precalculus

[viscid thistle]

But not sure if it’s right

[lilac storm]

how is that in y=mx+b form?

[dim rampart]

She is right though

[viscid thistle]

I got the first two derivatives

And from it I got that inflection points 2,2/e^2

[lilac storm]

yes i have that

[viscid thistle]

And then you have to find the derivative f'(2)

[dim rampart]

$y=-\frac{1}{e^2}x+\frac{4}{e^2}$ should be the answer

[obsidian monolith]

168995235206721

[viscid thistle]

Yeah that’s right

[dim rampart]

But seriously, how isn't this calculus? American education system is weird...

[viscid thistle]

It’s calculus tho

[dim rampart]

then why are we in pre calc

[viscid thistle]

I have it now in calculus lol

[dim rampart]

channel

[viscid thistle]

Idk

[lilac storm]

ok, thank you so much fro your time and help, i really appreciate it

[viscid thistle]

But I’m pretty sure it’s calc

[normal night]

why in the world would u name yourself pretty girl

😔

people these days

[viscid thistle]

Why lol??

[normal night]

im gonna name myself pretty boy now

Guys

i need help

how can i be good at math?

[viscid thistle]

Study well

[normal night]

didnt help at all

[final cypress]

can i get some help?

for vectors?

[dusk edge]

what do you need

[dusk edge]

read math books, you'll get interested, you'll get better

[viscid thistle]

hii im a junior at highschool and im wondering where should i start learning calculus? someone told me here so im asking for resources & guides

[dusk edge]

mit opencourseware

[viscid thistle]

thanks

[viscid thistle]

also

is there like something i need to understand before going to calculus

[dusk edge]

or you should look at khan acadam

acadamy

academy

[viscid thistle]

ok

[dusk edge]

you'll see what you don't understand and than you should watch a seperate vid on that topic

[viscid thistle]

no i was asking when do...like... change? idk how to say it, english is not my first language. but when do i like change from precalculus to calculus

[dusk edge]

im sorry idk know what precalc is im not from america

all i know is like calculus is just what they teach at universities

[viscid thistle]

they teach precalc in 12th grade

so im 4 years early

[dusk edge]

u just need to know your algebra and trigonometry

[viscid thistle]

i know that

[dusk edge]

then you should follow the calc course on mit ocw

its on yt

and for stuff you don't understand you should watch seperate vids on that

[viscid thistle]

oke

i get it now

is 3blue1brown a good source when i do all of that?

[dusk edge]

yeah

he is good

[viscid thistle]

ok then

see ya

[dusk edge]

good luck with your studies!

[solemn flame]

Partial fraction decomposition

[placid ledge]

is that just simplifying the fraction?

[solemn flame]

I have to make a new fraction and solve for A and B once I set it up

[dusk edge]

factor the denominator

[strong ermine]

help

[acoustic harbor]

sin x is always less than or equal to 1

[strong ermine]

thanks

how to do this one

[lilac storm]

can anyone please help me start off this question?

[strong ermine]

lol this is calculus

buddy

#calculus

[lilac storm]

oh, ok

[dusk edge]

U sub

[lilac storm]

yes i know but i just need helping starting it off?

[dusk edge]

Let u = 2x^3 -1

Du = 6x^2

*Du = 6x^2 de

De

Dx

[lilac storm]

do you mean dx= du/6x^2?

[dusk edge]

Yes

[lilac storm]

and by the way, i already have this

[dusk edge]

How do you mean?

[lilac storm]

yeah so let u = 2x^3 - 1, took the derivative of it which is 6x^2 and then rearranged it to get dx = du/6x^2, but now i am stuck?

[dusk edge]

So 3x^2 is half 6x^2

[lilac storm]

sorry, what?

[dusk edge]

3x^2 is the half of 6x^2 right?

[lilac storm]

yes

but why are you dividing it by 2?

[dusk edge]

To write the 3x^2 in terms of U

Cuz otherwise you can’t integrate

[lilac storm]

oh, ok

[dusk edge]

So it will be the integral of 1/2 u^4 du

If I’m correct

[lilac storm]

i have 1/2 du?

[dusk edge]

And then use powerrule

[dusk edge]

You forgot the u^4

[lilac storm]

so i have 3x^2 times du/6x^2, how do you get u^4 from this?

[dusk edge]

You set u = 2x^3 - 1 right

[lilac storm]

yes

[dusk edge]

And in the original problem it’s raised to the 4th power

So it’s u^4

[lilac storm]

so when setting it up, is it 3x(u)^4 times du/6x^2?

[dusk edge]

Yeah

But the 3x is actually 3x^2

[lilac storm]

yes, my bad

[dusk edge]

I suggest watching the organic chemistry teacher on yt

For u sub

[lilac storm]

so from here i have 1/2 (u)^4 du, is this correct?

[dusk edge]

Yes

Then use powerrule

Then use FTC1

And don’t forget to back substitute for u

[lilac storm]

yeah so do i have 1/2 times u^5/5?

[dusk edge]

Yes

[lilac storm]

then i substitute for u

[dusk edge]

Yes

Then use FTC1

[lilac storm]

sorry what is FTC1?

[dusk edge]

Look it up

[lilac storm]

then i just plug in 1 and -1 and subtract them to get the answer, right?

[dusk edge]

Yes

[lilac storm]

ok, thank you so much for your help, i really appreciate it

[dusk edge]

Don’t mind me asking but how old are you

[dusk edge]

Oh yeah no problem

[stiff yew]

wait why is the slant asymptotic wrong? how is it x+8?

[dim rampart]

don't forget about the +4, @stiff yew

your function is $\frac{x^2}{x-4}+4$

[obsidian monolith]

uosɹǝdʎddɐɥʎɹǝʌ

[stiff yew]

@dim rampart wait what do you do with the +4?

do you just add 4?

[dim rampart]

yes

[stiff yew]

alright

tysm :)

[stiff yew]

What type of function has no Vertical asymptote? Like what does it need to have?

Trying to do this

[unborn blade]

Hi am I doing this right?

[past meadow]

looks right, up until the eq in the last line

you got its slope right but the y intercept is wrong

[unborn blade]

oops

+4

guess i cant do algebra

[past meadow]

still seems wrong

[unborn blade]

tyty

oh

its not b = 4?

