precalculus

[ivory echo]

Yea I copied and pasted, thanks!

[dire bloom]

are the spaces required

[ivory echo]

Only in certain times. If you're typing something like less than x, you need a space between \leq and x

Otherwise it's just one thing \leqx, do \leq x

[dire bloom]

i see

[ivory echo]

That should be enough to get you started so you can use it to describe the problem.

or etc use.

[dire bloom]

can i put another $ in the \text{}\ so i can put <=

$$
\begin{cases}
(x^{2} - 2x) - \lvert 1 - 3x - x^{2} \rvert \text{if x <= 0 or x >= 2}
-(x^{2} - 2x) - \lvert 1 - 3x - x^{2} \rvert \text{if 0 < x < 2}
\end{cases}$$

[obsidian monolith]

seventown:

[dire bloom]

dangit

OH

[viscid thistle]

double \ for new line

and doing won't actually showcase the space

[dire bloom]

$$
\begin{cases}
(x^{2} - 2x) - \lvert 1 - 3x - x^{2} \rvert & \text{if $x \leq 0$ or $x \geq 2$ }\
-(x^{2} - 2x) - \lvert 1 - 3x - x^{2} \rvert & \text{if $0 < x < 2$}\
\end{cases}$$

[viscid thistle]

but i'm wondering why are we talking about this

[obsidian monolith]

seventown:

[viscid thistle]

we all understood your problem

[dire bloom]

okay

[viscid thistle]

and what you wrote initially

no need to latex it

[dire bloom]

lol

[viscid thistle]

hence, reading #resources file was a better way as we don't waste time, and you can read it whenever

but uh w/e

why not just list the 4 cases

positive positive
positive negative
negative positive
negative negative

[dire bloom]

oh damn, i should've divided instead of subtracted?

[viscid thistle]

??

[dire bloom]

no i was just being confused

[viscid thistle]

what i mean is why are abs bars in it yet when you can do

positive positive
positive negative
negative positive
negative negative

oh okay

[dire bloom]

okay so i'd just need to find the domains for the 1-3x-x^2 absolute value hmm

[viscid thistle]

?

just do the same you did for the other ones

where are you stuck

[dire bloom]

i dont think i'm stuck but i spent a little time trying to find the domains for |1 - 3x - x^2|

[umbral cloud]

Anyone know some books contain complex number exercices or calculs and analysis?

[viscid thistle]

it's all g as long as you got it

Anyone know some books contain complex number exercices or calculs and analysis?
@umbral cloud occupied channel, please move.

maybe ask in #books-old

[dire bloom]

okay so I think i've found the whole piecewise function with all 4 cases, now i just have to find when each is below 0

[viscid thistle]

gl ig

[dire bloom]

xd

[next scroll]

Can somebody help explain this to me?

[viscid thistle]

@next scroll factor out the bottom

[next scroll]

@next scroll factor out the bottom
@viscid thistle so do I multiply them together and cancel out common factors?

[viscid thistle]

no don't multiply them together

just find common factors

[next scroll]

x^2 is all I see that are common

[uncut mulch]

factors not terms

[viscid thistle]

Put it in factored form

Just do that first

[next scroll]

2x^2-5x-42?

[uncut mulch]

why are you adding them

factorise: $x^2+x-42$

[next scroll]

because I have no idea what I am doing

[obsidian monolith]

ramonov:

[next scroll]

but that is only one of the sides?

[uncut mulch]

do that first

1 step at a time

[next scroll]

(x-6)(x+7)

[uncut mulch]

similarly, factorise $x^2-6x$

[obsidian monolith]

ramonov:

[next scroll]

x(x-6)?

[uncut mulch]

yes

[next scroll]

so x-6 are common?

[uncut mulch]

and then what's the lowest common multiple of
(x-6)(x+7) and x(x-6)

(x-6) is a common factor so you don't need to multiply by it again

[next scroll]

so the answer is (x-7)?

[uncut mulch]

no

[next scroll]

😢

[uncut mulch]

do you know what the lowest common multiple is?

[next scroll]

so lcm of 6 and 7?

[uncut mulch]

what's the lowest common multiple of
(x-6)(x+7) and x(x-6)

[next scroll]

yes but not how to do it with variables

[uncut mulch]

exactly the same idea

eg what's the lcm of 2 and 3?

[next scroll]

6

[uncut mulch]

what's the lcm of 4 and 6?

[next scroll]

12

[uncut mulch]

how did you determine that it was 12

[next scroll]

my calculator 😂

[uncut mulch]

your calculator has a lcm button?

[next scroll]

but 3x4 is 12

and 2x6 is 12

nothing lower can use both terms for a common multiple

[uncut mulch]

use the exact same idea here

what's the smallest (non-0) expression divisible by both
(x-6)(x+7) and x(x-6)

[next scroll]

x^2-42?

i have no idea

[uncut mulch]

(x-6) is already a common factor of both terms so there's no need to repeat it

then the lcm would simply be
x(x-6)(x+7)

[next scroll]

so the second x-6 gets canceled out and the LCD is x(x-6)(x+7)?

[uncut mulch]

i wouldn't say cancelled

[next scroll]

factored out?

[uncut mulch]

as in there's no point in multiplying it again

similar to how you reach the lcm of 4 and 6 is 12

[next scroll]

and the numerator is irrelevant?

[uncut mulch]

least common denominator is only about the least common multiple of the denominators
everything else doesn't matter

[next scroll]

makes sense. I really appreciate your help. thank you.

[quaint mason]

How do I solve for B?

[astral mantle]

can you show the whole thing

[quaint mason]

uh yeah sure

👍

[astral mantle]

alright sorry i got side tracked

you are looking for something like the (starting price * change factor)^t

@quaint mason

[quaint mason]

huh

so its similar to the equation for a?

