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#help-10
warm shaleBOT
obtuse pebbleBOT
@covert lantern Has your question been resolved?
covert lantern
ok with further thought brute forcing might be fesablr
covert lantern
<@&286206848099549185>
somber thistle
what
glacial obsidian
ok with further thought brute forcing might be fesablr
Yes it is feasible
This solution is optimal
glacial obsidian
It is not unique, you can change (D,E,F) to (5,0,1), instead of (4,2,0). Only these two solutions are optimal
covert lantern
damn
what was your process?
arent there like 38k possible? (although there is a fair bit of overlap
glacial obsidian
what was your process?
I wrote a code
glacial obsidian
arent there like 38k possible? (although there is a fair bit of overlap
Yes
38584
covert lantern
oh thanks!!
is it possible you could send it?
also what was the total runtime anyways
glacial obsidian
also what was the total runtime anyways
0.43 seconds
robust hearth
covert lantern
related to this?
glacial obsidian
is it possible you could send it?
I wrote it in Haskell, do you know that language?
covert lantern
im not familiar no
glacial obsidian
im not familiar no
I will send it anyway
fun (xs,k)
| k == ' ' = (30+5*a+2*b+c)*(2200+3*d+2*e+f)^200
| k == 'x' = (30+5*a+2*b+c)*(2220+3*d+2*e+f)^200
| k == 'y' = (30+5*a+2*b+c)*(2210+3*d+2*e+f)^200
| k == 'z' = (30+5*a+2*b+c)*(2205+3*d+2*e+f)^200
| k == 'w' = (30+5*a+2*b+c)*(2215+3*d+2*e+f)^200
| otherwise = 0
where a = xs !! 0
b = xs !! 1
c = xs !! 2
d = xs !! 3
e = xs !! 4
f = xs !! 5
xss = zip (list 6 12) (repeat ' ') ++
zip yss (repeat 'x') ++
zip yss (repeat 'y') ++
zip yss (repeat 'z') ++
zip yss (repeat 'w')
where yss = list 6 9
zss = filter (\x -> fst x >= 20*2000^200*10^10) $ zip (map fun xss) xss
aux k [] = k
aux k (x:xs) = if fst k <= fst x then aux k xs else aux x xs
list 1 n = map (\x -> [x]) [0..n]
list k n = concat [map (i:) $ list (pred k) (n-i) | i <- [0..n]]
To get the solution I evaluated aux (10^700,([],' ')) zss
obtuse pebbleBOT
@covert lantern Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
dapper bobcat
I need to create a piecewise function with the following constraints:
Atleast 1 compound function ( 2 component functions)
Atleast 1 composite function ( inner and outer function must both be different types)
Atleast 1 local max and 1 local min
And must be continuous over its domain.
obtuse pebbleBOT
I need to create a piecewise function with the following constraints:
Atleast 1...
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
@dapper bobcat Has your question been resolved?
solemn crater
Can someone help me with some more mathematical induction?
I have tried this a couple of different ways already, but I keep boiling it down to n/n+1, but I believe I am supposed to be getting (n+1)/(n+2), am I not?
obtuse pebbleBOT
Can someone help me with some more mathematical induction?
I have tried this a...
solemn crater
and u got it too
wdym? like last step instead of cancelling out the (n+2) on the top and bottom?
spring trail
why would u need to show anything after this?
solemn crater
cause I have to prove that is the case, no?
spring trail
u showed that the statement is true for n=1?
and n=2?
letting either one of those be n, u also showed it to be true for n+1 in a general form?
does it not satisfy the conditions for mathematical induction...
tell me if im dumb
solemn crater
I think I am the dumb one here. I am trying to figure out what the heck I am doing, but I believe I am supposed to show them seperately, like prove p(1), then prove p(2), then prove p(n+1).And to prove p(n+1) I need to boil that stuff down to (n+1)/(n+1)+1 or (n+1)/(n+2)
spring trail
dont worry abt boiling down to whatever
just master the method
u confident in mathematical induction or not?
solemn crater
hell no! lol I am just starting to learn it
the process is pre simple, but the actual math to prove it is whats getting me lol
solemn crater
I think so yeah. Just proving a few arbitrary values for n, then going to the end of the sequence and proving that the last one should hold as well
if all 3 (and possibly more) hold, then the equation should be solid?
I think? lmao
spring trail
how abt this
u show its true for 0 or 1 or something
just one is enough actually
then show that regardless of what u proved, the relation will hold true for n+1 IF it holds true for n
spring trail
why would u need to show anything after this?
thats what it does
assuming that its true for n
the formula works for n+1
now since it has already worked for say 1
it wil work for 2
3
and so on
solemn crater
like if it works for 1, and it works for (n+1), then it must work for every values in between?
spring trail
so it worked for n+1 and 1
let n=1
it will work for 2
now let n be 2
it will work for 3
thats the principle
solemn crater
I still think I am supposed to prove that last bit though. Like this is the example from our slides:
I get what you are saying, technically this is correct. But I think I need to show how to actually get to that final point
spring trail
u took the sum till n steps
then added the next term
u got a result that obeyed the formula
it is kinda confusing; i suggest watching a khan acad vid in 2x speed
u can ask ur teach tomorrow but if u take my work stop where i told u to then pk implies pk+1
solemn crater
it is kinda confusing; i suggest watching a khan acad vid in 2x speed
yeah this def seems like a good idea. I shall see if I can find one on there.
I think for now imma take a break and come look at it again a lil later. Thank you for all your help @spring trail I really appreciate it
spring trail
k gl
bitter schooner
I am inquiring about a problem with simplification. I understand everything exept how (10x) got there and why? Can anyone help?
obtuse pebbleBOT
I am inquiring about a problem with simplification. I understand everything exep...
foggy tendon
I am inquiring about a problem with simplification. I understand everything exep...
