help-17

[rose raft]

yes

[open cave]

ye

[silk canopy]

ty

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[lime gorge]

(Yes)

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[silk canopy]

is the answer 1

[lime gorge]

U can use an online calculator for this kind of stuff

It’s just manual computation

Or just do it in bots

[silk canopy]

i used to use mathpapa, but then it resulted me in getting wrong answer through the calculator and me tearing correct answers doing it a million times and figuring out i was correct

[lime gorge]

Just use a normal calculator if online software doesn’t work

Unless u want us to check ur work

That’s something that would be more appropriate

[silk canopy]

alright

i will buy a calculator

but will u be willing to check this specific q?

[lime gorge]

There’s free ones online too

Here use this

https://www.mathway.com/Calculus

[silk canopy]

alr

ty

[lime gorge]

This is good for this kind of stuff

Just make sure to put the degrees symbol

[silk canopy]

yep

thanks a lot

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[gray yarrow]

For circumference we need the two side lengths

Area which is given is their product

Lets call the sides a and b

A=a*b

C=2a+2b

To get a and b, we have to factor the area

For example
2x^2+5x+3=(x+1)(2x+3)

Do you know how to divide a polynomial by another polynomial?

Oh lol

That is weird haha

Seems a bit advanced for 8th grade

We can start by guessing a root of the polynomial

Divisors of the constant are good ideas

For example 3

If we find such a root and call it p then we can factor out the term (x-p)
That will be what we are after

For that factoring we need polynomial division

But i doubt you are supposed to do that in 8th grade

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[young remnant]

9. Here are some timetable information from the local train station.

• Journey C lasts twice as long as Journey D.
• Taking Train A halves the duration of Journey B.
• Train C reduces the duration of Journey D by five hours.
• Journey A and Journey B last for eight hours.
• Taking Train B halves the duration of Journey A and Journey C.
• Journey B lasts half the duration as Journey D.
  Based on the information above, please answer the four questions listed below.

1. Taking Train B will reduce the duration of Journey C by twelve hours.
    a. True b. False c. Insufficient Information
2. Journey D lasts for 18 hours.
    a. True b. False c. Insufficient Information
3. Taking Train C will reduce the duration of Journey D to eleven hours.
    a. True b. False c. Insufficient Information
4. Taking appropriate Train, Journey D lasts longer than Journey A.
    a. True b. False c. Insufficient Information

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[lusty fox]

tf

whats your question

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[orchid hill]

anybody got ability to do 4x4 matrix multiplication

[orchid hill]

i struggle

[golden iron]

you can have a computer do this if you need to check your work

but if you're expected to be able to do it on paper then you just have to practice

[orchid hill]

did via calc but website i regularly use says differently

[golden iron]

can you show your work?

[orchid hill]

had to inverse t2 then mult by t1

[golden iron]

lol i thought you meant calculation

but thanks for the pic of your calculator haha

[orchid hill]

no calc did it but lmao unsure if righht

chatgpt shit at maths

[orchid hill]

can u verify result

[golden iron]

yeah im plugging it in

also chatgpt is not for math

https://www.symbolab.com/solver/matrix-multiply-calculator/\begin{pmatrix}1%260%260%2610\\ 0%26-1%260%2620\\ 0%260%26-1%2610\\ 0%260%260%261\end{pmatrix}\cdot\begin{pmatrix}-1%260%260%269\\ 0%261%260%2619\\ 0%260%26-1%269\\ 0%260%260%261\end{pmatrix}?or=input

[orchid hill]

ah u ty

i see issue thx

[orchid hill]

heard u could get add ons

[golden iron]

it's not the purpose of any "llm" to be correct in a sense

they've been taught to sound like they're using language

which has uses

but math is certainly not one of them

[orchid hill]

oh

[golden iron]

even though they use so much math to work haha

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[forest root]

can anyone help with the next steps with this u-subsitition problem

[forest root]

do i sub u = pi/3*ln(x) back into the equation

[full arch]

looks good so far

[full arch]

could you not just integrate sin(u) du?

the same way you would sin(x) dx

[forest root]

like -cos(u) ?

[full arch]

yep

and then it turns out quite nice

gj

[forest root]

something liek this?

then do i sub in u = pi/3*ln(x) or just plug pi/3 and 0 into u

like -cos(pi/3) + cos(0)

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[hearty storm]

Can someone explain this to me. Solution in green and my working is in purple

[hearty storm]

I don’t understand them using Y(s) instead of L(y(t)). Say I want to keep it as L(y(t)) then should our formula in red be d/dt? Instead of d/ds?

