help-17

[hidden kelp]

Is it the concept of a piecewise function that you don't get?

[vast shale]

correct

this is the first time ive ever heard of it

[hidden kelp]

Okay well, a piecewise function is just, different functions depending on the value of the input

[vast shale]

ah

[hidden kelp]

so in your case, if x < -1, then you will have 6x+5

[vast shale]

so like a value given to a function

is predetermined

[vast shale]

i see

that makes sense

[hidden kelp]

if you want to graph it, it's just going to be pieces (this is where the name comes from) of different functions

so since they ask you for f(7), find which interval 7 belongs to, and plug in the correct piece

[vast shale]

i see

[vast shale]

i dont see where f(7) is on the thing

it doesnt list it out for me

it just says find

[hidden kelp]

yeah I just noticed it too

[vast shale]

undefined?

[hidden kelp]

I suppose so? Since the function is not defined at x=7

[dreamy viper]

maybe the interval is supposed to be x>-1?

weird question

[vast shale]

it was undefined

[urban edge]

Yes, if a number is not in an interval then it is undefined

[vast shale]

anyways i got this new problem

from what i can see -1 is not listed

on the right side

[hidden kelp]

it is

[vast shale]

it is?

uhhh

[hidden kelp]

look carefully :)

[urban edge]

The function literally does not define what the value would be, so it would be undefined (in the case for f(7))

[hidden kelp]

Alright gotcha

[vast shale]

i still dont see anything

x<-3 can be anything

x=-3 isnt it

[hidden kelp]

What about the bottom one?

[vast shale]

what does the underlined < mean?

[hidden kelp]

smaller or equal to

no line means strictly greater/smaller

[vast shale]

that would still mean it would range from -2 to 2

right

it doesnt specifically identify that it is -1

only around that number

[hidden kelp]

-3 to 2*. But yeah. Is -1 in that interval?

[vast shale]

ytes

[hidden kelp]

then the function exists at x=-1 :D

[vast shale]

oh okay

random questionm sorta but

doesnt the -3<x mean that it cant be -3

since its only greater than -3

[hidden kelp]

it does mean that x can't be -3

stricly bigger

[vast shale]

i see

real quick could you give me a short explanation on teh difference between a underlined < and non underlined one

[hidden kelp]

$-4 \le x < 4$

wait I forgot the latex

give me a sec xD

[twin meteor]

imtyp0

[hidden kelp]

There that'll do

So this means that is in between -4 and 4, right?

[vast shale]

yes

[hidden kelp]

But the line on the left means that x can be equal to -4. However, since there is no line on the right, x cannot be equal to 4

[vast shale]

if the line was on the other side that mean it could be 4 too right

[hidden kelp]

yes

[vast shale]

ok col

i get it

now that i know that the bottom one is within the range

what would be my next step here?

[hidden kelp]

So, we agree that -1 is in the bottom one?

[vast shale]

yes

[hidden kelp]

then, f(-1) means replace x with -1

[vast shale]

-3<-1__<__2

[hidden kelp]

oh not in there sorry, in the function left of it.

[vast shale]

2(-1)

?

[hidden kelp]

yeah!

[vast shale]

so -2 would be my answer then i assume

[hidden kelp]

should be yes

[vast shale]

i see

thanks

i apprecaite it

[hidden kelp]

I can try and graph it for you if you'd like. Maybe that'll help you visualize it more?

[hidden kelp]

No problem :L

[vast shale]

i think its fine and for this i just do the bottom one right

so my answer would just be 4

?

[hidden kelp]

yeah

[vast shale]

i see

[hidden kelp]

since for the top one, x can't be 0

[vast shale]

and the crossed out = sign means it cant equal that

ok cool

[hidden kelp]

exactly

[vast shale]

lets say f(1) would that fit fo rthe top one?

since it can be anything

not 0

[hidden kelp]

correct

[vast shale]

thank you

i apprecaite the help

.close

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[flat wren]

Does anyone know the answer to this

[open spade]

!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

@flat wren Has your question been resolved?

[vocal sleet]

@flat wren Has your question been resolved?

[burnt nova]

Hi

@flat wren

its letter c

[flat wren]

Thank you 🙏

[burnt nova]

.close

write that

[flat wren]

Close

[burnt nova]

so the chat can close

with the .

.close

[flat wren]

.close

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[hearty delta]

Hi, I keep getting this problem wrong and I have no idea where I am messing up, can anyone take a look?

[hearty delta]

@flat wren ?

[flat wren]

I don't understand wich is the answer

I got d but was wrong

[hearty delta]

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[alpine notch]

Can someone pls explain where I’m going wrong

[flat whale]

the answer is supposed to be 0 i'm guessing?

[alpine notch]

nope lol

@flat whale

[flat whale]

oh you're asked to calculate the area between the curves

thought you were just calculating the integral

[alpine notch]

yeah

[flat whale]

,w int -1 to 0 (4x^3- 4x)

[twin meteor]

Results provided by WolframAlpha

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Requested by rie.mann#0

[flat whale]

,w int 0 to 1 (4x- 4x^3)

[twin meteor]

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[flat whale]

missing a square on the 2(-1)

[alpine notch]

ok so like this?

[flat whale]

right

[alpine notch]

k

tyy

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[hearty delta]

Does the order of intigration matter while setting the Jacobian?