(x-5)^2 = (x-5)(x-5)
apply foil. mutliply each term with eachother
= xx -5x -5x +25
= x^2 -10x +25
bitter schooner
(x-5)^2 = (x-5)(x-5)
apply foil. mutliply each term with eachother
= xx -5x -5x ...
thanks appreciate it forgot how to use that its been so long 😅
.close
obtuse pebbleBOT
foggy tendon
thanks appreciate it forgot how to use that its been so long 😅
lol nw, i call it the fishy method cos the lines you draw when multiplying out look like a fish
bitter schooner
lol nw, i call it the fishy method cos the lines you draw when multiplying out l...
😆 Ill make sure to use that name from now on haha
foggy tendon
😆 Ill make sure to use that name from now on haha
fishy method 🔛 🔝
bitter schooner
fishy method 🔛 🔝
frfr
bitter schooner
fishy method 🔛 🔝
also thanks for responding quikly i have an exam rn
foggy tendon
also thanks for responding quikly i have an exam rn
nw good luck!
obtuse pebbleBOT
Available help channel!
sage dagger
Do gradients of tangents differ from gradients of secants
obtuse pebbleBOT
Do gradients of tangents differ from gradients of secants
sage dagger
Secant is just (y2-y1)/(x2-x1) right
brave bramble
A gradient is a gradient. You can't say it's only half.
sage dagger
Huh
brave bramble
The only difference between a secant line and tangent line is the relationship to the curve
But they are still both lines
foggy tendon
sec(x) = 1/sinx
tan(x) = sinx/cosx
they are different.
i dont think i understand the question tho
sage dagger
The only difference between a secant line and tangent line is the relationship t...
But isn’t the gradient of a tangent the difference quotient
And the gradient of a secant y2-y1/x2-x1
native inlet
sec(x) = 1/sinx
tan(x) = sinx/cosx
they are different.
1/cosx?
foggy tendon
But isn’t the gradient of a tangent the difference quotient
oh the gradient of a tangent
solar trellis
in the derivative you take a limit of the difference quotient
which may not be a difference quotient
brave bramble
Is there a question you are working on, that's generating this question?
sage dagger
Is there a question you are working on, that's generating this question?
Well im studying for something but no question in specific
obtuse pebbleBOT
@sage dagger Has your question been resolved?
spare belfry
I need help to learn how to solve these
obtuse pebbleBOT
I need help to learn how to solve these
sage dagger
spicy
pure geode
I'm feelin pretty spicy
Use these equations
$a^2 = c^2 - b^2 \
b^2 = c^2 - a^2 \
c^2 = a^2 + b^2$
warm shaleBOT
WeAreIngram
pure geode
Putting your two knows in on the right to calculate the left
obtuse pebbleBOT
@spare belfry Has your question been resolved?
spare belfry
Putting your two knows in on the right to calculate the left
What about the ones with only one number given
obtuse pebbleBOT
pure geode
With 1,c and 2,b notice the line through the sides. This means that those two sides are the same length, meaning that only one of the values is written you actually know both
obtuse pebbleBOT
Available help channel!
candid iron
I'm taking Business Calculus-1, and I was wondering if my answers in this question were correct. My answer for each question was ii, and I was beginning to question whether I was right or not because I got ii every single time
obtuse pebbleBOT
I'm taking Business Calculus-1, and I was wondering if my answers in this questi...
pure geode
I would agree with you, I think the question writer just wanted to cause you pain.
candid iron
I would agree with you, I think the question writer just wanted to cause you pai...
Thank you LOL
dusky mauve
obtuse pebbleBOT
dusky mauve
I calculated the area as 53.2 but the volume as 49.6
That's not possible right?
What did I do wrong?
dusky mauve
In pretty sure I messed it up but in theory I just have to find the area of each face right?
And then add them?
gilded needle
right
gilded needle
well you have three side lengths, mayve call them length, width, height
L, W, H
two of the sides have area L x W
there are four other sides
what are their areas?
gilded needle
W x H
and L x H ?
i mean your work in orange is almost right
you did three calculations, one for each "type" of side
but for some reason you only counted one of each of the last two types
there are 6 sides total
gilded needle
nw
dusky mauve
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
uneven otter
Is this correct
obtuse pebbleBOT
Is this correct
brisk matrix
how do you negative an implication
uneven otter
$(\exists \epsilon > 0)(\forall \delta > 0)(\exists y \in (0 - \delta, 0 + \delta)) \wedge \lnot (|g(y) - g(0)| \geq \epsilon)$
brisk matrix
\lnot
uneven otter
oh so it was correct before thuis
uneven otter
ye
brisk matrix
looks good
brisk matrix
(if you remove the \lnot and fix the parentheses)
brisk matrix
if you keep the $(\exists y\in I)$ in it's own parenthesized section, then the statement following that doesn't make sense (y is not quantified or accounted for)
warm shaleBOT
maximo
uneven otter
oh wait a min i need to change that too
brisk matrix
an alternative would be $(\exists \varepsilon > 0)(\forall \delta > 0)(\exists y)(\dots )$
warm shaleBOT
maximo
drifting sun
just wanna say I like your name lol @uneven otter
brisk matrix
consider the original statement as
$$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall y\in\mathbb{R})(y\in (x_0 - \delta, x_0 + \delta) \implies |f(y) - f(x_0)| < \varepsilon )$$
warm shaleBOT
maximo
brisk matrix
why is this wallowing my parenthesis
brisk matrix
this goes back to what i said earlier
you're quantifying the y in the premise
and specifically only in the premise
so y has no meaning in the Q of the P -> Q
uneven otter
consider the original statement as
$$(\forall \varepsilon > 0)(\exists \delta > ...
why for all y in R?
