help-28

[charred raft]

you get

[swift bridge]

N? or like n -1?

[glossy valve]

Goëtia

[swift bridge]

what does this "N" represent? Is it the natural number such that for all n >= N we have s_n +1 /s_n - L < e?

[charred raft]

Yes

[torn jolt]

well i let u with this idea first

[swift bridge]

how is it that you got (L - e)^{n-N}?

im telescopign rn but i only get L -e

[charred raft]

try it

s_n/s_n-1 <= L-e

s_n-1/ s_n-2 <= L-e

....

s_N+1/s_N <= L-e

s_n <= (L-e)^(n-N) s_N

[swift bridge]

Wait bruh my internet just crashed but I had something like s_n+1 < e * s_n/s_N

[charred raft]

multiply the inequalities i gave ya

[swift bridge]

Yeah I see I think i multiplied wrong stuff

[swift bridge]

shouldnt it be to the (n-N+1) power?

[charred raft]

proof?

[swift bridge]

we start the numerator at n+1

then end at N+1

so then there are n-N + 1 terms

[charred raft]

no

[swift bridge]

i think?

[charred raft]

forget n+1

we start at s_n/s_n-1

[swift bridge]

ah i see okay, then ye

and we and at s_N+1/s_N?

[charred raft]

yes

[swift bridge]

then we presumibaly take everything to the 1/n power?

[charred raft]

yes

[swift bridge]

do i also assume natural log preseves inequality?

[charred raft]

u tell me

[swift bridge]

because we haventn relaly proved that

[charred raft]

prove it

[swift bridge]

i think natural log is increasing function

but then that uses derivatives and stuff

[swift bridge]

wait this doesnt seem to use lim sup/lim inf tho

like here is what am getting

[charred raft]

write

[swift bridge]

$(\frac{s_N}{c})^{\frac{1}{n}} (L - \epsilon) \leq (s_n)^{\frac{1}{n}} \leq (\frac{s_N}{c})^{\frac{1}{n}} (L + \epsilon)$ where c is a constant

[glossy valve]

LXDL

[charred raft]

huh?

whats c?

aight

u have lim sup X <= L + e

[swift bridge]

i simpliefed this: $(s_N)^{\frac{1}{n}}(L- \epsilon)^{1 - N / n}$

[glossy valve]

LXDL

[charred raft]

and lim inf X >= L-e

oke oke

[swift bridge]

yes

[charred raft]

use that

[swift bridge]

wait sry when you say lim sup X, shouldnt it be $\sup {s_{k+1}/s_k : k \geq N }$

[glossy valve]

LXDL

[swift bridge]

or am i missing smthing ehre

[charred raft]

no

lim sup s_n ^(1/n)

[swift bridge]

wait wha.. so what is X then?

[charred raft]

X = s_n ^(1/n)

[swift bridge]

ah okay hmm, but here how did we reach this conclusion? it seems we just got rid of the s_N/c term

or like the s_N term in general

[charred raft]

what is u talkin about?

[swift bridge]

im confused on how we get lim sup s_n^(1/n) < L + e

[charred raft]

$\frac{\ln(s_N)}{n} \rightarrow 0$

[glossy valve]

Goëtia

[swift bridge]

yes

[charred raft]

thus $(s_N)^{1/n} \rightarrow 1$

[glossy valve]

Goëtia

[swift bridge]

oh ok

okay so $\sup { {s_j}^{\frac{1}{j}} : j \geq n } \leq L + \epsilon$

[glossy valve]

LXDL

[swift bridge]

and we have $n \geq N$

[glossy valve]

LXDL

[swift bridge]

we need to show that the sup of that is less than the sup of somethign else tho

[charred raft]

bro...

why is u complicating stuff

[swift bridge]

im confsued as well

idk bro im trying but im just getting confused

im not sure what we are trying to do here

like did we just show that lim sup = L

or

and now we are done?

wait

oh wait

is that what we just did

we show that lim sup = L

and by same reasonign lim inf = L

so we are done?

[swift bridge]

but this part is correct right

[swift bridge]

not sure why we need to use this tho...

like im still trying to think how we just got L-e < s_n ^ (1/n)

[charred raft]

bcz u asked how did the s_N/c disappeared

u asked the obvious

[swift bridge]

but dont we need to show using epsilon delta?

[charred raft]

tf?

[swift bridge]

because i dont think i can just say "since limit of any x> 0 x^{1/n} -> 1"

[charred raft]

we are taking the limit both sides

[swift bridge]

so you are saying we have $L - \epsilon \leq\lim s_n^{1/n} \leq L + \epsilon$?

[glossy valve]

LXDL

[charred raft]

yes

thats the result

[swift bridge]

so then we dont use lim sup or lim inf then right

it seems we just directly show

[swift bridge]

yeah im confused from the second to the third step

[charred raft]

if lim inf X = lim sup X = L => lim X = L?

[swift bridge]

yea i get tat

[swift bridge]

but not sure how lim inf s_n ^(1/n) >= L - e

[charred raft]

$inf (s_n)^{1/n} \geq (s_N)^{1/n} (L- \varepsilon)^{1-N/n}$

[swift bridge]

yes i agree with that

[charred raft]

you take limit both sides as n ->infty

[glossy valve]

Goëtia

[charred raft]

can continue?

