precalculus

[terse ravine]

It's wrong

[uncut mulch]

which is bigger

[viscid thistle]

your answers are correct, but in the wrong place

[terse ravine]

Was a silly mistake.

[celest void]

@lusty falcon use in conjunction with sin^2 a + cos^2 a = 1, see if you can write cos (2a) purely in terms of cos

[echo pelican]

Can someone help me with this problem 22 here? Im helping a student with precalc and this is the only thing on their practice exam I cant figure out

I know there is a sec(pi/2 - x) identity but this one is (pi - x)

[viscid thistle]

Do you know cos(pi - x)

[echo pelican]

Know only sec(a)

Just what it says there

[viscid thistle]

do you know it's identity I mean

[echo pelican]

No i do not

[viscid thistle]

do you know cos(pi/2 +x ) ?

[echo pelican]

I know theres a pi/2 - x for each trig function

But i do not know how to invoke those for this question

I thought it must be a typo the precalc trig stuff usually comes directly from the identities but i was wondering if there is something simple i am overlookimg

[cerulean quarry]

ok 1 sec

u need 22 answered right/

?

[echo pelican]

Yes

That is all the given information there

[cerulean quarry]

f(a)=11/3 and f(theta)=sec(theta)

u want f(pi - a )

[echo pelican]

Yes basically

Sec(a) = 11/3

Find Sec(pi-a)

[cerulean quarry]

first u need to solve for A

yes

do u know how to do that?

[echo pelican]

Well i think this wants an exact value

And no calculator

[cerulean quarry]

that might be hard

[echo pelican]

My impression is that it was meant to be a trig identity problem

But there was a typo

[cerulean quarry]

i dont remember the identity but all u do is solve for A

does is explicitly say that u need a exact answer?

[echo pelican]

We can solve for A in terms of cosine inverse function, but based on the problems around it i assumed it was supposed to be a trig identity problem

I guess it just says "value"

[cerulean quarry]

well its cosine 3/11 that doesnt come out very pretty

and ur subtracting it from pi so its still not going to be pretty

[echo pelican]

There's an identity for pi/2 so my thought is that they meant to write that

Pi/2 - x

[cerulean quarry]

for secant or cosine?

duh both ><

[echo pelican]

For every trig function

Lol yeah

[cerulean quarry]

let me check some trig identities real quick

[echo pelican]

Every basic* trig function

Thank you

I just wanted to make sure i was not overlooking something simple here

[cerulean quarry]

yeah i dont think so

[echo pelican]

Or what if you combined double angle identity with the (pi/2 - x) ?

Factor 2 out and treat it as double angle

Well wed have a/2 tho...

[cerulean quarry]

the problem is the 11/3 isnt a nice point on the unit circke

so no matter how u rotate it by those angles its still going to spit out an ugly number

[echo pelican]

Ok thank you im gonna try to take a crack at it mixing and matching identities but i thought it must be a typo

[cerulean quarry]

ok

i wouldnt solve for an exact answer at all i'd just leave it as an expression imo

[echo pelican]

Yeah i think there is no choice as written

[cerulean quarry]

the question looks likes its testing to see if u know how to solve for a

[echo pelican]

Maybe that is the case but this one was with like 4 trig identity problens in a row

And it is one division by 2 away from being an identity

Thank you very much for taking a look with me

I was afraid i was missing something very obvious

[cerulean quarry]

actually i didnt check what would happen if u put them in fraction form

[echo pelican]

How do you mean?

[cerulean quarry]

actually no nm

nm that

[echo pelican]

Ah oj

[cerulean quarry]

i think its as i said it was

[echo pelican]

Ok*

[cerulean quarry]

i thought of it then in came out wrong

it

[echo pelican]

Yeah seems to be no other way to answer it

[viscid thistle]

cos (pi-x) = -cosx is a derivable thing

If you know what cos(pi/2 +x)

[echo pelican]

Is that a trig identity?

Cos(pi-x) = -cos(x) ?

That must be it then if it is!

[viscid thistle]

It is

[echo pelican]

Thank you so much! I checked like 10 trig identity tables and none of them had that

[viscid thistle]

Can someone help me with this problem?

[queen vapor]

@viscid thistle what's the problem?

[viscid thistle]

I’m supposed to verify the trig function

But I don’t understand the first step...

[queen vapor]

First of all, Tanx = Sinx/Cosx

[viscid thistle]

Yes

[queen vapor]

So, they used for the denominator

[viscid thistle]

Ok

Idk if you can see, but I sent a pic of the problem I’m working on

[queen vapor]

Yeah

[viscid thistle]

Ok

[queen vapor]

Yeah, best way is 2Cosx is common

[viscid thistle]

Ok

[queen vapor]

So, 2Cosx (1 - 1/1-Tanx)

Now simplify the portion inside Parenthesis

[viscid thistle]

Ok, and distribute right?

[queen vapor]

yep

[viscid thistle]

So would it look like this?

[queen vapor]

( 1 - 1/ 1-Tanx)

[queen vapor]

Lemme use paper

wait

[viscid thistle]

Ok, thank you so much.

[queen vapor]

You had everything right

[viscid thistle]

Ok

Thank you!

[queen vapor]

Welcome

[split epoch]

Can someone let me know if I’m correct

[uncut mulch]

Question 4:
incorrect range,
incorrect x-intercept

[queen vapor]

,rotate

[queen vapor]

See the end points on the X axis

[uncut mulch]

Question 5:
incorrect domain,

[queen vapor]

like (-2,4]

Since, the function is bounded

so we have fixed range also

oops

[silver matrix]

How can I approach this problems?

