b is 3 here
#help-26
crystal goblet
crystal goblet
Its not because I said there should be a negative, you should try to figure out why the negative is there
gleaming cape
well maybe it'[s because in the equation y = a(x+4)(x-1)
we just have to combine -4 and 1?
crystal goblet
Yes, when you expand it, you get ax^2+3ax-4a
gleaming cape
why is there an a after 3 and -4
crystal goblet
Yes, when you expand it, you get ax^2+3ax-4a
And we are told that this is equal to x^2+bx+c
If you compare the coefficients of x^2, a should be 1
gleaming cape
OH CUZ WE EXPLANDED IT WITHOUT KNOWING A
crystal goblet
Yes
gleaming cape
ok thanks a lot i understand now
i didn't realize we could just divide everything by a and completely get rid of it
wait a second
can we?
crystal goblet
No, we have deduced that ax^2+3ax-4a = x^2+bx+c for all x, which means the coefficients have to be the same
The coefficient of x^2 in the right hand side is 1 and in the left hand side is a, so a=1
gleaming cape
so in this case
No other case tho
no case when a =/ 1
anyway no worries i got it
thank
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gleaming cape
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lucid tiger
How would i start this?
topaz sinewBOT
How would i start this?
cursive patrol
lucid tiger
Actually i think i know how to do it correct me if im wrong. It's (6 choose 2) because you have 6 numbers to choose for for x1 and x2. then its (10 choose 2) because you have 8-17 numbers to choose from for x4 and x5
eternal wing
Yeah that's right
Do you understand why this is counting the numbers where x_1 < x_2 < x_3 < x_4 < x_5?
lucid tiger
yea thats how i arrived at my solution
eternal wing
Yeah but do you understand why (6 choose 2) corresponds to the number of ways to choose x_1, x_2 such that x_1 < x_2?
eternal wing
I don't really understand your explanation
But I'll just say, that when you choose x_1 and x_2 from {1, 2, ..., 6}
You can take x_1 to be the smaller one, and x_2 to be the bigger one
And the choose function doesn't care about the order
lucid tiger
yea sorry thats what i meant. coudnt write it down properly
and we have identical process for x_4 and x_5
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coarse tusk
https://math.stackexchange.com/questions/3943934/why-is-it-okay-to-take-out-the-constant-when-finding-asymptotes-or-hyperbola
in the answer given by Z Ahmed here, they said that $$L_1 L_2=\pm s^2$$ is the most general form of a hyperbola, where $L_1=0$ and $L_2=0$ are any two lines and s is any real number
topaz sinewBOT
https://math.stackexchange.com/questions/3943934/why-is-it-okay-to-take-out-the-...
thorny flameBOT
kheerii
coarse tusk
my question is, why does this form represent every possible hyperbola? how can we prove this?
I haven't been able to prove this myself. Something I tried is rawdogging the entire thing by taking general forms for L1 and L2 and trying to convert it to the form $$(x-a)^2+(y-b)^2=e^2\frac{(lx+my+n)^2}{l^2+m^2}, e>1$$ which obviously didn't work
thorny flameBOT
kheerii
coarse tusk
my next thought was that there might be some series of linear transformations (like shown in the famous Matt Parker video about parabolas) which may lead to this same result
I'm not very familiar with Linear Transformations though, which is why I need someone's help
coarse tusk
my next thought was that there might be some series of linear transformations (l...
https://www.youtube.com/watch?v=hoh4TmPzu1w this is the one I am talking about btw
topaz sinewBOT
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coarse tusk
<@&286206848099549185>
neon iron
ya
topaz sinewBOT
@coarse tusk Has your question been resolved?
topaz sinewBOT
@coarse tusk Has your question been resolved?
weary crystal
I'm not very familiar with Linear Transformations though, which is why I need so...
Okay, well if you want to show that that's the general form of any parabola, $L_1 = ax+by+c, L_2 = dx+ey+f$, then complete the square
thorny flameBOT
doaby
weary crystal
that should work
or something along those lines, im assuming you just want to reduce that to the form of a hyperbolic equation
topaz sinewBOT
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junior tundra
Hello
junior tundra
I'm working on this problem
but I'm not sure exactly where to start
when I look online, a lot of people are using the disjunctive and conjunctive syllogisms, but our class hasnt covered those topics yet
so far, all i have to work with are a table of common valid argument forms, and a table of some logical equivalencies
Is it possible to prove the conclusion "∴ u ∧ w" from these resources, or am i missing some necessary tools?