@astral mantle

[astral mantle]

similar yeah but not the same

you would use 1-0.02 instead of .02

because of exponential decay

[quaint mason]

are u referring that to a or b

[astral mantle]

the order doesnt matter

its multiplicaton

$312346(0.98)^t$

[obsidian monolith]

Star_:

[astral mantle]

i think

[quaint mason]

oh howd u do that

[astral mantle]

do what

[quaint mason]

$312346(.02)^t$

[obsidian monolith]

KO7S:

[quaint mason]

oh thats cool

[astral mantle]

its latex

[quaint mason]

ah

[astral mantle]

its a typsetting language

[quaint mason]

i see i see

[astral mantle]

i use it a lot in university

a lot of math papers are written with it

you can check #resources for the latex cheatsheet

it will help with your formatting

[quaint mason]

never heard of it b4

[astral mantle]

are you in highschool

[quaint mason]

no

[astral mantle]

i never heard of it before uni

[quaint mason]

college

[astral mantle]

oh

[quaint mason]

i go to a cc tho

and my math professor is terrible

she doesnt teach anything

all she does is hand hw out

[astral mantle]

its ok to go to a CC

[quaint mason]

and say gl

[astral mantle]

i think bad professors are a universal experience mate

[quaint mason]

and on test day she just toss things at u

and thats how she gets paid

u want my class? lemme toss a load of busy work and make money off of that

Professor who cant teach math. Smh

[astral mantle]

thats how my history prof are

i go to a STEM school so

the history profs suck

[quaint mason]

damn bro

wait a second

i got everything right but my first equation on a was wrong

[astral mantle]

yeah it says linearly

so its gonna be the starting price - price difference between both years(time)

$312346-6246t$

[obsidian monolith]

Star_:

[astral mantle]

idk if this is correct or not

but its what makes sense to me

i think this is equivalent to 312346(.02)^t anyways

but this is the linear expression of it

thats probably why b was right but not a

[elder charm]

the mindless and correct way to solve such problems is to write the general form of the function you want to fit, substitute known points into it, and obtain a system of equations for the coefficients

[silver matrix]

Hello. Could someone explain me what am I doing wrong?

I should get a positive k in order to have a t years greater. But I can't see the error in my logic.

Ok, I've seen it I think.

A(t) = 2 and A in this case is 1. I'm going to try that.

Well, that's not the correct answer.

I keep getting the same result.

[silver matrix]

I see answers in other channels of pre-university. Am I formulating wrong my question? The title of the question is in the photo. The formula is A(t) = Ae^{kt}.

https://math.la.asu.edu/~mohacsy/internet-mat-117/webtestinfo/example-webtest4-spring06.pdf - I've google and my result was correct indeed.

Thing is, why am I getting a smaller age of the wooden artifact if 60 % > 50 %?

[night fiber]

Because when the artefact just dies, it has 100% of carbon 14

[silver matrix]

I see. Thanks.

[night fiber]

And you know the proportion of carbon 14 exponentially decreases (A%=e^{-kt}) then the older the artefact is, the less carbon 14 there is because of radioactive decay

[silver matrix]

And you know the proportion of carbon 14 exponentially decreases (A%=e^{-kt}) then the older the artefact is, the less carbon 14 there is because of radioactive decay
@night fiber One moment. I still don't understand it. If we have 100 %, then it means the tree is 0 years old?

That's why I don't understand it. The logic should be that whenever he has more percentage of carbon 14 it means that it is more proximate to die, but it also means that for achieving that result it will pass more years.

So if the wooden artifact is 5783 years old (because it has 50 % of carbon-14), whenever it reaches 100 % or a percentage greater than 50 it should have more years. Not less.

Ok, I think I'm starting to get it. So it starts with 100 % carbon-14

And then it keeps decaying that percentage, whenever it grews up. I see. Thanks.

[night fiber]

I don't know for trees

But for animals for example, they have a constant proportion of carbon 14 into their body, until they die, and carbon 14 starts decaying when they die

"fun" fact: because of the huge amount of carbon released in the past 150 years, the proportion carbon 12/carbon 14 in the bodies of recent animals is then modified so all the correlation tables are wrong. So imagine scientists in 1000 years who will find inconsistencies within the proportion of carbon 14 into the bodies

[stuck lark]

$\trig$

[obsidian monolith]

RokettoJanpu:

[viscid thistle]

@scenic slate still need help?

[scenic slate]

Yes please!

[viscid thistle]

also it does not hurt to say "i need help on .." or "i'm stuck at .."

what have you tried so far

and where are you stuck

[wispy parcel]

Question; if you have a function f(x)=(x^2-1)/(x-1), we have to remove a part of the domain at x=1, but also that function could be simplified to f(x)=x+1... so which is the "correct" representation? If we can say that (x^2-1)/(x-1) = x+1, then why do they behave differently as function rules, and is there a way to reconcile this difference? I guess my thinking is that if two expressions are considered equal, but they represent two different functions, there's something more going on?

[willow bear]

that's because strictly speaking an expression alone does not define a function cause there is no statement of what domain x is allowed to be in

[wispy parcel]

Yeah, but they're both polynomials, yet the first one has discontinuous behaviour, which IIRC it shouldn't have?

I'm just confused why the domain has to change while the two things should be considered equal, their behaviour should be the same?

[elder charm]

That's because it's a removable discontinuity.

Different "representations" can't have serious, meaningful discontinuities, but they can have removable ones.

Like, suppose I give you a function $y = (x+1)$. But I can also multiply it by $\frac{x+2}{x+2}$. This doesn't change the values of the function anywhere except -2, but in -2 it makes it undefined.

[obsidian monolith]

ConfusedReptile:

[willow bear]

(x^2 - 1)/(x-1) is not a polynomial

[wispy parcel]

Ah, my book is being a bit misleading then. It's all good, that makes sense @elder charm

[elder charm]

well, it's correct here that for t !=1 it's a polynomial

but yeah, I can also see why you're confused

polynomials defined everywhere but in a point are pretty weird 😅

[wispy parcel]

Probably not the best way to start an introduction to polynomials but at least it got the ol' brain thinking I guess!

[round quest]

how can i simplify this

[elder charm]

multiply numerator and denominator by 4x, for example.

[round quest]

k

[viscid thistle]

if the inverse of a many-one function is not a proper function because the domain doesn't map onto unique elements, then what is it classified as?

it's a relation (not sure if there's some more specific term for this case)

okay thank you

[quaint mason]

How do i solve for T(m)

[elder charm]

calculate your T(m) for some m and you'll see very quickly why yours is wrong

[quaint mason]

wym??? @elder charm

[elder charm]

for m=1: 30.90388
for m=2: 0.0058717372

clearly either your formula isn't right, or something very bad is happening in Tempe 😛

[quaint mason]

LMAO

hmm

ye ppl are being dehydrated

and they are dying

@elder charm

thats why u need to drink water

[astral mantle]

1+growth rate

[quaint mason]

🙂

[astral mantle]

instead of just the growth rate

[quaint mason]

oh 100% thingy

?