I know VA: 3
HA: y = 2
x-intercept (y=0): 4

[willow bear]

i'd start with y = 2 + 1/(x-3) and then do some scaling and shifting to get the y-int right

[silver matrix]

ok thanks

why have you put 2 +1

instead of 3?

shifting and scaling i guess what I learnt on transformation functions, like multiplying x or summing to get a vertical or horizontal shift, etc

[willow bear]

i didn't say $y = \frac{2+1}{x-3}$ god forbid

[obsidian monolith]

Ann:

[silver matrix]

yeah, i mean u said start with that

not that it was the final solution i know

[willow bear]

i said $y = 2 + \frac{1}{x-3}$ and honestly i'm really upset you read the 2 as being in the num

[obsidian monolith]

Ann:

[silver matrix]

i'm sorry :(

[willow bear]

you can find a function of the form $y = 2 + \frac{k}{x-3}$ where $k$ is a constant

[obsidian monolith]

Ann:

[silver matrix]

any resources you could share with me?

I... i don't know how to do it

nor why start from that

I only understand the x - 3 for the VA

but not why doing the rest of it

ok

what is the meaning of scaling?

2 + 2/x-3 satisifies the VA and the x-intercept

ok, now I see my problem

HA is not y = 2

well yes

but not directly doing 2 = 2 + 1/x-3

ok, problem is i need another x

how am I going to get both y = 0 and y = 2

that's the problem i hav

[willow bear]

PARENTHESES

\verb|1/x-3| reads as $\frac{1}{x} - 3$, not as $\frac{1}{x-3}$. spacing is not a substitute for parentheses.

[obsidian monolith]

Ann:

[willow bear]

nor why start from that
it's a bit arbitrary but the form i gave you is relatively simple

just require that the curve pass through (4,0) and you'll get an equation in k which can then be solved

@silver matrix honest to god you're just kinda really overthinking this rn

[echo pelican]

@silver matrix

The horizontal asymptote occurs when you have the same degree in numerator and denominator

[silver matrix]

@silver matrix honest to god you're just kinda really overthinking this rn
nah, I'm not. I'm just dumb. I don't understand how someone can be so cool, that by giving some premises can build a function. I don't see a way that can find any function you want; that would be even cooler

my god

[willow bear]

i'm not looking for ALL rational functions which satisfy this

[silver matrix]

yes I know

[willow bear]

i'm just giving one of many possible methods

[silver matrix]

👍

[rare zephyr]

does a. means h(t) = 0?

does b. means do I have to differentiate it twice or once?

[viscid thistle]

does a. means h(t) = 0?
@rare zephyr looks like it

Though it could've been stated it more clearly

[rare zephyr]

alright

[neat birch]

does b. means do I have to differentiate it twice or once?
you need to find the maximum and minimum heights, meaning that the derivative of that formula will intersect with the x axis

[chrome thorn]

Is there an asymptote to a square root function?

If its a rational function inside the square root

[blissful ridge]

I didn't write out all the steps

But we can say that it holds from there, right?

@willow bear
Sorry for the ping, that channel was occupied

Ignore the quality of pages

[velvet blade]

@blissful ridge bro in the second page on RHS, isn't that supposed to be (3n+3)/2?

[blissful ridge]

Where??

[velvet blade]

https://discordapp.com/channels/268882317391429632/363224154469826562/765962534511050792

first line

[blissful ridge]

Actually in the first page I wrote divided by 3 by mistake

It should be 2

[velvet blade]

Oh okayy

[hearty vector]

can someone tell me how to graph rational functions like this ?

ik how to get the VA, HA, Hole, xint, y int

i just have trouble graphing it by hand

[viscid thistle]

same steps as every function

If you have all that info, i don't know where do you have troubles

Start by imagining an idea of how it'll look

[hearty vector]

i also dont understand how the line in the middle happens

when there is a H.A. at y= -1

my math is questionable

[viscid thistle]

What?

How the line appears???

[hearty vector]

i mean like

the third line in the middle

goes through y = -1

[viscid thistle]

Yeah so what

A horizontal asymptote can be perfectly crossed

do you agree that an asymptote is a line?

[hearty vector]

uhh sure?

[viscid thistle]

the HA is just there to indicate us end behavior

horizontal asymptote is a behaviour of the function for very big x

[hearty vector]

i havent learned what lim is yet

horizontal asymptote is a behaviour of the function for very big x
@viscid thistle as x goes to the extreme value of it right?

[viscid thistle]

yes

[hearty vector]

ok

so i guess now my main question is, how do would i graph this by hand? do i literally just graph the intercepts, and plug in test points?

[viscid thistle]

Start by imagining an idea of how it'll look
then, organise and represent your data obtained (HA, VA, x-intercept, y-intercept) and make sure you mark everything first

and then draw the line

[hearty vector]

what are some key things that i should consider by imagining how it'll look?

the degree?

[viscid thistle]

do you know about derivatives?

[hearty vector]

nope

[viscid thistle]

basically your data obtained, HA, VA, x-intercept, y-intercept, domain, maybe range

[hearty vector]

ik the difference quotient

do you know about derivatives?
@viscid thistle

Lul

basically your data obtained, HA, VA, x-intercept, y-intercept, domain, maybe range
@viscid thistle ok, i'll give it a go

thank you

[viscid thistle]

domain is very important

[hearty vector]

mhm

[viscid thistle]

also when f(x)>0 is a very useful one

yeah but ig they haven't learned derivatives so i didn't include it

[hearty vector]

thank you for the help!

