mossy latch
ahhhhh alr
#help-43
compact pewterBOT
molten badger
Soosss
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deft tangle
tlog2=32logt
karmic escarp
tlog2=32logt
change it to the same base
nvm what am i even talking ab
move it to the left side
and combine
tacit robin
karmic escarp
my bad i didnt actually do this question lol
theres a great video by bprp that talks ab lambert w
there's an entire playlist for this sake
hushed magnet
fwiw, they probably want you to guess the solution ||t=256||
||its somewhat reasonable to think that if there is a nice solution, then it might have the form t=2^k. plugging it in quickly gives k=8||
simple maple
are you still solving the original 2^t=t^32? cuz i'm getting a very different solution
molten badger
indeed, plot it to desmos give us more than 1 solution
deft tangle
Bro
simple maple
okay, sorry. I'm getting two very different solutions
deft tangle
Wolfram uncle and desmos aunty is not allowed to enter in exam hall
molten badger
Wolfram uncle and desmos aunty is not allowed to enter in exam hall
they must mean the integer one
simple maple
is a calculator allowed in the exam hall?
hushed magnet
!original
deft tangle
What?
hushed magnet
I guarantee that you are not asked for all solutions
deft tangle
I guarantee that you are not asked for all solutions
True
molten badger
So you can safely assume t=2^k
hushed magnet
or any of the ugly ones
compact pewterBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
simple maple
but... it has two. And neither is nice
karmic escarp
lambert w
karmic escarp
in fact if u plot this in there's no integer solutions
hushed magnet
in fact if u plot this in there's no integer solutions
there is....
simple maple
Dena literally wrote it
deft tangle
Why do you use calculators?
hushed magnet
your calculator cant handle big numbers like that
karmic escarp
your calculator cant handle big numbers like that
im using desmos
karmic escarp
that's not how that works vro
hushed magnet
well you need to zoom out further at least
karmic escarp
well you need to zoom out further at least
the trend tells me it's not gonna do a u-turn
molten badger
you didn't zoom out far enough
hushed magnet
it will
simple maple
the y value for x=256 is about 1.15*10^77
molten badger
2^256 is an incredibly large number
simple maple
you're not zooming far enough to see that
karmic escarp
you know what
my bad im wrong
i zoomed all the way up to 5x10^60 and i realised the gap was minimising
fucks sake is this fucking malarkey question
@deft tangle DROP THE CLASS
simple maple
honestly my first approach if with calculator would be numerical approximation for the small positive one
simple maple
<@1340556085467287603> DROP THE CLASS
Dude. No.
karmic escarp
For a bad joke
londoners 💔
hushed magnet
its entirely reasonable to guess t=2^k
karmic escarp
id assume half of the people here comes from either london or new york
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copper gazelle
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kind viper
,calc 3*33.25 + 0.25
boreal girderBOT
Result:
100
well that's the correct x value & i believe the only one
snow stone
compact pewterBOT
kind viper
$\frac{f(a-u)+f(u)-f(u)}{f(a-u)+f(u)} = 1 - \frac{f(u)}{f(a-u)+f(u)}$
boreal girderBOT
Ann
subtle helm
add both of the integrals tgt
with
u will get 2I = integral from 0 to a of smth dx
snow stone
ok no more hints lmao
subtle helm
[ \int_a^b f(x) \dd x +\int_a^b g(x) \dd x = \int_a^b f(x) + g(x) \dd x]
boreal girderBOT
k
boreal girderBOT
k
snow stone
@kind viper @subtle helm thx for both help, hope i did ok
subtle helm
works too ye
i would do
[ I + I = \int_0^a \frac{f(x) + f(a-x)}{f(x) + f(a-x)} \dd x = \int_0^a \dd x = a]
So, ${2I = a \implies I = \frac{a}{2}}$
boreal girderBOT
k
snow stone
since it is =I so I+I goes 2f(x)/f(x)+f(a-x)
how u got f(x)+f(a-x) in the numerator ?
subtle helm
wdym
we've shown that
[ I = \int_0^a \frac{f(x)}{f(x) + f(a-x)}\dd x = \int_0^a \frac{f(x-a)}{f(x) + f(a-x)}\dd x]
boreal girderBOT
k
compact pewterBOT
@snow stone Has your question been resolved?
novel narwhal
A company makes a mean monthly income of $20,300 with a standard deviation of $3,200. In one given month, the company makes $29,500. What percent of the time does the company make between $15,000 and $25,000? (i can't believe i'm asking for help on statistics, but here i am)
compact pewterBOT
A company makes a mean monthly income of $20,300 with a standard deviation of $3...
novel narwhal
do i use z-scores or something? idk
eager thunder
uhh let me see
eager thunder
do i use z-scores or something? idk
yes. can you calculate them?
novel narwhal
yea ik how to calcuate z-scores
but idk how to use them to make the probability thing
and i can't use the empirical rule because the z-scores aren't integers
eager thunder
oh uh the formulas are extremely complex and unless this is a high-level college statistics course you likely can use a chart or an online p-value calculator
novel narwhal
okay
eager thunder
wait actually
its not calculable by a formula at all what am i saying
the only way to do it is by computer approximation
novel narwhal
can you tell me how to use the p-value calcuator
i don't understand all the terms
nvm.
.close
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snow stone
is it true that this equation has 2 parallel tangents that has the same gradient (0.5)?
compact pewterBOT
is it true that this equation has 2 parallel tangents that has the same gradient...
snow stone
ive found two for (0,1) and (0,-1)
but somehow teacher says theres only one tangent with gradient of 0.5
rotund sphinx
Yeah your teacher is being a muppet
snow stone
cool
snow stone
,w tangent to x^2-xy+y^2=1 at (0,1)
is it works for pro wolfram alpha or also the free version?
rotund sphinx
Texit uses the free version as it’s a publicly available bot
Take that as you will
snow stone
k thx for ur help
rotund sphinx
k thx for ur help
compact pewterBOT
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snow stone
.close
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snow stone
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snow stone
hey, is it ok to say it using a theorem of "The composition of two continuous functions is continuous."
winged lion
Yes
compact pewterBOT
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paper moon
compact pewterBOT
kind viper
,rccw
boreal girderBOT
native shard
you can’t
kind viper
uhhhh you dont?