[vapid torrent]

is j hat dot k hat = 0?
since (0,1,0) dot (0,0,1) = 0 * 0 * 0 = 0

[quaint mason]

1+.0001900^m

but

is it cuz its increasing?

[elder charm]

@quaint mason Well, does a 0% growth rate means everyone dies on the spot? 😛

[quaint mason]

YEAHHHHHH

[elder charm]

@astral mantle and that's wrong, forgot to divide by 100 to convert from percents to actual values

[quaint mason]

lmao jk

[astral mantle]

my b

[quaint mason]

divide by 100?

oh

1.019 divide by 100??

or

did u mean

.019 divide by 100

bcuz of the percent

to decimal

[astral mantle]

extra 2 zeros

$162652(1.00019)^m$

[obsidian monolith]

Star_:

[quaint mason]

$162652(1.0001900)^m$

[obsidian monolith]

KO7S:

[astral mantle]

its the same thing

[quaint mason]

said to round to 7th decimal place

[astral mantle]

adding extra zeros doesnt do anything

bruh

[quaint mason]

xD

so since growth factor is approx 1, the growth rate in population would increase?

[astral mantle]

the closer it gets to 1 the slower it grows

the higher above 1 the faster

the lower the slower

well not slow but

it decays below 1

[quaint mason]

so since the growth factor is close to 1, theres a growth at higher rate, so when is there no growth at all? when its exactly 1?

[astral mantle]

no growth when its exactly 1

1 to any power is 1

so 1 * starting value

would be it

so it wont grow

[quaint mason]

ok

[quaint mason]

can u check that statement rq

[quaint mason]

whats the formula for calculating an annual percentage change?

[elder charm]

$$
\frac{A_{n+1}}{A_n} - 1
$$
?

[obsidian monolith]

ConfusedReptile:

[quaint mason]

huh

i dont think that would make sense in this context

well what is An?

[elder charm]

the value of the thing you're calculating the change of after nth year

[quaint mason]

so if year 1= like idk 900

and year 5=1500

nth would be 600?

well An??

[elder charm]

since several years have passed, you'd need to take a degree-4 root

because the formula of exponential growth is generally $A_n = A_0 k^n$.

[obsidian monolith]

ConfusedReptile:

[elder charm]

So if $A_5 = A_0 k^5 = 1500$ and $A_1 = A_0 k = 900$, then $k = (\frac{A_5}{A_1})^{\frac{1}{4}} = (\frac{1500}{900})^{\frac{1}{4}} ~= 1.1362$

[obsidian monolith]

ConfusedReptile:

[elder charm]

@quaint mason

so in this case, the growth rate per year is 13.62%

(I'm assuming you do mean exponential growth and not linear).

[quaint mason]

ye exponential

but

where did u get ^1/4 from @elder charm

[elder charm]

solving that system.

divide the first equation by the second:
$$
k^4 = \frac{A_5}{A_1}
$$

[obsidian monolith]

ConfusedReptile:

[elder charm]

and take the fourth root.

[quaint mason]

tyvm @elder charm

[turbid summit]

are basic non-algerbraic fractions included here?

[astral mantle]

do you mean functions

[turbid summit]

nope, just the "three fifths" kinda thing

the easiest stuff

[astral mantle]

if you have questions about fractions id probably go to #prealg-and-algebra

[turbid summit]

i'll do that, thanks

[viscid thistle]

The domain of f(x) is [0, infinity)
When x = 1, f(x) = (1/2)

When x = 2, f(x) = (1/5)

When x = 3, f(x) = (1/10)

When x = 4, f(x) = (1/17)

I have to figure out an equation that satisfies this

and unless I am using a piecewise funciton I have no idea how to do this

[mild wagon]

Hello

Can anyone help me with 2 logarithms

[tender sail]

-2cos(x) = 1

Solve for x

how should I do this?

I don't see any immediate way other than

arccos(1/-2) = x

but I was told I wasn't allowed to do this

please ping if anyone responds

[tardy ridge]

@tender sail you are supposed to know arccos(-1/2)

[gilded brook]

What am I missing? There is apparently one more answer

[uncut mulch]

you missed the trivial solution

if you managed to get pi, you should've got the missing solution too

[gilded brook]

Ahh

I see it was 0

How do you find two more solutions?

[uncut mulch]

general solutions, periodicity of sine

[minor pivot]

anybody know how to write sin(cos-1 2x) as an algebra expression?

[gilded brook]

And to do that you just do 2pi+those?

[uncut mulch]

introduce the general solution at:
pi*x/3 = ||n*pi + (-1)^n * (pi/6)||

@gilded brook

[gilded brook]

Ahh Alright I gotcha

[little valley]

Can anyone suggest me some tips or anything to improve in Combinatorics, I am really weak in it. Just can't get it no matter how hard I try

[viscid thistle]

It's hard maybe because there is not set pattern or a set way of thinking in it

A slight modification can make the problem from easy to ultra hard

So it only needs more practise on your thinking part. How do you think approach the problem

[edgy pasture]

Hi I need help

So for part b how do I put it in ti84

[edgy pasture]

Anybody?

<@&286206848099549185>

[formal iris]

we cant see lmao

[copper breach]

Hey guys

so I've been workin on my homework and I found some stuff on limits

and idk how to do them

can you guys show me what yo ugot for these

[chrome glen]

hi guys im from the UK and we dont have precalculus so can someone explain it to me?

[noble jay]

I think precalculus would be like as level maths

functions, trig etc

oh there's also this "anything in the us precalc curriculum goes here, examples include: trigonometry, logarithmic and exponential functions, function sketching, etc."

