help-13

[unkempt egret]

Ironic username

[pliant apex]

Don't care don't mind

[unkempt egret]

In any event, if you wish to receive help you are expected to be courteous and cooperative

[cedar kiln]

@pliant apex Has your question been resolved?

[cedar kiln]

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[scarlet folio]

im well aware that i need to use the product and/or quotient rule + chain rule but i dont even know like which parts get derived

[scarlet folio]

someone save me

[runic garnet]

Which question

[scarlet folio]

honestly any i mostly need help on what steps to do first

do i do chain or quotient/product

im so lost

[runic garnet]

Ok start with number 1

First u take the derivative of outer, then multiply it by derivative of inner

[hollow totem]

I think he confused abt when to use product rule and when to use chain rule

[scarlet folio]

would the first one just be 5(sec^2(4x))

i cant do this

[runic garnet]

Yes, but then multiply by derivative of 4x

[scarlet folio]

so just 4

[runic garnet]

Yea

[scarlet folio]

thanks so uh how do i do the hard ones

[runic garnet]

Look at number 2

How do u think u should start

[scarlet folio]

uhhh

(2x)(sec(x^3-1))+(sec(x^3-1))(tan(x^3-1))(3x^2)(6x)

if im being honest i replicated a practice problem if this is even close to right

like do i just derive everything ?

[runic garnet]

How’d u get 6x at the end

I’m getting x^2

[scarlet folio]

was i not supposed to derive the 3x^2

[runic garnet]

Which 3x^2

Where did that come from

The derivative of x^3 - 1 is 3x^2 and that’s it

[scarlet folio]

i derived the x^3

oh

burh

[runic garnet]

U don’t differentiate it again

But u forgot the x^2

[scarlet folio]

where do i get that from

wait nvm its from the x^2 in the beginning

my bad

[runic garnet]

Yes

[scarlet folio]

f(x)

[runic garnet]

Yea

[scarlet folio]

so its (2x)(sec(x^3-1))+(sec(x^3-1))(tan(x^3-1))(3x^2)(x^2)

[runic garnet]

Yea

Now just simplify

[scarlet folio]

woah what

i thought i didnt have to simplify in calculus

lord save me

[runic garnet]

It’s not that bad

Factor out the sec

Combine the x’s

[scarlet folio]

i still dont understand though like why do i derive the x^2 but leave the sec(x^3-1) alone

is it uhm product rule

where even is chain rule in this

[runic garnet]

U didn’t leave the sec(x^3 -1) alone

[scarlet folio]

no in the first part

before the addition

[runic garnet]

Oh yea, that’s cuz u took. The derivative of x^2 there

That’s just product rule

[scarlet folio]

oh ok

[runic garnet]

Chain rule is when u took the derivative of x^3 - 1

[scarlet folio]

second half was product/chain?

[runic garnet]

Yea

[scarlet folio]

ok, can u help with the other 4

[high knot]

I think the answer is 4

Your welcome

youre

[scarlet folio]

ok thanks

[high knot]

Yw

[scarlet folio]

this is like unpaid labor

[high knot]

Im a calculus god

[runic garnet]

Ok try the next one then

@scarlet folio

Got any clue? @scarlet folio

[scarlet folio]

im trying

dy/dx = (-csc(x^2)cot(x^2))(2x+1)-(csc(x^2))(2)(2x) all over (2x+1)^2

there

[runic garnet]

Can you resend the original problem down here so I don’t have to scroll

[scarlet folio]

does it matter what side the chain rule is on

like would it effect the answer if i did it like first

[runic garnet]

I’m not understanding what u mean by first

And speaking of chain rule, u forgot to chain the x^2

[scarlet folio]

i dont under stand how to do this

[high knot]

just put that in chatgpt

[runic garnet]

Bruh, clearly u do considering you’re getting reasonable answers

[scarlet folio]

i know the product rule and quotient rule

but where did i miss the chain

i thought the 2x was the chain

bless

g`(x) was 2 and f(x) was csc(X^2) and i coughed up the 2x for the chain rule

[scarlet folio]

please help me mak

[runic garnet]

U put the 2x in the wrong place

[high knot]

wow I got publicky shamed for that

[scarlet folio]

WHAT

[high knot]

Yeah if u want I can put it in chatgpt for u

its not cheating bc its not my work

[scarlet folio]

thats gonna help me on this test

thansk

[high knot]

Yw

[scarlet folio]

ill chat gpt in class

[high knot]

I took ap calc

[scarlet folio]

what do tou mean

[runic garnet]