paper moon
😭
native shard
that shit ain’t even
native shard
ok and f(1) = 0 but f(-1) ≠ 0
paper moon
Exactly
paper moon
Nope
native shard
was one side drawn already?
native shard
throw that shit in the bin
compact pewterBOT
native shard
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fair thunder
in this proof, how are we knowing that b - ak will contain every element, aka natural number.
dont we need another proof for that
compact pewterBOT
in this proof, how are we knowing that b - ak will contain every element, aka na...
kind viper
b - ak will contain every element, aka natural number.
wdym "contain every element"?
i think i remember seeing this exact proof posted here earlier, with this exact wording and all its flaws
fair thunder
as in how do i know that
lets say im taking 5
how will i know that b - ak = 5 for sure
kind viper
you don't
fair thunder
i think i remember seeing this exact proof posted here earlier, with this exact ...
yeah im the one with it, i was reviewing stuff lol
kind viper
it's not guaranteed that $5 \in S$
boreal girderBOT
Ann
fair thunder
then how is this working for every natural
boreal girderBOT
Ann
fair thunder
well, yeah
i mean then, how will i prove the main statement that
b = aq + r for integers
fair thunder
**where** are you seeing S=N
well i was thinking like that
now that you question it, it makes sense that it isnt
but then if, lets say 5 isnt in S
how is this proving that 5 = aq + r
or i mean
other way
kind viper
.....
i have almost 0 idea what the hell you're actually trying to ask here...
would you like me to give you a restructured version of the proof which doesn't have the original's notation fuckups and language weirdness?
fair thunder
yes please, and let me rephrase my question, just gimme a minute
okay so S is a subset of N right
kind viper
that it is
fair thunder
and we're trying to prove that b = aq + r
and b can be any integer
now we're taking the set S such that b - ak is an element and b - ak > 0
so just imagine that S is {2,3,4}
so taking 2 as least element we're proving this
but as you said
5 may not be in S
so then how do i ensure that
this works for every integer
like if a subset is {5,6,7}
how do i know that b - ak can also be equal 5 for some integer b,a,k
or how do i prove so
kind viper
\textbf{Theorem 2 (Division Algorithm).}
Given any integers $a, b$ with $a \neq 0$, there exist unique integers $q$ and $r$ such that $b = aq + r$ and $0 \leq r < |a|$. If $a$ doesn't divide $b$, we have $0 < r < |a|$ instead.
\textit{Proof.} The following is a step-by-step breakdown of the proof.
\begin{enumerate}[(1)]
\item If $a$ divides $b$, then the theorem is satisfied with $r=0$.
\item We thus assume $a$ doesn't divide $b$, and set out to prove the theorem using the well-ordering principle.
\item To this end, define a set $S$ as follows:
$$S := { b-ak \mid k \in \bZ, b-ak > 0 }.$$
\item To apply the well-ordering principle and claim $S$ has a smallest element, we need to establish the following:
\begin{enumerate}[(a)]
\item $S$ is a subset of the natural numbers.
\item $S$ is nonempty.
\end{enumerate}
After establishing that the smallest element of $S$ (which we'll call $r$) exists, we will need to do extra work to ensure that $r$ does what we want it to do.
\item Elements of $S$ are required to be in the form $b-ak$ (thus they are integers) and also positive. Thus is clear every element of $S$ is a natural number, and so $S \subseteq \bN$.
\item Taking $k$ such that $ak = -|ab|$ (indeed, this $k$ may be expressed more explicitly as $-\mathrm{sign}(a) \cdot |b|$), we get $$b - ak = b - (-|ab|) = b + |ab|.$$ Since $b + |ab|$ is positive, it belongs to $S$. Thus $S \neq \rien$.
\item Now that we proved that the well-ordering principle applies to $S$, let $r := \min(S)$.
\item We need to show three things:
\begin{enumerate}[(a)]
\item $0 < r < |a|$.
\item There exists an integer $q$ such that $b = aq+r$.
\item No other pair $(q,r)$ satisfies the above two conditions.
\end{enumerate}
\end{enumerate}
boreal girderBOT
Ann
kind viper
\begin{enumerate}[(1)]
\setcounter{enumi}{8}
\item To prove (8a), suppose towards a contradiction that $r > |a|$ (Clearly $r=|a|$ is not possible anyway as $a$ doesn't divide $b$). Then $r = b - ak$ for some integer $k$. But then $r - |a| > 0$ and also
$$r-|a| = b-ak-\mathrm{sign}(a)a = b - (k+\mathrm{sign}(a))a \in S,$$
and $r-|a| < r$. So $r$ is supposed to be the smallest element of $S$, yet we've found a smaller one than that. Contradiction!
\item To prove (8b), note that since $r \in S$, there exists an integer $q$ such that $b - aq = r$. From this $b = aq+r$ follows directly.
\item Now we prove (8c). To this end, suppose that there exists a different pair $(q', r')$ satisfying the conclusion of the theorem, i.e. that $b = aq+r = aq'+r'$, and $0 \leq r, r' < |a|$. We will aim to show that $q=q'$ and $r=r'$, as this will let us conclude that $(q,r)$ is indeed unique.
\item Since $aq+r = aq'+r'$, we get that $r - r' = aq' - aq = a(q'-q)$ and therefore $r-r'$ is a multiple of $a$.
\item However, as $r$ and $r'$ both lie between $0$ and $|a|-1$ inclusive, we get that $$-|a|+1 \leq r-r' \leq |a|-1.$$
\item The only multiple of $a$ in the range $[-|a|+1, |a|-1]$ is 0.
\item Thus $r-r'=0$ and so $r=r'$.
\item from $r=r'$ and $aq+r = aq'+r'$ and $a \neq 0$ we get that $q = q'$ too.
\item This proves uniqueness, and thus the theorem.