[past meadow]

it is not b=4

[unborn blade]

8

[past meadow]

yes

[unborn blade]

sorry im trolling

hard

thanks T_T

[past meadow]

april foolsing me

[unborn blade]

LOL legit

my brain

zzzz

[unborn blade]

can someone tell me where i'm going wrong?

[astral mantle]

should simplify to -2/(2+h)

redo your work

[unborn blade]

i know what the answer is

im just not sure where im going wrong

nvm im stupid

[fallow hinge]

hey guys, I'm learning de moivre's theorem, and I'm struggling on solving the solutions
when z^n = equation,
I understood up until the part where
the teacher wants us to substitute k values with -1, 0, 1

or like -2, -1, 0, 1 depending on n value

why do we do that?

[willow bear]

you see how you've expressed the angle inside the cis as 15+90k?

you will want all integer values of k that will make 15+90k lie between -180 and +180.

also it's "z^n = number" rather than z^n = equation

@fallow hinge

does this make sense to you?

. . .

[fallow hinge]

Uhh

but why specifically near 0?

sorry abt the late response I'm still working on that

[willow bear]

what do you mean by "near 0"?

[fallow hinge]

notice how the k values are near 0

-2, -1, 1, 2

so the values inputted into k are near 0 and 0

[willow bear]

well, that's only a consequence of your angles having to be between -180 and 180.

if your range for angles was 0 to 360, your k would always go from 0 to n-1

[fallow hinge]

I understand now, thank you

[dusk edge]

Btw how is this precancerous

Precalculus

[obsidian monolith]

datboi

[dark furnace]

the question is modified from (https://tutorial.math.lamar.edu/Classes/CalcI/TrigEquations_CalcI.aspx example 5).. sorry i kept deleting, doubting my question

wait... i think (a) is wrong, from the website he only added (pi)(n), instead of (2)(pi)(n).. but regardless.. is (b) still correct thoe

i did get the right solutions using (b) but it might be idiots luck

[uncut mulch]

b) should use pi * n too
also bad notation on the end

if you plan to use = like that, don't write those braces

or just write $n \in \bZ$

[obsidian monolith]

ℝamonov

[dark furnace]

i see, so im assuming b) is one of the solutions (but pi * n give all solutions), but i need to find the other angle also if i were to use 2*pi * n.. thanks!

[vague nymph]

hello there

[jolly raven]

yeah Z is better because it nice

and compact

[lilac storm]

so for b, the answer that i get by hand is 0.68 but when i check my answer for b in geogebra in c, the answer is showing 1.46, can anyone please help me out with like what mistake i have made?

[viscid thistle]

I'm pretty damn sure this is wrong, but I don't know where

Could anyone help?

[mild swan]

What have you tried?

Note: You can use Desmos to check your answer, but make sure you know how to do the problem.

[viscid thistle]

This. I think the error comes from the evaluation for 1 and 0 but I don't know where

[viscid thistle]

What's desmos?

[mild swan]

Oh rip - thought those were possible answers, but the possible answers are right below the question lol

[mild swan]

Probably the most broken online graphing calculator there is

[viscid thistle]

Never heard of it...

[mild swan]

The subtraction is very close 🙂

But -1/3 - 1/2 = ?

[mild swan]

https://www.desmos.com/calculator

[viscid thistle]

Is it like geogebra?

[mild swan]

yup

[viscid thistle]

Okay now I know for sure it is wrong, but I can't spot the error

[mild swan]

Error lies in simple subtraction

It's a small error

[viscid thistle]

🤣

[mild swan]

$-\frac{1}{3} - \frac{1}{2} = ?$

See it now?

[obsidian monolith]

Plant Man

[viscid thistle]

Not really no...no comments.

[mild swan]

ah gotcha

[viscid thistle]

Oh fuck

🤣

[mild swan]

welp - use desmos to check

in the future 🙂

[viscid thistle]

Casually forgets rules learnt like a year or 2 ago...

[mild swan]

yup - that'll happen

🙂

😅

it's happened to me more often than I'd like to admit.

[viscid thistle]

Definetly not ready for MAT then

[mild swan]

¯_(ツ)_/¯

Not necessarily

Simple mistakes are inevitable

[viscid thistle]

Mistakes like these are just... Especially when you have homework on these for the next day...

Imma have to go through what I wrote lmao

Man there were like 2 pages of the stuff too...

But thanks bro

Really appreciate it

[lilac storm]

anyone there to help?

[sick steppe]

no, no one's here. . .

[lilac storm]

can you help me out?

[viscid thistle]

Where?

[sick steppe]

I can totally help you with that well worded question you posted

Dont ask to ask

just ask.

[lilac storm]

ok

[viscid thistle]

Dw bro I managed to forget how adding fractions worked, no-one will make fun of you🤣

[lilac storm]

#precalculus message

[viscid thistle]

Oof I feel bad for missing that now

But sorry, I don't know, that's way beyond what I know how to do

[lilac storm]

are you there?

[lilac storm]

anyone there to help?

i posted a question a while ago and i still haven't received any help for it so is there anyone there that can please help me out?

[lilac storm]

anyone there?

[mild swan]

@lilac storm repost it - people probably won't want to scroll up or try to following your messages

[lilac storm]

oh, ok

#precalculus message

[lilac storm]

is it possible if you can help me?

[mild swan]

What's the question?

Got a specific question to ask?

Idk if I wanna look over all of your work for you - but if you send one problem at a time, then I may be willing to go through it with you @lilac storm

[lilac storm]

yeah so my specific question is that, for b when solving by hand, i get 0.68 but when checking my answer in geogebra, it is showing 1.46 as the answer so if you could just have a look at my hand written solution to see if i made a mistake?

[mild swan]

your work is a bit fuzzy

a bit difficult to see in the picture

[lilac storm]

yeah i can very quicky tell you what i have done

so i first used the delta x = b-a/n formula which gave me 0.5

then i solved for the integral using the trapezoidal rule

which gave me the answer of 0.68

but when i check in geogebra, the answer that it is giving is 1.46

[mild swan]

you got a using geogebra right?

