precalculus

[pale bison]

f(0) = 2x + 1
f(0) = y
f(1) = 3x
f(1) = y + 2
from this, you obtain
2x + 1 = y
3x = y + 2 => y = 3x - 2
this means
2x + 1 = 3x - 2 => x = 3
this means (from the equation 2x + 1 = y), y = 7

[full garden]

❤️ thank u so much

❤️ if it wasn't for you I would've not been able to solve it thank u so much

[pale bison]

ah fuck i made a mistake

[full garden]

oh

[pale bison]

just the last two lines

sorry

[full garden]

dont say sorry if it wasn't for u i would've not been able to solve that question thank u so much bro ❤️

[pale bison]

lol np

[full garden]

hey guys i have a question please

so it it correct if i do this ?

and then factor everything out?

can someone please tell me if thats the correct way of starting

[sour eagle]

The discriminant shouldn't include x

Only the coefficients

[full garden]

oh

[sour eagle]

so it would be (k^2+1)^2-4*k^2

[full garden]

so only (k^2+1)^2

[sour eagle]

Yes

[full garden]

okay thank u so much bro

thank u ❤️

[sour eagle]

Np 👍🏿

😄

[full garden]

guys how do I solve for this quadratic

i tried using like quadratic formula

and didn't get that recperical thing

[hasty magnet]

are u asking for a or b @full garden

[full garden]

b sir

thank u for asking

[hasty magnet]

what did u get for x

after doing the quadratic formula

[full garden]

this

[hasty magnet]

u sohuld get x=-k and x=-1/k

what is that

[full garden]

the simplest form

so like from the above question we got the dicrimnant

[hasty magnet]

k is a constant

[full garden]

that's how I set it up

how can i fix it 😅

[hasty magnet]

idk how u are getting those things

what are you doing?

[full garden]

like its impossible to get x = -k

yes bro I am jus trying to solve it using the quadratic formula

[hasty magnet]

ok first can u expand the equation

and tell me what you got

then we can apply the quadratic formula

[full garden]

oh

i thought expanding was going to make it wrong

[hasty magnet]

nah u have to expand it

[full garden]

because b is like (k^2+1)

oh

[hasty magnet]

before u apply it

[full garden]

okay

[hasty magnet]

ok good

[full garden]

thank u

[hasty magnet]

can u combine any like terms

[full garden]

yes

[hasty magnet]

combine em

[full garden]

now apply the quadratif formula bro?

quadratic*

[hasty magnet]

...

that's not combining like terms

do u seem anything u can combine

from the other thing u sent

[full garden]

i dont think i can combine them bro

I can just factor out the common x

and Ig use it for the b

[hasty magnet]

not factor

combine

oh wait

ok

i thought u could combine like terms but i was wrong

u actually have to take the quadratic formula of it lol

[full garden]

okay

[hasty magnet]

ok

idk im getting -1 @full garden

[full garden]

thank u @hasty magnet ❤️

u helped me a lot thank u so much ❤️

[hasty magnet]

i did?

@full garden

i got it if u need

did u get it?

if u still need help here's how u do it https://mathsolver.microsoft.com/en/solve-problem/k { x }^{ 2 } %2B `left( { k }^{ 2 } %2B1 `right) x%2Bk %3D 0

i kept doing it wrong because i forgot u needed absolute values

😅

[full garden]

@hasty magnet thank u so much ❤️ thank u ofc u did even if u didn't help me u still tried which means a lot to me thanks buddy

[rapid tangle]

How do i evaluate the fraction in problem 7 to apply limits

[lilac pier]

multiply numerator and denominator by x^3

@rapid tangle

[rapid tangle]

@lilac pier I tried that earlier put how do I apply limits to infinity on that

[lilac pier]

you get (x^6 - 6x^5 + 11x^4 - 6x^3) / - 6

what'd i do wrong

[stuck lark]

denom

[lilac pier]

numerator approaches +infinity

oh

LOL yeah

actually for the infinity part, write the denominator as 1/x^3 - 6/x^2 + 11/x - 6

so as x approaches infinity, the denominator approaches -6

the numerator approaches infinity since x^3 has the highest power and is positive

so it's something of the form infinity / -6 which is just -infinity

@rapid tangle

[rapid tangle]

Yes, the answer's correct

Thanks
Just got wound up at the algebra part cuz I'm relatively knew to this stuff

[tender bolt]

if i am given g(x) = 2x+1 f(g(x))= 3x+7 , how can i find f(x)?

[harsh smelt]

plug g(x) in

[tender bolt]

where

[harsh smelt]

if f(x)

what is stopping you from having g(x) as variable for f

[tender bolt]

but what can i do with that>

i dont know f(x)

i know g(x) and i know f(g(x))

[past meadow]

i think commander means sub 2x+1 for g(x) in f(g(x)) so that you have f(2x+1)=3x+7

[viscid thistle]

Zeus wants to know f(x) given g(x) and f(g(x))

What changed between g(x) and f(g(x))? @tender bolt

[tender bolt]

what do you mean

f(g(x)) means that if we put g(x)

inside of f(x) we get 3x+7

[viscid thistle]

$g(x)=2x+1$
$f(g(x))=3x+7$

[tender bolt]

that is f(2x+1) = 3x+7

but how can i solve that

[viscid thistle]

Ok so

What happened

The coefficient increased by 1

And the +1 turned to a +7

[tender bolt]

yes

oh i see

what yo uasked

but how does that help me

[viscid thistle]

Well the +1 in g(x) turned into a +7

So f(x) has a +6 somewhere

Follow?