@chrome glen

[viscid thistle]

check khan academy's playlist on pre calculus

[chrome glen]

@noble jay ahh thank you. where would i ask questions about integration

[noble jay]

there;s #calculus

there's*

[copper breach]

hello?

nvm i figured out the others

how would i do this

can someone please help me asap

[uncut mulch]

is this a test

[copper breach]

no

it's a review for my upcoming test

which is tomorrow

and i have no idea how to do it

[lament fiber]

(this is from Quora)

it was also in my book

this doesn't make sense to me
at this point we haven't been taught about monotonicity etc
how do I use it to solve this kind of stuff

these answers make no sense to me

(Please ping me when you answer)

[viscid thistle]

Oh this one

2^x + 3^x + 4^x = 5^x

Divide both sides by 5^x

(2/5)^x + (3/5)^x + (4/5)^x = 1

Left side is a strictly decreasing function

So there's only a unique solution. @lament fiber

I think sum of monotonic functions in general is not always a monotonic function

One can be monotonically increasing other can be decreasing so we can't generalise it I think

So I refrain from reading that solution

[lament fiber]

I don't get the strictly decreasing thing

as x increases, y decreases?

[viscid thistle]

uh yeah.

[lament fiber]

okay, stupid me

[viscid thistle]

It's fine

[eternal osprey]

in transformations, whats the order for applying stretches, reflections and translations?

[astral mantle]

horizontal shift to stretch/compress to reflection to vertical shift

basically follow order of operations

[eternal osprey]

so translations first?

because the way you wrote it, it seems to say "horizontal translation, stretch/compress, reflection, then vertical translation"

[noble jay]

@eternal osprey https://www.youtube.com/watch?v=lokEdLotXT4

this video should be perfect for you

just 6m

[copper breach]

how do I get help

[unique hill]

just to confirm, this is correct right?

https://cdn.discordapp.com/attachments/684683687186792468/762827660363300944/348756950936190976.png

[viscid thistle]

cis(π/2) = cos(π/2) + isin(π/2) = i

[elder charm]

$$
e^{i x} = \cos(x) + i \sin(x)
$$

so yes.

[obsidian monolith]

ConfusedReptile:

[unique hill]

so i can write i as $e^\frac{\pi (e^\frac{\pi i}{2})}{2}?$

[obsidian monolith]

nighty:

[unique hill]

and then i can keep going infinitely?

[elder charm]

sure

in a way, you could say that for the following function:
$$
f(z) = e^{\frac{\pi z}{2}}
$$

[obsidian monolith]

ConfusedReptile:

[elder charm]

i is a fixed point.

[unique hill]

ah ic

thank you

$e^{i\theta}=\cos(\theta)+i\sin(\theta)$

[obsidian monolith]

nighty:

[unique hill]

and for the restrictions of theta

is the angle for theta between -pi and pi inclusive or non inclusive?

@elder charm wait soz for ping but when i keep going infinitely for https://cdn.discordapp.com/attachments/363224154469826562/762839797139505182/524541200875782155.png

doesnt it diverges to infinity?

[elder charm]

Uhh, why'd it? It's just repeatedly applying the f(z) mentioned above to i, which is a fixed point.

[unique hill]

https://cdn.discordapp.com/attachments/684683687186792468/762842002655936522/348756950936190976.png

doesnt this give an infinitely large real no.?

but at the same time i is unreal

[elder charm]

AgaIn, why would it?

though I see what you mean

[forest cape]

oh hello @unique hill

oh hello @rough lotus

wait did i invite you here?

[rough lotus]

i think so

or else @unique hill

[forest cape]

i think its that f(f(...(f(z))) will still have the z there?

[rough lotus]

👀

[unique hill]

wait blair i think i invited u

[forest cape]

since my infinite fraction seems to suggest that its f(f(f(... where the z has become lost (which is not trye)

[rough lotus]

yea probs?

¯_(ツ)_/¯

uhh

where the z has become lost
z is still there at the "end" of the thing?

[unique hill]

im just confused as to why that pyramid is not an infinitely large real number

but yh it shouldnt be real

[rough lotus]

im just confused as to why that pyramid is not an infinitely large real number
wait wdym?

[forest cape]

well i = f(i)
i = f(f(i))
...

but if you say i = f(f(... infinitely, where did the i go?

i think thats the problem?

[unique hill]

basically yh

[rough lotus]

well u need the i at the end to make it simplify right?

uhh if this makes any sense

[forest cape]

ya

[rough lotus]

the base case is $i = e^{\frac{\pi i}{2}}$

[rough lotus]

right?

umm

[unique hill]

mhm

[rough lotus]

and then u have a bunch of recursion

$i = e^{\frac{\pi e^{\frac{\pi i}{2}}}{2}}$

[rough lotus]

umm

but it only works becus that i is there at the end?

[unique hill]

ye

[rough lotus]

well ig im kinda just repeating what u 2 r saying

[unique hill]

same thing with the pyramid tho

[rough lotus]

but like if u take out the i

yea

hmm

[unique hill]

wait actually i think i understand

cuz it's not actually taking out the i

so then https://cdn.discordapp.com/attachments/684683687186792468/762842002655936522/348756950936190976.png

is correct then ig

[forest cape]

i dont think so?

the problem is when approaching infinite recursion

for example: 0.000001, but if there are infinite 0's there isnt a 1 anymore

wait thats a bad example, but i think you get the point?

[unique hill]

yh kinda

but like

that pyramid should be true tho

[gritty axle]

hey, quick concept question: If a function is differentiable at a point, it must be continuous. Is this correct?...

[rough lotus]

not necessarily?

let $f(x)$ be a function such that
$$
f(x) =
\begin{cases}
0, & x < 0\
1, & x \ge 0
\end{cases}
$$

[rough lotus]

umm so f'(x) = 0 for all x, but its not continuous at x = 0?

does that make sense?

[viscid thistle]

What's the derivative at x=0

[rough lotus]

0?

wait is it?

idek

yea, cus f(x) = 1 at x = 0, diff 1 is 0

(someone correct me if im wrong)

[viscid thistle]

Use the limit definition of a derivative

[rough lotus]

oh tru

wait what

[viscid thistle]

You're talking differentiability and don't know how to define derivatives using limits ?

[rough lotus]

derivative approached from below is the same as derivative approached from above?

[viscid thistle]

Read it up don't make up things

[rough lotus]

does that only apply to continuity?

[viscid thistle]

Read it up don't make up things
.

The limits definition for a derivative

[rough lotus]

oh u meant 1st principles?