I didnt add that in my
original one
drifting sun
this is a function from R to R
uneven otter
ahh
brisk matrix
I didnt add that in my
i think you should address it
another way to write this statement is
$(\forall \varepsilon > 0)(\exists \delta > 0)(|x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon )$
warm shaleBOT
maximo
brisk matrix
the x is implicitly a "for all x"
and we want to quantify it over the whole implication
not just at the premise
uneven otter
consider the original statement as
$$(\forall \varepsilon > 0)(\exists \delta > ...
$$(\exists \varepsilon > 0)(\forall \delta > 0)(\exists y\in\mathbb{R})(y\in (x_0 - \delta, x_0 + \delta) \wedge |g(y) - g(0)| \geq \varepsilon )$$
warm shaleBOT
Calc II Victim
brisk matrix
ok
drifting sun
this comes from analysis and typically people write $\abs{x-x_0}<\delta$ to represent the set $(x_0 -\delta, x_0 + \delta)$ hence why maximo wrote it the way they did
brisk matrix
np
warm shaleBOT
Blue Guilmon
drifting sun
it makes it easier to read for sure
brisk matrix
yeah, i was trying to declutter to emphasize the quantifiers
uneven otter
tysm
drifting sun
whichever professor gave you this assignment
you really should thank them btw
not many people get to practice working with statements like this
uneven otter
my professor is such a goat
but ngl his assignments
be hard asf
compared to other semesters assignments
for this course
drifting sun
he's breaking it down quite well though and it'll save you a lot of headaches in the future if you decide to go into more theory and advanced topics
I would 100% thank him
uneven otter
I see
drifting sun
the typical way people learn this in the US is trial by fire
drifting sun
they just throw it at you and say "prove it" after only giving you a definition and doing maybe 1-2 examples on the board
that's how most introductory math classes are at the beginning theoretical level
then what ends up happening is most people can't even properly handle the statements like you're practicing doing now, let alone prove it
uneven otter
thats exactly the case for us. but he suggests shit ton of homework for practice
which helps alottttt
brisk matrix
calc 2, what is your goal with math
uneven otter
like if I try doing these questions with just the knowledge I gained from lectures. I would have struggled alot
no clue im takin math rn because of computer science
drifting sun
well this is quite advanced
brisk matrix
so this is for a discrete math course
uneven otter
I used to hate maths but idk university changed my mind
uneven otter
so this is for a discrete math course
intro to math proofs apparently
brisk matrix
yeah same difference
uneven otter
oh
brisk matrix
the name is synonymous albeit a misnomer
but i suggest you do familiiarize yourself with this content, especially if you do enjoy it. graph theory and combinatorics are really powerful as a computer scientist
and understanding those requires you understand what you're doing right now
drifting sun
you could easily jump into a math major from a course like this, it doesn't get much harder than this imo
uneven otter
graph theory seems so fucking cool
brisk matrix
graph theory seems so fucking cool
be careful what you wish for haha
uneven otter
😭
drifting sun
^
definitely recommend no matter what you end up doing taking a few linear algebra courses, proof-based and numerical analysis courses/ matrix analysis
if you have the opportunity
brisk matrix
you could easily jump into a math major from a course like this, it doesn't get ...
eh i wouldn't agree completely, but i agree with the sentiment
also it's surprising how much calc2 has been doing for just this course. they've done some group theory and now they're touching on analysis. it's basically baby proofs for each but still
drifting sun
it is used across almost all fields these days
uneven otter
I think next year i have to take calc III linear algebra and Algebra I
brisk matrix
damn i can't believe you're doing algebra 1 as cs
uneven otter
its a required course apparently
drifting sun
algebra 1? as in abstract algebra?
uneven otter
brisk matrix
pretty interesting class, and surprisingly useful
drifting sun
oh ok that's just a proof-based linear algebra course
drifting sun
yeah
take a numerical methods course too if you can fit it in
after you've taken these courses
brisk matrix
make sure you also sprinkle some cs classes here and there between all the math 
drifting sun
hey if you like it you like it, when I was doing my undergrad I took 5 math classes a semester my senior year xd
uneven otter
what the fuk
drifting sun
I basically took every math class my uni had to offer and started doing the grad ones
but that's because I loved it
anywho, does this answer all your questions?
uneven otter
thats insane I dont think i could handle even 3 math courses in one sem
yes thanks alot
drifting sun
math becomes a lot of the same after a certain point
uneven otter
Im ngl doe I hated calculus doe
this course is way more fun
and better
then the 2 calc courses I took so far
.close
empty salmon
How do I do the 17th question y'all
obtuse pebbleBOT
How do I do the 17th question y'all
royal basin
there is a stupid trick
empty salmon
what is it
royal basin
the left-hand side is a quadratic function of x, but at x=a, x=b and x=c its value is equal to 1
so the graph of y = (LHS) passes through (a, 1), (b, 1) and (c, 1)
desert sinew
how'd you get x=a and all anyway
By experimenting. It’s like trying your luck with small / "obvious" choice of number and check to see if they are the root(s).
empty salmon
ah okay
empty salmon
the left-hand side is a quadratic function of x, but at x=a, x=b and x=c its val...
so basically ax^2 + bx + c but all the coefficients equate to the root?
empty salmon
i see
empty salmon
my point is the left-hand side simplifies to 1 lol
since we got 3 roots
is the answer D
royal basin
indeed
empty salmon
alright
empty salmon
the left-hand side is a quadratic function of x, but at x=a, x=b and x=c its val...
sorry i didn't understand this properly but how'd you know the LHS is a quadratic function of x
cus like if x = b then (x-b)=0
wouldnt that convert everything to 0
royal basin
how'd you know the LHS is a quadratic function of x
it's the sum of three fractions each of which is a quadratic.
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
neat sparrow
@glacial obsidian #help-27 message I cannot argue for this assumption philosophically...
obtuse pebbleBOT
<@665690258456838176> https://discord.com/channels/268882317391429632/9034864800...
neat sparrow
is there any other assumption that would also be sufficient for making the conclusion true? (S(AK), AK(r) and r are just propositions and they are not independent, well idk if this leads to a contradiction, but I think I can assume that r and S(AK) are independent, because S(AK) increases the probability of AK(r), but it also increases the probability of $AK^{-1}(r) )$ and $AK^{-1}(r) )$ decreases the probability of r as much as r increases the probability of r, thus then S(AK) and r are independent. The $AK^{-1}(r) )$ is not relevant here I think, but I can explain more if you think it is relevant.