[mighty crater]

yo I need help with a question

For the following function, sketch the reciprocal. State all key points and properties. (App 8)

this is Rational Functions unit

[full pagoda]

find values for f(x) and flip 'em @mighty crater

for example, f(3) = 4 so the reciprocal is 1/f(3) which is 1/4

[mighty crater]

I need help with this question

[jagged orbit]

how can i construct a function such that lim f(x) x-> -4+ is 0 and lim f(f(x)) x -> 4+ DNE, is not +/- infinity
and also of domain r?

could anyone help :/

[jagged orbit]

<@&286206848099549185>

[mighty crater]

<@&286206848099549185>

[junior sable]

can someone please explain to me how 16^(3/4) = (16^(1/4))^3? I get that it's what I'm supposed to do, but I don't understand how it works.

[mighty crater]

Use photomath

@viscid thistle what grade u in

Yooooo

I need help with this

@viscid thistle bro

yea plz

these are marks lol

yea

those are just marks

like for the question

[wraith idol]

$\frac{(x-1)(x-2)}{(x-1)(x-2)}$

[obsidian monolith]

Whoever:

[mighty crater]

Man, yall are confusing the hell out of me. I just need this for my assignment due tmrw

So u don't know this... @viscid thistle

For the following function, sketch the reciprocal. State all key points and properties

how about this

@wraith idol how about this question

[mighty crater]

@wraith idol is the slant asymptote the same as an oblique asymptote?

[wraith idol]

same thing

[mighty crater]

ok thanks

[mighty crater]

when is it asking for Interval(s) when the graph is above the x-axis and below the x-axis for f(x)= 1/x^2 What does that mean, how do i do it @wraith idol @viscid thistle

bottom 2

@viscid thistle

@wraith idol

[echo pelican]

@junior sable

[bitter sandal]

hello i need some clarification regarding the sequence * 1 + 1/4 + 1/9 + 1/16 + ... + 1/n^2 *

does the n mean that it continues forever?

[harsh socket]

hello i need some clarification regarding the sequence * 1 + 1/4 + 1/9 + 1/16 + ... + 1/n^2 *
@bitter sandal la n significa que llega hasta un numero especifico

the n means that it reaches a specific number

[bitter sandal]

so the number terms of this sequence would also be "n" ?

hmm i starting to kinda get it now, thank you

[harsh socket]

so the number terms of this sequence would also be "n" ?
@bitter sandal Si hablas de como queda la sumatoria ya cuando converge, pues si deberia quedar en terminos de n

If you talk about how the sum is already when it converges, then if it should be in terms of n

[bitter sandal]

ahh i see i see, that makes sense now. thank you

[harsh socket]

🙂

[mighty crater]

Anyone can tell me if these are right or not, and what is the last one?

[steel venture]

your third one is wrong, your fourth justification could use some work

[solid juniper]

Guys help please

[steel venture]

@mighty crater

@solid juniper send the question

[mighty crater]

Can u tell me what is the 3rd on plz

[steel venture]

no

[solid juniper]

Its 5 questions so please guide me while solving it.

im dumb btw

[mighty crater]

sry bro, there is so much goin on and im stressin

[solid juniper]

So i wont understand anything right away

[steel venture]

no worries, the answer to the third one is much simpler than you think

your justification is perfect, build a simple funciton around that justification

[mighty crater]

ok

[steel venture]

flip go over to #help-9

[solid juniper]

Ok

[bitter sandal]

can this be expressed in sigma notation? >> " -1/2, 1 -3/2, 2 -5/2, 3

the only solution i have gives me the fractions only

[steel venture]

pleura can you post the question

[bitter sandal]

the instruction says: Express each sum using the summation notation

[steel venture]

start by doing a simpler case, multiply all the terms by 2

see what you get

[bitter sandal]

uhhh negative 1

[steel venture]

?

wait do you understand what the question is asking you to do

[bitter sandal]

wait if you mean the entire thing.. uh 9

[steel venture]

dont actually add them, just write out the sum

[uncut mulch]

note that you tried to consider :"1 - 3/2" and "2-5/2" as single summands

instead consider there to be 6 summands where the signs alternate

in which case you could include something like (-1)^n

[mighty crater]

I need help with this question

[willow bear]

@mighty crater still need help with this?

[solid juniper]

I need help

[chrome thorn]

How would I find the vertical and horizontal asymptotes of this function? The square root is confusing me

[placid delta]

How to find focus in this equation?

y^2+4x=0

[viscid thistle]

the focus

id change it so its based on x

y^2 + 4x = 0
x^2 + 4y = 0 (lol swap them, ill swap them back later)
x^2 = -4y
(x^2)/-4 = y

then complete the square, and find the focus

[dusk onyx]

$f(x) = x^2+lx +(\frac{l^2}{4} - \frac{9m^2}{4})$

[obsidian monolith]

ActiveChapter:

[dusk onyx]

range is the set of all real values?

[sour plinth]

that's domain

find the minimum value

of f(x)

[dusk onyx]

yikes

i see

thats alot of work

[sour plinth]

do you know how to find the minimum value of a parabola

[dusk onyx]

algebashing

i do

its just annoying to do the algebra on that

[sour plinth]

i'd differentiate it

wait nvm precalc

[dusk onyx]

youre right i could

[sour plinth]

you can use -b/2a

[dusk onyx]

much easier that way

im not on precalc level

american education system is weird to me

[sour plinth]

the x value of the min/max of a parabola is always the average of the two roots

so it would be -b/2a (for a parabola ax^2+bx+c)

[dusk onyx]

2x + l = 0

so x = -l/2

i put that in and find the y value

and then y >= answer

is the range right

[mighty crater]

@willow bear Yea i do

[willow bear]

what a delay wow

[mighty crater]

my bad bro

I had a lot of stuff that I had to do man

My bad

[willow bear]

please don't bro me

[grim canyon]

^^ I respect ya Ann I am trans too 🙂 sorry that some people don't understand it.. but I do 🙂

@willow bear Sorry about ColdKurama bae

[mighty crater]

ok I never knew tf, why are yall getting aggressive on me'

[willow bear]

i wasn't aggressive in the slightest

[mighty crater]

Im simply asking for help

[grim canyon]

no were not aggresive

[mighty crater]

Ok my bad

[willow bear]

anyway, i was busy when you replied

[grim canyon]

were just saying we are like this

[willow bear]

can you repost the problem

[mighty crater]

ok

[willow bear]

it was about sketching the graph of 1/f(x) when that of f itself is given i think?

