precalculus

[knotty spear]

so for |x| < pi

its 0

[viscid thistle]

floor(sinx/x) is not defined at x=0 because sinx/x is not defined at x=0. Neither function is 0 at x=0.

[knotty spear]

except for x=0

[viscid thistle]

The limit for the former is 0 and for the latter is 1 but both are only limits

[knotty spear]

yes

[ruby otter]

define former and latter here

[knotty spear]

former is lim floor(sinc)

latter is floor(lim sinc)

👀

[echo plaza]

the one before and the one after

[viscid thistle]

lol

@knotty spear you're wrong

[willow bear]

former = 1st in a pair

[knotty spear]

[willow bear]

latter = 2nd in a pair

[knotty spear]

lim x to 0 floor(sin(x)/x)) = 0

[viscid thistle]

@ruby otter former is first I said (floor(sinx/x)) and latter is second (sinx/x). This is always the case.

[knotty spear]

floor(lim x to 0 of sin(x)/x) =1

[ruby otter]

ya

[willow bear]

$0 = \lim_{x \to 0} \floor{\frac{\sin(x)}{x}} \neq \floor{\lim_{x \to 0} \frac{\sin(x)}{x}} = \floor{1} = 1$

[ruby otter]

but lim xto0 floor(sinx / x ) = 0

[obsidian monolith]

Ann:

[echo plaza]

}

[willow bear]

$0 = \lim_{x \to 0} \floor{\frac{\sin(x)}{x}} \neq \floor{\lim_{x \to 0} \frac{\sin(x)}{x}} = \floor{1} = 1$

[knotty spear]

thats what i said yes

[echo plaza]

modiwhat

[knotty spear]

modifé

[willow bear]

modifié

french for "edited"

[echo plaza]

🥖

[obsidian monolith]

Ann:

[willow bear]

there

i don't see why this is still being talked about

[ruby otter]

Ann

[knotty spear]

idk i got pungged

[limber bone]

can any1 teach me about those step functions

ceil and floor?

[ruby otter]

do you remember a nasty integral i asked before??

[viscid thistle]

Ann est en train d'écrire...

also what was the nasty integral

[willow bear]

$\floor{x} = $ largest integer $\leq x$ \ $\ceil{x} = $ lowest integer $\geq x$

[obsidian monolith]

Ann:

[ruby otter]

erircé'd niart ne tse nna

[echo plaza]

[limber bone]

what if u put an integer

its the same?

[viscid thistle]

yes

[limber bone]

ok

[ruby otter]

it is same as x

[limber bone]

so is floor 2.5 = 2

and ceil 2.5 = 3

[willow bear]

$x \in \bbZ$ iff $\floor{x} = \ceil{x} = x$

[viscid thistle]

yuh

[obsidian monolith]

Ann:

[willow bear]

and yes

[ruby otter]

floor 2 =2

[limber bone]

ceil 3.1 = 4?

[willow bear]

yes

[limber bone]

wow thats cool

[echo plaza]

good

[limber bone]

can u do calculus with these?

[knotty spear]

are we still talking about sinc or nah

[echo plaza]

yes

[willow bear]

wdym "can you do calculus"\

[echo plaza]

bit messy

[limber bone]

like can u differentiate them or like do stuff

[willow bear]

the functions are discontinuous at the integers, and locally constant everywhere else

[ruby otter]

it means can you do calculus !!

[echo plaza]

their derivatives are mostly 0

and sometimes not defined

[willow bear]

where they are differentiable, their derivative is 0

[limber bone]

okk

[ruby otter]

there's another like floor and ceil

[limber bone]

,wolf graph ceil(x)

,wolf graph floor(x)

[obsidian monolith]

Display results online and refine query

Requested by mo2men#4865

Display results online and refine query

Requested by mo2men#4865

[ruby otter]

its called round

another is fractional part

[echo plaza]

round is the shit one

[willow bear]

frac(x) is just x - floor(x)

[viscid thistle]

round is just floor(x + 1/2) though

[limber bone]

whats frac(x)

[willow bear]

i just said it

[viscid thistle]

fractional part of x

[limber bone]

x-floor(x)

[ruby otter]

no floor(x) = x- frac(x)

[viscid thistle]

frac pi = 0.1415...

[echo plaza]

does latex hav it

[viscid thistle]

lol radical

that's what ann said

[echo plaza]

it's that wobbly ting

[willow bear]

it's a comb

[ruby otter]

no oxide

[willow bear]

,w plot x - floor(x)

[obsidian monolith]

Display results online and refine query

Requested by Ann#8192

[viscid thistle]

wdym no
that's eggsactly what ann said

[willow bear]

floor(x) + frac(x) = x

for all x

[ruby otter]

floor(x) = x- frac(x) - Radical

frac(x) is just x - floor(x)- Ann

[willow bear]

i said frac(x) = x - floor(x)

you said floor(x) = x - frac(x)

these boil down to exactly the same thing

[ruby otter]

😂

just kidding

[willow bear]

💢

[viscid thistle]

[knotty spear]

[echo plaza]

[viscid thistle]

@willow bear I learned about why I was wrong, @thick raptor show me da wae

[willow bear]

prepare for a storm of rees

[viscid thistle]

cheety monker

[thick raptor]

[ruby otter]

[viscid thistle]

I never learned about the floor function

but makes sense now

y=/= 1 so round to 0

[ruby otter]

thats 2000iq

[thick raptor]

[ruby otter]

[willow bear]

[tawny nacelle]

[tropic fulcrum]

Any suggestions on where I begin?

