pi/2)?

[viscid thistle]

on

no

it doesn't

[fleet yew]

You sure about that?

[viscid thistle]

yea

[fleet yew]

Check again

[viscid thistle]

2pi =! 0

[fleet yew]

Lmao

[viscid thistle]

wait

what am i doing wrong?

[fleet yew]

0 doesnt equal 2pi but sin(0)=sin(2pi)

[viscid thistle]

😐

[fleet yew]

lol

Do you know what h(x) means

Or h(something)

It means you plug something in for x into the formula

So if h(x)=2x then h(3)=2(3)=6

[viscid thistle]

ok

[fleet yew]

So tell me

Does h(0+pi/2)=h(2pi+pi/2)?

[viscid thistle]

yes

:0

it does

[fleet yew]

So yeah 2pi is the period

[viscid thistle]

yes

[fleet yew]

Thats not entirely mathematically rigourous but its the right answer

So yeah good job

[viscid thistle]

ok

this

is giving me a headache

[fleet yew]

What is

@formal iris you still need help?

Pretty sure pepe is done

[viscid thistle]

uhh

lmao

[fleet yew]

Lol r u?

@viscid thistle

Its cool

[loud quartz]

Is this the correct way to solve this

[viscid thistle]

do i have the right answer?

[patent beacon]

3sin(2x)

[viscid thistle]

ok

[patent beacon]

Wait I derped

[viscid thistle]

yea

lmao

[patent beacon]

Don't trust me you had it haha

[viscid thistle]

i didnt

i got it wrong

idk

@patent beacon

lmao

[patent beacon]

Hmm it's a horizontal stretch by 2, no?

[viscid thistle]

@fleet yew any ideas?

yes

[fleet yew]

lol looks right

DUDE

[viscid thistle]

lmao

[fleet yew]

your equation is right

[viscid thistle]

im a clown

[formal iris]

AMD someone helped me out

[fleet yew]

but read the question again

[patent beacon]

Oh haha

[viscid thistle]

wait

[fleet yew]

this is the kind of shit that fucks you on standardized tests

[patent beacon]

At least someone here can read

[viscid thistle]

do i have to add 3 and 1/2?

[fleet yew]

yes

[viscid thistle]

u rkidding

[fleet yew]

bro ik

that's cheap

[viscid thistle]

WTFFFF

i hate this

[fleet yew]

lmao

[viscid thistle]

i cant with math no more

[fleet yew]

nah man it's cool

that got me too for a solid minute

they're trying to trick you

you just have to read carefully

[viscid thistle]

yea

a = 3; b = 1/2

[fleet yew]

bro now comes the hard part

add those numbers

[viscid thistle]

it's too hard for me. i think i'll let you solve it

lmao

its too hard

what is this

[fleet yew]

broo

it's exactly what i posted 30 min ago

something about the general eq of a sinusoidal

[viscid thistle]

hmm

ok

[viscid thistle]

to calculate a, you have to take the max y value and subtract it with the min y value then divide it by 2

[fleet yew]

wow

very helpful

[viscid thistle]

bruh i'm trying

lol

[viscid thistle]

to find d, you have to look at how much the graph has shifted up or down

https://www.desmos.com/calculator/klxhi5erc3

[keen sigil]

could someone help me out with this?

[willow bear]

what is giving you trouble here

[keen sigil]

im not sure how to compose the (fg^-1)(x)

[willow bear]

wdym

given that there's no circle there, i'd assume it stands for multiplication, not composition

[keen sigil]

oh yea sorry thats what i meant

[willow bear]

so if you have g^-1 written out it should pose an issue to you?

[keen sigil]

oh is that saying it's just g^-1 times f(x)?

[willow bear]

g^-1(x) times f(x) yes

[keen sigil]

ohh alright

that clears it up

thanks

i thought it was (fg)(x)^-1

[slim wigeon]

Well

You have one side and an angle

(there may be a different way to do this, but I'll explain how I remember)

Are you there @slate oracle

[slate oracle]

Yes.

[slim wigeon]

ok

I assume you know of SOH CAH TOA

[slate oracle]

Yes.

[slim wigeon]

And whats the area formula for a triangle?

[slate oracle]

area=hxb/2

[slim wigeon]

ok

since the rotation is arbitrary, we'll set b=14

so we want to find h

using soh cah toa we can find one of them that works for what we have and what we want

we have b

we want h

[slate oracle]

Yes.

[slim wigeon]

which one of the three works best/most easily?

[slate oracle]

toa

[slim wigeon]

ok

[slate oracle]

We know a and b.

Right?

[slim wigeon]

You only currently know b and B

if you call the angle B

[slate oracle]

Wait what?

Oh ok.\

Yeah, sorry.

[slim wigeon]

Whats the relationship that TOA describes?

[slate oracle]

tan=oppo/adj

[slim wigeon]

ok

so what is our opposite in this case?

[slate oracle]

14?

