ok let me see...
#help-17
small quarry
acoustic flower
small quarry
The concept of proportion is equal
small quarry
oh
acoustic flower
on how to solve it
acoustic flower
ya proportions are 2 ratios right?
small quarry
So let me continue to explain this
small quarry
so the 2 ratios need to be the same?
umm
let me explain more with the picture you sent
small quarry
So in this identity, what you can understand is: (5/2) / (3/2) = (5/6) / (1/2)
after you calcuate it, it should be the same amount
acoustic flower
soo it would be (15/4) = (5/12)???
small quarry
wtf???
acoustic flower
oh god im panicking
acoustic flower
maybe you can call a helper if you have any other questions
how to do that
vocal sleetBOT
@acoustic flower Has your question been resolved?
vast shale
how to do that
What's the question
small quarry
Oi am back lol
small quarry
how to do that
Ping helpers
small quarry
I explained to diva
vast shale
Do I help you doing it or solve it?
small quarry
Teach him about ratio and proportion
split wind
I am here, no worries
@acoustic flower , DarkWolfBG is here for rescue 
vast shale
Lol
Well, ratio is just basically fractions.
a:b = a/b
You can write (5/2)/(3/2)
And doing that you will get (5/3)/(1/1)
Multiply the denominator by 2
vocal sleetBOT
vast shale
Thus that becomes (5/6)/(1/2)
vocal sleetBOT
Available help channel!
fast quarry
vocal sleetBOT
fast quarry
i dont understand how they got the top x bound = 1 when in the question all it says is x>=0
so the bottom bound is 0
but then cant the top bount be infinity
signal drum
take the second equation where $0 \le y \le 1-x^2$
twin meteorBOT
MathLover
signal drum
if you ignore the y you get that $0 \le 1-x^2$. Solving it will get you the upper bound of 1
twin meteorBOT
MathLover
signal drum
Hope that helped
vocal sleetBOT
vocal sleetBOT
Available help channel!
whole gulch
vocal sleetBOT
whole gulch
Actually
Hello good morning
Could someone help me for number 7
Im not understanding what it is asking for
Here is my work for 7
Isnt the image
Hard to draw on my line paper
How do i describe that without having to draw the visual
<@&286206848099549185>
vocal sleetBOT
@whole gulch Has your question been resolved?
uneven wave
I just began learning about probability distributions for continuous random variables, noticing that most of the distributions were curves because there are an infinite number of values that the random variable can take on. What if the random variable only took on a finite number of values, such as just 0 and 1? Do we have a Bernoulli distribution here that can be represented as a histogram?
vocal sleetBOT
I just began learning about probability distributions for continuous random vari...
mild flower
yeah
curves are sometimes used for finitely many values as well when there are just so many that a bar graph would look dumb
vocal sleetBOT
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waxen hawk
vocal sleetBOT
waxen hawk
Why the blue and red are equivalent
paper depot
definition of a riemann integral
$\int_a^b f(x) \dd{x} = \lim_{n \to \infty} \frac{b-a}{n} \sum_{i=1}^n f \paren{a + \frac{(b-a)i}{n} }$
twin meteorBOT
Ann
paper depot
in particular when a = 0 and b = 1, b-a becomes 1 and the sum becomes the sum of f(i/n)
by the way the parentheses around i/n really want to be taller
waxen hawk
definition of a riemann integral
it is possible to prove the equal sign between the blue and red without the definition riemann integral? is there another way to think of it
waxen hawk
like through geometry
paper depot
ok so you're looking for geometric intuition, not a formal proof.
paper depot
ok sorry what
now you've confused me
first you ask if it's possible to explain the equality geometrically
now you say you already know how to do that
wtf
waxen hawk
now you say you already know how to do that
i just feeling it, but im not sure how to prove it
paper depot
?????
waxen hawk
**Ann**
by feeling, i can tell that is true
but i wanna know why it is true, why it is working
waxen hawk
Just draw the rectangles for small n, like n=4
flat whale
What about it
waxen hawk
besides the i
but im famillar with Riemann sum already
im just confused about the (b-a) that seems came out of nowhere
flat whale
but im famillar with Riemann sum already
Apparently not enough!
flat whale
im just confused about the (b-a) that seems came out of nowhere
This is very basic to Riemann sums
vocal sleetBOT
vocal sleetBOT
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real gale
Should eigenvalues of a positive semi-definite matrix always be non-negative?
real gale
Yes
This is an xyproblem 
What is the value of a, the function f:R->R, f(x,y) = ax^4+8y is a convex function?