[viscid thistle]

Yea ..

[rough lotus]

lol oops

[viscid thistle]

Use it you'll see right and left hand derivatives aren't the same

In this case

[rough lotus]

oh do u get like (0 - 1) / ( 0- ) = inf?

ok mb, ig my case doesnt work

[viscid thistle]

No case works

It has to be continuous in order to be differentiable

[inner gale]

hey, quick concept question: If a function is differentiable at a point, it must be continuous. Is this correct?...
@gritty axle yes

[supple meadow]

Hi

can someone help me

how to get arcs?

@obsidian monolith

[rough lotus]

$\begin{tikzpicture}
\draw (0, 0) arc (0 : 90 : 1);
\end{tikzpicture}$

[rough lotus]

\draw (co-ordinate of centre) arc (start angle : end angle : radius length);

@supple meadow is that what u meant?

[supple meadow]

yup

How can u find arc in inverse function

[blazing raven]

Arc length?

[rough lotus]

\draw (co-ordinate of centre) arc (start angle : end angle : radius length);
*\draw (co-ordinate the arc starts at) arc (start angle : end angle : radius length);

[supple meadow]

Thank u

[worn anvil]

Got a Pre-Cal question on rational functions

How does one write this? I've never seen a rational like this before

[echo wagon]

General form: y = a/(x-p) + q where p is the vertical asympyote and q is the horizontal asymptote

[worn anvil]

I got it, thx

@echo wagon

[candid sapphire]

Would points outside domain be considered to be discontinuous?

So like for ln(x) would the function be discontinuous for x<=0?

[viscid thistle]

What do you think?

Think for a second

[candid sapphire]

I thought it would definitely be discontinuous. But what type of discontinuity?

I'm leaning towards infinite discontinuity because of how the graph is at x=0

But that doesn't make much sense for points less than zero

[wispy ledge]

Can someone help me with this

[viscid thistle]

@candid sapphire it is nonsense to talk about continuity at points not in the domain of the function

@wispy ledge occupied channel, please proceed with a free one.

[candid sapphire]

I mean for this example I would agree, but for removable discontinuities, the point that is removed is not in the domain, but it can still be classified as a removable discontinuity no?

[viscid thistle]

Uh, are you asking me if there is a point not in the domain, it can be considered a removable discontinuity?

[candid sapphire]

yeah

Like if you have ((x-3)(x-2))/(x-2)

certainly undefined at x=2

[viscid thistle]

Of course, as i said, if a point is not on the domain of the function, it is nonsense again to talk about continuity

[candid sapphire]

What would a removable discontinuity be then?

would you have to define f(2) = 0 and then it would be a removable discontinuity?

[viscid thistle]

,w plot (x²-x-2)/(x-2)

Okay wolfram doesn't show the hole

My gosh is it that hard to notice that there is literal ongoing conversation @terse ravine

Not even 15s between the last message

would you have to define f(2) = 0 and then it would be a removable discontinuity?
The problem here is that i'm not exactly understanding what you are saying, a removable discontinuity is an undefined point that it is obviously not in the domain of the function and therefore, it is a discontinuity as the name suggests

Example of a removable discontinuity

[stuck lark]

a removable discontinuity is an undefined point that it is obviously not in the domain of the function
is this a defn you're giving or rejecting

[opal cosmos]

How to do sin(9pi/8)

[viscid thistle]

kinda giving

Why?

how does this statement make sense

its got me so confused

My fricking god

I've got interrupted like 4 times

sorry

[stuck lark]

it sounds off, can you clarify smth like

is an undefined point

[viscid thistle]

@viscid thistle @opal cosmos read #❓how-to-get-help , i believe it is not that hard to consider reading the rules of server you are gonna request a service from.

Like, yeah that sounds uh, "a point not in the domain of the function"

I was kinda referring there to the statement made by them

[stuck lark]

would you have to define f(2) = 0 and then it would be a removable discontinuity?
as long as lim x->2 f(x)!=0 this WILL make f have a removable disc at 2

also this https://cdn.discordapp.com/attachments/363224154469826562/763085717315518493/SAVE_20201006_191001.jpg

i can't say f has a removable disc at 2. if i define f by the above for x!=2 then further define f(2)=0 THEN i can say f has a removable disc at 2

[worn anvil]

Anyone got an equation for this? I can't solve it.

[viscid thistle]

This looks like a rational function.

Are you aware that the vertical asymptote is a value that which the function is undefined at?

[sick seal]

im not sure what to do with the lack of given information

[elder charm]

You don't need any more information than what the function's called 😛

[sick seal]

huh?

[elder charm]

average rate of change is just change divided by time (well, coordinate here).

and change is the difference in value.

in other words, just $\frac{f(7)-f(4)}{7-4} = \frac{f(7)-f(4)}{3}$

[obsidian monolith]

ConfusedReptile:

[sick seal]

ahh. now that i see it written out i remember this

thanks man

[gritty axle]

Could someone help me with question 4, part b?

[astral mantle]

for the tangent line you want to take the derivative of f(x) and substitute in the given x point into the derivative

this will give you the slope

you can then plug in the slope and the given point into the formula

$(y-y_1)=m(x-x_1)$

[obsidian monolith]

Star_:

[astral mantle]

where $(x_1,y_1)$ is the given point and m is the slope

[obsidian monolith]

Star_:

[astral mantle]

@gritty axle

[gritty axle]

So I have my equation on the left, does my solution/explanation explain it well?...

Would I plug my slope in actually? Cause I found it in part A to be 1/6

[astral mantle]

$f'(9) = 0.5(9)^{-0.5} = \frac{1}{6}$

[obsidian monolith]

Star_:

[astral mantle]

then you need to evaluate the y point of x = 9

which is 3

[gritty axle]

Where did the 0.5 come from

[astral mantle]

i took the derivative

of sqrt x

to find slope

[gritty axle]

I got the derivative to be 1/6

[astral mantle]

yeah thats the slope

1/6

of the line

[gritty axle]

Okay

I don’t think I’m following the process I guess

So from the derivative I put it into what formula

[astral mantle]

$(y-3)=\frac{1}{6}(x-9)$

[obsidian monolith]

Star_:

[gritty axle]

How did you get 3?