[hearty delta]

With u first im getting 1/7 but with v first I get -1/7 so I was wondering if the oerder mattered while chaninging bounds

[proven knot]

<@&268886789983436800>

[open spade]

<@&268886789983436800>

ur not even pinging anyone lol

[proven knot]

I unfortunately dont quite remember but i believe that the jacobian can be oriented either in the positive or negative direction so the order does matter depending on what you want to do

The result of integration should be the same under 2 different changes of variables

iirc

[hearty delta]

Ah crap

ok

@proven knot so if im switching the variables for this funtion

For region r

it does matter what i write first, even after the substitution

also how would I know which is the right order?

[proven knot]

That unfortunately i dont remember

[hearty delta]

I switched it around really quickly and the answer did not seem to change

so maybe not?

the negitive/pos J is throwing me off however

[proven knot]

If you do the algebra right the coordinate change should.get you the same integration

iirc if you switch the order it may cancel out

[hearty delta]

Yeah, my only question is what is the right order according to the basic funtion

[proven knot]

The extra negative cancels elsewhere in the algebra

[hearty delta]

Hmm ok

so not safe to assume it will always work out but it just did in this case?

This was kinda my work, but for the last final int where would i know what the right order is

or if J should be post or neg

[proven knot]

My vague recollection is that it should work out if you do the algebra right but like i said im just trying to offer something from what i remember of calc 3

[hearty delta]

ah ok no worries

Ill just keep this up incase someone else knows

Thank you so much tho

[proven knot]

Yw hope you get an answer

[proven knot]

https://math.stackexchange.com/questions/1028000/can-you-switch-the-order-of-the-determinants-when-changing-variables-using-the-j

So this explains that the jacobian is the magnitude of the determinant

So it's always positive

You just take the absolute value if yoi got it negative

And then order of integration doesnt matter

[hearty delta]

Ah ok perfect

So regardless the Jacobin is always positive

And for the order just follow the normal rules and it works

[proven knot]

Thats what they said in there and it corroborates my hazy memory of these problems lol

[blissful sentinel]

Yeah the Jacobian is always positive

[blissful sentinel]

It doesn't matter, but one order of integration might be more convenient than another

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@hearty delta Has your question been resolved?

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[crisp rampart]

can someone help me with this

[crisp rampart]

A sample 15 appliances with standard deviation of .15v. construct a 95% confidence interval for the standard deviation and the variance.

[vocal sleet]

@crisp rampart Has your question been resolved?

[crisp rampart]

no

<@&286206848099549185>

<@&286206848099549185>

...

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[bleak fractal]

An upscale resort has built its circular swimming pool around a central area that contains a restaurant. The central area is a right triangle with legs of 60 feet, 120 feet, and approximately 103.92 feet. The vertices of the triangle are points on the circle. The hypotenuse of the triangle is the diameter of the circle. The center of the circle is a point on the hypotenuse (longest side) of the triangle. The building permit that the resort obtained requires that the resort state how much water the pool will hold so the city can manage the resort’s water rights effectively.

The resort owner wants to line the largest circular edge of the pool with 24K gold leaf. Explain how to determine how many linear feet of gold leaf you need for this task.

[bleak fractal]

is the question asking for the circumference of the whole circle or just half the circle?

i'm confused on what the largest circular edge means

[vocal sleet]

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[bleak fractal]

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[plush shoal]

Is this what its asking me to do?

[open spade]

Yeah looks right

[plush shoal]

Is it just like that one answer?

Cus theres an a and b

[open spade]

Well you already showed x-2 is a factor by polynomial division right?

[plush shoal]

Yes

[open spade]

And you factored it further in the work you posted

You should be done with the question assuming you also showed work for the division

[plush shoal]

So its like that is the final answer and the complete way right?

[open spade]

yep

[plush shoal]

Okay then, thank you

[open spade]

np :D

[plush shoal]

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[spark cloak]

You invested your summer earnings into an annuity from which you can draw expenses while you are at university. If you need to withdraw $1200 each month for 9 months of university, how much do you need to invest in an account, earning 6% per year, compounded monthly, in order to cover your expenses?

[spark cloak]

I got an anwser of 855.21

[paper depot]

that sounds way too low

how are you gonna make even your first withdrawal of 1200 if you have under a thousand bucks in there

[spark cloak]

see that why I am asking

I used the correct formula

[spark cloak]

ill show you my work

also

bi monthly in 6 years

[paper depot]

,rccw

[spark cloak]

how do I insert that

in the formula

[paper depot]

bimonthly??

[paper depot]

did you perhaps mean monthly?

and 6 years????

[spark cloak]

once ever 2 moths

different question

[paper depot]

ffs

[spark cloak]

lets focus on this one

[paper depot]

dont try to ask about two different questions at once

[spark cloak]

alright

[paper depot]

$PV = R \cdot \frac{1 - (1+i)^{-n}}{i}$ is your formula?

[twin meteor]

Ann

[paper depot]

looks a little sus if you ask me

[spark cloak]

ya

Its the one the teacher provided

[paper depot]

i am not 100% sure if this is the right formula

[spark cloak]

Ill show you an example

that he did

[clear stirrup]

isn't this the same question from yesterday

[paper depot]

i am losing interest rather quickly in this

[clear stirrup]

🤨

[spark cloak]

How much money would be needed in order to withdraw \large $1000 at the end of every year, for \large 4 years, from an account that earns \large 8% interest per year, compounded annually?