I found the Hessian matrix and checked if the positive semi -definiteness
$H(x,y) = \begin{bmatrix} 12ax^2 & 0 \ 0 & 0 \end{bmatrix}$
twin meteorBOT
real gale
det(H(x,y)) = 0
paper depot
well duh of course its determinant is 0
paper depot
sure does
real gale
So for all a≥0 the function f(x,y) is a convex function?
But the answer key mentions a**>0, and not a≥**0
paper depot
Well I have been given four different definitions
show them all.
real gale
A twice differentiable function f : R^n → R is convex, if and only if the Hessian H(f(x)) is positive semi-definite for all x ∈ R^n.
paper depot
ok
the zero matrix is pos-semidef.
for a=0 you have a linear function, whose hessian matrix is zero everywhere.
so evidently the answer key must be wrong, but i'd like to see your other 3 definitions
real gale
so evidently the answer key must be wrong, but i'd like to see your other 3 defi...
-
If epigraph of a function f: R^d->R is a convex set, then f is a convex function.
-
A function f: R^d-> R is said to be a convex function iff for all x, y in R^d and c in [0,1], f(cx+(1-c)y) ≤ c f(x) + (1-c) f(y).
-
A differentiable function f : R^d->R is convex if and only if for all x,y in R^d, f(y) ≥ f(x) + (y-x)^T grad(f(x))
paper depot
great
but you're still told that for a=0 your function somehow isnt convex/
even though it fits definitions 1, 2 and 3 even then?
narrow forge
vocal sleetBOT
narrow forge
i dont under why this relation
produce a circle
i thought it would produce two points -1,0 and 1,0
rugged orchid
Well all the pairs of x y that satisfy those conditions form a circle
But idk where y is coming from and it’s not specified
noble current
substitute 0 in both x and at y once
this is the simple circle formulae
(x-h)2 + (y-k)2 = r2
past badge
I dont know how to do part ii
vocal sleetBOT
I dont know how to do part ii
chrome raptor
Pic of the whole exercise, please.
vocal sleetBOT
@past badge Has your question been resolved?
past badge
Pic of the whole exercise, please.
its from a paper
also i think its a mistake
so dw
.close
strong river
How to solve $xe^{-x}+\ln x=0$ for x?
vocal sleetBOT
How to solve $xe^{-x}+\ln x=0$ for x?
twin meteorBOT
YellowZ
river minnow
I think you can use lambert w function here
vocal sleetBOT
@strong river Has your question been resolved?
pastel ruin
i think u take lnx to the rhs and take log base e both sides
vocal sleetBOT
@strong river Has your question been resolved?
viral copper
,w solve xe^(-x) = ln x
twin meteorBOT
velvet plover
,w solve xe^(-x) = -ln x
twin meteorBOT
vocal sleetBOT
vocal sleetBOT
Available help channel!
vast shale
I tried simplifying this by multiplying (3+sqrt(x+9))/(3+sqrt(x+9)) but the denominator still adds up to be 0
vocal sleetBOT
I tried simplifying this by multiplying (3+sqrt(x+9))/(3+sqrt(x+9)) but the deno...
civic otter
Show your calculations, because your method is correct
There might be a silly mistake, maybe a wrong sign
peak matrix
simplify denominator
it's -x, right?
try factoring out x from numerator and then cancel
vast shale
oh, so I shouldn't substitute x = 0 at this stage?
peak matrix
It's completely fine that after doing conjugate, the denominator stays 0. The difference is that you can usually cancel something
vocal sleetBOT
vocal sleetBOT
Available help channel!
worthy lichen
vocal sleetBOT
hybrid flicker
!status
vocal sleetBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
@worthy lichen Has your question been resolved?
worthy lichen
2
hybrid flicker
Did you do the inequality chain using 3^2, 2^4...?
twin meteorBOT
rafilou2003
worthy lichen
yh thats the first thing i did
hybrid flicker
alright, did you apply log_3 to this whole chain ?
hybrid flicker
and so you should have found $2 < 4log_3(2) < 3 < 5log_3(2) < 4$
twin meteorBOT
rafilou2003
worthy lichen
yes
hybrid flicker
focus first on $2 < 4log_3(2) < 3$, what inequalities does this give us on $log_3(2)$ ?
twin meteorBOT
rafilou2003
worthy lichen
0.5< log_3(2) < 0.75
twin meteorBOT
rafilou2003
worthy lichen
0.6 < log_3(2) <0.8
hybrid flicker
yes
hybrid flicker
yes !
worthy lichen
thanks bro
hybrid flicker
np
worthy lichen
.close
rotund hornet
vocal sleetBOT
regal bane
In other words, given a binomial distribution of n tests, p = 0.3, what's the probability there's at least 0.25n successes?