[swift bridge]

uhh, yes i think this works

but it is (inf s_n)^(1/n)

not inf( (s_n)^(1/n) )

[charred raft]

no

its s_n^(1/n)

then i apply inf

[swift bridge]

oh okay yeah sry

yes continue

[charred raft]

u continue lol i aint doing ur hmwk

[swift bridge]

no i mean L - e cant jsut result from that right

[charred raft]

u figure it, u got the tools now

[swift bridge]

ok thanks

[swift bridge]

also this looks similar to gpt background but idk

[charred raft]

if gpt can do that, then go ask it

[swift bridge]

ok sry. i tried but gpt doesnt really explain

[charred raft]

good luck

i use gpt, to directly give me the rendered latex , without being correct in writing the latex code

[swift bridge]

i see alr

does L - e must alwayus be greater than 0?

[full forum]

@swift bridge Has your question been resolved?

[charred raft]

obv

e -> 0

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[chilly kindle]

Number 2

[chilly kindle]

,rccw

[chilly kindle]

i’m not sure what i’m doing

it asks to indicate the portion of the unit circle that satisfies the inequality, sketch a rough graph, then write in interval notation

[vague moss]

!show

[full forum]

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

[chilly kindle]

I don’t know where to start

other than me knowing where cos is -rt2/2

would the interval notation be like

<@&286206848099549185>

[leaden ermine]

looks correct

[chilly kindle]

okay tysmmmm

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[leaden ermine]

one thing, make the ] a ) at 2pi

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[past badge]

Hello,

I was wondering if someone could tell me the steps to answering this problem? I am having a bit of trouble comprehending how to solve it...

[past badge]

Any help is most appreciated

[keen bloom]

Try construct a line than connects the 2 points where the lines touch the circumfurence to create a triangle

[keen bloom]

Then find the arc

Then the angle

Then the other angle and finally x

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[past badge]

Thank you :)

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[frank plover]

I've been banging my head trying to prove this for a couple of hours now and Im pretty sure its something silly that Im missing

[frank plover]

$(A - B) \cup (B - A) = (A \cup B) - (A \cap B)$

[glossy valve]

Creeperarcade 2.0

[frank plover]

I applied the set difference identity and simplified it to the following

$(A - B) \cup (B - A) => (A \cap B^c) \cup (B \cap A^c)$

[glossy valve]

Creeperarcade 2.0

[frank plover]

and I cant use a distributive law because they dont both have the same letter in them

[charred raft]

prove double inclusion

X = Y <=> X subset of Y AND Y subset of X

[frank plover]

I'm not allowed to use that

[charred raft]

why not

[frank plover]

it has to be done via set algebra

teachers instructions

[charred raft]

ok

[frank plover]

I got a zero on my last assignment for using set algebra when we were practicing double inclusion

ik stupid but thats why its hurting my head

[rapid rain]

if you want to do it a similar way then

why not write "- (A cap B)" as "cap (A cap B)^c"

so start from the RHS and get to the LHS

[frank plover]

$- (A \cap B) = \cap (A \cap B)^c$

[glossy valve]

Creeperarcade 2.0

[frank plover]

this?

i didnt know that was a thing you could just do?

[rapid rain]

I mean coupled with the rest

$(A \cup B) - (A \cap B) = (A \cup B) \cap (A \cap B)^c$

[glossy valve]

rafilou is not not born in 2003

[frank plover]

so my understanding is that youre saying to do the RHS

and solve for the LHS

[rapid rain]

try to arrive at the LHS yes

[frank plover]

it took me like 5 minutes to process that sentence

let me try that rq

thats actually such a smart way to reverse engineer the question

thank you

so much

I got it

I'm got so zoned in on the LHS that I forgot about the RHS

[rapid rain]

that was fast

[frank plover]

I did part of it in my head

unless I made a illogical leap

I think its correct

[rapid rain]

I mean

[charred raft]

it doesnt matter, u can still apply it

[rapid rain]

after getting $(A \cup B) - (A \cap B) = (A \cup B) \cap (A^c \cup B^c)$

[glossy valve]

rafilou is not not born in 2003

[frank plover]

how so?

i thought they both had to be the exact same

either $A^c$ or $A$

[glossy valve]

Creeperarcade 2.0

[charred raft]

B^c U B = E

A U B intersect E = AUB

[frank plover]

hold on

so that is just the application of the distributive law for union?

i thought they had to match?

[charred raft]

basically ye

[frank plover]

so how is that law applicable

[charred raft]

let X = A intersect B^c

do X U (B intersect A^c)

X U B intersect X U A^c

expand X

[frank plover]

$(X \cup B) \cup (X \cup A^c)$

[glossy valve]

Creeperarcade 2.0

[frank plover]

oh I see

wait

omfg

I didnt entirely understand what you did I get it now

it took the variable analogy

I thought you mean reversing the distributive law

thats what I was looking at

so going from the RHS of the law and turning it into the LHS

[charred raft]

no i just answered ur first question, bout why u couldnt use the law

[frank plover]

yea i understand now

thank you for explaining that concept

I didnt know that was a proper way of applying that law

[charred raft]

but my idea to do the proof, is just endup with a similar expression for both the RHS & LHS

if they are equal to a similar expression both of them then they must be equal

a=c && b=c => a=b

[frank plover]

yea i understand that

the way we are required to do it

is start with one side and turn it into the other

discrete math is such a new concept to me

[charred raft]

after u apply the distributive law

u'll get this instantly

[frank plover]

this was super helpful I can finally get my assignment done

i knew it was something tiny

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[frank plover]

.close

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[spring bay]

how would I write this graph in four ways sin, -sin, cos, and -cos?