I'm also curious if there's a more systematic way to approach this problem, rather than guessing and checking
topaz sinewBOT
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sudden vapor
i know this is probably beyond the scope of a help channel but what does it mean if the kl divergence between two distributions is the same non-zero value forwards and backwards?
topaz sinewBOT
i know this is probably beyond the scope of a help channel but what does it mean...
sudden vapor
does it mean anything special?
topaz sinewBOT
@sudden vapor Has your question been resolved?
topaz sinewBOT
sudden vapor
fickle needle
it just wants to be drank
slate bronze
Close this channel
Don't waste time
restive inlet
correct me if i'm wrong but that's equal to
1 <@&268886789983436800> ping
sudden vapor
lol
urban grove
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sharp dew
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sharp dew
Sounds good and reasonable?
Isolate that shit
I might be tripping
Forget about what I had said
How to solve questions like this?
verbal crater
isolate t
verbal crater
yes
sharp dew
Why should we do this?
verbal crater
do what
sharp dew
Transferring that shit into base-10 log instead a base-3 or some
verbal crater
most calculators only have buttons for log 10
sharp dew
most calculators only have buttons for log 10
Ohh
sharp dew
most calculators only have buttons for log 10
May I ask why
verbal crater
it just means that i dont fucking know lol
sharp dew
it just means that i dont fucking know lol
Why because he’s ugly?
verbal crater
¯_(ツ)_/¯
sharp dew
What a cocky human, arrogant
verbal crater
but yeah base 10 logs is what most people use
sharp dew
I see
verbal crater
along with the natural log
sharp dew
I should have canceled out these common factors on both sides
Before I transferred them into log shits
For calculation convenience
Is that true?
verbal crater
yeah
verbal crater
because smaller numbers are easier to work with
sharp dew
because smaller numbers are easier to work with
That’s ambiguous
sharp dew
because smaller numbers are easier to work with
Why
Why more smaller, more easier to work with
verbal crater
you mean why we should cancel those common factors
sharp dew
Yes before we turn them into logs
verbal crater
i mean you dont have to
verbal crater
it is usually recommended to do that
sharp dew
it is usually recommended to do that
Why
sharp dew
it is usually recommended to do that
Why
sharp dew
do what makes you happy
No, I think it is about efficiency
verbal crater
and that
sharp dew
You have to prove that employing such a move can grant more efficiency In calculations
Than do not employing it
Is all of these canceling out common factors would be harder to (take more steps) achieve after we transferring it into logarithm
Be objective
verbal crater
hmm
technically yes because you would need to do an extra step of cancelling those common factors
sharp dew
technically yes because you would need to do an extra step of cancelling those c...
What kind of “extra steps”, would you mind making a example?
Like be specific
verbal crater
like 10000/22500
you first figure out what common factors are there
you can see 100 is a common factor
sharp dew
Yes
verbal crater
dividing that you get 100/225
then you have to figure out what the common factor of these two numbers are
spoilers: its 25
sharp dew
It is 5^2 I suppose
Yes
Then we compare that with log simplification
And judge them objectively
Log(10000/22500)
sharp dew
Log(10000/22500)
Cannot I do the exactly same thing?
sharp dew
Then why would you claim that simplification would be harder in log form?
sharp dew
Then why would you claim that simplification would be harder in log form?
^
Is it cognitive bias?
Why would you claim that
verbal crater
Then why would you claim that simplification would be harder in log form?
oh so you're questioning the order of the steps
?
sharp dew
oh so you're questioning the order of the steps
It seems like both ways are almost identical, but one is expressed in log form while the other is not
They are the same in terms of complexity, the simplification
Which poses a problem to your claim
verbal crater
well then i formally apologise for my logical error
sharp dew
As you states that we should cancel these common factors out before taking it into log form
And it would be easier to operate
sharp dew
well then i formally apologise for my logical error
What specific logical error is it? Would you categorize it as confirmation bias?
I mean you have been told/ recommended such manipulation can simplify the operation
You have been told and employing the thing repetitively.
That prompts the confirmation bias in your mind
Implant a bias in your mind
verbal crater
fucking hell what was that
anyways point is
i prefer simplifying first then take log
if you prefer first log then simplify
so be it, it works both ways
if i said one way is easier than the other
then im sorry
sharp dew
if i said one way is easier than the other
It is fine, because I have been told the exactly same thing when dealing with log questions in the past
verbal crater
also one reason i like simplify first then log: you might make some unneeded mistakes with the log, example you might add or multiply a 10 outside the log because you thought log(100) = 10 and you need to seperate that
but if you're careful, then go ahead and do what you like
not exactly which one is more efficient, its which one is less likely to make mistakes
unless you think that less mistakes = more efficient then it doesnt matter
sharp dew
Yes, as arguably most people are more familiar with non-log operations
That might be a reason to do this tho, to justify the move
To avoid mistakes, as you stated
verbal crater
but again if you want to just take log(10000/22500), so be it, but if you want to simplify that make sure anything that happens inside the log stays inside
sharp dew
I see
I will close the channel now
As my question is solved already
Thank you for the discussion
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on how i can go about this
vale ingot
What have you tried so far?
thorny heath
What have you tried so far?