You'll want to use a normal approximation to do this
vocal sleetBOT
@rotund hornet Has your question been resolved?
rotund hornet
In other words, given a binomial distribution of n tests, p = 0.3, what's the pr...
So the way I’ve written it is correct?
vocal sleetBOT
@rotund hornet Has your question been resolved?
rotund hornet
<@&286206848099549185>
vocal sleetBOT
@rotund hornet Has your question been resolved?
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@rotund hornet Has your question been resolved?
heady fjord
So the way I’ve written it is correct?
yepp
remote sorrel
log _3(log _4(x^2+1)^2+log _8(8x^3))=0
vocal sleetBOT
log _3(log _4(x^2+1)^2+log _8(8x^3))=0
merry python
$\log_{3}\left(\log_{4}\left(x^{2}+1\right)^{2}+\log_{8}\left(8x^{3}\right)\right)=0$?
twin meteorBOT
B-eard
merry python
Oh you wrote it
spiral inlet
merry python
are you aware of $a^{\log_{a}b}=b$
twin meteorBOT
B-eard
remote sorrel
**B\-eard**
ya
merry python
Apply that here
remote sorrel
pls tell me how
merry python
Do 3^ both sides
remote sorrel
im new to log problems
remote sorrel
Do 3^ both sides
hmm ok lemme try
twin meteorBOT
B-eard
remote sorrel
2log4(x^2 +1) + 3log8(x) =0
merry python
are you sure about the RHS?
remote sorrel
are you sure about the RHS?
ya maybe
merry python
Okay so you're saying 3^0=0
remote sorrel
a term log8(8) will separate out on lhs
so then 1-1 on rhs
hence 2log4(x^2 +1) + 3log8(x) =0
merry python
Man, 3^0 is 1
remote sorrel
Man, 3^0 is 1
ya i know
merry python
so RHS would be 1
remote sorrel
then this becomes 2log4(x^2 +1) + 3log8(x) =0
merry python
$\log_{4}\left(x^{2}+1\right)^{2}+\log_{8}8-1+\log_{8}x^{3}=0$
twin meteorBOT
B-eard
merry python
This is what you mean right
remote sorrel
**B\-eard**
nah
merry python
for number in base, use '_'
remote sorrel
2log4(x^2+1) + log8(8) +log8(x^3) = 1
then 2log(x^2 +1 ) + 1 + 3log8(x) = 1
=> 2log_4(x^2 +1 ) + 3log_8(x) = 0
remote sorrel
2log_4 (x^2+1)+3log_8 (x)=0
yes
remote sorrel
2log_4 (x^2+1)+3log_8 (x)=0
i am unable to solve after this point
merry python
are you aware of $\log_{a^{n}}b=\frac{1}{n}\log_{a}b$
twin meteorBOT
B-eard
elfin moon
vocal sleetBOT
elfin moon
My answer is is this correct?
- 6 appears on both dice+ 6 appears at least one time
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)(1,6)(2,6)(3,6)(4,6)(5,6)
11/36
outer warren
you wrote (6,3) twice
outer warren
wdym
outer warren
no
probabilities not summing to 1 is a red flag
they seemed to forgot that (6,6) has at least one six
outer warren
wdym by x=2
elfin moon
Probability of getting 2 six
X=0 probability of 0 six
X=1 probability of getting at least one six
Got it?
outer warren
well they used Y there,
and defined that as the success of rolling at least one six
and at least 1 six includes anything more than 1 six as well
Y=1 isn't the event that you get exactly one 6
you could set the distribution up like that,
but you'd define Y or whatever differently
elfin moon
well they used Y there,
and defined that as the success of rolling at least one ...