[slim wigeon]

yes

and the adj is the one we want to find

what other information do we currently have

[slate oracle]

That the angle is54

[slim wigeon]

ok

where does that go into TOA

[slate oracle]

Not sure.

[slim wigeon]

what do you usually put into sin cos and tan

[slate oracle]

Uh.. what?

[slim wigeon]

you have $sin(), cos(), and~tan()$ what goes into the paratheses?

[slate oracle]

x?

[obsidian monolith]

Sei:

[slim wigeon]

and x is usually what type of number? (integer, angle, other?)

[slate oracle]

Angle I suppose

[slim wigeon]

ok

so we have

$Tan(54º)=\frac{14}{height}$

[obsidian monolith]

Sei:

[slim wigeon]

right?

[slate oracle]

Yes.

[slim wigeon]

and you want the height, so how do you proceed?

[slate oracle]

How.. would I proceed? I'm sorry.

[slim wigeon]

You want to isolate height

[slate oracle]

So I would multiply each side with 14.

[slim wigeon]

no

that would give you $14*tan(54)=\frac{14^2}{height}$

[obsidian monolith]

Sei:

[slate oracle]

divide, sorry.

by 1/14.

I mean multiply by 1/14

[slim wigeon]

closer

but still not quite

height is on the bottom

right now

[slate oracle]

Oh.\

[slim wigeon]

if we do something to one side of an equation we have to do it to the other

so what can we do to get rid of height on the right

[slate oracle]

multiply each side with height?

[slim wigeon]

yes

now we have $height*tan(54)=14$

[obsidian monolith]

Sei:

[slate oracle]

Yes./

[slim wigeon]

and are a step closer to getting height isolated

[slate oracle]

divide each side with tan(54) now?

[slim wigeon]

yes

[slate oracle]

We're left with height=14/tan(54)

[slim wigeon]

yes

[slate oracle]

Sorry.

[slim wigeon]

its fine

[slate oracle]

So now, we know our height, right?

[slim wigeon]

just trying to help you understand this stuff

yes

you do

[slate oracle]

Thanks for all the help, btw.

[slim wigeon]

you can put 14/tan(54) in your calc to evaluate

[slate oracle]

I would get 20.777675732

[slim wigeon]

Not gonna check, but I believe you

[slate oracle]

Wait.

[slim wigeon]

?

[slate oracle]

I got 10.17159

[slim wigeon]

ok one sec

[slate oracle]

Ok.

[slim wigeon]

i am gonna check

10.17 is correct for degrees

ok so now you have b and h

and your area formula

plug in and solve

[slate oracle]

Ok.

[slim wigeon]

Lemme know what you get

[slate oracle]

So if I plug everything in, I'll get A=10.17(14)/2

Right?

And if I solve that, I'll get 71.19

[slim wigeon]

Looks good

[slate oracle]

What's next?

[slim wigeon]

You've found the area

I think you're done with that question

[slate oracle]

Oh, thank you!

[slim wigeon]

No problem

[vague minnow]

hey, I need to solve this equation system, how do I start? do I use graphs maybe, or some substitution?

[vague minnow]

nvm, I solved it :v

[gilded stirrup]

How do you find the interval for a function f(x) where the average rate of change is 0?

Is this correct?

[fleet yew]

yeah looks good

[proud jetty]

how to I do that one

Could someone help me step by step?

[viscid thistle]

factor out the cosx from each term

[proud jetty]

^ could u show me how to do it step by step sir?

[viscid thistle]

sure

[proud jetty]

thanks alot

[viscid thistle]

i'll try

[proud jetty]

pls do

[patent beacon]

Step 1:
cos(x)[√2 sin(x) - 1] = 0

And that's the step by step of factoring out the cos(x) from each term

[proud jetty]

Oh

so cos(x)=0?

[patent beacon]

Yes, that's one of the solutions. The other factor gets some solutions too

[proud jetty]

how would I factory them all probably?

ik all the answers

but I wanna know the setps to get them

@viscid thistle any luck?

[viscid thistle]

im working on it

[proud jetty]

if u need the answers i have them

I just need the steps

[viscid thistle]

alright so your second step is to set: sqrt2sin(x) - 1 =0

[proud jetty]

okay

wut happen to cosx?

[viscid thistle]

you have to calculate cos ^-1 (0) = x

[proud jetty]

😮

[viscid thistle]

forget what i said about square rooting

this is what u do next:

you set sqrt2sin(x)=1

then u divide both sides by sqrt2

[proud jetty]

ic

[viscid thistle]

then u find the sin inverse of 1/sqrt2

last step should look like : x = sin ^ -1 ( 1/sqrt2)

[proud jetty]

😮 yea i got it thx alot

[gilded stirrup]

I know it tells me to use guess and check, but is there a way to do this algebraically?

[viscid thistle]

I dont think so

[gilded stirrup]

F

[viscid thistle]

Might be wrong tho, but i have never heard of any other way of doing it

[gilded stirrup]

This is gonna take so long

[viscid thistle]

Lol

[gilded stirrup]

Ah ok

[fleet yew]

@gilded stirrup which problem specifically?