It’s supposed to be here on this term

Instead of at the end

[scarlet folio]

i thought even after deriving the inside thing was untouched

[high knot]

U can install ai on ur calculator the exam

youll be fine

[scarlet folio]

SHUT UP

NOW

[runic garnet]

Ok I’ll write it out

[high knot]

pm me and ill tell u how

[scarlet folio]

like -csc(u)cot(u) times u`

i didnt know i had to derive that

[scarlet folio]

please send

[runic garnet]

Don’t cheat

Not worth

[high knot]

Cheating got me a 5 ngl

on ap calc and chem

[scarlet folio]

what the hell is the prime notation on keyboadf

`

[high knot]

Its ok to do it sometimes

[runic garnet]

Nah

[high knot]

i was valedictorian too btw

trust me

[runic garnet]

I got a 5 on calc ab and bc and apush and macro without cheating

[scarlet folio]

oh my god bruh im gonna get a 0

BRUH

[runic garnet]

Doesn’t mean anything if u come from an easy district and cheat

[high knot]

Ok

Actually

[runic garnet]

Ok lemme show u

[high knot]

ngl u got scammed

Why would u not cheat

[scarlet folio]

can you do it on paper this format is messing me up

im adding like a million parenthesis

[runic garnet]

Many reasons

[high knot]

Labor

why the hell did u take ab and bc

💀💀

just take BC

[scarlet folio]

dude isnt this stuff supposed to be easy im going to get cooked

"youre gonna wish you are doing derivatives later on in the year"

mak save me

[high knot]

Ai

[scarlet folio]

with your cheats

[high knot]

U can jailbreak your calculator

[high knot]

Are you a woman

u have nice hand writing

[runic garnet]

We weren’t allowed to take BC as juniors

[runic garnet]

No

[high knot]

oh ok

rare occurence

ur above the average male handwriting level congratulations

[scarlet folio]

does the chain always go with f(x)

i need to find my notes wtf

[runic garnet]

The chain goes with anything that needs chain

Doesn’t matter where it is, if there’s a chain needed, do it there

[scarlet folio]

WAHT

[high knot]

when is ur math test

[scarlet folio]

tomorrow

[high knot]

wow

I have one too!!

[scarlet folio]

so ive come to my last resort of discrd math help

[high knot]

times r tough

[scarlet folio]

wait how do i know when it needs a chain

wait a minute

[scarlet folio]

that is just

[high knot]

end it

[scarlet folio]

increduble

[high knot]

why that take so long to write

[runic garnet]

I wasn’t writing the whole time…

[high knot]

……….

…

[runic garnet]

But generally I don’t write like that, I do more slanted stuff when I take notes

[high knot]

u write slantedv

?

[runic garnet]

Read what I just wrote then answer your own question

[high knot]

be fr

[runic garnet]

?

[high knot]

you write in italics or you can’t write in a straight line

Which one

[runic garnet]

I have no clue what ur talking abt

[high knot]

Wow

[runic garnet]

But @scarlet folio do u see how i did it

The chain rule

[high knot]

I explained that perfectly

[runic garnet]

Nice cursive

[high knot]

Treesh is in dire need

[scarlet folio]

im actually so screwed

[high knot]

If she doesn’t succeed she will have to contact her academic enemy, Corey Mills

[scarlet folio]

im on the verge of crying over stupid numbers

YOU

[high knot]

do you understand chains

[scarlet folio]

but how am i supposed to tell

isnt it when like f(g(x))

[runic garnet]

Yea, that’s what csc(x^2) is

[scarlet folio]

oh

[runic garnet]

Csc is f

x^2 is g

[high knot]

You’locmake me relapse

you’ll

[scarlet folio]

OH omg

wait help with 4

ill cook

[high knot]

We should all start a zoom call

your teacher must sleep at the school

To get in at 6am sharp

[runic garnet]

Who even are you

[high knot]

Maybe AP art is more of your speed

[runic garnet]

What are you doing here

[high knot]

she told me to help

For moral support

[scarlet folio]

so this would be (-csc(x^2)cot(x^2))(2x)(2x+1)-csc(x^2)(2)

clarifying

how is this not wrong the parenthesis seem so out of place

[runic garnet]

-csc(x^2)(2)

At the end

[scarlet folio]

oops

[high knot]

I’m having flashbacks

[scarlet folio]

flashbacks of derivatives

[runic garnet]

Good

[scarlet folio]

theres hope

i see a light

[runic garnet]

Flashbacks to cheating on AP exam

[high knot]

A New Hope

[high knot]

youre so right

I got a ghetto calculator just to cheat

[runic garnet]