[astral mantle]

used the formula

you use the point it gives you

[gritty axle]

which formula

[astral mantle]

they gave you x

[gritty axle]

yes

[astral mantle]

you sub x in to find y

so you get (9,3)

and use the formula i already sent

[gritty axle]

can you type out the formula to show me?

[astral mantle]

i already did

[gritty axle]

to get 3?

[astral mantle]

i already explained how to do the entire problem

to get 3 i plugged 9 into the given equation

you are calculating the equation of a line tangent to the graph at the point (9,3)

[gritty axle]

oh, so you put 9 in for sqrt (x) to get 3, which is the y value

[astral mantle]

$y=\frac{1}{6}x+\frac{9}{2}$

yeah

[obsidian monolith]

Star_:

[gritty axle]

$(y-3)=\frac{1}{6}(x-9)$

[obsidian monolith]

Mg Shay:

[astral mantle]

yes thats correct

[gritty axle]

so thats the equation of the tangant line at 9

[astral mantle]

yeah just simplify

[gritty axle]

in point slope form

[astral mantle]

you need to simplify to get it in point slope form

solve for y

$y=\frac{1}{6}x+\frac{9}{2}$

[obsidian monolith]

Star_:

[gritty axle]

how did you get to that point?

[astral mantle]

i just simplfied the equation

[gritty axle]

yes but how

[astral mantle]

distributed the 1/6 and added 3 to both sides

[gritty axle]

add 3 over?

okay

[astral mantle]

yeah

[gritty axle]

is that not y=mx+b form?

not point slope?

[astral mantle]

oh yeah yeah

you just dont have to simplify i guess

but theres really no point for point slope form

[obsidian monolith]

br0k3n:

[unkempt magnet]

is this invertible?

wait wrong function

$f: \mathbb{Q} \rightarrow \mathbb{Q} \quad f(x) = x³$

[obsidian monolith]

br0k3n:

[unkempt magnet]

is this function invertible?

i think its one to one so injective

as for surjectivity

thats the thing im wondering are there any rational inputs which have an irrational image

[gritty axle]

uhhh

what happened to the normal calculus channel

[blazing raven]

wait @unkempt magnet what's the domain you want in the inverse?

[stuck lark]

@unkempt magnet cbrt(2) is in Q*. there exists no x in Q where x^3=2

[blazing raven]

yeah but if he restricts the domain to only those numbers that were y-values of f(X) = x^3 he's okay.

[stuck lark]

you mean restrict f's codomain to f's image, then f can be made bijective. but no one asked that

[unkempt magnet]

@stuck lark what does Q* mean?

i wanted the domain of the inverse to be Q @blazing raven

oh shit i see wym

well

yeah seems like it would only work if you restrict the inverse's domain to the function's image but idk if that would work

the question basically just asked if it has an inverse so i guess the answer would be no

the image has to be equal to the codomain but it wouldnt be surjective because as you said numbers like 2 cant be an image

yeah thanks for that

[stuck lark]

@unkempt magnet Q*=irrationals. and no prob

[unkempt magnet]

@stuck lark

actually

i was wondering how to show that its not surjective

because for the original function cbrt(2) would not be in the domain even though 2 itself is in the domain

with the inverse you would cbrt it but still would there be any way to show it without calculating the inverse

i guess, that f(Q) = R, not Q

ok no the image would surely be Q

or would it

the image wouldnt be the entire set of all rational numbers

[willow bear]

f(Q) cannot be R no matter what f is

[unkempt magnet]

yeah i thought so

maybe verbally explaining would be enough

The image of f is not Q. There are rational numbers in the codomain which have no preimage, e.g 2, as there is no rational solution for x in f(x) = 2.

thus its not surjective which means theres no inverse

[stuck lark]

that's all

[mossy tiger]

how would I factor this

x^6 - 63x^3 - 64

[blissful ridge]

Let x³=t

[abstract folio]

Wondering how i would solve this

[viscid thistle]

Vieta

[viscid thistle]

for C) is the monthly case load for min average cost 750 and the cost would be 0?

[silent veldt]

@abstract folio 1) find p, q, r:
we know that f(r) = 0
by substituting r into f and solving for f we can find that r = 3
therefore we know that f(x) has a factor of (x-3)
next divide (x^3-3x^2-4x+12) by (x-3) to get x^2-4 which can be further factorised into (x-2)(x+2)
therefore f(x) = (x+2)(x-2)(x-3)
from here we can find that (p, q, r) = (-2, 2, 3)

2. find expression of g(x):
   we know that g(pqr) = g(p-q) = g(p+r) = 0
   bu substituting values we can find factorized form of g(x), which is g(x) = a(x-12)(x+4)(x-1), where a needs to be found
   we also know that g(0) = (r+1)! = 24
   by substituting 0 into g(x) we get:
   g(0) = a(-12 * 4 * -1) = 48a
   therefore
   48a = 24
   a = 0.5
   g(x) = 0.5(x-12)(x+4)(x-1)

[viscid thistle]

I know that the vertical asymptote is the denominator when it's equal to 0 but why is this the case

Because near x=2 function shoots up to +/- infinity

And it's around a vertical line x=2 so

That's it

Like try drawing it

It'll give you a better feel of it

[viscid thistle]

okay

[hollow silo]

Hi, sorry if I have the wrong channel, please direct it to me. I'm 16 in the uk and doing further maths as my a level atm. I just wanted to know if I could be directed into a website or book that could help me refine and practise the concepts of argand diagrams and complex numbers. Any help is great 😄

[iron violet]

hey quick question how does the top part become the bottom part?

you use the (x-3) to cancel out the numerator?

[viscid thistle]

Yes

Look

[willow bear]

-x+3 = -(x-3)

[iron violet]

ohh

[viscid thistle]

$-x+3={\color{green}{-(x-3)}}$ so then $$\frac{-x+3}{6(x+3)(x-3)}$$ $$\frac{{\color{green}{-(x-3)}}}{6(x+3)(x-3)}$$ $$\frac{{\color{green}{-\cancel{(x-3)}}}}{6(x+3)\cancel{(x-3)}}$$ $$\frac{-1}{6(x+3)}$$

Sniped

[obsidian monolith]

Al𝟛dium:

[iron violet]

I see

Thank you very much

both of you

[dawn void]

woah

ur good at texit

can u teach me

@viscid thistle

[viscid thistle]

I mean yeah but what do you need help on, there's a file at #resources to learn the basics i believe

[astral mantle]

the best way to learn tex is to practice

start writing assignments in tex

[surreal iron]

someone do this pls and tell me what they get

I got 0.7460... m /h

[patent beacon]

Have a picture of your work?