They it should be 11/36 no?
outer warren
yes
outer warren
they defined the event differently
if done properly, the correct result would've been reached
elfin moon
Nope
The event is not the issue
The main issue is the answer to check 6,6
Including or not
outer warren
i didn't say the issue was with defining the event itself
the issue is that they didn't consider the event properly
outer warren
yes...
the guy in the video included 6,6
and 11/36 is correct
AND what I'm saying is that the way the book defined the event was fine
but they fked up the calculations by ignoring (6,6)"
which as mentioned earlier
they seemed to forgot that (6,6) has at least one six
vocal sleetBOT
vocal sleetBOT
Available help channel!
viral copper
$\int_0^{\infty} \frac{2^{\cos x} \sin(\ln 2 \sin x)}{x} \dd x$
vocal sleetBOT
$\int_0^{\infty} \frac{2^{\cos x} \sin(\ln 2 \sin x)}{x} \dd x$
viral copper
yeah so my answer is π but I can't confirm it anywhere
twin meteorBOT
NEON
viral copper
Wolfram alpha doesn't seem to accept it
,w $\int_0^{\infty} \frac{2^{\cos x} \sin(\ln 2 \sin x)}{x} \dd x$
twin meteorBOT
Requested by neonperseus
viral copper
hmm
Idk where I'm going wrong
$\Im \int_0^{\infty} \frac 1x e^{\ln 2 e^{ix}} \dd x$
twin meteorBOT
NEON
viral copper
$\Im \sum_{k \geq 0} \frac{\ln^k 2}{k!} \int_0^{\infty} \frac{e^{kix}}{x} \dd x$
twin meteorBOT
NEON
viral copper
$\Im \sum_{k \geq 0} \frac{\ln^k 2}{k!} \int_0^{\infty} \frac{\cos kx + i\sin kx}{x} \dd x$
twin meteorBOT
NEON
viral copper
$\sum_{k \geq 0} \frac{\ln^k 2}{k!} \int_0^{\infty} \frac{\sin kx}{kx} \dd kx$
twin meteorBOT
NEON
viral copper
$\frac{\pi}{2}\sum_{k \geq 0} \frac{\ln^k 2}{k!}$
twin meteorBOT
NEON
viral copper
$\pi$
twin meteorBOT
NEON
twin meteorBOT
vocal sleetBOT
@viral copper Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
tall cosmos
tall cosmos
how do i get c
tall cosmos
.reopen
vocal sleetBOT
✅
vocal sleetBOT
@tall cosmos Has your question been resolved?
vocal sleetBOT
@tall cosmos Has your question been resolved?
inner latch
Hello! Suppose I have a list axis-aligned rectangles of arbitrary length. The width and height of each rectangle are also random. What would be an efficient algorithm to organize the positions of each rectangle around some point such that no rectangles overlap and the perimeter of the shape created by the center points of these rectangles is minimized?
vocal sleetBOT
Hello! Suppose I have a list axis-aligned rectangles of arbitrary length. The wi...
inner latch
please let me know if my question is not clear
vast shale
What are the operations you can perform on them?
inner latch
What are the operations you can perform on them?
If I understand you correctly, we can only translate them, no rotation or scaling
inner latch
Should the algorithm be heuristic or exact
no preference
inner latch
does inside perimiter count?
Could you further explain what you mean by this?
thank you
river wagon
if the resulting shape has a hole, do you include the perimiter of that hole?
inner latch
yes
Essentially the point around which we are organizing the rectangles should act as a "center of mass"
also there will be a set gap, say 5 pixels, between each adjacent rectangle, but that shouldn't be too hard to implement
I've found this problem rather difficult to approach
river wagon
Yeah
I don't get what the point of the point is, but perhaps you can put the rectangles on the top and bottom of a line.
like a city skyline with a reflection
inner latch
hmm
Would you like me to explain the actual issue I'm trying to solve rather than my interpretation of the solution? I want to avoid the xy problem
river wagon
Sure
inner latch
I'm tiling computer desktop windows automatically without altering their size so that they don't overlap and look appealing. The point refers to the center of the screen. The end result should resemble the windows "orbiting" the center of the screen with an even distribution.
here, let me show you an example of what 3 windows would look like:
Do you understand?
river wagon
Oh, nice visuals. Is the sizes of all windows given?
inner latch
yes, we know the sizes of each window
river wagon
and the size of the screen
inner latch
yes
river wagon
what if they dont fit?
inner latch
the point of this over traditional tiling is that we are respecting the initial configure size of each window
inner latch
what if they dont fit?
then they go off the screen
for now we don't care if they don't fit
inner latch
I'm not quite sure I understand exactly how the skyline approach would work for more complicated setups where there is not a complete horizontal line or there are more than two rows
if you are a programmer, could you convey this approach using pseudo-code?