"I need to prove $P(AK(r) & r | S(AK) ) > P(AK(r) & r | \neg S(AK) )$
with AK(r), r and S(AK) propositions.
And I know: $P(r | AK(r)) > P(r | \neg AK(r) ) $
And I know: $P(AK(r) | S(AK)) > P( AK(r) | \neg S(AK) ) $
is this possible?"
warm shaleBOT
Benne-girl
neat sparrow
"this is impossible because: We choose a random number x uniformly in [0,10]. Let r: x is in [5,6]; AK(r): x is in [0,6]; S(AK): x is in [0,5]. Then:
P(r | AK(r)) > P(r | not AK(r)) because P(r | AK(r)) = 1/6 and P(r | not AK(r)) = 0
P(AK(r) | S(AK)) > P(AK(r) | not S(AK)) because P(AK(r) | S(AK)) = 1 and P(AK(r) | not S(AK)) = 1/5
But P(AK(r) & r | S(AK)) < P(AK(r) & r | not S(AK)) because P(AK(r) & r | S(AK)) = 0 and P(AK(r) & r | not S(AK)) = 1/5"
but if you assume the following premise, it becomes possible, but im wondering if there is also another premise that would also make the conclusion true
"If you change the assumption P(r | AK(r)) > P(r | not AK(r)) by P(r | AK(r) and S(AK)) > P(r | AK(r) and not S(AK)) then it is true
Because of this https://en.wikipedia.org/wiki/Chain_rule_(probability)"
I can argue for: $P(r | AK(r) & S(AK)) > P(r | \neg AK(r) & \neg S(AK))$ but not sure if this does the job
warm shaleBOT
Benne-girl
neat sparrow
I have two other assumptions that I can make, idk if they follow from the previous assumptions, but $P(\neg AK(r) | \neg S(AK)) > P( \neg AK(r) | S(AK) )$ and $P(\neg r | \neg AK(r) ) > P( \neg r | AK(r) )$ , idk if this helps
warm shaleBOT
Benne-girl
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
golden spear
The points where the inscribed circle touches with the sides AC and BC make a ratio of 4:3. What is the perimeter?
obtuse pebbleBOT
The points where the inscribed circle touches with the sides AC and BC make a ra...
golden spear
<@&286206848099549185>
obtuse pebbleBOT
@golden spear Has your question been resolved?
obtuse pebbleBOT
@golden spear Has your question been resolved?
neat sparrow
is this valid?
it is indeed possible to prove that P(A & B | C) > P(A & B | ¬C). Let's proceed with the proof:
P(A | B) > P(A | ¬B) (Given)
P(B | C) > P(B | ¬C) (Given)
P(¬B | ¬C) > P(¬B | C) (Given)
P(¬A | ¬B) > P(¬A | B) (Given)
P(A | C) = P(A) (Given)
P(A | B & C) > P(A | ¬B & ¬C) (Given)
P(A | B & C) > P(A | ¬B & C) (Given)
P(¬B | C) > P(B | C) (Given)
P(¬B | ¬C) > P(B | ¬C) (Given)
Now, let's analyze the conditional probabilities involved:
P(A & B | C) = P(A | B & C) * P(B | C) (Multiplication rule of probability)
P(A & B | ¬C) = P(A | B & ¬C) * P(B | ¬C) (Multiplication rule of probability)
Using the given information, we can make the following comparisons:
From (1), (2), and (6):
P(A | B & C) > P(A | ¬B & ¬C) implies P(A | B & C) * P(B | C) > P(A | ¬B & ¬C) * P(B | C) (Multiplying both sides by P(B | C))
From (7), (8), and (9):
P(A | B & C) > P(A | ¬B & C) implies P(A | B & C) * P(B | C) > P(A | ¬B & C) * P(B | ¬C) (Multiplying both sides by P(B | C))
From (10) and (11):
P(A & B | C) = P(A | B & C) * P(B | C)
P(A & B | ¬C) = P(A | B & ¬C) * P(B | ¬C)
Combining the above inequalities, we have:
P(A | B & C) * P(B | C) > P(A | ¬B & C) * P(B | ¬C)
Multiplying both sides by P(A | B & ¬C), which is greater than zero:
P(A & B | C) > P(A & B | ¬C)
Thus, with the given premises, we have proven that P(A & B | C) is indeed greater than P(A & B | ¬C).
obtuse pebbleBOT
is this valid?
it is indeed possible to prove that P(A & B | C) > P(A & B | ¬C...
timid silo
is this valid?
it is indeed possible to prove that P(A & B | C) > P(A & B | ¬C...
I saw that your previous question got ignored. If this is an AI of any sort, this is probably why
The helpers in this server are allergic to AI
neat sparrow
The helpers in this server are allergic to AI
are you?
timid silo
Somewhat
timid silo
I'm not capable of helping you, I'm just here to tell you why you're not getting any answers
neat sparrow
alright
Any <@&286206848099549185> with some familiarity with logic and probabilty theory willing to help me out?
I want to prove: P(r & AK(r) | S(AK)) > P(r & AK(r) | ¬S(AK)).
This is all I know:
P(r | AK(r)) > P(r | ¬AK(r)) (Given)
P(AK(r) | S(AK)) > P(AK(r) | ¬S(AK)) (Given)
P(¬AK(r) | ¬S(AK)) > P(¬AK(r) | S(AK)) (Given)
P(¬r | ¬AK(r)) > P(¬r | AK(r)) (Given)
P(r | S(AK)) = P(A) (Given)
P(r | AK(r) & S(AK)) > P(r | ¬AK(r) & ¬S(AK)) (Given)
P(r | AK(r) & S(AK)) > P(r | ¬AK(r) & S(AK)) (Given)
P(¬AK(r) | S(AK)) > P(AK(r) | S(AK)) (Given)
P(¬AK(r) | ¬S(AK)) > P(AK(r) | ¬S(AK)) (Given)
can a proof be constructed?