Valedictorian for what 😂

[scarlet folio]

so for #4 4sin would be f(x) and (cot(3x)) is g(x)

[high knot]

Being the best

[runic garnet]

Cheater

Yea ik

[runic garnet]

Yea

[high knot]

Work smarter not harder

[scarlet folio]

im going to hit oyu

[high knot]

student debt paid

the heartbreak emoji

💀

[scarlet folio]

that was me

hold on let me try 3

4*

[runic garnet]

Ultimately it comes down to a fundamental disagreement 🤷‍♂️

[high knot]

I took the high road

sorry for you

[runic garnet]

have fun

[scarlet folio]

wait are you leaving me i need help

STOP

[high knot]

I think i ticked hom off

him

[scarlet folio]

lol nooo

[runic garnet]

No, I was talking to mak lol

[high knot]

Oh ok

Thanks

[scarlet folio]

i sounded like a desperate ex but please i cannot pass this class

[high knot]

I was joking i did not cheat i thought it was obvious

💀💀

[high knot]

just ask Corey Mills

[runic garnet]

@scarlet folio what’d u get

After this I gotta go

[scarlet folio]

wait is this even the product rule

[runic garnet]

No

[scarlet folio]

wait n

4cos(cos(3x))??

times uh

3

[runic garnet]

Is that ur answer?

[scarlet folio]

yes

[runic garnet]

How did u get cos on the inside

[scarlet folio]

i thought i had to leave that alone

oh wait

am i supposed to find the derivative of cos too

[runic garnet]

U should be getting:

[scarlet folio]

i thought this was just f(g(x))

[runic garnet]

$4\cos (\cot (3x)) \cdot \dv {x} \cot (3x)$

[scarlet folio]

omg

COT

where did the cotangent comr rfrom

nvm im screwed whatever thanks for the help you did provide lol

upaid labor

[runic garnet]

Cot is on the inside

[scarlet folio]

im embarassed i read it wrong

i thought it said cos

so for the second part i just derive (cot(3x))

which is -csc^2(3x)

[runic garnet]

Yes then chain rule again

Chain rule for the 3x

[scarlet folio]

OH OK thanks

imma just try to figure out the rest on my own, thanks for the help now i know how to at least start it

.close

[cedar kiln]

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[cedar kiln]

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[rapid lance]

i need a little help on this precalc stuff

[nimble mountain]

there are no nice roots :(

[rapid lance]

huh

that’s not good to hear

nvm chat i gave up

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mystic rapids]

yo

[mystic rapids]

If I got this

[crimson sedge]

yes ?

[mystic rapids]

it can be solved by this?

or in parts

multiply that twice

[crimson sedge]

no u must developp

[mystic rapids]

and then integrate it

so by part?

[crimson sedge]

oh wait

i didnt understand what u meant

[crimson sedge]

u mean that this is a primitive ?

[mystic rapids]

Its after integrate

[crimson sedge]

no it is wrong

[mystic rapids]

so

like this?

I hope u understand

[crimson sedge]

like what ?

[mystic rapids]

The plus thing was wrong

[crimson sedge]

thats the good idea yes

[mystic rapids]

So the integration is x^7/7

[crimson sedge]

and it is 9x^2 at the end of devloppement

[mystic rapids]

And bla bla

Yea

My bad

[crimson sedge]

yes and bla bla

[mystic rapids]

K thanks so much bro

[crimson sedge]

np

[cedar kiln]

@mystic rapids Has your question been resolved?

[mystic rapids]

No I got an other question

With this one I can do

yeah no?

[cedar kiln]

@mystic rapids Has your question been resolved?

[dire geode]

no that's the wrong antiderivative

you need to use chain rule when doing u sub

or maybe you just forgot a constant

u = z^2 => du = 2z dz

[mystic rapids]

so it just ends in

U/22

u 11 / 22

I dont know how to put that arrow

[cedar kiln]

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[lyric locust]

How do I do my test points like this?

<@&286206848099549185>

[cedar kiln]

@lyric locust Has your question been resolved?

[twin heron]

solve for Xs

factor poly

plug in x and see if the condition is true

[lyric locust]

Likes this?

[cedar kiln]

@lyric locust Has your question been resolved?

[cedar kiln]

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[junior marsh]

.reopen

[fathom star]

Is this the equation I should be doing for this?

ay=2.9m/s^3*10

I assumed I was supposed to use a derivative of 2.9^3 since its not m/s^2 but idk that.