thanks for your help btw
inner latch
correct
however, I'm not sure if rows and columns is the best way of looking at this
consider the following arrangement:
river wagon
I think the biggest disadvantage of the city-skyline is that its very wide.
inner latch
yeah
river wagon
The circumference is roughly optimal i think
so its not a math question really
sorry bud, I think you have to use your own artistic skills here
inner latch
What if we were to instead say that the windows have to fit into a circle of minimum radius?
inner latch
so its not a math question really
sorry, I wasn't sure where else to ask this
inner latch
sorry bud, I think you have to use your own artistic skills here
I still think there is an actual solution to this
river wagon
I'll return to this soon
inner latch
ok, thank you for your time
vocal sleetBOT
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vocal sleetBOT
@inner latch Has your question been resolved?
vocal sleetBOT
inner latch
.reopen
vocal sleetBOT
✅
vocal sleetBOT
@inner latch Has your question been resolved?
river wagon
Here is a greedy O(n^3) algorithm:
class placed_rectangle
vector2 posA
vector2 posB
class rectangle
vector2 dims
class placement_option
vector2 origin
vector2 dims
vector2 direction # either [1, -1], [1, 1], [-1, -1], or [-1, 1]
bool taken
function pack_rects
input rectangles (array of absract_rectangle)
rectangles.sort(by -height*width)
rect1 = popfirst(rectangles)
placed_rectangles = [new placed_rectangle(
posA = rect1.dims/2,
posB = - rect1.dims/2)]
positions = array(placement_options) #here, you need to specify 8 starting placement options
for rect in recangles
maxscore = 0
bestp = none
placed = new placed_rectangle(
posA = p.origin,
posB = p.origin + rect.dims * p.dir
)
for p in positions
if (!p.taken and !overlap(rect, rect2) for all rect2 in placed_rectangles)
score = min(p.dims.x, rect.dims.x) + min(p.dims.y, rect.dims.y)
if score > maxscore
maxscore = score
bestp = p
end
end
end
if bestoption != none
bestp.taken = true
placed_rectangles.push(new placed_rectangle( posA = bestp.origin,
posB = bestp.origin + rect.dims * bestp.dir ))
positions.push(new placement_option(
origin = [bestp.origin.x, bestp.origin.y + bestp.dir.y * rect.dims.y],
dir = bestp.dir,
taken = false))
positions.push(new placement_option(
origin = [bestp.origin.x + bestp.dir.x * rect.width, bestp.origin.y],
dir = bestp.dir,
taken = false))
end
@inner latch It does the perimiter minimization, but hopefully in a more space efficient way 😛
So it uses these placement options which I have colored red here. They are possible places to have corners
It's not completely done.
inner latch
oh wow thank you. I'll study this and translate it into my project to test
inner latch
ah, ok
inner latch
i realised it has some flaws
what would those be?
river wagon
If it places a rectangle here, it does not create any corners in the top left
Don't worry about me. It's a fun excercise.
river wagon
Wanna use dm instead?
inner latch
sure!
inner latch
.close
worthy lichen
vocal sleetBOT
worthy lichen
how do i do this without a calculator?
minor sandal
Don’t quote me on this but That’s definitely cosine
You want to get X by itself
Do for each equation get X by itself
karmic imp
Don’t quote me on this but That’s definitely cosine
Or it could be sine
hazy helm
worthy lichen
wait so what do i do when i get X by itself?
karmic imp
Sine looks more wavy
No
minor sandal
If the answer is positive that means you go to the right and if it’s negative you go to the left
pine moss
just inspect where the line goes off to infinity
karmic imp
Sine and cosine are identical, just horizontally shifted
pine moss
you can see that the line isnt at infinity at multiples of pi/2
karmic imp
Cos always has that u factor
And no, that's secant and cosecant
pine moss
now you just go and see which equations are solved for the denominator 0 at pi/2
and you can eliminate those
worthy lichen
Ok I’ll go try this out now
unique swan
I've been told that the definition of sin/cos/tan changes when you move away from the context of right triangles but I don't understand how it can since it's the same function button on a calculator...please explain that.
vocal sleetBOT
I've been told that the definition of sin/cos/tan changes when you move away fro...
unique swan
There will still always be a leg opposite any angle in a triangle whether it's a right triangle or not.
unique swan
Well, sides then?
paper depot
any angle has a side across from it yes
heavy yoke
you can make definitions based on the circle that both have the same values for angles less than 90 degrees and are still defined for larger angles
unique swan
I mean, okay.