In this problem, we know R=1000, i=0.08, and n=4.

Substituting into the present value formula gives PV=\frac{1000\left [ 1-(1+0.08)^{-4}) \right ]}{0.08}.

This simplifies to give PV=\frac{1000(0.265)}{0.08}.

PV=$3,312.50

This means $3,312.50 would be needed in order to withdraw $1,000 annually for 4 years.

[twin meteor]

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

[paper depot]

is that really your teacher or is that chatGPT

it really looks like chatGPT

[spark cloak]

i swear to god its the teachers

ill even send the ss

[paper depot]

i have a feeling i'm going to spit venom if i continue

[spark cloak]

caught me lackin

[paper depot]

so im out

[clear stirrup]

It is the right formula

PV of an ordinary annuity

[spark cloak]

ok I was too lazy to do It yesterday and just took the pictures for our conversation

but im bugging out

how is the amount so low

i mean it can be right I guess

[paper depot]

R = 1200, i = 0.005 and n = 9 sounds like what it should be to me

9 compounding periods and interest 0.5% per compounding period (month)

[spark cloak]

0.005

?

oh an error

[paper depot]

0.5%/mo

[spark cloak]

mb

[clear stirrup]

this is not it

[spark cloak]

mm i plugged in the wrong things

[clear stirrup]

This is correct

[spark cloak]

i need to divide the rate by the amount of payments

like compounded weekly

monthly

and etc

?

[clear stirrup]

Bro did I not give you the formula with the numbers plugged in

💀

yesterday

[spark cloak]

I dont think

Im not 100% sure

[clear stirrup]

Yes divide by the number it is compounded in a year

[spark cloak]

oh you did but you plugged in a difference question

[spark cloak]

0.05/12 in this instance

im assuming

[clear stirrup]

Yes because it's compounded monthly

[spark cloak]

ok

[spark cloak]

Do i do 9/12

when I insert it

in the formula?

ya for sure

no nvm

[clear stirrup]

It depends on if you are using 'n' as periods or 'n' as number of times compounded

[spark cloak]

mmm

[clear stirrup]

If it's the latter, then you use nt

If the former, you just use periods

which is 9

[spark cloak]

so what should i do for this one

[clear stirrup]

one period = 1 month

For which

[spark cloak]

$1200\frac{\left(1-\left(1+0.005\right)^{-\frac{9}{12}}\right)}{0.005}=896.0804627$

[twin meteor]

deviousglxy

[clear stirrup]

It should just be 9

[spark cloak]

$1200\frac{\left(1-\left(1+0.005\right)^{-9}\right)}{0.005}=10534.8767$

[twin meteor]

deviousglxy

[spark cloak]

this is the one I copied down

it seems right

[clear stirrup]

Just by looking you know it's impossible for it to be 896

[spark cloak]

yep

4. If $25, 000 is invested into an account paying 4.5% per year, compounded bi-monthly, how much money can be withdrawn from the account every two months for the next 6 years?

this uses same formula

only difference

is trying to find

the r

also how is this compounded

[clear stirrup]

We are trying to find PMT

[spark cloak]

mmm so it is different

[clear stirrup]

Well you can use the same formula, just rearrange to solve for PMT

[spark cloak]

what is the better alternative

[clear stirrup]

Use a financial calculator/excel

[spark cloak]

I have to do it on paper

what does the financial calculator do

?

[clear stirrup]

It's just an easy way of solving time value questiosn

questions

You feed it the known values and it calculates the value you're looking for

[spark cloak]

ya I used one yesterday to check for anwsers

I dont get the every 2 months

for the next 6 years

so

1/6 multiplied by 6

which is just 1

[clear stirrup]

If it's every 2 months

How many times is that in a year

[spark cloak]

6

[clear stirrup]

That's what you use

[spark cloak]

for the next 6 years

where do I plug that

[clear stirrup]

You do it the same way as if it was compounded monthly

but using 6 instead of 12

[spark cloak]

so

6 multiplied by 6

[clear stirrup]

Yes

[spark cloak]

which is 36

I dont know why I tried to divide

so what is going to be the rate

I have 0.045

divide by the 36?

which would be

[clear stirrup]

No

36 is number of periods

[spark cloak]

then im going to have to use 6

[clear stirrup]

For rate you always divide it in the context of how many times it's compounded in a year

[spark cloak]

ok

0.045/6

[clear stirrup]

Mhm

[spark cloak]

mmm

$25000=x\left[\frac{1-\left(1+0.0075\right)^{-36}}{0.0075}\right]$

[twin meteor]

deviousglxy

[spark cloak]

I got a good anwser

of

794.99

[clear stirrup]

Yeah that seems correct

example of financial calculator

[spark cloak]

oh thats what you use

are there actual

calculator

that are meant for finance that arent online?

[clear stirrup]

Yeah

TI-83 or TI-84 also can do TVM calculations

in apps

[spark cloak]

are those also used for calc

or seperate calculators

[clear stirrup]

ti?