[willow bear]

yeah ok so what have you done so far

[mighty crater]

No given equation

[willow bear]

you don't need the equation

you just need to know how to visualize going from (x, y) to (x, 1/y)

[mighty crater]

i have a very rough sketch

[willow bear]

oof no

way off

[mighty crater]

Yea i figured

[willow bear]

ok so first off draw the lines y = ±1

the points where your graph meets either of those lines? those are the ones that stay in place

[mighty crater]

The problem is with the curve right?

The lines are ok??

[willow bear]

everything is bad in your sketch

[mighty crater]

ok 😆

[willow bear]

zeros, as i said above, turn into asymptotes. there are no asymptotes in the original graph but if there were any they would become zeros

there are several key points on the graph which can be transformed explicitly

(1/2, -9/4)
(3, 4)
(4, 2)
(5, 3)

[mighty crater]

i may be wrong but when we transform these points, do we add a negative in front of the coordinates in order to make the reciprocal?

@willow bear ?

[willow bear]

why negative

you're not multiplying by -1

you're going from y to 1/y

[mighty crater]

Ok i see now

Thank u

Ok i might have one more question later, if that doesnt bother u

[grim canyon]

I mean Ann isn't the only one that could help in this channel so dw

[mighty crater]

ok yall are saving my ass, I appreciate yall

[grim canyon]

xD

[mighty crater]

Can someone check is these are correct or not

I think the 2nd one is wrong

[random cloud]

I just have to isolate y in terms of x right?

And then do the same for x in terms of y again?

[grim canyon]

,rccw

[random cloud]

Yes?

[lime bolt]

lol what else would u do

[random cloud]

How would you even solve this with compelteing the square

Cuz I've seen some weird sifusruons involving fractions where you can find an exact amount

[lime bolt]

note that 2(y^2+xy)=2(y+x/2)^2 -x^2/2

so u can just rearrange, square root and minus x

[random cloud]

Where did the 6 go

[lime bolt]

what?

[viscid thistle]

@sweet cliff PLEASE

[lime bolt]

16 consectuive messages spamming this one guy in different channels

[grim canyon]

^^ yep..

[mighty crater]

Im still kinda struggling with this question

[cedar pawn]

Whats a reciprocal

[quaint mason]

?

in what

[sick seal]

i think of it as an opposite. think of the number 2 written out as 2/1. 1/2 is the reciprocal of 2 because multiplied together it = 1

@cedar pawn

[quaint mason]

just swap the numbers.

are u sure u asked the question in the right topic

[sick seal]

hit it with the good ol uno reverse to make it = 1

i gotta question i just need a second to word it

[quaint mason]

i couldnt tell if he was kidding xD

[sick seal]

nvm i figured it out. rubber duck for the win.

[quaint mason]

lmao

aight

[viscid thistle]

how do I do question b?

i've been stuck on it for a while now

[blissful ridge]

What have you tried?

[uncut mulch]

the same way you did a)
just more algebra

^

[viscid thistle]

i think I got it

basically that (fog) (x) = x

right?

and then I try to simplify in order to try and get x

[blissful ridge]

Yeah

[zinc chasm]

https://drive.google.com/file/d/1vQyZiPhDxMqGwcAnkRy6Ob8Hja0BhnLe/view

Heya im in 8th grade

and this is my hw lol

im pretty bad at graphs so i need some explanations

any help

<@&286206848099549185> ?

[idle moat]

how do you solve this

[full pagoda]

what do you think your first step will be?

@idle moat

[idle moat]

divide by 3

after that, im a bit lost

[full pagoda]

Do you know what the opposite of euler's number is?

[idle moat]

no

are you asking for the value of e?

[full pagoda]

the natural log is the opposite of e

to get a variable out of the exponent, you take the natural log of both sides.

[idle moat]

so lne^x = ln (10/3)

[full pagoda]

correct

and since ln is the opposite of e, they "cancel" out

well they equal 1, at least

[idle moat]

i got it, thank you

[full pagoda]

no problem :)

[idle moat]

what about for

i got down to the step

log (6x - 24) = 2 log x

actually

log (6x - 24) = log x^2

nvm lol, i got it now

[full pagoda]

one of those is wrong

you wouldn't get log(6x-24)

[idle moat]

-8*

[full pagoda]

there you go

[burnt owl]

help

idk how to do d, f, c, and c

help for any of them would be much appreciated

[full pagoda]

@burnt owl d is the same as f(x) + h(x)

so x^2-3x-28 + sqrt(5-7x)

f is just plugging g(x) into itself for x

so whenever you see x in g(x), plug in the entire g(x) equation

c is the same thing except whenever you see an x in g(x), plug in h(x)

and then you put c twice and i'm not sure you meant to

[burnt owl]

oh

[viscid thistle]

@umbral haven remove this from here

And dont double post in future

[umbral haven]

yes sir

where does that question belong

calc or precalc

[viscid thistle]

Calc

Idk much about this precal and calc division

[hallow bison]

Can anyone help me solve this?

[odd cloud]

Apply linear velocity = angular velocity * radius, here in your problem liner velocity and radius are given.