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[hearty storm]

<@&286206848099549185>

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[low seal]

approximate area under the graph of f(x)=x*(2^x), x∈[-1,2], six subintervals, right endpoints

[golden iron]

do you have any work?

you can check against here https://www.symbolab.com/solver/riemann-sum-calculator/riemann \int_{-1}^{2 }x\cdot2^{x}dx%2C n%3D6 ?or=input

make sure you have right-point type selected in the dropdown

[low seal]

does that make it -1 and 2 inclusive?

it gave the same answer i had before, (13(sqrt2)/8)+5, or 7.298

which was wrong

[golden iron]

the right-sum shouldn't include -1

[low seal]

it does include 2

[golden iron]

yes

[low seal]

hmm

[golden iron]

x1, ..., x6 = -1/2, 0, 1/2, ..., 2

is there an answer key, how do you know this is incorrect?

[low seal]

its an online portal

[golden iron]

does it expect an exact answer? or does it say to round decimal points

[low seal]

i tried both

[golden iron]

if you could get a picture of the og question that would be great

[low seal]

i think its decimal

[golden iron]

yeah i would expect decimal

is there a rounding convention from class or the portal expects?

[low seal]

usually 2 or 3 decimal places but i just tried more in case

[golden iron]

yeah i don't know then. hopefully it's just a bug :/

[low seal]

thanks

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[vast shale]

How does this function have a limit

[flat whale]

As x approaches what

[vast shale]

infinity

Horizontal asymptote

But that's y so what about x

[flat whale]

The first term goes to zero because the denominator's degree is higher than the numerator's

[vast shale]

wowee

That's so cool

If the degree of denominator is higher than the numerator it always has a horizontal asymptote at y = 0

Messed around with desmos

Alright thanks

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[gaunt panther]

How can I calculate this when I don’t have the value of x1?

[rose raft]

what is the top

[gaunt panther]

sry

[rose raft]

oh

are you solving a or b

[gaunt panther]

b

look at my workings

[dull bear]

Well of course you can see where the stationary point is “about”, I’d say use that to make an educated choice for your x1, no?

[gaunt panther]

Could we just ignore the Lnx

And set 2x-3 = 0

And that would give an estimate of x1?

So x = 3/2

[dull bear]

I mean, I guess you could choose x1 as 3/2, but then, ergh

[dull bear]

They literally give you a plot anyway for it

[gaunt panther]

OH MY

[plain minnow]

Plots are friends not foes

[gaunt panther]

I didn’t see the graph

Mb

Yh calm that makes sense

Thank u chartbit

[dull bear]

[just so happens that 3/2 is pretty good of a choice anyway ]

[wraith python]

Use Picard's Method.

[gaunt panther]

Newton raphson method very underrated imo

It’s cool

I’m acc learning about it now for the first time

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[exotic peak]

Hello is 2pi/4 in 90 degrees in the unit circle?

[unreal heart]

pi/2 is 90 degrees yes

[exotic peak]

okay thank youuu!

[unreal heart]

if you ever need a conversion the formula is radians*180/pi

[exotic peak]

okay ill take note of that thankss

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[quiet sand]

Howdy

[quiet sand]

I just need some help with like 3 questions. All geometry related

Let ABCD be an isosceles trapezoid with bases AD = 21 and BC = 9. Given
that the altitude of the trapezoid is 8, find its circumradius.

This is what I have so far

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[quiet sand]

<@&286206848099549185>

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[charred echo]

i've been trying to figure this out for like an hour. im not even sure where to begin as theres literally not a single problem thats similar to that in my textbook. tried doing something with harmonic mean, geometric mean etc but didnt get to the answer

if a+b=12 then find the max value of a^2+3ab+b^2

[charred echo]

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[boreal remnant]

This is way simpler than you're making it lmao

b = 12-a

Then you just have to maximize:
a² + 3a(12-a) + (12-a)²

[paper depot]

.reopen

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[charred echo]

yes i realised that and felt stupid af

[paper depot]

ah.

[charred echo]

it was just in the section with geometric, harmonic etc means and thought i had to use those

but yes thanks

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[hushed atlas]

to integrate ln x all squared is the best way by parts

[hushed atlas]

nvm

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[hushed atlas]

would u set u as lnx squared or just lnx

if u do the latter u have to integrate lnx which is by parts again

idk im seeing different methods in diff places so not sure which is better

nvm again lol

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[sudden compass]

!status

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[polar ivy]

1

[sudden compass]

which part?

[polar ivy]

everything

[rapid swift]

asking someone to do your exam for you

while having absolutely nothing to go off

I mean cmon

[polar ivy]

its nit my test

it was a q one of my friends gave me

but idk how to do it

and if its a test, i would do it in class in exam conditions

<@&286206848099549185>

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[orchid hill]

What is the volume of the work envelope of a Cartesian
robot whose x,y,z axes each have maximum reach 1.2m
and maximum stroke 80cm?

