precalculus

[frosty folio]

Hey I have a question if anyone's free to help

I'm trying to find the area of a quadrilateral with missing sides

[hallow sluice]

Yeah same here. Being self taught is fun but often gets rocky at times

[civic moon]

Oh well do yk the expansion of (x+h)^n

[gleaming walrus]

yeah that's what I mean

this problem doesn't required you to know a lot about limits or derivatives or whatnot

if you know the needed algebra theorems inside, you can solve it

[civic moon]

Uhm have u heard of
$(x+h)^n = x^n + \sum\limits_{k=1}^{n} \binom{n}{k} x^{n-k} h^k$

[obsidian monolith]

Scholar Of Mysteries

[civic moon]

@gleaming walrus

[gleaming walrus]

yes, I learned it to solve that problem

[civic moon]

Yea so what do u need help with? 😭

[gleaming walrus]

I don't need help 😭

I was talking in the discussion channel but I can't send images there

[civic moon]

Oh

Alr mb!

[median vapor]

what is that Man

[orchid urchin]

binomial theorem

very usefukl

[bronze geyser]

what the helly

too many unknown signs and their uses

https://klipy.com/gifs/its-time-to-stop-7

[thorny remnant]

Ohh I love these, though I'm not always good with them, idk why.

[tropic river]

Why is this incorrect?

I used :

https://cdn.discordapp.com/attachments/779196945448304640/1364708570355990538/togif.gif

[orchid urchin]

the 6x - 3 should be a 6x + 3

[tropic river]

why?

I'm 🕊️

[orchid urchin]

I am not too sure what method you used but I divided the fraction, or the first term by x^2 on top and bottom and got (12x + 6 - 3/x^2)/(2 + 7/x^2) now the -3/x^2 and 7/x^2 as x approaches infinity reaches zero so those are negligble giving you (12x + 6)/2 which is just 6x + 3

[tropic river]

I used ruffini

12x3+6x2-3 = 3 . (2x-1) ( 2x^2+2x+1)

and

[tropic river]

( 2x^2+2x+1)/ 2x^2+7

[orchid urchin]

that doens't equate though

[tropic river]

[orchid urchin]

h wait it does

didn't see the 3

[tropic river]

k

[tropic river]

and I used this

[tropic river]

lim x -> inf

2/2

so

[tropic river]

3 . (2x-1)

x --> inf

[tropic river]

[orchid urchin]

your work makes sense but your taking the limit of not just a fraction but 3 things, namely the fraction, cube root and the 7x

[tropic river]

hm

[orchid urchin]

all terms must be compared together before concluding anything

[tropic river]

isn't it x. 1

?

[orchid urchin]

the dominant behavior of this cube root is x + 1

[tropic river]

why why

[orchid urchin]

giving (6x + 3) + (x + 1) - 7x

[tropic river]

[orchid urchin]

that gives 7x + 4 - 7x which is just 4

depending on how u do it if you get a 1/3x^2 lets say as x approaches infinity that fraction will be 0

so it won't even matter you can just get rid of it

[tropic river]

but i have 6x-3 instead 6x+3

[orchid urchin]

try factoring out a x^3 inside the original cube root, which does give x + cbrt(1 + 3/x + 1/x^3), and the idea here is that the cuberoot of (1 + 3/x + 1/x^3) approaches 1 + 1/3 * (3/x + 1/x^3), which after multiply out and taking the lim gives x + 1

[orchid urchin]

you probably got that because you tried to cancel something you can't

like the quadratic with the denominator

[orchid urchin]

oh I see you did do that, then you should have gotten x * (something small), which is x * cuberoot(1 + tiny) now you can use linearization which I did use to figure out what that cube root would end up being, but there is an algebraic way if u want to know

its just painful

[tropic river]

I have exam of calculus tomorrow

i'm 🕊️

[orchid urchin]

study up on linearization man

or actually just those type of cube root limits

[dark hollow]

is ap precalc worth or shld i jst take normal precalc

i have the opportunity to take precalc in 10th and take ab/bc in 11th

[round geyser]

How would we solve this

[undone pumice]

I’m in AP rn I mean I think it’s a good bridge to take ap calc and get into ap exam math structures with frqs and all. and more rigor to colleges/weighting possibly

[echo sinew]

Take square root to both sides (you can do it without worries because everything is non negative)

[round geyser]

Btw

I remembered a^4-b^4

That’s what I got

[neat ledge]

you can do the fourth power square root on both sides so you get 2x+1 = |3x+4| and that working out is quite easy

i checked on my calculator and that should work

[round geyser]

Yep

[neat ledge]

Looks like a fun one

[high sequoia]

U need help with this or just answer?

[warm ibex]

Guys in a differential equation, What exactly is x and y?

y is a function of x right? so like y=y(x) or h(x) whatnot. And x in an differential equation is the variable, input for the function y.
and dy/dx is just the y derivative, y'(x).

and in the image when we wrote g(y) this y in g(y) is the y in the dy/dx? so g(y) is a function which has an input of y=y(x)/h(x), and f(x) is just a function which has the input x, which is the same input y depends on?

um sorry if i dont make sense at all.

[spark anchor]

2

[round geyser]

I get bc 4 is 2^2 and 8 is 2^3

[neat ledge]

it took me like 3 pages and 5 tries cause i tried stupid shit but i got it

[round geyser]

I used nth root of(x^m) = x^m/n

And get same base

Then that cancels and left with 2x-1=x+1

Subtract over to get x=2

[neat ledge]

i simplified to (2^4x-2)^3=2^3x+3)^2 then just work it out to 2^12x-6=2^6x+6 and then you remove the 2 and simplify

[round geyser]

Yea

That works

[rough leaf]

Bro

Is ts your TikTok account😭😭

[pure sage]

hey guys im new and i really need help with calculus...can anyone help me...

[valid topaz]

#calculus

[exotic barn]

You are free to ask your questions in #calculus like plens suggested ^^

[bronze geyser]

looks like it

it has the same hieroglyphs on it

[round geyser]

Yea why?

I go to all the mathematicians and place requests

[round geyser]

Is this what u said “if u want all solutions even complex then let LHS=a⁴,RHS=b⁴, so u get a⁴-b⁴=0, (a-b)(a+b)(a²+b²)=0, so a=b,a=-b and the last is just the quadratic so finding ex is trivial”

[sly narwhal]

Anyone know a place to find a good and quick explanation of why the properties of conics hold. All the hs curriculums just dumb it down and don’t explain crap and I’d rather not read 40 pages for this.