[flint pulsar]

The size of a population of rabbits is determined by the rule P = $6400*30^{0.2t}-400$ where P is the size of the population t years after January 2006. Sketch the graph of P against t.

[obsidian monolith]

steph:

[flint pulsar]

i was just wondering what does the e represent for the x-int?

[willow bear]

e is e

yknow, the base of the natural logarithm

[delicate rivet]

wait but what is a natural base of a logarithm? like what does it represent?

[full pagoda]

euler's number comes up everywhere. it's mostly related to exponential growth or rate of change

[flint pulsar]

okay, so how would it be applied in these types of questions? like when i solve for the x-int, how would i know to use e?

i also solved for the x-int on my calculator and it came up with a slightly different answer to the x-int provided

[full pagoda]

e is mentioned or can be derived in just about every growth formula there is

at least of the ones i can think of

so you'd know to use it based on a given formula

i know population growth formula has e in it

[delicate rivet]

ohhh, so you'd choose whether or not to use e depending on the situation in the problem?

[full pagoda]

pretty much

but when talking about population growth, it's in the formula

[delicate rivet]

wait what's your population growth formula?

[full pagoda]

The one that comes to mind for me is the Pert formula

P_0 * e^rt

where p_0 is initial population, r is rate, and t is time

[delicate rivet]

ohhh, so instead of a numerical value we use it, (for population growth)

[full pagoda]

when applying that formula, yes

[flint pulsar]

ohh that makes more sense thank you sm for your help!

[delicate rivet]

last question to double check; you'd basically use e for situations where we need a 'smoother' growth?

[full pagoda]

honestly i wish i knew more about it. i'm not sure about 'smoother'. most of what i'm basing this on is the fact that it is an essential constant when talking about growth/rate of change. if you're studying those topics, you will be seeing euler's number a lot

[willow bear]

e is just a number lol

the function e^x has some properties which set it apart from other exponential functions but those don't really come up until calculus

[full pagoda]

comes up in alg 2

[delicate rivet]

ooo okie, tysm for you help!

[full pagoda]

👍

[bitter sandal]

is this true? apparently my book doesnt say something about it, and i've searching in google but idk how do i word it out on the search bar

[blissful ridge]

Yes that is true

[bitter sandal]

thank you

[cedar parcel]

can some help me?

The problem I have is due tommorow

[viscid thistle]

@cedar parcel i can't help if you don't post the problem

[fossil crown]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[cedar parcel]

the bottom two scteenshots are the mls

*mLs

[viscid thistle]

Yeah no don't multipost just realised

[fossil crown]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[cedar parcel]

can someone jsut help me?

*just

And i posted in multiple channels because one one is helping me

*I

[viscid thistle]

What kind of argument is that

[cedar parcel]

its not

its a fact

[viscid thistle]

So you are telling me you break the rules of the server just because you can't be patient?

[cedar parcel]

its due in an hour

[viscid thistle]

Well that's definitely not our fault.

[mighty crater]

My man is desperate

just help the man yo

[viscid thistle]

?

[onyx saffron]

我在试做数学作业，可是一问题得难。问题是“A Ferris wheel 50 ft in diameter makes on revolution every 40 seconds. If the center of the wheel is 30 feet above the ground, how long after reaching the low point is a rider 50 ft above the ground“。正确的答案是”55.903秒钟“，为什么？

[viscid thistle]

@onyx saffron still stuck?

[onyx saffron]

i got an answer but it isn’t 55.903, it’s 15.903, i don’t understand why 40 was added

[vapid blaze]

At what points does this graph have a constant rate of change.

[viscid thistle]

At points of inflexion

[blissful ridge]

Rate of change is zero at that very instant but can we say it's constant??

[onyx saffron]

since it’s an odd power greater than one, the derivative is an even power and therefore it is never constant

i think

[vapid blaze]

This is the equation

It has an Odd power, so I was thinking the same

[viscid thistle]

Rate of change is zero at that very instant but can we say it's constant??
Hmmmm

[onyx saffron]

the derivative never remains constant, so there are no points where it has a constant rate of change

[viscid thistle]

Hmmm right

Mb

i got an answer but it isn’t 55.903, it’s 15.903, i don’t understand why 40 was added
@onyx saffron can i see what you did

[vapid blaze]

Thanks guys

[viscid thistle]

If you are still looking for help

[viscid thistle]

That is literally ilegible

[onyx saffron]

ohh the watermark

give me a second

so i set up a triangle, used sign to find the angle, added 90 to that, divided by 360, multiplied by 40, and that gave me my answer

[viscid thistle]

Okay hold up

@onyx saffron okay

You there?

yo?

[onyx saffron]

sorry i was running

what did i do wrong ?

[viscid thistle]

everything until "divided 360" is correct

excluding diving by 360

so you have your angle arcsin(4/5)*pi/2

do you know the relevant formulas for time of revolution or angular velocity?

@onyx saffron

[onyx saffron]

i think so

[viscid thistle]

can you be more active?

if possible

if not it's okay but let me know

[onyx saffron]

i'm here

[viscid thistle]

okay

[onyx saffron]

i'm trying to get the angle in degrees, not radians

[viscid thistle]

is this familiar to you $\omega=\frac{2\pi}{T}$

[obsidian monolith]

Al𝟛dium:

[onyx saffron]

nope

wait maybe

does that have to do with the period of a function?

[viscid thistle]

yes

[onyx saffron]

then yes, i have

[viscid thistle]

T is the time spent to complete one revolution

which you have

[onyx saffron]

ohh, okay

[viscid thistle]

so when we get the angular velocity we can finally solve for the time with $\theta=\omega t$

[obsidian monolith]

Al𝟛dium:

[onyx saffron]

so i use that instead of /360

[viscid thistle]

with theta being arcsin(3/4)+pi/2

yeah working with angles works too obviously

but you need to be consistent

be careful with units

[onyx saffron]

that makes sense

thanks

[graceful bear]

how to solve circles?