[hot lily]

Lets start with sin

[honest hill]

is there any extra info given about the point?

[spring bay]

no

[hot lily]

Do you know about $y = a\sin(b(x - h)) + k$

[glossy valve]

King Leo

[hot lily]

What more info is needed?

[spring bay]

yea

[hot lily]

So whats the amplitude of this function

[honest hill]

that its a minimum?

[hot lily]

Oh i see

[spring bay]

2?

[hot lily]

So whats a

[spring bay]

2

[honest hill]

if its not a minimum, the amplitude is greater than 2

[hot lily]

Fair point

But i think its safe to assume thats a minimum point

$y = 2\sin(b(x - h)) + k$

[glossy valve]

King Leo

[hot lily]

@spring bay do you know what b does

[spring bay]

is that like the dilation?

no, how would I find that

[hot lily]

Close, its like a horizontal stretch. You can find it using pb = 2pi, where p is the period of the graph

[spring bay]

wait how did u get the period?

[hot lily]

You can look at the graph

It requires a little bit of extra thinking

Do you need help?

[spring bay]

yea

is it because one full revolution on a unit circle is 2pi so for sin one full revolution it’s also 2pi?

[hot lily]

Hint: the horizontal distance between an extreme (minimum or maximum) and the closest midline is 1/4 of the period

In other words:

Each line segment that i drew on the x-axis spans about 1/4 of the period

[spring bay]

ohh alr

wait can we also just assume that the graph is in the standard shape of sin without compression or stretching in the x-direction?

[hot lily]

No you cant. Thats the point of finding a, b, h, and k

[hot lily]

But for the purposes of the drawing, it works on any sin or cos function
#help-28 message

[spring bay]

alr

so if the period is 2pi, then b would be 1?

[hot lily]

That's why one revolution of the default y = sin(x) has a period of 2pi

[spring bay]

how would I find the horizontal shift

oh wait is it 3pi

.close

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[runic heath]

why is this wrong?

[runic heath]

to find the function i took the points (1,4) and (3,0)

slope -2

[full forum]

@runic heath Has your question been resolved?

[gritty rose]

(3,0) is not on the line

(4,1) is

Use that instead

[runic heath]

ohhh the only line i can use is the adj?

like the line that touches the blue area?

can i use (3,2) then?

[full forum]

@runic heath Has your question been resolved?

[runic heath]

<@&268886789983436800> anyone knows?

[vague moss]

You can decompose this area into two parts

A square and a triangle

How would you do that?

[gritty rose]

Yes that's on the line

[full forum]

@runic heath Has your question been resolved?

[runic heath]

bottom part square and top triangle? but what does that give me

[vague moss]

Oh wait it's integration lol

You know how to find the equation of a straight line from 2 points on it?

[runic heath]

yes, i just didnt know what points to use

i assueme its the slanted line?

[vague moss]

yes

[runic heath]

i see lemme try with 1,4 and 3,2

[vague moss]

You could take any two from (1,4), (2,3), (3, 2), (4,1)

[runic heath]

4,1 would work even if it's outside of the area?

[vague moss]

Yes because we just want the line for now

[runic heath]

thank you i solved it the area is 6

is this one also about just finding two points of the red line?

[vague moss]

Yes

But you must be careful

The line has breaks

You should be careful to only select points which are on the same slope

[runic heath]

so i only pay attention to x=0 and x=1?

[vague moss]

For the first question

Yes

[runic heath]

the slope is the entire red line?

[vague moss]

The function is piecewise

[runic heath]

how so

[vague moss]

It has different equations for various parts

For example

[runic heath]

ohh right cuz it's a sharp turn

[vague moss]

If you find two points between x=1 and x=6

And find the line connecting them

and then substitute x=0

you wouldn't get the right answer

[vague moss]

^^

[runic heath]

ohh

[vague moss]

yes

So you must be careful

The first part, you can go as normal

What is the equation of the line for x=0 to x=1?

[runic heath]

first part two points can be
(0,-3) and (1,-3)

second part two points can be hmm

it all has to be either below or above the x axis? not mixed?

or not mixed with the straight line afterwards

like what if i pick (2,-2) and (5,1)

for the second one

[vague moss]

It can be mixed

[runic heath]

i see so this should do

[vague moss]

Since they are on the same line it wont matter

yes

[runic heath]

thank you ill calculate and come back with answers haha

first equation is f(x)=-3
integrated is -3x
area=-3

how can the area be negative

[placid vale]

It's under the x axis so it's negative

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[stuck fiber]

I got the right answer butttttt

Im noticing ive developed a bad habit 😅

I dont really read the questions, I kinda just guess what I need to do, like I see the word 'distance' and I just think oh put it inside sqrt(a^2 + b^2) and see what happens'

And yeah I got lucky this time

But I need to stop this habit, because even though I 'know' how to do it, I still dont really know what the question is asking

Like what does it mean by 'as a function of t'?

And 'Your answer should only contain only one sine or cosine and one t', this is supposed to give me a hint to suggest a certain way of how to approach the question right? What is that hint?