Which led me to do
3 * 2.9m/s^2 = 8.7m/s^2

8.7m/s^2 x 10 for time in the original acceleration formula since we're at t = 10
So ay=87m/s^2?

[plucky owl]

It might be a typo, since acceleration is m/s^2 and that's what the statement given says

[fathom star]

I wondered if that could be the case after discovering jerk but wasnt sure.

[mellow haven]

the given unit is not wrong

m/s^3 multiplied by the s from t makes it m/s²

and yes, 2.9 m/s^3 is called the jerk lol

anyways, dont think too much about it

take the acceleration and integrate twice

[plucky owl]

Then why does it say, vertical acceleration, ay? If it's acceleration, it should be units in m/s^2 not m/s^3

[mellow haven]

it is m/s² lol

[mellow haven]

ay = 2.9 t

ay = 2.9 t

2.9 has a unit of m/s^3

and t has a unit of seconds

multiply them, you get m/s²

m/s^3 is not the unit of ay, its the unit of the 2.9

$\frac{m}{s^3} , s = \frac{m}{s^2}$

[wraith dagger]

Emily

[fathom star]

So shouldnt it be: 0+0+14.5*100?
29/2 x (10)^2

[plucky owl]

They did not write that as clearly as I can read lol it's 2.9 m/s^3 times t

Emily was right

[mellow haven]

i encountered this notation a lot before

😛

[mellow haven]

why are you hurrying to substitute

first find the expression of y(t)

and then substitute the jerk and the time!

start with ay = 2.9 t

integrate to find vy

then integrate to find y

[fathom star]

Im trying to learn integration so ill see if I can figure that out I guess

ay = 29m/s^2

[mellow haven]

no, ay = 2.9 t

linear function

like in math y=mx

[fathom star]

but we're looking for the height at t=10?

[mellow haven]

yeah, so we first derive the expression of the height

integrating the acceleration gives the velocity

and then integrating the velocity gives the height

what function has a derivative of 2.9 t

?

[fathom star]

Trying to figure it out

All I know atm is the power rule

[mellow haven]

yeah exactly, use it

[mellow haven]

yes, thats when differentiating

for integraing, its

$\int x^n , dx = \frac{x^{n+1}}{n+1}$

[fathom star]

t^2.9 + c? (nvm)

[wraith dagger]

Emily

[mellow haven]

$v_y(t)=\int a_y(t) , dt = \int 2.9 , t , dt$

[wraith dagger]

Emily

[mellow haven]

just apply the formula!

[fathom star]

um

Im doing this

so it would be 2.9tdt, does the t after integration become t^2?

[mellow haven]

according to this formula, it becomes t²/2

[mellow haven]

no, thats not how we do it

you cant take t outside of the integral

you can only take 2.9 out, yes

$v_y(t)=\int a_y(t) , dt = \int 2.9 , t , dt = 2.9 , \int t , dt$

[wraith dagger]

Emily

[mellow haven]

$v_y(t)= 2.9 , \frac{t^2}{2}$

[wraith dagger]

Emily

[mellow haven]

i only applied this formula

$\int t^1 , dt = \frac{t^{1+1}}{1+1}=\frac{t^2}{2}$

[wraith dagger]

Emily

[fathom star]

the ay in ay(t)dt is technically (2.9m/s^3 x t) (t) dt, for the formula?

[mellow haven]

no, the (t) in ay(t) is just to show that the acceleration is a function of time

like f(x)

$v_y=\int a_y , dt = \int 2.9 ,t , dt$

[wraith dagger]

Emily

[fathom star]

ah ok

[mellow haven]

okay, so

starting from ay = 2.9 t

what would vy = ?

[fathom star]

I feel bad but im honestly still lost. I think its that I dont understand this on any level

[mellow haven]

okay, lets start over

[fathom star]

wait

maybe actually

[mellow haven]

its really simple

integrate acceleration, which is 2.9 t

and it becomes velocity

[mellow haven]

and how to integrate acceleration, using this formula

[fathom star]

Ok so am I plugging in variables at this point then multiplying it by ay which equals 29m/s^2?

[fathom star]

Oh, this basically. plug in 10 for t

[mellow haven]

NO

we plug in nothing yet

we are only deriving the expression of the velocity

not the height!

[fathom star]

so vy = ay(t^2/2)

[mellow haven]

vy = 2.9 t²/2

not ay (t²/2)

because ay = 2.9 t

[fathom star]

so what am I supposed to do with this?