[lilac storm]

yes

[mild swan]

can you resend your work?

it's a bit hard to see - not much light

[lilac storm]

yes of course

just a moment

[mild swan]

👍

Trapezoidal Rule - having trouble remembering it

it's basically creating trapezoids with x = -1, -0.5, 0, 0.5, 1, and the bases of y = 0 and a slant of the segment that connects f(-1) with f(-0.5), f(-0.5) and f(0), etc.?

[mild swan]

are you sure this is correct

0.25 * 2e is already greater than 0.68

[lilac storm]

but i am still getting 0.68?

[harsh cipher]

Hi

I have question about cubic regression function.

I'm given cubic regression function how do I find the value of dependent variable when I know the value of independent variable is 75?

Ill take a screenshot of the equation.

[jolly raven]

would independent variable be x and dependent variable be f(x)

[harsh cipher]

tbh not sure for cubic regression.

[jolly raven]

thats what i would assume it is since in science that typically what they are

input vs output, x vs f(x)

well i see one thing you could do

[harsh cipher]

so then all I need to do is plug in those numbers

what can I do?

[jolly raven]

whats weird to me here is how this is just algebra

not really calculus but whatever

[harsh cipher]

It is

[jolly raven]

id set f(x) to 75 like instructed

only issue is that it wants one value of x

and there should be 3

is there a parameter for what x should be? like above 0 or something

[harsh cipher]

No

I think dependent variable is y ?

😦

[jolly raven]

yeah... yeah it is

so i was stupid

and u should set x to 75

and solve

so that should answer the whole thing

sorry for my mix up about what a dependent variable is

[harsh cipher]

it's okay

thanks for your help 🙂

[jolly raven]

no problem!

[harsh cipher]

I think I got it. I just plug in the equation in my calculator and trace x = 75

dependent variable is x. you were correct.

[jolly raven]

hm

[dark sky]

no it dont, right?

[pliant locust]

Correct

[dark sky]

you would just bring the k down to get: klogb(x) right?

[willow bear]

yes

[velvet blade]

I can't find this log division proof

dividing two logarithms

$\frac{\log_a b}{\log_a c} = \log_c b$

[obsidian monolith]

Researcher in Pre-algebra

[velvet blade]

Can't find the proof on khan either

Wait, I got it

[lean thistle]

if f(x) = 2^x + x^2 - 2x

[obsidian monolith]