proof by contradiction
so basically
assume that there exists a positive integer k such that:
k is not a divisor of (k - 1)!,
k is not a perfect square, and
k is not a prime number
but i got stuck after
vale ingot
Hmm, contradiction is a solid first thing to try but I'm not sure if it's the best approach here
thorny heath
tru
vale ingot
Well you have 3 cases, k is prime, k is a perfect square, and k | (k-1)!
thorny heath
Yea
vale ingot
But in another way, you could say that if k is not prime and k is not a perfect square, then k must divide (k-1)!
weary crystal
induction could also be fun :p
thorny heath
😭
vale ingot
Hmm, idk divisibility proofs don't play nice with induction
thorny heath
yeah kinda tricky
vale ingot
But do you understand how saying that "k not being prime and not being a perfect square implies that k | (k-1)!" is essentially the same as the question above?
weary crystal
ah so you're saying to assume 2 are false and check the last is true, then you have 3 cases
vale ingot
Basically yeah
weary crystal
I'm sure that would work, that's a good way of looking at it
thorny heath
But do you understand how saying that "k not being prime and not being a perfect...
i think so, cause i can start by assuming $k$ is a positive integer that is not prime and not a perfect square. Then $k$ has at least two distinct prime factors.
then if i let $p$ and $q$ be two distinct prime factors of $k$. Since $k$ is not a perfect square, both $p$ and $q$ must appear in the prime factorization of $k$ with an odd power
if i consider the factorization of $(k - 1)!$:
[
(k - 1)! = 2^{a_1} \cdot 3^{a_2} \cdot 5^{a_3} \cdot \ldots \cdot p^{a_p} \cdot q^{a_q} \cdot \ldots
]
thorny flameBOT
q1z
thorny heath
up until this point
the blueprint is sorta set
but id have to do this case assuming 2 are false
vale ingot
I think you've got the idea, think you're good to grind away at this for a bit?
weary crystal
trying to use the fact that k does not divide (k-1)! to prove 2 things seems awfully tricky
I bet there's a much faster way
vale ingot
Is this due tonight?
thorny heath
Is this due tonight?
yea
weary crystal
proof by contradiction?
maybe? assume none of the above hold and see what happens
but still would be very icky
the first condition is an issue
vale ingot
Hmmm, doaby do you think you can handle this?
I don't think I can give any more good advice without just giving the answer
weary crystal
I've got no clue, at this point I'm brainstorming with them lol
thorny heath
😭
vale ingot
Alright lemme think...
thorny heath
what if we use [
M = k \cdot (k - 1)! = 2^{b_1} \cdot 3^{b_2} \cdot 5^{b_3} \cdot \ldots \cdot p^{b_p} \cdot q^{b_q} \cdot \ldots \cdot (k - 1)!
]
thorny flameBOT
q1z
weary crystal
seems fairly trivial actually for even k, maybe you can do something more if you know it's odd?
vale ingot
So a way to prove "A or B" is true is to provide a proof that one of them is always true, or that one being false implies the other is true
thorny heath
seems fairly trivial actually for even k, maybe you can do something more if you...
if ks odd
then k - 1
is
even
WAIT
weary crystal
oh
vale ingot
seems fairly trivial actually for even k, maybe you can do something more if you...
Elaborate on this
thorny heath
NO WAYA
vale ingot
Oh did you get it?
thorny heath
nah my eyes decieved me
weary crystal
lmao
thorny heath
false hope
vale ingot
lmao fair enough
Okay, but doaby can you be explicit for me?
Why is it trivial if k is even?
thorny heath
doabys cooking
weary crystal
Elaborate on this
for even k, it's trivial if k = 2 since 2 is prime. then if k is even and larger than 2, i'm pretty sure you can check k|(k-1)! is always true
use the first few even numbers, it def works
just gotta prove that
vale ingot
Show me how it works for the first few even numbers
q1z do you think you can guess why it's trivial if k is even and greater than 2?
thorny heath
ok so if $k$ is even and larger than $2$, it can be expressed as $k = 2m$ for some positive integer $m \geq 2$. but then if we consider $(k-1)!$:
[
(k-1)! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2m-2) \cdot (2m-1)
]
Since $k = 2m$, the factors $2$ and $m$ would just be this no?
[
(k-1)! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2m-2) \cdot (2m-1) = 2 \cdot m \cdot \text{{other terms}}
]
which would imply that $k \mid (k-1)!$ for even $k$ larger than $2$
statement basically holds for both odd and even $k$. If $k$ is odd, $k \mid (k-1)!$ due to the even factor in $(k-1)!$, and if $k$ is even and larger than $2$, $k \mid (k-1)!$.
thorny flameBOT
q1z
weary crystal
there ya go
vale ingot
Fantastic!
weary crystal
we cooked
vale ingot
Now do you think you can extend this to any k that isn't prime or a perfect square?
thorny heath
YOU GUYS COOKED 🔥 🧑
weary crystal
now I can sleep lmao
thorny heath
Now do you think you can extend this to any k that isn't prime or a perfect squa...