I mean like all of the connections between all the ways to define conic sections and why cones connect to general 2nd degree equations. Figured out a lot on my own but I’m bound to miss some interesting things I’d rather not misss

[round geyser]

Were u in NotMathClub’s live

[orchid urchin]

x and y are nothing special, dy is just a little bit of y (lets say y is an hour a tiny bit of an hour would be 1 second for example), and dx is a little bit of x, so the ratio dy/dx is simply saying how much does y change when we change x (dx) by a little bit. Speaking precisely if you have x, then dx is the derivative of x, same with the y. In the image you show its a bit more complicated though because we have a function f, where x is the input and a function g where y is the input, and theres no derivate weird stuff happening there but we equate it to dy/dx. Depending on how far you are in calc you may realize you can seperate the variables and get (1/(g(y)) * dy = f(x)dx which you get by just algebraic manipulation, and the right side is the derivative of f(x) and the left side is the derivative of 1/g(y), or the reciprocal of g(y).

dy/dx is read as "the derivative of y with respect to x"

now you would want to integrate after rearranging

you typically do differential equations after completing calc 2, or 3 as well

and next time use #calculus

[exotic barn]

<@&268886789983436800>

[sly narwhal]

No, haven’t heard of it

[round geyser]

Ok

[rough leaf]

Nah I’m js asking cuz I saw it

Anf it looked familiar

[rough leaf]

Nope

[round geyser]

Okay

[round geyser]

But anyway

Come pop up his live

Really fun

[sly narwhal]

OH THAT GUY. I don’t think I was in any livestreams but I’ve commented on a few shorts of his

[round geyser]

Yep

[bronze geyser]

im first

[distant kelp]

@gentle vault Hi

[gentle vault]

Hi

[viscid thistle]

Hii can someone help me with this? Im not sure if this is correct

[uncut mulch]

Is incorrect

why are you multiplying 3 to the 8x

what is the derivative of 8x

[viscid thistle]

I thought thats what your supposed to do i think i got it mixed up

Im new to this stuff i just started learning today im learning by myself i dont go to school 💔

[viscid thistle]

you guys got any youtube video recommendations for this one?

[velvet wyvern]

,rccw

[uncut mulch]

linearity and
power rule
check out stuff from organic chem tutor

also use your space,
write stuff on new lines

+- can also be simplfied down to -
making the final result
9x^2 - 8

[velvet wyvern]

this one too https://youtube.com/@professorleonard?si=KEfipJGZKmpVYwOd

[viscid thistle]

ohhh okay i see thank you!!

[gentle vault]

organic chem tutor is goat

[tribal wind]

any good vid for continuity and differenciability,
and tricks for integration, highschool level to solve multiple choice questions?

[urban kernel]

Just differentiate the options if there is an MCQ for integration.

Far easier.

Also, continuity is not that hard imo.
A function is continuous if for all c in its domain, lim_{x->c} f(x) = f(c).

A function is differentiable at a point if it is continuous at it, and also smooth.
Extend it to all points for a differentiable function.

The smoothness is given by the limit of the difference quotinet.

[rugged sleet]

I see you mention 'smooth' but I think that this definition is a bit dangeroous to operate with. Would be better to prove that a function is differentiable at a certain point and hence continuous at it. Smoothness is just a vague term, in several cases you can't declare a curve to be smooth at certain point without checking if it is differentiable at it.

[urban kernel]

I see.

I will take better care.

[tribal wind]

ohhh thank youuu @urban kernel @rugged sleet 🛐🛐🛐

do u have any videos? a short one, just for explanation works

[urban kernel]

Khan Academy (Calculus AB/BC) Playlist is good imo for beginners.

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-4/v/differentiability

Remember, i am not recommending this for rigour.

It's for intuitive understanding.

[orchid cedar]

hmm, i dont know calc, but are u saying that the continuity of a function depends on how it changes its outputs as u change the input

[urban kernel]

No.

[orchid cedar]

like when your input approaches a value, the function's output also reflects that

its output gets closer to the output of f(that value)?/

[urban kernel]

Yeah..., you can say that

[orchid cedar]

why though?

why does continuity depend on that?/

like isn't it obvious?

does every function not support that?

why?

[rugged sleet]

A function is continuous at x=c if it attains the same value at every point in the neighbourhood of c.

[orchid cedar]

ahh ok

[rugged sleet]

Piecewise functions don't.

[orchid cedar]

but if it breaks, there will be atleast one value that makes the function undefined

[rugged sleet]

Check this example:
f(x) = x^2 if x>=2 and f(x)=2x-1 if x<2

To determine if f is continuous at x=2, you will have to check if lim x->2 f(x) = f(2)

[orchid cedar]

hmm

ok

what about the second function

[rugged sleet]

So, for the provided function, can you comment on its continuity at x=2

[orchid cedar]

oh, nvm

ok

[orchid cedar]

its the same func srry

[rugged sleet]

f is the same function defined differently for different points of domain

[orchid cedar]

yea yeah srry

hmm ok

[rugged sleet]

I hope you know how to check if limit of a function exists at a point?

[orchid cedar]

uhh....... about that

[orchid cedar]

you make dx, smaller and smaller right?

[rugged sleet]

if lim f(x) exists as x->c then this applies that its right and left hand limit exist and are equal.

[orchid cedar]

hmm

wait, how do u know which direction to approach fro

m

do u do both

what if the limit exists on one side

what would that entail.

it would be dis-continuos right

[velvet wyvern]

what do you mean

[orchid cedar]

like x + dx or x- dx

or is that not how that works?

[velvet wyvern]

you take both limits

[orchid cedar]

ok

but : "You must take both limits"?

otherwise, it would be incomplete

right?

[velvet wyvern]

elaborate

[orchid cedar]

i digress

[velvet wyvern]

?

[orchid cedar]

what i mean is that, if we take the limit from one side, that does not like fully define the function cuz it may be dis-continuos from the other side

[velvet wyvern]

for a function to be continuous at a point, the left hand limit and right hand limit as it approaches that point should be same and equal to the value of function at that point

[orchid cedar]

yes, so one must approach from both sides, before declaring that its continuous

[velvet wyvern]

right

[orchid cedar]

thanks!

[hardy venture]

$\sin^2(x)+\tan^2(x)=\cos^2(x)-1$

[obsidian monolith]

YeetusDeletus5

[round geyser]

x=?

[valid topaz]

newton's method?

i don't rly see a nice way to do this

[nimble wolf]

try rrt first

[round geyser]

I did

I get P/Q is 1,1/2,1/4,3,3/2,3/4

6x2=12 possible solutions

[valid topaz]

wait ||3|| is a solu

[nimble wolf]

-1 works

[round geyser]

x=3,-1,-1/2

[valid topaz]

hmmmmm

oh i was being dumb

[round geyser]

Two repeated roots; -1 and -1/2

[nimble wolf]

yea

[round geyser]

According to desmos

When I did rtt, my teacher let us use desmos even on quizzes and tests

[foggy rampart]

There is a freshman in my pre-cal class

I am so jealous of him

[heavy ridge]

im new here what do i do

[hushed sphinx]

That depends on what you want to achieve.

[surreal iron]

i know someone taking a2 in middle school

[undone pumice]

im a freshman in precal

ik a few other kids in my school doing it (like... 3, 4 that i know of)

[foggy rampart]

I wish i was

[undone pumice]

🥀

[foggy rampart]

How i be feeling

No one at my school likes math like i do

[undone pumice]

damn

[round geyser]

I’m a senior

[undone pumice]

ik u told me before

[round geyser]

Yep

[novel lantern]

Where do I go from here?

[exotic barn]

,rccw

[exotic barn]

I would go back to line 1 and instead bring -log|x| to the left

and then apply on both sides the exp func

[novel lantern]

That fixed that. Thanks!

Went ahead and did that @exotic barn still kinda rough on this

[exotic barn]

Apply e^x on both sides

bryh

[novel lantern]

Why would applying e^x to both sides work in this case?