[tender bolt]

so f(x)= kx + 6

[viscid thistle]

Yes

If im correct

So try to think of g(x) without the +1 now

[tender bolt]

so there is not really any method

[viscid thistle]

Wait no

[tender bolt]

it is just trying to find it

[viscid thistle]

Hold on

[uncut mulch]

let u = 2x + 1

do some rearranging and subs

[viscid thistle]

Cant you do something with inverse

[past meadow]

you're overthinking it. perform the substitution ramonov mentioned

[viscid thistle]

Im pretty sure theres a legit method with inverse

[tender bolt]

i just thought it in my had and i got f(x)= 3/2x + 11/2

[past meadow]

thats the correct answer zeus

[viscid thistle]

^

[tender bolt]

oh so i can do 2x+1= u

[past meadow]

yes

that would be the best method here

[tender bolt]

f(u) = 3x+7

y=3x+7

[viscid thistle]

Eeee

[uncut mulch]

isolate x to get it in terms of u

[viscid thistle]

^

If you do $f(g^{-1}(x))$

[obsidian monolith]

AASlavoj:

[viscid thistle]

Shouldnt that be f(x)

[tender bolt]

i did it like this f(u) , where u=2x+1 f(u)=3x+7

u= (x-7)/3

f(x)=2*u+1

[uncut mulch]

uh no

[tender bolt]

f(x)= 2*(x-7)/3 +1

[uncut mulch]

u=2x+1
isolate x from that equation

[tender bolt]

i did that

obv

thats how i got (x-7)/3

i any case, i got what i needed, i checked also

thanks mans

man*

[uncut mulch]

wait

[tender bolt]

@uncut mulch its correcti checked it

thanks

[uncut mulch]

that notation looks messed up

[viscid thistle]

How can u be two different things

[uncut mulch]

^

[viscid thistle]

Whatever

Lol

[uncut mulch]

extremely hard to follow what's actually being done

[harsh smelt]

How can u be two different things

[viscid thistle]

Animal abuse inside that box

[uncut mulch]

who the fk is Schrodinger?

[harsh smelt]

it is Schrödinger minimized

[sour eagle]

ö looks like 😮

[tender bolt]

wtf

[sour eagle]

[hexed ermine]

It's the pogchamp

[warm siren]

lol

[rapid tangle]

How do i solve this

[serene heath]

what have u tried

[rapid tangle]

I don't know how to expand xⁿ

The formula wasn't given

Like i solved the prev one where it was a^x

[serene heath]

you could make a sub here

u=x^1/2

or multiply by (x-1)/(x-1)

[rapid tangle]

Ok imma try, thanks

Hey..uhh so still not getting it
Sorry, if it's basic stuff but I'm really new to this 😅@serene heath

I could solve it using L'Hopital rule but I wanna solve it algebraically cuz they didn't teach that rule in this book

[serene heath]

what did you get after subbing

[rapid tangle]

😶oops sorry i took the wrong limit
Thanks for your help, got it

This is correct right?

[viscid thistle]

Yup

Big brain

My man

I miss u

@rapid tangle good job

Me too lol

How u been

#chill :)

[silver matrix]

I don't know how to do it.

Any help would be appreciated.

[hasty magnet]

the increments are by 7 @silver matrix

so one line is counted as 7 numbers until that line

it's asking what would y be if x is 2

[sour eagle]

x-intercept = intersection between the graph of the function and x axis
y-intercept = intersection between the graph of the function and y axis
so f(14) = 0
f(0) = 28
You can get the equation of the "line" between points (0,28) and (14,0) by calculating the slope
m = -28/14 = -2
y = -2x+28
So f(2) = -2*2+28 = 24

@silver matrix

[hasty magnet]

bruh ur not suppose to give them the answer @sour eagle

[silver matrix]

Why not?

[hasty magnet]

that's not how you teach

thats y

[sour eagle]

😔

[hasty magnet]

we're here to guide you, not do it for you

[silver matrix]

Now, here the answer is A. right?

Thanks to his explanation I know how to mechanize this exercises

[sour eagle]

Yes

[full garden]

hey guys I just have a very silly question

for this one I just plugged 1/x in f

and I got this

but I feel like it's wrong to do that because the question can't be that simple 😅

can someone please just quickly tell me if that's correct please

[smoky pagoda]

im not familiar with the notation but if Im assuming correctly, id say it's -2*(1/(-3(x-4)))-5

[full garden]

oh

[smoky pagoda]

or when simplifying: 2/(3x-4)-5

cross check with someone else though

it's the way I think the notation is being used

[full garden]

okay thanks buddy I'll check thank u

[smoky pagoda]

i havent actually seen this before

[full garden]

sameee

[smoky pagoda]

but the way I understand it what's in the square brackets becomes the new input for f

[full garden]

i thought we just had to sub 1/x into f

oh

yes i think ur answer is correct bro thank u

[rapid tangle]

Where did I go wrong?

[signal herald]

@rapid tangle 5th line

You factor out an x, but take out the negative sign as well

That’s why you end up with -1/2 instead of 1/2

[rapid tangle]

@signal herald ah yes, a rather silly mistake
Thanks tho

[signal herald]

@rapid tangle Where do you go to school? It's interesting that you're learning limits in precalculus, especially using series expansions

[drowsy karma]

So I'm looking at the answer key of my homework but I don't understand where that 3 came from in the final portion of the question.

The 3 that came into 3y

[harsh smelt]

he moved everything to right

[drowsy karma]

wym

[uncut mulch]

multiplied both sides of the equation by 3 to get rid of the fraction

[drowsy karma]

Ahh

thanks

[rapid tangle]

@signal herald heyy so m currently in 10th grade moving to 11th
I'm really just trynna teach myself calculus
Rn I'm following this book called Differential Calculus for Beginners by Joseph Edwards

[drowsy karma]

how does how does it go from that to that?

Idk if its flying over my head or??

[harsh smelt]

well if a/b = -1

then a obviously is -b

[uncut mulch]

is that a 9, a or q

[drowsy karma]

Its a q

Im just gonna question my teacher tmr

[harsh smelt]

-q-3=-4 btw

q+3=4

q = 1

[uncut mulch]

multiplied both sides by 12

[harsh smelt]

look
$\frac{(-q-3)}{6} \frac{3}{2} = -1$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

$\frac{3(-q-3)}{12}= -1$

[obsidian monolith]

Commander Vimes:

[uncut mulch]

though it would be more efficient to multiply by the lcm of 2,6 (which was 6)

[harsh smelt]

$\frac{3(-q-3)}{-1}= 12$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

which is the same as $\frac{3(-q-3)}{1}= -12$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

do you understand why?

"the cosine of an angle in the unit circle just equals cosine, "

hm

indeed!

well your idea is correct

look

sin is opposite leg divided by hypothenuse

(for right triangle)

and cos is near leg by hypothenuse

so in case of unit circle we have hypothenuse 1

since it is radius

and because of that

x and y components are cos and sin respectively

we define unit circle in terms of circle btw

x^2+y^2=1

[viscid thistle]

hey guys! How do i find the two fundamental solutions for cos(5x+3x)=0.03?

i found one, but not sure how to find the other

[harsh smelt]

hey guys! How do i find the two fundamental solutions for cos(5x+3x)=0.03?
@viscid thistle cos(-x)=cosx

[viscid thistle]

@harsh smelt i appreciate that tip, but im not quite sure how to apply it

i know that my first answer is in QI and my second should be in Q4

[harsh smelt]

suppose cos(10)=0.3

[viscid thistle]

is that where thats heading?