[orchid hill]

unsure whats l x b x h in this

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[placid cradle]

I have a question

[placid cradle]

is Un∼An => |Un|∼|An|

[paper depot]

yes

|U_n|/|A_n| = |U_n/A_n| -> |1| = 1

[placid cradle]

ah yes

aight cool thanks

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[molten rain]

how do i do questin 9

[scenic zephyr]

Is this how you do it?

i know the max and min are 8 and 1, respectively

[molten rain]

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.close

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✅

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[sage burrow]

I'm trying to understand generally how to show a polynomial contained in Z_p is irreducible

I understand fundamentally you first substitute integers 0 up to p-1 and show that f(x), our polynomial, is not equal to 0 for any of these values, but its the next stage which confuses me,

that being "showing a polynomial is not the product of irreducible polynomials"

i.e, say deg( f(x) ) = 7. we have to show that there is no possible irreducible pair of polynomials with respective degrees 6 and 1, 5 and 2, or 4 and 3. while 6 and 1 can be trivial by looking at coefficients, the remaining have always had 'tricks' to them that ive never understood

i.e, when we specifically work in Z_2, i have the corollary "In Z_2, x^2 and (x^2 + x + 1) are the ONLY irreducible quadratics in Z_2". Which is fine, but I don't really have anything else for, say, Z_7 or Z_191. Is there a way I can approach this stage more generally, rather than using corollaries such as the one above?

[half imp]

Over a field every linear polynomial has a root

Thus, any reducible polynomial of degree 2 or 3 has a root

[sage burrow]

oh, so you can very easily deduce deg2 and deg3 polynomials with just that?

would you then, like

apply that idea to polynomials of higher degrees

rearrange them into other polynomials and see if they are reducible

[half imp]

It only works to show they don't have degree 1 factors

so for degree 7 you'd still need to check 2+5 and 3+4

Which doing that manually sounds

really horrible

[sage burrow]

it is and i'm expected to do it

i'll show you a past-paper question i was given with this idea in mind

[half imp]

maybe Eisenstein's criterion?

[sage burrow]

f(x) = x^6 + x^5 + x^4 + x^3 + 1 in Z_2

so you'd be testing pairs of 3 and 3, or 4 and 2

[half imp]

yeah

considering all irreducible deg 2 polys is the start

(if they were reducible f would have a root)

[sage burrow]

and you're supposed to rearrange to like

f(x) = x^4 (x^2 + x + 1) + x^3 + 1

x^3 + 1 = x(x^2 +x + 1) + (x^2 + x + 1)

so f = (x^2 + x + 1)(x^4 + x + 1)

x^4 + x + 1 is irreducible as
(x^2 + x + 1)^2, (the only non-x^2 irreducible quadratic in Z_2 squared) =/= (x^4 + x + 1)

and -that's- how they wanted me to show 'there is no pair of 4 and 2'

which i understand despite how baffling the rearranging was at first glance

its the "the only non-x^2 irreducible quadratic ... " part that irks me

is there a list of these i'm supposed to be able to apply

[half imp]

there is a pair of 4 and 2, isn't there?

didn't you just give it?

[sage burrow]

oh sorry i just showed they were both irreducible lol

yes

f = (x^2 + x + 1)(x^4 + x + 1) so that is your pair, rather

[half imp]

But yeah there are 4 degree 2 polynomials over Z2

Only one of them is irreducible: x^2+x+1

not sure why x^2 is claimed to be irreducible, it's x*x

[sage burrow]

that is weird actually

i just copied what was in my notes 1:1 there

but it does go against what i understand irreducible to even mean

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[open burrow]

Hello, any clues on how to do this?

[full hatch]

Induction

Don't forget a >= 0

[vocal sleet]

@open burrow Has your question been resolved?

[open burrow]

Ok thanks I’ll try that

[vocal sleet]

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[wooden lance]

What would the area of this little valve opening be

[wooden lance]

The diameter measures 8mm

[lime gorge]

Do u know the formula for area of a circle

[sly glade]

Blue = 8mm and you want the area of the green circle?

[wooden lance]

Yea it should be 50mm^2

?

But that doesn't seem right because it working with my model

It isnt*

[lime gorge]

50.266 mm^2?

[wooden lance]

Yea

[lime gorge]

What’s the model

[wooden lance]

I'm modeling the drainage of water out of a container

[lime gorge]

The speed at which it drains?

Mass flow rate, volumetric?

[wooden lance]

Volumetric flow rate

Q=Av

I did the whole model with differential equations and stuff but it just doesn't work

I tried graphing it and realized if there was just one more zero (as in if the valve was smaller by a factor of 10) it would be perfect

[merry dragon]

unit conversion somewhere maybe?

[wooden lance]

I've double checked my calculations a zillion times man

[boreal remnant]

what do you mean doesn't work

[wooden lance]

I can send yall screenshoots

[boreal remnant]

are you saying that this valve has broken the laws of physics?

[wooden lance]

Plugging in a value for t yields an absurd value

I have post up on the help forum with the screenshots

I can tag yall if you'd like

[merry dragon]

yeah, a link would help please

[wooden lance]

.close

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[vocal sleet]

Please don't occupy multiple help channels.