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
timid silo
can't help you here but maybe this could be asked in #probability-statistics or even #advanced-probability if you dont get help here
neat sparrow
hmmm alright
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
neat sparrow
if you prove that A > B, because you know A>C and C>B can you then say that you deduced this from these premises? Or should one say that one proved it from one premises or like what is the verb for this??
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
north quarry
if you prove that A > B, because you know A>C and C>B can you then say that you ...
i would say its a deduction ? maybe idk
neat sparrow
i would say its a deduction ? maybe idk
that's what I was thinking as well, but not sure
neat sparrow
if you prove that A > B, because you know A>C and C>B can you then say that you ...
<@&286206848099549185> maybe somoene who is a bit more experienced with maths might know?
hidden garnet
Okay holy shit i couldn't make the slighest sense of the notations
hidden garnet
if you prove that A > B, because you know A>C and C>B can you then say that you ...
This is fine, the greater than sign is a transitive relation
neat sparrow
This is fine, the greater than sign is a transitive relation
I understand
neat sparrow
This is fine, the greater than sign is a transitive relation
but like what is the verb that we generally use for such sentences? Is deduce fine?
neat sparrow
May I know what's AK?
um yes, thanks a lot that you are willing to help, but I am now doubting, thus how about I ping you again when I got everything clear for myself? Otherwise I might waste your time
hidden garnet
No problems, thanks for the consideration even
neat sparrow
No problems, thanks for the consideration even
might I ask you another question before having figured this out, in order to figure this out:
let's say I want to defined a set 'F(q)' which includes all propositions 'f(q)' that entail 'K(q)', i.e. that one knows that q. Could I say something like this then? I would like to know if one can use universal quantification over a 'proposition-function' like f(q) in which the argument q is the q over which we are doing universal quantification just before that:
$\forall q \forall f(q) { [ f(q) \in F(q) ] \rightarrow [ f(q) \models K(q) ] }$
warm shaleBOT
Benne-girl
hidden garnet
<@&286206848099549185>
I believe someone else might be better in this case, I'm not really versed in this one
neat sparrow
let's say I have a set of propositions {s_1, s_2, … , s_n} and I would like to express that the elements of this set have a certain property, for example that if s_1 or s_2 or ... is true, then if A then B (A --> B). How would one express this formally?
<@&286206848099549185> anyone around with some familiarity with set theory (notation) / formal logic (notation) ?
hidden garnet
By the way, if someone doesn't see your question here, you might try #proofs-and-logic
brisk matrix
that’s an odd way to write that @neat sparrow
is your goal to express the set {f(q) | f(q) -> K(q)}?
or {f(q) | f(q) entails K(q)}
neat sparrow
that’s an odd way to write that <@286265557743828993>
Absolutely im kinda asking a weird thing
neat sparrow
By the way, if someone doesn't see your question here, you might try <#328208536...
Alright thanks!!
neat sparrow
is your goal to express the set {f(q) | f(q) -> K(q)}?
What I want to do is quantify over an open formula
brisk matrix
i would say forall f
neat sparrow
What I want to do is quantify over an open formula
The example is made up to ask about that
brisk matrix
it’s like a distinction between f(x) and f
neat sparrow
it’s like a distinction between f(x) and f
What is the difference?
brisk matrix
ok i am at my computer now
@neat sparrow did you figure out your question or do you still want to talk about it
timid silo
Latin
obtuse pebbleBOT
@neat sparrow Has your question been resolved?
fallen needle
how would I solve c and d
obtuse pebbleBOT
how would I solve c and d
timid silo
do you know that $\tan\theta = \frac{perpendicular}{adjacent}$ and similarly, $\\cos\theta = \frac{adjacent}{hypotenuse}$
warm shaleBOT
physicsIsThicc
nimble iris
im stuck
royal basin
consider: multiplying both sides by r
restive acorn
Actually hang on one sec
nimble iris
oki
nimble iris
mhm
restive acorn
6costheta
nimble iris
maybe using cost=x/r before multiplying both sides by r??
restive acorn
Makes a circle with diameter 6
restive acorn
You not only need the eq for circle
You also have to be familiar with
r = asintheta
r = acostheta forms
Those guys all make circles
With diameter a
nimble iris
woa
restive acorn
Center would be
royal basin
You also have to be familiar with
r = asintheta
r = acostheta forms
Those guys all make circles
you do not actually need to know this
restive acorn
I said it wrong eatlier btw
royal basin
you need to know how to convert between cartesian and polar
x = r cos(θ), y = r sin(θ)
it is for that reason that i suggested multiplying both sides by r to get r^2 = 6r cos(θ)
restive acorn
They want it in the form x^2 + y^2
nimble iris
maybe using cost=x/r before multiplying both sides by r??
i was actually able to go thru w this and get 6x = x^2+y^2
royal basin
i was actually able to go thru w this and get 6x = x^2+y^2
yes exactly
empty salmon
What am I doing wrong here
obtuse pebbleBOT
What am I doing wrong here
empty salmon
<@&286206848099549185>
high lily
try expand
gleaming ridge
not very helpful smh
empty salmon
can't we just substitute α + β with -b/a and all
empty salmon
is that a hollow knighht pfp i see
also i just realised i wrote alpha and beta as a and b in the screenshot instead damn
gleaming ridge
$\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$
warm shaleBOT
numbpy (anti-glomed)
gleaming ridge
is that a hollow knighht pfp i see
also i just realised i wrote alpha and beta a...