[mellow haven]

thats velocity!!

and the integral of velocity gives the height, what we're looking for

so, you gotta integrate the velocity now

$y=\int \frac{2.9}{2} t^2 , dt$

[wraith dagger]

Emily

[fathom star]

I have to go get food. 😭

Wait are you integrating dt again for velocity?

@mellow haven

[mellow haven]

we are integrating velocity to get the height

and then once we have the expression of the height

we will substitute t=10 in it

[fathom star]

So for the integration symbol, are you putting v of X at the top and v initial at the bottom?

[mellow haven]

im not putting anything

the rule is that

$v_y=\int a_y , dt , + c$

[wraith dagger]

Emily

[mellow haven]

$y = \int v_y , dt + c$

[wraith dagger]

Emily

[mellow haven]

the c's are called constants of integration

(they are both 0 in our case, since the rocket was at rest)

[cedar kiln]

@fathom star Has your question been resolved?

[fathom star]

For now I guess xP

[cedar kiln]

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[small nacelle]

i need help graphing slope. how do I graph an improper fraction? my equation was in point-slope form which was y + 2 = 2/3(x-2) and i converted it to slope-intercept form which turns out to be y = 2/3x - 10/3.

[stark plinth]

Slope is generally represented as:
(change in y)/(change in x)
Since you've written it in slope-intercept form, start from the intercept and plot that.
Then move (upwards) in y units, and (right) in x units and plot the next point.

You've calculated the slope to be (2) / (3), so y and x will move accordingly. Every time y moves by 2, x will move by 3.

If it helps to connect with prior examples, when you've seen line equations in the past where the slope is an proper integer, you can also represent them as fractions as well.
For example: y = 2x + 3
Has a slope of '2' or '(2)/(1)'. Likewise, it means everytime y changes by 2, x will change by 1.

[cedar kiln]

@small nacelle Has your question been resolved?

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[queen sundial]

74

[mellow haven]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@queen sundial Has your question been resolved?

[queen sundial]

Okay I guess I can use discrimination

Discriminant*

[queen sundial]

But how would I solve using derivatives

Ik question doesn’t say that but the lesson I’m on is about that

[mellow haven]

using the general equation of the tangent line to a function f

do you know it?

[cedar kiln]

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[cedar kiln]

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[wild patio]

A is a mixture and B is a compound?

[fast lotus]

Yep

But this is not math 🙄

[wild patio]

STEM

[wild patio]

And is it a physical change

Since it’s the same elements

Just different form

[fast lotus]

Chemical

The thing changed

From hydrogen to water

[wild patio]

It’s the same elements

[fast lotus]

New product is formed

[wild patio]

OHHH

[fast lotus]

Ok so

[wild patio]

H2O

[fast lotus]

Yes

[wild patio]

So if it was 2 oxygen and one H bonded together

Would it still be chemical

[fast lotus]

Yes

[wild patio]

Why that isn’t a element

02H

So what would make it a physical change

[fast lotus]

Because Hydrogen cannot

Form 2 bonds

It has only 1 electron to form a bond

[wild patio]

So what would make it a physical

[fast lotus]

Physical change hmm

It's not

Formation of water is chemical

For physical

[wild patio]

And B is more stable because there closer together

[fast lotus]

U can basically deform stuff and say it is physical

[wild patio]

Or is it not

[fast lotus]

Water is more stable yes

Reason being that the reaction is exothermic

[wild patio]

Yea but

Since there closer together

That’s how we determine

What’s more stable

[fast lotus]

Not exactly

[wild patio]

So how would I determine

[fast lotus]

If u go closer and closer, u would see repulsions

[wild patio]

Well can’t do that

[fast lotus]

My reason is that reaction is exothermic

[wild patio]

So a chemical is more stable then a physical

[fast lotus]

Since it is exothermic, h2o has lower energy

[fast lotus]

Um wot?

Didn't get you

[wild patio]

Idek

[fast lotus]

What's ur doubt exactly

[wild patio]

Nvm I see it

[fast lotus]

Mhm

[wild patio]

So movement differs in a solid liquid gas

[fast lotus]

Yes

[wild patio]

Solid barley moves and less free

[fast lotus]

Yes

[wild patio]

Liquid moved past eachother and a little more free

[fast lotus]

Yes

[wild patio]

And gas Particles move the fastest and don’t interact with eachother

[fast lotus]

Yes

But they can

[wild patio]

Or do they

[fast lotus]

Interact

[wild patio]

So I shouldn’t say that

[fast lotus]

Yes they can collide

[wild patio]

They move the fastest and in random motions and can collide

And are the most free

[fast lotus]

Yes

They move faster yes

They are free and collide yes

Correct

[wild patio]

21A is solid

B is liquid

C is solid?