\end{enumerate}
boreal girderBOT
Ann
kind viper
so just imagine that S is {2,3,4}
{2,3,4} is not something that S can ever be.
you're definitely overthinking it somehow.
i don't know how to address your overthinking
so im instead gonna tell you to read the 2-page, broken-down-as-much-as-i-could proof
and tell me the number of the earliest step that confuses you or demands more explanation
@fair thunder
i.e. i want you to apply the well-ordering principle on the set of all labels of the steps that confuse you if said set is nonempty
fair thunder
okay so every element of S is a natural number
but every natural number is not in S
kind viper
and if it is empty, i.e. if you understand everything i wrote without any hitch, then tell me so
fair thunder
and we're proving this for set S
then how does this apply to all natural numbers then
kind viper
did you read my proof y/n
fair thunder
yes
kind viper
did you read all the steps
fair thunder
i did yes
kind viper
ok then can you tell me the lowest step number at which you get confused
fair thunder
5
kind viper
ok
yes, S is a subset of N and in general is a proper subset of N.
this is not an impediment to the fact that a and b are arbitrary.
"we're proving this for set S" is kind of misleading imo btw
S is a tool we use in the proof, but it doesn't dictate the scope of the proof in any way
fair thunder
see, im confused in this specific part
b - ak can be any integer, this is very easy to understand
but, arent we proving this exact same thing that any integer can be written in the form b - ak
so my confusion arises that how exactly are we assuming b - ak to be some integer
and even if it is, how is it following to be correct for every integer
integer as in positive integers
kind viper
b - ak can be any integer, this is very easy to understand
idk how it's very easy to understand when "b-ak can be any integer" is quite false without a lot of qualifiers
arent we proving this exact same thing that any integer can be written in the form b - ak
no we aren't.
fair thunder
> b - ak can be any integer, this is very easy to understand
idk how it's very e...
i mean we're considering b,a,k to be integers
kind viper
you say the earliest step number which confuses you is 5. this implies you understand steps 1-4 inclusive, correct?
fair thunder
yes
kind viper
yes so
we're planning to use the WOP
WOP says "any nonempty subset of N has a smallest element"
our goal then will be to introduce some kind of subset of N that has something to do w our problem
and then do some shit to its smallest element
fair thunder
i understand the whole proof what youre doing
im confused as to how is this proof following over to all integers
kind viper
step 3 introduces the subset we're working with
step 4 sets out to prove that it is the kind of set WOP deals with
fair thunder
and then do some shit to its smallest element
right
kind viper
steps 5 and 6 prove the two sub-goals outlined in 4
fair thunder
right
kind viper
step 8 outlines the goals for showing this r really is the r that we want
im confused as to how is this proof following over to all integers
i have no idea what this is supposed to mean, i'm sorry.
copper sierra
theyre probably still hung up on "b-ak can be any integer"
fair thunder
this proof, its supposed to work for, lets take N for now
so this statement will work for
1,2,3,4,....
right?
now we're proving for the set S, this is true
BUT, set S may not contain 5?
so thats where im stuck
or is it that, for proving it for 5
we take some integers so that it IS 5
and then continue showing it works
am i correct?
like proving it for 5
we take the set to be {5,6,7,...}
and then show it works
and as a whole this proof works for every set like this
copper sierra
i think ur either still hung up on "b-ak can be any integer" or confused on how S is defined
fair thunder
oh wait
i think youre right
im having confusion with the set S definition
i was thinking set S is some integers which can be written in the form b - ak
but its b - ak elements only
god😭
okay but then how are we knowing that b - ak will be containing every integer from 1,2,3,4,....
copper sierra
S contains exactly all positive b-ak
fair thunder
like it is trivial to understand but, shouldnt there be like a, statemnt for that as well or something
or it doesnt matter if it has all natural numbers
copper sierra
b-ak does NOT cover all integers. u should check this by examples
copper sierra
u might be misunderstanding the statement of the theorem too
it says for all a,b there exist q,r
fair thunder
right
copper sierra
so in the proof a,b are arbitrary and q,r are to be constructed
copper sierra
lets take it slow
fair thunder
the remainder, which is the set S
that doesnt contain all integers
and it doesnt matter if the remainder contains all integers
cuz thats not the point
only a and b are supposed to be all integers
is that right
copper sierra
if b=10, a=3 then S=?
fair thunder
1,4,7,...
ohhhhh
im getting it now
set S depends on a and b
and for that a and b, this holds
which then proves the statement riiighhhttt
copper sierra
all i did is show u that S != N
upper bane
sneaking in to add that if you have access to Rosen's ENT, there is an alternate version of the proof that may help you understand better
fair thunder
whats Rosens ENT lol
upper bane
Rosen's Elementary Number Theory
fair thunder
THANK YOU SO MUCH THO yall
really helped a lot
thank you
fair thunder
Rosen's Elementary Number Theory
oh
copper sierra
hopefully that got u a lil unstuck
fair thunder
im gonna give it a read then
copper sierra
now u should go thru anns outline again
fair thunder
hopefully that got u a lil unstuck
yeah i got it completely now
fair thunder
now u should go thru anns outline again
i did get the proof before, i wasnt getting the S not equal N part thats why
but yes that is a cleaner version
copper sierra
btw in the future its a good habit to do lil examples to get unstuck and back on track
copper sierra
np 
fair thunder
.close
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kindred hemlock
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kindred hemlock
how to solve this question
subtle helm
[ E(x) = \sum xP(x)]
boreal girderBOT
k
subtle helm
x becomes x-1
kind viper
how to solve this question
"all payouts are reduced by $1" is equivalent to "after you play the game, the dealer takes $1 from you"
kindred hemlock
kind viper
think about it that way
kindred hemlock
i tried this but i got the wrong question
kind viper
btw your -1 and 0 should be -2 and -1
kindred hemlock
"all payouts are reduced by \$1" is equivalent to "after you play the game, the ...
oh thats a nice way to think about it
kindred hemlock
btw your -1 and 0 should be -2 and -1
ye i dont understand why tho, because -1 and 0 arent payouts, the dealer dont pay anything tho
kindred hemlock
oo dam payouts can be negative?
kind viper
or from breaking even to losing 1
kindred hemlock
or from breaking even to losing 1
what does payout mean
is payout the amount the dealer give to the player
compact pewterBOT
kindred hemlock
.reopen
compact pewterBOT
✅
kindred hemlock
oh wait sorry i have one quick question
what does doubling a payout mean, it means x2 for all payouts even the negative ones? doubling negative payouts means the dealer gain double?
wanton ferry
yes
kind viper
what does doubling a payout mean, it means x2 for all payouts even the negative ...