[gilded stirrup]

b and d? They took so long to guess and check

[fleet yew]

If you dont mind going into a bit higher level math

[gilded stirrup]

sure I might not understand fully tho lol

[fleet yew]

You dont need exact answers do you?

[gilded stirrup]

I think answers close to the solutions are fine

so no

[fleet yew]

Look up the taylor series for cosine

You only need the first 2 or 3 terms

[gilded stirrup]

hmm ok gimme a second

[fleet yew]

Basically all these functions can be represented using an infinite sum of terms in x

[gilded stirrup]

oooooohh I see

[fleet yew]

So like adding together a bunch of polynomial terms

[gilded stirrup]

ok Okok I understand lol

Thank you!

[fleet yew]

@gilded stirrup what do you mean by log(x)

e or 10?

[gilded stirrup]

10

it’s a grade 12 math book so I think it just assumes it’s 10

[acoustic harbor]

im in 12th grade

only thing we use is ln

[gilded stirrup]

ooof

In is convenient tho

[acoustic harbor]

what class are you taking?

my math book is some calculus 6th edition book

[gilded stirrup]

grade 9 fundamentals

my book oh

my book is grade 12 advanced functions and calculus and vectors

[uncut mulch]

depends on the book. some use log for ln

[gilded stirrup]

makes sense

but I’m pretty sure it’s base 10

[acoustic harbor]

yeah if it doesnt have a base

its 10

[tardy ridge]

feng is my father

[uncut mulch]

yeah if it doesnt have a base
it's 10
depends on context

[gilded stirrup]

feng is my father
@tardy ridge no

[smoky needle]

I know this is a dumb question, but will this have a vertical asymptote? I’m getting confused.

[willow bear]

yes, this thing has vertical asymptotes at ±1

[smoky needle]

@willow bear What happens if it was x = -1?

[willow bear]

what do you mean

[gilded stirrup]

@smoky needle divide all by x, then you have 1/x^2-1 solve x^2-1=0 to find VA and take the limit as x approaches infinity to get horizontal asymptote

[lilac pier]

@viscid thistle you can find the first factor by guessing

then others come from long division

[viscid thistle]

This is what I did

I’m confused

[fleet yew]

Integral coefficients?

Lmao thatd be fun

[lilac pier]

@viscid thistle yeah so try plugging in the integer values, see what factor u get

[stable pasture]

does vectors count as precalculus?

uh for some unknown reason my teacher says my answer to this problem is wrong (problem 4)

this is the figure

this is what I got, but he says it doesnt match with the answer in his solution book

can someone point out what I'm doing wrong or if it is right?
I just took $\vec{n} = \vec{AC} \times \vec{AB}$ and then found $\hat{n}$

[obsidian monolith]

PhysicsMonster:

[viscid thistle]

@lilac pier none of the possible rational roots equal zero when I plug them in

this is the question again for reference

[gilded stirrup]

If you’ve tried all of them they either have no integer roots or there are imaginary roots

[muted steeple]

Hi guys, can a quadratic have 3 solutions?

[pure dagger]

nuh

[muted steeple]

https://www.slader.com/textbook/9781305071759-precalculus-mathematics-for-calculus-7th-edition/57/exercises/109/

then how this person got it?

I can follow what he's doing at it seems correct.

In my own answer, i simply divide out the last factor.

he uses it to get -1

[pure dagger]

that ain't a quadratic equation

[muted steeple]

how is it not?

[pure dagger]

the (.)^(1/2) factor makes it non-quadratic

if you don't literally have a x^2 + b x + c, then it's not quadratic

[muted steeple]

with substitution it is.

[pure dagger]

what are you substituting

also, substitution of an equation can make it so that the new equation has a different amount of solutions

in any case; if you're just dividing out the (x + 1)^(1/2) factor, you're potentially dividing by zero, so you're making a mistake there

[muted steeple]

if you say $y = (x+1) $ then you have something along the lines of $4y^{1/2}-5y^{3/2}+y^{5/2}=0$ then some factoring and you get $y^{1/2}\left(y^2-5y+4\right)$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

oh

[pure dagger]

and that substituted term you create is still not a quadratic expression

it's a quadratic times a square root

[muted steeple]

oke, it makes sense now.

[pure dagger]

cool!

[muted steeple]

thanks.

[jade iris]

You invest an income stream of 𝑓(𝑡) = 𝑃𝑒0.02𝑡 dollars at a rate of 6% compounded continuously for 43 years into a retirement account. Find the value of 𝑃 needed for you to have 4 million dollars at retirement. Please round to the nearest dollar. How would I solve this and if this is in the wrong chat tell me!

[willow bear]

do you mean $f(t) = Pe^{0.02t}$?

[obsidian monolith]

Ann:

[jade iris]

yes with a rate of 6% compounded continuously for 43 years

[willow bear]

oh an income stream huh.