[fast lotus]

Yes

[wild patio]

And for 23

Physical
Physical
Chemical
Chemical
Physical

[fast lotus]

Yes

[wild patio]

Deadass?

[fast lotus]

Mhm

[wild patio]

Physical
Chemical
Chemical

[fast lotus]

Third is physical

Change in state is physical

[wild patio]

I thought melting or burning

Is chemical

[fast lotus]

Burning is chemical

If it forms new compounds

But melting is not

Ice, water, steam

Interconvert

[wild patio]

Ahhh

Got it

So had it said burn

Chemical

[fast lotus]

Yep

Yes

Burning includes reaction with oxygen

[wild patio]

And 27

[fast lotus]

But converting into gas is not

[wild patio]

Is this physical

[fast lotus]

Yep

[wild patio]

Because it just took a new form

What would make it chemical

[fast lotus]

Imagine

U have wood

And u burnt it

[wild patio]

Chemicals

[fast lotus]

CO2 will be formed

So chemical

But what happens if u heat water?

[wild patio]

So why is 27 physical

[fast lotus]

Water vapour

[fast lotus]

Burning is different from heating

[wild patio]

How do you know those elements is water

[fast lotus]

No no

[wild patio]

So heating is always physical

[fast lotus]

In that figure

[wild patio]

No matter what the second image is

[fast lotus]

Do u see any new compounds?

[wild patio]

So if something is heating

No matter what it is

[fast lotus]

Mhm

[wild patio]

It’s physical

Always

[fast lotus]

Yes

Wait always yes is not accurate

Look

If a substance is converted from solid to liquid to gas and vice versa

It's physical

But if some new compounds are formed, then chemical

[wild patio]

So for 31

The answer is

The atoms get faster due to the heat

Or do I gotta add something else

[fast lotus]

Yes atoms get faster

[wild patio]

And for 33

[fast lotus]

But if enough heat is supplied

Change in state is also possible

[wild patio]

Rock at top is potential

Bottom is kinetic

Wood is potential Ashes is Kinetic

[fast lotus]

Nop

[wild patio]

Tight spring is potential spring relaxed is Is kinetic

[fast lotus]

They didn't mention that they left the rock from top

[wild patio]

It says bottom of hill

[fast lotus]

Yes so basically at rest

On the ground

No energy

[wild patio]

So that’s more stable?

[fast lotus]

Yes

[wild patio]

So wood is potential

Ashes is Stable

[fast lotus]

Yes

[wild patio]

Tight is potential

Relaxed is stable

[fast lotus]

Yes

[wild patio]

So for 37

What’s endothermic and what’s exothermic

[fast lotus]

It says energy is absorbed

[wild patio]

So

[fast lotus]

Hm?

[wild patio]

?

[fast lotus]

Are u asking the definition?

[wild patio]

So this is endothermic

[fast lotus]

Yes

[wild patio]

Because endothermic absorbers

[fast lotus]

Yes

[wild patio]

Ok and 39

That shii confuses me

[fast lotus]

🙄🙄

What kind of question is that

Hypothesis is like a guess using the known stuff

[wild patio]

A hypothesis is a guess based off data

[fast lotus]

Theory is something which is proven

[wild patio]

Ahhh got it

[fast lotus]

Yes

[wild patio]

So 43

How does that relate

[fast lotus]

Idk lol

[wild patio]

So how would she test this

.close

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[nimble mountain]

im lost trying to understand correspondence theorem

[raven shard]

which part are you having trouble with?

[cedar kiln]

@nimble mountain Has your question been resolved?

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[nimble mountain]

i shall come back stronger

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[blazing zephyr]

find all possible pairs of (a,b,c) where a,b,c are non negative integers that are less than 10 and they sum to a number thats divisible by 5

[solid juniper]

well there's always this. for each nonnegative integer $z$, the number of ways to make a sum of $z$ is the coefficient of $x^z$ in $$(1 + x + x^2 + \ldots + x^9)^3$$

[wraith dagger]

:bending_skull:

[dire geode]

How did you type that so fast

[solid juniper]

are you making fun of me

i was typing something else and changed my mind ok

[dire geode]

No I'm actually just stupid. I thought you sent it right after skill sent the question but you sent your message after I finished reading

[solid juniper]

lol

i thought you were making fun of me typing slow

[raven shard]

do you have to count the number that exist, or do you have to "find all" of them

[dire geode]

Never

[solid juniper]

lol i didn't see that. i hope not

[raven shard]

it doesn't actually take long to write out

but ye slightly non-ideal

[solid juniper]