Ramboo

how would i find $(f^{-1})’’ (4)$
```Compilation error:```! Double superscript.
<recently read> ^
                 
l.55 how would i find $(f^{-1})’’
                                      (4)$
I treat `x^1^2' essentially like `x^1{}^2'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2020/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./354585240515248128.aux) ) 
Here is how much of TeX's memory you used:```

[willow bear]

write g(x) = f^-1(x) for simplicity...

then you have $f(g(x)) = 1$

[obsidian monolith]

Ann

[willow bear]

try to differentiate that twice with the chain and product rules

and youll have an expression for g''(x)

@lean thistle

[lean thistle]

f(g(x))=1?

why is that

[willow bear]

bah

not 1 sorry

x

[lean thistle]

yeah

[willow bear]

i was thinking of the derivative at the same time and i was thinking about how x differentiates to 1 under what i suggested

sorry. cant think 100% straight rn.

[lean thistle]

its ok

you gave me a hint thats enough

thanks

[stark fulcrum]

hey guys, what’s the derivative of -2xe^-0.1x^2 ?

[viscid thistle]

do you know the chain rule?

[stark fulcrum]

yes am I supposed to use that ?

[sick steppe]

it's product rule

and as written very ambiguous

[mild swan]

$-2xe^{-0.1x^2}$

[obsidian monolith]

Plant Man

[mild swan]

?

[lilac storm]

can anyone please help me with this question?

[lilac storm]

anyone there to help?

[gray path]

ill send in a moment

[lilac storm]

ok, thank you

[gray path]

its pretty simple, just keep track of variable names

[lilac storm]

so that's it

[lilac storm]

ok, thank you so much for your help and time, i really appreciate it

[gray path]

only took a couple seconds

[lilac storm]

so that's it for the question?

[gray path]

yeah

fairly certain

[lilac storm]

ok, thank you so much once again

[flint bloom]

Im just lost on what the problem wants.

I know it will take a little over 45 years for $850 to compound t0 $5,000 with r= 0.039, n=4, and solving for t

but like it says use years 1-5 which isnt gonna get me anywhere close

[flint bloom]

<@&286206848099549185>

[viscid thistle]

Could someone please help me out with this one? I dont understand what to do with the square root

[opaque current]

Are you supposed to use integration?

[sick steppe]

u sub

[opaque current]

or reverse chain rule

[steel flare]

Would someone be able to give me an example of when a cubic function will have a global maximum but no minimum point

I'm trying to solve one of the communication questions from my homework but I don't understand how this situation can work out

[willow bear]

@steel flare i have a feeling they're expecting you to give a cubic function w/ a restricted domain

such as f(x) = x - x^3 for x in [0,infty)

if you don't restrict the domain you won't have a global max NOR a global min

[steel flare]

Okay thanks

So the main concept is to restrict the domain correct?

[steel flare]

will this give me a situation where the function will have no global min but will have a global maximum?

@willow bear

[willow bear]

you didn't need to ping me twice

anyway, try it out

graph it

[lilac storm]

find the domain of the function.

so is the answer for this question just f(s,t) is defined for s^2+t^2 is greater than equal to 0, therefore the Domain=all points (s,t) in the st-plane such that s^2+t^2 is greater than or equal to 0?

[mild coral]

The domain for this function would be All Real Numbers since s and t are being squared, any negative numbers will become positive.

You will never get a negative number in that function because each variable is being squared

Now, if one of the variables was not being squared, say the 't' then the domain would be >=0

The range of the function is >=0 however

For part b, use the equation: Vf=Vi+at, so it will be: Vf=0-9.8(2)

It is negative since going down

For part A, it should be the same since it is just falling and no other forces presumably are acting on the ball.

Should be yeah, least for part b

I would assume the same answer

You don't really have any sort of distance so we have to assume it is at free fall, started with 0 velocity and that there is no air resistance

So for part A, I would put the same answer as for part b

Do this:

[f(x)f-f(x)i]/xf-xi

Taking the final function - the initial function and dividing that by the final-initial

@woeful yoke

No pm

But I can help sure

Take the derivative of that function and input for x

Have you done derivatives yet?

Sounds right

You've done derivatives in Pre-calc? Interesting

Write it as: f'(4)=-24

Since we are finding the instant rate of change, we want to use the derivative of that function since f(4) does not equal -24

Yes

That little quote is showing it is the first derivative

So f''(x) is the second derivative

after 4, we just use numerical values

Indeed

The domain should be [-5,2]

Since in this piece wise function, the lowest number it goes is -5 and the heighest is 2

Any number smaller than -5 or larger than 2 can't be counted for in this since there is no function represented for those values

If it is asking for the domain then yes

Shouldn't be

Then yeah, [-5,2]

You know why it is closed brackets?

Good

Yes

Closed means included, open means excluded

Good

Get the X and Y values for each integer and draw the graph

Plot the points then I would say make it look like a right rectangle as best as possible

Like this

But like

Only from the lowest Y to highest Y

So don't go lower 0

Since 3 and 0x values are both 0

Highest Y value will be 2

So, something like this

Yeah

I assume it means calculate the area of it

So just do 2*3

yup

ok

No clue about integrals

We begin that this week

Why the hell is precalc learning integrals? XD

Not your teacher

US Private school?

Makes sense

It is your school

[echo meadow]

guys did they just forget about the pi

[mild coral]

Yeah my precalc book doesn't even go through integrals

Allow me to look, I would post the integral question in #calculus

For this problem, since it is an absolute value problem, you will need to calculate 2 areas and add them. So 1/2bh

Sounds probable

Same thing as the previous problem except with a cubic function instead of an absolute value function

Just sounds like you need to make a lot of small rectangles

I would think that since they are teaching you integrals, they would just make you guys use those for the area under the curve

yeah

Graph the function

It will make sense

that is the function

That entire area from x=0 to x=2

and make 20 rectangles to approximate the area

calculate it.

yup

of course

Nahh

🙂

[viscid thistle]

@echo meadow no they didn't "forget" about anything.

[obsidian monolith]

Al𝟛dium

[echo meadow]

how did sin (2x) get canceled out

@viscid thistle

[viscid thistle]

sin(π)=0. the whole term will be 0 as well.

there's a typo.

\begin{align} y'&=12\sec²(π-2x)\ &=12\frac{1²}{\cos²(π-2x)}\ &=12\frac{1}{\underbrace{(\cos(π-2x))²}{\cos({\color{green}{\alpha}}-{\color{blue}{\beta}})=\cos({\color{green}{\alpha}})\cos({\color{blue}{\beta}})+\sin({\color{green}{\alpha}})\sin({\color{blue}{\beta}})}} \&=12\frac{1}{(\ \cancelto{-1}{\cos(π)}\cos(2x)+\cancelto{0}{\underbrace{\sin(π)}{0}\sin(2x)}\ )²} \ &=12\frac{1}{(-\cos(2x)+0)²} \ &=12\frac{1}{(-\cos(2x))²}\ &=12\frac{1}{\cos²(2x)} \ &=12\sec²(2x)\end{align}