paper depot
but "leg" implies adjacency to a 90° angle
unique swan
Ah, okay.
paper depot
I've been told that the definition of sin/cos/tan changes when you move away fro...
the new definition extends the old
unique swan
So, is sin/cos/tan still a ratio between 2 chosen sides?
paper depot
in general no.
unique swan
bitch slaps himself
heavy yoke
you can use trigonometry for non-right triangles (law of sines, law of cosines), but it’s not as simple as a ratio between sides
paper depot
sin, cos and tan are just functions that take an angle and return a number
unique swan
And what does that number represent?
paper depot
as far as an all-encompassing representation goes, (cos(θ), sin(θ)) are the coordinates of the point you get by going θ radians ccw around the unit circle.
unique swan
So the number simply converts degrees to radians then?
paper depot
no.
this has nothing to do with angle unit conversions AT ALL.
forget about that.
in practice when doing geometry you wouldn't care much for what the outputs of the trig functions represent and instead kind of blackbox them away in things like the law of sines or the law of cosines
right triangle trig isn't obsolete despite this, of course.
unique swan
Okay, but if (cos(theta), sin(theta)) represent the x, y coordinate of a point on a unit circle then why couldn't we say that translates to a specific angle? That's bizarre to me that we can't.
paper depot
idk what you're talking about and i don't want to risk opening a third can of worms
unique swan
Okay, thanks anyway, I'll figure this one out later. It's just not clicking right now.
.close
unique swan
PISS ON IT!
vocal sleetBOT
PISS ON IT!
unique swan
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
covert salmon
Where did the 5/32 come from?
vocal sleetBOT
Where did the 5/32 come from?
paper depot
it's $\binom{-1/4}{2}$
twin meteorBOT
Ann
covert salmon
and what does that become?
paper depot
$\binom{\alpha}{n} = \frac{\alpha(\alpha-1)(\alpha-2) \dots (\alpha-n+1)}{n!}$
twin meteorBOT
Ann
paper depot
so $\binom{-1/4}{2} = \frac{-\frac{1}{4} \cdot \paren{-\frac{5}{4}}}{2}$
twin meteorBOT
Ann
paper depot
$\binom{-1/4}{3}$...
twin meteorBOT
Ann
paper depot
you know how to read sigma notation, do you not?
covert salmon
no
paper depot
well that's where your issue is
you should learn/review sigma notation
i've seen you struggle with it earlier today i believe
covert salmon
Did they get the 15/125 by doing ( -1/4 x -5/4 x -9/4 ) / 3! ?
paper depot
15/128 not 15/125
covert salmon
ye
ok makes sense
I have another question tho
where did the 1 in the beginning come from?
because for n = 0
wouldn't we have -1/4 / 0!
which isn't 1?
twin meteorBOT
Ann
covert salmon
Ah that is standard for all bionomials?
paper depot
the numerator in $\binom{\alpha}{n}$ is a product of $n$ terms starting with $\alpha$, but for $n=0$ the numerator is the empty product (the product of nothing), and the empty product is 1
twin meteorBOT
Ann
covert salmon
tysm!
paper depot
Ah that is standard for all bionomials?
yes, look up generalized binomial coefficient
cold flint
Dd
cold flint
No
blissful sentinel
Uh
Please read #❓how-to-get-help
If you don't read it, then please don't use the help channels
And stop spamming in other people's help channels
@cold flint
vocal sleetBOT
@cold flint Has your question been resolved?
blissful sentinel
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vocal sleetBOT
vocal sleetBOT
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vast shale
For the property of Laplace transforms: [
\map g t = u_c \map f {t-c} \implies \map G s = e^{-ct} \map F s
]
I understand the procedure on how to go from the inverse laplace transform to the Laplace transform, but not in the opposite way. What I'm asking is how should I deduce [
\map G s = e^{-ct} \map F s \implies \map g t = u_c \map f{t-c} ]
vocal sleetBOT
For the property of Laplace transforms: \[
\map g t = u_c \map f {t-c} \implies ...
twin meteorBOT
vast shale
Hm, I guess they are inverses so the first implication is sufficient
somber yew
Yes, generally you prove stuff from time to frequency since the inverse laplace is not easy to work with if you haven't had complex analysis
vast shale
I see. My book did mention that there is a way to calculate inverse laplace transforms, but it was relating to complex analysis thus outside of the scope of the book
Well fair enough. I guess I'll just stick to the first kind of implication from hereon
Thanks
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vocal sleetBOT
vast shale
vocal sleetBOT
Available help channel!
vast shale
Hi
vocal sleetBOT
@vast shale Has your question been resolved?
sudden fox
vocal sleetBOT
dull maple
No. It shouldn't.
sudden fox
how
dull maple
How is it -3?
sudden fox
look at desmos
dull maple
You just plotted points there.
sly sierra
is there a trick here somewhere? why wouldn't it be -3?
sudden fox
im confused too
dull maple
is there a trick here somewhere? why wouldn't it be -3?