..oh, forget about that
vale ingot
You are legitimately on the right track!
weary crystal
if k is odd, you get to throw away a prime factor of 2, which will almost def make things easier
vale ingot
Think about multiples of 3 larger than 9
Also it should've been even k greater than 4, that was my bad 😛
thorny heath
Also it should've been even k greater than 4, that was my bad 😛
sorry where
thorny heath
if k is odd, you get to throw away a prime factor of 2, which will almost def ma...
well if its odd k = 2m+ 1
[
(k-1)! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2m) \cdot (2m + 1)
]
thorny flameBOT
q1z
vale ingot
Oh we were talking about even k greater than 2, but 4 doesn't divide 3!
So it has to be even k greater than 4
weary crystal
touche
vale ingot
But 4 is a perfect square so we're still good
weary crystal
yep
thorny heath
Since $k = 2m + 1$, the factor $2$ is present in the product:
[
(k-1)! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (2m) \cdot (2m + 1) = 2 \cdot \text{{other terms}}
]
thorny flameBOT
q1z
thorny flameBOT
q1z
vale ingot
I don't think that holds, since it would imply that a prime k divides (k-1)!
vale ingot
ok so if $k$ is even and larger than $2$, it can be expressed as $k = 2m$ for so...
But this is legit on the right track
thorny heath
But this is legit on the right track
and that would be for the case
where we said
2 false, 1 true
weary crystal
I sleep now, gl!
weary crystal
any time :p
vale ingot
Yeah, we want to show that k is not prime, and k is not a perfect square implies that k | (k-1)!
Good night 🙂
vale ingot
You are legit super close!
weary crystal
noooo just finish it tmrw!!
thorny heath
ill try 🥲
vale ingot
I'm still down to help if you're up for it
thorny heath
i would but its getting rlly late 😭
vale ingot
Well in that case I'll tell you this
Your proof that an even k that is greater than 2 is 99% right, it just needs to be greater than 4 (again, sorry about that)
thorny heath
no worries
vale ingot
Because you said k = 2m, and 2 and m are parts of (k-1)!
I would say to try to extend that to 3, 4, and any other number
thorny heath
thanks
vale ingot
No prob 🙂
thorny heath
so for the other 2 cases
i shoudl aim to
like
show k is not prime, and k is not a
oh wait
vale ingot
Actually, the beauty of showing 2 being false implies that the 3rd is true means that you don't have to worry about showing anything else is true
Because it would mean that if the 3rd were false, the 1st or 2nd must be true
vale ingot
A => B is the same as ~B => ~A
weary crystal
actually i just thought, and if k is any odd number that isn't prime, k | (k-1)!
even squares have this property
vale ingot
9 doesn't divide 8!
weary crystal
9 | 8! since 8! has 3 and 6
vale ingot
Oh wait you're right lmao
weary crystal
yeah lol
vale ingot
But it still matters for 4
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unkempt ridge
unkempt ridge
3x-5 + 13/x+3 is what I got which is wrong
cyan mesa
3x-5 + 13/x+3 is what I got which is wrong
It is right
Reread the last bit of the question
unkempt ridge
3x-5 + 13/x+3 is what I got which is wrong
I got to this ^ but what do I do after?
cyan mesa
Multiply by (x+3)
And whatever is left is the answer
Don’t distribute
(3x-5)(x+3) + 13
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fervent terrace
Hi, i want to check that the answer to this question may be 6x/2x-5. Or if it is something else please let me know
topaz sinewBOT
Hi, i want to check that the answer to this question may be 6x/2x-5. Or if it is...
fervent terrace
@haughty drum
boreal mulch
Hi, i want to check that the answer to this question may be 6x/2x-5. Or if it is...
I got that too
topaz sinewBOT
@fervent terrace Has your question been resolved?
gray jungle
Hi, i want to check that the answer to this question may be 6x/2x-5. Or if it is...
yes the answer is indeed 6x/2x-5
boreal mulch
can someone briefly explain why tan(x) cycles every pi? im kinda tired and can't really think it through, thanks
topaz sinewBOT
can someone briefly explain why tan(x) cycles every pi? im kinda tired and can't...
lapis raft
Do you know the visualisation of tan?
boreal mulch
Do you know the visualisation of tan?
yes
but that's a result of sinx/cosx
you mean the graph or
lapis raft
I mean no, in circles
boreal mulch
I mean no, in circles
unit circle?
lapis raft
Yes
lapis raft
By adding pi, you rotate the point by 180 degrees, which basically doesn't change anything
boreal mulch
what does it look like when rotated by pi
lapis raft
From x,y it goes to -x, -y
Or, you can suppose signs of cos(x) and sin(x) changing
Try deriving tan(x) = tan(x+pi)
boreal mulch
so -sinx/-cosx = tanx
lapis raft
Yes
lapis raft
Oh yes, and you missed division symbol
lapis raft
Yep
boreal mulch
Yep
ah i see thanks
i didnt know tangent could actually be drawn haha
tysm
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north cloud
easy
umbral canyon
Good day, I assume that I can ask a question here.