[obsidian monolith]

Al𝟛dium

[viscid thistle]

@echo meadow now it's fixed. any more doubts? there are more methods to simplify this, but this may be more natural to you as it simply uses the addition/substraction formulas.

also please avoid asking in 2 channels at the same time

[mild swan]

I wish I was this good at latex

[echo meadow]

thank you !

[echo meadow]

oh stupid past me

[ancient horizon]

@viscid thistle Are you able to help me with a binomial theory question?

[mild coral]

What's the question?

[dense oar]

How do I find the rectangular form of
r = 4 csc(θ)

[ancient horizon]

Sorry I forgot to ping @mild coral

[mild coral]

It was easier to explain it this way

Follow this, use Pascal's triangle

[uncut mulch]

look up binomial theorem

[ancient horizon]

What do I do with the x^4

Is there somewhere in the equation to plug it in or

[mild coral]

You perform the binomial theorem

Look at the image I sent and view the (a+b)^4

I don't like the way the question is worded

[ancient horizon]

OHH I SEE I SEE

What do I plug in for a

[mild coral]

Granted, I do have comprehension issues sometimes

[ancient horizon]

Or where does the (2x-11)^9 go

[mild coral]

Well, (2x-11)=(a+b)

[ancient horizon]

So 2x is a and -11 is b?

[mild coral]

Yes

[ancient horizon]

Did you mean to leave out the superscript of 9

Do I do anything with that

[mild coral]

(a+b)^n

You can expand this using Pascals triangle

[mild coral]

^

[ancient horizon]

Yes I’m using that and I’m plugging everything in now but I didn’t add the superscript of 9 anywhere

[mild coral]

Cause you don't need to

That exponent is signifying it is, say, the 9th expansion of it using the theorem

In the question, you replace the 9 for the 4

Just focus on (a+b)^4

[ancient horizon]

Okay wonderful tysm!! Are u familiar with arithmetic and geometric sequences? I’ve done my work for them I and I know what I did wrong, I just need a little guidance to fix them

[uncut mulch]

whut no

you don't just replace the 9 for 4

that is NOT what the question is asking for

[ancient horizon]

Oh? What shall I do then

I was just working through I’ll start over

[uncut mulch]

look up binomial theorem

[ancient horizon]

Ok. I was still a little confused but I have a different question that’s in another format so I don’t have to replace any superscripts

[uncut mulch]

$(a+b)^n = \sum_{k=0}^{n} \binom n k a^{n-k}b^k$

[obsidian monolith]

ℝamonov

[mild coral]

I was trying to explain what to do in a simple way, when I meant replace the 4 and 9, I meant like perform the expansion in terms of the exponent 4 for the theorem

[uncut mulch]

k=0 gets you the first term
k =1 gets you the second term
etc

[mild coral]

I am not the best with explaining but I try

[uncut mulch]

it was far from what you're supposed to do

[mild coral]

Sounds false but I disliked the question format anyways

[uncut mulch]

there's nothing wrong with the question

[mild coral]

Sounds false again

but I will agree to disagree

Perhaps I had misinterpreted the question.

[sick steppe]

you did

[mild coral]

https://tenor.com/view/montypython-draw-flesh-wound-sword-call-it-a-draw-gif-5447899

[uncut mulch]

they were being asked about something being raised to the power of 9 specifically
doesn't make sense to ignore that number completely and replace it with a number of your choice

[mild coral]

I was suggesting the bottom

[sick steppe]

You said do (a+b)^4

[mild coral]

No

That's not what I meant

[sick steppe]

[mild coral]

I was trying to relate to the theorem

[sick steppe]

It might not have been what you meant, but it's what you said

[mild coral]

I was referencing this the whole time.

[sick steppe]

yeah, that means jack shit for (a+b)^9

since it's a completely different row of pascal's

[mild coral]

They were asking for the 4th expansion of it however

Again, like stated previously, perhaps I had misinterpreted the question.

[sick steppe]

yeah.. they wanted the Ax^4 term of the expansion

[mild coral]

Alright.

That is the first part of the expansion I referenced, I unfortunately forgot to mention it though

[solemn flame]

Hi can someone help me clear the first and second pivot I already wrote out the matrix

[mild coral]

Are you solving for each variable? If so, do systems of equations

[solemn flame]

@mild coral my teacher specifically wants us to do it in matrix form but o really don’t understand it

[mild coral]

Show me the matrix you created

[solemn flame]

@mild coral

[mild swan]

should be 5 -2 1 7 for the bottom row

[lofty mulch]

/frac{sec(x)tan(x)+csc(x)}{sec(x)+cot(x)(csc(x)}

$/frac{sec(x)tan(x)+csc(x)}{sec(x)+cot(x)(csc(x)}

how do you use the bot on here?

anyone could help out in getting the answer tan(x) as the answer?

[mild coral]

It should be 5 -2 1 7 for the bottom, I believe that is a Z

[mild swan]

i would start with rewriting everything in terms of sin and cos, then clearing the complex fraction

[lofty mulch]

I have a picture of it but I get cot(x)

I forget my arithmetic, multiplying by 1/cos(x) * sin(x)/cos(x) should equal sin(x)/cos^2(x)

[mild swan]

Are you able to send a larger picture?

[lofty mulch]

I could attempt

[mild swan]

actually wait

just gotta move the img to google

[lofty mulch]

oh yea

if anything seems hard to see let me know

[mild swan]

gotcha

i think you have to multiply the leftmost term by sin(x)

otherwise the common denominator isn't cos^2(x) sin(x)

@lofty mulch

[lofty mulch]

hmm

damn

[mild swan]

so here the numerator should be 1

sin^2(x) + cos^2(x)

[lofty mulch]

arithmetic gets bad bad

me*

[mild swan]

it gets everyone

promise

[lofty mulch]

thanks I'm gonna retry it

[mild swan]

👍

[lofty mulch]

Yeppity yep 👍

@mild swan Thanks, got tan(x)

[mild swan]

😄

👍

[lofty mulch]

I needed a hero

fr

[mild swan]

well - me is not where you're going to find it

XD

other people here are much better at just about everything than i am

i'm only in high school - and in this server lie college professors

so

i'll let you decide who's more reliable

[hollow moat]

anyone know quadratics?

[mild swan]

[mild swan]

#❓how-to-get-help message

[hushed vine]

https://dontasktoask.com

[placid ledge]

https://youtu.be/phaIklEphSM?t=36

[mild swan]

@hushed vine @placid ledge 👏 👏 👏 👏 👏 👏 👏 👏 👏

I commend you both

[faint nest]

hi does anyone know if this is right

i put another answer last time and it was wrong

[sick steppe]

$\frac{11}{12}=\frac{3}{12}+\frac{8}{12}$

[obsidian monolith]

moshill1

[sick steppe]

@faint nest

[hushed vine]

I don't think it is.

Do you know what $\tan{\frac{11\pi}{12}}$ is?

[obsidian monolith]

Spaceship

[mild coral]

Use the values from the unit circle: 3pi/4 and pi/6. They both add up to be 11pi/12

[mild swan]

@faint nest you got it?

[faint nest]

Yes thank u

[unborn blade]

im a bit confused about the variables