[harsh smelt]

then cos(-10)=0.3

[viscid thistle]

Im following that

cos is an even function, i agree

[harsh smelt]

suppose you found solution which is A

then because cos is even

-A is also solution

[viscid thistle]

what if i wanted the solution to only by between 0 and 360

[harsh smelt]

then transorm -A

-90 = 270

360-A btw

[viscid thistle]

i tried that as well!

[harsh smelt]

A and 360-A will give the same cos

,w cos(8x)=0.03

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[harsh smelt]

basically what i said

[viscid thistle]

how did they come up with 33.965?

Sorry if i'm not understanding by the way! im trying my best!

[drowsy karma]

@harsh smelt ya i think i get it

[lilac pier]

@viscid thistle You have cos(8x) = 0.03, that means cos(8x) is positive

For simplicity, let 8x = alpha. So you have cos(alpha) = 0.03, which means cos(alpha) is positive. You should know cos is positive in the first and fourth quadrants.

For the first quadrant, you get alpha = 88.28. For the fourth quadrant, it's 360 - 88.28. So you have alpha = 88.28, 271.72. Now use alpha = 8x.

[fleet yew]

@lilac pier dont just give out answers

[viscid thistle]

^

that way he won't learn it nor reason through it

[willow bear]

no, the domain of tan is not $\mathbb{R} - { n \pi/2 \mid n \in \mathbb{N} }$.

[obsidian monolith]

Ann:

[silver matrix]

Ok..

and what would be it?

[willow bear]

$\mathbb{R} - { (2n+1)\pi/2 \mid n \in \mathbb{Z} }$

[obsidian monolith]

Ann:

[silver matrix]

Uf.. I do not understand that 2n + 1

Indeed, I do not understand all this mathematical demonstrations or whatever is called when they add that + 1

[willow bear]

it can be rewritten as $R - \curly{n\pi + \frac\pi2 \mid n \in \bZ}$

[silver matrix]

it's annoying for me

[obsidian monolith]

Ann:

[willow bear]

if you wish

does that make it clearer

[silver matrix]

Yeah

Thanks

Now I understand why they add that 1

it's a better way the first way but I needed to know where it came from

[silver matrix]

Is this true? At first I thought it was, but then I said that the domain could be (in interval notation) (-\infty, 2] U [3, \infty)

As the x can have all that values except the ones between 2 and 3

This ones included

But then he says that the domain has to be the intersection between the roots.

What do you guys think?

[viscid thistle]

Yes

The domain of a function is all the values that can produce a result passing through the formula

sqrt x-3 is only defined in R when x>=3

[willow bear]

sqrt**(x-3)**

[silver matrix]

So?

What is the solution?

There's no solution then?

He's right?

[viscid thistle]

and sqrt 2-x only when x<=2

[willow bear]

sqrt**(2-x)**

[viscid thistle]

so the function as a whole has no real domain

[willow bear]

if you write \verb|sqrt x - 3| then it is not clear whether you are talking about $\sqrt{x}-3$ or about $\sqrt{x-3}$

[obsidian monolith]

Ann:

[viscid thistle]

indeed, sorry

[silver matrix]

What is hte difference between that example and this one?

I mean the values don't have an intersection either

But here there's a solution

In the other case, it wasn't

Why

This is not solved algebraically, I HATE this explanations

[viscid thistle]

the square root doesn't exist if what is being squared is negative

[willow bear]

in either case, you don't want negative numbers under your square roots

[viscid thistle]

so in essence, what you're looking to answer is

when is (x+4)(x-3) negative

[willow bear]

^

and then, once you figured that out, take everything BUT those points

[silver matrix]

Is a circle an even, odd or neither function?

The professor says that techinically a circle is not a function

[tawny nacelle]

yes it isnt

[harsh smelt]

it is.

it takes pair (x,y) and maps it to point (x,y) what's the problem

[patent beacon]

It's not a function R → R, which is usually where we talk about odd/even

[silver matrix]

Could you tell me, even so, if it's an even odd or neither function? It's symmetrical both on the y-axis and on the origin

Oh ok

so it's anything then right?

If it's not a function

then can be neither, nor odd nor even I guess

[serene heath]

the question is nonsense

[tawny nacelle]

its a relation on the set R

[serene heath]

since its not a function

[silver matrix]

https://www.youtube.com/watch?v=fKyBOLsqRlo&list=PL0o_zxa4K1BXUHcQIvKx0Y5KdWIw18suz&index=10
Min: 17:34

I don't like how he explain things, but 🤷‍♂️ this guy makes ace exams

He taughts you how to mechanize things

So

Left as an exercise I guess. The thing is that I've heard about R but I don't even know what that is

They are like planes? Dimensions? Real numbers?

No idea

[serene heath]

the set of all real numbers

[silver matrix]

Ok

I've seen sets in a very basic way

so

So R1, R2 and R3 are then differente thins right?

[willow bear]

yes
R^1, or just R, is the number line

R^2 is the x-y plane

R^3 is three-dimensional space in the everyday sense of the word

[silver matrix]

Can there be R\infty?

[harsh smelt]

Hilbert space

[silver matrix]

or R0?

[harsh smelt]

or R0?
zero vector?

[willow bear]

no to both of these

[silver matrix]

yeah i guess so

oh

[willow bear]

R^∞ could be understood as the set of all infinite sequences of real numbers

tho it never really comes up bc there's not much you could do with it

R^0 is just... well, you could say it's just {0} but i don't see a use for that except maybe to plug a hole in some definition

[silver matrix]

So if there are higher dimensions than 11, why the physicists try to invent a theory that explain all in the world with x dimensions?

[willow bear]

i.e. a technicality that is almost surely never comes up in practice

uh

...what

[silver matrix]

yeah

[willow bear]

what are you even talking about

[silver matrix]

how can you explain more than 11 dimensions

[willow bear]

this went metaphysical really quickly

...

...