[echo plaza]

factor theorem

[tropic fulcrum]

so would i basically plug stuff in for K and then see if substituting 1 for x = 0

[torn swift]

well you can divide the polynomial by that factor

that's something you can do with it, but I have not worked that out, so not so sure what you end up with

[echo plaza]

find f(1)

[tropic fulcrum]

so k =3 or 1

[echo plaza]

ye

[tropic fulcrum]

awesome, thx for the help guys really appreciate it 😃

[echo plaza]

np

[sinful sky]

hey can anyone help me solve this question without a calculator

there we go its bigger

[willow bear]

ok so what's giving you trouble here

[sinful sky]

ok so heres what i know first

uh lemme draw it real quick

so heres what i need to do

then i multiply

[willow bear]

so far so good

[sinful sky]

1+9999e^-0.8t on both sides right

[willow bear]

that's possible

[sinful sky]

so that gives me 4000(1+9999e^-0.8t)=10000

then i divide 4000

[willow bear]

ok

you get 1 + 9999exp(-0.8t) = 2.5

[sinful sky]

yea

subtract 1

on both sides

then divide 9999

[willow bear]

yes

you get exp(-0.8t) = 1/6666

[sinful sky]

then u can turn it into ln(e^-0.8t)=ln(1/6666)

[willow bear]

yes

[sinful sky]

simplify it into

[willow bear]

that's called applying ln to both sides

[sinful sky]

uhh

shoot

[willow bear]

no

i mean that what you did there

going from exp(-0.8t) = 1/6666 to log(exp(-0.8t)) = log(1/6666)

you applied log to both sides

you're making perfectly reasonable steps so far

what are you having trouble with now?

[sinful sky]

so basically ive been just cancelling and simplifying it

then it turns into -0.8t = -ln(6666) bc fraction or w/e

right

so the issue is ive always plugged this into my calc

5ln(6666)/4

how would i solve this

hehe

.

it has to be over 4000 students

so 5ln(6666)/4 = 4000

[willow bear]

no

[sinful sky]

right...?

[willow bear]

5 ln(6666)/4 is the number of days

not the number of students

[sinful sky]

oh

[willow bear]

all that's left to do is evaluate 5 ln(6666)/4

which i admit is kinda hard to do without a calculator

[sinful sky]

yea im not allowed to use a calculator

and im trying to learn how

but i think i might just have to lose that question

[willow bear]

,w 5 log(6666)/4

[obsidian monolith]

Display results online and refine query

Requested by Ann#8192

[willow bear]

yeah honestly idk if i were computing this by hand i'd leave it be

[sinful sky]

all i figured out was how to simplify it and stuff xd

[willow bear]

idk why you're forced to not use a calculator

[sinful sky]

it goes all the way to calc

this is only precalc

feelsbadman

and no calculator for anything

i dont know how im gonna survive

3 hour test

i dont think thats enough time for me to try calculating

since its an exam

[torn swift]

no one will be able to calculate ln of that without at the very least, a table of logs

[sinful sky]

welp hopefully it doesnt have it on the exam

these r practice questions for the upcoming exam i have

[torn swift]

if it was log base 10 then, the maybe we would have a chance at trying to figure it out but ln is difficult

[willow bear]

if you're asked to calculate the log of a number like this and not given a calculator it's just evil

[sinful sky]

ok i have my final option

im gonna draw on a pencil and roll it

best of luck to me

thanks guys

[torn swift]

if it was base 10 then we could do it using a log table

[opal epoch]

gamer

[torn swift]

or use change of base to get it to log 10 but that's way tedious

[sinful sky]

i only have 3 hours and theres a bunch of questions

PENCIL ROLLING TIME

[mellow violet]

Can someone please walk me through the steps on this, because it makes no sense to me.

[short sorrel]

ok so

$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$

[obsidian monolith]

Namington:

[short sorrel]

we want to clear up those fractions

since they're annoying to deal with

so, lets multiply the entire fraction by sqrt(3)/sqrt(3)

$\left(\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\right) \left(\frac{\sqrt{3}}{\sqrt{3}}\right)$

[obsidian monolith]

Namington:

[short sorrel]

we're allowed to do this because sqrt(3)/sqrt(3) is just 1

so we're multiplying by 1

which doesnt change the value

now, distribute the sqrt(3) in

$\frac{\sqrt{3}\left(1-\frac{1}{\sqrt{3}}\right)}{\sqrt{3}\left(1+\frac{1}{\sqrt{3}}\right)} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

[obsidian monolith]

Namington:

[mellow violet]

ok i'm with you so far

[short sorrel]

now, we have $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$, but generally we don't like having radicals in denominators - they're troublesome

[obsidian monolith]

Namington:

[short sorrel]

so, we do a process called 'rationalizing the denominator'

to do this, we multiply by the conjugate of the denominator

the denominator is sqrt(3) + 1, so its conjugate is sqrt(3) - 1

same numbers, the sign is just flipped

but again, we cant just multiply willy-nilly; instead, we multiply by (sqrt(3) - 1) / (sqrt(3) - 1), because that's the same as multiplying by 1

$\left(\frac{\sqrt{3} - 1}{\sqrt{3} + 1}\right) \left(\frac{\sqrt{3} - 1}{\sqrt{3} - 1}\right)$

[obsidian monolith]

Namington:

[short sorrel]

the conjugate multiplication might seem like a random choice, but itll make more sense once we expand

$\frac{(\sqrt{3} - 1)(\sqrt{3} -1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - \sqrt{3} - \sqrt{3} + 1}{3 - \sqrt{3} + \sqrt{3} - 1}$

[obsidian monolith]

Namington:

[short sorrel]

do you follow that expansion?

