oh i am usually good at these
#help-19
main creek
yes
mystic saffron
well yeah okay
i will just open another one if i am still stuck
thank you tho @main creek
i really appreciate it
.close
sudden mauve
²
sudden mauve
sorry my thing kept going to wrong tab
quasi sparrow
sorry my thing kept going to wrong tab
.close
plush idol
Can somebody explain to me how to use the general formula of simpson's 3/8 in a simple way? Like why did we put y3 there and why the rest there
plush idol
Here
quasi sparrow
Can somebody explain to me how to use the general formula of simpson's 3/8 in a ...
You're asking for a proof?
quasi sparrow
Can somebody explain to me how to use the general formula of simpson's 3/8 in a ...
Then what are you looking for
plush idol
I want to know the explination of the formula...
Like why did we put the y1 y2 y4 y5 there and then put y3 there
See the image please
quasi sparrow
I want to know the explination of the formula...
Like why did we put the y1 y2 ...
That comes from the formula
plush idol
Yes. Explain it please
quasi sparrow
If you want to know why the formula works, that's what the proof is for
But you just said you don't want the proof
So 
plush idol
I just want a simple explination
plush idol
Like in the trapezoidal rule we took the first and the last then multiplyed the rest by 2
Here what did we do?
plush idol
Look
plush idol
In simpson's 1/3 rule we took the first and the last then multiplied the odd ones by 4 and the even ones by 2
Explain 3/8 like that if you can
quasi sparrow
In simpson's 1/3 rule we took the first and the last then multiplied the odd on...
Did you read the wiki yet
plush idol
Yes
quasi sparrow
What don't you understand about it
quasi sparrow
In simpson's 1/3 rule we took the first and the last then multiplied the odd on...
They give a similar formula
plush idol
Continue
quasi sparrow
What don't you understand about it
.
quasi sparrow
I understood 1/3 and trapezoidal but not 3/8
???????
They're just fractions
What are you looking for man
You said you don't want the proof
You said you read the formula
quasi sparrow
https://en.m.wikipedia.org/wiki/Simpson%27s_rule
Yes I told you to read that section here
plush idol
Here is its formula.
Now explain to me how to USE it. I do not want to know how did it come into existence
quasi sparrow
It's exactly the same as the other Simpsons rule except with 3/8 and 3 in some places
Is it the 3/8 you're having trouble with?
quasi sparrow
Just till me how to use it
You evaluate the function at the interior points then add them
Then multiply by 3
Then add the endpoints
Then multiply by 3/8
plush idol
We multiply 3/8 by the step-size then we multiply the Answer by:
The first point + the last point plus 3* what?
- 2*what
This is what I want
odd edgeBOT
@plush idol Has your question been resolved?
dawn siren
Trying to find exact value of this function
odd edgeBOT
Trying to find exact value of this function
dawn siren
noble palm
what have you tried
dawn siren
I tried converting 5π/8 to degrees and looking for a special triangle
I get 112.5 degrees which doesn't have a special triangle
noble palm
theres a handy formula to help you
it's double angle formula for sin
you kinda just need to memorize and get used to using it
but it's sin(2x) = 2sin(x)cos(x)
here you have that x = 5pi/8
and you want to find sin(x)cos(x)
well that is sin(2x)/2, or sin(5pi/4)/2
noble palm
sure
dawn siren
OMG because (sin2A)/2=sinAcosA
noble palm
👍
noble palm
yup, no problem
dawn siren
.close
gleaming hedge
in calculating a limit, our teacher put x^x-1 like e^(xln(x)-1)
wooden python
$x^{x - 1}$ се е превърнало в $e^{x \ln(x) - 1}$, нали?
gleaming hedge
това не се ли получава e^(xln(x))/e
clever fjordBOT
Ann
wooden python
$x^x - 1$?
clever fjordBOT
Ann
gleaming hedge
да
wooden python
$x^x - 1 = e^{x \ln(x) - 1}$?
clever fjordBOT
Ann
gleaming hedge
да
wooden python
но може би само изглежда така поради лошия почерк...
gleaming hedge
аз бях доста учуден
тогава е просто e^(xln(x))-1
аз питах асистента ни след като приключи неща от поредицата на
дали му плащат добре за допълнителни семинари
дали му харесва
той си държеше на аргумента, че не е толкова до плащане, колкото е до да помага
но честно казано...
не знам, смисъл аз го гледам като някакви глупави грешки, които не трябва да ги има
защото все пак е преподавател във висше
както и да е
благодаря ти @wooden python
.close
tiny wind
odd edgeBOT
tiny wind
I dont understand how to solve this
lethal spoke
do you understand gradient intercept form of a line and conditions for perpendicularity
tiny wind
do you understand gradient intercept form of a line and conditions for perpendic...
