.close
#help-4
proper raft
vale dockBOT
vale dockBOT
โ
crimson yoke
what happened?
crimson yoke
first simplify the inner part
proper raft
oh ok
crimson yoke
thats a repeating number
doesn't matter
the exponent can be a fraction
$\sqrt[a]{b^c}=b^\frac{c}{a}$
rocky lotusBOT
Suika
proper raft
x^16-x^7 = x^(9)y^(3). because dividing exponents is subtracting, i believe.
crimson yoke
yes
proper raft
the same way if you multiply exponents by distribution, you're adding them
ok let me try this
so now its ^3radx^9y^3
that works
crimson yoke
yes
crimson yoke
correct
crimson yoke
hahah yeah
proper raft
i love figuring things out so when i use ai and stuff like that it makes me feel dumb
so i approached the help channel for a more
i guess holistic approach?
ok
crimson yoke
oki
proper raft
i think from here on ill try to figure out and maybe use tools to guide me to the right answer
i wont be bothering your day anymore
tysm
.close
vale dockBOT
crimson yoke
i wont be bothering your day anymore
don't mind it
feel free to make another help channel
crimson yoke
i think from here on ill try to figure out and maybe use tools to guide me to th...
good luck!
manic hatch
U can use wolfram to check answers
https://www.wolframalpha.com/
proper raft
U can use wolfram to check answers
https://www.wolframalpha.com/
oh ty
forgot about wolfram
vale dockBOT
Available help channel!
mild jewel
vale dockBOT
mild jewel
i feel like f is constant
i tried f(x)=0, 1, -1 but it isn't equal to the second part
and i don't wanna try more constant functions
snow zealot
You can't possibly gain any certain value by giving random inputs in the function
mortal temple
there is a constant solution
let f(x) = k for all x
then solve k = 1/2 + sqrt(k-k^2) for k
light saddle
is that the CG maths 2025?
dire cloud
let f(x) = k for all x
then solve k = 1/2 + sqrt(k-k^2) for k
theres also the domain of the sqrt
k-k^2>=0
mortal temple
yes, if k does not satisfy k-k^2 >= 0 then it cannot be a solution
dire cloud
so k belongs to [0,1]?
mortal temple
yes
dire cloud
let f(x) = k for all x
then solve k = 1/2 + sqrt(k-k^2) for k
yeah this is giving 2 solutions of k in the given range for me
mortal temple
i think one is an extraneous solution
dire cloud
i think one is an extraneous solution
simplify thy english please
mortal temple
there is only 1 solution
mortal temple
(2+sqrt(2))/4 is in that range but it isn't a solution
mortal temple
squaring an equation introduces extraneous (false) solutions
dire cloud
oh yeah it doesnt work in the original equation
mortal temple
exactly
dire cloud
gives a negative value inside the square root
ah i get it know
but what if f(x) isnt a constant?
dire cloud
yeah i think i miscalculated
dire cloud
(2+sqrt2)/4 works and (2-sqrt2)/4 doesn't work
this seems right
dire cloud
but what if f(x) isnt a constant?
?
mortal temple
part 2 asks us to prove that f is periodic
vale dockBOT
@mild jewel Has your question been resolved?
light saddle
I think you can show that || f is 2-periodic ||
mortal temple
but i got this
$f(x+2)=\begin{cases}f(x) &f(x)\geq 1/2\1-f(x) &f(x) < 1/2\end{cases}$
rocky lotusBOT
Axe
light saddle
I found that since || f(x+1) ^2 - f(x+1) + 1/4 = f(x) - f(x) ^2, we can let g(x) = f(x+1)^2 - f(x) + 1/8. then g(x+1) = -g(x) and g(x+2) = g(x) . Then, you can solve a quadratic to find f, and since f >= 1/2, there is only one value for f(x) ||
f is determined by g so f is periodic
did I go wrong somewhere?
mortal temple
why is f >= 1/2 ?
light saddle
f(x) = 1/2 + sqrt (...)
mortal temple
oh yeah, then f(x+2) = f(x) right?
light saddle
I think yea
using g maybe you can make any 1-periodic function satisfy property E as long as you define it properly on [0, 2]
vale dockBOT
vale dockBOT
Available help channel!
tidal citrus
How to do this?
x = 500
y = 1000
(5(2x) + 2x) + y - 300 + 2(8(1.25 * 2)) + 0.5 * 2
hardy coral
How to do what?
tidal citrus
That equation
hardy coral
That's not an equation
hardy coral
If you want to calculate the result of the expression given those values of x and y, then just replace all the x's by 500 and all the y's by 1000
tidal citrus
like what is the answer to that
red tulip
brackets first, then multiplication, then addition or subtraction
tidal citrus
and explain how you did it so i know how to do it
red tulip
adding and subtracting doesnt matter the order
but u need to do brackets and multiplication first
tidal citrus
can you help
hardy coral
Take (5(2x) + 2x) + y - 300 + 2(8(1.25 * 2)) + 0.5 * 2
Replace x by 500 and y by 1000:
(5(2*500) + 2*500) + 1000 - 300 + 2(8(1.25 * 2)) + 0.5 * 2
Then calculate...