[exotic barn]

The roots of the quadratic are 0 and 10 and since the leading coeffcient is positive, it's open upwards, so within the roots it would be negative

because e^x is the inverse of ln (I am assuming log means here with base e)

but you can take any base

[novel lantern]

Right… Knew that just need to make cheat sheets.

[novel lantern]

Thank you for your help!

[exotic barn]

Np!

[hallow flicker]

what does exp mean here

[uncut mulch]

exponential function

e^

[hallow flicker]

oh

[orchid urchin]

can you show me more of this problem

specifically the base of the log

and when u write a straight bar is that abs value?

[orchid urchin]

ah nvm this was yesterday, just take away that e^x is specifically the inverse of ln(x), NOT log, but in your problem assuming the straight bars are abs symbols then you can just change log(24) and change it to log(24/|x|) giving you:

log(|x - 10|) = log(24/|x|)
And since there is no notation provided for the base it is assumed to be 10, with that being said they have the same base here so you can remove the logs giving you |x - 10| = 24/|x| where x cannot be 0 (otherwise division by zero).
Solve to get:
|x(x-10)| = 24
And split into cases:
x(x-10) = 24
x(x-10) = -24
You should know what to do from there

Also for the step where you just remove the logs, note that a log is injective making it one to one, and if you want more intuitive reasoning think of the situation where you raise a constant b to the power of log_{b}(x), which is just x, and instead of applying "e^x" you would just apply the base of the log to get rid of them.

[orchid urchin]

It's important to note that because of what @exotic barn said here about e^x is the inverse of ln, and regardless they have the same base meaning its okay to remove the logs no matter the assumed base

[obsidian slate]

easy question to try removed

[gentle vault]

um

Isnt that meant to be in #calculus ?

Idk tho

[obsidian slate]

oh it wasnt on my channel list for some reason

[echo sinew]

!status

[round geyser]

Do I have to show steps

[echo sinew]

I guess so

But what's the issue with this exercise? Which part are you having trouble with?

[dry verge]

Did Khan Academy mess up?

[raw hill]

where did x go

[dry verge]

Oh wait, I realized that they talked about the co-vertices and not the vertices.

The answer isn't an imaginary number, whoops.

[round geyser]

Reasoning

The first method is synthetic divison

I think

Then I got R(x)=6

Will it be (x+1)(x-4)+6

Since we know (x+1)=0 then
0(x-4)+6

Then it’s 6

[uncut mulch]

what does this have to your previous question

also this isn't what the question is asking for

[round geyser]

This one

If it that I have to solve the first requirement

To get the second

[uncut mulch]

but what does that have to do with the f(x+1) = x^2 - 3x + 2 image you posted

[round geyser]

That was the second one while waiting for them to respond

To the first

[uncut mulch]

they responded
they were waiting for you to say which part you had trouble with

[round geyser]

Then I did

It was the reasoning

The wordings are confusing to me

[uncut mulch]

did you do part i)

[round geyser]

I thought I have to do a then b/i

a is easy

[uncut mulch]

i mean b i),
there's only one i) there so i ditched the a

[round geyser]

It’s subtracting the “log_2 x^2

[round geyser]

Okay

[round geyser]

Do we just use the given and solve or I have to proof the first equation to a

[uncut mulch]

wdym

you said a) was easy, so i'm assuming you did that already

b) i) is a continuation of that

solving the equation you just got

[round geyser]

Okay

First one was condensing

[round geyser]

That’s what I got

[uncut mulch]

continue solving, you're not done with that yet

you've only factorised,
haven't explicitly stated the roots yet

[round geyser]

We get x=-5/3 and 6

[round geyser]

I see that -5/3 doesn’t work bc

log (-x)= undefined

[uncut mulch]

poor wording

log(negative value) is undefined.

[round geyser]

When I plugged in -5/3

I get 2 + 2 log (-5/3)

On RHS

Which is undefined

[uncut mulch]

yes

[round geyser]

Like this

[round geyser]

^

I assume x=6 works

[uncut mulch]

didn't really need to eval the whole thing,

just starting domain and/or log(-5/3) is undefined is sufficient

[round geyser]

Guess what 6 works

[uncut mulch]

you've also dropped the 2 in front of the log on the right side when plugging your values in

and the way you're evaluating when x=-5/3 is also bad

because you're essentially doing algebra that ignores the restriction that disqualifies it from being a solution

[round geyser]

Okay

What would the domain be

[uncut mulch]

intersection of all arguments of the logs being greater than 0

[round geyser]

Bc when I saw the graph

They touch at (1,0)

Do I write x>0

[uncut mulch]

x>0 would be the domain, yes

the way you're evaulating it

,w log(2,(-5/3+3)(-5/3+10)) = 2 + log(2,(-5/3)^2)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by ramonov

[uncut mulch]

would actually be true

[orchid urchin]

hey -5/3 doesn't work tho

because the origional equation on the RHS is not squared

[uncut mulch]

yes..

that's the point i'm making

[orchid urchin]

oh okay

[orchid urchin]

also do u say your pau part as ram?

[uncut mulch]

yes

first letter is the greek rho

[orchid urchin]

thats cool

[young spire]

@tired garden
hello, can i ask you a question?

[round geyser]

Okay

Is there a formula that has that Greek letter that measures density

[echo sinew]

λ: line density
σ: surface density
ρ: volumic density

These are usually used in physics/engineering

I remember them appearing on electrostatics, and also a little bit in mechanics

[round geyser]

p=mv

[echo sinew]

Huh, that's not a density

That's linear momentum

(Sometimes I've found it called q)

[echo sinew]

And this doesn't have greek letters

[round geyser]

With has volume

ρ= something/volume

You know what I mean

[echo sinew]

Now yes 😅

[round geyser]

What is it

[echo sinew]

ρ = mass/volume

[round geyser]

Yess

[echo sinew]

But often this is simply called d, not ρ
(at least in my country)

[round geyser]

g/cm^3

[round geyser]

That’s the unit for density

[round geyser]

But d=vt is distance

[echo sinew]

That's just one of infinite units

[echo sinew]

??? No 🙈

[round geyser]

Distance = speed x time

[echo sinew]

Oh yeah you've changed it my bad

That's only for uniform rectilinear motion though

[round geyser]

It’s just (mass’s unit) per (v^3)

Is this right

[echo sinew]

It's simply "whatever unit for mass" / "whatever unit for volume"

[round geyser]

Okay

[echo sinew]

The international systems uses kg/m³, but often you can encounter g/cm³, g/L, mg/mL and so on and so forth

[round geyser]

Okay

[echo sinew]

But this is physics, not calculus

[round geyser]

We can convert them easily if I remeber

Okay

[echo sinew]

Of course

[round geyser]

Okay

[round geyser]

Is centi means 100

Ik cent is

[echo sinew]

"centi" means a hundredth (aka 1/100 or 0.01)

[round geyser]

Ok

[true geode]

jeez, vary your responses at least

you come across as a redstone repeater, just outputting the same thing over and over again

...yes I know that's not technically what they do just deal with it

[orchid urchin]

dayum

[winged nexus]

he seems interactive enough to me

[dire aspen]

okay

[tired garden]

sup

[round geyser]

Who

[stark solstice]

fk precalc

[amber narwhal]

nooo

[orchid urchin]

i think someone had a rough study session

[orchid urchin]

any1 know the most elegant part of precalc?