I am learning about person's r and t-test.
The formula for the t-test is on the attached image.
My question is, what would happen if r equals to 1 or -1?
Solving it normally would lead to a math error. I have searched the internet for answers but I can't find any. In that case, can I assume that 1 or -1 to be +-0.9999999...?
thank you
topaz sinewBOT
Good day, I assume that I can ask a question here.
I am learning about person's ...
fair thorn
if r = -1 or 1, the function is undefined
however, if you take the limit, it'll b smth else
hope that helps
wait hang on'
shrewd horizon
if r is 1 or -1 then surely the limit would just be inf or -inf, unless n = 2
fair thorn
yep
shrewd horizon
considering that r < -1 or r > 1 also doesn't really work, unless you're allowing complex numbers, it might just be that only -1 < r < 1 makes sense
umbral canyon
considering that r < -1 or r > 1 *also* doesn't really work, unless you're allow...
so i cannot proceed to t-test if the value of the person's r has a perfect positive/negative linear relationship?
topaz sinewBOT
@umbral canyon Has your question been resolved?
warped zealot
topaz sinewBOT
warped zealot
I need help understanding this solution
specifically on the 'when the original distance separation is doubled'
I dont see any step on the solution where the seperation distance was doubled
topaz sinewBOT
@warped zealot Has your question been resolved?
topaz sinewBOT
@warped zealot Has your question been resolved?
thorny heath
@weary crystal thoughts?
warped zealot
wrong channel dude but whatever
weary crystal
<@297564071794835457> thoughts?
okay you're missing a lot of stuff here but you have the right idea
thorny heath
okay you're missing a lot of stuff here but you have the right idea
sorry in which case
weary crystal
for part (b), your first sentence just doesn't make sense lol so I'd delete it
weary crystal
wanna start a new channel tho? you kinda yoinked someone else's lol
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outer carbon
Hi
outer carbon
So I had an exam and they took the question paper back so I cant show the question
but it is
tan x = alpha tan3x
for what value of alpha will there be no real solution for x
I think
mellow arrow
$\tan\left(x\right)=α\tan\left(3x\right)?$
thorny flameBOT
B-eard
outer carbon
**B\-eard**
ye
mellow arrow
do you know the formula for tan(3x) in terms of tan(x)?
outer carbon
do you know the formula for tan(3x) in terms of tan(x)?
wdym by in terms of tan x
I just know it like
that triple angle formula
outer carbon
3 tanx + tan^3x / 1 - 3tan^2x
mellow arrow
that triple angle formula
write it out just to make sure
outer carbon
thats it
outer carbon
So, do you know where tangent is undefined?
idk what that would mean
mellow arrow
no
mellow arrow
maiden cipher
You're going to have a rough time of it if you don't know what you're aiming for
outer carbon
thats what I wrote bro
abstract wadi
There's a plus in what you wrote.
maiden cipher
punch in tan 90 on your calculator
abstract wadi
Apple pi, you're not helping.
maiden cipher
No
abstract wadi
And you're confused yourself about the problem.
maiden cipher
tan 90 is not infinity.. tan 90 is undefined
abstract wadi
Apple pi, this is not necessary for the problem, at ALL.
mellow arrow
$\tan\left(x\right)=α\left(\frac{3\tan\left(x\right)-\tan^{3}\left(x\right)}{1-3\tan^{2}x}\right)$
outer carbon
Idk where he saw 90
thorny flameBOT
B-eard
mellow arrow
now factor out a tan(x) from RHS
outer carbon
alpha = 2
mellow arrow
and cancel
outer carbon
now factor out a tan(x) from RHS
[ 3 - tan^2x ] * alpha = 1 - 3tan^2x
mellow arrow
yep
maiden cipher
it's a secondary way of solving, since you're given that the desire is a solution for a where there is no real solution for x, but.. you guys have fun
mellow arrow
now can you solve for tan^2(x) in terms of alpha?
tranquil grail
@abstract wadi dude I like your pfp 😂😂
topaz sinewBOT
outer carbon
<@748029603087384657> dude I like your pfp 😂😂
ok I think hes just imaginary as well
topaz sinewBOT
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sweet roost
hello can somebody please help?
topaz sinewBOT
hello can somebody please help?
sweet roost
Suppose that the car license plate must be a sequence V1/V2/C3/C4/V5/V6, where each Vi represents any vowel (taken in A, E, I, O, U) and each Cj any digit . Given the hypothesis that if the license plate contains some letter "A" then it cannot contain any letter "U", how many different license plates are possible?