tidal citrus
what now
red tulip
write it out
tidal citrus
Take (5(2x) + 2x) + y - 300 + 2(8(1.25 * 2)) + 0.5 * 2
Replace x by 500 and y by 1000:
(5(2500) + 2500) + 1000 - 300 + 2(8(1.25 * 2)) + 0.5 * 2
Then calculate...
what now
red tulip
go through each part
tidal citrus
Which one
red tulip
starting with whats in brackets
hardy coral
Calculate the insides of parenthses, then the multiplications, then the additions, ...
tidal citrus
Can u show me
hardy coral
No
tidal citrus
there are 2 brackets
red tulip
take the 5(2(500))
hardy coral
You can try though, I'll correct you if you want
tidal citrus
hmm
red tulip
whats 2 times 500
tidal citrus
1000
red tulip
and whats 5 times 1000
red tulip
so you have (5000 + 2(500)) + 1000 - 300 + 2(8(1.25(2))) + 0.5(2)
red tulip
finish off the first bracket
red tulip
i said the first one
tidal citrus
( ?
red tulip
i meant whats inside the brackets
tidal citrus
Ohhh
tidal citrus
so 6000 right?
red tulip
do you see how the process works?
tidal citrus
Kind of
red tulip
it technically doesnt matter if you do the first set of brackets or the last one as long as you are doing the brackets first
tidal citrus
Ok
red tulip
and you just simplify it as you go along
tidal citrus
What now
red tulip
its now 6000 + 1000 - 300 + 2(8(1.25(2))) + 0.5(2)
try doing the second set of brackets
tidal citrus
2(8(1.25(2)))?
red tulip
yes
tidal citrus
Ok what do i do first
red tulip
start with whats the deepest into the brackets
tidal citrus
So 1.25?
red tulip
think about the brackets as layers, you do the bottom layer first
yeah
1.25(2)
work your way outwards
red tulip
why 8?
tidal citrus
you said work your way outwards
red tulip
yeah start with the very inside
tidal citrus
i did
red tulip
again work your way outwards
tidal citrus
so 2.5 x 8
red tulip
ye
red tulip
20
tidal citrus
ok so what now
red tulip
times by 2
tidal citrus
ok so 40 right?
red tulip
because theres that 2 on the left side of the bracket
yes
now its 6000 + 1000 - 300 + 40 + 0.5(2)
its pretty close
tidal citrus
0.5 x 2 is 1 right?
red tulip
yes
tidal citrus
So 6000 + 1000 - 300 + 40 + 1
tidal citrus
So its 7741?
red tulip
wrong
tidal citrus
What is it then
tidal citrus
so then its
red tulip
-300 is 6700
tidal citrus
what now
tidal citrus
red tulip
you used the right method at the end there was just a small calculation error
you said 7741
vale dockBOT
@tidal citrus Has your question been resolved?
spring juniper
The problem I have been attempting to solve is calculating the Kelly fraction of a 5 leg parlay with legs with corresponding odds of winning: 62.65% 59.58% 59.49% 59.49% and 58.18% and the entire parlay has payouts such that 5 correct pays 10x 4 correct pays 2x 3 correct pays 0.4x. I lack an understanding of why exactly taking the log is what yields the best long term growth so an explanation of that would be great or a reference to something. I'm not very experienced when it comes to probability theory especially when its sequential such as it is here so that's where I'm stuck. Also it would be nice if someone could check my work thus far to ensure its right
vale dockBOT
The problem I have been attempting to solve is calculating the Kelly fraction of...
@spring juniper Has your question been resolved?
spring juniper
<@&286206848099549185>
vale dockBOT
@spring juniper Has your question been resolved?
spring juniper
covert lance
@spring juniper well, there are several levels at which you can understand why taking the log is optimal. Are you familiar with a term called ergodicity?
spring juniper
no
vale dockBOT
@spring juniper Has your question been resolved?
vale dockBOT
@spring juniper Has your question been resolved?
vale dockBOT
@spring juniper Has your question been resolved?
lucid goblet
???
vale dockBOT
@spring juniper Has your question been resolved?
sharp hinge
i think bot doesnt know if ur question is resolved
lyric sundial
i think bot doesnt know if ur question is resolved
Yeah you have to click @spring juniper
pure forge
desert knot
it works because the amount of money you have after a series of bets is a bunch of independent random variables multiplied together
like if i bet 40% of my money, i'm multiplying it by a random variable that's either 0.6 or 1.4
the EV of final money is the product of the individual EVs, so if you just want to optimize that then you optimize each individual EV
But the problem with that is you'll almost always end up falling short of that EV. We want to know roughly what the median outcome should be, not the mean
from the central limit theorem we can notice if we're adding a bunch of RVs together, the median should about equal the mean
maximizing median outcome is the same as maximizing the median log of the outcome
and log(X_1 * ... * X_n) = log(X_1) + ... + log(X_n)
so the best way to obtain the median log is to maximize each individual E[log(X_i)]
and that corresponds to the best way to obtain the median value
vale dockBOT
vale dockBOT
Available help channel!