[round geyser]

What

[low yoke]

Wdym

[echo sinew]

Indeed, how do you measure elegance? 😅

[smoky schooner]

Guys i made fun notes of calc like the idea of it in a fun way does anyone want to have a grasp

[inner moon]

I'd like to see

[round geyser]

Is bearing same as angles

[uncut mulch]

Not quite

bearings use angles to describe direction

[cerulean phoenix]

Can somebody explain that question marked line only?

ignore the language cuz there is no english translation for me

[potent slate]

It’s upper limit - lower limit

At x = 5 (upper limit) it will be
(5sqrt(25 - 25))/2 + 25/2 sin^(-1) (1) = 0 + 25/2 sin^(-1) (1)

At x = 3 (lower limit) it will be (3sqrt(25 - 9))/2 + 25/2 sin^(-1) (3/5)

[atomic violet]

ive done too many area under the curve to forget ab this 😭 thermo fluids is killing mee

i know of exponent rules like 1/x becomes x^-1

however how do i integrate that exactly? you add 1 to the power and divide by the new power

am i missing something there

you cant divide by zero

[smoky schooner]

Am a send u on dm later just leave me a reminder cuz am busy rn so sry

[cerulean phoenix]

alright understood

[willow wraith]

Integral of 1/x is not the same as normal integration

It’s ln|x|

[thorny lily]

hello guys, I’m new here and just starting out with calculus. Could anyone suggest some good resources to help me get started?

[willow wraith]

looking into conceptual problems first

usually starting with physics actually

[thorny lily]

which physics problems do you think are best for beginners to grasp the core concepts???

any resources you’d suggest for learning?/

[willow wraith]

classical mechanics problems will may help you learn WAY more why calculus is needed and you’ll start to see relationships that kind of usher in calculus

[valid topaz]

<@&268886789983436800>

[proven pagoda]

i get into this chat and i feel like "dang, ap precalculus failed me 💔"

[hushed sphinx]

There are often people who for some reason or another try to talk about calculus questions here instead of in #calculus.

So don't assume everything being spoken about here is something a precalculus class should have taught you.

[proven pagoda]

that channel isn't even showing up on my server channel list, i guess I didn't dd all of them lol

[proven pagoda]

like this, i'm just confused on the syntax lol

[hushed sphinx]

Yeah, the server uses a Discord feature that aims at not overwhelming newcommers with a bazilion channels, but it's somewhat wobbly. You can right click the server name (above the channel list) and click "show all channels" in the menu.

[proven pagoda]

yes yes i did

lol i thought i did that already

[hushed sphinx]

Is it writing e^blah as exp(blah) that confuses you? Or leaving out the parentheses?

[proven pagoda]

yeah the exp and my curriculum never used abs value

and the arrow is that just like simplification or

i can see the math it's just stuff ive never approached i guess

[potent slate]

The arrow is an implication arrow
A ⇒ B (B follows from A, or A implies B)

[potent slate]

Well, do you know what modulus (absolute value) is?

[proven pagoda]

like do i know what absolute value is? yes, i've just never used it graphically and i haven't done any solving with it since 9th grade lol

[potent slate]

As you can see, the output for a modulus is always non-negative

[proven pagoda]

so can an absolute value's function never have a y value of -a?

[potent slate]

^

[proven pagoda]

alright

what cabn transformations look like

[fossil kindle]

when is |x| even used

alg 2?

[proven pagoda]

it was introduced to me in alg 1

and i think i used it in geometry a little bit in real world problems?

[fossil kindle]

they introduce it but i dont think they actually do anything w/ it

[proven pagoda]

idrk i don't remember lol

[fossil kindle]

besides distance but thats not really using the parent function tbh

[proven pagoda]

all it really is is just the distance from 0 so

[potent slate]

f(x) = a|x - h| + k
h : horizontal shift
k : vertical shift
a : vertical compression or stretch
if a is negative, the graph reflects about the x axis

[proven pagoda]

and then b(x-h) i'm guessing is horizontal compression or stretch?

or slope

lol

[proven pagoda]

but this doesn't matter cause you can just multiply the number by a so don't mind me 🙏

[orchid urchin]

its a subjective question, which part of precalc delivers maximum results with minimal, surprisingly simple formulas or methods, like eulers identity.

[novel lantern]

Getting stuck again. Simple algebra failing me

Nvm forgot can factor it

[viscid thistle]

Istg if I have to do this STUPID axes rotation by angle transformation again in calc I will die

Actual nightmare

[raw hill]

Me when eigenvalues of $\begin{bmatrix} a & b/2 \ b/2 & c \end{bmatrix}$

[obsidian monolith]

Civil Service Pigeon

[viscid thistle]

Anddd there’s a 10 second method

25 minutes of my life down the drain

[orchid urchin]

what u typin

[loud fossil]

me?

[orchid urchin]

@loud fossil

[loud fossil]

-.-

[orchid urchin]

I mean getting help doesn't mean your bad at math

just means your improving

[smoky schooner]

I hope this will help,p

So let me wrap up everything I said. So basically the derivative and the limit is is like something that is widely misunderstood upon others. For example, the derivative is mostly known as the accurate change like the accurate change that's not like like the slope. But actually the derivative is how much the velocity or the thing we're measuring changes as we as it approaches and this approaches zero meaning it's more than the accurate change. It's something related to the physics behind of the thing and not only the numbers. Like it's not just the slope it's literally the physics everything behind it. For example, so we have a car. So the car is basically transferring energy from chemical to mechanical to kinetic. The chemical part is when the car is not even moving is when the car is literally at zero zero constant velocity or the unit we're measuring it in and as it moves and this approaches zero when this mechanical and chemical energy turns into kinetic it increases and gets away from the zero. But as the friction of the ground acts as a external nerving uh force that's like how Newton Law said that any force second law actually that any force acted by another force is the only way the force can change. It decreases and makes it approach zero. So when we solve something like 3x^2 the derivative is basically when

we manage the power we're basically measuring how much it approaches zero and then when we are multiplying the constant it's basically how much like it's getting away of zero. So it's basically widely just connected to physics more than just numbers and abstract proofs. It's basically the way we should look at look in it when we actually solving real life problems instead of just looking for how much is the rate of change. We should look of how much does the thing change as it approaches and this approaches zero. And this gives us more accurate and more reliable answer that can be used for any operation then if we're using it for only the slope we're basically just like we're basically just like um finding an answer for the temporary state. And when we say it's an accurate estimation we're just saying an accurate estimation for for example a two points and then that we combine but it's actually a way to solve this problem forever like without actually having to measure every single part of the of the graph. So the misconception is that if the thing that we actually believe is that derivatives that means we need to actually measure every single small part while we're only measuring how much the thing changes as it approaches and this approaches zero. And when we zoom in into something we're just seeing the little slope of the thing instead of seeing how much is the change over certain time for a more reliable answer we use the derivative that is actually the real versus the misconception which is just a dumb idea of what we see that's only used for hard like thinking cuz if what it was actually true we'd sit years trying to find the derivative. And basically the the thing changes as the the the the certain type of graph changes is what changes the way and method we use for example power rule, chain rule, etc. And this gives us an accurate estimation that helps us understand everything very well. I hope you enjoyed what I said.