I mean yeah

Common calculator to use

even in college

[spark cloak]

def should get one

they will make my life much easier

thanks a lot

for all the help

have a great night

.close

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[tacit hill]

Okay, so I've been working on this problem for a couple days now and I've hit a roadblock and now I'm starting to get desperate. If you haven't seen the meme below, it's a false proof that π = 4. This is obviously wrong; however, folding in the corners actually does more closely approximate the area of the circle, so I was wondering if I could use this to get a new series definition for π.

The graph depicts each of the corners of each iteration of the "square" on a quarter circle. If I can find the coordinates for all the infinite coordinates for each iteration of the "square", and then find the area of the rectangle between that point and where it ends up when you fold it, I can simply subtract the sum of all those rectangles' areas from the unit square to get the area of the quarter circle (π/4). Using a bit of basic geometry, you can get a sort of recursive formula for the corners and the area removed when a corner is folded up (both are shown at the top of the notebook paper). However, this isn't quite good enough to immediately form a series so I've been chasing after a more general form.

I decided to organize the coordinates in a pascal-esque triangle where each corner produces 2 new corners: the one one the left keeps the x coordinate and the one on the right keeps the y coordinate. I've also denoted them as [n, k] where n represents the nth row and k represents the kth column (both indexed at 1 rather than 0). While I have been able to spot that half of the points are just reflections of the other half and I've written down a couple general forms for specific values of k, I still haven't found a general form for [n, k] or noticed any possible patterns. It is here that I ask for help if you have the time.

[vocal sleet]

@tacit hill Has your question been resolved?

[tacit hill]

<@&286206848099549185>

[tacit hill]

Just an acknowledgment of seeing the question would be nice if possible 👍

[plucky spear]

https://www.blogs.unicamp.br/zero/964/

i found this on the internet

i can't help you with the problem though just thought the meme was interesting

[frozen bobcat]

the problem is it doesn't actually converge to the circumference of the circle.

[molten willow]

^

[frozen bobcat]

there's always intervals where the rectangle is off the circle only scattered points that are on the circle. So the perimeter of the rectangles will always be strictly larger than the circumference of the circle with a measurable difference

[tacit hill]

I wasn’t asking about the perimeter? I was asking about the area

And I have tested it experimentally, it genuinely does approximate pi/4

[frozen bobcat]

XD I was just looking at the meme, I didn't scroll up to see the question lol

[tacit hill]

Bruh moment, I don’t blame ya though lol

[frozen bobcat]

the area should work if you can write that area as a series.

[tacit hill]

I need the general form for [n, k] to make a series for that though

So far I’ve only got a recursive form

[frozen bobcat]

that... is not my specialty, you'll have to get someone else's help for that.

[tacit hill]

Alright, back to my dwelling chamber

[frozen bobcat]

it wouldn't surprise me if this is a known series though. so some deep googling might be fruitful.

[tacit hill]

Don’t really know what to search for though

And it doesn’t look like anything I’ve seen on Wikipedia (granted, I’m famously very bad at googling things)

[vocal sleet]

@tacit hill Has your question been resolved?

[tacit hill]

<@&286206848099549185> I don’t mean to push y’all but I haven’t gotten too much help

[pallid forge]

found this video about it

https://www.youtube.com/watch?v=Rv0c7R8brjE

[tacit hill]

Again, I’m talking about the area under our curve

Not the perimeter

[lyric fossil]

please ping helpers only once

what you’re doing would converge to the area of a circle, yes

it’s essentially integration

[tacit hill]

Oh sorry sorry

[lyric fossil]

whether this is novel or not, i don’t know

but probably not

that’s not a diss

it’s just that these things have been looked at for centuries so it’s unlikely it’s new

[tacit hill]

I understand but I have no idea how I would begin to look for that

[lyric fossil]

(in particular this is the limit of a riemann sum on sqrt(1-x^2))

well what’s your goal with this?

is it just to prove to yourself that no one has done it before?

[tacit hill]

I just wanna know if there is a general form for a given corner

[lyric fossil]

then this is just a riemann sum

i can give you the general form if you give me a minute

i'll write (S_n) for the area after folding ((n-1))-times
[
S_n = \sum_{i=0}^n \frac{\sqrt{1-\frac{i^2}{n^2}}}{n}
]

let me double check this with a calculator, but i believe it should be it

ok there's a slight issue give me a sec

[tacit hill]

Aren’t riemann sums equally spaced along the x axis though?

[lyric fossil]

they don't need to be

but this one i made is

and does your folding not follow an equally-spaced folding pattern?

[twin meteor]

maximo

[tacit hill]

No they are not equally spaced

[lyric fossil]

then this is still a riemann sum but of the following form

[
S_n = \sum_{i=0}^n\sqrt{1-x_i^2}\Delta x_i
]

[twin meteor]

maximo

[lyric fossil]

where (\Delta x_i = x_{i+1} - x_i)

[twin meteor]

maximo

[lyric fossil]

there is an obvious off-by-one error here but i can't be bothered to find it

and note that this is only for the first quadrant, but you can just multiply this by 4

[tacit hill]

I mean this isn’t exactly the same as finding the coordinates for a given corner

[lyric fossil]

how are you deciding when to fold?

[tacit hill]

Like sure you get the same area in the end but like plenty of formulas exist for pi

[tacit hill]

So you start with a given corner, fold it in such that it makes contact with the normalized point on the circle (keeps same angle), and the two new coordinates keep either the x coordinate or the y coordinate of the original corner

[lyric fossil]

what is a "normalized point on the circle"

do you have a drawing of this or similar?