yes, it is. It's the little ghost
empty salmon
yes, it is. It's the little ghost
nice
empty salmon
just sub all values
mhm
okay so
i got the roots as
-a/b
and -b/c
ill substitute those
gleaming ridge
,w expand bc(x + a/b) (x + b/c)
warm shaleBOT
gleaming ridge
ill substitute those
yep
empty salmon
,w expand bc(x + a/b) (x + b/c)
wait how'd you get that
gleaming ridge
new roots are -a/b and -b/c just find an equation that has these roots
gleaming ridge
yee
empty salmon
ill just multiply all that
gleaming ridge
yes
empty salmon
yea i got bcx + (b^2 + ac)x + abc
empty salmon
oh yes i did
gleaming ridge
only issue is that, none of the answer match
gleaming ridge
looks like we're supposed to have (a+b) instead of (b^2 + ac) as coefficient of x
I don't think they're same, are they?
empty salmon
nah idts
gleaming ridge
yeah so unless there's some trick involved that's the answer
empty salmon
so the options themselves are wrong?
empty salmon
aight
gleaming ridge
I really hope so
lime jewel
obtuse pebbleBOT
lime jewel
Can i please get some help heree
royal basin
@lime jewel any progress so far or are you stuck
lime jewel
Im stuck heree
royal basin
ok
the remainders of our number mod 3, 6 and 9 must each lie in set A = {2, 5, 7}
tell me: which members of A are possible values for the remainder mod 3?
@lime jewel
obtuse pebbleBOT
@lime jewel Has your question been resolved?
dawn trout
How can I simplify this equation?
(a, b and s are constants?)
obtuse pebbleBOT
How can I simplify this equation?
(a, b and s are constants?)
royal basin
dawn trout
Yes
restive jungle
How can I simplify this equation?
(a, b and s are constants?)
i laughed sorry 💀
would it be possible to take 1-(cos(st) out and do stuff with it ?
obtuse pebbleBOT
@dawn trout Has your question been resolved?
dreamy forum
ok lets be honest this cant be simplified
dawn trout
ok lets be honest this cant be simplified
Okay
dreamy forum
maybe i am missing some rules for constants let me quickly check
nah
I mean saying a, b and s are constants does nothing if no value for a, b and s are supplied
At least for simplification
Because here they could be variables for all I care 
Anyways you may shrink the term down in size and make it more ugly in the process
perhaps even harder to work with who knows
I mean you can expand the usual way
Put it all into their respective fractions and get something along the line of $\sqrt{\frac{2a^2 + 2abt + b^2t^2 - 2a^2 \cos(st) + 2abt \cos(st)}{s^2} - \frac{2b^2t \sin(st)}{s^3} + \frac{2b^2 - 2b^2 \cos(st)}{s^4}}$
warm shaleBOT
cyrolx
dreamy forum
and put it all in one fraction
the usual way but this alone already looks disgusting
obtuse pebbleBOT
@dawn trout Has your question been resolved?
dawn trout
Alrighty~ thank you!
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
little lynx
obtuse pebbleBOT
little lynx
How should I proceed to solve this question?
rich plume
assume the initial amount to be a variable let's say x
little lynx
ok
rich plume
Now find the total interest in terms of x
little lynx
(x * 5 * 7)/ 100
which is 35x/100
the total amount they're asking for would then be x+35x/100
rich plume
Yes
little lynx
which is equal to 6615
rich plume
Yes
rich plume
I can take the 100 to the rhs
First make the denominator same
rich plume
Noo
little lynx
so 36x=661500
rich plume
$a+\frac{b}{c}=d$ and $a+b=cd$ are different
little lynx
I did that only
warm shaleBOT
B-eard
rich plume
Hint:||multiply both sides by 100||
little lynx
**B-eard**
it is
rich plume
$x+\frac{35}{100}x=6615⇒\frac{100x}{100}+\frac{35x}{100}=6615⇒\frac{100x+35x}{100}=6615⇒135x=661500$
warm shaleBOT
B-eard
rich plume
or you can do like this also
$x+\frac{35x}{100}=6615⇒100\left(x+\frac{35x}{100}\right)=6615\cdot100⇒100x+100\cdot\frac{35x}{100}=661500⇒100x+35x=661500$'
warm shaleBOT
B-eard
little lynx
oh, right
proper kelp
how would you do this question? any help is appreciated
obtuse pebbleBOT
how would you do this question? any help is appreciated
proper kelp
Oh, the question is , which of the following differential equations yields the direction field below
obtuse pebbleBOT
@proper kelp Has your question been resolved?
obtuse pebbleBOT
Available help channel!
tame vale
How do we solve $\frac{(x-2)(x+3)}{x-3}=0$? Can we just multiply by x-3 and then solve the equation (x+3)=0 V (x-2)=0?
obtuse pebbleBOT
How do we solve $\frac{(x-2)(x+3)}{x-3}=0$? Can we just multiply by x-3 and then...
warm shaleBOT
Bennxy
dusk ruin
More or less, solve the numerator for 0 directly
dusk ruin
Multiplying by the denominator is bad practice generally though, bc it removes part of your equation or expression
high lily
generally speaking you should ensure that the values you obtain don't make the denominator zero
dusk ruin
that’s part of it yeah though not applicable in this case
high lily
i.e. you should be careful if you instead had something like
$$\frac{(x-2)(x-3)}{x-3} = 0$$
and not blindly multiply both sides by $\red{x-3}$ here
warm shaleBOT
ℝamonov
tame vale
i.e. you should be careful if you instead had something like
$$\frac{(x-2)(x-3)}...