[viscid thistle]

?

[blissful ridge]

We humans and even you hobo cannot understand such high level of communication

[viscid thistle]

This reaction plus HoboSas's pfp made me laugh so bad

?
I'm talking about this

Plus their dp

how to solve life problems?

Wait are you serious

no

how to solve circles?
@graceful bear

be more specific

I guess we will never find out

[graceful bear]

Conic section of circles is what I meant to ask

[waxen creek]

?
@viscid thistle @marsh crag

[viscid thistle]

it's still very vague

[waxen creek]

WE found a MATH GORILLA

OMG

YESS

HI

[marsh crag]

OMG

[viscid thistle]

yo

[marsh crag]

YO

You like maths?

[viscid thistle]

#chill

[marsh crag]

O damn

[quaint mason]

im confused on what it means by smaller and larger values

[viscid thistle]

n < your number < n + 1

[quartz oxide]

can someone help

[quaint mason]

@viscid thistle wait so

what is n??

[viscid thistle]

An integer

[quaint mason]

um

[viscid thistle]

Do you even understand the problem?

[quaint mason]

ah i got it

log4 (1/50) which is 1/50=4^y

=.02=4^y

solve for y for where .02 is between the values

ty @viscid thistle

i have a question, in that case, what happens if you do not have a base? @viscid thistle

[viscid thistle]

Wdym?

[quaint mason]

so one of the question asked the same thing but doesnt give a base

its just log(.009)

[viscid thistle]

It's probably either base 10 or e

[quaint mason]

is it usually just 10 by default?

[uncut mulch]

depends on context/whether the book has predefined the notation/level of math

[quaint mason]

i see

thank u

[uncut mulch]

at lower levels it's usually 10
as you get higher, e

[quaint mason]

aka 2.71

approx

[loud marsh]

how the fuck do you do this

wait should this go in trig

[uncut mulch]

there's a fine line between certain trig questions and pre-calc

so its ok

what have you tried?

[viscid thistle]

Are we assuming he's asking to prove it?

[uncut mulch]

yes

[viscid thistle]

Lol alr

[uncut mulch]

you could do:
$$LHS = \br{\frac{\tan(x) + \tan(y)}{\cot(x) + \cot(y)} \cdot \frac{1}{\tan(x)\tan(y)}} \cdot \tan(x)\tan(y)$$

[obsidian monolith]

ramonov:

[viscid thistle]

@loud marsh ^

[uncut mulch]

and the problem reduces to simplifying the stuff in the big set of to 1

technically not mixing sides 😄

[novel pecan]

Y'all are smart asf wtf

[torn seal]

Helloo Im new to this server and I was wondering if someone could help me out

for my first question Id just like to check my work

Question: Find the Inverse of f(x)=2x^3+1.

My answer: f(x)=2x^3+1
y=2x^3+1
x=2y^3+1
x+1= 2y^3
y= x+1/-4
f^-1(x) x+1/-4

[echo pelican]

@torn seal

[torn seal]

Ohhhhh- Im so bad at thisss

Thank you for that.. @echo pelican

[echo pelican]

Yw good luck with the other problems

[torn seal]

thank you

Can you maybe show me how to solve something like this

[echo pelican]

Yes hold on

Im assuming when they right fg

They mean f of g of x

[torn seal]

basically f times g

[echo pelican]

For this composition of functions its like function done to a function

fg

Means do to g(x) what f says

So replace the x in f(x) with whatever g(x) is

@torn seal

Then wherever x appears in f you are instead replacing it with g(x)

[torn seal]

Ohh you're really smart, hopefuly Ill understand this. SO basically you just replace f and g

[echo pelican]

If its

fg

That is saying

Go into f(x) and wherever x appears we are going to instead replace it with g(x)

If i said gf it would be the other way around

[torn seal]

That makes so much more sense actually

[echo pelican]

We can do even

F(g(h(x)))

3, 4, 5 composition of any number of functions

The inner most one gets done first and you work your way outward to the outside most

[torn seal]

wait Im talking about the product

[echo pelican]

I am 90% sure that fg notation is composition

If it means multiply f and g

It would be like this:

[torn seal]

ive gotten this f(x) = 2x - 5
g(x) = 2 - x

(2x - 5)(2 - x)
4x - 2x^2 - 10 + 5x
-2x^2 + 9x - 10

[echo pelican]

Yeag

Perfect

Check about the notation with your prof

To be sure

[torn seal]

yay thank you

Im sorry if Im bothering you but is it okay if I ask another question..?

[echo pelican]

Lol go ahead

It doesnt hurt to ask

[torn seal]

aw thank you so much

[echo pelican]

Ok for this one

P gets you pound of apples

C gets you cost if you already know pounds

So you can do a composition of functions that will get you

Cost of P pounds

C will be the outside function since Pounds needs to be inside of the cost function

Similar to the first way i showed you the fg problem

Basically you want

C of P

This will get you your cost in terms of hours

Then you can substitute 32 in for h

Just @ me if you have trouble

@torn seal

[torn seal]

how does this look

P(n) = 375(32)
P(n) = 12,000

C(n) = 0.35(12,000) + 1000

cost of operating the cannery for 32 hours: $5,200.00

[echo pelican]

Yeah that seems about right

Good stuff

[torn seal]

:)

[sand prairie]

i dont know where to really put this nor do i know if im allowed to ask for ideas but can anyone help me with ideas of fractal art

[solar cradle]

sierpinski triangle, menger sponge, the dragon fractal

hilbert's curve

[sand prairie]

sry for not responding, but I want to draw something that isn't one of those

i have no ideas

[solar cradle]