[tacit hill]

So if you have a line going from the origin to a corner, it’s where the line intersects the circle

[lyric fossil]

i see give me a sec to think about it

i think it may have a nice sequence to it

[tacit hill]

I have some specific values and a recursive formula (very top of notebook paper) here if they are of any value

[lyric fossil]

yeah ok nvm i don't see this having a nice explicit sequence

i mean it might but i don't think i could calculate it

[tacit hill]

Are there semi-similar problems that I can look to for inspiration?

[lyric fossil]

probably, but i dont know of any off the top of my head

the good thing here is that we just want to find a sequence of sequences for the x values

the heights will be always solvable via sqrt(1-x^2) from there

[tacit hill]

Right and also because it’s reflected across y=x

[lyric fossil]

you can use that to your advantage too yeah

what was your recursive formula?

[tacit hill]

(x, y) -> (x/sqrt(x^2+y^2), y) and (x, y/sqrt(x^2+y^2))

[lyric fossil]

it also looks like rather than x and y, we just want to find the sequence for the angles

which should make this a whole lot easier

[tacit hill]

Ah that’s true that could help

(Don’t know how I didn’t think of that 💀)

[lyric fossil]

im struggling here since it's a 2d sequence but once you start with angles it should be more straightforward

[tacit hill]

Oh I mean the angles don’t look pretty though

[lyric fossil]

yeah but it should be simpler to make explicit

something like arctan(something with n and m here)

oh my

there's a very nice sequence going on here

[tacit hill]

Hmm, I’m not sure if I see anything

[lyric fossil]

let me work on it for a little longer and i'll share

ok so there is a very nice subsequence going on here

the furthermost line will have the form y = x/sqrt(n)

[tacit hill]

yesyes, that's the k=1 case for [n, k]

[tacit hill]

see the middle section: I've calculated them up until k = 4, but I haven't been able to string them together

also, more often than not the denominators are starting to look like a sum of (n-k)^k but 1. I don't know how to prove that and 2. it probably isn't true

[lyric fossil]

i am inclined to agree with you. it looks like some sort of sum going on, but i can't put my finger on it
it does look like a very nice sequence though

like not completely dubious

[tacit hill]

yeah, feels just out of reach though unfortunately

I'm gonna look back at this and be so mad at myself lol

from the future: definitely am lmao, this was so much easier than I was making it out to be

it's almost 5 AM here, Imma probably call it a day now. let me know if you find anything!

[lyric fossil]

there is this mse post about it https://math.stackexchange.com/questions/4485663/calculating-pi-by-subtracting-squares-from-a-circumsquare

the comments shared a recurrence relation

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[white coral]

Can anyone possibly confirm my answer, just telling me if it's right or wrong

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[white coral]

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[digital shell]

i'm annoyed rn

i don't understand this math

the lim(x-9) with x at -3 is limx-lim9

but then they have written -lim-1 here??? isn't this incorrect

[whole oasis]

it's limit of -1

(-1 is part of the limit, it's inside it)

[paper depot]

i think the issue is they typoed an extra minus sign in

which is a mistake on their part

[digital shell]

is the extra minus on the -1

[whole oasis]

ah ye

[digital shell]

shouldn't it b positive

doesn't it change our answer significantly then

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[dim quail]

$\int_0 ^\pi \frac{x \sin x}{1 + \cos ^2 x}$

[twin meteor]

Bettim

[dim quail]

How to do this

[viral copper]

what have you tried

[hasty pulsar]

what have you

.QDSNBHUILFOQEJNKFDV

[astral shadow]

The same idea as last question should take you somewhere.

[paper depot]

missing dx btw

[viral copper]

This looks like Kings rule

[scenic ravine]

use I think you asked a very simlar question just now, use the same idea

[dim quail]

Nothing, u sub did't work

Oh okay

The denominator changes

I'll try wait

[viral copper]

cos(π - x) is -cos(x)

So it actually doesn't

[vast shale]

Are we sure this has an elementary antiderivative?

Or a real valued one

[viral copper]

We don't need the anti derivative

Of this specific function atleast

Kings rule and ||adding the integral to itself|| simplifies our job massively, I've spoilered it because I want OP to figure it out.

[astral shadow]

We just had the op do these exact steps 5 mins ago.

[tacit hill]

||can't you just use IBP?||

[scenic ravine]

I don't think so, and even if you can, it's unnecessary

[tacit hill]

I mean x is an easy derivative, and sin(x)/(1+cos^2(x)) is also a pretty straightforward integral

[dim quail]

How do I proceed

[tacit hill]

simplifies pretty nice too

[viral copper]

ah

[hasty pulsar]

it doesnt have an elementary antidervative

i thought ibp first too

[viral copper]

Wait where did the sine go

[astral shadow]

sin(x)/(1+cos^2(x)) is doable. But not afterwards.

[viral copper]

After $2I = \int_0^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} \dd x$

[twin meteor]

NEON

[dim quail]

Good question lol wait I'll bring sine back

[scenic ravine]

yeah, that makes it an easy u-sub from here

[dim quail]

What will my u be?

[viral copper]

think about it

what should your u be

[dim quail]

If I take the dr, i get sin2x

So it can't be thst

Oh wait

Cosx

[astral shadow]

there you go.