[mellow violet]

yes, makes sense so far

[short sorrel]

alright, now we simplify

$\frac{3 - \sqrt{3} - \sqrt{3} + 1}{3 - \sqrt{3} + \sqrt{3} - 1} = \frac{4 - 2\sqrt{3}}{2}$

[obsidian monolith]

Namington:

[short sorrel]

finally, we simplify by distributing in division by 2

$\frac{4 - 2\sqrt{3}}{2} = \frac{4}{2} - \frac{2\sqrt{3}}{2} = 2 - \sqrt{3}$

[obsidian monolith]

Namington:

[mellow violet]

holy smokes, it was the rationalizing the denominator part i had no clue on, just when I think I got a handle on it..hah. Thanks a ton, I would have never figured that out.

[short sorrel]

(you dont really have to show the distribution step, you can skip straight from the (4-2sqrt3)/2 to 2 - sqrt(3), but its nice to show why it works)

yeah, its a bit of an algebra "trick"

multiplying by a conjugate gets rid of the middle term

which is quite helpful

thats the same reason $a^2 - b^2$ factors to $(a+b)(a-b)$, for example

[obsidian monolith]

Namington:

[short sorrel]

so, by multiplying by the conjugate, we were able to get rid of lingering sqrt(3)s

which rationalized the denominator

[mellow violet]

makes sense

alright with that knowledge, lemme go back at it and see if i can work the problem

thanks

[fiery spruce]

How do you find the angles of this wothout law of sines/law of cosines

[willow bear]

the triangle is isosceles

so it is possible to draw its altitude/median/bisector and obtain two congruent right triangles

[fiery spruce]

How would I find x?

[swift wagon]

Can you clarify what all the numbers mean and why there is a o next to 180

[fiery spruce]

1800m across an open stretch of road, both of the angles are angles of elevation to a helicopter somewhere in the sky between the two people

The goal is to find the altitude of the heli

[swift wagon]

Oh

Is that 51 degrees?

[fiery spruce]

52

[swift wagon]

One sec

Is that a right angle?

Must be right

[fiery spruce]

Your only given the two angles and the side

[swift wagon]

So I would just use the sine rule to find on of the other sides

Then use SohCahToa assuming that is a right angle

[earnest wedge]

Finding power of a complex number (in rectangular form) convert to polar, applying devoires theorem, then convert back to rectangular
Is anyone able to help me with this?

[serene heath]

post ur question

[earnest wedge]

I don’t have specific examples but like I don’t get it

More of the finding powers/roots

[eternal folio]

do you know de Moivre's theorem?

[valid vector]

Is this correct?

And also, do Pythagorean Identities only work if they are to the 2nd power, or can 1 + tan(x) = sec(x)?

[serene heath]

that equality isnt always true

[valid vector]

Wdym?

[torn swift]

What have you tried so far in terms of simplifying the left hand side?

[wet hare]

Can someone help me out with this, please? I solved it, but it still marks it wrong. It tells me to use n as an integer constent, and to enter my response in radians. I only have 2 tries and I used 1 of them.

I don't know what else to add.

[limber bone]

sin(x)(sin(x)+1)=0

so either sin(x) = 0

or sin(x)+1=0

there are infintely many cotermina nagles

• sin is periodic too

so just write ur answer +2*pi *n

[wet hare]

I wrote this

[patent beacon]

You're forgetting the cases where sin(x) = 0

[wet hare]

For this one I don't really get it either, I wrote this down, but I don't know if I'm right since it says to write it in a comma-separated list. 😕

Its still marks it wrong in the previous one, because I haven't submitted it yet. Too nervous to do so.

[patent beacon]

sin(x)[sin(x)) + 1] = 0
sin(x) = 0 or sin(x) = -1
x = πn or 3π/2 + 2πn

[spring thunder]

x=pi*n tho

[patent beacon]

Oh yeah

[wet hare]

So it should be 3(pi)/2 + 2n(pi), right?

[viscid thistle]

@valid vector
Factor out 5 in the numerator and it becomes 5(secx-1)(secx+1)

[patent beacon]

@wet hare
No

x = πn or 3π/2 + 2πn

[wet hare]

3(pi)/2 + 2(pi)n? @patent beacon

But that's what I wrote, right?

[patent beacon]

I think it's pretty clear that we're not writing the same thing, you must think they're equivalent. Why?

[wet hare]

So it should be 3(pi) + 2(pi)n then

[patent beacon]

No, because that doesn't cover 0

[wet hare]

I am sorry but I am confused..

(Pi)n or 3(pi)n+2(pi)n

[patent beacon]

To get rid of the n, and list a few examples:
x = 0, π, 3π/2, 2π...

[wet hare]

It says I should write it as a comma-separated list so maybe a comma in between those two.

So the answer should be ×= 0,(pi),3(pi)/2, or 2(pi)? Whichever two I pick of those you mentioned?

[valid vector]

Why is this wrong? I've revised it like 20 times now

Pls tag me

[vast karma]

@valid vector dont you need to write brackets?

[swift wagon]

@valid vector It’s only 2pi/3

If you put in 4pi/3 then you get -sqrt3/3

[valid vector]

Oh OK, thanks!

[ruby otter]

Any ideas

Question 7

,rotate

[obsidian monolith]

Something went wrong while running your command. The error has been logged and will be fixed soon!

[ruby otter]

,rotate

[thick raptor]

Lots of simplification can be done if you take the log of it

[serene heath]

looks like e

[ruby otter]

I'm haveing one power one by sin

[thick raptor]

??

[ruby otter]

I did the inner one first

[thick raptor]

That's not how limits work?

[ruby otter]

Then?

[thick raptor]

Didn't I just tell u wut to do?

[ruby otter]

Log for what?

[thick raptor]

The problem?