yes
kindred hemlock
yes
oh i see, if the question was rephrased like "if all the outcomes that can be won by the player is reduced by 1, we still reduce 1 for the negative wins?"
deft tangle
dy/dx+y/x=x^2
compact pewterBOT
dy/dx+y/x=x^2
deft tangle
hints please
covert crater
Use integrating factor
forest token
compact pewterBOT
forest token
how to do this integration
i first substituted x-->tan^2 theta
but then i got lots of rubbish which i couldnt simplify
kind viper
kind viper
well at least $\int \arctan(\sqrt{x}) \dd{x}$ looks like it might be less ass
boreal girderBOT
Ann
kind viper
or if you really want, $\int \mathrm{arc} ; ; \tan ; x \dd{x}$
forest token
yeah thats true
boreal girderBOT
Ann
forest token
how to proceed from here?
forest token
oh lmao
cold widget
oh this was a good one , yeah converting to arctan is the way, then ibp or sub,
or directly ibp works but its quite lenghty that route
kind viper
i would substitute x = u^2 first and then ibp
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dusky quartz
don't quite understand the question
compact pewterBOT
don't quite understand the question
ionic python
so that means $-1<x_i<1$
boreal girderBOT
Allen
dusky quartz
can't I just set $x_i=-x_{i+1}$
boreal girderBOT
laestia
dusky quartz
which would imply n=2562
ionic python
but ther might be a smaller value (idk, ask someone else im bad at these)
dusky quartz
pretty damn sure n=2561 is not allowed
weak cobalt
$x_i^2 < 1 \implies n > 2561$
boreal girderBOT
Nel
weak cobalt
So yes if you find a solution with n = 2562, you've solved it
dusky quartz
ig I'll just set x_odd num = sqrt(2561/2562) and x_even to be additive inverse
k thx
.close
sullen gorge
3 orders are
1° > 2° > 3°
3° > 2° > 1°
1° > 2° > 3°
Combining these three orders together, what is the final order?
compact pewterBOT
3 orders are
1° > 2° > 3°
3° > 2° > 1°
1° > 2° > 3°
Combining these three order...
upper bane
do you have the original question? i don't think i understand what this question even wants judging from the 1, 2, 3
covert crater
you can send a screenshot of the question
sullen gorge
If there are 3 orders as such, then if we consider all three, what would the final order be as a summary of the 3 orders
What will come first second and third
sullen gorge
Primary, Secondary and Tertiary amines
upper bane
this lowkey sounds like a biology/chemistry question
stuck haven
Yes it is
sullen gorge
And this is sort of a math question
stuck haven
sullen gorge
There is mathe evrywhere
stuck haven
well said
upper bane
in that case, could you please show the original problem as it is
stuck haven
but wheres the context
upper bane
maybe there's extra context behind the question that could help
stuck haven
mhm
sullen gorge
Context as in?
stuck haven
wdym by the order of amines, what kind of reaction is that
upper bane
cuz i don't think we can just rank these amines as if we're ranking racers or smth
sullen gorge
An order represents how many carbons an aniline is attached to
stuck haven
there should be order of acidity/basicity/sn1-sn2 reaction
upper bane
why not just post the original question though
sullen gorge
It's not a reaction
stuck haven
oh
sullen gorge
why not just post the original question though
That is the original question
upper bane
that's it? no paper, no website?
stuck haven
and how can three orders be combined?
wide moss
It's not a reaction
It gets kinda difficult to answer your question if you don't define what you mean by those terms beforehand doesn't it
covert crater
if it's a chemistry question, you can go ask in the chemistry server #old-network
stuck haven
Yeah im lost
covert crater
if it's a chemistry question, you can go ask in the chemistry server <#406129911...
most likely gonna get better help than in a maths server
upper bane
if you say this is math, then i assume you're asking us to somehow rank these amines, even though i have no idea how amines can be ranked
but idw to assume, so this question is super ambiguous as it is formulated here
covert crater
I think they are talking about primary amines, secondary amines and tertiary amines
sullen gorge
Yes it's about ranking
upper bane
i'm aware they are
stuck haven
if you say this is math, then i assume you're asking us to somehow rank these am...
Yeah well he said as in the number of carbons attatched to aniline, then again how can three inequality hold at the same time
upper bane
but ranking in what sense
stuck haven
Yes it's about ranking
in the basis of what
upper bane
and why are the first and third orders the same
the second order is just a reverse of the other two
so what kind of ranking is this, if it is a ranking?
sullen gorge
My teacher made me believe that they are ranked, but I belive that basicity rankings are not calculated but rather proven experimentally
sullen gorge
so what kind of ranking is this, if it is a ranking?
Just literally join then together, don't confuse the question
stuck haven
join?
sullen gorge
Say the 3 amine are given marks
upper bane
uhuh
sullen gorge
And are arranged in 3 descending orders
stuck haven
sullen gorge
Then statistically calculate what the final order might be
upper bane
so in the first and third tests, we get an order of 1 > 2 > 3
wide moss
Just literally join then together, don't confuse the question
None of this is clear from your question
upper bane
and in the second test we get an order of 3 > 2 > 1
stuck haven
so in the first and third tests, we get an order of 1 > 2 > 3
so if we take a mode that should be the final order
upper bane
assuming by "just join" that we are to rank them by score, then i suppose the final order is 1 > 2 > 3?
stuck haven
Yes
upper bane
1 tops the scoreboard twice and loses once
3 does the opposite
2 is just in the middle all the way
sullen gorge
I believe range average and all those things are from statistics so this is technically a math question
stuck haven
stuck haven
im crying
upper bane
other than the 1, 2, 3
stuck haven
we need inputs
stuck haven
Yeah so where are the basicity datas?
upper bane
in first run it could be pH 8 > 7 > 6
then in the second run it could be pH 14 > 7 > 1
sullen gorge
2 is just in the middle all the way
OK so what are your thoughts
upper bane
well, since your question is imprecise, my answer is also gonna be as imprecise
and if you accept imprecise answers, 1 > 2 > 3
stuck haven
Yea thats the final order I assume
sullen gorge
Yeah the final order can really just be anything (not really) which leads me to belive that this ain't calculated
stuck haven
upper bane
i have no idea, this is literally your question
stuck haven
Im really confused
upper bane
there's no data for this to be statistics, there's no numbers for this to be algebra
sullen gorge
But there is an order that we can use to assume the numbers
stuck haven
and the orders arent defined properly
sullen gorge
OK look at this for example
upper bane
you already gave the orders
unless now order means something different
but sure, send it over
sullen gorge
90 > 89 > 0
90 > 89 > 10
90 > 89 > 0
upper bane
isn't that what i said
but that is falling into the territory of assumptions
if we wanna assume, we can literally make the final order to be anything we want
stuck haven
I mean what do you even mean by final order
sullen gorge
So, Primary gets 180, secondary gets 267 and tertiary gets 100.