You can come up with your own fractals then

you can do it with any arbitrary recursive pattern

[craggy hamlet]

Can someone give me a short definition of cosine? thank you

*and explanation

[solar cradle]

Imagine a right triangle, with one of the (non-right) angles as being x

The cosine of x is the ratio of the length of the adjacent side to the length of the longest side

The longest side is also known as the hypotenuse

[craggy hamlet]

ok thanks. then what is tangent then

[sand prairie]

do you think a simple pyramid/triangle with increasing blocks per layer work

im really overthinking this

[solar cradle]

ok thanks. then what is tangent then
@craggy hamlet the opposite side over the adjacent side

do you think a simple pyramid/triangle with increasing blocks per layer work
@sand prairie not really a fractal then

[sand prairie]

wait, im confused again xD

oh

its not repeating

[viscid thistle]

i need help with basic calculus

[willow bear]

@viscid thistle ok do you have a problem youre doing rn?

...

[viscid thistle]

yes

sorry

had to go to toilet

whats the derivative of

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

does power rule applies over function ??? i learnt it only applies over x not like (x+k) k is constant

@willow bear

[willow bear]

have you heard of the chain rule?

[viscid thistle]

yes

ik it

[willow bear]

cause that's what you're gonna need in order to differentiate this

$\dv{x} (x^2 + k^2)^{-3/2} = -3x (x^2 + k^2)^{-5/2}$ is what it's gonna work out to.

[obsidian monolith]

Ann:

[viscid thistle]

WHAT

the heck

-3x

[willow bear]

ok can you show me the chain rule as it's stated in your textbook

so that i don't confuse you with notation you're not familiar with

[viscid thistle]

where did that -3x jumped from

shouldnt it be

-3/2

power rule

[willow bear]

-3/2 from differentiating the outer function, 2x from differentiating the inner

[viscid thistle]

u performed 2 steps

wait there is an inner and an outer function ?

[willow bear]

yes? (x^2 + k^2)^{-3/2} is a composition of two functions

it's f(g(x)) where f(t) = t^(-3/2) and g(x) = x^2+k^2

[viscid thistle]

OH MY GOD

you considered that power as a function

while i thought its the power in power rules and i can bring it down and thats it the derivative

wont that work ?

[willow bear]

the power rule technically only applies to functions of the form x^p

[viscid thistle]

ok

[willow bear]

it can also apply to (x+k)^p because technically you need the chain rule here but the derivative of (x+k) is 1 so it's ok

but if it's (something more complicated)^p you will need the chain rule and it'll give you a factor that you can't ignore

[viscid thistle]

no it doesnt applies here

you wrote -3x

while power rule

-3/2

[willow bear]

no, i am not using the product rule here

no it doesnt applies here
what doesn't apply

[viscid thistle]

power rule doesnt apply here

you just jumped 3 not 3/2

and u wrote an x there

power rule should have given -3/2

[trim hemlock]

Yes, it did

[willow bear]

-3/2 from differentiating the outer function, 2x from differentiating the inner

[trim hemlock]

But the chain rule also wants to join in

[willow bear]

did you like... read what i said, cause i thought i explained clearly what came from where

i can do it in more detail if you want

also you still havent shown me what notation you use for the chain rule. what does it take to get a simple answer to a simple question out of you?

[viscid thistle]

left d(right) + right d (left)

what i am asking is you cant apply power rule in (X+k)^p

you have to consider it a composite function

[trim hemlock]

I think you should reread what ann said

[willow bear]

yeah

cause it feels like you're flat out refusing to do that

also i said CHAIN rule not product rule

[viscid thistle]

fog =(f’ o g) × g’

[willow bear]

okay.

f(g(x)) where f(t) = t^(-3/2) and g(x) = x^2+k^2
your function is this

f'(t) = -3/2 t^(-5/2)

g'(x) = 2x

(f' o g)(x) = -3/2 (x^2+k^2)^(-5/2)

[viscid thistle]

this is so hard to imagine

but i learned it nvm

thanks

[trim hemlock]

fog =(f’ o g) × g’
Try not to use this notation

It can get very confusing sometimes

[viscid thistle]

when i saw this function

i thought that

[willow bear]

(f o g)' on the left hand side.

[viscid thistle]

-3/2 will jump down and -5/2 remains at top thats it the answer inside bracket remains unchanged

[trim hemlock]

What?

[viscid thistle]

but as he said

it the power rule only works for x^p

not for (x+k)^p

so i get the point

[trim hemlock]

No, the power rule, or any rule, works for all of the function

Not just x or any variable alone

But then you have to also remember the chain rule

Chain rule is very important

[viscid thistle]

then the answer should be -3/2(x+k)^5/2

if the power rule works then that should be answer

[trim hemlock]

But where is your chain rule?

And also thats not correctly derived

The outtermost function is not derived properly

[viscid thistle]

chain rule isnt used in power rule what i have learnt is that power rules says ( jump the power and subtract the power 1 )

[trim hemlock]

Nah, chain rule is used everywhere

[viscid thistle]

please kill me

[trim hemlock]

You just dont see it in functions like x^2

[viscid thistle]

but its the same here

[trim hemlock]

Its confusing at first, just do more exercises

[viscid thistle]

its just that

instead of x there is (x+k)

me poor physics guy

: (

the question asks me to calculate maxima

me dont know diffrentiation

me dead

[trim hemlock]

Well if it doesnt specify which method to use, you could solve it in other ways

[viscid thistle]

ur talk doesnt make sense

chain rule isnt used in power rule

they are diffrent rules

[trim hemlock]

Ehhh, lets try this f(g(x)) = x^2, g(x) = x, then f'(g(x)) = 2x.g'(x)

And g'(x) is 1 isnt it

Hmmm

But that is too complicated tho, so why not just write f(x) = x^2 and f'(x) = 2x by this rule called power rule

As i said, chain rule is everywhere

[viscid thistle]

ok

i learned

forget logic

can you find maxima of this

kqz/(z^2+r^2)^3/2

z is variable

[trim hemlock]

No, unless you specify what the output and input in as well as what all of that letters mean

[viscid thistle]

they are constants

z ranges from 0 to infinity

its a odd function

[trim hemlock]

Well then firstly use your quotient rule and then apply the chain rule afterwards

Do it slowly

As to not cause any confusion

[viscid thistle]

ok

is quotient rule derived from product rule ??