[smoky needle]

Hello would this function: f(x)= - | x-4 | +4 have a max or a min? Or neither?

[patent beacon]

If you know what |x| looks like, which is like a big V with minimum at (0,0)

You can get f(x) by taking |x| and

• Flipping over x axis
• Moving up 4 units
• Moving right 4 units

Should help visualize that there's a max at (4,4)

[smoky needle]

Ahh I see, so since we flip it over the X, it would be a downward V; therefore there’s a max?

[patent beacon]

Yaya

[jade iris]

@willow bear yep its been a while since i have done one

[smoky needle]

@patent beacon thank you!!

[patent beacon]

Np, feel free to ask if you have anything else!

[smoky needle]

@patent beacon I finished my quiz last night, but I went back to my work and see if I made sure I did them right. So far, I got those. But I made small mistakes on my other ones.😓

Thank you once again

[jade iris]

Can someone explain 8B?

[acoustic harbor]

set the equation to equal 0

[jade iris]

and solve for B? @acoustic harbor

[acoustic harbor]

wait why do you have Y and B

[jade iris]

i had y as the rate per month with was 1000 but i didnt know if that was relevent to use

[acoustic harbor]

im pretty sure everything is in terms of B

not y

just set the differential equation to equal to 0 and find what B values make it 0

[jade iris]

@acoustic harbor would that make it stable then if it is 16,666.67?

[blazing raven]

x^2 + y^2 + z^2 = 27 ... jk

[acoustic harbor]

?

[civic sierra]

Where x=y=z, taadaa

[acoustic harbor]

3

they dont have to be the same

[civic sierra]

I wasn't being serious :)

[gilded stirrup]

Is there a good website for practice mhf4u exams

The ones I found online are either just the exam no answers or are really shady

[viscid thistle]

Openstax.

Has lessons, a shit ton of problems and answers.

[gilded stirrup]

ty sm!

[viscid thistle]

You can start from there.

https://openstax.org/subjects/math

[gilded stirrup]

This is so much better than just searching for exams in google

ty

[tardy ridge]

@gilded stirrup is my father.

[fleet yew]

@tardy ridge prove it

[gilded stirrup]

@fleet yew hahahaha hes not

[viscid thistle]

Anyone know which one is the answer

been stuck on solving this for like 20 minutes

[patent beacon]

2T + 0C + 2F = 6
1T + 5C + 0F = 11
100T + 120C + 60F = 460

[viscid thistle]

?

[patent beacon]

First line is to get the correct fiber
Second line is to get the correct fat
Third is for calories

[tribal wharf]

process of elimination should also be pretty easy?

only the first and last answers come out to 6 grams of fiber for example

[viscid thistle]

im still stuck

how would process of elimiation work

[tribal wharf]

2 toast, 1 cottage cheese, 2 fruit is how many grams of fiber

2 toast * 2 grams per toast + 1 cheese * 0 grams per cheese + 2 fruit * 2 grams per fruit = 8 grams in total

since there are 6 grams of fiber in the meal, clearly the second answer is not correct

[viscid thistle]

so is the asnwer A?

@tribal wharf

[tribal wharf]

i mean yes

is it the way the chart's presented thats throwing you off?

seems pretty straightforward to me

[viscid thistle]

dam it makes sense

how did I not catch that

I was making matrix charts

but the asnwer is like right there

lmao

Thanks @tribal wharf

[gilded stirrup]

On khan academy they teach you about multiplying polar coordinates in polar form by transforming it into e^phi*i+another but isn’t it just possible to add the angles and multiply r by the other r to get the polar coordinate?

So it would be like z1=r1cis theta z2=r2cis phi and z1 x z2 would be r1*r2(cis(theta+phi)

Actually nvm I found it on YouTube

Khan academy does it the really long way

[muted steeple]

Hi guys, where am i going wrong solving this

$x^2\sqrt[]{x+3} = (x+3)^{3/2}$

So i rewrite this with fractional exponents

$x^2(x+3)^{1/2} = (x+3)^{3/2}$

I can see that i can get rid of these fractional exponents by squaring the both sides, so i get.

$x^4(x+3) = (x+3)^3$

Now if i divide the (x+3) from both sides i get

$x^4 = (x+3)^2$

If i expand, i get

$x^4 = x^2+6x+9$

Then moving everything to the left

$x^4 - x^2 - 6x - 9 = 0$

Now it seems to be in a standard form, so i group and factor

$x^2(x^2-1)-3(2x+3) = 0$ which doesn't make sense to me.

[obsidian monolith]

Publius:

[pale bison]

texed it for you

ok

so from

[muted steeple]

thanks, didn't know what happened there

[pale bison]

you can divide both sides by (x+3)^(1/2)

that way you don't have to deal with issue of factoring probably

[muted steeple]

oh, ofc. Though i'm still interested in where i went wrong.

surely, it should have factored either way?

[pale bison]

i don't see why it should factor

certainly not by grouping it seems

you can try the rational root test

[muted steeple]

what is that?