,w expand (\sum_{i=0}^9 x^i)^3

[wraith dagger]

Results provided by WolframAlpha

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

Requested by slayyla

[solid juniper]

eh that's gonna be a lot?

it would take a while to write out

[raven shard]

we'll see about that

[solid juniper]

,calc 6 + 36 + 73 + 63 + 21 + 1

[wraith dagger]

Result:

200

[solid juniper]

bs problem if you need to "find" them all

[solid juniper]

if you just need to find the amount i think i can outline a not amazing but probably feasible way following this

there would be a lot of binomial coefficient computing though

[solid juniper]

or maybe nvm on that

[raven shard]

ok well

so i guess that was roughly 9 minutes

[solid juniper]

you wrote out all 200?

[raven shard]

yes

[solid juniper]

lmao

ok but like

if that's intended, imagine grading this

[raven shard]

lmao

[cedar kiln]

@blazing zephyr Has your question been resolved?

[blazing zephyr]

sorry, the first

[raven shard]

ok perfect

[raven shard]

if a = 0 and b = 0 how many possible values of c are there?

[blazing zephyr]

2?

[raven shard]

correct. now if a = 0 and b = 1 how many possible values of c are there?

[blazing zephyr]

2

[raven shard]

correct. (you might see where this is going)

[blazing zephyr]

so if we lock a and b we get 2 possibilities for c?

so is it like 10×10×2=200?

[raven shard]

yes

[blazing zephyr]

oh1

.close ty

[cedar kiln]

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[marsh horizon]

Need help

[marsh horizon]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[cedar kiln]

@marsh horizon Has your question been resolved?

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[manic hedge]

i want to arrive at LHS = RHS

can help guide me?

this is proving

[cerulean sail]

You have, haven’t you?

Also small note, RHS is sec(2x) + tan(2x), you wrote tan(x) there

If you want to write it as a “chain”, say LHS = … = RHS, then you can always just do this:

[cerulean sail]

(I.e. wrote the steps in order to the bottom for the left, then write the steps you did for the right backwards from the same thing)

[cedar kiln]

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[oak leaf]

Can anyone Help me with this?

[cedar kiln]

@oak leaf Has your question been resolved?

[mint moss]

equilibrium, right?

[oak leaf]

yes

actually, i solved it but answer was not matching

so, solution was wrong in book

🙂

[mint moss]

is it 5/8

?

[cedar kiln]

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[brisk linden]

how to differentiate this?

[pseudo pilot]

it is $(2-x)^{-2}$ isn't it?

[wraith dagger]

MetuMortis

[brisk linden]

yes

indeed

[pseudo pilot]

use chain rule

[brisk linden]

my bad did i say differentiate?

i Meant INTEGRATE

[pseudo pilot]

substitue u = 2-x then

[brisk linden]

ok and then>

what rule do i have to use?

[pseudo pilot]

no special rules, just basic integral substitution:
$\ u = 2-x \ du = -dx$ thing

[wraith dagger]

MetuMortis

[dreamy void]

Yea do what griffin said

[cedar kiln]

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[odd sierra]

help plz

i got 0.24286896 from my calc

for mutltiplying part

3 sf only

how does does that work if we have a leading 0?

0.243?

[hollow trail]

the leading 0 is not significant

[odd sierra]

yes

[hollow trail]

but you should not round until the end of the calculation

[odd sierra]

oh

Ok

wait but

oh

becuz they are same rules

but what if we had addition with division u have to round early right?

[hollow trail]

no, you round at the end. you keep track of how many significant figures the intermediate result should have without actually rounding it

[odd sierra]

Ok perfect

thnx

[cedar kiln]

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[dire geode]

why are you setting x=1

[crimson sedge]

x^2 cancels out right?

or x^2 / x^2 =1

[pseudo pilot]

it does not

[crimson sedge]

if it doesnt work where do i get it wrong

[dire geode]

,tex .wrong cancel

[wraith dagger]

riemann

[crimson sedge]

show me logic please

[dire geode]

just take x = 2

what's the first term?

[pseudo pilot]

did not know it is that common to have a tag lol

[fallow apex]

[mellow belfry]

you can't simplify that way with the denominator

[crimson sedge]

i know it doesnt work but i want to know why

is it already simplified

[runic garnet]

$\frac x{x+y} = \frac {x+y-y}{x+y} = 1- \frac y{x+y}$

[runic garnet]

heres something u can do

idk if thats even related to what u asked, i just wanted to latex

[dire geode]

it doesn't work because you ignored the -1 term

[crimson sedge]

oh so simplifying by division only works when the whole thing is multiplication on top and bottom?