[ruby otter]

I mean which one in the prob

[thick raptor]

Are you implying that you want to work with one part of the problem and ignore the rest of it like I just told you not to?

[ruby otter]

At least u must give me one step

Just log what does it mean??

I said I got like 1^csc(x)

[thick raptor]

Okay then

That's 1

Is that an answer choice?

[ruby otter]

No

e. Root e
1/e. Root 1/e

[thick raptor]

Then why are you getting 1^csc(x)?

[ruby otter]

These are the four options

I got one but it's wrong however

[thick raptor]

how are you getting 1^csc(x)?

[ruby otter]

I did limir for inner one

[thick raptor]

And which limit rule says you can just take the limit like that?

If you're going there you may as well've taken the limit all the way in

$\lim_{x\to0}1+x=1$\$\lim_{x\to0}\left(\frac{(1+x)^{1/x}}e\right)^{\frac1{\sin(x)}}=\lim_{x\to0}\left(\frac{1^{1/x}}e\right)^{\frac1{\sin(x)}}$

[obsidian monolith]

SAckalope:

[thick raptor]

But that gives me 1/e inside, not 1

[ruby otter]

What is the final answer

[thick raptor]

We're not here to give you answers. You should already know that.

And as I told you already, if you take the log of the whole limit, it simplifies a lot

[ruby otter]

Well if u r not here to give answera

You could explain what u say

[thick raptor]

I'm honestly unsure what more I can explain without just doing it for you

[serene heath]

do it for him, but do it in more code

[thick raptor]

[ruby otter]

For which function you want me to take log

[thick raptor]

What's the limit?

What is the problem you are working on

[ruby otter]

That's what we must find

[thick raptor]

Write down the problem you are working on

[ruby otter]

That pic

It's hard to write so I sent pic

[thick raptor]

Just write it here for me

[ruby otter]

https://cdn.discordapp.com/attachments/363224154469826562/573967026423332911/573965397934735371.png

7th one

[thick raptor]

Okay

[ruby otter]

Hey I'm not in pc

[thick raptor]

Honestly not sure what you can't understand about taking the log of that

[ruby otter]

Log of what

What do u mean by that

[thick raptor]

...

[ruby otter]

that

"THAT"

[thick raptor]

Someone should help this man figure out what problem they're working on

[ruby otter]

,w that

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[ruby otter]

,w Take log of that

[obsidian monolith]

Display results online and refine query

Requested by Radical Ninja#7581

[ruby otter]

U mean this

[thick raptor]

[ruby otter]

Write that step

[thick raptor]

How about trying to actually write out the problem you're working on

[ruby otter]

I said I'm on mobile

I don't know why you have this kinda attitude

[serene heath]

take log of ur expression

[ruby otter]

Like e power log!

[thick raptor]

It's because I'm trying to help you, yet you don't seem to know what problem you're working on?

[ruby otter]

I still don't know what does take log mean

It makes me think what is log

[thick raptor]

You don't know what a logarithm is?

[ruby otter]

Yes

[thick raptor]

Yes as in you don't know what a logarithm is?

[ruby otter]

Yes as in I know

[thick raptor]

But you don't know how to take the log of something?

[ruby otter]

Do u ever see some 1 solving limits without knowing log

Maybe except you

[thick raptor]

Given how hard it is to help you

[ruby otter]

Like x -> lnx ??

[thick raptor]

I have to ask these basic questions

Do you know how to take the log of something or not?

Does the concept of something like ln(1) = 0 make sense?

[ruby otter]

I think I'm not from your universe

Maybe explain

[thick raptor]

[ruby otter]

Maybe in your Universe it's different

What does log mean to u

[thick raptor]

[gilded matrix]

can u pls listen to weeby

[thick raptor]

Personally it means an integral

[gilded matrix]

You cannot compete with weeby's divine intellect.

[thick raptor]

Or the inverse of an exponential function

[ruby otter]

Fine can u write that step

Like what u mean

[thick raptor]

No I am not going to, because as you've implied, logarithms should be something you know before doing these sorts of limits

[ruby otter]

Like log (e^f(x))

[thick raptor]

And I don't have the energy to teach you an entire thing for the sake of one problem

[ruby otter]

I never asked you to

I asked that one step

[thick raptor]

Yes, and I did, but I'm not willing to explain every minute detail that should've been clear, given what you're working with

[ruby otter]

only you can understand what you speak

[thick raptor]

<@&286206848099549185> need help with a limit, they can't seem to understand what I mean by taking the log of the problem they're working on

[ruby otter]

They?

[torn swift]

is it not of some sort of special form you think?

$\lim_{x \to 0} \bigg {(}\frac{(1+x)^{\frac{1}{x}}}{e}\bigg{)}^{\frac{1}{\sin(x)}}$

thank

[obsidian monolith]

⚡Amphy⚡:

[ruby otter]

Just say what does taking log mean

[torn swift]

most likely ln(x)

[ruby otter]

Like $ x \to lnx $

[torn swift]

why is it multiplied by x?

[obsidian monolith]

Radical Ninja:

[torn swift]

why is x approaching ln(x) now?

[ruby otter]

Doesn't make sense right??

The same here

[torn swift]

he did not say that though

take the log of both sides is literally just do this:
\
$x+5=2 \to \ln(x+5)=\ln(2)$

[obsidian monolith]

⚡Amphy⚡:

[ruby otter]

But what is both sides here in my problem mean??

[torn swift]

the limit equals some number just call it whatever you want, perhaps Q?