odd pelican
how do i prove something is bijective
vale dockBOT
how do i prove something is bijective
steady charm
you can prove that it is injective and surjective separately
mortal temple
someone pointed out the other day that it's strictly increasing, and that theta(0)=0 and lim x->infinity theta(x)=1
the first part implies injective, and the second part implies surjective
odd pelican
is that all the proof i need?
mortal temple
i don't know, perhaps you want to justify this with theorems from your textbook
odd pelican
my notes never mention limits in this class
sour needle
my notes never mention limits in this class
I suspect you are expected to use the definitions of injectivity and surjectivity
odd pelican
yea i think so
sour needle
i'll go through the injectivity proof with you, then you do surjectivity
Let's start with the def
odd pelican
thank you but im kinda clueless tbh my prof is not good at explaining
sour needle
a function f is injective if and only if whenever f(x_0) = f(x_1), x_0 = x_1
essentially that just means that unique inputs have unique outputs so they are one-to-one
like any other iff proof, you need to do both sides
so we can start with f(x_0) = f(x_1)
charred fulcrum
you don't need both sides
odd pelican
a function f is injective if and only if whenever f(x_0) = f(x_1), x_0 = x_1
is x_1 just x is included in 1
sour needle
x_1 just denotes another input
charred fulcrum
it's either f(x1)=f(x2)=>x1=x2
or x1!=x2 => f(x1)!=f(x2)
sour needle
x sub 1
charred fulcrum
for every x1,x2 of the domain of theta
sour needle
it's either f(x1)=f(x2)=>x1=x2
or x1!=x2 => f(x1)!=f(x2)
you do need to prove both sides
anyway, let's choose two arbitrary inputs from R^+ x_0 and x_1
plug those in to theta
we get
x_0 / (1 + x_0) = x_1 / (1 + x_1)
charred fulcrum
you really don't need both sides
the definition of injective is just a "then"
it's a one way implication
sour needle
is it not an iff?
mortal temple
Thunder is right
charred fulcrum
same with surjective
sour needle
didn't realise it's a one way implication, that makes things easier
mortal temple
Yeah it's f(a)=f(b) => a=b, that's a one-way implication
sour needle
we get
x_0 / (1 + x_0) = x_1 / (1 + x_1)
from here we can cross multiply to get rid of the denoms and then group all the variables on one side
x_0 ( 1 + x_1 ) = x_1 ( 1 + x_0 )
x_0 + x_1 x_0 - x_1 - x_1 x_0 = 0
then we can look to factor here
x_0 ( 1 + x_1 - x_1 ) - x_1 = 0
and simplify
x_0 - x_1 = 0
clearly this final statement can only be true when x_0 = x_1
which means that any arbitrary element of the domain maps to a unique element in the range under theta
odd pelican
from here we can cross multiply to get rid of the denoms and then group all the ...
is there a way to write this in like actual math terms like that one bot on the server does cause im lost just looking a this
sour needle
sure, give me a sec
odd pelican
ty
sour needle
[ x_0 ( 1 + x_1 ) = x_1 ( 1 + x_0 ) ]
[ x_0 + x_1 x_0 - x_1 - x_1 x_0 = 0 ]
then we can look to factor here
[ x_0 ( 1 + x_1 - x_1 ) - x_1 = 0 ]
and simplify
[ x_0 - x_1 = 0 ]
rocky lotusBOT
shsgd
sour needle
this is just a rough draft, you will need to refine it in your final answer. But hopefully this helps you understand how you use the definition for a proof
sour needle
anyway, let's choose two arbitrary inputs from R^+ x_0 and x_1
you'll want to start with a sentence like this
mortal temple
Then if $\theta(x_1)=\theta(x_2)$...
rocky lotusBOT
Axe
sour needle
Consider $x_0, x_1 \in R_{\geq 0}$, then if $\theta (x_0) = \theta (x_1)...$
is \greq not >=?
mortal temple
\geq
rocky lotusBOT
shsgd
sour needle
good call
odd pelican
I genuinely just dont understand it might be cause of the concerta wearing off but like i just dont get it
sour needle
what part don't you get?
odd pelican
Everything bro i look at the question and just dont know where to start/what to do
mortal temple
Start: $\theta(x_0)=\theta(x_1)$
Goal: $x_0=x_1$
rocky lotusBOT
Axe
odd pelican
This is for injective?
mortal temple
Yes
sour needle
Have you had a class on writing proofs?
like different kinds of proofs, common language to use, structure etc.
odd pelican
This for the start?
odd pelican
like different kinds of proofs, common language to use, structure etc.