[dim quail]

I get tan inverse?

[viral copper]

indeed

[tacit hill]

ah wait arctan(cos(x)) ≠ cos(arctan(x)) 💀

[dim quail]

Pi by 2

Is that what I should get?

No sait

Got it

Pipi\4

[viral copper]

,w intergate (sin x)/(1 + cos^2 x) from 0 to pi

[twin meteor]

Results provided by WolframAlpha

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

Requested by neonperseus

[viral copper]

pi^2/4?

[dim quail]

Yes

I get pi square by 4

[viral copper]

right that seems good

[scenic ravine]

yup, that's correct

[dim quail]

Thanks i go for dinner byee

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[fast quarry]

i still cant seem to do this part of the question

im struggling with the partial differentiation of arctan

[vocal sleet]

@fast quarry Has your question been resolved?

[proven garden]

calculate d/dx tan^-1(x)

y = tan^-1(x) then tany = x and do implicit differentiation

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[fast quarry]

.reopen

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✅

[fast quarry]

sorry just seen

[fast quarry]

but im doing partial differentiation so the tan y on the left just goes to 0 as you treat it just like a number or something right?

i dont put a dy/dx in there as its not treated as a variable

[proven garden]

y is a function of x

d/dx tany dont go to 0

[fast quarry]

oh

i dont really understand this part to be honest

i thought it goes to 0

[proven garden]

$\frac{d}{dx}tan(y) =sec^2(y)\frac{dy}{dx}$

[twin meteor]

WhereWolf

[proven garden]

$sec^2(y)\frac{dy}{dx}=1$ and solve for dy/dx

[twin meteor]

WhereWolf

[fast quarry]

yes i understand that, but when doing partial derivatives, with respect to x, doesnt tan(y) go to 0?

[proven garden]

when x and y are independet we do partial derivative

[fast quarry]

ok so dy/dx = 1/sec^2(y)

i dont think i gert the correct answer

[proven garden]

from tan(y) = x we get sec(y) = sqrt(1+x^2)

so d/dx tan^-1(x) = 1/1+x^2

with that info do your partial derivatives

[fast quarry]

okay

wait do u mind if i restart from this part, im getting very lost, sorry

im here now, definately something wrong right

hello

@proven garden

sorry to ping

[proven garden]

it's fine

just do the partial derivative?

what's wrong

[fast quarry]

so i need to put the derivative of arctan(yx^-1) into my answer for dh/dx.
but in my answer i get dy/dx = 0? is that the correct answer?

is that correct then?

[proven garden]

no that's not how you do that

that's what I did to find d/dx tan^-1(x)

that's not applicable here cause y is not a function of x

do the partial derivative normally

[fast quarry]

is the pink part wrong then?

which part is messed up

[proven garden]

dont set y = tan^-1yx^-1

why are you doing that

[fast quarry]

i thought that is how to differentiate arctan

[proven garden]

h = xtan^-1(yx^-1) and find h_x

I told you d/dx tan^-1(x) = 1/(1+x^2)

treat y as a constant and find d/dx h

[fast quarry]

is this ok now

[proven garden]

I think you differentiated tan^-1(yx^-1) wrong

[fast quarry]

oh wait i see

[proven garden]

btw d/dx tan^-1(x) = 1/(1+x^2) not x/(1+x^2)

[fast quarry]

that is u prime just place holder if the x is actually a function of x not just x

right?

meaning derivative of u

u'

better now?

this is where i got that formula from

[proven garden]

bruh

that's using chain rule

[modest aurora]

lmao

[proven garden]

when u = sqrt(x)

[fast quarry]

in my case i replaced it with u = yx^-1

and then continued to do the partial differentiation

yeah u have to use chain rule after the product rule right? sorry im trying this is kinda hard for me icl

[proven garden]

oh that's fine

[fast quarry]

i thin im correct now

[proven garden]

still wrong

[fast quarry]

damn

[proven garden]

you differentiated yx^-1 wrong

and you forgot a

x

[fast quarry]

let me try again lol

surely this is correct now, pls circle the parts that are wrong if there are

i really cant spot

[proven garden]

that seem's fine

[fast quarry]

okok thank u sm

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[pearl dragon]

i need help with finding the surface area of a cylinder. i know how to find the area of a circle but not the virtical part.

[daring willow]

like do u want the formula for it or do u need help proving the formula?

[pearl dragon]

i need the formula

[daring willow]

oh o

ok

2piradiusheight + 2pi*r^2

[pearl dragon]

like in this idk how to get the part of the rectangle

[pearl dragon]

ty

[daring willow]

np

[pearl dragon]

how could i find the height?

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[marsh coyote]

6.5.1 I'm not too sure what the hint is trying to give me:

Corollary 6.5.1 tells me that (1/n^(1/k)) converges

Theorem 6.1.19 gives me the standard limit laws

How would showing that (n^q) does not converge help me

Ah okay nvm I misread the question oops

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[half sail]

I need help solving 47.4 divided by 0.003 on paper

[rugged vortex]

if you need to , turn them each into fractions

then use basic fraction division tricks

[half sail]

when i try to do it i get 158 as my answer and itss supposedly 15,800

i need to solve it just using division not fractions

[rugged vortex]

each decimal can be represented as a fraction; its a way to eliminate decimal points and you'd be dealing with intergers for the most part

[half sail]

I am really confused as well because i thought there was supposed to be a decimal point

[half sail]

im not up to that stuff yet lol

[rugged vortex]

youre telling me you dont know how to represent 0.003 as a fraction?