[orchid hill]

.close

[vocal sleet]

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[split delta]

i need help but this is vietnamese math

[split delta]

its vietnamese 8th grade

[vocal sleet]

@split delta Has your question been resolved?

[split delta]

not yet

[dapper condor]

what have you tried

[split delta]

i think i will convert it to a constant and then apply the cauchy inequality

[split delta]

do u have an idea??

[dapper condor]

ill be honest, i have no idea

also doesn't the cauchy inequality only apply to vectors

[cobalt crypt]

cauchy-schwarz should work

[dapper condor]

sorry i know this not my help channel, but do you mind explaining to me how this works as i am currently learning something similar to this

[cobalt crypt]

[split delta]

hey my friend hints that the schwarz is the fastest way

[cobalt crypt]

theyre probably right about that

do you know the corollary to CS called titu's lemma

[split delta]

ugh nope

[cobalt crypt]

just use CS then

[covert granite]

in 8TH GRADE!!!

[split delta]

yes

[split delta]

nah i havent learnt about that yet

[cobalt crypt]

you havent learnt cauchy-schwarz?

[covert granite]

8th grade...

[cobalt crypt]

[split delta]

yes but i have no idea for this exercise

[cobalt crypt]

well a hint is

[echo zodiac]

ez

[cobalt crypt]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[cobalt crypt]

can you not

[split delta]

sorry he is my friend

[echo zodiac]

🙂

no coppy 🙂

[split delta]

just say it!

[cobalt crypt]

this is the hint

[covert granite]

is this actually 8th gradr

[split delta]

ofc

[echo zodiac]

y

[covert granite]

im in 10th grade

and have no idea

[split delta]

what nationality are u

[covert granite]

10th grade in us

11 in uk

[echo zodiac]

:))

[split delta]

just ignore him

[echo zodiac]

🙂

[split delta]

oh really:) thanks for ur help but i still dont know how to do this exercise:(

[covert granite]

ask ur friend???

[cobalt crypt]

well in the context of the exercise

1 has another face

[split delta]

he doesnt help me how to do it instead he sends the ans so i dont understand

[echo zodiac]

để t dạy m

=cauchy

[split delta]

english pls

[cobalt crypt]

do you know how to apply cauchy-schwarz to this?

[split delta]

this is english server not vietnamese

[echo zodiac]

messs

[cobalt crypt]

there arent any rules on what language you have to use here so you're free to speak vietnamese if you'd like

[split delta]

ye i think so

[cobalt crypt]

give it a try

[split delta]

ahh okayy

[covert granite]

oh ik what this is

the thing u can prove via induction

uh

OH WAIT

they are all positive

[split delta]

yes they r

[covert granite]

so u can apply Cauchy schwarz

[twin meteor]

beo
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[split delta]

so complicated...

[echo zodiac]

for m @split delta

[cobalt crypt]

😬

yikes

thats a lotta work

[split delta]

yes ofc

[echo zodiac]

;-;

[cobalt crypt]

if you apply C-S to what i wrote before

you get

[echo zodiac]

I have the answer,anybody needs it?

[covert granite]

~0.266?

[twin meteor]

beo

[split delta]

1

[covert granite]

welp

[echo zodiac]

😄

[covert granite]

i got 1/3

wtf

[cobalt crypt]

mmm 1

,w a^3/b + b^3/c + c^3/a where (a,b,c) = (1/3, 1/3, 1/3)

[twin meteor]

Results provided by WolframAlpha

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

Requested by 夢雪#2250

[cobalt crypt]

thats definitely 1

[echo zodiac]

;-;

[covert granite]

wtf

[cobalt crypt]

5(1/3)^3 is also definitely 1

[covert granite]

so how

....

[split delta]

min=1 when a=b=c=1/3 right Khánh?

[cobalt crypt]

well this is quite easy to minimise

under the condition that a + b + c = 1

[echo zodiac]

yes sir

[cobalt crypt]

if you sum this inequality you just get the minimal value

[echo zodiac]

@split delta wake up

[split delta]

nahh

im so sleepy

i need to sleep

[echo zodiac]

NO

[cobalt crypt]

,ti karin#0305

[twin meteor]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[cobalt crypt]

[echo zodiac]

go learn grandson

[split delta]

its 12 am u know

[cobalt crypt]

,ti beo#2475

[twin meteor]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[cobalt crypt]

[echo zodiac]

:d

[cobalt crypt]

come back after you've slept

this problem wont solve itself when you're dead tired

[echo zodiac]

@split delta wake up grandson

[split delta]

ugh hey can i ask u a question?

[cobalt crypt]

dont ask to ask, just ask

[split delta]

what nationality are u

[cobalt crypt]

🤐

[echo zodiac]

china ,chingchong

[split delta]

excuse me?