I saw some types of proofs
Its only my third week
Second*
mortal temple
I think you're misusing $\in$
rocky lotusBOT
Axe
mortal temple
No, 0 and 1 are not sets
sour needle
that symbol means whatever is to the left of it is an element of the set to the right of it. Not sure if you have been introduced to sets yet
sour needle
ok, in the og question theta is defined as a map from R^+ (positive real numbers) to [0, 1)
so we chose two elements of the domain, R^+, and called them x_0 and x_1
in other words, x_0 and x_1 are both positive real numbers
odd pelican
Can i just use x_1 and x_2
sour needle
you can use whatever two variable names you want
they just represent arbitrary elements of R^+
right, I need to go, im sure Axe will answer any other questions you have. Good luck
odd pelican
So this is ok?
odd pelican
Okok and now i do like: replace x by a real number?
mortal temple
x1 and x2 are nonnegative real numbers
odd pelican
What is better 1 or 2
mortal temple
You can just swap those
odd pelican
Swap what
mortal temple
$\theta(x_1)=\theta(x_2)$. On veut montrer que $x_1=x_2$
rocky lotusBOT
Axe
mortal temple
Now use the definition of theta
odd pelican
the x\1+x?
mortal temple
Yeah
mortal temple
No
odd pelican
๐ช
odd pelican
where did the division go
mortal temple
we get
x_0 / (1 + x_0) = x_1 / (1 + x_1)
^
odd pelican
mortal temple
Yep ๐
odd pelican
and now what
odd pelican
[ x_0 ( 1 + x_1 ) = x_1 ( 1 + x_0 ) ]
[ x_0 + x_1 x_0 - x_1 - x_1 x_0 = 0 ]
then we can look to factor here
[ x_0 ( 1 + x_1 - x_1 ) - x_1 = 0 ]
and simplify
[ x_0 - x_1 = 0 ]
rocky lotusBOT
Noah
odd pelican
exactly this?
mortal temple
The variable names are wrong
odd pelican
why is the second parentesis negative
am i dumb
?
oh wait nvm
u want it to be = 0
Now i have this @mortal temple
I think this is good
vale dockBOT
@odd pelican Has your question been resolved?
odd pelican
how do i show its also surjective
odd pelican
?
odd pelican
<@&286206848099549185>
vale dockBOT
@odd pelican Has your question been resolved?
vale dockBOT
@odd pelican Has your question been resolved?
indigo condor
i cant do this im not smart help me please
vale dockBOT
i cant do this im not smart help me please
flint phoenix
i cant do this im not smart help me please
which question?
indigo condor
1 b and d and 2 b
indigo condor
Okay, do do ou know pythogoras' theorm?
tbh this is the first year im trying in school
and its late
so if it not the a2+b2=c2 then no
rocky lotusBOT
wai
indigo condor
ah
woeful oxide
i cant do this im not smart help me please
why are you solving for slope
OHH
shit bro i cant read
mb
wait no
i can read
why are you solving for slope??
indigo condor
dude idk what im doing im lost
woeful oxide
dude idk what im doing im lost
dyk what angle of elevation and depression is
woeful oxide
i did that like 2 days ago
yeah man calculating tangent is a calc conept its 100% not talking abt tangent lines
indigo condor
idk this is the first time im trying in school
i did the work before but its been so long
woeful oxide
what grade r u in
indigo condor
11
woeful oxide
uhh for starters
angle of elevation/depression is the angle that is made relative to two pointss
if you had a pole thats 20m tall
and you looked at the top (assuming you are both at ground level)
the angle that you look up at from is the angle of elevation
or if the pole was 20m into the ground
ykw look at a diagram
if normal eye level was the horizontal line
and i looked down to j
the angle that forms inbetween the line and j is the angle of depression
or if i was at j
looking at T
the angle made there is the angle of elevation
ay @indigo condor u still w me
woeful oxide
bro ๐
anyways
watch a video on sohcahtoa
or like
theres prolly one by oct
hes useful
woeful oxide
in still going straight into welding
underwater welding makes money ig ๐ญ
indigo condor
i can oxy stick and mig weld rn
indigo condor
i can
i get good marks in ir
i passed a class last year with a 51%
i get good grades in welding tho
woeful oxide
51.. ๐
indigo condor
yes
woeful oxide
yeah man trades r going to save you
indigo condor
they better
woeful oxide
bros going to be making 150k a year with 0 life expectancy ๐ญ
ngl that shit is impressive tho
indigo condor
im just trying to get good grades this year becuase my dad will help me with a camaro
woeful oxide
well shit
indigo condor
bros going to be making 150k a year with 0 life expectancy ๐ญ
life spand at 45 max
woeful oxide
id reccomend joining a peer tutoring club if you have one
and then grinding everyday
atleast an hour a day
indigo condor
i have a tutorial becuase i need help with classes
woeful oxide
tutor?
indigo condor
tutor?