Can u say that in english 
lethal spoke
Can u say that in english <:joobisweat:919975534006517871>
to get gradient intercept divide through by coefficient of y
tiny wind
u are using a lot of fancy words that i dont understand
wooden python
do you know either of the words "slope" and "gradient"?
also that guy's speaking in a way that's bordering on gibberish so not your fault if you don't gete him
get him*
@tiny wind
tiny wind
also that guy's speaking in a way that's bordering on gibberish so not your faul...
okay ty i thought i was going crazy
wooden python
oh lmfao the question even mentions the word "slope"
tiny wind
im on a different question now
tiny wind
yeah i got it wrong so it gave me a new one
wooden python
same format different numbers?
tiny wind
yes
wooden python
ok show us
tiny wind
wooden python
im still gonna ask you this: are you familiar with the concept of slope
tiny wind
yes
tiny wind
wooden python
if i gave you the equation y = 8x + 11 and told you this defines a line,
and asked you for the slope of this line,
would you be able to answer
tiny wind
no
wooden python
do you know slope-intercept form
tiny wind
no
ok
tiny fulcrum
Hey I am stuck with this and dont really know what to do. I started with
sqrt( (x-x1)^2 +(y-y1)^2 ) + sqrt( (x-x2)^2 +(y-y2)^2 ) = k. I would appreciate help so much. Thanks
red pendant
can smb explain why
odd edgeBOT
can smb explain why
odd edgeBOT
@red pendant Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
tidal barn
How to calculate determinant of this matrix
odd edgeBOT
How to calculate determinant of this matrix
late quartz
👀
odd edgeBOT
@tidal barn Has your question been resolved?
ivory raven
you can decompose it into a block matrix
one 1x1 zero block in the top left, (n-1)x1 and 1x(n-1) blocks full of as, and an (n-1)x(n-1) identity matrix
$$\operatorname{det}\begin{bmatrix} A & B \ C & D \end{bmatrix} = \operatorname{det}(A-BD^{-1}C)\operatorname{det}(D)$$
clever fjordBOT
Desync
stray token
hello
ocean knot
Let f:I->R and g:J->R be convex functions such that f(I) is contained in J and g monotonically increasing. Show that g•f is convex
ocean knot
In the image is the deffinition of convex functions
I dont know how to start. I have to show that inequality with g(f(x))
<@&286206848099549185>
.close
odd edgeBOT
odd edgeBOT
Available help channel!
formal iris
Hi, I am a Undergrad Student taking Calc 2. Can someone help me understand how to integrate this function using U - Subsituiton?
amber schooner
notice how the derivative of sqrt x is 1/(2sqrt(x))
and how we also have sqrt(x) in the denominator
it’s just missing the 2 in the denominator
so u can multiply by 2/2 to get it to look like sin(u)du
or u can think of it like du=1/2sqrtx dx
dx=2sqrtx du
and the sqrtx will cancel
then u have 3sin(u) * 2 du
=6sin(u) du
and this is easy to integrate
just -6cos(u)
which is -6cos(sqrtx)
amber schooner
and don’t forget the +C
amber schooner
I was making the mistake of saying dx = du/2sqrt(x)
yea no no that would be if u was our original variable and we did a substitution with the variable x
formal iris
Ah ok thanks !!!
odd edgeBOT
odd edgeBOT
Available help channel!
fathom talon
odd edgeBOT
fathom talon
can someone help with this?
quasi sparrow
do you know the circumference of the wheel?
odd edgeBOT
@fathom talon Has your question been resolved?
mystic saffron
odd edgeBOT
mystic saffron
someone show me how to do this pls
quasi sparrow
,rotate
clever fjordBOT
quasi sparrow
did you try plugging each of the options in and see if there's an integer k that makes the equation zero
mystic saffron
did you try plugging each of the options in and see if there's an integer k that...
i did not
i will try and lyk
mystic saffron
did you try plugging each of the options in and see if there's an integer k that...
what do you mean
fluid tundra
put kx on the right hand side
divide everything by x
you will see
plugging in answer choices is not necessary with this in mind
fluid tundra
i mean subtract kx from both sides
fluid tundra
then divide both sides by x
mystic saffron
ill have 7/x
fluid tundra
that is correct
mystic saffron
okay
fluid tundra
now you can observe the answer choices
mystic saffron
observe? 😭
fluid tundra
observe? 😭
you have a 7/x term so you can intuit that x should have a factor of 7 in it
mystic saffron
what is blud saying
odd edgeBOT
@mystic saffron Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
simple ether
odd edgeBOT
simple ether
not sure how they subbed it in and simplified it to that
toxic monolith
not sure how they subbed it in and simplified it to that
$\text{look here:}\\\int_{0}^{4}\frac{x}{\sqrt{x}+1}dx\overset{\ast }{=}\\t=\sqrt{x}\Leftrightarrow x=t^{2}\Leftrightarrow dx=2tdt\\\overset{\ast }{=}\int_{\sqrt{0}}^{\sqrt{4}}\frac{t^{2}}{t+1}\cdot 2tdt=2\int_{0}^{2}\frac{t^{3}dt}{t+1}=_{\cdots }etc$
clever fjordBOT
Joanna Angel
zealous barn
i have a series from n=2 to infinity : 2/(n^3-n) and I am trying to express it as a telescoping series but cant seem to figure it out
zealous barn
i tried partial fraction decomp but it didnt seem to work
would that be the only approach?
leaden karma
what did you get for your partial fraction decomposition?
obsidian patio
Answer?
leaden karma
I don't see why that wouldn't work
odd edgeBOT
Answer?