[onyx glen]

it means that your answer has to have t in it, as opposed to just being a number.

which would be a natural consequence of doing it the correct way anyway

[onyx glen]

no, it's just a goal for how much you should simplify your answer -- so that you cannot just pop that shit into the distance formula, decide not to simplify it at all, and call it a day

[stuck fiber]

Ahh I see that makes sense

Ok well I understand the question a lot more now haha

Thank you!!

Also welcome back!! Where have you been you were the most active person on the server then you just dissappeared for like 2 years 😝

[onyx glen]

1

[stuck fiber]

Dont pick up the phone, you know hes only calling cause hes drunk and alone

Ok well... good to see you

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[orchid lake]

Could anyone help me solve this? I don’t understand what to do

[hollow wharf]

Think of some number a such that $6a^2$ or $60a^2$ is the number inside the square root

[glossy valve]

CST (please ping when replying)

[orchid lake]

Sorry I don’t understand do u think u could explain a little more? 😢

[hollow wharf]

Start with 600

Can you think of a square that can be multiplied to 6 to make 600?

[orchid lake]

10^2 x 6

Ohh

[torn jolt]

Linearization L(x) = f(a) + f'(a)(x - a) (oh... not calculus yet)

[orchid lake]

And then after that how can I use the decimals provided?

[past venture]

I don't think shes doing that xd

[past venture]

you can use that fact sqrt(ab) = sqrt(a) * sqrt(b) !

if you notice that 600 = 6 x 100

u can split it up

[orchid lake]

Ohhh and then turn it into a mixed radical?

Like 10 root 6

[past venture]

yea!

[orchid lake]

Then would I have to divide for questions like d)?

[past venture]

mhm

[orchid lake]

Ok so for d) would it be like root 6 over root 100

Or root 6 over 10

[past venture]

yep

exactly

[orchid lake]

Can it be further simplified?

[past venture]

well i think ur meant to use the number given for sqrt(6)

so sqrt(6) / 10 ~= 0.245

[orchid lake]

Ohhh I see

So then I would just have to plug in the given value

[past venture]

yes

the kinda theme of all the questions is like

how can u change it to k sqrt(6) or k sqrt(60)

[orchid lake]

Ohhh I see now

[past venture]

so u can just plug in the value

yeah!

[orchid lake]

Tysm!

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[past venture]

np!!

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[random lichen]

how do i draw this

i got this so far

[toxic nebula]

What you have to do with this?

[random lichen]

nvm i got it

[random lichen]

make venn diagram with that info

then find probability of spanish which i can do

[toxic nebula]

Ok

If you have done it, then .close ir

[random lichen]

. ckise

.ckise

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[nimble juniper]

Could someone help me solve for c and d

[nimble juniper]

Pretty sure PBS and QCS are similar but I don't know how to prove them

[nocturne shard]

BSC and QSP are similar

thats it

[nimble juniper]

Ahh okay

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[keen shell]

i need help

[keen shell]

determine the slope of the tangent line at point (-2,2) on the curve f(x)=2x^2+3x

i’m using limits

just started calc

and that’s all my teacher taught

grade 12

is this wrong then gang 💔

icl if she seen me use this i don’t think she’d pass my ass

it’s an assignment that was due like an hour ago i just woke up 😭

oh word

bettt

thanks bro

yeah my teachers super good she like gives us the super long and boring way first

for us to understand it

then she’s gonna teach us all the shortcuts

i’m hyped

cyaa

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[wintry geode]

How do I get the equation of a parabola given two points (x1, y1), (x2, y2), and the y-coordinate of the vertex k?

[wintry geode]

This is what I've already tried but solving for h looks like a cubic (I don't know how to get solutions for cubics with variable coefficients)

Should I continue from here or try a different method?

Oops

I made a mistake

I accidentally wrote y1-h instead of y1-k somewhere

imma see if i can solve it now

I think i got it
nvm sry abt that

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[feral quest]

5x^2+3x-5 identify the Like terms

[feral quest]

!status

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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

[feral quest]

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[random glacier]

how do i find out quartile 1, quartile 3, ect? im confused af im dumb 😭

[wet marsh]

do you know how to find the mean?

[random glacier]

no i dontt :(

wait i think i do

[wet marsh]

Okay try finding that for this data set

[random glacier]

u eliminate the same amount of numbers from both sides?

then u get the mean?

[wet marsh]

Not sure if I follow but kinda?

[random glacier]

okayy

so i put them in order?

[wet marsh]

so like what's the mean here

yes

[random glacier]

ohh okay

[random glacier]

wait a sec

[wet marsh]

wait @random glacier I'm sorry I don't know the english terms but what I was thinking of is the median

[random glacier]

i found it

ohh

😭

[wet marsh]

okay well what it is

[random glacier]

the mean and median r different wait a sec

[wet marsh]

yeah you might as well say the mean though to make sure you understand how you get it

[random glacier]

the median is

72.5

[wet marsh]

you mean the, mean right if so yeah!

but what about the median

[random glacier]

72 i think 😭

[wet marsh]

what does your ordered data set look like

[random glacier]

wdym?

[wet marsh]

like

69, (...), 78

[random glacier]

69, 70, 70, 71, 72, 74, 76, 78

[wet marsh]

okay and the median is the central element right?