Which means the final order can be 2° > 1° > 3°
But on the contrary
stuck haven
ok ig you need inputs for that
upper bane
ok, i see what you mean
primary gets 190 btw
but without data
we can't tell what is what
sullen gorge
If it was
30 > 20 > 10
40 > 10 > 5
10 > 8> 2
Primary gets 45, secondary gets 38 and tertiary gets 52
Making it 3° > 1° > 2°
stuck haven
so basically
upper bane
again, that's exactly what i said
stuck haven
my question is
sullen gorge
So this just isn't right to assume like this
upper bane
but that's what you're having us do, you know
stuck haven
Do you have the data set when youre asked that?
upper bane
if we don't assume, there is no answer to your question. period.
stuck haven
otherwise the answer will vary
sullen gorge
So maybe my teacher is wrong to tell me that this final value is due to the other 3 orders, or maybe the order she gave us that 2° > 1° > 3° is the most general order that has the most outcomes
upper bane
i have no comment here
stuck haven
same
upper bane
you gotta ask your teacher
sullen gorge
😢
upper bane
this is basically trying to do statistics but without the data
every single statistician will boo us out of the lecture hall if we try
stuck haven
this is basically trying to do statistics but without the data
arbitary statistics😭🙏
compact pewterBOT
stuck haven
Yeah nws
young raft
i had a stroke reading all of that xddd
upper bane
statistics 101: please use data
compact pewterBOT
Available help channel!
winged dock
Let $f(x) = x^2 + 6x + c$ for all real number s$x$, where $c$ is some real number. For what values of $c$ does $f(f(x))$ have exactly $3$ distinct real roots?
compact pewterBOT
Let $f(x) = x^2 + 6x + c$ for all real number s$x$, where $c$ is some real numbe...
boreal girderBOT
Triaengle
cerulean steeple
you would presume 3 distinct real roots would be 2 distinct and 1 double root
kind viper
yeah given that deg f(f(x)) = 4
compact pewterBOT
@winged dock Has your question been resolved?
winged dock
Will differentiating this get me anywhere ??
hushed magnet
if g has a double root, then it is also a root of g'. so then gcd(g, g') is not 1. the gcd is easy to compute with the euclidean algorithm. tho whether its easy (and helpful) here with the c is a different question
compact pewterBOT
@winged dock Has your question been resolved?
winged dock
,w divide x^4 + 12x3 + (42+2c)x2 + x(36+12c) +c2 + 7c by x3 + 9x2 + (21+c)x + 3c+9
Hmmm
The value of c given by this is somehow the answer
<@&286206848099549185>
molten badger
Let $f(x) = x^2 + 6x + c$ for all real number s$x$, where $c$ is some real numbe...
,w (x^2+6x+(11-sqrt(13))\2)^2 + 6(x^2+6x+(11-sqrt(13))\2) + (11-sqrt(13))\2 =0
boreal girderBOT
molten badger
@winged dock It seems like $c=\frac{11-\sqrt{13}}{2}$ is the only solution here
boreal girderBOT
Alexis_Fx
molten badger
please correct me if I'm wrong, I'm not so sure about this
vast shoal
**Triaengle**
Basic function iteration?
This feels like our secondary school function question.
wait
let me think deeply
a, b, and c are all positive real numbers. Prove the following expression.
molten badger
uhh, what
molten badger
what's going on here
molten badger
does this suppose to relate to triaengle original question?
vast shoal
does this suppose to relate to triaengle original question?
Forgive me for not understanding the specific role of this group QAQ，so what the word triaengle actually means
vast shoal
emm i suppose i don't actually understand awa
molten badger
<@1261742170835128382> currently hold this channel
uh, so, he holds this channel, that mean you can't post anything unrelated to the topic he want to get help with
winged dock
<@1261742170835128382> It seems like $c=\frac{11-\sqrt{13}}{2}$ is the only sol...
Yes this is the answer
vast shoal
so what the topic()
molten badger
Yes this is the answer
yeah, have you solved it?
vast shoal
emm what about the problem i just posted above
winged dock
wolfram alpha solved the question but I don't now how
molten badger
emm what about the problem i just posted above
you can open another channel to get help
!help
compact pewterBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
molten badger
wolfram alpha solved the question but I don't now how
okay, so
a hint is $x^2 + 6x +c =0$, when does this has 1 root?
boreal girderBOT
Alexis_Fx
winged dock
,w GCD x^4 + 12x3 + (42+2c)x2 + x(36+12c) +c2 + 7c by x3 + 9x2 + (21+c)x + 3c+9
boreal girderBOT
molten badger
you won't have to deal with differentiate here
winged dock
a hint is $x^2 + 6x +c =0$, when does this has 1 root?
Hmmm but wait how does that help
molten badger
let u=x^2+6x+c then f(f(x))=u^2+6u+c, then it would have 2 root let say a and b then u=a and u=b one must has 1 roots and other one has 2
so we want to know how we can add -a to u so u-a=0 has 1 root
sick shard
x=-3±√(6-c-+√(9-c))
by quadratic formula
we want one root to vanish, let inner root be 0 → c=9
x=-3±√(6-9±0) complex roots
let outer root be 0
6-c-+√(9-c) = 0
-+√(9-c) = c-6
9-c = c²+36-12c
c²-11c+27 = 0
c=(11±√13)/2
which gives x that you got here
molten badger
you're cooking here haha, that's why you was typing for so long
sick shard
1st thing, you don't need to simplify
f(x)=x²+6x+c
f(f(x)) = 0
f²(x) + 6f(x) + c = 0
f(x) = ... (by formula)
x²+6x+c = ...
again quadratic
x = ...