[short sorrel]

lim(stuff) as x->0 = y

[torn swift]

$\lim_{x \to 0} \bigg {(}\frac{(1+x)^{\frac{1}{x}}}{e}\bigg{)}^{\frac{1}{\sin(x)}}=Q$

[obsidian monolith]

⚡Amphy⚡:

[short sorrel]

huh

i have never seen Q used as a variable

[torn swift]

so then take the log of both sides

[thick raptor]

Don't even need to do that

[short sorrel]

outside of physics

[torn swift]

Q is quantity to me, so I use that lol

[ruby otter]

Yeah now you guys start a fight

[thick raptor]

Just take the log of the thing

[ruby otter]

Each and everyone is gonna come up with a new idea

[thick raptor]

Taking the log of both sides is simply taking the log of the left side and taking the log of the right side

[ruby otter]

And take log means take log right??

[thick raptor]

Not sure what you're trying to ask

Eating apples means eating apples

[ruby otter]

Exactly

[thick raptor]

So from what I gather

You know how to take the log of both sides of an equation

Yet you don't know how to take the log of a single thing?

[ruby otter]

Coz it doesn't make sense to me

Not the math but what you say

[thick raptor]

But am I right?

That is what you are struggling with?

[ruby otter]

Maybe ya acc to you

But TBH I was struggling to do that prob

[thick raptor]

It's not about me, it's about what you're struggling to understand

[ruby otter]

Its not about me it's about what you're struggling to explain

Maybe you don't wanna explain

[thick raptor]

I'm not explaining at all. If someone asked you how to compute the limit of x+2 as x -> 4, would you try to explain why 4+2 = 6?

[gilded matrix]

are you guys gonna solve something or just argue with each other and be unproductive?

[thick raptor]

There are parts that should be understood and parts that shouldn't

I tried asking them if they understood what it meant to take the log of both sides

[ruby otter]

There are parts which should be understood by you not be me

Maybe to you what take log means is different to me

[thick raptor]

But yes, this is fairly unproductive

[gilded matrix]

Radical Ninja stop being ignorant

if you dont know how to take the log of it go learn it , I don't think we can teach you here to that level

It's hard to teach you by only text

[ruby otter]

Lol you are leaving your fan to fight with me??

[gilded matrix]

nobody is fighting no one

You are the one who came here asking for help, either participate or good luck on your own!

[thick raptor]

If someone asked you how to compute the limit of x+2 as x -> 4, and they didn't understand why 4+2 = 6, what would you do?

That's the position we're in right now

[ruby otter]

I would explain

TBH that's true

I would explain or show them what's what not like you being here instead texting this much you could have showed me that one step

And some1 else pops outta nowhere calls me ignorant

[gilded matrix]

It's inefficient to try and explain it here

[thick raptor]

Well that's fine if you want to try and do that, but it's really not worth my time to travel down a rabbit hole like that

[serene heath]

wow well this.... escalated

[thick raptor]

could've been worse

[torn swift]

the other day some guy could not differentiate a function to save his life

explained how to use the power rule like three times and he still could not do it

[native sequoia]

Taking the log of an expression (provided that it is positive) is like using f(x)=exp(log(f(x))) and ignoring the exp until the very end. @ruby otter

and taking the log of your limit seems to be in the right direction.

[sinful sky]

what am i doing wrong

did i mess something up

[torn swift]

what does f(x+t) mean you think?

[sinful sky]

i'd imagine it would just be using the original f(x) then + t

[torn swift]

well f(x+t) means, where ever you see and x, plug in x+t

[sinful sky]

oh so it would be (x+t)^3+1

?

[torn swift]

you'll have to expand (x+t)^3

[sinful sky]

ah

[torn swift]

annoying lol

[sinful sky]

ok ty

[torn swift]

let me work that one out as well

this should just end up giving the derivative

difference quotient stuff

[willow bear]

i'd imagine it would just be using the original f(x) then + t

very much no

that is f(x)+t

not f(x+t)

the two are very different and should not be confused

[sinful sky]

yes ty i understand

replace the x instead

uh what am i looking at for this question

[torn swift]

what is the stuff at the beginning lol?

[sinful sky]

yea

idk

this website glitches i guess

[torn swift]

lol let's just pretend it wasn't there and actually move on with the problem though

[sinful sky]

i can do the right side

yea i just wanted to know what the left side was

[willow bear]

uhh

okay no this is definitely some sort of glitch

[sinful sky]

oh right whats the || and little o

[torn swift]

ok then so you already know how to do it, just some weird rendering issue lol

[willow bear]

can you reload the page

[sinful sky]

o is for comparison right

it doesnt fix it so its w/e

if i refresh the questions disappear

[willow bear]

those sticks and the circle are weird

[sinful sky]

ah ok

[willow bear]

best guess is that the intended expression was $\sqrt{\frac{a^{1/2}cb^{1/3}}{b^{-5/3}a^{5/2}}}$

[obsidian monolith]

Ann:

[round eagle]

Description

A new bridge is to be constructed over the East River in New York City. Two structural possibilities exist: the support could be in the shape of a parabola or the support could be in the shape of a semi-ellipse.

The space between the supports needs to be 1050 feet. The height at the center of the arch needs to be 350 feet.

could i get some help with this?

[torn swift]

so basically choose what shape it has to be?

[round eagle]

id probably go w ellipse since its elongated

but i have to do both

i know how to do parabolic equations, but semi elliptical equations are quite a lot

@torn swift

y=(-0.001)x^2+350

thats my parabolic equation right now

[torn swift]

well it's not all that bad actually since you have the distance of both the major and minor axis

[round eagle]

true

525 and 175

[torn swift]

choose your ellipse center to be at the middle of the bridge

[round eagle]

well i gotta find out both equations anyways, so at least i have the parabolic one doen

so (0,350)

since the bridges apex has to be at 350 ft

[torn swift]

yes, the distance from the origin to apex is 350 and the distance from the origin to any given side is then 525

[round eagle]

ive never done elliptical equations, so ill need some help if you dont mind

[torn swift]

is there is picture to go along with the problem?