[pale bison]

in your case, it'll be $\pm 1, \pm3$ or $\pm9$

[obsidian monolith]

Publius:

[pale bison]

,w factor x^4 - x^2 - 6x - 9

[obsidian monolith]

Display results online and refine query

Requested by Publius#2314

[muted steeple]

not sure which part you are referring to.

-9/1?

that's what i get from the rational root test.

[pale bison]

factors of 9

[muted steeple]

1,9,3,3?

[pale bison]

$\pm\frac{\text{factors of 9}}{\text{factors of 1}}$

[obsidian monolith]

Publius:

[muted steeple]

oh

and what does that tell me sorry?

[pale bison]

so 1, -1, 3, -3, 9, and -9

it tells you that those are the possible rational roots

[muted steeple]

that it can be factored?

[pale bison]

so if it factors, it'll be one or more of these:
(x-1)
(x+1)
(x-3)
(x+3)
(x-9)
(x+9)

but like

dividing or rising both sides by something can lead to troubles

[muted steeple]

how come?

i know why dividing can

not sure why raising can

[pale bison]

if you divide by (x+3)^(1/2) as i suggested, you are assuming x != -3, meaning you might lost this root if it happens to be one

squaring both sides will get rid of the signs

[muted steeple]

oh

[pale bison]

so you need to break it into cases

[muted steeple]

what do you mean when you say break it into cases?

like you did above?

[pale bison]

assume x < 0, and square and see where that gets you
assume x > 0, do the same thing
x = 0, same thing again

but dividing doesn't give you that many cases

it only wants you to assume x != -3

[muted steeple]

i guess that assumption is unfounded

so i can't divide

[pale bison]

plus, dealing with fourth degree polynomials is generally bad unless it factors nicely

[muted steeple]

well, this is part of my quadratic equation section

[pale bison]

this is nothing really profound, in all cases you get solutions by running through all cases

and like here:

you generally don't want to multiply by something and then divide

[muted steeple]

i haven't been told any of this in the book 😄

i thought it's a standard way to manipulate an identity

regardless whether it's quadratic or linear

[pale bison]

once you get used to dealing with polynomials you'll have some idea on how to find roots

or quadratic rather, anything else is annoying unless it is made purposely good

[muted steeple]

okay, so let's say i want to redo this again properly.

assuming i can't divide

or whatever

then i need to cover test cases x < 0, x > 0 and x = 0

so i have three equations like this?

for x = 0 $\sqrt[]{3}=\sqrt[]{0^3}$?

[pale bison]

no, that's the worst way you can go about it

less case is better

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

okay

then what's the best way to go about it

[muted steeple]

i'm just curious

[pale bison]

$x^2(x+3)^{1/2}-(x+3)^{3/2}=0$

[obsidian monolith]

Publius:

[pale bison]

$(x+3)^{1/2}(x^2-(x+3))=0$

[obsidian monolith]

Publius:

[muted steeple]

okay, this looks like what i've done before.

[pale bison]

$\sqrt{x+3}(x^2-x-3)=0$

[obsidian monolith]

Publius:

[muted steeple]

not sure why i didn't realise this ...

[pale bison]

now go from here

the way i recognized this is because i have something with (x+3)

[muted steeple]

then, i can use the zero product rule to work out the two products right?

wait, that middle term is troublesome

in the parenthesis

[pale bison]

idk what that rule means

[muted steeple]

if you have products ab=0 then a=0 or b=0

[pale bison]

but either $\sqrt{x+3}=0$ or $x^2-x-3=0$

[obsidian monolith]

Publius:

[pale bison]

yes precisely

it's a quadratic so nothing bad really

compared to the degree 4 polynomial you managed to cook up

[muted steeple]

yes 😄

I was hoping that it would factor nicely

okay, but how would i find $x^2-x-3=0?$

[obsidian monolith]

꧁༺Vocal༻꧂:

[muted steeple]

that x would get in the way no?

[pale bison]

the quadratic formula since you can't factor it

[muted steeple]

oh

jeez

i'm really not on point.

thank you for your time.

really informative.

[pale bison]

np

[gilded stirrup]

could someone explain the complex roots of z^3=-512

the khan academy video doesn’t help

like why is there a random variable k*theta

[fleet yew]

@gilded stirrup what is the real third root of -512

[viscid thistle]

like why is there a random variable k*theta
@gilded stirrup k usually is the integer

[gilded stirrup]

It’s -8

Ohh

[fleet yew]

@gilded stirrup -8 times the third roots of unity

[gilded stirrup]

third roots of unity?

[fleet yew]

@gilded stirrup third roots of 1

Unity=1

e^0, e^i2pi/3, e^i4pi/3

[gilded stirrup]

OOOOOOOOHHHHHHHHH

IT ALL MAKES SENSE NOW

OK THAT WAS THE LAST THING I NEEDED FOR COMPLEX NUMBERS TY

[proud jetty]

https://gyazo.com/2c65b49457597d9c527418f7e07f6f11.png

how do u get 1.14?