Here we should get x-2=0 right?
high lily
yes
tame vale
also use `\vee` next time
Could you explain why and how I should use \vee?
high lily
that's the proper latex symbol for or instead of your makeshift capiltal V
$\vee, \text{V}$
tame vale
Ohhh ok got it, thanks
warm shaleBOT
ℝamonov
tame vale
.close
solid heron
how do I go about solving this if I expend it it looks like it would just be (1)(1)...(1)
obtuse pebbleBOT
how do I go about solving this if I expend it it looks like it would just be (1)...
oak bolt
You can't bring the lim inside the product because one of the limits of the product contains n
pure geode
Did you try the infinite series expansion of sin
solid heron
we are not even thought that here is highschool
pure geode
Since when do you do this in highschool
vague bloom
I dunno how you were supposed to solve it in high school but the answer is B
obtuse pebbleBOT
@solid heron Has your question been resolved?
restive pendant
how do I go about solving this if I expend it it looks like it would just be (1)...
what tools do u know?
vivid creek
Can someone give me some tips/ help answer this question
obtuse pebbleBOT
Can someone give me some tips/ help answer this question
royal basin
a or b?
vivid creek
A
royal basin
notation could have been better
in particular don't use the letter x as a multiplication symbol
but thatll do
now, you have here two things you need as input for the formula:
the radius and the angle
which of these do you already have and which are you missing
royal basin
the correct answer to the question i just asked was "I already have the angle, but I am missing the radius."
not any complaints about how you dont know how to do something.
anyway, ok.
would you like me to guide you along in your thinking, or would you like me to give you one pointer and have you try to do it yourself?
vivid creek
Guide along the thinking please
I tried think what I could do but I haven’t gotten anywhere
I was thinking about circle theorem
But not sure about that
royal basin
"circle theorem", singular?
vivid creek
Yea
royal basin
to you, there is only one circle theorem in existence?
royal basin
me: "circle theorem", singular?
you: yea
me: so to you there's only one?
you: no
anyway ok let's leave that aside
vivid creek
The angle OED looks like a right angle but I’m not sure if it’s drawn to scale
royal basin
let's look into that further
what angles on the diagram can you see that you know for certain are right angles?
vivid creek
Assuming it drawn to scale. I would say OED OFD BCA and OAC
They all look like right angle atleast
royal basin
no, do not assume the drawing is to scale.
vivid creek
Well I would say OED is a right angle since it perpendicular to OB
royal basin
how do you know "it" (ED) is perpendicular to OB?
vivid creek
Well it because it looks like it is
royal basin
no, do not assume the drawing is to scale.
royal basin
what about the fact that AC and BC are tangent to the circle?
they are explicitly stated as such, too.
royal basin
"circle theorems", what a vague reference.
OBC and OAC are right angles because they're the angle between a tangent and its corresponding radius.
royal basin
royal basin
"paddle"?
vivid creek
Auto correct mb
royal basin
since ED is parallel to BC.
vivid creek
How do the right angles help me tho
royal basin
again, consider triangle OED.
consider also the side OD that i just drew in.
what can you tell me about side OD?
vivid creek
OD would be 13.1cm?
vivid creek
Well I use 10/cos 0.7
royal basin
you are jumping ahead.
i mean, sure, angle EOD is 0.7 radians, so OD = OE/cos(0.7) = 10/cos(0.7)
but there is something else you should be able to tell me about OD
vivid creek
Yes thank you for the help
Could you help me with the next question?
No b
A different one
This one 8
vivid creek
Thank you
obtuse pebbleBOT
@vivid creek Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
sour oriole
I have to find the area in the colored space, using the integration topic.
bold bane
What have you tried?
sour oriole
I already wrote the integral that I think raises the problem.
Can I use latex notation here?
bold bane
Yes.
grizzled shore
$5x+2$
warm shaleBOT
frosst
grizzled shore
Just like normal
sour oriole
$\int_{a}^{b} (f(x) - g(x)) , dx$
warm shaleBOT
Daniel H.
sour oriole
Yeah, I already did that.
And the values that I found was x_1 = 5, x_2 = -2 and x_3 = 2
fierce vale
How did u get 3 solutions??
sour oriole
I took -2 for the a, but I don't know which value fits for the b
sour oriole
How did u get 3 solutions??
Yes
bold bane
How did u get 3 solutions??
One of the functions is cubic.
bold bane
I took -2 for the a, but I don't know which value fits for the b
So what are you having difficulty with?
grizzled shore
Yeah, I already did that.
And the values that I found was x_1 = 5, x_2 = -2 and...
The left point (a) should be negative by the diagram
bold bane
grizzled shore
Then, you can observe that the diagram contains only 2 intersections
You can easily deduce was the right point must be
fierce vale
He doesn't have the full graph though
sour oriole
$\int_{-2}^{2} \left[ \left(10 - 4x - 3x^2 \right) - \left(2x^2 - 10 - x^3)\right) \right] dx$
warm shaleBOT
Daniel H.
fierce vale
b=min(2,5) cause both functions are decreasing when x approaches infinity
bold bane
,w integral ((10-4x-3x^2) - (2x^2 - 10 -x^3)) from x=-2 to 2
warm shaleBOT
sour oriole
Is it possible that this is the area? With decimals?
bold bane
Your answer is correct.
sour oriole
Okay, thank u!
Btw, why the positive value for "b" should be 2 and not 5?
Is it because is a quadratic function?
bold bane
So you found three intersections; x = -2, x = 2, and x = 5. The graph shows the area between two intersections, one of which is in the 2nd Quadrant which means that the left bound is at x = -2. The next intersection after x = -2 is at x = 2.