[surreal iron]

yep 1 sec

[patent beacon]

Why 0.4×1.2×1.2?

[surreal iron]

r = o.4h

[patent beacon]

Ah right

Well, yeah this looks good!

[surreal iron]

k

ty

[iron violet]

could someone show me how to solve this?

[burnt sonnet]

I can walk through the operations with you if you want

[iron violet]

please

[reef halo]

query detected
parse query
catch exception
punctuation
0.4×1.2×1.2
0.4 times 1.2×1.2
try
catch exception

[burnt sonnet]

Do you know in which order you should prioritize. Addition, subtraction, multiplication and division?

[iron violet]

Yes PEMDAS right?

[reef halo]

store query

[burnt sonnet]

what does that stand for? 🙂

[reef halo]

0.4 times 1.2 times 1.2
try
catch exception

[iron violet]

Parentheses, Exponents, Multiplication and Division

[burnt sonnet]

mdas is multiplication , division , addition and subtraction right?

ok cool

Yes that works fine 🙂

[reef halo]

parse string
0 point 4 times 1.2 times 1.2
try
catch exception

[burnt sonnet]

So you have: $(4+1)(\sqrt{(3)(4)+52}+8)$

[obsidian monolith]

egocarpo:

[lime bolt]

guys you are getting trolled lol

[burnt sonnet]

@iron violet Where would you start?

[reef halo]

parse string
0 point 4 times 1 point 2 times 1.2
try
catch exception

[burnt sonnet]

using your rule

[iron violet]

the 4+1?

[burnt sonnet]

sure that works fine 🙂

$5(\sqrt{(3)(4)+52}+8)$ so we have this?

[obsidian monolith]

egocarpo:

[iron violet]

after that I dont know

[reef halo]

push

[burnt sonnet]

Okay so you would want to just have a number in the square root so you could evauluate that

$(3)(4)+52$ so try to make this into a number

[obsidian monolith]

egocarpo:

[iron violet]

so 64 +8 now?

[burnt sonnet]

almost

the 64 you got is inside the square root

[iron violet]

oh right

[burnt sonnet]

Now we have $5(\sqrt{64}+8)$

[obsidian monolith]

egocarpo:

[burnt sonnet]

Do you know what the square root of 64 is?

[iron violet]

8?

yeah 8

[burnt sonnet]

yes

so then we have $5(8+8)$

[iron violet]

5(16) now'\

[obsidian monolith]

egocarpo:

[burnt sonnet]

yes good

[iron violet]

ok I see now

thank you I appreciate your help

[burnt sonnet]

So I would say that you want to get the the expression inside a function to be a number so you can evaluate it

or maybe you havn't talked about functions yet. The square root thing

[iron violet]

Yeah I forgot a bunch of stuff from pre calc and I am now in calc so I have to refresh

[burnt sonnet]

oki cool!

Good to refresh, Good luck!

[iron violet]

Thank you again

[burnt sonnet]

yw

[rough loom]

How could I go about solving this?

I see vids that say solve the inner summation first, but what if the inner sum uses variables from an outer sum?

[terse ravine]

@patent beacon I think this is right.

[patent beacon]

Ye

[terse ravine]

🙂

[patent beacon]

Ye

[minor pivot]

sin(cos-1 2x) Does anybody know how to write that in an algebraic expression?

[quick mirage]

lets assume a = cos^-1(2x)

so cos(a) = 2x

cos(a) = 2x/1 (adjacent / hypoteneuse)

sin(a) would have to be opposite / hypoteneuse, but we have no opposite

but actually, we do!

since we know the adjacent side and hypoteneuse of this imaginary triangle, we can use pythagoreans theorm to find the opposide side
adj ^ 2 + opp ^ 2 = hyp ^ 2
opp ^ 2 = hyp ^ 2 - adj ^ 2
opp = sqrt( hyp ^ 2 - adj ^ 2)
opp = sqrt(1^2 - (2x)^2)
opp = sqrt(1 - 4x^2)

remember, a = cos^-1(2x), so sin(a) = sin(cos^-1(2x))

sin(a) = opp/hyp
sin(a) = sqrt(1 - 4x^2)/1
sin(a) = sqrt(1 - 4x^2)

Therefore, sin(cos^-1(2x)) = sqrt(1 - 4x^2)

@minor pivot

[minor pivot]

hey thanks

so it's the same as cos(sin-1 2x)

[quick mirage]

yep

[minor pivot]

thanks!

[supple meadow]

Hi help

[terse ravine]

-31=-31(-3)+b

-31=93+b

[quick mirage]

y = 59

59 = -31(-3) + b

[terse ravine]

oh yea

-34=b

[supple meadow]

Help

what is the area

the device can sense movement up to 30 m away through the angle of 85°

so what is the area that their device can monitor?

[quick mirage]

The range in this case would be a part circle

so think of the radius of the circle as 30 m and there's only 85 degrees of the total 360 degrees of the circle

hint: ||Area = pi*r^2||

@supple meadow

[supple meadow]

Thanks

[wicked star]

https://gyazo.com/e1c11c382f2559b48e8be33b0cc594e5

can someone explain how i got this wrong?

@violet tide

@torn oriole

@wind igloo

[torn oriole]

dont tag staff.

or honorable or helpers

read rules

[wind igloo]

Looks like the software sucks. Not much anyone here can do about that.

Unless you put 40 and they wanted 4 theta. idk

Oh, yeah, that's what it looks like.

Also, don't ping specific people, especially moderators unless they're already helping you.

[blazing raven]

the theta, catital O, zero conundrum.