its like a period i go into a room with other and there is like 6 teatchers
to help
woeful oxide
ahh gotchu
indigo condor
whith any work thats needs to be done
indigo condor
im trying
i have 3 thing 1 is 2 sides and 1 im basicly done and the other is 3 pages 2 sided
and its 11:45
woeful oxide
id reccomend watching a video on trig
and then if you get stuck ask us
we cant rlly teach you an entire unit
(well we can but not many people would)
woeful oxide
thats what i thought but we all start somewhere
like naturally im very good at learning maths
indigo condor
i aint
indigo condor
i have verrrrry bad adhd
indigo condor
7-9 hours a day sometimes
same if hours where like muinuits
indigo condor
cya\
vale dockBOT
@indigo condor Has your question been resolved?
vale dockBOT
wintry raptor
You made a mistake when evaluating the definite integrals
vale dockBOT
vale dockBOT
Available help channel!
blazing goblet
vale dockBOT
blazing goblet
Don't have an idea on where to go from this
jaunty flame
Don't have an idea on where to go from this
use euclidean distance
vale dockBOT
@blazing goblet Has your question been resolved?
blazing goblet
Yeah I get that however, how would I go about it since I don't have the other coordinate
<@&286206848099549185>
dark pewter
So the set of points where Q could be is a circle with radius 17 and center at P, right?
blazing goblet
Yeah I think so
(not a clue what you said, but I searched it up and it checks out)
dark pewter
have you learned about equations of circles?
blazing goblet
No I only just started geometry as a 9th grader
I'm 2.5 weeks in
<@&286206848099549185>
vale dockBOT
@blazing goblet Has your question been resolved?
jovial edge
Pythagorean triple
midnight pier
Hello. I am trying to work part a and part b now
vale dockBOT
Hello. I am trying to work part a and part b now
midnight pier
What I have is this: {11, 13, 17, 19} - prime
{12,, 14, 16, 18, 20} - even
15 - odd, not prime and of the card is numbered 15, Silena gets another chance
For Silena to win, she has to get a prime, so I have 4/10, but I guess it's not that simple
vale dockBOT
@midnight pier Has your question been resolved?
midnight pier
I guess I may have to work out if she picks 15 and the second pick is prime
I multiplied 1/10 and 4/9 to find if she chose 15 and gets a second pick
final adder
the event that silena wins in round 1, is mutually exclusive to the event that she wins in round 2, correct?
midnight pier
Oh yeah
final adder
so what can you do about two favorable cases that are mutually exclusive?
final adder
wdym?
midnight pier
Oh, I think I misunderstood
final adder
so what can you do about two favorable cases that are mutually exclusive?
your answer should be an operator c:
midnight pier
I misread
midnight pier
so what can you do about two favorable cases that are mutually exclusive?
Add them alone
There is no intersection to subtract
final adder
Add them alone
yes!
oki so i see that you have worked out each probability for win in round 1 and round 2
final adder
yep!
midnight pier
Wait, I am trying to understand the mutually exclusive part again
Is it considered two rounds
or just 2 tries
A second try if she picked 15
final adder
or just 2 tries
same thing, just diff terminology
we can use tries if you want though!
midnight pier
Alright no prob
So, it's not mutually exclusive because when she picked out the 15, she did not put it back in the bag?
So there would be nothing in common for the intersection?
final adder
it is mutually exclusive, no?
final adder
So there would be nothing in common for the intersection?
ye
midnight pier
it *is* mutually exclusive, no?
Yes
I think I get it now
I am going to look at part b
(b) only one card is taken from the bag, given that Silena loses.
Oh, is this conditional probability?
I was just thinking even/total
I was just thinking like 5/10
one moment
I am writing it
final adder
ok!
midnight pier
in my book
P(one card| loses) = P(one card n loses)/P(loses)
If for part a, the proability of her winning is 4/9
midnight pier
ok!
Can the probability of her losing be 1 - 4/9?
I got 5/9 if I do the long working out similarly to when I got 4/9 or if I subtract 4/9 from 1, I get 5/9
midnight pier
Can the probability of her losing be 1 - 4/9?
Initially, I wrote winning. I meant losing
Now to find the intersection P(one card n loses)
I think that would just be the 5/10
I got 9/10
I am going to look at part c
final adder
Can the probability of her losing be 1 - 4/9?
yes
midnight pier
Nice
Alright
This looks like a Union thing
Like some addition
"either" , "or"
If she takes two cards, is it that when she takes up the first card for it to be prime, it would be 4/10 and the second card to also be 4/10 ? or 3/9
I believe someone asked this in class. Hmm. Let me check if I can find what they said
midnight pier
I believe we are to assume that the card is not replaced
Hii. I lost internet connection
final adder
Hmm, I would approach it by calculating the probability that she loses
midnight pier
ohh
final adder
In general, when you see โat leastโ, try to calculate the opposite case
weary pilot
Hello. I am trying to work part a and part b now
what are you using to do the problem
like whats your general framework
id recommend using 'states' to solve these kinds of problems
midnight pier
In general, when you see โat leastโ, try to calculate the opposite case
Ah, thanks
midnight pier
id recommend using 'states' to solve these kinds of problems
What are states?