[trim hemlock]

Yep

[past meadow]

it certainly can be

you'll need chain rule too though

[viscid thistle]

Well then firstly use your quotient rule and then apply the chain rule afterwards

wont it make double diffrentiation

[trim hemlock]

Please not

And only apply the chain rule to the part where you differentiate a composite function

Or more specifically, (z^2 + r^2)^(3/2)

[viscid thistle]

first i used quotient rule

i got the expression

now its already diffrenciated 1 time

[trim hemlock]

Can you show it?

[viscid thistle]

i dont know how to use texit bot

[trim hemlock]

No need

[past meadow]

take a picture

[trim hemlock]

Use a camera

[viscid thistle]

=0

[trim hemlock]

I think we arent doing the same function, can you post the original function?

[viscid thistle]

yes we arent

[trim hemlock]

kqz/(z^2+r^2)^3/2
So you are telling me we arent differentiating this?

[trim hemlock]

Ok then, why is x present in your derivative?

There is no x in the original function

[viscid thistle]

its multipication sign

XD

sorry

[trim hemlock]

Ok, next time use a dot, or dont use anything for multiplication sign

That can make a lot of people misunderstand

[viscid thistle]

my x are curvy

[trim hemlock]

Secondly, can you tell me the two functions we are using the quotient rule for?

[viscid thistle]

yes z and

[trim hemlock]

Curvy or not, it can be misinterpretated

[viscid thistle]

bottom

one

[trim hemlock]

So f(z) = kqz
g(z) = (z^2 + r^2)^(3/2)

[viscid thistle]

yes

[trim hemlock]

Do you agree?

Ok

So lets apply the quotient rule

The numerator is f'(z)g(z) - g'(z)f(z)

Right?

[viscid thistle]

yes

[trim hemlock]

so tell me what is f'(z)

[viscid thistle]

1

if u took constants out

[trim hemlock]

is it?

thats a bit odd

the rule states differently

so 1 is wrong

[viscid thistle]

its kq

[trim hemlock]

yes

exactly

and g(z)?

so f'(z)g(z) is?

[viscid thistle]

i did this thing and reached here

[trim hemlock]

no, just follow me

[viscid thistle]

ok

[trim hemlock]

what is f'(z)g(z)?

[willow bear]

but as he said
it the power rule only works for x^p
not for (x+k)^p

does that "he" refer to me

[trim hemlock]

probably the teacher

nvm then

[viscid thistle]

kq(x^2+k^2)^3/2

[trim hemlock]

good

[willow bear]

i'm not a he.

idk why people keep pretending i'm a guy when i'm so clearly not

[trim hemlock]

ok now what is g'(z)

[viscid thistle]

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

[trim hemlock]

what is x?

[viscid thistle]

x is z

[trim hemlock]

we are differentiating it with respect to z

try not to change the variable

[willow bear]

gamer science, am i entitled to an acknowledgment that you fucked up

[viscid thistle]

yes from where have you gain such knowledge

[trim hemlock]

and also 3/2 - 1 is definitely not -5/2

[viscid thistle]

HBfasdufbsdibsdaiufbasdofndsnfhua

ok its

0.5

[trim hemlock]

and the derivative of x^2 is not 3x

i think im with ann on this......

[viscid thistle]

i was seeing from here

[trim hemlock]

but we arent differentiating from that

bruh

did you really remember anything ann told you

like i said, just do it slowly, and you wont get confused

[willow bear]

i can't even get them to answer the simplest questions i think they insist on not listening to me at all

[viscid thistle]

i must not waste your time i am sorry i owe u a big time i m dumb

[trim hemlock]

d/dz[(z^2 + r^2)^(3/2) ] = 2z(z^2 + r^2)^(1/2).(3/2)

this should be g'(z)

[stuck lark]

@viscid thistle it was pointed out before. the main issue is you referred to her as “he”

[willow bear]

2z**^2**?

[trim hemlock]

wait oops

[viscid thistle]

@viscid thistle it was pointed out before. the main issue is you referred to her as “he”
@stuck lark so thats the main issu

[willow bear]

@viscid thistle it's not the main issue and it has nothing to do with the math, it just makes me really upset when people call me "he".

[viscid thistle]

u hate bois

[willow bear]

?????

bruh did you even read what i said

or are you a gd misogynist

[viscid thistle]

yes i read

d/dz[(z^2 + r^2)^(3/2) ] = 2z(z^2 + r^2)^(1/2).(3/2)
@trim hemlock

yes

[willow bear]

where did i say i "hate bois"

[viscid thistle]

u dont want urself to be called a boi

[stuck lark]

what logic is that

[willow bear]

yknow what i don't need to explain myself to someone so intentionally thick

[viscid thistle]

why siddhart and axazeel are spectators

if u hate something u will hate called by that name

simple logic

[stuck lark]

that was a rhetorical question

[viscid thistle]

i am thick ?

[willow bear]

yes you are. you just committed the fallacy of the converse, too.