[sick seal]

im at a roadblock here, ive somehow managed to get the correct answer ||17/2||, but i honestly have no clue how i got there or why its correct. the only thing i know about this problem is that t =/= -7.
even after putting this on a graph i am still stumped on how to correctly solve it

think i figured it out, i multiply t+7 both sides then solve for x

imean t

[solar heron]

@sick seal should its be -17/2?

shouldnt*

[sick seal]

:) yes... that is my one weakness in math, remembering to put the lil - in the answer box\

lol

[solar heron]

lol well hope that helps

[sick seal]

thank you mate

[sick seal]

ayy yo, back at it again but with a box this time, im told to find the domain and range as intervals, but when i think of domain and range i think of x and y. also
not sure how to find the intervals of a function.

my best guess for this would be that i use the length and with to find the domain, (-oo,5)U(5,oo), but just looking at that i can tell thats not the right answer

i guess im having a problem understanding what the domain and range represent in this function.

[willow bear]

the domain is the set of all values of x which make sense in the context of this problem

for example, x=1,000,000 does not make sense because you can't cut squares that large out of a 5 by 7 rectangle at every corner

[sick seal]

OH

[willow bear]

so this will have to be excluded from your domain

[sick seal]

i get it now

so domain would look like (0,5) and range would be (min volume, max volume)

or am i still missunderstanding?

bc im assuming the range would be the volume

[willow bear]

no, the domain would be (0, 5/2)

the range would be (0, max possible volume]

[sick seal]

oh yeah theres 2

i see. that clears that up

thanks mate

[viscid thistle]

lmfao

my sister's precalc class

makes me mad

[sick seal]

what am i missing here, there must be a key point i didnt pick up because i was willing to put my life on the line that these were the right answers.

[uncut mulch]

you didn't fully read the instructions

actually its a bit unclear

but i think they just something like
f(4) for a

[sick seal]

i should probably ad that this is at the bottom of the page

[uncut mulch]

f(r) represents the area for radius r cm,
f(4) will be the area for radius 4 cm

which is the function representation of it

[sick seal]

OOOOOH because f(4) is the area and 4 is radius

isee

lemme try that rq

dingdingding

that makes sense. i also got a wacky question i know ill need help with coming up in a few.

wait i applied the same thing and its all clear now

thanks man

[viscid thistle]

i need help, ping me when available

[viscid thistle]

Just post your question

nvm its g

[vapid folio]

has anyone here heard of anything called "cho-cho-chal"?

[willow bear]

what

[vapid folio]

my teacher keeps saying it. no clue what it means

[viscid thistle]

are you in US?

[vapid folio]

yeah

[willow bear]

any context?

[paper plume]

uh quick question, when asked to find all values where a function is continuous, is this the same as finding the domain?

[patent beacon]

No, since something like
f(x) = 0 for x ≤ 0
f(x) = 1 for x > 0
Has a domain R, but is not continuous at x = 0

,w graph step function

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[patent beacon]

As you can see the blue line jumps there

[paper plume]

ah i see, that makes a lot of sense

thank you

[quaint mason]

tryna find the exponential function.

[opaque idol]

Can someone help please

[elder charm]

,rotate -90

[elder charm]

... just substitute? Or do you mean 8?

[quaint mason]

reptile

how do u set up exponential function for this

??

[proud raven]

look up exponential decay equation

[quaint mason]

uhuh so like P(1-n)^x

[proud raven]

usually $A_0 e^{rt}$

[obsidian monolith]

jan Niku:

[quaint mason]

rt?

[proud raven]

A_0 comes from t=0

yea, rate times time

so rate is a constant

t is independent variable

[quaint mason]

i thought that

[proud raven]

first solve for A_0 where t=0

[quaint mason]

exponential equation was like

[opaque idol]

,rotate -90
@elder charm wait what do u mean?

[quaint mason]

y=ab^x

[proud raven]

if its a differential form its uhh

y'=ay

but im assuming precalc so the other form

a_0 comes from intial condition (here altitude 0) and then solve for r using any other point

[quaint mason]

what a_0?

the initial point?

[proud raven]

so instead of t use x

let x be altitude

so where altitude is 0

A_0e^(rt) is just A_0, right?

since its e to the 0

so you can solve for that constant since youre given pressure at x=0

[quaint mason]

so e=exponential right?

[proud raven]

sorta

[quaint mason]

well like

um

[proud raven]

its more the variable in the exponent

that creates the behavior

(e to the x)

no reason you have touse e as the base

its just convention, among other reasons youll see later

[quaint mason]

ah i already know

so like percentage growth and percentage decay

where its like 1-n for decay and 1+n for percentage growth and thats basically what e is

[proud raven]

ohhhh

[quaint mason]

for example

if u wanted a idk

[proud raven]

yea, i knowwhat youre talking aboue

[random cloud]

Guys I need help

[quaint mason]

5% increase annually

ye

[proud raven]

thats compound interest

exponential growth is just continuous compound interest

sorta

theyre really similar, exponential growth is easier to solve for and deal with i think

[quaint mason]

bcuz it doesnt stop regardless of what number u enter?

for example

16(1.007)^t

[proud raven]

more uhh, the growth is happening at every instant

[quaint mason]

u can enter whatever value for t

[proud raven]

instead of like, once a month

[quaint mason]

say t=time

[proud raven]

yea i think i get you

so exponential is no 1-n or any of that

simpler form

[opaque idol]

Can someone help?

[quaint mason]

what i was thinkin when solving for this

cant i plug it into my calculator and do like i believe it was...ExpReg?

and there get equation for it

[random cloud]

Is this channel ever quiet for a hot moment

[quaint mason]

and then plug 7000 in

[proud raven]

they tell you not to use regression

but you can

[quaint mason]

ah shoot

ugh

[proud raven]

or solve for constants

which is easy

[quaint mason]

sup @random cloud

[proud raven]

'fitting' i think

[quaint mason]

fitting?

[proud raven]

yea

[random cloud]

I need to factor this

[proud raven]

idk thats what id call it

[random cloud]

The answer on the bottom is wrong

[proud raven]

@random cloud difference of squares :p

[random cloud]

So I decided to do algebraic long division to find out

[quaint mason]

@opaque idol cant u plug in 4 for m?

[random cloud]

And it didnt work out

@proud raven BRUH

[opaque idol]

But how about the T @quaint mason

[random cloud]

And I thought I was being clever

Fuck me

[quaint mason]

substitute that in for t @opaque idol

[proud raven]

x^6-64=(x^3)^2-8²=...