[random glacier]

yeah i think so

if that is it i got 72

as central

[wet marsh]

you have 8 elements

[random glacier]

ohh

wait

so do i add it all together

then divide

it by 8

?

[wet marsh]

no you add the center two

The median is the central element, since we don't have a middle one we take the average of the closest two

[random glacier]

ohh

okay

i got 71.5

[wet marsh]

yeah so

that's our quartile 2

do you know what the quartiles mean?

[random glacier]

not rlly 😭

[wet marsh]

okay so like,
the median is the center of our data set

[random glacier]

okayy

and thats quartile 2?

[wet marsh]

yes

[random glacier]

okay that makes sense

[wet marsh]

so then what do you think quartile 1/3 is?

if median is the quartile 2

[random glacier]

idk how to figure it out 😭

[wet marsh]

Okay so like

[wet marsh]

let's take this

the median is the center of our data set

the quartile 1 is the center of the lefthand side of our data set basically

so what's the median of 69,70,70,71?

[random glacier]

70

[wet marsh]

yeah

so then what's quartile 3

These are the quartiles
69, 70,(70) 70, 71,(71.5), 72, 74, 76, 78

[random glacier]

75? 😭

[wet marsh]

yeah!

[wet marsh]

so then which one of these is right

[random glacier]

4

i think

😭

[wet marsh]

Q2=71.5

[wet marsh]

I said this badly but like

basically the quartile 2 divides the data set into 2 sub-sets(what's left and right to it)

and the quartile 1 is just the median of the left-hand side while the quartile 3 is the median of the right-hand side

[random glacier]

OHHH

3

is the answer

[wet marsh]

yes!

[wet marsh]

because our 1st bit is the minimum of the set, 69
the left side of the rectangle is Q1, so 70
the middle bit of the rectangle is Q2, so 71.5
the right side of the rectangle is Q3, so 75
the final bit is the maximum of our set so 78

[random glacier]

oh tysmtysm

this helped alott

[wet marsh]

by the way you NEVER include the Q2 in the two sides, so if we for example added 71.5 then our minimum,Q1,Q2,Q3 and maximum would be the same

[random glacier]

tyty

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[torn jolt]

Hello

[torn jolt]

Is everything here correct?

Btw Idk why the first one is 3 marks i feel like im missing something

[keen bloom]

First one is wrong

[torn jolt]

If the first one is wrong how is this

[keen bloom]

You cant solve inequalities like that

You have to move everything to one side

[keen bloom]

And factor

Yh but not the equal sign

[torn jolt]

So what negative did i multiply

Ohhh ok

Is the second image wrong as well

👍

Hows rhis

[keen bloom]

This?

[torn jolt]

How is this

[keen bloom]

Yh thats right

Wait

X is not bigger than 0 i think

[torn jolt]

What sign do i use than

[keen bloom]

Or nvm

No ur right mb

[torn jolt]

So is questuon 4 correct?

[keen bloom]

You can always check on desmos.com

[keen bloom]

The one you wrote on the test no

Lol

I thought the same

[torn jolt]

Im so confused 💀

@neat thicket @keen bloom

[keen bloom]

Sup

[keen bloom]

Yh its right

[torn jolt]

Ok so did i write the right thing?

[keen bloom]

When you have a quadratic like that its good to put it on a grapsh

[keen bloom]

Yea

Not on the test tho

You would lose points if u divided like that

[torn jolt]

What do i do then

[keen bloom]

This is right

I was talking about the original image

[torn jolt]

Oh ok

And question 5

A and b

Are these correct?

The completign the squayre

@keen bloom

[keen bloom]

Uh sorry

Bruh

You were supposed to find the turning point

Factor out the 3

Then complete the square

[copper zenith]

yh

[keen bloom]

Thanks for the affirmation

😭

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[bold hound]

Guys, it is urgent
I need to know how to convert mg/L/h into ml/h
According to my situation, I have 0.735

And I am lost since I can't get any information that helps me going further welppp

[umbral dome]

those do not appear to be compatible units

[bold hound]

Dang
I am helping my friends with a scientific project and they asked me how to do it

Is there any way for me to make them compatible? With a litrage or smth

(if I end up with mg/H)

Ah yeah I guess I will just need Co2's density?

[umbral dome]

the first one appears to be density per hour(?) and the second is volume per hour(?)

[bold hound]

If you need more context, It is a plant's Co2 absorption

They found out the plant could absorb 0.735mg/L/H

[unique wagon]

could you show the specific question

and why you need to convert it into ml/h

[bold hound]

There is no specific questions
They just wanna convert it to compare it to other plants

Which have values surely said in ml/h

[unique wagon]

ah I see now

[unique wagon]

I think measuring co2 absorption in density/hour is completely wrong

it should be expressed in ml/h or g/h

[bold hound]

Oh
Internet told me it was possible

[unique wagon]

could you send me the source you used?