topaz sinewBOT
Suppose that the car license plate must be a sequence V1/V2/C3/C4/V5/V6, where e...
sweet roost
i got that the total possibilities are 62500 (5 x 5 x 10 x 10 x 5 x 5)
and we subtract the cases where if V1=A or V2=A or V5=A or V6=A then the other V's can't be "U"
so 4 x (10^2 x 4^3)
because if V1=A for example, then the possibilities for V2, V5 and V6 are not 4 anymore but 3.
so 62500 - 4 x 6400 = 36900
i dont have the solution so i dont know if this is right, can somebody check please?
slim maple
because if V1=A for example, then the possibilities for V2, V5 and V6 are not 4 ...
wouldnt it be "not 5 anymore but 4"?
slim maple
so 4 x (10^2 x 4^3)
also it seems like this might slightly over count
since this would count AA11OO twice for example
sweet roost
we consider the cases where V1 = A or V2=A or V5=A or V6=A, then the cases where V1 and V2=A, V1 and V5=A, etc... then the cases where V1, V2, V5 = A, etc... and finally the cases where all V's= A
sweet roost
we still consider 62500?
slim maple
yeah
sweet roost
62500 - 4 x (4^3 x 10^2) - 6 x (10^2 x 4^2) - 4 * (10^2 x 4) - 10^2 ?
sweet roost
4 x 4^3 x 10^2 is the cases where only 1 V = A
slim maple
should it be 4^3?
sweet roost
should it be 4^3?
where?
slim maple
where?
in the second term
but the point im making applies to the 4^2 in the third term and the 4 in the fourth
ah wait
no this is right
yeah youre done gj
sweet roost
62500 - 4 x (4^3 x 10^2) - 6 x (10^2 x 4^2) - 4 * (10^2 x 4) - 10^2 ?
So this?
slim maple
should be yeah
sweet roost
should be yeah
Okay thank you very much!!
slim maple
Okay thank you very much!!
wait
you applied pie wrong
should be alternating + and -
because the first term is the number of cases with at least one A
so the second is the number with at least 2 etc
sweet roost
Oh yea im dumb
slim maple
if you wanted to do it with only - it should be 3^3,3^2,etc
because then it would be the cases with exactly 1 A, exactly 2 As, etc
sweet roost
because then it would be the cases with exactly 1 A, exactly 2 As, etc
If this is the formula
A U B U C U D in our case
Then what is 62500?
We shouldnt consider 62500
slim maple
A U B U C U D in our case
yeah after further consideration PIE isnt the fastest way to do this
because all of the sets are subsets of each other
slim maple
because then it would be the cases with exactly 1 A, exactly 2 As, etc
this would be faster
sweet roost
if you wanted to do it with only - it should be 3^3,3^2,etc
Uh i don't get it sorry
slim maple
Uh i don't get it sorry
well ok so if we have exactly one A
then there are 4 ways to place it and the rest must be E,I,O
so 4*(10^2)*(3^3)
and you can generalize
sweet roost
so 4*(10^2)\*(3^3)
Oh okay because if we have exactly one A, the other V's can't be A nor U
sweet roost
62500 - 4 x (4^3 x 10^2) - 6 x (10^2 x 4^2) - 4 * (10^2 x 4) - 10^2 ?
62500 - 4 x (3^3 x 10^2) - 6 x (10^2 x 3^2) - 4 * (10^2 x 3) - 10^2
Seems right
sweet roost
right
Sorry to bother you, if we wanted to solve it using inclusion/exclusion, how would we do it?
Alternating +/-
topaz sinewBOT
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tranquil grail
ok I think hes just imaginary as well
Fr
obtuse widget
Hello, I need help with conditional probability
there are 3 boxes: A, B, C. in box A there are 6 white balls and 2 black balls. There are 5 white balls in box B. In box C there is 1 white and 1 black. A ball is drawn from a randomly selected box and it turns out to be white. Then another ball was drawn from the same box. What is the probability that a white ball was drawn in the second draw?
topaz sinewBOT
Hello, I need help with conditional probability
there are 3 boxes: A, B, C. in b...
@obtuse widget Has your question been resolved?
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@obtuse widget Has your question been resolved?
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@obtuse widget Has your question been resolved?