[echo zodiac]

sure

:))

[cobalt crypt]

you might not want to say that here, lest you get muted or banned

[split delta]

sorry about that

[cobalt crypt]

you are not the one that needs to apologise

[split delta]

my friend has a bad behavior so hope u forgive him

on the whole, thank u so much for helping me a lot, see u tmr!!

[cobalt crypt]

[split delta]

.close

[vocal sleet]

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[vocal sleet]

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[obsidian wave]

I'm stuck with this. It's not clear to me why the author uses g(a,c) = g(b,d) to then prove that a = b.

[vocal sleet]

@obsidian wave Has your question been resolved?

[obsidian wave]

hello, not yet

[fluid obsidian]

What?

I don't understand your doubt

You have to show that g is not 1-1, so you just consider g as 1-1 and get a contradiction

[obsidian wave]

My doubt was why is g(a,c) equal to g(b,d). It's supposed to be g(a,d) and g(b,d), right?

[vocal sleet]

@obsidian wave Has your question been resolved?

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[fleet salmon]

how is Fa right

shouldnt skater one Fa be on the right

bec East

[lime gorge]

equal and opposite forces right? because she pushes the guy to the right with a force of 70 N, she herself is propelled backwards with the equal and opposite force of 70 N

at least, i think so

its not what forces they exert, but what forces are exerted on them that's shown on their individual fbd's

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@fleet salmon Has your question been resolved?

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[robust heron]

could someone point out how theta value is wrong?

[vocal sleet]

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[scenic zephyr]

Where did that 8.75 come from?

[scenic zephyr]

Ohhh...

Got it

Thanks

[vocal sleet]

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[vast shale]

how do u solve [
x^x = 27^{x+27}
]

[karmic imp]

,w Solve[x^x = 27^(x+27), x]

[twin meteor]

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Requested by dldh06#2659

[karmic imp]

So wolfram gave the complex solutions first before the real one? Wack

[flat whale]

If x is a positive integer, you know x=27^k for some positive integer k

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[topaz wagon]

help

[topaz wagon]

i have no idea where to start

ive swear ive done these before and got them ez but i just need to activate a neuron

[nova ingot]

Maybe start with 2 equations, 1 going up the stream and 1 going down, then maybe youll spot it

Usually when this happens to me i start using the longer way until it clicks

[topaz wagon]

ok

ill try

[topaz wagon]

i still dont mnow

[rapid swift]

make some variables

write some equations

that is how you start almost every word problem

there is no excuse for having absolutely nothing

at the very least, you can identify some quantities that will be relevant

[vocal sleet]

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[thorny spear]

How do we create a quantitative answer the question above in terms of vectors n (unit length) and w (a vector pointed towards the sun

This is what I got so far:

The amount of solar radiation captured by the sun appears to be at the highest when the vector n and vector w are co-inciding and pointing in the same direction.

Hence, the amount of solar radiation captured seems to be proportional to the angle between the vector n (unit vector) and the vector w pointing towards the sun. Hence, to quantify the reduction in solar radiation captured, we can use the dot product formula:

n·w = |n||w|cos(theta)

Since n is a unit vector, |n| = 1. Therefore, we can simplify the equation as:

n·w = |w|cos(theta)

Not sure how to proceed

Ohhhhh I got something

Part 2:
The amount of solar radiation captured by the sun appears to be at the highest when the vector n and vector w are co-inciding and pointing in the same direction. Hence, when the angle between the normal vector and the vector pointing towards the sun is zero (i.e., the solar panel is directly facing the sun), the cosine of the angle is 1, which means that the dot product of the two vectors is equal to the product of their magnitudes.

Hence, the amount of solar radiation captured seems to be proportional to the angle between the vector n (unit vector) and the vector w pointing towards the sun. As the angle between the normal vector and the vector pointing towards the sun increases, the cosine of the angle decreases. The reduction in solar radiation captured is given by 1 - cos(theta), where theta is the angle between the normal vector and the vector pointing towards the sun.

Taking the dot product formula,
n.w = |n|*|w|*cos(theta
) Since n is a unit vector, |n| is 1, hence giving
n.w=|w|*cos(theta)
Equating for cos(theta) gives us
Cos(theta)=(n.w)/|w|

Hence, to quantify the reduction in solar radiation captured, we can use the following formula:

Reduction = 1-cos(theta)
Reduction = 1-(n.w)/|w|

Can anyone verifiy is this correct?

[thin vale]

is this chatGPT

[thorny spear]

Do I really sound that robotic

[thin vale]

no, just the spacing and formatting seems chatGPT-esque

[thorny spear]

Fair enough 😭

Can you please check it out though, I tried to make it explanatory asf and I wanna make sure its right before submitting

[thin vale]

I am a little busy, I think the long text is what might be discouraging people from committing to reading it all since we are all volunteers. But I bet that someone will help you out in a reasonable amount of time since it isn't too complex. Good luck!