it says h represents the height of the cone

but then dh/dt is the change in water level

same with V, it says V is the volume of the cone

but dV/dt is the volume of water

i cant see the connection for some reason

[crystal osprey]

Do you understand what d/dt actually means?

Well technically it already gives you the definitions there

Basically, they're all variables that are rates in respect to time.

h is a variable that represents the height of the cone in cm. So we've established that the unit for distance in this case is centimetres, which has to be consistent throughout all the variables if you want a related rate. We've also established that the unit for time here is in minutes.

Now we're not actually given any values for time. But that's fine, because the point here is that if all of these variables actually relate to each other, then their values will change in a specific rate after a certain amount of time passes. This means that given the equation for the volume of a cone and finding the derivative in respect to time, you can actually relate all the variables with each other to find their values.

[unborn blade]

yeah I think I understand most of that

[crystal osprey]

Going back to the the variable 'h' and its relation to dh/dt as your original question was. In this case, dh/dt is the change in rate (or growth) that 'h' goes through at a specific point in time. The units is cm/min.

[unborn blade]

but the height of the cone doesnt change right?

[crystal osprey]

The height of the cone during let's say 5 minutes will always be a fixed value if that's what you're asking

But dh/dt at 5 minutes shows how fast the height is changing at that instantaneous moment

[unborn blade]

I think that parts confusing to me just because of the wording

because if we know the height of the cone cant change, and h represents the height of the cone, how does dh/dt give the change in water level with respect to time

im still really new to calculus

so i probably got some fundamental concepts not fully cemented in my brain yet

which is why im confused

[crystal osprey]

No worries, it's normal to be really confused especially when it comes to related rates

Let's imagine an easier scenario. When you have a car accelerating at a random constant value, you'll have a total distance traveled over time... correct?

Let's say that it's been 20 seconds since your car started accelerating. You would have traveled a certain distance from your starting point. But that distance changes constantly, because your car has a different speed at any place in time due to acceleration.

But with related rates, what we're saying is that all these variables actually relate to each other. The distance traveled by the car, the speed of the car. But what are they in relation to? In this case, time. Time is the independent variable that defines the distance traveled at that minute, or the speed at that minute.

It might be difficult to imagine dh/dt to give the change in water level, but it's no different with the speed of the car. In fact, if I wrote 'h' as the variable for distance covered by the car, then dh/dt would be the speed of the car.

Because distance over time is exactly what speed is. If distance in this case is measured in miles, and time in this case is measured in hour, then my speed would be miles per hour, or mph as you're probably more used to seeing.

So let's say you've been driving at a certain acceleration for this amount of time. If you look at your speedometer, you'll see the speed that you're going at in that very instant. Let's say 60mph. But maybe 1 minute later, it becomes 75mph. That's how dh/dt can be different. This is of course, assuming that there's something that actually made the rate change in the first place. In this case, it's your car's acceleration that is making your speed change at every instant.

Now in the case for the water rising in the cone specifically, you've got realize WHY the rate of the height changes. If it was a cube, then the rate of the water rising won't ever change, because the volume of the cube will always be uniform throughout. But you're given a cone. The volume of the tip of the cone is smaller than the volume of the bottom, since that's where it's widest. So when you're finding the dh/dt or the rate of change in the height of the cone, it will change at every instant. If you were to pour the water into the cone uniformly, it'll take faster to fill the bottom since it's narrow, and slower to fill near the top since it's larger in volume. This is why the change in height will gradually get lower as you go higher and higher, does that make sense?

[unborn blade]

yes that makes perfect sense

[crystal osprey]

aight that's great!

[unborn blade]

i think that initial car acceleration example was a moment of clarity for me

that was really useful

[crystal osprey]

That's good to hear!

[modest palm]

,w number of primes below 100000

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Wind0₹#9519

[willow bear]

if you want to just use the bot go to #bots

oh this was 4h ago

joy

[civic sierra]

So when we have a fraction with a polynomial in the numerator and denominator, then if the degree of the polynomial in the numerator is bigger by 1 than the degree of the polynomial in the denominator that means that there exists a slant/oblique asymptote?

Or so I think... my question is: Is there a rule for the existence of a slant asymptote in rational functions?

(ping me)

[willow bear]

@civic sierra yes it's what you just said

deg(num) - deg(denom) = 1

[civic sierra]

Thanks for confirming

[viscid thistle]

aaa okok.

how do i solve this, analytically?

i tried substituting in β/2=π-α in the trig part but it just spits out a big mess

it goes like $$\frac{\cos ^2 (\alpha/2)+\cos(\frac{\pi}{3})}{\sin^2(\alpha/2)}=\frac{\cos^2(\alpha)+\cos(\frac{2\pi}{5})}{\sin^2(\alpha)}$$

[obsidian monolith]

asteris !!

[viscid thistle]

which looks clean, but idk how i should go about it

[viscid thistle]

nvm nvm i figured it out

[earnest barn]

Earlier, I had to sketch the following curve

Two properties about the curve:

• It is symmetrical about y = x
• 2 <= x + y <= 4 since (x + y)^2 - 2^(x + y) >= 0 and (x+y)^2 and 2^(x+y) intersect at 2 and 4, by considering the graphs of each function

I tried drawing in where the curve was bounded so that I could sketch it, but I found it very hard to sketch this curve

When I looked into the answers, the curve was a closed loop

I did not deduce that the curve was a closed loop

Questions:

• How could I have deduced that the curve was a closed loop given the information I worked out?
• How could I approach sketching this curve?

[earnest barn]

<@&286206848099549185>

[dusky spindle]

based on the degree what kind of polynomial is this? (-x^2)+(2x)-5

[sick steppe]

what's the degree of it?

[dusky spindle]

2

[sick steppe]

so what type of polynomial has degree 2?

[dusky spindle]

parabola?

[sick steppe]

yes

[dusky spindle]

it was quadratic. rip

[jolly raven]

Same thing???

Parabola is the output of a quadratic function

On a real graph

[gritty wing]