[bold hound]

Let me ask 'hem

Bruh I just said it was me

Let's say I just asked chat got and got this

Oh wait

Tf
I am so lost
I'm tired af

'kay so
I am theorising

If we divide what we got by the liter of the box
Do we get rid of the liter thing ? ( I do not know all the steps they have done so far)

[gritty rose]

!nogpt

[full forum]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[bold hound]

Ah yeah but it wasn't to answer to my own question
I answered people where the mistake could be

I am better in maths than I am in actual chemistry (I haven't worked on this during my chemistry classes) and I can't find where to get infos

Usually, he redirects me on websites where I can verify the infos, but here I can't
That's why I am asking you

But I'll be careful with showing my mistakes (when I helped myself with Chat gpt) here

[gritty rose]

you need to just form your question properly

[bold hound]

Yeah but in that case
I am definitely lost
I don't even know what to ask properly so chat gpt won't be a useful tool

[gritty rose]

no one knows what you're asking man. go ask your friends for more information

[bold hound]

Nevermind they gave up
It's late so we all are very tired

Sorry for bothering you all

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[woeful anvil]

Am i able to do problem 39 without using a triple angle identity?

[gritty rose]

yes

[woeful anvil]

how would i set it up?

i know cos theta is x/r

but how would i do it with the

cos(3theta)

[gritty rose]

if you know what r=cos(theta) looks like, then r = cos(3 theta) is 3 of those

https://jwilson.coe.uga.edu/EMAT6680Fa2013/Thurston/Write-Ups/Write-up 11/Polar_Petals.html

[woeful anvil]

ah

[gritty rose]

just trace r = cos(3theta) for theta = 0 to pi/6 first, then pi/6 to pi/3

[woeful anvil]

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[woeful anvil]

thx

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[unborn acorn]

how do i solve a² + b² - ab = 1 in Z

[unborn acorn]

and then

[leaden ermine]

a³+b³ = a+b is maybe better to work with

for integer sols

[real barn]

Diophantine equation

maybe

[unborn acorn]

ehh wait

[real barn]

one set of solutions is 1, 1

for sure

-1, -1 as well

given that this is the graph

therefore -1, 0 is also a solution

1, 0 as well

[unborn acorn]

the og question was : for all a, b in Z, show that N(a+bj) = 1 if and only if the pair (a,b) is (+ or - 1, 0), (0, + or - 1), + or -(1, 1)

[leaden ermine]

thanks for hiding it

[unborn acorn]

and for z complex number, N(z) = |z|²

j^3 = 1 obv

[real barn]

we have solutions: {a,b} = {1, 1}, {1, 0}, {-1, -1}, {-1, 0}, {0, -1}, {0, 1}

but that's only from graph

there are many solutions as you can see on the graph

[real barn]

ohhhh

then N(a+bi) = ab + 1

therefore to satisfy N(a+bi) = 1

ab + 1 = 1

ab = 0

[kindred grove]

||
substract ab on both sides, you get (a-b)^2 = 1 - ab, so you need 1-ab>=0, i.e. ab<=1 to have a solution
now if you add 3ab instead, you get (a+b)^2 = 1+3ab, i.e. 1+3ab >= 0, i.e. 3ab>=-1 is required
that limits your options pretty quickly
||

[rapid rain]

My recommendation was to distinguish cases ab <= 0 and ab > 0

in the latter case it helps rewriting a^2 + b^2 - ab = (a-b)^2 + ab

[unborn acorn]

so since we know that we need ab <= 1 to have a solution, if ab>0, we need to have a = 1 and b = 1 or they're both -1 or they're both 0 or one of them is 0

now if ab<=0

we cannot use the first condition given by aplatypus

if we use the second one

3ab>=-1, with ab<=0, ...

ab has to be 0

i dont know

[unborn acorn]

can we use the previous question? The one that we just solved with U(A)?

[kindred grove]

got a link to it ?

[unborn acorn]

this

[unborn acorn]

this

[kindred grove]

well do you have another way of finding the units of A without using that norm argument?

[rapid rain]

no, in fact it's kinda the opposite

we rather find all units of A

by solving N(z) = 1

[kindred grove]

yeah the norm is here to help you

instead of working with A directly which might be pretty ugly, you just work in Z

[unborn acorn]

ok

so we use ab<=1 and 3ab>=-1?

orr

[kindred grove]

yeah -1 <= 3ab <= 3, combining the two inequalities

so indeed as you said a=0 or b=0 works

if both aren't 0 you have almost no leeway at all

[unborn acorn]

the first condition gave more solutions, but a or b dont need to be 1 as long as the other is 0

so is there a way to find all of the solutions?

@kindred grove

[kindred grove]

yes, if you push the reasoning with these inequalities just a little further

[kindred grove]

what a,b actually satisfy these inequalities if both are non-zero ?

@unborn acorn

[unborn acorn]

1 and -1 but for the first condition, if we assume ab>0 and we know ab <= 1 if we want a solution, if we take 0, a can be anything

[kindred grove]

yes i know

I'm deliberately excluding the case where either is zero, we already covered that

[unborn acorn]

ok but we also need to study that case?

[kindred grove]

which one ?

• either one of a,b is zero, then (a,b) satisfies the inequalities
• or a and b are both not zero, and this forces (a,b) to be ???? to satisfy the inequalities

my question is what do you put in the ????

[unborn acorn]

(1,1) or (-1, -1)

[kindred grove]

yes indeed

so with these two inequalities we managed to restrict a lot which points (a,b) we need to look at

now you can plug in these cases in the original equality

and see which (a, b) are actually solutions to the equality

[unborn acorn]

ok, so normally i just need to plug in the first case where one is 0 and then it should force the other one to be 1 or -1?