[sweet wyvern]

cos^-1(.42) is approximately equal to 1.14

[proud jetty]

hmmm

nvm ic thx

[sweet wyvern]

yw

[next kernel]

Her guys

How to write this in rectangular coordinates

r = 5/(1+cos(theta))

[sweet wyvern]

x = r cos(theta), y = r sin(theta)

[next kernel]

Mmk so multiplty r

Get r² = 5r/(r+x)

Does it become 5 + 5r/x

[sweet wyvern]

x = ( 5/(1+cos(theta)) cos(theta) --> 5 cos(theta) /1 + cos(theta)

[next kernel]

Ηow did you get all that?

Step by step

[sweet wyvern]

so we know r = 5/(1+cos(theta) right? We use the formula r cos(theta) = x but we plug in 5/(1+cos(theta) into r cos(theta) = x instead of r. We are substituting the value r with 5/(1+cos(theta) instead

hence why I wrote x = ( 5/(1+cos(theta)) cos(theta)

[next kernel]

Is that legal? My brain hurts

I have no idea but my assumption is that the answer is 1.

[patent beacon]

The multiplicative identity, by definition

Which is a fancy way of saying the identity matrix

No?

Note that this is the way we define an inverse matrix. If ' means inverse,
AA' = I

[next kernel]

@sweet wyvern

That gives us x = 5/costheta

What would we do with that information?

[sweet wyvern]

you simplified incorrectly

(5 cos(θ))/(cos(θ) + 1) = x

[next kernel]

That makes no sense because you end up with x=(5costheta)/(costheta+cos²theta)

[sweet wyvern]

no you only multiply the numerator by cos(theta)

if you multiplied both the numerator and denominator by cos(theta) then you'd effectively be multiplying the radius by one

[next kernel]

you mean you multiply the fraction by cos(theta)/one

[sweet wyvern]

yes

[next kernel]

And then?

[sweet wyvern]

then you get what I showed you earlier: the x-coordinate of the rectangular coordinates

so now you need to solve for y

and the equation you use for conversion to y is y=r sin(theta)

[next kernel]

I would get y= 5sin(theta)/(cos(theta) + 1)

[sweet wyvern]

yes

[next kernel]

How does that...work out for the answer?

[sweet wyvern]

((5 cos(θ))/(cos(θ) + 1), (5sin(theta)/(cos(theta) + 1)) should be your rectangular coordinates

[next kernel]

Except those aren’t coordinates on a point.

Coordinates on a point would be like 2 +/- √5

[sweet wyvern]

do you have a specific value for theta?

[next kernel]

No

The answer simply takes the sides of the equation and trades parts to rewrite it using √ (x² + y² ). I think you took the wrong method for procedure.

[sweet wyvern]

then I don't think you can do much else

oh

[next kernel]

The professor made a mistake when asking for coordinates, because those were not included in the answer.

She meant to say conversion.

This is all a mess.

[sweet wyvern]

yeah sorry man

[fleet yew]

@sweet wyvern are you applying for mod

[sweet wyvern]

no it was just a joke

[fleet yew]

I think you should lol

[viscid thistle]

i cant seem to wrap my head around how to solve this

any tips?

[fleet yew]

Uh

Just find cot(45°)

Lol

[viscid thistle]

thanks ive mastered it!

[fleet yew]

Really?

[gilded stirrup]

lol

[fleet yew]

What did you get?

This is a hard problem don't worry too much if you get it wrong

[viscid thistle]

i keep trying to convert cot into sin/cos and squaring the unit circle values

but that doesnt seem to be the way to solve it

so im not to sure what to try now

[fleet yew]

first of all

Define cotangent

[viscid thistle]

1/tan

[fleet yew]

Ok

Whats tangent

[viscid thistle]

sin/cos

[fleet yew]

So whats 1/(sin/cos)

[viscid thistle]

cos/sin

[fleet yew]

Yeah you said that cot=sin/cos lol

Probably just a typo but you should know

[viscid thistle]

oh i meant i was converting cotangent into terms of sin and cos!

i see the confusion now

[acoustic harbor]

amd i got a question in calculus channel

real easy

[viscid thistle]

oh

i solved it

nice

[gilded stirrup]

sun so cot= cos/sin and use the trig ratios

Ok nice

[viscid thistle]

it seems i was on the right path

[violet nest]

Hello

I'm having trouble converting a logarithmic equation to exponentials

So this is my graph and equation pulled up from Excel

But this is apparently what my friend is getting

Our values are based on the same table, and I need the equation in the second picture's form to find my K values

Can anyone help me convert my top equation to the second?

[sharp marsh]

Find all solutions of the equation in the interval [0, 2π). (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
4 sec^2 x + 2 tan^2 x − 6 = 0

Help?