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
obsidian patio
Hurry up
obsidian patio
K
zealous barn
I don't see why that wouldn't work
leaden karma
you can factor n^2-1 as (n+1)(n-1) and decompose further
zealous barn
hmm ok
toxic monolith
hmm ok
$\text{note that:}\\S_{n}=\sum_{k=2}^{n}\frac{2}{k^{3}-k}=\sum_{k=2}^{n}\frac{2}{k\left( k-1 \right)\left( k+1 \right)}=\\=\sum_{k=2}^{n}\left( \frac{1}{k+1}+\frac{1}{k-1}-\frac{2}{k} \right)=\\=\sum_{k=2}^{n}\left( \frac{1}{k+1}-\frac{1}{k} \right)-\sum_{k=2}^{n}\left( \frac{1}{k}-\frac{1}{k-1} \right)=_{\cdots }etc$
clever fjordBOT
Joanna Angel
zealous barn
wait what
toxic monolith
remember to use partial sums of a series when using definition to test the convergence of a series
zealous barn
i dont understand
toxic monolith
i dont understand
have you ever investigated of the convergence the series due to the definition?
zealous barn
no
toxic monolith
no
so why are you dealing with such a problem now?
We use the telescopic methods shown above to examine the convergence of a series based on the definition
toxic monolith
ok
so I gave you a hint, how to continue your problem, please carefully read what I've written for you
zealous barn
ty
toxic monolith
yw
zealous barn
.close
wise lava
i need help understanding this. why is step one multiplying by a fraction with a denominator of 1-cosx?
latent scaffold
The first step is multiplying the numerator and denominator by the conjugate of the denominator.
wise lava
is that to get a common denominator?
latent scaffold
It's not the immediate goal no. Multiplying by the conjugate like that usually helps simplifying stuff. For instance, doing it here gives you a (1-cos^2 (x)) in the denominator, which can be simplified.
wise lava
to sin^2(x)
latent scaffold
Yes
wise lava
is c = sinx?
that gives them the common denominator of sin^2x
ok i got it thanks 😊
.close
vestal island
odd edgeBOT
hoary marsh
!status
odd edgeBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
green elm
try doing a change of variables to $\int_{-a}^0 f(x)dx$
hoary marsh
But not sure
clever fjordBOT
Bungo
green elm
see what you end up with
green elm
what did you get?
fervent hound
Any function can be written as the sum of an even and an odd function
clever fjordBOT
Adam Chebil
fervent hound
Integral of the odd fct is 0 and
the integral of the even fct from -a to a is 2 times the integral from 0 to a
mystic saffron
suppose we have:
x+y=1
-7=0
is this system consistent or inconsistent
mystic saffron
like suppose the augmented matrix gives 0=-7 after reduction
wooden python
well
what do you think
can it be consistent
or is the unignorable -7=0 gonna fuck us up?
mystic saffron
i always thought it would make the system inconsistent (no solutions) BUT wolframalpha apparently does not agree with me
wooden python
show exact WA input and output
sonic nova
The presence of a single false statement in the system renders the system inconsistent
wooden python
scroll down?
sonic nova
It's pretty much the same way you use that word normally:
The information is inconsistent with truth
mystic saffron
mystic saffron
The presence of a single false statement in the system renders the system incons...
yeah that's what i have on my textbook too
sonic nova
Just seems like wolfram is confused about how to present the answer (or maybe the way you entered it)
tender carbon
odd edgeBOT
tender carbon
cant find a way to change bix sizes here in wolfram
i want to increase z
now it is 0.2 max
odd edgeBOT
@tender carbon Has your question been resolved?
tender carbon
.close
odd edgeBOT
Plot 60sinxsiny/(xy) <@782319730160959508>
frigid canopy
How's $\lim_{x\to0}x\left[\frac{1}{x}\right]$=1?
odd edgeBOT
How's $\lim_{x\to0}x\left[\frac{1}{x}\right]$=1?
clever fjordBOT
Why am. I here
frigid canopy
where [x] is the floor function
My attempt:- I can re-write[1/x] as 1/x-{1/x} where {x} is the fractional part function
so I'll have 1-x{1/x}
now what?