,w (x^2+6x+(11+sqrt(13))\2)^2 + 6(x^2+6x+(11+sqrt(13))\2) + (11+sqrt(13))\2 =0
boreal girderBOT
sick shard
yea that's the only lol
molten badger
using Vieta, you could rule out (11+\sqrt13)\2
oh wait nvm
it can't
just make sure -(the roots) + c =< 9
sick shard
x=-3±√(6-c-+√(9-c))
by quadratic formula
we want one root to vanish, let inner ...
you didn't understand this?
compact pewterBOT
compact pewterBOT
Available help channel!
rocky sinew
need help w/ this, i'm pretty sure the answer is 4 but i'm not good with graphs
covert crater
need help w/ this, i'm pretty sure the answer is 4 but i'm not good with graphs
seems right but it's an estimate
so the answer varies
rocky sinew
yeah im not rlly sure tbh
covert crater
yeah im not rlly sure tbh
Generally they have a range for acceptable answers
rocky sinew
my school is insanely specfic tbh..
im rlly confused on this aswell but i kinda just guessed and i think its 2.75??
vivid breach
im rlly confused on this aswell but i kinda just guessed and i think its 2.75??
tamika's time is 2.75 in the second meet
ashlyn's time is 3.0 in the second meet
rocky sinew
tamika's time is 2.75 in the second meet
ashlyn's time is 3.0 in the second meet
where would they have the same time in the meet then.. it feels like a trick question cause i have like 3 questions with that same graph
vivid breach
which meet do you think that is
late mica
Does anyone know a good algorithm for checking if there are roots in a specific interval of a non-linear function?
covert crater
Does anyone know a good algorithm for checking if there are roots in a specific ...
!occupied
compact pewterBOT
Does anyone know a good algorithm for checking if there are roots in a specific ...
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
rocky sinew
i cld be wrong tho
vivid breach
tamika's time is 2.75 in the second meet
ashlyn's time is 3.0 in the second meet
well you can see in the second one their times are different
vivid breach
isnt it the 5th meet actually??
yes
wheat pasture
Generally they have a range for acceptable answers
(to be fair, as they draw a line of best fit for you here, I can see why they'd only accept 4 in this question (or, given you draw your own line, a value that's based on said line) 
)
compact pewterBOT
vivid breach
np
compact pewterBOT
Available help channel!
covert crater
(to be fair, as they draw a line of best fit for you here, I can see why they'd ...
Oh I initally thought they were the one who drawed it lol
compact pewterBOT
Oh I initally thought they were the one who drawed it lol
covert crater
.close
strange pendant
compact pewterBOT
strange pendant
how do I prove this implication
does the implication is a synonym for subseteq?
verbal rover
hushed magnet
you are supposed to show that Bn C subset (A symm diff B)^c. and somewhere along the way you are allowed to use that C subset A
its just a normal inclusion proof
strange pendant
so how do I do it?
hushed magnet
where you are allowed to use an extra property
strange pendant
let x in BnC
strange pendant
its just a normal inclusion proof
I need handhold
turbid surge
Hii renato
turbid surge
I managed to enter university somehow
Omg LETSGOOOO congrats
turbid surge
now I am trying to survive it
Welcome to the club
magic wren
Take an element x from B intersection C. Note that x is in C. Thus by assumption x is in A.
strange pendant
x in BnCnA
strange pendant
You need to show that x is not in A\B and that x is not in B\A
? the complement of the symdiff?
A symdiff B = (A - B) U (B - A)
magic wren
yeah to show that x is in the complement of that set, you need to show that it is not in A-B and not in B-A
magic wren
To show that x is in (A symdiff B)^C, you need to show that x is not in A symdiff B.
To show that x is not in A symdiff B, you need to show that x is not in A - B and not in B - A
strange pendant
strange pendant
To show that x is in (A symdiff B)^C, you need to show that x is *not* in A symd...
you are so right, im so stupid jajjaa
I was overcomplicating everything
strange pendant
You need to show that x is not in A\B and that x is not in B\A
how?
full moat
With respect to what are you taking your complement?
full moat
with respect to the universal set
With respect to V?
strange pendant
yes
full moat
Welp, ok
full moat
V is the reference set for A,B,C
That's not the universal "set" then, though
Y por cierto, hablo Español si lo prefieres
full moat
Bueno, primero, el que no sea necesariamente cierto ya es un cambio
strange pendant
That's not the universal "set" then, though
it is
@full moat @magic wren
@full moat @full moat
full moat
https://es.wikipedia.org/wiki/Conjunto_universal
Bueno, está bien, pero por lo general V es la clase propia de los conjuntos, V={x:x=x}
magic wren
To show that x is in (A symdiff B)^C, you need to show that x is *not* in A symd...
Assume x is in A-B. Since x is in A, we see that x is not in B. This is a contradiction. Thus x is not in A-B.
full moat
Igualmente, dame un segundo para que intente tu problema, no debería ser terrible
strange pendant
Ok gracias gordo
strange pendant
Assume x is in A-B. Since x is in A, we see that x is not in B. This is a contra...
wdym?
magic wren
If you have an element of A-B, and that same element is an element of A, then it cannot be an element of B
because if it were, we would have an element in A-B that is in both A and B
strange pendant
To show that x is in (A symdiff B)^C, you need to show that x is not in A symdiff B.