[restive pasture]

what is the probability that two numbers (real numbers) from 0 to 10 have an absolute difference of less than 3

[torn swift]

the final equation will depend on where they want to origin to be, but most likely it will be at the middle since the parabola equation uses the midpoint as the origin I assume

[round eagle]

theres no picture for me to pass off

base off of*

and likely, since the apex is at 0,350

[torn swift]

well for the ellipse it's simple enough really

we set the origin to be the middle of the bridge

[round eagle]

so the final parabola ended up being y=(-0.001)x^2+350

[torn swift]

so we don't need to consider (h,k) since the ellipse is now centered at the origin

[round eagle]

satisfying the 1050 wide and 350 tall

and okay sorry to cut you off

[torn swift]

We know that the major axis is the x, as that distance is the longer one

[round eagle]

yes

it makes the ellipse sideways right

"0" but sideways

[torn swift]

oval, yes

that's an interesting way to describe it lol

so now we have to write down the general form of the equation

[round eagle]

me rn

[torn swift]

so for an ellipse centered at the origin with it's major axis along the x, the equation will be...?

[round eagle]

(x-h)^2/a^2 + (y-k)^2/b^2=1

i have no idea where to start

x is useless tho since were at the origin

and y is

in that case

[torn swift]

this is directly from your book, so let's go through all the parts of it

what is the center of the ellipse?

[round eagle]

the origin right

[torn swift]

yes the origin so that means that (h,k) is...?

[round eagle]

itll simplify calculations aswll

0 and 0

[torn swift]

good

so now the major axis is horizontal or vertical?

[round eagle]

horizontal since the sideways oval right

its looooooong oval

[torn swift]

yes, it is horizontal

[void quiver]

so many smart people

[torn swift]

now what is a?
a is the distance from the origin to either end of the ellipse along the major axis

[round eagle]

needs to be 1050

[torn swift]

so the distance to one side of the bridge to its center is...?

[round eagle]

525

[torn swift]

now we have to find b, which is...?

[round eagle]

1050 wide and 350 tall

the other axis right?

so 175

[torn swift]

b is the distance from the origin to either end of the minor axis, so it would be the distance from the origin to the top of the bridge

[round eagle]

so 350/2

so we have 525 and 175 for the halves

[torn swift]

pretty sure 350 is measured from the ground up

[round eagle]

o

yeah

[torn swift]

thought it was 350 from the ground to the top

[round eagle]

the apex of the arch is 350

[torn swift]

ok so it is 350 then

[round eagle]

but wait

this is a semi ellipse and a parabola

so wouldnt it be 175 since its only half?

mb for not stating that its a semi ellipse

[torn swift]

the fact is is semi won't alter the values of a and b, we'll just have to restrict the values so that we only get the upper half of the ellipse

[round eagle]

oh okay

so stick w 350

[torn swift]

so we have a,b and (h,k) and you know which equation form to use so put it all together and write the equation for me

[round eagle]

okay brb

(x-0)^2/(525)^2

• (y-0)^2/(350)^2

=1

[torn swift]

perfect, and you can get rid of the zeros lol

[round eagle]

okok

[torn swift]

$\frac{x^{2}}{525^{2}} + \frac{y^{2}}{350^{2}} =1$

[obsidian monolith]

⚡Amphy⚡:

[round eagle]

ooo thats fancy

[torn swift]

here's what it looks like in a graph calc in case you were wondering

[round eagle]

yeah im required to graph it as well

[torn swift]

graph by hand?

[round eagle]

oh no just a screenshoot. picture

shot/ picture

[torn swift]

ok lol, well graph you own please

[round eagle]

yeah i just did it

so my ellipse satisfies the equation and delivers the 350ft height as well as the 1050 ft width

but my parabola satisfies the height, but over shoots the width

because y=(-0.001)x^2+350

gets me a 591.61 width

i dont think it matters nevermind lol

but thank you so much ❤

u really helped me out alot @torn swift

preciate you

[torn swift]

you have to revise that parabola equation then lol

[round eagle]

oof

dont worry i will figure it out, i dont wanna burden you anymore

[viscid thistle]

(x^2)/((x-3)^2(x+5))

keep messing this up

doing partial decomposition

(A/(x+5))+(B/(x-3))+(C/(x-3)^2) = x^2

A = 25/64

C=9/8

B=39

(25/64)/(x+5) + 39/(x-3) + (9/8)/(x-3)^2

says I'm wrong still

[torn swift]

$\frac{x^{2}}{(x-3)^{2}(x+5)} =\frac{A}{x+5} + \frac{B}{x-3}+\frac{C}{(x-3)^{2}}$

[obsidian monolith]

⚡Amphy⚡:

[torn swift]

@viscid thistle this is what you did correct?

[viscid thistle]

yes

[torn swift]

so what is the next step you did?

[viscid thistle]

multiplied the whole right side of the equation by (x-3)(x-3)(x+5)

resulting in (x-3)(x-3)A + (x-3)(x+5)B + (x+5)C

[thick raptor]

= ?

[viscid thistle]

x^2

[torn swift]

that is correct so far

[viscid thistle]

I then let x = 3

and let x = -5

then arbitrarily chose x = -10 to find B

[thick raptor]

x=0 tbh

[torn swift]

I would actually multiply that whole thing out and then equate the powers of x to find the coefficients that way

[thick raptor]

God no why

[torn swift]

I guess it's not fast but it always works lol

[thick raptor]

That's 🤢

[viscid thistle]

I know C = 9/4 is correct

I am kind of sure A = 25/64

B is probably where I messed up

[thick raptor]

But plugging in numbres always works

And it avoids systems of equations

[torn swift]

eh I always did it equating coefficients lol

[thick raptor]

How are you getting C = 9/4?