lavish kite
{1/x} will always be between 0 and 1, and x approaches 0, so x{1/x} approaches 0
restive delta
Nvm
frigid canopy
as x appraochs zero, 1/x tends to infty
lavish kite
it does, but {1/x} doesn't, because it's the fractional part
fluid tundra
as x appraochs zero, 1/x tends to infty
1/x is unbounded, but {1/x} itself is bounded
{} is a function that always outputs between 0 and 1, so when defining limits, it's a constant value
frigid canopy
yes I understand but that would just mean the limit is bounded between 0 and 1 , right?
fluid tundra
as bee said, the x term goes to 0
lavish kite
well {1/x} is always between 0 and 1
but then you multiply it by x, which is going to 0
high tusk
Help
rotund dawn
one moment
high tusk
high tusk
one moment
oh shit can u buy me nitro
rotund dawn
I can't even buy food
yeah I made the mistake of paying yearly
and then i lost my job xd
high tusk
how much
rotund dawn
i could add 10 more dollars to it
this
fluid tundra
i could add 10 more dollars to it
LOL
high tusk
aight aight
high tusk
this is wrong
thanks dad
rotund dawn
$6\sqrt{x^2y^3}\left(\frac{x^{\frac{1}{3}}}{y^{\frac{1}{2}}}\right)$
high tusk
yes
clever fjordBOT
Juke | ping me if no response
high tusk
how do u type that jesus
rotund dawn
latex
high tusk
it would make me crazy
high tusk
yes
rotund dawn
that simplifies to just $6xy^{\frac{3}{2}}$
clever fjordBOT
Juke | ping me if no response
rotund dawn
does that make sense so far?
high tusk
how
high tusk
uhuh
rotund dawn
I just did that to the radical
high tusk
ooo
okay
Oh uh
one mistake...
um
its actually not 6 x sqrt x²y³
its
6sqrtx²y³
i js saw that
rotund dawn
oh
high tusk
yea...
rotund dawn
is that supposed to be the 6th root?
rotund dawn
breh
rotund dawn
u wrote it so poorly bro 
anways
okay
same idea
x^(2/6)y^3 and then
idk what the right side says
high tusk
u wrote it so poorly bro <:kekw:752021756650455061>
shhh
rotund dawn
is that n?
high tusk
idk what the right side says
its
multiplied by
X^1/3 / Y^(-1/2)
ill js show the question in the tb
clever fjordBOT
rotund dawn
oh
high tusk
this
rotund dawn
perfect
high tusk
thanks dad (he never said that part)
rotund dawn
this is what u worte
it's all wrote, mainly bc u wrote it wrong to start
anyways
make this into exponents
what do u get
I'll type it as u answer
also bro
the y^3 IS inside
are u trolling me? 😭
rotund dawn
okay
my sleep deprived ass is struggling a bit xd
anwyays
$x^{\frac{2}{6}}y^{\frac{3}{6}}$
clever fjordBOT
Juke | ping me if no response
rotund dawn
u see how i got this?
high tusk
nuh uh
high tusk
mhm
high tusk
yes
high tusk
ph
rotund dawn
and the root is 6
high tusk
yes
rotund dawn
$x^{\frac{2}{6}}y^{\frac{3}{6}} = x^{\frac{1}{3}}y^{\frac{1}{2}}$
do u see how this simpified
wait typo
clever fjordBOT
Juke | ping me if no response
rotund dawn
there, tpyo fixed
do u see that
is that step clear?
( i'm gonna assume so bc it should be )
high tusk
yes
rotund dawn
anwyays
now we have the multiplying term on the right
which is.......
that thing
high tusk
yep
rotund dawn
so $x^{\frac{1}{3}}y^{\frac{1}{2}}\frac{x^{\frac{1}{3}}}{y^{\frac{-1}{2}}}$
clever fjordBOT
Juke | ping me if no response
high tusk
y will become 1/2 in numerator
high tusk
is that right
rotund dawn
I mean
yeah
sure, let's go with it
I wasn't gonna do that
but it's the same thing
high tusk
yep
rotund dawn
$x^{\frac{1}{3}}y^{\frac{1}{2}}x^{\frac{1}{3}}}y^{\frac{1}{2}}$
clever fjordBOT
Juke | ping me if no response
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
rotund dawn
now just simplify
COMPILE ERROR MY ASS, IT LOOKS FINE U BOT
sigh
anyways
just combine em now
and ur good
rotund dawn
do u know how to muliply exponents
rotund dawn
idk what u wrote
high tusk
$2x^1/3 x 2y^1/2$
clever fjordBOT
sin city ☆
rotund dawn
just write it by hand
high tusk
$2x^{\frac{1}{3}} X 2y^{\frac{1}{2}}$
rotund dawn
use {
high tusk
oh
clever fjordBOT
sin city ☆
high tusk
Uwu
high tusk
oh
rotund dawn
so don't have any 2 of anything
we are multiplying
so we add exponents
so we just have $x^{\frac{1}{3}+\frac{1}{3}}y^{\frac{1}{2}+\frac{1}{2}}$
clever fjordBOT
Juke | ping me if no response
high tusk
how do u type that so quick
rotund dawn
bc ik \frac wants two {} so I always do \frac{}{} and then fill in
that way i don't have to track the braces
same for exponent
technically I don't need the braces in the exponent code part here
but dw about the nuance, just always use it tbh
anyways does this make sense?