[thorny spear]

Ah, okay, thank you regardless 🙂

[vocal sleet]

@thorny spear Has your question been resolved?

[vocal sleet]

@thorny spear Has your question been resolved?

[thorny spear]

!close

.close

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[sterile lotus]

Can someone please help me on this question here. It is given that f(x) = m + nx where m and n are constants. Given also that f(0) = 1 and f(4) = 21, find the value of m and n.

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[pale perch]

do you understand what it means for f(0) to equal 1?

[sterile lotus]

Yes

[pale perch]

what would it give you

[sterile lotus]

If im correcti it should be

1 = m + nx

[pale perch]

not quite, remember its f(0) so x=0

[sterile lotus]

Hmm

0 = 1?

[pale perch]

how did you come to that?

[sterile lotus]

Not sure

Sorry let me try again

[pale perch]

no worries

[sterile lotus]

Would it be

m + n0 = 1?

[pale perch]

thats right, then what can you get from that

[sterile lotus]

That m would be equa to 1

[pale perch]

correct

now you know that, you can do a similar process for f(4)=21

to find n

[sterile lotus]

So it would be

1 + 4n = 21?

[pale perch]

exactly

[sterile lotus]

Ohh

Man thanks

[pale perch]

np

[sterile lotus]

But can I ask you 1 more question

[pale perch]

i do have to go now however im sure someone will see it if you post it here

[sterile lotus]

Oh okay no worries

It is given f(x) = 3x - 5

Find the value of t, given that f(t) = 10.

Would it be x is now equal to t?

.close

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[devout urchin]

Hello, I have this equation in the form of ln(y) = ...

I need this in the form of y = ...

So I removed the ln in the second line. Is it correct?

Just need confirmation

[paper depot]

no

you did not "remove the ln", you raised e to the power of both sides, except you did it improperly on the right.

[devout urchin]

@paper depot what's the proper way?

[paper depot]

e^ln(y) = e^(ln(a) + b ln(x))

simplify this properly

[devout urchin]

I know about it, I just don't understand what to do with e^b.ln(x)

I'm pretty sure the y = e^ln(a) is already correct, no?

[paper depot]

no, it is not

you cannot rewrite e^ln(y) = e^(ln(a) + b ln(x)) as just y = e^(ln(a))

[devout urchin]

I don't know man, I googled about natural log properties and it doesn' have these kind of stuff

I can't find ln(x + y) = ...

[paper depot]

why would you need to look at the log of a sum here

[devout urchin]

which part did I got wrong?

e^bln(x) right?

[paper depot]

you're overthinking this massively or just being blind

[devout urchin]

man I give up

[paper depot]

[devout urchin]

what do u expect me to do lol, I'm absolutely clueless here. back in hs they didn't get in depth about these kind of stuff either

[devout urchin]

for this conversion

does it have any name?

this is all I could find

[paper depot]

it's an exponent law.

you cannot forget exponent laws just because there are logs in the picture.

[devout urchin]

ahh found it

thank you

.close

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[silent tusk]

Hi, please can someone help me with this?

[silent tusk]

This is what I've got so far:
Q11:
Radius: 10cm
Diameter = 20cm
Circumfernece : 3.14 * 20 = 62.8

25% of the circumference of each circle overlaps. And since their are two circles in total 25 + 25 = 50%

Circumference of shaded area: 62.8/2 = 50 50 % of 62.8 = 31.4
C = 3.14 * D
31.4 = 3.14 * D
D = 31.4 / 3.14
R = 5

Area of shaded area: 25 * 3.14 = 15.7

[silent tusk]

My answer matches none of the answers above, so I'm not so sure what to do.

[silent tusk]

Ohk, I understand. Tysm

.close

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[vocal sleet]

@polar ivy Has your question been resolved?

[vocal sleet]

@polar ivy Has your question been resolved?

[polar ivy]

<@&286206848099549185>

[vocal sleet]

@polar ivy Has your question been resolved?

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@polar ivy Has your question been resolved?

[polar ivy]

<@&286206848099549185>

[misty hamlet]

That image is hard to read, could you let us know what part you are struggling with and what you have tried already?

[polar ivy]

im struggling with part i

i know that a is at its max so that mean that theta should be the max on the cos graph

but im wondering whether or not theta could be 0, pi or 2pi

and im not too sure which one it is

@misty hamlet

[misty hamlet]

I don't know what the question is asking. Could you take a clearer image

[lone niche]

a(1 - cos(theta)) is at its max
a is positive
so what theta value would maximize that

[vocal sleet]

@polar ivy Has your question been resolved?

[velvet sparrow]

@vast shale

[vast shale]

ur right

[vocal sleet]

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[icy knoll]

can someone help me with this?