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@obtuse widget Has your question been resolved?
lusty frigate
idk
weary crystal
Hello, I need help with conditional probability
there are 3 boxes: A, B, C. in b...
this seems like a situation you'd use the formula for total probability
in this case, something like
$P(W) = P(A) P(W | A) + P(B) P(W | B) + P(C) P(W | C)$
thorny flameBOT
doaby
weary crystal
or if you don't like formulas, a probability tree is a very good idea
topaz sinewBOT
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frail python
how do I know which one I should do -(x-3) or +(x-3) for the absolute value
frail python
for question e
topaz sinewBOT
for question e
frail python
when x approaches -2 from either side it's the same answer? I'm confused
tranquil grail
how do I know which one I should do -(x-3) or +(x-3) for the absolute value
For x < 3 do -(x-3)
fast shore
if both sides give the same answer then the limit exists
if its not the same number then there is no real limit
tranquil grail
< 3 and it became a heart
tranquil grail
so it's -(x-3) when it's approaching -2 from either side
Yes
frail python
if both sides give the same answer then the limit exists
yes ik
tranquil grail
It would change a bit if you were finding the limit at x = 3
glad sierra
in an additive group $\mathbb{Z}_{n}$, if we get a generator a, then (n-a) is also a generator.
topaz sinewBOT
in an additive group $\mathbb{Z}_{n}$, if we get a generator a, then (n-a) is al...
glad sierra
similarly if I get a primitive root of U(p), can i find the other primitive root?
thorny flameBOT
Dubs
whole geode
@glad sierra https://math.stackexchange.com/a/133720/25445
ruby jungle
can someone help me with this i forgot, this was a long time ago and the teacher wanted to see it now💀
topaz sinewBOT
can someone help me with this i forgot, this was a long time ago and the teacher...
ruby jungle
I need what property was used
topaz sinewBOT
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abstract lily
Number 3 is that the halves of a line seperated by the midpoint are equal
Number 4 I would call symmetrical
topaz sinewBOT
@ruby jungle Has your question been resolved?
ruby jungle
ok
molten crystal
solution to second oreder homogeneous diff equations with complex roots looks like this
topaz sinewBOT
solution to second oreder homogeneous diff equations with complex roots looks li...
molten crystal
why are we just using one of the roots?
and not also the conjugate?
nevermind i've got it
.close
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strong mural
topaz sinewBOT
https://media.discordapp.net/attachments/1018703538517454919/1206302854168453200...
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strong mural
.close
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obsidian plover
I need some help with fourier transforms, specifically fourier transform of a circle (spatially) to the frequency domain, i get this result
topaz sinewBOT
I need some help with fourier transforms, specifically fourier transform of a ci...
obsidian plover
I get this, but im absolutely clueless as to why this happens, i understand the FFT of a block function, but i feel as if i cant apply that knowledge here
I would be incredibly thankful if any kind person could help me
topaz sinewBOT
@obsidian plover Has your question been resolved?
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@obsidian plover Has your question been resolved?
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@obsidian plover Has your question been resolved?
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@obsidian plover Has your question been resolved?
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@obsidian plover Has your question been resolved?
topaz sinewBOT
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maiden star
topaz sinewBOT
maiden star
where did the 1/2 go from line 1 to lione 2
woven creek
wow
abstract wadi
It's still there.
1/2 -1/2cos^2(x) = 1/2(1-cos^2(x))
haughty mural
distributive rule: ab+ac=a(b+c) or in this case ab-ac=a(b-c)
topaz sinewBOT
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humble glacier
Sanity check:
The Fréchet derivative is defined for normed spaces $(V , | \cdot |{V} ) \also ( W , | \cdot |{W} ) \also \mathcal{U} \subseteq V \ open \ subset$ as:
[ \lim_{| h |{V} \to 0} \frac{| f(u + h) - f(u) - Jh |{W}}{| h |_{V}} = 0 ]
Now choosing Rn and Rm as normed spaces and changing h = x - a we have the limit as
[ \lim_{x \to a} \frac{| f(x) - f(a) - Jx - Ja |}{| x -a |} = 0 ]
And J is the jacobian
thorny flameBOT
Tobi
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
humble glacier
How can the jacobian move to the other side?
topaz sinewBOT
@humble glacier Has your question been resolved?
steep void
I'm stuck on some math homework because I don't understand how I'm supposed to use one variable to get another one
topaz sinewBOT
I'm stuck on some math homework because I don't understand how I'm supposed to u...
steep void
The question is "deduce the roots of the function P, written in algebraic form"
I have the figured out the expression in the second picture, but I don't know how I'm supposed to work with both variables, and I don't see how switching back to z from u would help
opal vault
so what happened was
we know that 0 is not a root of P
so P(z) = 0 is equivalent to P(z)/z^2 = 0
so it means we have to solve u^2 + 2u - 3 = 0
opal vault
no
steep void
Oh
opal vault
u = z + 1/z
opal vault
no need to isolate z
solve for u first
that's the reason why we consider u in the first place
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molten vine
in the jacobian matrix, how do i know which variable to partial differentiate first ??
topaz sinewBOT
in the jacobian matrix, how do i know which variable to partial differentiate fi...