[jagged lintel]

do you live in breaking bad lmao

why is your room so orange

[fossil kindle]

its prolly a light

[crimson onyx]

I’m an algebra 2 student and I was curious what’s ahead, I looked at this, and ngl im kinda scared now..

[novel lantern]

lol it’s a reduced blue light led

[willow wraith]

derivative is a graph of instantaneous slope

you’re overcomplicating it a lot

In physics for position functions your input is usually a time

so the derivative of that position function throughout a time interval would just be dp/dt

meters / second

which is just velocity

Velocity with respect to a time is acceleration

and instantaneous slope is just the slope over an extremely small interval as the length of that interval approaches a finitely small number

So just this

[smoky schooner]

What ur saying is technically kinda the same as what i said

[smoky schooner]

Thats what i said

But am saying

How do we find this conctusie sle

Slope(

[smoky schooner]

This is not what ur gonna take its just a theory

[willow wraith]

<@&268886789983436800>

[dry verge]

Did I do something wrong here?

[rough leaf]

I don’t think so

Qait

It’s gonna be 13sqrt2/2 - 13sqrt2/2 i

[crimson onyx]

Oh ok, thanks phew

[dry verge]

Ohhh, that's how you do it.

Thank you so much! That was much simpler than I thought it was going to be.

[viscid thistle]

DONE WITH CONICS YAYAYYAYAY I HOPE I NEVER SEE THESE EVER AGAIN OMLLLL time for sequences and series hope this ain't harder😛

[undone pumice]

gulp wow

[viscid thistle]

once I do s&s I will be DONE with precalc🎉

[undone pumice]

🙏

what is s&s?

[viscid thistle]

Sequences and series

[undone pumice]

ohh

[viscid thistle]

I acronymed it cuz I said it msg before

[undone pumice]

ah

oops i can't read fah

[viscid thistle]

how hard would you guys say sequences and series is compared to conics and paras

[orchid urchin]

sequences and series can be so much harder, oh boy the sigma notation, and learning how double sums work, and triple sums, and recursive sums, and deriving all the equations for a{n} and S{s} and then even more, its a lot, way harder than conics imo.

[orchid urchin]

its so much more difficult

although if your not self study it will probably be around the same difficulty

[viscid thistle]

oh great 😭

im self studying so im cooked lol

[orchid urchin]

im self study so I had to learn it in so much depth, and rederive where all the formulas come from and develop new techniques to invent new ones

because you have to learn all the basics which go by fast, for the single sum, and nail the notation down with practice and the rules for it and why they work.

and then move onto double sums

then triple

until you get the pattern and you can do quadruple on your own

thats how I did it atleast

and then taking a sequence of numbers and thinking about it as a function and finding first differences, second, and third, and so on

and then finding the degree of the polynomial based off of that

for f(x)

and rederiving the equation using the form Ax^3 + Bx^2 + Cx + D for example

and you would do so by setting up a system of equations and setting the polynomial to the sum of the first term, first two terms, first three terms, and first four

so @viscid thistle it can get kinda crazy if you mean to go deep

definitely worth it though because now I look at it and be like, oh this stuff is easy

as when I first saw it I couldnt do it at all

[viscid thistle]

alr, wish me luck then😭 so far I litteraly just started 2 minutes ago so

imma stuydy for 4 hours straight

[orchid urchin]

the start should be rather easy, the formulas for a linear sequence are just simple and the sum is simple as well and creative depending on how you want to intepret it if you do research that, same for geometric, once u get into sigma notation and that type of stuff though then you will be able to invent those types of formulas

the weirdest thing I do till this day is factor out a sigma

[orchid urchin]

this basically

since the sigma literally just means add and its all linear u can just do that

[viscid thistle]

so if the sum of x's was 10 then would it be a(10) +b(10) = (a+b)(10)?

[orchid urchin]

yeah

because if u set 10 to u so 10 = u you will have au + bu = u(a+b) @viscid thistle

thats one way u can think about it

[viscid thistle]

nice, thank you

[viscid thistle]

@orchid urchin yo arithmetic sequences lowk rlly fun

nvm just sequences

im aboutta do arithmetic

[viscid thistle]

wait I think I can do this now

let me try it

im not sure if I got this right at all since I litteraly had no idea what was going on so I made up a thing on how to do it lol

51?

[raw hill]

,w \sum^{3}{i=1} \sum^{4}{j=2} (i+j)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[viscid thistle]

whooop

[raw hill]

You seem to be off by $6=1+2+3$, so maybe recheck somewhere there

[obsidian monolith]

Civil Service Pigeon

[orchid urchin]

oh also ill mention if u can do sequences and series on a good level u can begin discrete calculus

which is lowk easy

and calculus too if u can do that

[orchid urchin]

Now try to find an explicit formula for the sum of (-2)^i from i = 0 to n

i personally would recommend not using the geoemetric series formula

and learn about telescoping series to solve that

[smoky schooner]

Its also wrong to so dw

[gentle vault]

$pi$

[obsidian monolith]

🗣

[gentle vault]

Uh idk

[crimson onyx]

Wat

[fossil kindle]

3.14sqrt(-1) or psqrt(-1)

[silver robin]

Can someone help me with understanding the use of tangents and normals in conics?

[shell current]

The ones done were in class, didnt get it someone mind explaining 7, 8, 10, 12 my dms are open anytime.

[haughty crown]

THE ALGEBROS

[haughty crown]

Ok but do you want it explained here or in DM’s?

[shell current]

Im In precal honors bro

[shell current]

Sure dms would be great

[undone pumice]

Precal reviews a lot of algebra

[shell current]

ive noticed that, i was never great at algebra bcs i never focused in class lol

[haughty crown]

I sent you a message R

[undone pumice]

Oh… that’s… not great

[lunar pumice]

What do you think for 12)

[young spire]

hope you're doing well, here is the question:
i struggle to understand the change of base rule of logarithms. even though i made a proof for it myself. i understand the proofs but they don't really explain anything. they just show you why the rules work algebraically but i want to know the logic behind the rule.
i don't understand that why you get a completely different base when you divide two logs, with the same base.
i asked the same question when I tried to make sense of adding up rule. I asked the question of why you multiply the input values when you add up logarithms with the same base? How does this actually work when all you're doing is summing exponents up? and why doesn't this work when you add up logs with the different bases? I was desperate by answers, so i asked that question to ChatGPT and it told me this:
"You are trying to mix two different counting systems into one system."
that made kinda sense but my gut says that it didn't explain anything mathematically, thus this is not really reliable or understandable.
can you tell me the logic of base of logarithms?

there are literally too many rules in logarithms and it feels kind of frustrating to make sense of all them.
i could just use them as tools which will save me a lot of time and would be actually smart since I need to pass the universty entrance exam but i have a lots of time ahead of me plus that's not really how i study.

[hushed sphinx]

It's a bit hard to know what will help when you already have a proof and that doesn't convince your gut ...

But ... suppose the only way you know to calculate powers is b^y = exp(y·ln(b)).
Then, instead of looking for log_b(x) as a solution to b^y = x, you're solving x = exp(y·ln(b)) which is the same as ln(x) = y·ln(b).