[odd verge]

kekw

[crimson sedge]

i have no idea what that is

[mellow belfry]

yas, you have to factor

[crimson sedge]

is the original question already simplified

dont mind the does this work

i want to know if the original is already simplified

[mellow belfry]

I think it can't be simplified

[crimson sedge]

so its already at its most simple form isnt it?

[mellow belfry]

because below there is a difference of squares

[crimson sedge]

there is something missing in my algebra logic

this is something im trying to do

lol

i feel like all of them are

i want to know the logic behind it if it is the case

[mellow belfry]

Well, to simplify an equation it's necessary that it be factored

[crimson sedge]

ahhh

so for B

its -1(6-x) / (6-x)

can be factored

c

[mellow belfry]

yep, this is factored

[crimson sedge]

i dont know how to factor c

ill try to factor D

[mellow belfry]

3

[crimson sedge]

3(x) / 3(x-2) ?

x / (x-2) ?

[mellow belfry]

yep

[crimson sedge]

yeah i just dont understand the question i guess

im thinking "if there is an x term in the numerator and an x term in the denominator the expression can always be simplified" means youre canceling the x on top and bottom to be 1

but according to you were just finding if an expression with Xes on top and bottom can be simplified

okay i think i got it now

okay guys

i understand that you cant cancel the x's in (x-5) / (x+1)

but i dont understand it enough to be able to form the logical articulation of it

is it because in the order of operations

(x-5) is basically its own set?

and (x+1) is its own set

and its seperate from division and multiplication

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

holy crap

every time there is a question to simplify rational expressions

they never gave me a question like that

so im extremely unfamiliar with it

[cedar kiln]

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[craggy hazel]

Hi I am very confused on how to find the domain of combined functions

composite*

.close

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@eager hemlock Has your question been resolved?

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[zinc jolt]

is this correct

[vernal lichen]

is this proofs?

@zinc jolt

[glacial field]

You can’t conclude 5 with the given info, use SAS

Such that RQ = SQ, m<RQT = m<SQT, and TQ = TQ

[zinc jolt]

yes

[zinc jolt]

oh ok

thanks

.close

[cedar kiln]

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[vapid otter]

What is the domain of $f(x) = ln(\frac x {x - 1})$ ?

[wraith dagger]

Devil Wears Prada

[vapid otter]

do i set the denominator greater than 0?

for ln

cuz i know sqrt's are greater than or equal than

ln are just greater than

right

[dim sage]

no becuase -/- is positive

[vapid otter]

?

what

[dim sage]

the denominator can be negitive but cant be 0

[vapid otter]

yes

[dim sage]

and it can only be negitive when the top is also negitive

[vapid otter]

what

how do i find the domain of this

what is the first step

[dim sage]

set the demoator so it cant = 0

[vapid otter]

yes

and

[undone star]

we can simplify this question to x/(x−1)>0

[vapid otter]

oh

okay

[dim sage]

set the function so it is greater then 0

[undone star]

this can be split into 2 trivial case

the most obvious one is if x is bigger than 1

so we know for sure that x>1 is part of the domain

[vapid otter]

$\frac 1 {x - 1} > x$

[wraith dagger]

Devil Wears Prada

[undone star]

what isthis for?

[vapid otter]

im solving

[undone star]

not quite

[vapid otter]

the inequality

how do i do it thebn

oh

is it

1/x on the right

[undone star]

yeah

[vapid otter]

$1 > \frac 1 x \cdot (x - 1)$

[wraith dagger]

Devil Wears Prada

[undone star]

where does 1 come from

we are solving x/(x-1)>0

it is best to solve it grpahically if u struggle to understand it visually or intuitively

[vapid otter]

no i cant use a graph

not allowed

[undone star]

u can try to graph it urself to convince urself

like not with desmos

but just plotting a few points if u are confused

[vapid otter]

no

im telling you i need to understand it algebraicly

how do i find the domain algebraicly

[undone star]

your algebra is wrong currently

[vapid otter]

well correct it please

[undone star]

i cant really explain it besides just doing it analytically as it is

x/(x-1) > 0 has 2 cases

[vapid otter]

okay and

[undone star]

either x and x-1 are both positive

or x and x-1 are both negative

like before, obvious case is if x>1

and also x < 0

if x < 0, x is negative

x-1 will also be negative

[vapid otter]

can i just solve this algebraicly

[undone star]

then the inequality also holds

[vapid otter]

i dont know what ur doing