[silver matrix]

No, it would be a non-technical discussion about what people are studying today

[pale kettle]

You can just work with these things algebraically

[harsh smelt]

dimension in math != dimension physically

[willow bear]

i am not sure what kind of answer you want to hear

because like

[pale kettle]

You don't need to be able to visualize them to be able to work with them

[harsh smelt]

dimension in math is just like roughly standard lenght of an array

[willow bear]

the space of all possible configurations of a system is a meaningful thing physicists talk about

and the dimension of THAT can go as high as you want with complex enough systems

[silver matrix]

I'm just saying that I do not understand why physics try to say that the world is x dimensions when in math you see higher dimensions than that

[willow bear]

why does it matter what you see in math

[pale kettle]

Because the world isn't that high of a dimension?

[willow bear]

what you see in math need not have direct relation to anything in the real world

or like

[pale kettle]

Just because we can think of R^(10000) doesn't mean the world can have that many dimensions?

[silver matrix]

so math is apart from the world/universe?

[willow bear]

direct relation in specifically the way you're thinking of

[pale kettle]

I mean, there are things in the world that have high dimension in some sense

[silver matrix]

that explodes my mind tbh

[pale kettle]

Like data sets can have high dimension

[silver matrix]

and doesn't makes sense in some way

[pale kettle]

This isn't the standard "physical dimension of space"

[silver matrix]

then you are not explainging all

[willow bear]

because again, R^n for high n is the natural setting for modeling complex systems with many components and many degrees of freedom

[pale kettle]

I can have a dataset that takes values in R^(10000)

[patent beacon]

There's different ways to interpret a vector space. Literally representing "space" is one of them, for which R⁴ could never work

[willow bear]

like if you have three bodies interacting gravitationally in three dimensional "physical" space, that's already three position and three velocity vectors - one for each body
and the state of the system as a whole could be expressed as a single point in R^18

up to technicalities as to what that point would ac be

add more components and interactions and the dimension of the state space can be cranked as high as you wish

[silver matrix]

I have to study more to comprehend these type of things. From a premature perspective I see incoherent listening to a physicist talking about M-theory or whatever theory (obviously in a divulgative way) and then see the group monster with 256 dimensions or whatever that thing is (again, in a divulgative way)

[pale kettle]

Dimension just can have different meanings?

[willow bear]

of course you have to study more to comprehend these types of things

[pale kettle]

Dimension in these situations usually means dimension as a vector space

So like R^(10) has dimension 10

[patent beacon]

Self-teaching AIs are really just spicy functions R^10 to R^5 or whatever

[slow roost]

i did the math but not sure if its right.

[maiden pebble]

is tan(5x) = sin(5x)/cos(5x)

[past meadow]

Yes

[maiden pebble]

awesome

https://i.imgur.com/aCYvzOV.png I got a very long answer, could someone tell me if im close?

[viscid thistle]

https://www.symbolab.com/solver/derivative-calculator/\frac{d}{dx}\left(csc\left(5x^{5}\right)\right)

[maiden pebble]

$ csc(5x^5)cot(5x^5)(25x^4)cot(5x^5) + (-csc(5x^5))(-sin(5x^5))(25x^4) $

[obsidian monolith]

Bridge:

[maiden pebble]

thanks rent free i see where i went wrong

[viscid thistle]

np

[maiden pebble]

Could someone help me solve the derivative of sqrt(2t)

[empty wolf]

Do you know how to write square root as an exponent?

[maiden pebble]

Yes

[empty wolf]

Okay, so once you've done that, do you have an idea for derivative rules you know that would apply to (2t) raised to some power?

[maiden pebble]

power rule i believe its called

$ \frac{1}{2}(2t)^{-1/2}$

[obsidian monolith]

Bridge:

[empty wolf]

Yeah, nice, that's exactly it. There's just one step left. The power rule would be for $t^{1/2}$, but you have 2t inside

[maiden pebble]

finally...

[obsidian monolith]

flormulaic:

[empty wolf]

So is there any other rule you can use that takes care of when you replace e.g. x with 2x or a f(x)?

[maiden pebble]

what?

[empty wolf]

Do you remember learning about the chain rule?

[maiden pebble]

Yes

I just didnt understand your question

[empty wolf]

Sorry, that's what I was trying to ask about

So, can you try applying the chain rule to the steps you've already worked on?

[maiden pebble]

is my equation correct?

[past meadow]

Wait why do you need chain rule here

[empty wolf]

The equation you put above would be for just the power rule. You also need to handle the "2t" inside somehow (using the chain rule)

[past meadow]

...

Constant multiple rule

Much easier than chain rule

Original expression simplifies as Sqrt(2)*sqrt(t)

[maiden pebble]

I dont think ive heard of the Constant Multiple rule before

[empty wolf]

I did not know that was a thing, but it sounds like giving a name to the chain rule of a linear function

[maiden pebble]

Wait, flormulaic do you want me to apply to chain rule to my original equation or my most recent one?

[empty wolf]

I was thinking about just applying it to the most recent one. The only problem is in the most recent equation you sent, you're just skipping the fact that when you apply the power rule to (2t)^{1/2}, there is a 2t (a function) that you need to pull the derivative from afterwards

Sorry if that's unclear

[maiden pebble]

So you just want me to mutliply $ \frac{1}{2}(2t)^{-1/2}$ by 2

[obsidian monolith]

Bridge:

[empty wolf]

Exactly like that, yes

[maiden pebble]

So the fraction obv cancels

[empty wolf]

Yes

[maiden pebble]

Ooh ok, so i get 1/sqrt(2t)

[empty wolf]

I think so, yes

[maiden pebble]

Awesome thanks

pretty sure those are the same thing

[empty wolf]

That's 1/sqrt(2t), it just separated the two

[past meadow]

Sorry i misread the message

My bad

[empty wolf]

Sneaky, I think the rule you're talking about would not pass through a nonlinear function (like square root)

[maiden pebble]

, calc 2/3/4

[obsidian monolith]

Result:

0.16666666666667

[maiden pebble]

, calc 2/1 * 4/3

[obsidian monolith]

Result:

2.6666666666667

[maiden pebble]

, calc 2/3 * 4/1

[obsidian monolith]

Result:

2.6666666666667

[maiden pebble]

, calc 2/3 * 1/4

[obsidian monolith]

Result:

0.16666666666667

[maiden pebble]

thank god

[prime crypt]

. Write a polynomial inequality in factored form with integer linear factors who’s solution is:

𝑥 ∈ [−3]∪[−2,0]∪ [􏰇1/2,2/3]

[willow bear]

what's that box meant to be before the 1/2

also do you have a screenshot of the problem just in case

[prime crypt]

Yep

[] includes the value, so less than or equal to.

just said in set notation

[stuck lark]

i heartily dislike [] used for both closed intervals & set roster

[prime crypt]

Yep they force it up your mouth in ib lol

[harsh smelt]

well

@prime crypt

if this borders are solutions for inequailty

they are roots of underlying equality

(sounds a bit communistly lol)

[prime crypt]

hahaha

you’re right, just tried that and got something

just used one of those sign charts to see if the solutions fit for either < or >

[willow bear]

aight well ok like

you're gonna want a nonstrict inequality

because (polynomial) > 0 will only ever give you open intervals

[viscid thistle]

Tag helper after 15 mins

[stark vale]

guys where did this 3 come from

[harsh smelt]

just to get rid of radical

to get 2^(5/5)

[viscid thistle]

so 2^(2/5) * 2^(3/5)

add exponents

2^(5/5)

[stark vale]

so we see that when raise it to 2 power wont work

thats why we put 3?

[harsh smelt]

like if he wish he could've put 2^7138138183 under root

but ^3 cancels root

[stark vale]

okey thanks

[patent beacon]

I'm thinking something has gone wrong

It looks like whatever goes off the top, comes back at the bottom. With that being said, the amplitude would then be 5

Periodic functions should never look like this lol

[silver matrix]

Thanks for the answer. I guess the period is 10pi, no?

And what about the formula?

[patent beacon]

,w graph 5cos(1/5 x)

[obsidian monolith]

Display results online and refine query

Requested by Kaynex#7520

[patent beacon]

That would be it. Note that 1/5 comes from 2π/10π

[silver matrix]

Amazing, thanks.

what the actual fuck

[viscid thistle]

bruh

[silver matrix]

And this is the Mathematical Association of America

arggg

I'm gonna tell my advisor

[viscid thistle]

very poorly made

[silver matrix]

My answer?

[viscid thistle]

no the software

and the question

[silver matrix]

Yeah, agree

[still meadow]

isn't that Webwork as a whole?

[viscid thistle]

Anyone here ?

[pale bison]

no

[viscid thistle]

we're all ghosts

Q.29

I have solved this auestion

But i just need someone to check if my logic is correct or not

[pale bison]

sure, post your solution

[viscid thistle]

Okay

My handwriting is trash

Just let me explain what i did ?

[pale bison]

sure

[viscid thistle]

So first i checked for what values of x sqrt(1-|x|) would be valid

As its a square root

It should always be positive for all real values of x

So sqrt(1-|x|) >=0

Solving this inequality we get

x must belong to the interval [-1,1]

[pale bison]

good, keep going

[viscid thistle]

And then i just took the intersection of the above 2 sets

And got the answer

Have i considered all the appropriate cases ?

[pale bison]

yes

you actually don't need to consider the -sqrt 5 case, since |-1 - sqrt 5| > 1

so looking exclusively at [-1, (-1 + sqrt5)/2] will work

oh wops nvm

i missed the 2 at the bottom

yeah, you did everything correctly, discard what i said above

[viscid thistle]

Thanks 🙂

[pale bison]

oh wait wops,
|(-1 - sqrt 5)/2| > 1

[viscid thistle]

so looking exclusively at [-1, (-1 + sqrt5)/2] will work
@pale bison Shouldnt it be a closed like this ) ?

Since the roots arent considered

[pale bison]

yes, since you have strict equality, i'm just giving a domain for which they're defined on

[viscid thistle]

Yes 🙂

[sour eagle]

@viscid thistle but the solution ((-1-sqrt5)/2, (-1+sqrt5)/2) is not an answer in that list. Or maybe im just dumb. Sorry for the ping!

[viscid thistle]

@sour eagle Its alright 🙂 , actually the solution will be obtained after intersection of 2 sets

Thats just one of the sets

[sour eagle]

Oh ok! Wasn't sure if that was the final solution or not 😅 Sorry again for the ping

[gloomy mortar]

Okay I’m absolutely going nuts with Precalc. I have a few assignments left and I just can’t figure them out. I asked some family too. Can someone please dm me so we can talk about these calculus parts of the class (since this server doesn’t like giving me notifications in here 🙈)? I’m just being driven nuts with it and only have a few more days till the school year is over.

[rich flint]

@gloomy mortar go ahead and post your work so someone can see if they can help you

@gloomy mortar we have rules regarding dm make sure you review those but your better off uploading a screenshot on here so people can help you. If I can’t see the work I don’t feel confident that I would be able to assist. Good luck 👍

[gloomy mortar]

That’s all

[viscid thistle]

That's all eh

Wow great

No one will have time to walk through all

@gloomy mortar you gotta choose

[past meadow]

the general prolem solving method for geometry problems like those in the second pic onward goes like this: Draw a diagram. list or think of relevant formulas(cosine rule, sine rule, etc are going to be especially important here). See if you have enough values to use any of those. If you don;t, look for values that you can find that you would be able to plug into those formulas. Then, do your best to solve.

[raven brook]

Now how to solve this ?

[pale bison]

,rotate

[pale bison]

what have you tried?

[viscid thistle]

@raven brook dont multipost.

[sterile notch]

So I had this question:
Show: 2 * ln(y) = √(ln(y))

2 * ln(y) - √(ln(y)) = 0

2 * a^2 - a = 0
a( 2a - 1 ) = 0

2 * ln(y) - √(ln(y)) = 0
√(ln(y)) ( 2 * √(ln(y)) - 1) = 0
√(ln(y)) = 0, ln(y) = 0, y = e^0 = 1
or
2 * √(ln(y)) - 1 = 0
2 * √(ln(y)) = 1
√(ln(y)) = 1/2
ln(y) = 1/4
e^(1/4) = y

Though I was wondering, why I don't need to put ± next to ln(y), after squaring 1/2, to 1/4

[uncut mulch]

definition of the square root

$(\sqrt{x})^2 = x$

[obsidian monolith]

ramonov:

[sterile notch]

Looks like abs value tbh

[uncut mulch]

$(\sqrt{x})^2$ is not the same as $\sqrt{x^2}$ for all values of x

[obsidian monolith]

ramonov:

[sterile notch]

Hmm, what values?