[young spire]

would you like to learn my proof

[hushed sphinx]

Another way to look at it would be to say: Logarithm functions are just arbitrary multiples of ln -- after all, these all satisfy f(xy) = f(x)+f(y), by linearity.
And we define the "base" of one of these functions to mean the input that happens to map to 1.

[tired garden]

feel free to share it

also, are you comfortable with exponent rules? because if you are, log rules are really not far behind

[lunar pumice]

do you understand that logs are basically reverse exponents

[young spire]

i am

[young spire]

i do

[lunar pumice]

ok so as tropo said, try to turn your log rules to use exponents

like, what does log(xy) = log x + log y mean?
what does that look like if you had to write it with exponents instead

[lunar pumice]

a^??? = xy
a^? = x
a^? = y
a^??? = xy = a^? * a^? =
i know this is far from valid math but you would think about that rule like this

the ? is why logs are helpful

anyways with exponent rules you would have a^(? + ?) = a^??? = xy
so ??? = ? + ?

thats why a log xy = log x + log y

[young spire]

yeah, actually that's why these rules are called as "the properties of logarithms" because you can prove them by the very own definition of logarithm and exponent's basic properties.

loga(x)=y <=> a to the power y = x
i assume that all the properties are basically the interpretation of this definition

[lunar pumice]

well yeah log rules are just what happens when you use exponent rules

and write them with logs

[tired garden]

just checking, are you saying that angel's explanation is one you already know but doesn't satisfy you?

[hushed sphinx]

Yeah, that goes through so many steps that it's understandable if it doesn't produce much intuition. I would instead start just with the definition of log_b(a) try to derive the fraction at the end of the calculation:

log_b(a) is the solution to b^x = a.
But b is c^(log_c(b)), so this is the same as c^(log_c(b)·x) = a.
Now take log_c on both sides again, giving log_c(b)·x = log_c(a).
And now you only need to divide through by log_c(b) to isolate x.

[lapis shuttle]

sorry for intruding guys but uh... are ALL of these answer choices wrong or am i just not understanding

[young spire]

yeah

[lunar pumice]

ok i don't really understand what you are asking

you know log rules exist because they are just exponent rules written in a different way and you ask why that is?

[tired garden]

uhhh looks very wrong yeah, these look like 3-3sin t, not 3sin4t

[lapis shuttle]

ok thank you

I thought i was tweaking

[tired garden]

so would you like me to try explaining a different way, that helps visualize these properties intuitively?

is that your ask?

[young spire]

i think my problem is about the logic between the base

[tired garden]

you mean like the change of base stuff?

here's how i like to think about it:

but before i proceed, let me clarify some vocabulary to be rigorous

if we say "three fives make fifteen", the "make" here implies that we are adding the 5's together

i am going to use the same language, but i will use multiplication instead, since we are working with exponents and logs

so in this case it would be "three 5's make 125"

make sure to always think about multiplication as our "building action", not addition

when we say log_b c = n (aka b^n = c), we are essentially saying that an "n number of b's make c"

does this make sense so far?

to "visualize" it you can think of the b's as objects, grouped into boxes, with n of them in each box, and c is the value of the entire box

@young spire ping me if you're comfortable with this visualization and ill proceed

[young spire]

i was already comfortable with these actions

[tired garden]

just checking

ok so

lets say we have as in your proof log_c a and log_c b

picture a as a giant box, picture b as a small jar

log_c a is how many objects fill the box

log_c b is how many objects fill the jar

so log_c a / log_c b is basically the number of objects in the box, divided by the number of objects in the jar

what is log_b a? the number of jars that go into the box

[tired garden]

isnt that just this?

[young spire]

i didn't quite grasp the situation

chatgpt told me something similar to this, refering to the ratio

[tired garden]

what in particular do you struggle with? can you point to something more specific?

[young spire]

the thing that i struggle with is that how do we get to the base b

like, why does this ratio lead us to log_b(a)

[lunar pumice]

maybe write it as exponents

[tired garden]

lets try a specific example

so log_2 512 / log_2 8

imagine 2 as a ball

i can fit 9 balls into a box, which i will use to represent 512, as 2^9 = 512

similarly, i can fit 3 balls into a jar, so the jar is 8

log_2 512 is how many balls fit in the box

log_2 8 is how many balls fit in the jar

so what is log_8 512?

that would be the number of jars that fit in the box

does this help?

[young spire]

when you divide logs, dont you divide the exponents

i dont understand how the base just come in to the solution, i guess the base matters only if you try to write it as another log

[tired garden]

edit: ah, i think so? depends on what exactly you mean. that is what we are doing here in this example

do you at least follow my intuition

perhaps we can figure out how to explain away the confusion through that example

[young spire]

how did we directly come to this

the problem is here

[tired garden]

well, we said log_8 512

8 is the jar

512 is the box

since 8 x 8 x 8 = 512, 3 jars fit into the box

and that is log_8 512

so log_8 512, in our example, can be visualized as the number of jars that fit in the box

[young spire]

why do we find how many jars would fit in the box

[tired garden]

because the entire point of this was to help with visualizing the concept

you dont have to do this at all

[young spire]

no no, like

loga(x)/loga(y)=log?(???)

when i first saw the rule, i asked to myself that if i was finding this rule, what would i do to divide two logs? as it turned out, i was struggling to determine the base.
its easy to prove the rule when the base is determined and all you need to explain is why ... is equal to ...

[tired garden]

so we can either pretend we know the left side and figure out the right side or vice versa, which way do you wanna approach first?

[young spire]

we should try to find the left side (base) and pretend to know the right side (input)

[tired garden]

ok

so let's pretend that, once again, 8 represents the jar, and 512 represents the box

so log_8 512 represents the number of jars that fit in the box, which we know is 3

but let's say we don't measure this directly, we don't just shove the jars into the box and count

let's say we have to use some kind of measure, like volume

we say, for instance, the jar is 1 liter, the box is 3 liters, or something like that

except instead of measuring in liters, we measure in balls

what is a ball? doesn't matter, literally any amount

so hypothetically, if we set the ball to be 2, as in our previous example

[hushed sphinx]

I wonder if some of the problem might be that you expect the rule to work like "we must always rewrite this to that when we're doing calculations". That wouldn't be a helpful way to look at it; it is more "these two different calculations always yield the same result" -- a plain fact that you can then do with what you want. It can sometimes be useful to replace one with the other or vice versa when that happens to make something you want to do simpler.

[tired garden]

we can find log_8 512 by first measuring the amount of balls that fit in a jar, then the number of balls that fit in a box, and this can be a proxy to find the amount of jars that fit into the box without directly measuring jars inside the box

the ball could be any other value and this would still work

you can try this with the ball being sqrt2

and that's why the general base c just appears, and can be anything

[hybrid leaf]

Does anyone have AP Precalc exam prep materials that are either legit or leaked? I need resources for final exam and especially help for FRQ#2 part Bii and C

[potent slate]

Just do past papers

[hybrid leaf]

there's only so many and FRQ 2 i dont understand

[potent slate]

Well, do as much as you can

[potent slate]

What FRQ#2? What year?

[haughty crown]

Try to find a video on YouTube explaining it, instead of just doing it @hybrid leaf

[potent slate]

?