grizzled shore
^^
wise dragon
$$\text{I need to find the work done by the field: } \
F(x,y,z)=(yze^{xz}-2y,;;e^{xz},;;xye^{xz}+3z^{2})\
\text{along the curve} \newline
\gamma :\begin{bmatrix} cos(t) \ sin(t) \ 0 \end{bmatrix}\;\;\;\; t\in[0,2\pi]
\
\
\text{I found the curl of } \overline{F}:\
\nabla \cross F = (0,0,-2)\
\text{and } \gamma':\
\gamma':\begin{bmatrix} -sin(t) \ cos(t) \ 0 \end{bmatrix};;;; t\in[0,2\pi]\
\text{but if I try to use Stokes theorem I get:}\
\oint_{\gamma}^{}\overline{F}\cdot d\overline{s} =\int_{0}^{2\pi}\int_{0}^{1}(0,0,-2)\cdot(0,0,1)\rho ;;d\rho d\theta = -2\pi\
\text{If instead I just calculate the work by the definition:}\
\oint_{\gamma}^{}\overline{F}\cdot d\overline{s} =\int_{0}^{2\pi}(-2sin(t),;;e,;;cos(t)sint(t)e)\cdot(-sin(t),cos(t),0)dt =\int_{0}^{2\pi}(2sin^2(t)+ecos(t))dt=2\pi
\
\
\text{Now, the correct answer is }2\pi \text{ but I don't understand why Stoke isn't working. Why is there a minus? I think I oriented something wrongly, I'm not quite sure how to transform the };;\hat{n}dS;; \text{ in flux integrals most times, so any help with that also would be amazing}$$
obtuse pebbleBOT
$$\text{I need to find the work done by the field: } \\\
F(x,y,z)=(yze^{xz}-2y,...
wise dragon
Now, the correct answer is 2pi but I don't understand why Stoke isn't working. Why is there a minus? I think I oriented something wrongly, I'm not quite sure how to transform the ndS in flux integrals most times, so any help with that also would be amazing
polar fossil
very anticlimactic for all that work you put in to typing it but that's how it goes sometimes 
wise dragon
lol
thanks
any tips on the ndS with flux integrals?
usually i do the cross product between the partial derivatives of the parametrization for n
polar fossil
i'll be honest it's been like five years since i've done that
fierce vale
How to find the roots of a cubic without trial and error ?
obtuse pebbleBOT
How to find the roots of a cubic without trial and error ?
fierce vale
I mean without guessing a root and then doing long division
fierce vale
Which formula?
ruby path
Tbf Tartaliga's algorithm isn't that hard to learn
ruby path
Which formula?
Just look it up, it's impractical though
I wonder
,w solve ax^3 + bx^2 + cx + d = 0
warm shaleBOT
fierce vale
It looks ugly
native inlet
Which formula?
$\small{x=\sqrt[3]{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)+\sqrt{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}+$\$\sqrt[3]{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)-\sqrt{\left(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\right)^2+\left(\frac{c}{3a}-\frac{b^2}{9a^2}\right)^3}}-\frac{b}{3a}}$
native inlet
couldn't even do the LaTeX
the formula is pretty garbage
and that's only one solution
to get the other two you need to use roots of unity
ruby path
if you need the roots then you're just better off using Newton's method
fierce vale
I guess I'm sticking with trial and error fot now 😅
fierce vale
if you need the roots then you're just better off using Newton's method
Couldn't find it
warm shaleBOT
XxMrFancyu2xX
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
native inlet
probably should learn it tbh
fierce vale
Is it the same as tartaliga's algo?
ruby path
nope
native inlet
idek wht Tartalgia's Algorithm is?
ruby path
It's a substitution to make a standard cubic polynomial into a depressed one, and then theres a formula that gives you the roots of a depressed cubic
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
crude dawn
helo
crude dawn
i have a question
i have function
and i have to find all asymptotes
and im vondering
its looks like that
and
plucky rivet
do you have work?
crude dawn
are there asymptotes near to 1 and 3?
like that
sorry for that awesome drawing
but u know the idea...
like
the domain ends at 1
but the 1 is in domain
if i calculate limit to 1 from risght side it will be an symptote or not
i know there are oblique ones but vertical too?
pseudo swift
nah they're not asymptotes
there should be at least one coord going to infinity
not the case here
flat aurora
is this correct
obtuse pebbleBOT
is this correct
plucky rivet
if you are asking if these blue lines are equal, then no
flat aurora
if you are asking if these blue lines are equal, then no
why
plucky rivet
now its true
you should be able to see the curve on right side of the vertical orange line
pure geode
yes it is
pure geode
I don't think you can with just that info
flat aurora
I don't think you can with just that info
thats what is given only
pure geode
Well without knowing how much it was shifted up by you can't find that length
because it changes
flat aurora
Well without knowing how much it was shifted up by you can't find that length
could u help me with this?
.close
quasi walrus
can someone help me finish this problem please
https://docs.google.com/document/d/1EeodbtF_Dh3bDzxUWhUe6AgnWJ9WFcUSalWTvXy6PsU/edit
obtuse pebbleBOT
can someone help me finish this problem please
https://docs.google.com/document/...
fierce vale
can someone help me finish this problem please
https://docs.google.com/document/...
Take a screenshot the doc is not readable
royal basin
the doc just says
aiuna98n091j0328jh019j09j0iamosmaosmoamomsaosinf09j09j0129j-19-21210-10kimn0
over and over...
quasi walrus
Take a screenshot the doc is not readable
sorry i fixed it
accidently left edit mode on
ofc someone has to be like that 🙄
tardy epoch
Take a screenshot the doc is not readable
Do this
obtuse pebbleBOT
@quasi walrus Has your question been resolved?
obtuse pebbleBOT
@quasi walrus Has your question been resolved?
obtuse pebbleBOT
@quasi walrus Has your question been resolved?
timid silo
obtuse pebbleBOT
timid silo
How would I go about solving this?
rigid lintel
you could draw it
timid silo
you could draw it
wym?
rigid lintel
draw the lines |x| and |2x - 1|
timid silo
draw the lines |x| and |2x - 1|
you mean graph it? i cant rlly understand what your trying to say
rigid lintel
pen and paper
timid silo
dude im not gonna graph it
rigid lintel
the alternative is algebra
timid silo
it would take too much time
rigid lintel
lots of it
timid silo
the alternative is algebra
yes please
rigid lintel
alright well
divide by |x| on both sides
should make it pretty easy
and also check the case that x = 0
timid silo
its okay i found a solution
dapper silo
it would take too much time
graphing it would take literally 5 seconds if you know what to look for
timid silo
its alr
i just found out
i could set 2x-1
to the positive
and negative
of abs value of x
.close
long hedge
find difference quotient of f(x)=x^2-6