[torn swift]

I hardly do partial fractions anyways so it was never a huge time loss

[thick raptor]

smh

[torn swift]

now that SA points it out, you should check your value of C

[viscid thistle]

let x = 3*

8C = 9

C = 9/8

oh fuck

[thick raptor]

[torn swift]

[viscid thistle]

[thick raptor]

my eyes

[torn swift]

[viscid thistle]

is C = 9/8 wrong

[thick raptor]

You just showed C = 9/8

[viscid thistle]

A = 25/64

now i'll let x =0

[thick raptor]

Mhm

x = 0 => small numbres

Usuw6hbs7mogtaply works out nicer is all

[viscid thistle]

585/64 = 3B

(25/64)(-3)(-3)+(9/8)(5)=3B

[thick raptor]

Actually mb

[viscid thistle]

585/192 = B

[thick raptor]

It's easier here to partially multiply it out

You get x^2 = Ax^2 + stuff + Bx^2 + stuff + stuff

1 = A + B

[torn swift]

equating coefficients I see 👀

[thick raptor]

Generally speaking I only do it for the highest and lowest powers

Cus it's messy in between

[viscid thistle]

oh

I see where I messed up B

[thick raptor]

now if u wanna be big bren, you would take the residue of both sides of the original thing at x=3

[viscid thistle]

b = 39/64?

[thick raptor]

Mhm

[viscid thistle]

so

(25/64)/(x+5) + (39/64)/(x-3) + (9/8)/((x-3)(x-3)) = x^2

[thick raptor]

= x^2/[(x-3)^2(x+5)]

[viscid thistle]

oh yeah

[torn swift]

$\frac{x^{2}}{(x-3)^{2}(x+5)} =\frac{25}{64(x+5)} + \frac{39}{64(x-3)}+\frac{9}{8(x-3)^{2}}$

[obsidian monolith]

⚡Amphy⚡:

[viscid thistle]

yeah sorry guys I think I'm just messing up the arithmetic

[thick raptor]

[torn swift]

partial fractions tend to get messy sometimes lol

[thick raptor]

||assuming our gender||

@torn swift only if u do it inefficiently

[viscid thistle]

holy shit that sucked

[thick raptor]

[torn swift]

if it works it works

[viscid thistle]

but I kept writing the wrong numbers

like a tard

[thick raptor]

Only took u 15 minutes in this one go

[torn swift]

math test be like: you have 5 minutes left

[viscid thistle]

first time doing these

we actually get too much time on our tests

[torn swift]

time to make the fraction larger then

[thick raptor]

[hybrid charm]

Partial fractions difficult

[torn swift]

irreducible repeated quadratic factors in denominator then

[thick raptor]

Larger thonkers

lel

[viscid thistle]

yeah

that's the next question on this hw

[thick raptor]

For those I usually tackle with sub in 0 and take leading coefficient

[torn swift]

multiply everything and equate coefficients

[thick raptor]

bruh why wouldn't you plug in points that make the problem trivial tho

[torn swift]

because I forget lol

[thick raptor]

[hybrid charm]

I would prefer subbing values of x

[thick raptor]

Who the hecc u forget

[torn swift]

because out of practice so I use the method that I remember the most lol

[thick raptor]

When u multiply 'em the factored form is literally staring u in the face

Then learn to plug in more

[viscid thistle]

this is fun though

compared to just applying trig identities

[thick raptor]

Oof

[viscid thistle]

even though it pissed me off because I can't remember the numbers for more than two seconds

[torn swift]

method of undetermined coefficients for differential equations be like...equate powers please

[thick raptor]

[torn swift]

that's what my book said to do lol

[thick raptor]

Bet u were also taught to just straight up multiply

[hybrid charm]

2 seconds not enough

[thick raptor]

er differentiate

[viscid thistle]

1/((x-1)(x+1)(x))

[torn swift]

that's what the book said to do, but I suspected that was very bad method so was looking for a better one

[viscid thistle]

= A/(x-1) + B/(x+1) + C/x

[thick raptor]

That one is particularly ez

[viscid thistle]

yeah

[thick raptor]

tbh memorize that one due to telescoping sums

[viscid thistle]

eh

I won't be taking maths anymore till fall

[torn swift]

where repeated irreducible quadratic factors at?

[viscid thistle]

question after this

[torn swift]

lol time to die then

[velvet ether]

partial fractions?

[hybrid charm]

Yeah

[velvet ether]

i agree, it gets a little messy sometimes but once you get it into a matrix you should be good from there

just don't screw up the numbers lol

[viscid thistle]

lol I fucked this one up I think

[velvet ether]

what was it?

[viscid thistle]

holy shit I'm dyslexic

1/((x-1)(x+1)(x))

[hybrid charm]

What do you get in matrix?

[viscid thistle]

(x+1)(x)A + (x-1)(x)B + (x-1)(x+1)C = 1

a = 1/2
c = -1
b = 1/2

says im wrong though

[velvet ether]

substitute random numbers in for x (0, 1, 2 for simplicity) and simplify the equations

ill give you an example

[viscid thistle]

I did -1 and 0

[hybrid charm]

x= 1 get A

x=0 get C

[thick raptor]

That's definitely right

[velvet ether]

you can do that too

[thick raptor]

Er wait

[velvet ether]

but a matrix is more consistent so you don't have to figure out the numbers

[thick raptor]

@viscid thistle you probably switched B and C

[viscid thistle]

yeah that's what happened

[thick raptor]

Literally dyslexic

[hybrid charm]

Matrix you mean solving system of equations?