weary pilot
oh nvm its not a game with an indeterminate amount of rounds
states is helpful for questions where you dont know how many rounds take place
the 'canonical way' to do them is with an infinite gp
midnight pier
what are you using to do the problem
I think this would be for part d
weary pilot
anyway thats irrelevant
midnight pier
Sure!
weary pilot
midnight pier
In general, when you see โat leastโ, try to calculate the opposite case
I didn't see any "at least"
midnight pier
Thank you. I am reading through it now
final adder
โEither number or bothโ = โat least oneโ
midnight pier
Ohh
I was using this question b as an example with the either or both. This is how I worked part b
midnight pier
โEither number or bothโ = โat least oneโ
I was wondering if it could be worked like that
midnight pier
Would it be possible to explain how I could go about implementing it for part d?
midnight pier
I was wondering if it could be worked like that
How do I know when to use the at least or when to use the Union like the example above with the 24 prizes? @final adder
weary pilot
lemme see if i can do it my prob is rusty
midnight pier
Sure, that's fine
midnight pier
lemme see if i can do it my prob is rusty
How is it going?
midnight pier
Sure no prob
midnight pier
Umm. I actually didn't get to finish look through it
weary pilot
did you understand the given examples
.
weary pilot
yeah elaborate
midnight pier
Why multiply 1/2 * 1
like (1/2)1 - (1/2)(1-p) .....I get the 1-p part
but why (1/2)*1
weary pilot
half the time, you straight up win.
midnight pier
If Alice gets head first, wouldn't it be p?
weary pilot
so the probability of winning half the time is 1
weary pilot
because if you win, you win 
weary pilot
what's the probability of getting heads if you flip a coin?
half the time, you'll get a head
midnight pier
1/2
weary pilot
so there it's $\frac12 \cdot 1$
rocky lotusBOT
ฮ ฮตฯฯฯ
weary pilot
the rest, you wont get a head, so there its $\frac12 \cdot 0$
rocky lotusBOT
ฮ ฮตฯฯฯ
weary pilot
overall, its 1/2.
midnight pier
Alright
weary pilot
...okay
weary pilot
Why multiply 1/2 * 1
do you geddit
i understand its a little weird to write it that way
midnight pier
the rest, you wont get a head, so there its $\frac12 \cdot 0$
Which rest?
I am trying to process it. I guess because I am not used to using the word "time"
weary pilot
alright valid
weary pilot
Which rest?
this just says 'if you get a tail, you wont get a head.'
midnight pier
Because suppose in 10 tries, I get heads 6 times. It wouldn;t be half of the time
midnight pier
do you geddit
Yes!
midnight pier
Let me see again
midnight pier
so did you get the example given
Yes
midnight pier
Sure
I'll start by saying the probability of Silena choosing a prime is 4/10...
multiplied by...
4/10(something) + (5/10)(something)
but then, what if she picks 15, would we factor in that? numbers from 11-20
weary pilot
yes
midnight pier
Okay
weary pilot
so we're finding the probability of her winning
4/10, she wins
so thats $\frac4{10} \cdot 1$
rocky lotusBOT
ฮ ฮตฯฯฯ
weary pilot
see?
west cloud
Is it always multiplied by 1
when it's the probability of winning then it's multiplied by 1
when it's the probability of not winning then it's multiplied by 0
midnight pier
when it's the **probability of winning** then it's multiplied by 1
4/10
weary pilot
okay so what do we have so far
weary pilot
yes
what's the probability of winning if 15 is chosen?
more specifically, what happens is 15 is chosen
midnight pier
more specifically, what *happens* is 15 is chosen
You pick again
weary pilot
yes
so the probability of winning if 15 is chosen is the same as the initial probability
okay
whats our final equation
midnight pier
So 4/10 + 0 + 1/10p? = 2/5 + 1/10p
weary pilot
yea
midnight pier
Nice
weary pilot
now you can just solve for p
south sorrel
Good morning everyone.
How do I integrate:
vale dockBOT
Good morning everyone.
How do I integrate:
south sorrel
$x^2\1+x$
weary pilot
sub 1+x as t
rocky lotusBOT
Blake
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
weary pilot
not that hard to figure that out
weary pilot
yes
south sorrel
$\frac[x^2][1+x]$
stark wedge
$\frac{x^2}{1+x}$
stark wedge
$\frac[x^2][1+x]$
LaTeX command arguments go in {braces}
weary pilot
ykw, almost
haughty fiber
t^2 - 2x-1? lol
well now you have to get rid of the new x you made, right?
south sorrel
well now you have to get rid of the new x you made, right?
t^2 is x^2 + 2x+ 1
and I need x^2
so isn't it t^2 -2x -1?
haughty fiber
it's more direct if you consider that x=t-1
south sorrel
t^2 -2t +1
haughty fiber
it gets you the same value anyways lol
haughty fiber
t^2 -2t +1
yes
south sorrel
$\int\frac{t^2-2t-1}{t}dt$
haughty fiber
this also lines up with the value of x^2=(t-1)^2
rocky lotusBOT
Blake
haughty fiber
yep. now it's all nice and linear, so you can split it up
south sorrel
is that partial fraction decomposition?
haughty fiber
no
south sorrel
I can't remember much from Calc 2 XD
haughty fiber
note that $\frac{a+b}{c}=\frac ac + \frac bc$
south sorrel
$\int\frac{t^2}{t}-\frac{2t}{2}-\frac{1}{t}$
haughty fiber
what is the 2 in the denom coming from
vale dockBOT
vale dockBOT
Available help channel!
tranquil bay
So im arguing with this guy on a youtube short of John Carter swinging a rock around on a chain. The premise of the movie is that heโs significantly stronger on Mars compared to Earth due to gravity.