u can then add and get ur final answer
which is $x^{\frac{2}{3}}y$
clever fjordBOT
Juke | ping me if no response
rotund dawn
that's the correct simplified form of this
high tusk
this made my brain hurt
rotund dawn
it's not hard, just annoying
just deal with it for whatever unit ur learning this for
and then you'll never interact with such weird shit again tbh
or at least, rarely lol
high tusk
just deal with it for whatever unit ur learning this for
yeah
im doing this for
fuckin
logs
high tusk
rotund dawn
anyways, I gotta get ready for class
high tusk
ive gotten my ass beat
near wedge
odd edgeBOT
near wedge
does this pairwise disjoint mean like a set of singletons
near wedge
yes, for example if you did the hiene borel on a set [0,1] would fit this?
wooden python
the what
near wedge
Heine–Borel theorem
im just thinking of an example of this sorry
would {2,3} U {4,5} be a better one
tall veldt
I'm not sure what Heine borel has to do with this
wooden python
{2,3} and {4,5} are disjoint yes
near wedge
i was just thinking like if you have a closed set from 0 to 1, you could make a finite covering of it no, and would it be disjoint if you did so?
wooden python
coverings don't need to be pairwise-disjoint.
wintry viper
Can someone help me with question 2 please
odd edgeBOT
Can someone help me with question 2 please
clever fjordBOT
odd magnet
!status
odd edgeBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
@wintry viper Has your question been resolved?
wintry viper
!status
I don’t know where to begin
Like what the mean or SD is
Usually it’s more clear
odd edgeBOT
@wintry viper Has your question been resolved?
mystic saffron
I dont think I understand cantors thrm, specifically why the powerset of a countably infinite set is uncountable
hasty dome
Which argument of Cantor's are you referring to?
The diagonal argument?
The theorem that shows it for more than just countable sets?
mystic saffron
I guess?
like i think this one is to do with the P(N)
but i dont think i understand
hasty dome
OK, but which of those two?
Usually the diagonal argument is used when first introducing the idea that the powerset of a countably infinite set is uncountable.
mystic saffron
Can you go over the diagnonal argument with me because ngl that sort of confuses me, like if you can just show there is no surjection from f:X -> Y, then it is uncountable?
is that. literally it
i mean it intuitively makes sense, i just dont know if my definition is percise enough
hasty dome
Yes, if there's no surjection from a countable set to another, the other is uncountable.
mystic saffron
I also recognize that this implies that the cardinality of X is less than Y
hasty dome
With all pairs of countable sets, you can get a surjection from one to the other.
Countable means you can label them with the naturals.
mystic saffron
right
hasty dome
So, let's say that f turns a natural number into a member of set A and g turns a natural number into a member of set B.
You can just do f(n) |-> g(n) to get a surjection.
mystic saffron
right
hasty dome
So, if you can't possibly get a surjection, they're not both countably infinite.
Does that make sense so far?
mystic saffron
yes
hasty dome
OK, so the powerset will be subsets of the original set.
So, you'll have like g(2) = {f(1), f(2), f(3)} or something like that.
Note that g(2) contains f(2).
mystic saffron
ok so because there is no surjection from N to any powerset unless it is a finite set, then it is uncountable
hasty dome
Well, it's more complicated to prove there's no surjection.
mystic saffron
hmm
hasty dome
There are also things like g(3) = {f(1), f(2)} where 3 is on the left but not on the right.
For example, g(something) is the empty set.
So, take all the numbers where g(n) = some set without f(n) in it.
Does that make sense so far?
Like let's say that g(27) = {}.
Well, there's a 27 on the left and no 27 on the right.
g(3) = {f(1), f(2)} has a 3 on the left and no 3 on the right.
So, we take all the natural numbers that are on the left but not on the right of one of those pairs.
We put them into a set, like {3, 27, ...}.
Then, g(something) = {f(3), f(27), ...}.
The question is whether something is in the set on the right.
Do you understand the setup here?
If not, do you have any questions?
@mystic saffron
hasty dome
No problem.
OK, so let's say that g(1000) = {f(3), f(27), ...}.
Will f(1000) be in that set?
mystic saffron
i guess not based on the previous examples?
hasty dome
Well, it might be, it might not.
The thing we do is to try both.
Let's say 1000 is on both sides.