is there a way to simplify this into an expression without the root?

[placid ledge]

yes

[gritty wing]

can ya break it down for me?

[opaque olive]

maybe its nicer to keep the square root

[gritty wing]

if I keep it in the root, I can't do the u sub

if you want context, here's what a computer thinks it comes out to be

but uh

[opaque olive]

$\frac{(x^4 + \sqrt{2}x^2 + 1)(x^4 - \sqrt{2}x^2 + 1) }{x^4}$

[obsidian monolith]

Yes

[gritty wing]

yea...

[opaque olive]

is that what you wanted

[gritty wing]

see the pic ^

those two are supposed to equal each other

[gritty wing]

I could theoretically integrate that but boi what a mess

[opaque olive]

one sec

so yeah just initially expand, and keep x^8 on the bottom to get that form

$\sqrt{(x^4 + \frac{1}{x^4})^2 + 4} = \sqrt{x^8 + \frac{1}{x^8} + 2} = \sqrt{\frac{x^{16} + 2x^8 +1}{x^8}} = \frac{\sqrt{(x^8 + 1)^2}}{\sqrt{(x^4)^2}} = \frac{x^8 + 1}{x^4}$

[obsidian monolith]

Yes

[gritty wing]

that's much nicer :D

[charred rock]

Okay so I went and taught myself set and interval notation, anyone mind telling me if I am understanding it? 🙂

[rigid beacon]

Everything is correct @charred rock

just some style things you'll see more often than the way you've written it

wait no there is one mistake

your range on the last thing

it should be

$$( -\infty, 0 ]$$

[obsidian monolith]

Spamakin🎷

[rigid beacon]

that order

[charred rock]

oh really? I was always confused as to the order of the range

[rigid beacon]

smallest to the left, larger to the right

[charred rock]

ahhhh okay

is that the same for x?

domain*

[rigid beacon]

yea it doesn't matter

interval notation in general

smallest goes on the left

ok next just style things

rather than

$$ { x | x \in \mathbb{R}} $$

[obsidian monolith]

Spamakin🎷

[rigid beacon]

more often we write

$$ x \in \mathbb{R} $$

[obsidian monolith]

Spamakin🎷

[charred rock]

oh i suppose mine is redundant

[rigid beacon]

we only use set notation with | if we need to be more specific

yea

so then for example with the range in the first problem

rather than

$$ { y | y \in \mathbb{R},~ y \geq 0}$$

[obsidian monolith]

Spamakin🎷

[rigid beacon]

people write

$${ y \in \mathbb{R} | y \geq 0 }$$

[obsidian monolith]

Spamakin🎷

[rigid beacon]

so like if you need to do specification (like y >= 0)

[charred rock]

does that then apply for range in c as well?

[rigid beacon]

you'll write the more general over arching set and then after the | (which is read as "such that") you'll add the specification

yea

what you wrote isn't wrong tho

anyone reading it would understand it

[charred rock]

thank you so much for telling me this, it is great to learn these things early

i really appreciate your help!

i went ahead and fixed that error and updated my style

[restive bridge]

I am having trouble understanding this proof.

https://i.imgur.com/b0CpKBN.png

This is the formula cited. What's giving trouble is when they use formula 6.

Why is it:

N(N-1)/2?

I can see that as opposed to formula 6, the value ends at 1 less than N, as opposed to ending at N itself - but I'm confused as to why that changes the outcome for the formula really.

[viscid thistle]

If you want a written proof and you have studied AP , then notice that 1,2,3...n is an AP .
Now Write $$S_n = 1 + 2+ 3 + 4 + ... + n$$
and again flip it to write $$ S_n = n + (n - 1) + .... + 1$$

Now for an AP , the formula for a sum with the first and last element is $ Sum = \frac{n}{2} (a + t_n)$

write this for both of our $S_n$ and then add them to get $$ 2 S_n = [\frac{n}{2} (1 + n)] + [\frac{n}{2}(n+1)] $$
Solving for $S_n$ gives what we need , namely $$S_n = \frac{n}{2}(n +1)$$ @restive bridge

[obsidian monolith]

xi64

[cursive birch]

could someone help me with this im confused :<

[viscid thistle]

i would actually suggest giving a re-read for the definition of a one sided limit / a limit in general if you are that confused

[hushed vine]

you cna ping helpers after 15 minutes right

[cursive birch]

like I think the limit as x approaches three from the right of f of x is 3 but im not sure if thats correct or not

[hushed vine]

i think yes but isn't the problem looking for the left side limit?

[cursive birch]

ah

im smart

[hushed vine]

been almost 30 minutes so <@&286206848099549185>

[fresh breach]

do you know in general how to take roots of complex numbers

what techniques do you already know and what did you try

[cursive birch]

<@&286206848099549185>

[hushed vine]

Nevermind I got it. I wasn't using the 2nd smallest

stupid me

[fresh breach]

yeah nilla what is your specific questions

question

so it got cleared up that the limit should be coming from the left

what does that mean to you

[cursive birch]

since its asking for the limit from the left side would that make it 1

[fresh breach]

yes

when it's asking for a limit from a specific direction

the value at the point is irrelevant

and the behavior elsewhere is irrelevant

[cursive birch]

ooh ok thanks

[fresh breach]

np

[strange pewter]

I have a problem

Do I ask here? It's calc

[stuck lark]

go ahead

[strange pewter]

Idk how to do that

[blissful ridge]

Exploit odd and even properties of definite integrals

[sick steppe]

Meme math

[dense oar]

How do I find the polar equation for
x - 4y + 2 = 0 ?

[dense oar]

<@&286206848099549185>

[fresh breach]

what do you already know about converting to polar form

do you know how x and y each convert individually

[dense oar]

X = rcosθ
Y = rsinθ

I’m pretty sure I figured it out now

[fresh breach]

ok thats great

[restive bridge]

I apologize, Idon't quite follow. Would it be possible to explain just this part?

in your explanation, it shows n+1. I'm just not getting why we have n-1 in the end result

[sick steppe]

cause the sum is n(n+1)/2

where n is just n-1 in this instance

[restive bridge]

OOOOHH

I got it, thanks bro.

So originally formula 6 is: (n +1)n/2

You substitute the formula in this case: [(n-1+1)(n-1)]/2 ---> n(n-1)/2

@sick steppe

TY

[sick steppe]

yes

[white idol]

can someone please help me with this

[willow bear]

what's giving you trouble here?

[fleet yew]

Polar?

x(θ) = f(θ)*cos(θ), y(θ) = f(θ)*sin(θ)

Find (dy/dθ)/(dx/dθ) at π/6

[willow bear]

@white idol seems to have disappeared

[white idol]

oh sorr

i dont get notified for stuff

[white idol]

im not exactly sure where to start

I first tried finding the derivative of the equation and realized that I don't know what to do with that

[willow bear]

maybe converting the curve to cartesian parametric equations would help

[white idol]

then i tried converting it to rectangular and realized that didnt work

[willow bear]

as AMD suggested

wym didnt work

why didn't it work

[white idol]

i got it in terms of x and y

if i reformat it as amd did, would i get (3+8sintheta)sintheta?

[willow bear]

that's your y

x = (3 + 8sin(theta))cos(theta)
y = (3 + 8sin(theta))sin(theta)

[ashen wren]

so we just finished our verifying identities lesson and I gotta say I'm really having a hard time figuring out where to begin on some of the questions the teacher assigned to us