[icy knoll]

like how one would i solve problem?

ik its product rule but i keep getting a dif answer everytime and its nothing like the actual answer

[hushed pewter]

!show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[icy knoll]

at the first derivative (im using quotient rule) to get the first part and i keep getting 1+3x/3x as 3/3x^2

and that parts apparently wrong

[paper depot]

deduct?

[icy knoll]

mb meant solve

[paper depot]

uh now it's even more vague

[icy knoll]

oh

[paper depot]

what are you trying to do, take the derivative?

[icy knoll]

yes

[marsh spear]

what do you wanna solve for?

Oh

[paper depot]

is the 4. part of the problem or only the problem number

[icy knoll]

only problem number

[marsh spear]

i think its just the problem number

[paper depot]

wanted to make sure

anyway uh

[marsh spear]

give me a minute i think i got this

[paper depot]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[marsh spear]

Oh

[paper depot]

anyway

[icy knoll]

yeah thats what i plan to do

[marsh spear]

myb myb

[icy knoll]

i need help solving it and understanding specifically

[marsh spear]

Ok

[paper depot]

(1+3x)/(3x) does not differentiate to 3/x^2

using the quot rule on this is overkill

[icy knoll]

oh

should i go straight into product?

[marsh spear]

I would distribute if i were u

[paper depot]

you should do some simplification here

[icy knoll]

simplification how?

[paper depot]

distribute first

[icy knoll]

multiply both?

ok

[paper depot]

then break the resulting big fraction into a sum

[icy knoll]

gimme a second

[marsh spear]

yeah

[paper depot]

and show us all your work obviously

[marsh spear]

i mean product rule isnt an overkill

but it will be difficult to like simplify from there

[icy knoll]

wait how would i simply these?

would i multiply 3-x to both top and bottom?

[paper depot]

no, you would not.

[paper depot]

bad handwriting.

[marsh spear]

im sorry

im using a bad mouse

[paper depot]

those x's look closer to n's or u's

[icy knoll]

oh so just the top..?

[marsh spear]

yeah

[paper depot]

of course

you know how to work with fractions do you not

[marsh spear]

also an x is supposed to look smth like that

wait lemme try again

[paper depot]

an x is supposed to look like an x and is not supposed to be confusable with other letters

[marsh spear]

yeah but it shouldnt be confused for the multiplcation symbol as well

unless u use a dot which in that case wont matter then

[paper depot]

you should use a dot, and when a cross is required, clearly raise it above the baseline.

[paper depot]

this is better but not ideal imo

[marsh spear]

on paper its much more recognizable ig

[paper depot]

ive seen many more piss-poor x's with this shape than i've seen good ones.

[marsh spear]

well im using a mouse so its not good

[marsh spear]

but on paper its decent

[icy knoll]

so like that

[marsh spear]

yes

now u can choose to do a few things ig

1

acc

the only thing

is

bring each thing on the numererator over 3x

yeah

[icy knoll]

wait uh

i need to change 9x to 8x cause i didnt see the -1x

so it looks like

(-3x^2 + 8x + 3)/3x

[marsh spear]

oh yeah

ok

[icy knoll]

whats the next step?

[marsh spear]

since

theres a single denominator

[icy knoll]

uhm..

[marsh spear]

wait

thats wrong

[icy knoll]

could you explain why we do that

[marsh spear]

wait

wait

[icy knoll]

ok

[marsh spear]

give me a second

[paper depot]

no, this is correct.

[icy knoll]

wait

wouldnt i just do quotient rule now?

[paper depot]

to make it easier for ourselves to take the derivative

and no

[marsh spear]

oh dont do quotient rule

please

dont

thats just

like

so much work

[paper depot]

the constant 3 is positive on the numerator tho

[marsh spear]

every thing works out really well

[paper depot]

using the quotient rule now would be masochism

[marsh spear]

^^

[icy knoll]

oh ok cause im rn looking over the answer key they gave and they did quotient

[paper depot]

[icy knoll]

💀

[paper depot]

,w simplify (1+3x)(3-x)/(3x)

[twin meteor]

Results provided by WolframAlpha

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

Requested by Ann#0413

[marsh spear]

thats a terrible answer key

unless

u were supposed to practice that rule

and only that rule

[paper depot]

just to make sure we're on the same page simplification wise
this here is correct despite @marsh spear's self-correction

[icy knoll]

ok wait what does this mean? like why do we do this

[paper depot]

there are better examples to practice quot rule on

[icy knoll]

its just simplification?

[marsh spear]

yeah

because

[paper depot]

we do this to make it easier for ourselves to take the derivative.

[marsh spear]

everything happens to workout lol

[icy knoll]

oh ok

[marsh spear]

try to avoid doing the quotient rule in this scnerio

[paper depot]

taking the derivative at this point is a matter of applying just the power rule.

[icy knoll]

ok

im still kinda confused..