Why should it be -3? I'm getting -7.
sudden fox
because if it's a parallelogram
sly sierra
oh wait, there are of course multiple parallelograms you could form that include those three points
sly sierra
but ABCD probably wants them to be traversed in that order as you go around the parallelogram
the one you drew would be ABDC
dull maple
we need to make the distances equal
Sure, that works. Although there are easier ways.
dull maple
we need to make the distances equal
You need more. That alone won't suffice.
sudden fox
alright thanks
i wasn't aware of the orders
lol
thank you for the help
i have another question tho
im wondering why it is not Friday
it says By Friday
so it will be done by 11:59 PM, Thursday, no?
aka By Friday
yeah i keep thinking how i got it wrong
but i just don't know
maybe the question has an error
<@&286206848099549185>
dull maple
so it will be done by 11:59 PM, Thursday, no?
Yeah. So, that's thursday night only.
By friday night, it'll rain another 0.75 inches.
vocal sleetBOT
@sudden fox Has your question been resolved?
sudden fox
By friday night, it'll rain another 0.75 inches.
no when they say Friday
vocal sleetBOT
no when they say Friday
sudden fox
it means
Friday non inclusive
no?
like in English
by is used to say
before something
like lets meet up by 7 oclock
it means
only until before
6:59
so when i put friday
i meant
until
Thursday 11:59
yes?
dull maple
Well, i think what matters is that you had it right. It's a semantic issue now. Lol
dull maple
I think that question just meant that by friday includes the day of friday as well.
sudden fox
i see
yes since they didn't specify what time of Friday
it's sketchy
but regardless i got a passing grade
just a weird question
anyways thanks for your help enemagneto!
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pallid zenith
twin meteor
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whole gulch
vocal sleetBOT
whole gulch
vocal sleetBOT
@whole gulch Has your question been resolved?
whole gulch
@spice suneres
<@&286206848099549185>
I’m trying to think of how else to do this problem
flat whale
Your integration order is confusing.
The set up to cylindrical is right but you didn't change the bounds on the integral. Then everything after that looks wrong.
Follow the worked examples here and try again
https://tutorial.math.lamar.edu/classes/calciii/DivergenceTheorem.aspx
Maybe do cylindrical surfaces first
vast shale
vocal sleetBOT
vast shale
I get [
\map u{x,t} = \sum_{n=1}^\infty c_n \map \cos{\f{n\pi \alpha}L t} \map \sin{\f{n\pi}Lx}
]
as a general solution but this seems wrong to me
twin meteorBOT
bleak oak
looks good to me whats wrong about it
at least the shape looks good not sure bout the constants
coarse sequoia
I completely forgot how to do slope equations where X is already given in this fashion
vocal sleetBOT
I completely forgot how to do slope equations where X is already given in this f...
coarse sequoia
From the top, what do I do? Find a way to isolate y?
I don't know how I'd isolate this with half of something
mild flower
y is already isolated
you just plug in the value they give you for x and calculate the corresponding y
for example for the first one it's $-\f12(-2) - 2$
twin meteorBOT
Hayley
coarse sequoia
Hold on lemme try it
Tried the table function on my calculator and now its staying stupid stuff like -1x/-3/2y
OH
I SEE WHAT YOU MEAN
I'm just so stupid
mild flower
Tried the table function on my calculator and now its staying stupid stuff like ...
you can try the table function if you want, i'd recommend just doing each line manually though
coarse sequoia
Much easier manually
Yeah for the -1 it gives me y = 3/2
Well -3/2 but still I don't get it
That's not a functioning coordinate
What am I supposed to do now?
coarse sequoia
Frankly, what am I supposed to do with that? Make a slope inside of a slope because that's only adjusting for y
And y can only be a solid number
We aren't dealing with 3 dimensions here
mild flower
why can't y be a fraction? 
coarse sequoia
Because I'm supposed to graph it too?