I can help you out for Bii. What did you get for Bi??

Your answer in Bi is your rate of change per month (slope) for the equation you have to compose in Bii

By what is given, 25 is your intercept (at t = 0) soooo you can make up a slope-intercept form equation and solve for x = 1,5

sorry, I meant to say t, not x lol

[undone pumice]

Honestly I might be the only one who does this but lets say a problem has me run a logarithimic regression then for the rate of change question i just switch to linear for the same data and take the rate of change there

cus frq 1 and 2 r both calc so i can do that

[hybrid leaf]

My bad.. I meand Bii 😅
It's the explanation ones - I don't know what they are looking for. For part C as well, I understand the concepts but its the specific phrases I don't state that lose me the points

[potent slate]

Yeah. I am talking about Bii

[hybrid leaf]

For example when it says determine the domain, technically the question doesn't say you have to find the domain but i believe that you are expected to find it

[hybrid leaf]

gosh typo for the second time Biii

[potent slate]

Oh 😭

[hybrid leaf]

mb

[potent slate]

@hybrid leaf D is a quadratic that opens down and increases on [0; 4]. A_t is a secant line to the quadratic

So concluding from that, why is A_t always less than D(t)?

Example of a concave-down quadratic and a secant line

[potent slate]

Okay so it’s asking you to connect the idea of a vertex of the graph to the domain. In the context of plays, it makes no sense for it to decrease. The maximum in a concave-down parabola is when the rate of change changes from increasing to decreasing soooo do with that as you will

[hybrid leaf]

Yeah... I can do the explanation part but I guess my main issue is not not kowing how to solve the problem... it's writing what they want

should I just write everything I know in hopes that one of those is what they want

[potent slate]

Well, if you know the explanation, write it down??

I basically gave you the entire answer

[hybrid leaf]

that the function has a maximum so the domain is from 0 to where that maximum is?

[potent slate]

Exactly

[hybrid leaf]

right. But now should I do the work to acc find the domain

or is that enough

[potent slate]

In part A, you found D(t)

Just use maximum = -b/2a

[potent slate]

I’d say the explanation is already enough, but if you want to include saying the maximum, go for it. Better to be safe than sorry loll

Anyway, happy to help)

If you have any more questions, feel free to reach out here

[zenith sigil]

Me and @crimson wraith need help with math

[tender quest]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[zenith sigil]

Uh

[potent slate]

Just post the question.

[echo sinew]

Read the message...
#precalculus message

[round geyser]

Why do we use this

For finding distance of one point

[subtle flicker]

isnt the answer ? 4/9 ?

[floral gorge]

do Eqn 1 ÷ Eqn 2
(x^2y^2)/(xy^2) = 36/18
x = 2

sub x = 2 in Eqn 2
2y^2 = 18
y^2 = 9
y = +- 3

therefore x/y is +- 2/3

[floral gorge]

thats the perpendicular distance formula

you can prove it with vector geometry later on

[round geyser]

Correct

[round geyser]

I got 50

[round geyser]

a=4, b=3, c=-196

h=12, k=34

[floral gorge]

correct

[round geyser]

The long way is

Make a slope formula

And use the distance formula with sqrt((sum of x)^2)+(sum of y)^2)

[round geyser]

How would u make an equation of this parabola

[hushed sphinx]

Presumably it is a parabola with vertical axis (without assuming that there isn't a unique solution), so plug those three points into y=ax²+bx+c, giving you three equations in a, b, and c that you can solve.

You can also wing it less systematically by writing down a function for the line through (8,0) and (14,18) and then adding an appropriate multiple of (x-8)(x-14) to make it pass through (12,16). That's a simpler calculation but perhaps more conceptually confusing.

[bold crag]

It's just like by quadratic general equation y=ax^2+bx+c

[round geyser]

It’s y=-ax²+bx+c

[hushed sphinx]

<@&268886789983436800> MrBeast

[round geyser]

Am I right

I also noticed (8,0) is on x-axis

[hushed sphinx]

You can write it in that form too.

[round geyser]

When I graphed it

(14,18) is the vertex

So then it’s y=-(1/2)(x-14)^2+18

[round geyser]

And I got x=20 also

Or y=-(1/2)(x-8)(x-20)

Even I can use mean value theorem on (12,16) and (14,18)

[potent slate]

Case 1: x = x - 1 => 0 = -1. Doesn’t work
Case 2: x = -x + 1 => x = 1/2 (only real solution)

[hushed sphinx]

1/2.
1/2 ± i/2
1/2 ± (1+sqrt2)i/2

[brisk raft]

,w x^8 = (x-1)^8

[obsidian monolith]

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Requested by altanis_

[round geyser]

Just got that

[brisk raft]

use the complex sqrt perhaps

$$\left(\frac{x}{x-1}\right)^8 = 1$$

[round geyser]

Set that to u

u^8=1

[hushed sphinx]

Wolfram writes the imaginary parts weirdly, but sqrt(3+2sqrt(2))/2 is the same as (1+sqrt(2))/2.

[round geyser]

Okay

Thanks

[brisk raft]

then immediately you have that $x = \left{ \frac{\zeta}{1 - \zeta} \right}$ with $\zeta = \exp(2\pi i k / 8)$ for each $k = 0, 1, 2, 3, 4, 5, 6, 7$

[hushed sphinx]

And there's a pair of solutions I missed with imaginary part ±(1-sqrt2)i/2, which I really should have exepected -- they correspond to an 8-pointed star with edge length 1 rather than a regular octagon with that edge length.

[brisk raft]

exclude 0

cuz 1-1=0 so x/1-x =undef

🙃

[round geyser]

Btw

x≠1 is the domain

[brisk raft]

howd you solve it if not with the roots of unity

[hushed sphinx]

Since x^8 = (x-1)^8, I know in particular that |x|=|x-1| which is only possible when the real part is 1/2.

[brisk raft]

very nice

[hushed sphinx]

So I was looking for regular polygons with number of sides dividing 8, and side length 1, where one of the sides could be the line between x and x-1.

[round geyser]

x^5-1 has complex solutions that is with radians

[brisk raft]

yes you can use roots of unity

$x^5 - 1 = 0$ has solutions $x = {\exp(2 \pi i k / 5)}_{k = 0, 1, 2, 3, 4}$

[brisk raft]

,w x^5 = 1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by altanis_

[brisk raft]

did i mess up somewhere

oh well

i will ping Moderatrosand you will be banned.

LMFAO did he get banned 😭

[west peak]

Practice makes perfect!

[brisk raft]

i was correct

idk why wolfram alpha decided to omit a solution

[haughty crown]

https://tenor.com/view/lol-rdj-juggtok-giftok-tonystark-facials-gif-3958302751287261929

[valid topaz]

||square (1+i) first||

[hushed sphinx]

Add then square it once more.