I'm struggling with this one

I'm having trouble determining which side to start in

[blissful ridge]

I mean it's not very intuitive to start on the RHS

[ashen wren]

alrighty then

[sick steppe]

1+tan^2(x)=sec^2(x) would be where I start probably

[somber yew]

Or if you're familiar with compound angle identities

Maybe this would be a major hint so 😛

[viscid thistle]

heyy, can someone please help me with this problem

[sick steppe]

slope of the secant on [-pi/6,pi/6]

[viscid thistle]

thank you!

[plush scroll]

can someone explain how this works?

I dont want a direct answer, i just need an explanation on how to get the answer...

[sick steppe]

is it (x+1)^3 in the integrand?

[plush scroll]

no ^2

[somber yew]

Can you evaluate (x+1)^2?

[plush scroll]

wdym? @somber yew

[somber yew]

Can you expand (x+1)^2 to another expression?

wink wink, (a+b)^2=...

[plush scroll]

I dont think we do that

[somber yew]

Hmmm?

Can you not write $$(x+1)^2=x^2+2x+1?$$

[obsidian monolith]

Manan.

[plush scroll]

the teacher never expanded a question when he was explaining it to me, this is the first question in this form i got

[somber yew]

I mean

You're not always constrained to do what has been taught in class

Also, this is precalc knowledge so it is assumed in a calculus course

[plush scroll]

i know..its why i came here, am a bit troubled with focusing in class, i am working on that though

[somber yew]

No worries.

[plush scroll]

lemme try working on it, and send a picture to where i end, then maybe u can correct after

[somber yew]

Sure!

I wouldn't be around here for long, it's getting late for me. But you can send your work here.

[plush scroll]

its okay, thanks

[somber yew]

Take a look at the original expression again.

Cross-check against the given options. Do you notice something?

[hushed vine]

@viscid thistle When else is this denominator 0?

[plush scroll]

lets do ur way...with expanding
(x+1) (x+1)
x^2+x+1
integration of (x^2 + 2x + 1) is going to be (X^3/3 + 2x^2/2 + 1x) = 7/3
(X^3/3 + 2x^2/2 + 1x) = 7/3 {we need to remove the denominators so we multiply it all by 3}
[2x^2/2] the denominator cancels the 2 from 2x^2, and 3 is multiplied to the x making it 3x^2
then its gonna be (X^3 + 3x^2 + 3x) = 7

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
This is what i came up with...i am like 80% sure its wrong so could u help me out if it is wrong

[somber yew]

This sounds right. Good job!

[plush scroll]

Wait...fr? that works???

[somber yew]

Yes haha.

That is how it was meant to be done.

At least in a reasonable way. 😛

[plush scroll]

omg..wow ...alr..ig i do pay attention in class a bit

thanks for verifying, i'd be stuck on it for another hour if i wasnt sure

[somber yew]

No worries. Goodluck!

[fleet ferry]

ello

does anyone know where to start learning calculus?

i have a basic understanding of limits, and its applications on physics

but thats it

[hollow hatch]

Books and Youtube video ofc

[rose gull]

khan academy

[scenic hollow]

Can somebody help me

[onyx wharf]

Equations that show a proportional relationship are of the form y = kx

There is no constant added

[scenic hollow]

Can you example more

[viscid thistle]

@scenic hollow please read #❓how-to-get-help , rule 3.

[tender quest]

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.

[viscid thistle]

do you know what proportionality is?

[unborn blade]

Can someone help me figure out where I'm going wrong?

I'm trying to find the second derivative using chain rule and quotient rule

omg im so stupid

nevermind

[kindred bluff]

It’s the very last part mate, you forgot to multiply the -3x^2 by 4

[tender lotus]

can someone tell me what i'm doing wrong?

for y i multiplied 5 by 2 over the square root of 3

[dim rampart]

cos(30)=y/10

,w 10*cos(30°)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by veryhappyperson#1152

[kindred bluff]

It should be square root of 3 over 2

That’s the cosine of 30 degrees

[tender lotus]

so I did put the cos of 30 as square root of 3 over 2 but when I cross multiply I am still getting the wrong answer

[kindred bluff]

cos(30)=y/10
Sqrt(3)/2=y/10
10*sqrt(3)/2=y

Try that

I crossed multiplied, as you did

[tender lotus]

I got 8.66 again and its still being marked as wrong

[kindred bluff]

Weird, maybe try and type in 5sqrt(3)?

[tender lotus]

I know its very weird and I was overthinking it

It worked!

[kindred bluff]

This worked?

[tender lotus]

but why did I have to change it back if it never asked for that

yes the 5sqrt3

so its just simplified

[kindred bluff]

I have no clue whatsoever

[tender lotus]

I guess if it can be simplified I have to leave it that way, thank you

[kindred bluff]

You’re welcome!

[dark sky]

help i dont know how to do 2

my guess is 17(0.7)^t

is that right ?

when i plugged in 37 to calculator, i dont end up with 4.5

[viscid thistle]

I see Koshen Lu

Chen Lu

[harsh cipher]

Hello

I'm curious about something which is related questions on a exam.

I just took a test that had 7-8 questions that were not taught in the chapter lessons.

Should I contact my instructor and explain this to him? He doesn't seem to care very much since he has not bothered to email me back about a different question.

[jolly raven]

hm

thats annoying to deal with

well i would but idk how he would respond

best of luck to you mirrion

[clear sundial]

How would I write the trig form of a complex number like -16i? I said that -16i=0-16i, meaning that a=0 and b=-16 but then solving for theta doesn't work because you'd be dividing by zero (since tan(theta)=b/a

[jolly raven]

maybe by definition of e^ix?

idk

[clear sundial]

only thing i got is that if arctan(b/a) is undefined it must be 3pi/2 since the arctan of that is also undefined?

but idk if that's true

[jolly raven]

isnt arctan of pi/2 + pi*n where n is an integer undefined for tan or arctan?

[willow bear]

arctan has the whole real number line as its domain

you're thinking of tan.

[leaden jewel]

can it be differentiable at any point though?

[willow bear]

there is one point where it's differentiable.

[leaden jewel]

how so?

[viscid thistle]

https://cdn.discordapp.com/emojis/767775988494762004.png?size=40

[willow bear]

zero

your function is differentiable at zero.

and in fact, f'(0) = 0

[viscid thistle]

@leaden jewel how old are you

which grade are you in

[leaden jewel]

just completed my school

[viscid thistle]

so?>

you're 19?

[leaden jewel]

but it needs to be continuous in the epsilon neighbourhood to be differentiable right?

[viscid thistle]

phi

put it as phi

ig

idk

[leaden jewel]

18

[willow bear]

it is continuous at 0

[viscid thistle]

i am 15

[leaden jewel]

What does it even mean to be continous at a point

[leaden jewel]

??

[willow bear]

you haven't seen the definition?

[stuck lark]

@viscid thistle keep offtopic away from the help channels

[viscid thistle]

sorry

[leaden jewel]

I did see the epsilon delta definition but they aren't zero are they?

[willow bear]

who are "they"

[leaden jewel]

epsilon and delta

[echo wagon]

Look harder