Well, the way we got {3, 27, ...} is that we said that g(3) doesn't equal something with f(3) in it.
g(27) doesn't equal something with f(27) in it.
So, if f(1000) is in g(1000), that's a problem.
Do you see why?
mystic saffron
its just not possible?
hasty dome
If 1000 is in {3, 27, ...}, then g(1000) doesn't contain f(1000).
mystic saffron
okay
hasty dome
Just like if 3 is in {3, 27, ...}, then g(3) doesn't contain f(3).
But we just said that f(1000) is in g(1000).
So, that can't happen.
Does it make sense why?
This can be hard to understand at first.
hasty dome
Or, to get rid of the functions, we can do:
3 |-> {1, 2}
27 |-> {}
Let's get the set of all numbers that don't produce a set with themselves in it.
3 doesn't produce a set with 3 in it.
27 doesn't produce a set with 27 in it.
So, something like {3, 27, ...}.
That set is in the powerset, so let's say 1000 |-> that set.
1000 can't produce a set with 1000 in it because that set doesn't contain anything where a number (here, 1000) produces a set with the same number in it.
But that means that 1000 doesn't produce itself, so it is in that set.
So, 1000 is both in that set and not in that set.
mystic saffron
i think that makes sense
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
Or, to get rid of the functions, we can do:
3 |-> {1, 2}
27 |-> {}
Let's get t...
thank. you for helping! i get it
odd edgeBOT
hasty dome
You're welcome.
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rocky edge
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rocky edge
Any help on this
shy smelt
Does omega go throug the origin?
chilly epoch
v = ω × r
odd edgeBOT
@rocky edge Has your question been resolved?
wind lion
If a= -2 then 3a/5 + 1/2 = 5
odd edgeBOT
If a= -2 then 3a/5 + 1/2 = 5
odd edgeBOT
@wind lion Has your question been resolved?
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@wind lion Has your question been resolved?
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@wind lion Has your question been resolved?
wind lion
Nope
wind lion
.close
rare glade
When it comes to vertical asymptotes, if an expression cannot be simplified further, for example x+1/x+2, does that mean there's automatically a vertical asymptote at x=-2? Or do you need to take a limit as it approaches -2 to verify that its a vertical asymptote (ik you have to have a limit to determine if its positive or negative infinity)
ember oak
When it comes to vertical asymptotes, if an expression cannot be simplified furt...
Do you mean $x+\frac{1}{x}+2$, $x+\frac{1}{x+2}$, or $\frac{x+1}{x+2}$?
clever fjordBOT
SWR
rare glade
bascially any rational function that can't be simplified further
the example gave was (x+1)/(x+2)
ember oak
the example gave was (x+1)/(x+2)
Yes. Parentheses are helpful
I was really thrown off, as I thought you meant x+(1/x)+2
rare glade
ok
ember oak
When it comes to vertical asymptotes, if an expression cannot be simplified furt...
A vertical asymptote is never guaranteed
Most trivial example is $y=\frac{x+2}{x+2}$
clever fjordBOT
SWR
ember oak
yes
rare glade
so it can be simplified
rare glade
yea ik
zenith jasper
I believe a non-removable discontinuity for a rational function is a vertical asymptote
rare glade
but im talking functions like (x^2)/(2x+1)
you can't simplify
so there would have to be a VA at -1/2 right?
rare glade
I believe a non-removable discontinuity for a rational function is a vertical as...
yea,
that what I was tryna say
ember oak
I believe a non-removable discontinuity for a rational function is a vertical as...
Not quite. A vertical asymptote is defined as an x coordinate where the graph approaches infinity in either direction
zenith jasper
Not quite. A vertical asymptote is defined as an x coordinate where the graph ap...
wdym?
the two-sided limit is +infinity?
(or -infinity?)
ember oak
the two-sided limit is +infinity?
I mean either side of the limit will go to either + or - infinity
zenith jasper
so 1/x would or would not have a vertical asymptote at x = 0?
slightly confused by your definition
zenith jasper
ok, sure
ember oak
zenith jasper
so, do you have a counterexample to the previous statement?
rare glade
But my question was to make sure that what I thought was correct
So I can IDENTIFY vertical asymptotes by looking at a rational function
ember oak
I believe a non-removable discontinuity for a rational function is a vertical as...