Assuming John Carter is swinging a rock that has more mass than himself on a chain 3 meters long, is it possible for him to stay standing in a fixed position.
mortal temple
not really, he'd have to dig in his heels, but even then it would be difficult
tranquil bay
If the mass of the rock is greater than that of the thing moving it, wouldnt he move before the rock??
mortal temple
ye
modest kettle
assuming the stone to weigh around 130 kgs
tranquil bay
Unless the normal force of him digging his heels in can close the gap
Which theres no damn way
Unless its the lightest rock known to man
marble furnace
So im arguing with this guy on a youtube short of John Carter swinging a rock ar...
It's more of a biology/physics question, yes we as humans on earth can lift heavier "weights(mass times gravity)" here on earth and since the gravity on mars is lesser that "large" rock with a high "mass" will have a lower weight, so it will be like swinging an ordinary rock on earth for him
tranquil bay
But when we lift those weights its overhead usually using the ground as a normal force because we dont have the mass to lift such heavy objects without pushing off of the earth
So to SWING such a thing is impossible
marble furnace
So to SWING such a thing is impossible
Yes
tranquil bay
It's more of a biology/physics question, yes we as humans on earth can lift heav...
But John will weigh proportionatly less as well
marble furnace
But John will weigh proportionatly less as well
His strength won't decrease
tranquil bay
Yes but his ability to move the rock will decrease because the force binding him to the ground is proportionatly weaker than earth
So he still cant move things with more mass than himself without a normal force
marble furnace
Yes but his ability to move the rock will decrease because the force binding him...
Kinda
Depends on the direction
tranquil bay
As seen in the clip he almost horizontally swings it
I mean the clip is an exaggeration of course
mortal temple
he would have to lean away from the rock until he was almost touching the ground
then his feet would slip
tranquil bay
Right
Ok so im not crazy and the math checks out that he cant swing the damn rock right
modest kettle
hoold on
tranquil bay
Im not tryna nitpick the movie but this douche in the comments basically said im wrong
And that he could totally realistically do that
mortal temple
idk i've never swung a rock on mars but i tend to agree with you ๐
marble furnace
he would have to lean away from the rock until he was almost touching the ground
I don't get which force does this
mortal temple
he's swinging from shoulder height. the chain pulling on him would topple him
tranquil bay
I think blurple may be running the numbers on it
Since he assumed the weight earlier
mortal temple
tranquil bay
And this is a 15kg ball and chain if im not mistaken they compete with when they do shot put
No wait this isnt shot put
Its the other thing
Hammer throw
mortal temple
yeah hammer throw
midnight pier
Im not tryna nitpick the movie but this douche in the comments basically said im...
Why do you even bother lol, debating with people in the comments is such a waste of time
tranquil bay
Well its an interesting question and i may have been wrong
midnight pier
I only troll people in the comments
tranquil bay
yeah hammer throw
But anyways theyre moving their whole body to counter the force of this ball that weighs maybe 20% of their body weight
Even with the gravity on mars being ~40% that of earths
To move such a large object should be impossible especially considering the lack of weight keeping him in place
Im curious what blurple will say
@modest kettle hey are you actually running numbers or should i close the channel
Well blurple if youre still calculating i admire and appreciate your dedication and you can dm me
.close
vale dockBOT
vale dockBOT
โ
tranquil bay
.close
vale dockBOT
modest kettle
Well blurple if youre still calculating i admire and appreciate your dedication ...
not much
the person would have to exert 17700 newtons to push the ground, assuming the stone to be 150 kg making 1 rps
10 times the force needed to jump on earth
and the gravity doesn't seem to affect this aspect
only the pressure exerted on the man's joints change
mortal temple
wouldn't his arms feel the same tension regardless of gravity?
plush timber
wouldn't his arms feel the same tension regardless of gravity?
Nope ig
vale dockBOT
Available help channel!
signal chasm
question
opaque stratus
so do people discuss olympiad problems here
weary pilot
so do people discuss olympiad problems here
flint phoenix
guys so do you have a link for the otis discord server
yeah, this channel is now in use. Probably ask further questions at #discussion #serious-discussion or check out #old-network
flint phoenix
like it cancels out
Becasue the area on the LHS is beneath the x-axis
signal chasm
Becasue the area on the LHS is beneath the x-axis
yeah
flint phoenix
You cannot integrate two areas together if they're not adjacent
signal chasm
wdym adjacent
flint phoenix
wdym adjacent
Like they're enclosed in the same field
flint phoenix
like as in same quadrant?