mild flower
$-\frac32 = -1\f12$ if that helps
twin meteorBOT
Hayley
mild flower
so midway between -1 and -2
mild flower
coarse sequoia
Wait no I'm a space off but still
mild flower
what did you get for -2?
wait your -1 value isn't right either, what did you do to compute that?
coarse sequoia
Yeah my first answer was wrong
y = -3 on -2
I told you -1 doesn't make any sense
Its literally asking for a slope within a slope
mild flower
and then redo that -1 calculation because -3/2 isn't right
mild flower
yeah
mild flower
perf
coarse sequoia
So 0, -2 for the next pair
mild flower
👍
coarse sequoia
Yep, thank you
Formatting issue
Thank yo uvery much for your patience with my stupidity
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vast shale
Hello can someone help me with this problem, idk how to start?
vocal sleetBOT
Hello can someone help me with this problem, idk how to start?
@vast shale Has your question been resolved?
vast shale
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pallid forge
are you struggling with how to multiply matrices together?
eager dirge
vocal sleetBOT
eager dirge
Everytime I plug the sides into sin, cos, or tan they don't match. Any ideas?
wraith venture
eager dirge
oh i wasn't aware cos sine and tan didnt work on non right triangles I thought it was only the pythagorean theorem that stopped working
pallid forge
Everytime I plug the sides into sin, cos, or tan they don't match. Any ideas?
this would only work if the triangle was a right triangle
eager dirge
I will try cosine rule
vocal sleetBOT
@eager dirge Has your question been resolved?
eager dirge
It’s my first time using the cossine rule, how did I mess this up?
twin meteorBOT
Fossil
small quarry
a = 16, b = 14, c = 6
16^2 = 14^2 + 6^2 - 2(6)(24)cos(A)
-12cos(A) = 1
cos(A) = -1 / 12
umm
eager dirge
now you do cos^-1(-1/12) = A correct?
small quarry
umm
not really...
one of the angle should be like 94...
idk
can't help
BRUH
I was wrong
cos(A) = 13/14
sorry for that I alr corrected
Cos A = 21.8
Done
eager dirge
Cos A = 21.8
wouldnt that be: A = 21.8 ?
eager dirge
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vocal sleetBOT
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Available help channel!
naive sable
This is false, right ?
vocal sleetBOT
This is false, right ?
lofty bridge
looks like it
sharp fractal
I need help
vocal sleetBOT
I need help
sharp fractal
if the ratio of two roots of equation lx^2 + nx + n = 0 is p : q. Then prove that ROOT(p/q) + ROOT(q/p) +ROOT(n/l) = 0
whole oasis
are there square roots on the LHS?
if so, then notice that square root is nonnegative, hence all the terms should be 0
and so root(p/q) = 0 and root(q/p) = 0 at the same time which is impossible
paper depot
ah wait no
whole oasis
mhm I've found a solution but
how can I explain why my previous reasoning is wrong then
Okay mb, they can be complex
vocal sleetBOT
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vocal sleetBOT
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Available help channel!
dull scaffold
question 3 help pls
vocal sleetBOT
question 3 help pls
dull scaffold
yea
dull scaffold
which is 48 right?
dreamy viper
5 6 8 and 9 can b3 first place no?
quick estuary
Multiply by 5
quick estuary
5 choices for first
2 for last
4! In the middle
Are we missing anything
Idts
quick estuary
Should be 240
dull scaffold
wrong
quick estuary
Ahm
dull scaffold
yes sur
quick estuary
Cant be 1 at first place
If 8 is ending
Number of permutations
4 x 4!
I see
Is it 192
What our mistake is
quick estuary
We considered
6 and 8 in the
Permutations
Did not fix it
Fixing 8
Only 4 can lead
dull scaffold
soooo?
quick estuary
192 is correct?
dull scaffold
yes sur
dull scaffold
ok
quick estuary
Then 4!
dull scaffold
bc 1 is taken
quick estuary
And likewise
Same cases if 6 ends
8 *4!
244!
2 x 4 x 4!
Sorry for causing confusion
dull scaffold
np
vocal sleetBOT
@dull scaffold Has your question been resolved?
safe nova
Hi. This is a calculus question from khan academy. I looked at the how to solve section for these types of questions but still confused.
vocal sleetBOT
Hi. This is a calculus question from khan academy. I looked at the how to solve ...
safe nova
Limits at infinity of quotients with square roots is the section
The answer is -3/4 but don’t understand where the negative comes from
safe nova
safe nova
For which x
viral copper
The ones in the denominator