[kindred grove]

yeah exactly

and then (1, 1) and (-1, -1) also work in the equality

so we won

[unborn acorn]

ok i see

thankss

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[kindred grove]

@unborn acorn the ellipse & the inequalities, looks cute

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[wicked zodiac]

how can you tell if a fujnctions is decreasing by find the dervaivtre

[wicked zodiac]

for intergarwl testg

for finding the series

conmverging or not

[tulip oriole]

if dy/dx < 0 then it's decreasing

[wicked zodiac]

yes

but there would be a x or something
how would i know if its less than 0

[high fjord]

dude what?

u wanna explain what ur question is?

[wicked zodiac]

how would i know if its less tahn 0

sometimes its not just a constant

[high fjord]

if whats less than 0?

the y value?, the integral? the derivative?

[tulip oriole]

you can find the critical points and test

[wicked zodiac]

i see

[high fjord]

ok im so confused but whatever

[tulip oriole]

depends where your serie starts

[wicked zodiac]

for this one

how would i find the critical points

[tulip oriole]

where dy/dx = 0

[high fjord]

oh

[wicked zodiac]

so 0

and then is it something like where i pick a numvber greater tahn 0 and less than 0

and if the side is postive i put+

[tulip oriole]

desmos gives me sqrt(e)

but yeah with integral test you only have to show that the function is eventually decreasing

can you show the original serie this is about

[tulip oriole]

what's the question rn

[wicked zodiac]

if it converfes or not

by using the test

[tulip oriole]

ok so back to finding the critical points

should give you sqrt(e)

[wicked zodiac]

ive done this calc1 but its only for ones not in fractions

how wouyld i do this

would i rewrite it?
or is there another way

without rewriting it

to find the critical point

these are he ones i see

bnefore

@tulip oriole

[hot lily]

Wdym without rewriting it

You need to differentiate it though

[wicked zodiac]

to find the cricitcal point

[hot lily]

And to see where its decreasing/increasing

[tulip oriole]

once you factor x out it's just 3-6ln(x) = 0

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[upbeat venture]

can someone explain how this works to me

[steel solar]

$g(f(x))=g(y)$, where $y=f(x)$

[glossy valve]

;(

[steel solar]

So essentially, f(x) is the input for g(x)

In math notation this would be $x\to f(x)$ roughly

[glossy valve]

;(

[upbeat venture]

im still a little confused

[craggy tapir]

\mapsto

[upbeat venture]

like how does g of f become (0,3)

[steel solar]

I didn't choose that since notational stuff, I chose the more lenient notation

[steel solar]

Because of the original x-input for f(x) which would be f(0)

[tulip oriole]

for example, if f maps 0 to 5 and g maps 5 to 3, then g o f maps 0 to 3

[steel solar]

f(0)=5, so g(5)=3 is what we want

[tulip oriole]

g takes the output of f as an input

[upbeat venture]

OH OKAY

WAIT THAT MAKES SO MUCH MORE SENSE

I WAS DOING IT BACKWARDS 😭

thank you!!!

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[thick dove]

How do I show that (-inf,2) isn't bounded below ?

[thick dove]

I understand that I let a∈ℝ and construct an x∈(-inf,2) such that x>a so the set has no lower bound but how do I construct it

If I just say "oh x>a" then I can argue that for any set and say anything has no bounds

[trim sun]

you cant do this by contradiction? like suppose (-inf, 2) has a lower bound B, prove B-1 is in (-inf, 2), that's a contradiction

[thick dove]

There's a way the course taught that I want to do

But I don't get it

[thick dove]

I can but I'd rather understand the way the course mentions

rather than just go around the issue and ignore it

[simple ridge]

Do you mean such that x<a?

[thick dove]

Yea mb

[simple ridge]

Np

[thick dove]

But still

Idk how to do it at all

Other than say "yea its smaller"

[simple ridge]

Can you show me an example of the way you are trying to use

Like an example from the course or something like that

[thick dove]

I don't at all understanding the courses example but its showing that (2,inf) has no upper bound:

Plan for the proof:

Show there exists no a∈ℝ:x<=a for all x in the set

Given a∈ℝ, pick x'∈(2,inf), x>a

e.g x'=max{a+1,3}

Proof:

Let a∈ℝ, write x'=max{a+1,3}
Then x'>3, so x>a+1>a

So a is not an upper bound

None of that makes sense to me I'm ngl, unless you assume a isn't in the set but thats just assuming there's no number bigger than infinity then by defining x'=max{a+1,3} to be 3

Idfk

I'm gonna go look at the lecture slides to see if that is how they did it bc that's what I have in my notes and it makes 0 sense

[simple ridge]

If the way you are talking about is just taking any a in (-∞,2) and showing that there exists x in (-∞,2) such that x<a then just take x=a-1 for example

For any in a in (-∞,2), a-1<a

So that a is not a lower bound

[simple ridge]

Are you sure this proof is exactly a proof written in your course ?