[half sail]

idk

1/3?

[rugged vortex]

no

[copper solstice]

3/1000

[rugged vortex]

show me your work.

[copper solstice]

Can i help ?

I wanna see this problem

[rugged vortex]

as long as you dont majorly interfere, sure

[copper solstice]

Okey

[half sail]

its easy im just dumb

[copper solstice]

That's 0.003 divided by 158 your problem is the other way around

[rugged vortex]

youre neglecting the decimal point and youre treating the divisor as 3 instead of 0.003

[half sail]

no 158 is the answer i get

[half sail]

i thought i had to move the decimal to the right 3 times or something

[copper solstice]

The problem is 158 divided by 0.003 right ?

[half sail]

and 47.4 move it one time

should i show what my book is telling me

[copper solstice]

Sorry 47.4 divided by 0.003

[half sail]

yea

thats it

[rugged vortex]

i mean 0.003 = 3 * 1/1000. If youre gonna divide 47.4 by 0.003 its gonna be the same as 47.4/(3 * 1/1000) = (47.4/3) * 1000

[half sail]

how do i show it on the paper though

it wants me to solve the problem in the way i was doing

[rugged vortex]

you could.

[half sail]

this is what im going off of

[rugged vortex]

yeah and this is why you gotta multiply your quotient by 1000 at the end

what youre doing is 474/3

you neglected the decimal point

[rugged vortex]

I broke it down for you

so essentially the real work here is for 47.4/3

you multiply that quotient by 1000

since 1 divided by 1/1000 is the same as multiplying by 1000

[half sail]

im so confused by all this

158 x 1000 is 158,000

the answer is 15800

why is 0.003 3/1000?

[copper solstice]

You add 0.003 one thousand times to itself to get 3

So 0.003 can be written as 3/1000

Hence the 1000 as the denominator

And 3 as the numerator

[half sail]

i dont think im supposed to solve it like that i never learned this

[copper solstice]

Where are you from ?

[half sail]

all the other questions just required me to move decimal points and just divide normally pretty much

[half sail]

usa

[copper solstice]

Well this is the correct answer

Wait a minute

[half sail]

what

[copper solstice]

I'm gonna do what you said

Move the decimal points

[half sail]

oh ok thank you

[half sail]

im trying to solve it like this though, nothing about multiplying by 1000

[copper solstice]

If you myltiply by 1000 you remove the decimals

There is no other way you can remove the decimals exept by multiplying by 1000

[half sail]

i thought i was supposed to move the decimal in 0.003 to the right 3 times, 47.4 i would move the decimal one time so thats 4 decimal places. I divide 474 by 3 and then move 4 places to the left and put a decimal

[half sail]

how do i know when to multiply by 1000

on questions like this

[copper solstice]

Tbh i just know when i can't exactly explain it,

Sorry but i just know how to do math but not teach it

[half sail]

i think i did it finally

its ok i appreciate you taking the time to help me

i just had to add zeros in order to move the decimal three places

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[hearty delta]

Hi, so for this question how do we pick the bounds?

[hearty delta]

like why is it 0 and 2 and then 0 and y

[sharp lynx]

the bounds from 0 to y would be the standard way of computing P(X < y) for a fixed y, then the outer integral considers all possible values of Y

[hearty delta]

so if it was y<x would it be 0 and 9 and then 0 and x?

[sharp lynx]

probably, 1s

[hearty delta]

hm?

[sharp lynx]

actually no it wouldn't

[hearty delta]

oh ok

[sharp lynx]

perhaps it would help to draw the joint support of X and Y and then look at the particular region X>Y

the geometry is slightly more complicated

[hearty delta]

yeah how would you graph it

is it just a straigt line at y=x

straight*

[sharp lynx]

the joint support is a rectangle

[hearty delta]

ok give me a second

[sharp lynx]

the line y=x would be included though yes

[hearty delta]

ohhh

ok so its a rectangle bounded at x=0,9 and y=0,2

and with y=x going through it

and the area is the funtion of F

[sharp lynx]

the region Y>X is above that line and is just a triangle

the region X>Y is below that line and is a trapezoid

[hearty delta]

hmm?

wont it be another triangle

[sharp lynx]

nope

[hearty delta]

oh it doesnt go directly through the center

so the geomatry would be more difficult but i see how it would look graphically

tysm

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[vast shale]

can someone expalin to me why the final answer isnt 36^11/10

i got everything the same except that its 36^11/10 because you multiply the 6s as well

[open spade]

What makes you think it's 36?

[vast shale]

add the exponents and multiply the 6s

[mild flower]

if you had like $3^2 \cdot 3^2$ do you think that would result in $9^4$?

[twin meteor]

Hayley

[open spade]

You don't multiply the 6's

[vast shale]

oh

no i wouldnt

looks similar to when you simplify variables with exponents x^2 * x^2 = x^4

[mild flower]

yes, it's exactly the same

[vast shale]

so as a general rule if it has an exponent and its the same, dont multiply those 2 together

right?

[mild flower]

correct, $x^a\cdot x^b = x^{a+b}$

[twin meteor]

Hayley

[vast shale]

ok thank you!