[icy knoll]

so i did this

and then i simply that further

[paper depot]

have you or have you not arrived at -x + 1/x + 8/3 with your simplification

[icy knoll]

i have

wait no

[marsh spear]

now just take the derivative

[icy knoll]

i didnt get -1x

i got 1x

without negative

[marsh spear]

yeah

u should come to that

[icy knoll]

ok yeah

[paper depot]

wiener

[marsh spear]

hi

[paper depot]

are you sure YOU aren't fucking up

because i think you are

[marsh spear]

potentially

[paper depot]

you had it right the first time round then corrected yourself into wrongness

WA is reliable for algebraic simplification

[marsh spear]

(3x+1)(x-3) = 3x^2 - 8x - 3

where am i wrong

[paper depot]

yeah but the second factor is supposed to be 3-x

[marsh spear]

Oh

shit

Oh ok

yeah

the first one is correct

myb

[icy knoll]

wait so we cant change the negatives to positives?

[icy knoll]

-3x^2 + 3 + 8x to 3x^2 -3 -8x?

[marsh spear]

no no no

i did it wrong

i multipled (3x+1) with x-3 not 3-x

[icy knoll]

oh

[marsh spear]

yeah sorry about that

so you should get that picture

[icy knoll]

wait uh

wouldnt the - x just be -1

wait

nvm

thats the wrong one

[paper depot]

the -x would not be -1

[icy knoll]

i meant 1/x

[paper depot]

but the -x would differentiate to -1.

[icy knoll]

nvm

i wrote it wrong

fixed what i did wrong

[marsh spear]

right

derivative of a constant is 0

derivative of -x is -1

and the derivative of 1/x is -1/x^2

[icy knoll]

ok wait

lemme try this part

oh wait

its just -2/x^2?

nvm

[paper depot]

no

[marsh spear]

yeah glad u found the issue there lo

[paper depot]

it's -1 - 1/x^2

$-1 - \frac{1}{x^2} \neq -\frac{2}{x^2}$

[twin meteor]

Ann

[icy knoll]

yeah i figured that out 😓

[marsh spear]

its alright

[paper depot]

if you wish to write this as one fraction for whatever reason you can do $-\frac{x^2+1}{x^2}$ but this is not really necessary

[twin meteor]

Ann

[icy knoll]

ok uhm..

wait how would i write that as a fraction..?

[marsh spear]

common denominator

[icy knoll]

is it possible to take the derivate of this answer in the same way we just did this problem?

cause this isnt product rule anymore

but if i change this to a fraction

i could possibly do quotient?

[paper depot]

that'd be the 2nd derivative of the original. it is not what you were asked for.

[paper depot]

no need for that NO NEED FOR THAT

[icy knoll]

oh wait so theres always a way around quotient??

[marsh spear]

well

that depends

if u can simplify like we did in here

then theres no need

but in some scnerios

quotient rule may be the only rule that can be applied

[paper depot]

no

[icy knoll]

ok

[icy knoll]

i tried that rn

but got the wrong answer again

i dont rly get where im going wrong

lemme send my work

[marsh spear]

r u tryna find the 2nd derivative

[icy knoll]

oh wait

nvm i got it

[marsh spear]

oh ok gud

[icy knoll]

yeah i just wanted to make sure i understood what you guys taught

[marsh spear]

are you in america?

[icy knoll]

yeah

[marsh spear]

oh

so im assuming ur in calc rn right

[icy knoll]

pretty much

[marsh spear]

should u have done like algebra 2 n stuff

[icy knoll]

yeah we did that previously

[marsh spear]

oh ok

yeah

[icy knoll]

im just not the brightest with understanding stuff till i actually do it

[marsh spear]

the problem with calclus is

[icy knoll]

and our prof only rly teaches theories

without practice problems

[marsh spear]

if you have a weak base in algebra then calculus will be a struggle

[icy knoll]

probably why im kinda struggling atm

[marsh spear]

yeah

dw tho

with more practice u will naturally start to get the hang of it

but if u have the choice to like just review some algebra or go over some things yk then u should

just so u can suceed with calc

[icy knoll]

yeah hopefully i will have time to

do you mind if you can help me understand another problem?

[marsh spear]

well its 1 am for me but i can try ig

[icy knoll]

o its 1 am here too

[marsh spear]

yeah

alr send the problem rq

[marsh spear]

oo

that

[icy knoll]

chain rule

[marsh spear]

is just chain rule

yeah

[icy knoll]

yeah

but i still keep messing up

[marsh spear]

derivative of ln is what

[icy knoll]

1/x

?*

[marsh spear]

right

derivative of cotx

[icy knoll]

uhmm

is it -secx?

wait no

[marsh spear]

-csc^2x

[icy knoll]

-secxtan?

oh.

ok wait

lemme write that down

ok

[marsh spear]

that

yeah

[icy knoll]

then what do i do about 10^x^5/3?

[marsh spear]

wait thats another exponent