No, I'm not sure how to explain the concept, so I'll pass it to the next helper
signal chasm
alr
vale dockBOT
@signal chasm Has your question been resolved?
vale dockBOT
vale dockBOT
Available help channel!
severe lodge
vale dockBOT
severe lodge
when it says vdue to vs it wants it where v is shorted right ?
oh or is that the one for Is
hollow rune
severe lodge
they dont really help
severe lodge
k thanks
fair wing
you may close this channel
tribal goblet
Yoooo can someone help me?
severe lodge
you may close this channel
im going to leave it open in case someone cans till help here
craggy mist
im going to leave it open in case someone cans till help here
Hey, it tells you to use superposition so do that, when you consider the voltage source open the current source, and when you consider the current source, short the voltage and add up the effects vectorially.
severe lodge
Hey, it tells you to use superposition so do that, when you consider the voltage...
so i have shorted the voltage source
And simplified to this
craggy mist
so i have shorted the voltage source
So, now you may find the effect due to current source, but be beware it's a supermesh.
craggy mist
do i need to know that for this problem ?
Ig you should
severe lodge
can i do the problem without it ?
craggy mist
Maybe there's a way, but you need to think for that, also if you visualise properly it's a wheatstone bridge with a voltage source inbetween
Sometimes redrawing ckts in a symmetric form, helps in visualization
severe lodge
Maybe there's a way, but you need to think for that, also if you visualise prope...
with the way i have it was going to use current div
but i cant right since the current splits so idk the
I in
craggy mist
but i cant right since the current splits so idk the
I in
Yes we can't do that, i would suggest you to learn supermesh first !
severe lodge
Yes we can't do that, i would suggest you to learn supermesh first !
i have to use superposition
craggy mist
i have to use superposition
We re using superposition, after that we're applying supermesh for one such case of superposition
Well, we can do it without supermesh too, nvm i overlooked
severe lodge
kk
craggy mist
So, this would be a equivalent ckt, so if we short the voltage source (R1| |R3)+(R2| |R4) will be rthe resistance
And so, you may calculate the current
maiden sandal
How many integral solutions to x1+x2+x3+x4=a are there if a=12
maiden sandal
i tried 12-4=8 (8+4-1,8) (11,8)=165 and it was wrong
astral flower
i tried 12-4=8 (8+4-1,8) (11,8)=165 and it was wrong
infinite?
vale dockBOT
@maiden sandal Has your question been resolved?
astral flower
do you mean positive integral solutions?
wild linden
@maiden sandal you're looking for integer partitions of 12 into exactly 4 parts. https://en.wikipedia.org/wiki/Triangle_of_partition_numbers
vale dockBOT
wild linden
(assuming that xi are positive integers)
astral flower
yes
vale dockBOT
โ
astral flower
you can use the stars and bars method
maiden sandal
but isn't that what I did let me think
ok theres another
x1>=0 x2>=-2 x3>=-1 x4=>4 x1+x2+x3+x4=15
<@&286206848099549185>
ashen prawn
x1>=0 x2>=-2 x3>=-1 x4=>4 x1+x2+x3+x4=15
whats the doubt
use pnc or multinomial ig
itd give quicker results
ashen prawn
ok somehow I got it right
type .close if doubt is over
vale dockBOT
@maiden sandal Has your question been resolved?
maiden sandal
.close
quick pewter
vale dockBOT
quick pewter
idk how to start to be honest
i figured the frame would just be 10*10 since 100 square feet but the key was way different
and idk the mural if 10 is wrong
hazy pivot
Well let's call the width of the mural x
quick pewter
okie
hazy pivot
Now what's the height of the mural
quick pewter
length or height
hazy pivot
Whatever you wanna call it, it's a 2d thing
quick pewter
okay if width is x and length is 5ft more then length is 5 + x
hazy pivot
Good
Now what's the width and length of the frame + mural
If it's 2ft wide on all sides
quick pewter
wait how do we know the width is x
tawny moon
Well let's call the width of the mural x
right here
hazy pivot
We assume
hazy pivot
We don't know anything, so we gotta start with something
quick pewter
Now what's the width and length of the frame + mural
is it x+9 for length
hazy pivot
Good
quick pewter
and x+4 for width
hazy pivot
Good
hazy pivot
So now what's the area of the mural
hazy pivot
had to draw this out lowk
That's fine lol
errant gyro
it's better to draw anyway
errant gyro
approaching a geometry question blind is often not recommended for basic geo
quick pewter
area is x^2 + 13x +36?
quick pewter
x^2 +5x
quick pewter
im confused, isnt the area of frame the same as frame + mural
since like
the mural fits inside
hazy pivot
No
errant gyro
no
quick pewter
๐ญ
hazy pivot
Yes but we want just the frame