[acoustic harbor]

use a trig id

change sec^2x to 1-tan^2x

or change tan^2x

and then factor

@sharp marsh

[sharp marsh]

Oh it's

The X is seperated from the 2

@acoustic harbor

Also is it not pi/6 and 5pi/6?

[acoustic harbor]

im not sure what you mean?

[sharp marsh]

Um it's not sec^2x but sec^2 (x)

[pseudo lance]

hello

when looking for a hole, should u do numerator =0 and denominator = 0 OR should u find an x value for which the graph doesnt exist

for example, in y=ln(x^2)/ln(x^2)+2. If u do numerator and denominator = 0 then u end up without a hole, but if you use x=0 then u do have a hole

this equation is confusing me

[fleet yew]

@pseudo lance holes are common divisors (roots) of the num and denom

[pseudo lance]

whats a divisor?

oh nvm

how come x=0 is a hole then? since its not a common divisor

[acoustic harbor]

a hole occurs when the limit is 0/0

[next kernel]

I wait.

[acoustic harbor]

i forgot everything about ellipses

[viscid thistle]

Ok what the fuck

How is this even done

[acoustic harbor]

1-1

[willow bear]

$1 - (1 - 2 \sin^2(x)) \neq 1 + 2 \sin^2(x)$

[obsidian monolith]

Ann:

[acoustic harbor]

is 0

[viscid thistle]

Oh wait its -1 + 2sin^2 x

[willow bear]

$1 - (1 - 2 \sin^2(x)) \neq -1 + 2 \sin^2(x)$ either

[obsidian monolith]

Ann:

[acoustic harbor]

what?

[viscid thistle]

Welp I’m in danger

[acoustic harbor]

do you know what 1-1 is

1-1 is not 1

[viscid thistle]

Yes it’s 0 but

[acoustic harbor]

Make it 0

[viscid thistle]

Wait it’s not distribution multiplacston oooooo

[acoustic harbor]

its straight forward from there

[willow bear]

you say $1 - 1 = 0$ yet simplified $1 - 1 + 2\sin^2(x)$ as something other than $2\sin^2(x)$ two times now

[obsidian monolith]

Ann:

[willow bear]

curious indeed

[viscid thistle]

Ok does that look a lil better

[acoustic harbor]

no

what do you not get

1-1 is 0

[viscid thistle]

I’m sorry I’m brain dead

There

[acoustic harbor]

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

which is equal to 2sin^2x

[willow bear]

$1 - (1-A) = 1 - 1 + A = (1 - 1) + A = 0 + A = A$

[obsidian monolith]

Ann:

[willow bear]

[acoustic harbor]

im a little worried

[viscid thistle]

My brain has just shut off since the quarantine started

Im so not ready for ap Calc

[acoustic harbor]

dont really use trig ids in ap calc

[viscid thistle]

Good I hate this stuff

But uh what do I do now

[acoustic harbor]

isnt it straight forward?

you have 2sin^2x/2sinxcosx

start canceling out stuff

[viscid thistle]

Oh huh so that would leave sinx/cosx

[acoustic harbor]

yes

[viscid thistle]

Now how exactly do this one work

[willow bear]

1/sec^2(x) = cos^2(x)

and tan(x) = sin(x)/cos(x)

so......

[viscid thistle]

So there’s this

[acoustic harbor]

1/cos^2x

not cos^2x

[willow bear]

**1/**sec^2(x) = cos^2(x)

[viscid thistle]

Would I cancel out the denominator this way?

No no

Never mind that’s not right

[acoustic harbor]

no

just multiply the reciprocal

[viscid thistle]

So the denominator cancels

While the numerator

Uh this?

[acoustic harbor]

yes

where'd ur 2 go

[viscid thistle]

Forgot to put it in front of sin

Or wait

Does it go between the sin and x while the cos numerator and denominator cancel out

[uncut mulch]

wdym

[viscid thistle]

I don’t know anymore

[uncut mulch]

currently you should have: $\frac{2\sin(x)\cos^2(x)}{\cos(x)}$

[obsidian monolith]

ramonov:

[uncut mulch]

cancelling a factor of $\cos(x)$ will leave you with: $2\sin(x)\cos(x)$

[obsidian monolith]

ramonov:

[jaunty mason]

I have a function that transforms a polar coordinate: $f(r, \theta) = (a sin \frac{r}{a}, \theta)$

[obsidian monolith]

Tāhā:

[jaunty mason]

What is the function that has the same effect, but in Cartesian coordinates?

[smoky needle]

Hello, is someone available to help me on a question?

[jaunty mason]

don't ask to ask!