Oh you said non-removable. My mistake. I thought you were referring to holes.
zenith jasper
I think you are correct, unless SWR points out a mistake in my reasoning (I don't think there is one)
ember oak
But my question was to make sure that what I thought was correct
Yes. You are correct
ember oak
So I can IDENTIFY vertical asymptotes by looking at a rational function
Tushar and I just wanted to be sure we agreed on the definition so as to not give you bad advice
rare glade
Ok
Yea I know the definition
It’s just this was just identifying VA by looking at a rational function
zenith jasper
another way of phrasing this is that a rational function only has two types of discontinuities: removable and infinite
cannot be a jump, because rational functions are continuous wherever they are defined
cannot be an "endpoint" discontinuity because rational functions are only "undefined" at finitely many points (the roots of the denominator)
zenith jasper
It’s just this was just identifying VA by looking at a rational function
yeah, just need to be a bit more precise about "cannot be simplified further"
but I think you mean non-removable discontinuity, so it works
odd edgeBOT
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thorny mulch
A student takes two exams, each of which is graded either positive or negative. The examinations are held on different days and the student knows the grade of the first examination before taking the second examination. The probability of passing the first examination is 60%. If the student passes the first exam, their confidence rises and they have an 80% chance of passing the second exam, but if they fail, their confidence drops and the probability of passing the second exam is 30%. What is the probability of passing one of the two exams?
odd edgeBOT
A student takes two exams, each of which is graded either positive or negative. ...
@thorny mulch Has your question been resolved?
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@thorny mulch Has your question been resolved?
half wave
Am I going crazy? Don't I get 17 not 18?
odd edgeBOT
Am I going crazy? Don't I get 17 not 18?
burnt scarab
85/5 becomes 17 then theres the -7/14 times -2 which adds to become 18
half wave
bruh
I'm honestly inputting this into Symbolab so I don't know why it's incorrect as well
Does anyone know what I'm doing wrong here? I'm getting 250 when I input the values, am I suppose to divide by 2?
odd edgeBOT
@half wave Has your question been resolved?
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@half wave Has your question been resolved?
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@half wave Has your question been resolved?
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Available help channel!
wintry glen
need help with this, put it in desmos and still confused
wintry glen
cinder elk
what exactly is it asking to put into the blank spaces?
odd edgeBOT
@wintry glen Has your question been resolved?
warm umbra
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@warm umbra Has your question been resolved?
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✅
warm umbra
Please help me <@&286206848099549185>
odd edgeBOT
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formal oxide
Might sound dumb
but how does this definition of prime work:
p is said to be prime if whenever p|ab either p|a or p|b
wooden python
p is prime if the divisibility of ab by p implies that one of a and b is itself divisible by p
here is how this fails with a non-prime:
4*9 is divisible by 6, but 4 and 9 themselves aren't
formal oxide
ok thanks
sick sigil
This is kind of physics but i figured a lot of people studied this in school anyway and should know
sick sigil
shouldn't the arrow be biimplication?
because lambda vacuum is the same as saying c
and lambda material is the same as saying v
and why are there 3 equal signs after the n
wtf is going on here
it's not a question it's a lecture slide
but they're saying that the definition of refractive index is n = c/v
and n = lambda vacuum / lambda material
so shouldn't this actually be written as
sick sigil
so what are the 3 equal sign thing
never seen this ever
i so dont deserve to study what im studying lmao
low locust
without knowing anything about the topic, I just want to point out that just because <=> is true that does not mean that just writing => is false
sick sigil
i thought that i would be getting negative points for using => instead of <=> if it goes both ways
low locust
no
sick sigil
oh
low locust
its still a true statement
frigid canopy
sick sigil
f me i'll start doing that in the future then
low locust
just like 4>=4 is still a true statement
sick sigil
ohhh i see i see
thank you both so much!
guess i could've googled, but it feels so much better to ask here because instead of arbitrary wikipedia sites i get input from "normal" people like myself :p
low locust
of course if you want to use the other direction then you better write that <=> holds
sick sigil
of course if you want to use the other direction then you better write that <=> ...
sorry could you explain that in other words?
you mean if i want to go from statement B back to statement A at a later stage in the solution?
but what does the "holds" part mean?
low locust
if you want to use that n=lambda_vac/lambda_mat implies n=c/v, then you need to write <= or <=>
holds just as in normal english. "equivalence holds".
sick sigil
oh ok yeye that makes sense then
low locust
or isnt that normal english? at this point I cant tell anymore
sick sigil
english isn't my native tongue
so while i can speak fluent "daily english" the sciency part isn't really applicable ._.
anyway, got it, thank you so much!