To show that x is not in A symdiff B, you need to show that x is not in A - B and not in B - A
magic wren
yes
strange pendant
yes
To show that x is not in A symdiff B, you need to show that x is not in A - B and not in B - A
how does that follow?
strange pendant
,, A \triangle B = (A-B) \cup (B-A)
boreal girderBOT
Renato
strange pendant
well I just showed the first part
how did the and appear
it should be an or
@magic wren
full moat
Ok, ya está
magic wren
A - B U B - A <=> x is an element of that set if it is in A-B or B-A
Negating that statement gives
x is an element of that set if [neg (it is in A-B or B-A) ]
<=> x is an element of that set if it is not in A-B and not in B-A
de morgan's laws
full moat
Notemos que (AΔB)^c=V\ ((AUB)\ (A∩B))
strange pendant
,, A \triangle B = (A-B) \cup (B-A)
\ (A \triangle B)^c = (A - B)^c \cap (B-A)^c
boreal girderBOT
Renato
magic wren
yes
full moat
Lo que a su vez implica (AΔB)^c=(V(AUB))U(A∩B)
strange pendant
then what stipdendi
<=> x is an element of that set if it is not in A-B and not in B-A
ok I get that
magic wren
Assume x is in A-B. Since x is in A, we see that x is not in B. This is a contra...
then this
recall that x is in A B and C
strange pendant
Assume x is in A-B. Since x is in A, we see that x is not in B. This is a contra...
so what?
contradiction
full moat
Luego sea x∈B∩C, se tiene que C⊆A, luego (x∈B)^(x∈C), mas esto implica (x∈B)^(x∈A), luego x∈A∩B⊆(AΔB)^c
strange pendant
?
full moat
Luego sea x∈B∩C, se tiene que C⊆A, luego (x∈B)^(x∈C), mas esto implica (x∈B)^(x∈...
Y ya está
magic wren
x in BnCnA
The contradiction comes from this
magic wren
The assumption that x is in A-B was false
strange pendant
can you guys put the proof in a full message and not in between messages
I am getting lost for a sec
though I appreciate the help I am being provided
full moat
Alright, I'll condense what I said in one message
Note that: (AΔB)^c=V\ ((AUB)\ (A∩B))
This then implies that: (AΔB)^c=(V \ (AUB))U(A∩B)
Thus let x∈B∩C, note that (x∈B)^(x∈C), but C⊆A, however, this implies (x∈B)^(x∈A), so finally we get x∈A∩B⊆(AΔB)^c
In case the notation is a little too messy, the key here is realizing that B∩C⊆B∩A, and that necessarily B∩A is in the complement
magic wren
Given: C subset A
Show: B n C subset (A symdiff B)^c
Take x in B n C.
x is in C.
By assumption x is in A.
(A symdiff B)^c = ( (A-B) u (B-A) )^c
= (A-B)^c n (B-A)^c
Show: (1) x is not in A-B and (2) x is not in B-A.
(1) x is in A and x is in B (we deduced in the beginning). Thus x in A-B is impossible, and therefore x not in A-B.
(2) x is in A and x is in B (we deduced in the beginning). Thus x in B-A is impossible, and therefore x not in B-A.
Therefore x in (A-B)^c n (B-A)^c.
Therefore x in (A symdiff B)^c.
Therefore B n C subset (A symdiff B)^c, which was to be shown.
strange pendant
Assume x is in A-B. Since x is in A, we see that x is not in B. This is a contra...
IT IS STILL POSSIBLE THAT X IS IN B - A
magic wren
perfect
compact pewterBOT
@strange pendant Has your question been resolved?
shell meteor
help! i factored the denominator and got it correct. but i expanded x^2-9 into x^+6x-9. i then factored it into (x-3)^2. where did i go worng
compact pewterBOT
help! i factored the denominator and got it correct. but i expanded x^2-9 into x...
cold mural
did you mean the denominator? x^2 - 9 is the numerator
shell meteor
oh yea my mistake
cold mural
yeah then the numerator is a difference of squares if you recall the formula for that
fathom crest
If you wanted to expand $x^2-9$ to $x^2+6x-9$, there are two mistakes. 1. you missed the $-6x$, since $$x^2-9=x^2-9+6x-6x=x^2+6x-9-6x$$ 2. you still can't factorize, since $$x^2+6x-9 \neq (x-3)^2=x^2-6x+9$$
cold mural
when you turned it into x^2 + 6x - 9 you added a 6x but didn't subtract it
boreal girderBOT
Good
fathom crest
Recall $$a^2-b^2=(a+b)(a-b)$$ which is what the solution did.
boreal girderBOT
Good
compact pewterBOT
Available help channel!
shell meteor
i understand how the numerator is converted to a/1 but im confused how the denominator is able to be rearranged like that?
compact pewterBOT
i understand how the numerator is converted to a/1 but im confused how the denom...
quartz yoke
as in what way?
shell meteor
how does (1/b)+c be rewritten as 1+bc/b
keen granite
how does (1/b)+c be rewritten as 1+bc/b
(\frac1b+c=\frac1b+\frac{bc}b=\frac{1+bc}b)
boreal girderBOT
PajamaMamaLlama
shell meteor
no sorry could you explain it a little more. are you saying since the c is multiplied by b in the numerator the b in the denominator gets cancelled out
keen granite
yes, but we're doing it backwards
because we can only add fractions if they have the same denominator
we introduce the extra b term on top and bottom
shell meteor
ohhhhhhh ok
keen granite
so the last step is algebraically valid
shell meteor
im sorry im a little fuzzy on how equations can be manipulated in these manners
keen granite
all good part of the learning process 
shell meteor
thx
strange pendant
compact pewterBOT
weak cobalt
Can you rewrite X - Y using only union, intersection, and complement?
weak cobalt
Yeah that's nice
weak cobalt
It's pretty much exactly what you're asked to do
strange pendant
can u help me derive the left side from the right side
weak cobalt
Just do the same thing you did but from bottom to top?
strange pendant
Just do the same thing you did but from bottom to top?
oh you are right dude
xDDDDDD
.solved
compact pewterBOT
compact pewterBOT
Available help channel!
sterile oracle
would u say i solved this correctly?
compact pewterBOT
would u say i solved this correctly?
sterile oracle
oops ss so small the problem is f(t) = 1/t, formula is f(t) - f(1) over t+2 and t cannot be equal to 1
molten badger
would u say i solved this correctly?
sterile oracle
stop 💔
molten badger
we simplify $\frac{f(t)-f(1)}{t-1}$ ?
boreal girderBOT
Alexis_Fx
bleak flower
sturdy reef
agreed!
subtle helm
how did the denom turn into t+2?
$\frac{\frac{1}{t} - 1}{t-1}$ not $\frac{\frac{1}{t}-1}{t+2}$, no???
boreal girderBOT
k
subtle helm
@sterile oracle
sterile oracle
how did the denom turn into t+2?
i noticed afterwards im dumb ye its supposed to be 1 not 2
subtle helm
also
i understand that u are trying to simplify the numerator
but using the equal sign is kinda wrong here
exotic prism
Hello!