I guess negative values wont work.

For the first one.

Since it'd be undefined I assume?

[uncut mulch]

depends if you are considering complex numbers

[sterile notch]

Ah right

[uncut mulch]

eg (sqrt(-1))^2 would be -1
but sqrt( (-1)^2) would be |-1| = 1

[sterile notch]

Yep, that makes sense

[raven brook]

@pale bison tried doin f'(x) = f (x) - f (1/3) / x - 1/3

@Al𝟛dium#4084 oki

[gloomy mortar]

@viscid thistle I still need this much help because people weren’t really helping before, so this has all built up. I guess now you realize why I can’t do it myself. People are refusing to help me

[viscid thistle]

You sent 13 problems

13...

[gloomy mortar]

Before I didn’t send many. I sent 13 problems now because it’s built up since people weren’t helping

[viscid thistle]

Aight

[gloomy mortar]

I get that it’s a lot but it’s gotten to that much because people haven’t helped. That’s why I really just need someone to help

And I’m coming to here because my teacher doesn’t ever answer her emails

And no one I know irl knows more than the second problem

Or if they have, haven’t replied

[viscid thistle]

3. Note that angular speed's formula is $$\omega=\frac{\theta}{t}$$ and when we talk about time that it takes to rotate once we talk about this formula $$\omega=\frac{2π}{T}$$ where T is the time it takes to rotate once (which you are given)

4. Is basically asking you for your lineal velocity, V. Whose formula is $$V=\omega \cdot r$$ whose r is given on the previous problem.

Next pic: 1) Note that Sine's law is $$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$$ where A is an angle and a is the side length located on the opposite of the angle (same for letters b and c), example given on the pic. While cosine's law: $$a²=b²+c²-2bc\cdot \cos(A)$$. So for 1 and 2, two lengths are given and one angle, choose which law is more convinient.

From 3) to 6) you gotta draw these.
3) 2 angles and one length given, calculate another length.

4. 2 lengths are given (while 90 being the length side) (use geometry here).

5. 2 angles and 1 side, pretty obvious to which law you should use to get the other side.

6. Again pretty obvious, 2 sides and 1 angle, calculate the missing angle.

Next pic:

1. 2 angles 1 side, calculate side.

2. 2 sides 1 angle, calculate angle.

3. 3 sides, calculate angle.

Last pic:

4. Again, 2 angles and 1 side, calculate the other side

5. Motion problem combined with 2 angles given, to calculate 3 sides, 2 or them are calculated by the motion equation (you gotta transform the units first) $$x=v\cdot t$$ where v is velocity which is given and time which is given as well, when both x are calculated, use cosine's law for the other length.

Bonus: 2 side lengths, 1 angle, missing 1 side, calculate it by one of the laws. And to find the area, i would use Heron's formula for the area of any triangle: $$A_{\text{triangle}}=\sqrt{s(s-a)(s-b)(s-c)}$$ where s is the semi-perimeter so $$s=\frac{a+b+c}{2}$$ and a, b and c are all the side lengths of the triangle

@gloomy mortar

[obsidian monolith]

Al𝟛dium:

[sour eagle]

HOW

🤯

[gloomy mortar]

Oh god let me try to break all that down and do it

Thanks for the info I’m just trying to not have my brain explode rn

Is 3 just 2pi/45 rpm then?

On the first page

Is this right for those two?

@viscid thistle

[viscid thistle]

The second one is not

And the first one must be expressed in rad/s not in rpm

For the second one look better at the formula

Its $V=\omega\cdot r$ not the other one

[obsidian monolith]

Al𝟛dium:

[gloomy mortar]

Wait what did I do wrong then in the second

[viscid thistle]

Its $V=\omega\cdot r$ not the other one

[obsidian monolith]

Al𝟛dium:

[gloomy mortar]

What do you mean not the other one

I plugged in the w and r is 125 right?

[viscid thistle]

You clearly didnt used that one

For the 4

I see a fraction

[gloomy mortar]

Yeah cause w=2pi/45

[viscid thistle]

Oh my srry its so small

I couldn't appreciate the equal signs

[gloomy mortar]

I multiplied that by 125/1

[viscid thistle]

Neither the dots

[gloomy mortar]

No worries

[viscid thistle]

So yeah its correct

And the first one must be expressed in rad/s not in rpm
@viscid thistle dont forget this one thing tho

[gloomy mortar]

Wait it says radians per minute though so wouldn’t it be rad/m not rad/s?

(I put rpm and derp, that’s revolutions per minute not radians per minute lol)

[viscid thistle]

Wait it says radians per minute though so wouldn’t it be rad/m not rad/s?
@gloomy mortar you are given 45s

Not minutes

Therefore it must be seconds

So rad/s

Or whatever notation that its equivalent to that

[gloomy mortar]

Yeah but it says calculate in radians per minute which I just now realized

Unless that’s a typo

[viscid thistle]

Where?

[gloomy mortar]

Right on number 3

[viscid thistle]

Oml how did i skipped that lol

So yeah

Basically from rad/s to rad/min

[gloomy mortar]

So just times it by 60?

[viscid thistle]

Look

[gloomy mortar]

I mean to get rad/s to rad/min do I times 2pi/45 by 60?

[viscid thistle]

,ask (2pi rad)/45s to rpm

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[viscid thistle]

^

[gloomy mortar]

It’s not rpm it’s radians per minute

[viscid thistle]

BRUH oml i need to eat urgently lol

[gloomy mortar]

Lol

[viscid thistle]

I mean to get rad/s to rad/min do I times 2pi/45 by 60?
@gloomy mortar yeah

[gloomy mortar]

So this for the last two?

@viscid thistle (When you’re back, if you’re eating)

[viscid thistle]

Totally correct

[gloomy mortar]

So I’m taking a look at the last two pages, and I have this so far. I feel like I should use the sines law, but with the info I have I’m confused...

Cause I have sides b and c but angle A...

If I use the Cosines law I get stuck

[viscid thistle]

Yah for the tree one you gotta use cosine law

It has 3 forms

Use the most convenient one @gloomy mortar

[gloomy mortar]

Wait so I don’t need that to be angle A then

Does it matter what’s angle A, B, and C?