[orchid urchin]

just binomial expansion

for all terms except first and last

and you find every part that is imaginary

[round geyser]

I get [(1+i)^2]^14

(1+i)^2=2i

[orchid urchin]

oh wait I lied its 0

[idle field]

use the polar form

[round geyser]

Is that right

2^14 * i^14

According to this

We use -1

[orchid urchin]

no its 0

this is propaganda

[round geyser]

Yes

[orchid urchin]

chinese

[round geyser]

It’s -16,384+0i

So it’s 0

I like solving this

I get log_2^a (log_2^b (2^1000)) = 1

[orchid urchin]

seems about right to me

[round geyser]

Then

Now it’s (2^a)^1

Or log_2^b (2^1000) = (2^a)^1

[round geyser]

^

[orchid urchin]

yeah, and then 1000/b = 2^a

[round geyser]

I thought it’s 1000*[log_2^b^2]=2^a

[orchid urchin]

well you can keep going in layers, like the first layer which is log 2a and all that stuff = 2^0 = 1, and then the second layer you get the log2b(2^1000) = (2^a)^1 = 2^1, then third layer you get 2^1000 = (2^b)^(2^a) = 2^(b * (2^a))
It's a bit odd here but equating the exponents from the last step we get, 1000 = b * 2^a and then since a na db are positive you just identify all powers of 2 that divide 1000, and sum em up to get 881

the only hard part is actually finding all the pairs and summing them

[orchid urchin]

which this would also get you there, just because it comes from the law property log_{a^b}(c^d) = d/b

[round geyser]

I got 128

[orchid urchin]

sjpw ,e tje wprl

show me the work

massive typo

[round geyser]

Okay so

I’ll start with log_2^b (2^1000) = 2^a

Move the base under 2^a

Now it’s 2^1000 = (2^b)^(2^a)

Same base can cancel

1000=b^(2^a)

We know 1000 is 10^3

So it’s 10^3=b*(2^a)

[orchid urchin]

im kinda confused on this step, it would have to be 1000 = b * (2^a)

[round geyser]

Yea

It is

I’ll continue

[round geyser]

10^3 can be 5^3 * 2^3

[round geyser]

Then it’s
5^3 * 2^3 = b*(2^a)

You see

b = 5^3=125 and a=3

[orchid urchin]

yeah so far I agree

[round geyser]

You add them

3+125=128

[orchid urchin]

its for all possible values of a+b though

thats just one pair

a = 3 and b = 125 which sums to 128

a = 2 and b = 250

a = 1 and b = 500

501 + 252 + 128 = 881

neat relationship btw

[round geyser]

Thanks

[round geyser]

How would u derive a=2 and b=250

[orchid urchin]

I just checked a = 1, a = 2, a = 3, etc

lol

[orchid urchin]

im surprised thats a comp math problem, ig it makes sense though since u have to find pairs of numbers and sum them up which is unusual but I feel like it is very straightforward

[round geyser]

Yep

And I notice a pattern

[acoustic galleon]

very cool pattern

[round geyser]

When a is increased b will be decreased

When a is 1 to 2, b is 500 to 250

[keen delta]

de moivre's theorem
1+i=sqrt2cis45
so sqrt2^28 * cis(45*28mod360)
and then u get there

[round geyser]

What’s cis

And what’s mod

[orchid urchin]

mod is modulus

[keen delta]

cos+isin

[orchid urchin]

like 5 % 3 = 2

it shows the remainder

[keen delta]

NOOOOO COMPUTER SCIENCE explodes

[round geyser]

Oh

[orchid urchin]

if x % 2 == 0 then "even" end

except for 0 = x

[keen delta]

you can also take everyone's advice and square it once, (1+i)^28=((1+i)^2)^14=(2i)^14 which as you know is not going to be imaginary

[round geyser]

Yep

[keen delta]

forgot one more rule, x>=0

I never remember where the = and > goes

[orchid urchin]

it can't be 0 because 0 is odd nor even

and it can't be negative

[keen delta]

in general u can do their tactics but for more weird problems on that I suggest doing de moivre's theorem

[orchid urchin]

if u cehcking if its a even or odd num

[keen delta]

0 is even trust

because

[orchid urchin]

how

[keen delta]

even numbers - even numbers = even number and 2 and 2 are both even

frfr

[orchid urchin]

show proof

[keen delta]

idk I don't do discrete meth

[keen delta]

hmmm

[orchid urchin]

what about -0

[keen delta]

let m, n be any positive integer where m>n
2m-2n=2(m-n)
since it is divisible by 2 and m-n is positive, it's a thing yeah
2n-2n=0

[keen delta]

-0 isn't real

[orchid urchin]

oh i see what I did, I assumed that it couldn't be even or odd because it was neither pos or neg

[orchid urchin]

it was satire

[keen delta]

discrete math is fake

wait

[haughty crown]

oh yeah? Then what’s (0i)^2, huh?

[keen delta]

0^2*-1 btw

[haughty crown]

I had a headache trying to read this what

[potent slate]

Modulo* The modulus is also called “absolute value,” while the modulo is the one that finds the remainder

Or I might be wrong

The modulus might also be the specific number you divide by to find the remainder, I guess

Or something like that

[orchid urchin]

pop quiz, whats the modulus of a complex numbert

[potent slate]

sqrt(a^2 + b^2)???

[orchid urchin]

nice

[potent slate]

for a complex number a + bi

[orchid urchin]

yeah and if u think about it its the same for x,y coords

non complex or rect coords

[potent slate]

True

[orchid urchin]

whatever u want to refer to them as

makes it easy to convert them to polar as well

[potent slate]

Mhm

[haughty crown]

He/he

https://tenor.com/view/haha-mickael-jackson-queendugif-ahah-hehe-gif-13642339

[drowsy trout]

yall tryna read an exploratory note?

[fossil kindle]

what is this

[drowsy trout]

idk bro

a dyadic kinda polynomial and quadratic sequence

[barren horizon]

Plot the points on a graph of the factored polynomial $p(x)=4(x-6)(x+3)(6x-1)(1x+2)$

[obsidian monolith]

emiya

[uncut mulch]

how much detail is required?

[barren horizon]

Thats easy stuff

Its called precalculus but its easy

Just algebra at that point

[eager meteor]

Yo

Have we solved the dream integral?

And if yes, how?

[orchid urchin]

taylor series expansion

according to the internet

[eager meteor]

K thx

Also…

Is there a proof for squeeze theorem?

[limber junco]

Yes

it is left as an exercise for you to prove

Just use epsilon delta definition

[eager meteor]

But I am in 9th grade

But I’ll try

Thx

[robust terrace]

It would js be annoying to expand

[last palm]

Guys, is there any website kinda thing where i can find calculus questions regularly??

[ancient inlet]

Daily Integral

[fossil kindle]

what is a squeeze theorem

[echo sinew]

https://letmegooglethat.com/?q=squeeze+theorem

[fossil kindle]

ig bro

not falling for ts again

[hushed sphinx]

What do you mean "falling for"?

The embed looks like the site is probably more than a bit snarky, but just seeing the embed itself should give you more than enough clue to understand Alberto's point without actually following the link ...

[jolly karma]

Hey chat how does one determine the direction of a parabola (or which variable changes) from this state?

[last palm]

Um write the equation with x square and its term on side side while keeping y on other side then its something like
Y=1/2(x)^2+5x+59
This proves the parabola is opening upwards and u can find its vetex by using (-b/2a,-D/4a)

Then find the y intercept and trace the graph

Also if x is squared term parabola opens up or down and if y is squared it opens left or right depending on conditions , here it opens up as x square is positive