So I have a function
f(x) = x^2 - 6
And got reflected by
y = x
The reflection is g(x)
Both f(x) and g(x) intercept at A and B
The question is asking A and B coordinate
since y = x
I just need to change
y = x^2 - 6
Into
x = x^2 - 6
And then factor it
X = x^2 - 6
0 = x^2 - x - 6
(x-3)(x+2)
So It's
A ( 3, 3 )
B ( -2, -2 )
-
Why i can do that? Why i can change y = x^2 - 6 Into x = x^2 - 6
-
Or maybe, this is actually false?
-
I need better understanding of this y = x reflection intercept. Maybe drawing or make another example could help
Thank you!
compact pewterBOT
Hello!
So I have a function
f(x) = x^2 - 6
And got reflected by
y = x
The refle...
spark folio
Why i can do that? Why i can change y = x^2 - 6 Into x = x^2 - 6
Simple enough to see, you already know that the specific points that you are looking for are on the line y=x
so for evey such point, you have the same x and y co-ordinates, so to find the specific values, you can easily replace y with x (since both are equal)
However, do note that it wont be true in general. This is possible only in this case since the line is y=x. If it were some different line, say 3y = 2x + 4, then you have to change the y appropriately (to (2x+4)/3 for this case).
Try to fiddle with the plots on some website like desmos
compact pewterBOT
compact pewterBOT
Available help channel!
vast shoal
how to get the max and min of the function
compact pewterBOT
how to get the max and min of the function
vast shoal
one method is that using a=cos2x,b=cos2y
i wonder is there any other methods
<@&286206848099549185>
compact pewterBOT
@vast shoal Has your question been resolved?
wind iron
one method is that using a=cos2x,b=cos2y
you can easily prove that f(x,y) <= 3*sqrt(8) and you have equality for f(0,0)
maybe you can do something similar for the minimum
compact pewterBOT
@vast shoal Has your question been resolved?
compact pewterBOT
@vast shoal Has your question been resolved?
vast shoal
.close
compact pewterBOT
compact pewterBOT
Available help channel!
fresh ginkgo
how come this legal for linear trans. T and Id?
compact pewterBOT
how come this legal for linear trans. T and Id?
fresh ginkgo
it seems wrong
this is what he used it to do
what he did means that i can just switch the order of any "combination" of linear trans.'s as long as i take out T
and get polinom form
compact pewterBOT
@fresh ginkgo Has your question been resolved?
compact pewterBOT
compact pewterBOT
Available help channel!
thorny kestrel
how come this legal for linear trans. T and Id?
because T and its powers commute
compact pewterBOT
because T and its powers commute
thorny kestrel
for example
$X (X+1) (T) = (X^2 + X) (T) = T^2 + T = (T + id)T = (X+1)X (T) $
, $X (X+1) (T) = (X^2 + X) (T) = T^2 + T = (T + id)T = (X+1)X (T) $
hopefully you get what I mean
.close
compact pewterBOT
compact pewterBOT
Available help channel!
lofty mountain
Teacher did it by taking option..
i want to understand solution with basic
compact pewterBOT
Teacher did it by taking option..
i want to understand solution with basic
upper bane
btw i think this is the third time i have seen this exact image
bleak dock
btw i think this is the third time i have seen this exact image
indeed
upper bane
#help-46 message
once here
bleak dock
https://discord.com/channels/268882317391429632/1018700692027883620/140483631391...
the helper there messed up cause they didn't realise it was simple interest
upper bane
true that
lofty mountain
i don't get the solution..😔
upper bane
#help-8 message
did you see this message?
bleak dock
exactly
which part of:
Since he paid 3000 (after three years)+5450(after five years)=8450.
Interest he has given is, 8450-7000=1450..
Which is interest on 7000 for 3 years and on 4000 for 2 years combined. Equate both and you'll get R.
don't you understand?
shadow burrow
6 maybe
bleak dock
the key is that the interest is calculated on the principal
at first the principal is Rs 7000
but after he pays off Rs 3000, he has Rs 4000 to pay off, and then the new principal is Rs 4000
so the 1450 is a combination of 7000 * (r%) for 3 years
and 4000 * (r%) for 5 - 3 = 2 years
compact pewterBOT
@lofty mountain Has your question been resolved?
upper bane
OP you seem to have marked it as incomplete, but have not provided any questions or input
lofty mountain
which part of:
> Since he paid 3000 (after three years)+5450(after five years)=8...
hana..
i dont get the last 7000 for 3 years and 4000 for 2 year one
bleak dock
hana..
i dont get the last 7000 for 3 years and 4000 for 2 year one
he pays everything off in 5 years
so he pays off the 7000 loan for 3 years
that means he pays off the 7000 - 3000 = 4000 loan for 2 years
he reduces his loan from 7000 to 4000, by paying off 3000
lofty mountain
he reduces his loan from 7000 to 4000, by paying off 3000
is there any int. given on 3000?
bleak dock
is there any int. given on 3000?
of course
it's the same percentage interest as the one for 7000
bleak dock
so the 1450 is a combination of 7000 * (r%) for 3 years
*and* 4000 * (r%) for 5...
which is why you can use the same variable, r, twice
upper bane
ok so
when you pay off part of a loan
you reduce the principal, but not the interest rate
you pay less on interest now because your principal has reduced
lofty mountain
so the 1450 is a combination of 7000 * (r%) for 3 years
*and* 4000 * (r%) for 5...
ohok
now i get it
i just have to equal the 7000 int. and 4000int to 1450
bleak dock
so the 1450 is a combination of 7000 * (r%) for 3 years
*and* 4000 * (r%) for 5...
yes then I've given you the algebraic equation in r, if you use your brain a bit
lofty mountain
let me do another question using this method
bleak dock
this question suffers from a lack of punctuation
lofty mountain
there are total 4 questions by this language
lofty mountain
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lofty mountain
.reopen
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lofty mountain
this question suffers from a lack of punctuation
.close
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Available help channel!
placid spruce
is there a faster way to solve without manually factoring each choice
compact pewterBOT
is there a faster way to solve without manually factoring each choice
vivid breach
plug in -2b into each answer
placid spruce
-2b as x?