dense rain
It doesn't really matter, changing the rows is only going to change the determinant by a negative sign and you are taking the absolute value anyway
topaz sinewBOT
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ashen glen
how do i find the max of $y=3-x^{1/2}-2x^{-1/2}$
topaz sinewBOT
how do i find the max of $y=3-x^{1/2}-2x^{-1/2}$
thorny flameBOT
yossi0
willow chasm
derivate
ashen glen
i have made ${dy/dx} = -1/2x^{-1/2}+x^{-3/2}$
thorny flameBOT
yossi0
ashen glen
i made that = 0 how do i solve
willow chasm
square both sides
maiden pelican
Take x to the - 1 /2 as a common factor first
mild hearth
i have made ${dy/dx} = -1/2x^{-1/2}+x^{-3/2}$
multiply through by x^(-3/2)
maiden pelican
U mean multiply through by x^(3/2)
mild hearth
yes
topaz sinewBOT
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sonic bone
why wrong
icy sky
derivative of cot is also negative
topaz sinewBOT
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willow helm
topaz sinewBOT
willow helm
is this right
topaz sinewBOT
is this right
lucid lagoon
yes
lucid lagoon
oh right
glacial adder
looks good
topaz sinewBOT
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neon iron
Can you explain how to get the answer? My teacher didnt explain it in a way I could understand
topaz sinewBOT
Can you explain how to get the answer? My teacher didnt explain it in a way I co...
mild hearth
Can you explain how to get the answer? My teacher didnt explain it in a way I co...
do you know sohcahtoa
void crow
do you know sohcahtoa
help
mild hearth
opposite to the angle, adjacent to the angle or hypotenuse
void crow
sohcahtoa
void crow
HAHAHHA
mild hearth
go away
neon iron
which side is JH here
is it opp
neon iron
hypotenuse
neon iron
siin
neon iron
wait why do u put 9 before sin
void crow
multiplies
neon iron
y
neon iron
Yeah
void crow
well
neon iron
idk why its that tho or how to get there
neon iron
i've figured sin out
void crow
sin?
neon iron
but why is ok
void crow
yeah, what is it
neon iron
hUH
neon iron
JH
neon iron
GH
neon iron
ya
neon iron
9
void crow
well yes
at the end
but if u replace GH with 9
itd become
sin 35 = JH/9
correct?
neon iron
ya
neon iron
wait what
void crow
$Sin (35) = \frac{JH}{9}$
thorny flameBOT
sin
void crow
its right infront of u
neon iron
wait but how come its written like this : 9 sin (35)
void crow
wait but how come its written like this : 9 sin (35)
okay
well
LOOK AT IT
ONCE AGAIN
neon iron
💀💀
neon iron
WHY IS IT WRITTEN IN THAT ORDER😭
thorny flameBOT
sin
neon iron
ya
neon iron
9
neon iron
ya
void crow
so that'd give us
neon iron
ok
neon iron
I get how u get everything
neon iron
But the question was "what expression can you use to find JH?" I just dont get how u make a proper expression, the answer was 9 sin (35) which was diff from the one u put up
void crow
But the question was "what expression can you use to find JH?" I just dont get h...
okay
what is an expression?
neon iron
a math statement
void crow
correct!
neon iron
yay!
void crow
so basically
an expression is used
to find the value of something
for eg
2x^2 can be used to find the value of x if its equal to 0
or 2x^2+4x+8 can be used to find the value of x if its equal to -29302
doesn't matter
now, sin cos and tan
they are ratios
you're aware of that, correct?
neon iron
yea
void crow
alright, good
you're doing great so far
now, if you know
angle and the value of a side are codependent on each other
thats why 90 degree angles are perpendicular to each other
and 180 degree angles are straight lines
sounds clear so far?
neon iron
yes
void crow
perfect
now, what sin cos and tan basically have
are fixed ratios
sin will never be adjacent/opposite
it'd always be opp/hyp
similarly to cos and tan with their ratios
sin 35 helps us to find its value, by considering the ratio of two sides
ie. opposite and hypotenuse
okay?
neon iron
ya
void crow
if the angle grows in measure
the ratio will grow bigger, until it approaches a certain value, ie. 1
sin 40 > sin 35
and sin 90 > sin 40 > sin 35
sounds good so far?
neon iron
yea
void crow
okay
so now you know that
sin 35 equals to the ratio of opposite side and hypotenuse
neon iron
okay that makes more sense
void crow
$sin 35 = \frac{Opp}{Hyp}$
thorny flameBOT
sin
void crow
lets replace opp and hyp together, okay?
neon iron
ok
neon iron
JH
neon iron
GH
thorny flameBOT
sin
void crow
correct?
neon iron
ya
neon iron
9 sin (35)
void crow
why 9?
neon iron
what abt JH
neon iron
Cuz GH is hyp and 9 was given for GH
void crow
Cuz GH is hyp and 9 was given for GH
correct
void crow
got it?
neon iron
ya