[thick dove]

From the lecture slides

The bit at the bottom in black is the "proof"

Hes assumed all a∈ℝ that aren't in the set are smaller than 2, which defeats the point of a proof

Cuz if you can just assume a is smaller than infinity, just say "no number is bigger than infinity" so there's no upper bound

[thick dove]

Do you understand this at all @simple ridge
It just makes 0 sense to me

[simple ridge]

Tbh I think this is making the proof longer than needed

Not to say that it is incorrect

[simple ridge]

So let me tell you what is going on

So the claim is that given any a in R , a isn't be an upper bound of (-2,∞)

So first of all choose any a in R

Then either a=< 2 or a>2

[thick dove]

Right

[simple ridge]

If a=< 2 then you can take x=3>a(which is in (2,∞)) so that x>a and hence a is not an upper bound of (2,∞)

You are left with the case of a>2

If a>2

Then a+1>3 so that a+1 is in (2,∞) and a+1>a so that a is not an upper bound of (2,∞)

This proves the claim

[thick dove]

ah I see

Thank you

[simple ridge]

So the purpose of x=max{3,a+1} is :

1. to make sure that x is in (2,∞).
2. to make sure that x>a

[thick dove]

Hmm I see

I still don't quite get the reasoning I'm sorry

Ik I'm being a dumbass rn

[simple ridge]

No you are not

Tell me what is still bothering you

[thick dove]

let x=max{a+1,3},
a>2 so a+1<3, so x>a+1>a
ahhh and x is in the set so an x in the set will always be bigger than a∈ℝ?

And for the case a<2, x=3 which is in the set

[simple ridge]

This is not quite right

[thick dove]

And bigger than a

So its not an upper bound

Ohh

[simple ridge]

a>2 doesn't mean that a+1<3

[thick dove]

Oops I meant a+1>3

[simple ridge]

Ohh in this case what you said is correct

[thick dove]

Nice thank you

[simple ridge]

Everything you just said is correct and exactly describes the proof that you sent

[thick dove]

The question on my assignment is showing (-inf,5) has no lower bound but same argument with x=min{a-1,some number smaller than 5}

[thick dove]

🙏

[simple ridge]

Exactly

Great job

[thick dove]

Ty ty

It can be any number smaller than 5 right

Or does it have to be 4

[simple ridge]

Np but I didn't do anything, you did the work

[simple ridge]

What do you think

[thick dove]

for min{a+1, something}

[simple ridge]

I would like to hear your opinion

[thick dove]

I think if x=min{a-1,5}

If a>=5 then for example 3 is in the set and smaller so a isn't a lower bound

If a<5, then a-1<4 so if x=min{a-1,4}, x<a-1<a and x is in the set so x<a

Right?

Wait x<a-1 is wrong? x=a-1 moreover

a-1<a
So x<a

?

[torn jolt]

I need help. It 24 I need a positive and negative answer. The numbers are 7,-7,-5,-2

[simple ridge]

Did you mean to write x=min{a-1,4} in the first line ?

[thick dove]

Yh mb

[simple ridge]

#❓how-to-get-help

[simple ridge]

Np I was just checking

[thick dove]

Its x=a-1<a right? Not x<a-1<a

[simple ridge]

Yes

[simple ridge]

But I have a comment here

[thick dove]

Ofc

[simple ridge]

Just to make your proof look better

[simple ridge]

Ah wait no

mb

You don't know for sure that x<a-1

For example take x=4.5 then x<a but x>a-1

[thick dove]

Yea so if a<5, x=min{a-1,4}=a-1<a
so x<a

?

And that means there's no lower bound bc if x=min{a-1,some number smaller than 4}=a-1, x must be in the set

right?

[simple ridge]

Ah wait

I am tired rn so forgive me for being stupid

[thick dove]

It's all good its 4am and my flatmate won't stfu talking to me in the library

[simple ridge]

Its 1 am for me xD

[thick dove]

So I can't think either

[simple ridge]

Ohh then your condition is worse than me

[thick dove]

is there an issue with my maths?

[simple ridge]

Ok so x=<a-1 is guaranteed because x=min{a-1,4} so if a<5 that means a-1<4 which means that x=a-1 . If a>=5 then a-1>=4 so that x=4

[simple ridge]

So this is right

(when a<5)

[simple ridge]

other than this you have no real issue. Just a slight fix , if x<5 then x=a-1<a and x is in (-∞,5) so that (-∞,5) is not bounded below (you shouldn't repeat that x<a, especially in the way you did it)

That's just for a better structure of the proof

But the ideas you wrote are correct and complete

Forgive me if I caused a bit of confusion above

[thick dove]

@simple ridge

Good or nah?

[simple ridge]

All of the ideas are correct but I suggest that you define x hat in case 2 instead of defining it at the beginning since you didnt use it in case 1

Also in case 2, x hat=min{a-1,1}=a-1 and a-1<a do x hat < a

You shouldn't mention that x hat in E when you say x hat=a-1 and a-1<a

Only after you say x hat<a that you mention x hat is in E

You can even make the structure better if you want but that is not a big deal rn

What matters the most is that your proof contains all of the necessary points

Just fix the structure of the proof slightly as I told you and then you will be done

[full forum]

@thick dove Has your question been resolved?

[full forum]

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[thick dove]

Again I have a query

[thick dove]

How do I show that {n∈ℕ:(1+1/n⁴)} has no maximum element?

[oak ruin]

it does

do you mean minimum

[thick dove]

Uhhh

[oak ruin]

It's a decreasing sequence

so the first term is the max

2 at n=1

[thick dove]

Oh wait

1-1/n⁴ oops

prove {(1-1/n⁴):n∈ℕ} has no max