[harsh smelt]

it is opposite to side

[gloomy mortar]

I mean does it matter if that is angle A, B, or C?

[viscid thistle]

Yeah

[gloomy mortar]

Because it doesn’t say what side is a, b, or c

[viscid thistle]

You can just name them

But respect the angle

Like if you put angle A, the side a must be on the opposite

[gloomy mortar]

Okay I still don’t get what I use though. Because the different versions of the cosine law only have one version for each angle

[viscid thistle]

Okay

For this one i'll tell you

Just so you get used to it

$$a²=b²+c²-2bc\cdot \cos(A)$$ so $$a²=30²+45²-2(30)(45) \cdot \cos(86)$$

Thats it

[obsidian monolith]

Al𝟛dium:

[gloomy mortar]

Oh wait wait yeah I know what to plug in for that but I still don’t get the info I need I mean

I need angle C

[viscid thistle]

Yeah true

But then you can do

$c²=a²+b²-2ab\cdot \cos(C)$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Bc now you know a, b and c

[gloomy mortar]

😳 I didn’t think of that

[viscid thistle]

Its all good

Playing with cosine law pretty much

[gloomy mortar]

I’m confusing myself

I feel like I need to take cos^-1 to cancel cosine and find <C but how do I do that here

Cause I can’t find cos^-1(4.07)

@viscid thistle

[willow bear]

4.07 = cos(C)

[gloomy mortar]

Yeah

[willow bear]

the fact that you got cos(something) to be a value above 1 is

kinda suspicious ngl

hoooold up

[gloomy mortar]

What did I do wrong then I’m confusing myself

[willow bear]

,calc 2 * 52.31 * 30

[obsidian monolith]

Result:

3138.6

[willow bear]

,calc 52.31^2 + 30^2

[obsidian monolith]

Result:

3636.3361

[willow bear]

so you had 2025 = 3636.3361 - 3138.6cos(C)

that's what you should've had, anyway

and then you... subtracted the 3636.3361 and the 3138.6?

yeah, that's where you went wrong

[viscid thistle]

@gloomy mortar ^

Its a lil bit tedious not to mess up the numbers

Be careful

[gloomy mortar]

No that’s the same number

497.7361

2025=497.7361cos(C)

That’s what I have

,calc 3636.3361 - 3138.6

[obsidian monolith]

Result:

497.7361

[harsh smelt]

2025=497.7361cos(C)

wtf is that

,calc 2025/498

[obsidian monolith]

Result:

4.066265060241

[gloomy mortar]

,calc 2025/497.7361

[obsidian monolith]

Result:

4.0684209965884

[gloomy mortar]

4.07, approximately

[viscid thistle]

,calc 30^2+45^2-[(2(30)(45)*cos(86)]

[obsidian monolith]

The following error occured while calculating:
Error: Parenthesis ) expected (char 30)

[harsh smelt]

,w range cos(x)

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[harsh smelt]

,calc 30^2+45^2-((2(30)(45)*cos(86))

[viscid thistle]

,calc 30^2+45^2-(2(30)(45)*cos(86))

[obsidian monolith]

The following error occured while calculating:
Error: Parenthesis ) expected (char 31)

Result:

3960.9858013643

[viscid thistle]

,calc sqrt(3960.9858)

[obsidian monolith]

Result:

62.936363097974

[viscid thistle]

Look

[willow bear]

@gloomy mortar you are making an algebraic fuckup

to demonstrate...

by your logic, the equation 35 = 50 - 2x would have x = 35/48 as a solution

which it does not

3636.3361 - 3138.6 * cos(C) is not (3636.3361 - 3138.6) * cos(C)

[gloomy mortar]

I used a calculator for that too

[viscid thistle]

@gloomy mortar redo it

[willow bear]

mythicalsheep

[gloomy mortar]

I just did

[willow bear]

i am not saying you did your arithmetic incorrectly

[gloomy mortar]

And got the same answer

[willow bear]

i am saying THAT THIS IS AN INVALID ALGEBRAIC MOVE

[harsh smelt]

Aldium i suppose you are needed in #help-2

[viscid thistle]

Aldium i suppose you are needed in #help-2
@harsh smelt busy day lol

[gloomy mortar]

Wdym Ann

[willow bear]

ok so like

can you show your work again

redone if possible

so that i can pinpoint the mistake more easily

[gloomy mortar]

Well I just showed the first part

Anything not shown was because calculator lol

[viscid thistle]

redone

[gloomy mortar]

Ik

[harsh smelt]

,w arccos(0.51339326451)

[gloomy mortar]

Why’d you subtract 2025 and put everything over 3138.6

[obsidian monolith]

Display results online and refine query

Requested by Commander Vimes#7532

[gloomy mortar]

That’s the opposite

[willow bear]

aight so like

you know what really is not helping

the fact that you're choosing to work with big messy numbers

[harsh smelt]

Why’d you subtract 2025 and put everything over 3138.6
because
a^2=b^2+c^2-2ab(cosC)

[gloomy mortar]

I’m using the scientific calculator to make my life easier so I don’t have to work with them

[willow bear]

algebra

[remote veldt]

sometimes it can be easier to make everything a variable, then put numbers in at the end

[willow bear]

$c^2 = a^2 + b^2 - 2ab \cos(C)$ can be rearranged as-is to $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$

[obsidian monolith]

Ann:

[viscid thistle]

sometimes it can be easier to make everything a variable, then put numbers in at the end
@remote veldt always lmao

[harsh smelt]

^

[remote veldt]

I mean like with coefficients of 1 or whatever it usually doesn't matter lmao

[harsh smelt]

also avoids round off errors

[gloomy mortar]

I literally did that I just didn’t put that equation in my work

[willow bear]

no you didn't.

you chose to plug in all the numbers and confused yourself.

[gloomy mortar]

To the equation yes

[willow bear]

and subtracted things that weren't meant to be subtracted.

[harsh smelt]

,calc 252.3130

[obsidian monolith]

Result:

3138.6

[willow bear]

$c^2 = a^2 + b^2 - 2ab \cos(C)$ can be rearranged as-is, \textbf{without plugging anything into anything}, to $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$

[obsidian monolith]

Ann:

[viscid thistle]

To the equation yes
@gloomy mortar that led to algebraic messings

[harsh smelt]

wtf

[willow bear]

ugh

too many cooks in the kitchen

im out

[gloomy mortar]

a²=b²+c²-2bc•cos(A)

That’s the first part

I did exactly that

Commander that was me dividing it on both sides to cancel it out

[harsh smelt]

where from 497

[gloomy mortar]

I already explained and calculated

[harsh smelt]

and i cannot see it

what here results in 497?