[willow bear]

a bunch of people, probably

just post your question

[jaunty mason]

,$ f(x, y) = \left(a sin \left(\frac{\sqrt{x^2+y^2}}{a}\right)\cdot\frac{1}{\sqrt{1+{(y/x)}^2}},a sin \left(\frac{\sqrt{x^2+y^2}}{a}\right)\cdot\frac{y/x}{\sqrt{1+{(y/x)}^2}})\right)

[obsidian monolith]

Tāhā:

[jaunty mason]

this is what i tried, please tell me there's a simpler version for it

oof, doesn't fit in the image

Well the x-component is

,$ a sin \left(\frac{\sqrt{x^2+y^2}}{a}\right)\cdot\frac{1}{\sqrt{1+{(y/x)}^2}}

[obsidian monolith]

Tāhā:

[jaunty mason]

and the y-component is

,$ a sin \left(\frac{\sqrt{x^2+y^2}}{a}\right)\cdot\frac{y/x}{\sqrt{1+{(y/x)}^2}}

[obsidian monolith]

Tāhā:

[smoky needle]

I am having trouble with #7. I don’t quite understand what it is asking me.

[willow bear]

it's asking you for the end behavior of f

as x -> +∞ and x -> -∞ respectively

[jaunty mason]

aaah can someone help me with my polar conundrum

[smoky needle]

@willow bear so will I just have to put any real numbers in for x to see if it is a neg or positive?

[jaunty mason]

nop

you're supposed to know that the term with highest degree will grow the greatest

and recognize that it has a negative coefficient in this case

[smoky needle]

So if it a positive value for that, would negative be the answer since -x^7 has the highest power?

[jaunty mason]

correct

maybe, i'll have more luck in questions?

[smoky needle]

@jaunty mason thank you, I appreciate it!

[jaunty mason]

np 😄

[fierce slate]

could someone help me with solving this? https://prnt.sc/sarabi

i dont understand how to find the equation without a given directrix

[sweet wyvern]

instead try finding the distance b/w the vertex of the parabola and its focus

call this distance p

the general formula that we want to use to determine our parabola (when it opens or down and we are given the focus or directrix) is x^2 = 4py

[viscid thistle]

you could also use process of elimination given estimates of points... say, it can't be the first one, because one of the points is (1,1) and that doesn't appear remotely true on the graph. this isn't reliable in the long term, but it works for tricky questions you don't know the traditional process of.

[sweet wyvern]

yeah lol, you can also instantly rule out the second option b/c of the y^2 (this applies when the parabola opens to the right or to the left)

[viscid thistle]

indeed.

the only remaining options are the third and last, and we can rule out one of them with a point estimation.

[leaden stratus]

Hello. Today we tried to solve this, but even our teacher was unsure about the steps. How would you solve this? Thanks.

The result is pi/4 + kpi.

[barren stream]

can someone assist me with this question please?

[willow bear]

• how many marbles are there in total?
• how many fit the description "the marble is green or has a number > 3"?

[barren stream]

alright there are 10 marbles in total

6 marbles are green, and 4 of them are larger than 3

@willow bear im not really sure what to do next.

or even if im correct so far.

[willow bear]

you have 10 marbles:
O1, O2, O3, O4, G1, G2, G3, G4, G5, G6

[barren stream]

yes.

[willow bear]

O for orange, G for green

how many fit the description "the marble is green or has a number > 3"?

[barren stream]

7

sigh.

thank you

i appreciate it a lot.

[muted steeple]

how do i go about solving c?

[rare pagoda]

you have a parabola opening up downwards, what do you know about where its absolute maximum will be?

[muted steeple]

"opening up downwards" ?

[rare pagoda]

aka it's growing towards -infinity

hint:find the vertex

[muted steeple]

i know nothing of that sort at this stage. I'm doing a chapter on quadratic equations.

[rare pagoda]

okay you know the equation y = x^2 right?

[muted steeple]

that's a circle?

oh

nvm

ye it goes U

[rare pagoda]

and -x^2 would make an upside down U

[muted steeple]

yes.

[rare pagoda]

that's what I mean by growing up/down

[muted steeple]

so i assume that negative -16x^2 is the down

[rare pagoda]

yep

[muted steeple]

oke

[rare pagoda]

and when it is growing down, you know there is going to be a global max

aka the vertex

[muted steeple]

yes, so i need to find the global max

but i have not learnt that, i know how to find roots etc.

wait, can i set the roots somehow to global max?

[rare pagoda]

if you get the roots, then the t value for the vertex would be the average of those 2 roots

but you can use the formula $t = \frac{-b}{2a}$

[obsidian monolith]

magnusChadson:

[rare pagoda]

make sure to plug the t value that you get back into h(t)

[muted steeple]

oooh i think i can see it.

so the arithmetic average of the two roots would give me the time for the global max right?

wait, but i need to find the height.

ahh...

[rare pagoda]

kinda weird that problem c is find the height while problem d is find the time

since you need time to find height

[muted steeple]

https://cdn.discordapp.com/attachments/363224154469826562/706907252249854484/195274489901219849.png this resembles the square completion formula

a and b are the coefficients for the quadratics?

[rare pagoda]

yeah

the formula I gave you tells you where the vertex of a parabola is

im surprised they haven't taught you that

[muted steeple]

they have not, i have a feeling there's another way of doing this.

i'm certain they wouldn't give me this problem if they didn't expect me to solve with with tools i've learnt so far.