.close
mystic saffron
\dm
\usetikzlibrary{circuits.logic.US, positioning}
\begin{tikzpicture}[circuit logic US]
\node [nand gate, inputs={nn}] (A) at (0,0) {};
\node [nand gate, inputs={nnn}] (B) at (0,-2) {};
\node [nand gate, inputs={nn}] (C) at (0,-4) {};
\node [nand gate, inputs={nn}] (D) at (0,2) {};
\node [nand gate, inputs={nn}] (E) at (3.5,-0.15) {};
\node [nand gate, inputs={nn}] (F) at (3.5,-1.85) {};
\draw (A.input 2) -- ++(-2,0) -- ++(0,-1) coordinate (wire1) -- ++(-1,0) node[left] {$Clk$};
\draw (B.input 2) -- ++(-2,0) -- ++(0,1);
\draw (C.input 2) -- ++(-3,0) node[left]{$D$};
\draw (C.output) -- ++(0.8,0) coordinate (wire2);
\draw (D.input 1) -- ++(-1,0) -- ++(0,-7) -- ++(2,0) -| (wire2);
\draw (A.output) -- ++(2.6,0);
\draw (B.output) -- ++(2.6,0);
\draw (F.output) -- ++(1.5,0) coordinate (outputq') node[right] {$Q'$};
\draw (E.output) -- ++(1.5,0) coordinate (outputq) node[right] {$Q$};
\draw (F.input 1) -- ++(-0.5,0) -- ++(0,0.3) -- ($(E.output)!0.5!(outputq) + (0,-0.5)$) -- ($(E.output)!0.5!(outputq)$) coordinate (wire3);
\draw (E.input 2) -- ++(-0.5,0) -- ++(0,-0.3) -- ($(F.output)!0.5!(outputq') + (0,0.5)$) -- ($(F.output)!0.5!(outputq')$) coordinate (wire4);
\fill (wire1) circle [radius=2pt];
\fill (wire2) circle [radius=2pt];
\fill (wire3) circle [radius=2pt];
\fill (wire4) circle [radius=2pt];
\draw (A.input 1) -- ++(-0.5, 0) -- ++(0,0.5) coordinate (wire5);
\draw (B.input 3) -- ++(-0.5, 0) -- ++(0,-0.5) coordinate (wire6);
\draw (B.input 1) -- ++(-0.5, 0) -- ++(0, 0.5) coordinate (wire9);
\draw (C.input 1) -- ++(-0.5, 0) -- ++(0, 0.5) coordinate (wire8);
\draw (D.input 2) -- ++(-0.5, 0) -- ++(0, -0.5) coordinate (wire11);
\draw (D.output) -- ++(0.8, 0) -- ++(0, -0.7) -- (wire5);
\draw (wire2) -- ++(0,0.7) -- (wire6);
\path (B.output) -- ++(0.8, 0) coordinate (wire7);
\path (A.output) -- ++(0.8,0) coordinate (wire10);
\fill (wire7) circle [radius=2pt];
\fill (wire10) circle [radius=2pt];
\draw (wire8) -- ($(wire7) + (0, -0.7)$) -- (wire7);
\draw (wire9) -- ($(wire10) + (0, -0.7)$) -- (wire10);
\draw (wire11) -- ($(wire10) + (0, 0.7)$) -- (wire10);
\end{tikzpicture}
mystic saffron
okay so i am trying to understand the behaviour of this circuit
not much context is needed other than to know that Q and Q' must be complementary to each other, and there can never be an instance where both are 1
so, when Clk = 0, the second level gates experience no change
so the circuit pretty much turns off at that point
the rest is pretty hard to grasp so i would appreciate like some sort of guidance with this
.close
odd edgeBOT
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Available help channel!
leaden light
Did I solve it right?
odd edgeBOT
Did I solve it right?
vernal pilot
Did I solve it right?
No, there's a mistake in going from 3rd to 4th line
It should have been $\frac{dy}{y} = \frac{2x \text{ dx }}{2x^{2}+2}$
clever fjordBOT
smidgin
odd edgeBOT
@leaden light Has your question been resolved?
leaden light
No, there's a mistake in going from 3rd to 4th line
Am I right with this one ?
vernal pilot
Am I right with this one ?
No, $ln(2x^{2}+1) \neq 2ln(2x+1)$
clever fjordBOT
smidgin
leaden light
ok, I need to get rid with ln. But 1/2 ruins everything
vernal pilot
$1/2ln(2x^2+1) = ln(\sqrt{2x^2+1})$
clever fjordBOT
smidgin
vernal pilot
You are incorrectly using the fact that $ln(a^{b})=bln(a)$
clever fjordBOT
smidgin
vernal pilot
**smidgin**
Use this to get rid of the ln
vernal pilot
$ln(a)+ln(b)=ln(ab)$
clever fjordBOT
smidgin
vernal pilot
$ln(y)=ln(\sqrt{x^2+1})+ln(C) \implies
ln(y)=ln(C\sqrt{x^2+1}) \implies y=C\sqrt{x^2+1}$
clever fjordBOT
smidgin
vernal pilot
You are mistaken with the properties, you are only allowed to say $ ln(a)=ln(b) \implies a=b$, not $ln(a)=ln(b)+ln(c) \implies a=b+c$ .The first statement is true because the natural logarithm is an injective function. The second statement can be easily proved false if you take a=6, b=3 and c=2
leaden light
well now I really get it.
clever fjordBOT
smidgin
leaden light
but should I write it this way, or it's unnecessary F(x, y) = C ?
vernal pilot
Depends on how you ask. Some people will be fine if you just left it without getting rid of the ln, some people want you to write it that way