[viscid thistle]

Better $a=\sqrt{b²+c²-2bc\cdot \cos(A)}$

[gloomy mortar]

The whole equation before cos(C)

[harsh smelt]

[gloomy mortar]

Al3dium I did nothing different

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Look

[gloomy mortar]

Yeah I know

[remote veldt]

why have you already plugged numbers in mythical sheep. Why are there numbers in front of the cos(x)? why are there numbers anywhere?

this is the issue that ann was stating

[harsh smelt]

$a^2=b^2+c^2-2ab\cos A$

[gloomy mortar]

You guys are ignoring what I said

Those are the values plugged into that equation

The first one

[obsidian monolith]

Commander Vimes:

[gloomy mortar]

The one with cos(A)

[harsh smelt]

yes

[remote veldt]

do not plug values in directly

[harsh smelt]

A = C now

[remote veldt]

wait

[gloomy mortar]

I know the equation

[harsh smelt]

it is just sumbols

$c^2=a^2+b^2-2ab\cos C$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

if you wish so

[gloomy mortar]

I did that for the second one

Look at the triangle

That’s where the values come from

I found 52.31 in the first equation

[harsh smelt]

$c^2=a^2+b^2-2ab\cos C$ hence \
$2ab\cos C = a^2+b^2-c^2$

[viscid thistle]

$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$

[obsidian monolith]

Al𝟛dium:

[viscid thistle]

Just redo it

[gloomy mortar]

I already know the equations

[obsidian monolith]

Commander Vimes:

[inner karma]

AlKhashi's

[gloomy mortar]

I did and you guys won’t believe me

[inner karma]

theorem

[harsh smelt]

$\cos C = \frac{a^2+b^2-c^2}{2ab}$

if you did

then

cosC=?

[obsidian monolith]

Commander Vimes:

[gloomy mortar]

4.07

[harsh smelt]

wrong

you did it wrong

[harsh smelt]

you did $\
\cos C = \frac{c^2}{a^2+b^2-2ab}$

[obsidian monolith]

Commander Vimes:

[harsh smelt]

instead of

if you got cos = 4.07 you by definition cannot do it correctly

[viscid thistle]

,calc arccos((52.31^2+30^2-45^2)/(251.3130))

[remote veldt]

use arccos

[harsh smelt]

and wolfram

[obsidian monolith]

The following error occured while calculating:
Error: (intermediate value)(intermediate value)(intermediate value) is not a function

[harsh smelt]

,w

[obsidian monolith]

Please submit a valid query! For example, ,ask differentiate x+y^2 with respect to x.

[viscid thistle]

,w arccos((52.31^2+30^2-45^2)/(251.3130))

Ik lol

[obsidian monolith]

Display results online and refine query

Requested by Al𝟛dium#4084

[harsh smelt]

here is correct answer

[gloomy mortar]

That’s not the equation

[harsh smelt]

instead of

[viscid thistle]

^

[gloomy mortar]

Stop talking

This is what I did

The equation in red is what I did

Which is what I was told to do

[harsh smelt]

here is correct

here shit happens

[gloomy mortar]

Stop for a second

,calc 52.31²+30²

[obsidian monolith]

The following error occured while calculating:
Error: Syntax error in part "²+30²" (char 6)

[viscid thistle]

^2

[gloomy mortar]

Ik, my phone automatically autocorrects it

,calc 53.31^2+30^2

[obsidian monolith]

Result:

3741.9561

[gloomy mortar]

,calc -2(52.31)(30)

[obsidian monolith]

Result:

-3138.6

[harsh smelt]

52.31 btw

[gloomy mortar]

Yep mistyped

,calc 52.31^2+30^2

[obsidian monolith]

Result:

3636.3361

[gloomy mortar]

,calc 3636.3361-3138.6

[obsidian monolith]

Result:

497.7361

[harsh smelt]

this operation is illegal

[viscid thistle]

Yep

[harsh smelt]

2025=3636.3361-3138.6cosC

[viscid thistle]

You didn't respect the order of operations

[gloomy mortar]

I didn’t do the exponents first because I mistyped

The first one

[viscid thistle]

What

[gloomy mortar]

I did them originally but I mistyped it so I had to recalculate

[harsh smelt]

are 3633 and 3138cosC similar terms of the equation?

can you add bananas to chairs?

meh

[remote veldt]

@gloomy mortar The point I have been making is that your fundamental error came in in this step (red arrow). Do you understand why I say that you shouldn't have done that

[gloomy mortar]

Everything is plugged in the exact same as the equation

[remote veldt]

yes. But you should not be plugging in anything that early

you should isolate for the variable, and THEN plug in

[harsh smelt]

AND ANYWAY

[remote veldt]

you made algebraic errors because there were numbers there

[harsh smelt]

are 3633 and 3138cosC similar terms of the equation?
@harsh smelt

[remote veldt]

you subtracted non-similar terms, for example

isolate first

before there are ANY numbers

and then plug in

[harsh smelt]

can i add bananas and chairs?

[remote veldt]

it will minimize these silly algebraic errors

[viscid thistle]

yes. But you should not be plugging in anything that early
Exactly, which is what everyone has been telling you @gloomy mortar

[gloomy mortar]

The equation is c²=a²+b²-2ab•cosC

[harsh smelt]

advantages of operating with symbols:
easier to read
easier to parse
mimnal round-off-errors

The equation is c²=a²+b²-2ab•cosC
yes

we need cosC

[gloomy mortar]

I plugged into that because that’s the equation I was given

[viscid thistle]

...

[remote veldt]

and we are saying to wait to plug in until AFTER you have isolated for the thing you want

right now, that equation has c^2 isolated

manipulate it until the isolated thing is C (or cos(C) if you prefer)

[harsh smelt]

or. at least AcosC

[gloomy mortar]

Okay well I wasn’t told that at the start I went with the equations I was given

[viscid thistle]

We know

Neither no one told you to plug those in

But lesson learned, okay?

[gloomy mortar]

I literally barely even know how this works how am I supposed to know not to plug it in from the start