[velvet ether]

nope it works for partial fractions

just ignore the linear dynamic system part thing. this was part of something else i was doing

but that also works

[hybrid charm]

Oic

[velvet ether]

😄

[viscid thistle]

huh got same answers

i'll try putting it in again

-1/x + (1/2)/(x-1) + (1/2)/(x+1) = 1/((x+1)(x-1)(x))

wow I'm retarded

says I'm wrong still

[velvet ether]

hmm

what did you get?

[viscid thistle]

just says I'm wrong

[velvet ether]

oh wait nvm lol

so the original problem was A/x + B/(x-1) + C/(x+1) = 1/((x+1)(x-1)(x))?

[viscid thistle]

(1)/((x)(x-1)(x+1))

yes

[velvet ether]

kk gimme a sec

i got the same as you did

what did the solutions say?

@viscid thistle

[velvet ether]

oh wait a sec

nvm thats what i got lol. can i see your work?

[viscid thistle]

yeah

(1)/((x)(x-1)(x+1))

A/x + B/(x-1) + C/(x+1)

[velvet ether]

i did mine the matrix way

[viscid thistle]

(x-1)(x+1)A + (x)(x+1)B + (x-1)(x)C

let x = 0

A = -1

[willow bear]

,w partial fractions 1/(x(x-1)(x+1))

[obsidian monolith]

Display results online and refine query

Requested by Ann#8192

[viscid thistle]

u wot

[velvet ether]

yep thats what it says in the key

[willow bear]

oh but the thing is x^3 - 1 is not x(x-1)(x+1) lmao

[velvet ether]

i don't see anything wrong

[willow bear]

are you sure the denom wasn't x^3 - x

[viscid thistle]

1/(x^3-1)

did I fuck this up

[velvet ether]

LOL

[willow bear]

bc x^3 - 1 is not x(x-1)(x+1)

[hybrid charm]

[viscid thistle]

i want to die

[velvet ether]

the partial fractions are right

so thats a plus

not like it matters though lol

[viscid thistle]

oh I see

my bad

[hybrid charm]

Let's see it as practice of extra problem

[viscid thistle]

,w partial fractions 1/(x^3-1)

[obsidian monolith]

Display results online and refine query

Requested by Caesar#9438

[willow bear]

did i not answer this already in #help-1

also, in response to the previous problem

[willow bear]

the -1 is just what they multiply both sides by

the 0 is because you're looking for x intercepts aka roots

no, it's there to get rid of the non-unity leading coefficient

if you know what an x-intercept is, then it should be obvious why the function is equated to 0 and not something else, and when this is appropriate to do

[velvet ether]

if the graph is an upside down u, it has reached its highest point on the vertex, hence its maximum value

if the graph is just a regular u, it has reached its lowest point at the vertex, hence the name minimum value

np!

[torn swift]

+x^2 will open up

so it will have min value

[velvet ether]

that too ^^

[torn swift]

-x^2 means it will open down so it will have max value

[velvet ether]

the axis of symmetry is always gonna be the x-value of the vertex

[torn swift]

h is x coordinate of vertex

[velvet ether]

and axis of symmetry

😄

[viscid thistle]

Anyone have a good resource for inequalities? I'm really struggling to grasp them.
Specifically solving them.

I've got a question, what's the difference between Precalculus and Calculus?

Is Calculus harder?

Precalculus is the stuff taught before calculus.

Oh so it's built on top of it?

yes.

You pretty much need to know the stuff before calculus to do calculus

Interesting
Thank you

[serene heath]

who would've guessed that precalc is previous to calc

[viscid thistle]

lmao

[frigid vale]

hmmmmmm

[placid tusk]

I'm guessing this goes here

[obsidian monolith]

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

[placid tusk]

Is this a valid proof of the triangle inequality

?

[frozen needle]

looks alright

[idle dust]

heyi have 3 trig questiongs

can someone help me

1. sin(arctan(sqrt(x)/(1+x)))

im not sure how to remove the function composition

[tawny nacelle]

wtf

[olive star]

wtf

is that even precalc

college level precalc

?

i barely learned from my precalc class

it was online

it was hard

[tawny nacelle]

[olive star]

the whole reason i joined this server was bc that precalc class was too hard

[idle dust]

mniip

[cerulean wharf]

you could try replacing sin and arctan by the complex formulas

[idle dust]

BRUh

[tawny nacelle]

like

f(x) = sin(x)

g(x) = arctan(x)

it kinda goes like that

[idle dust]

f(g( BUT THAT DOESN t change anuthing

[tawny nacelle]

y du wanna change anything ??

[idle dust]

it says simplify the following to remove the trigonometric and inverse trigonometric funtions

[tawny nacelle]

bRUh

[idle dust]

how u do dis

no comprnedo

[tawny nacelle]

bro

draw some triangles

[idle dust]

OK

[tawny nacelle]

i got $\sqrt{\frac{x}{x^2 + 3x + 1}}$

[obsidian monolith]

soap:

[tawny nacelle]

now check the answer and tell me if its rite

imma watch anime meanwhile

ping when u do

[idle dust]

wadafuk

[tawny nacelle]

is it rite tho ?

[idle dust]

NANDESCA

i have plunged into deep confusion

[tawny nacelle]

u know what nandesca means huh

[languid crane]

FOIL

[tawny nacelle]

[languid crane]

first outer inner last

right?

[tawny nacelle]

thats gay

[languid crane]

eyy

ok

im doing algebra

:(

[tawny nacelle]

more like

algebRUH

amirite