[mild flower]

,tex .exp rules

[twin meteor]

Hayley

[vast shale]

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[vast shale]

I need help with this question. I feel like its poorly written

[little fossil]

why do you think that?

you need to transform these infos in equations

[vast shale]

ehh nvm

yeah i just need help

[little fossil]

ok

[paper depot]

!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

[vast shale]

1

[vast shale]

ok

[paper depot]

before you do that, you need to clearly define your variables

here it will be most helpful to make variables to represent the cost per minute and cost per mile.

[vast shale]

c for cost per minute

m for cost per mile

[paper depot]

as you say.

ok, so:

Ryan's first taxi ride was 3.0 miles, took 7 minutes, and cost $8.50.

can you write this as an equation?

(don't forget the $2 base rate)

[vast shale]

so it would be c+r+2=8.50?

[paper depot]

no, on two counts

one, you don't have any variable named r

[vast shale]

oh

2+3m+7c=8.50

[paper depot]

you still don't have any variable named r

[vast shale]

my apologies

[paper depot]

and now you've got your own variables backward.

[vast shale]

done

[paper depot]

would you like to rename your variables at this point so as to cause less confusion afterward?

or should we stick with these names

@vast shale ?

[vast shale]

we should stick with those names

[paper depot]

ok, as you say.

so the equation for the first taxi ride is 2 + 7c + 3m = 8.50
the second taxi ride

was 7.0 miles, took 14 minutes, and cost $16.00.

[vast shale]

7m+14c+2=16

[paper depot]

aight

from these two equations you can find both rates

[vast shale]

im assuming we solver for m and c

and then we use it to solve for the third equation now

yup

will do now

got my answer

its B

thanks

.close

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[vast shale]

studying for math placement test and i wont be allowed a calculator, dont know how id solve 7 root 128 without a calculator

where would i go from here

[solid sage]

Well it’s kind of intuitive

Just 2, not really a method but just identifying that 128 is a number that is frequent

[vast shale]

is there no method i can use to find the answer

i wont have time to experiment on the placement test 😂

nvm lol i found a yt vid

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[elfin moon]

So here in the last line....pi/4 +pi = 3pi/4
Because x,y>0 but x+y/(1-x.y) <0

But my doubt is 3 pi/4 is not in range?

[obtuse sierra]

answer is wrong

site is toppr?

[elfin moon]

Yes

[timid shell]

Never trust toppr

[elfin moon]

U indian

[obtuse sierra]

yes

[elfin moon]

Most answers given by them

Answer should be pi/4 ?

[obtuse sierra]

-pi/4

[elfin moon]

See

[elfin moon]

Show me how? Why

[obtuse sierra]

um tan^-1(-1) = -pi/4

isnt it

[elfin moon]

I know this

[timid shell]

What if x and y are <1 then your condition is false

[elfin moon]

But here x,y so we need to check as we discussed earlier

[obtuse sierra]

but no condition is given

about x and y

[timid shell]

He made the condition

[obtuse sierra]

just positive

[obtuse sierra]

oh

[timid shell]

From x,y>0

[obtuse sierra]

dont do that

if the condition is specifically given, then apply that

cuz most of the time xy is given less than 1

[timid shell]

.close

Kar bro

[elfin moon]

Ni kar rha yar

[elfin moon]

Is this generally true?

[timid shell]

Konsi condition ki baat kar rha hai likhke Dede

[obtuse sierra]

the product xy

[timid shell]

<1?

[obtuse sierra]

yes ig

[elfin moon]

X,y<1 bola h tumne

[timid shell]

Hn ho skta hai kyunki ques mein bas x>y>0 diya hai

Toh ek ko 1/3 dusre ko 1/2 maanle

[elfin moon]

Ye question me x,y >0 hai and x+y/(1+x,y) <0 hone par pi + karne par

Pi+pi/4 =5pi/4
This is wrong because (-pi/4) hoga so 3pi/4

[timid shell]

pi+pi/4=5pi/4???

Pi-pi/4=3pi/4

[elfin moon]

Ohh haaan edited

5 pi/4 range me ni h

Pi- pi/4 = 3 pi/4 hoga toh isme pi/4 ni ho sakta?

@timid shell @obtuse sierra

[vocal sleet]

@elfin moon Has your question been resolved?

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[narrow olive]

Find the first and second derivative of:

[obtuse sierra]

for first derivative apply chain rule

[narrow olive]

for the chain rule, i'm not sure what to put as the separate functions if that makes sense

[paper depot]

you've got a composition of three functions here

[obtuse sierra]

yes

[paper depot]

and they're all exp

[narrow olive]

oh

how would i do the chain rule on three functions

[obtuse sierra]

the first function is the question one

the e^e^e^x

2nd function is e^e^x

3rd is e^x

[wary mantle]

The e-pyramid

[obtuse sierra]

differentiate first one

then 2nd

then 3rd

[narrow olive]

ok

ohh i see

thank you 🙏

.close

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[quartz mauve]

is |x-1| the same as saying "x-1 or x+1"?

[hard atlas]

no

[scenic ravine]

no. x-1 or 1-x

[lethal mauve]

It doesn't mean "or", it's a value that depends on x

[quartz mauve]

i see

so for |x-1|=2, for example, how would i solve for x

do i check for both -(x-1) and x-1