I run to this step, what should I do next
#help-10
wicked crown
well if you wanted to turn a -3 into a 1 you could for example divide the row in a way that the -3 would turn into a 1
wicked crown
what would your matrix then look like ?
I dont think it should cause an error if you do it right
at least thats what I would do
because you know
you cant really do any further subtraction with any of the two other rows
spiral bluff
Yes I know
wicked crown
because then you would destroy the leading zeros
so you are left with mulitplying or dividing
and thats totally legitimate
have a try
wicked crown
what was it ?
spiral bluff
Thanks a lot!
x=6 , y=-3, z=-2
And I have checked it by substitution into the three equation
wicked crown
and also a tip, we also actually had to do it. Write down the operation you are doing, so your way of solving the system becomes simpler to follow and you will profit from it too, thats why I wrote those comments over my arrows
your book also does that, in a slightly different way
spiral bluff
I think they are aggravating since I'm not sure whether my steps are correct or not
But I will take your advice
I learn a lot from you guys today, thanks from the bottom of my heart
wicked crown
I think they are aggravating since I'm not sure whether my steps are correct or ...
its really just
- getting the first leading zero row by subtraction or addition
- using that row to get the row with two leading zeros
- "padding" those rows now by dividing or multiplying until you only have ones after the leading zeros
- Swapping the row so that you have this triangle
wicked crown
I learn a lot from you guys today, thanks from the bottom of my heart
np, good luck for the future!
obtuse pebbleBOT
@spiral bluff Has your question been resolved?
dry pumice
can someone explain implicit differentiation to me? i don’t understand it
obtuse pebbleBOT
obtuse pebbleBOT
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south inlet
obtuse pebbleBOT
south inlet
I am doing u-substitution, I don't understand where the tan(u) went?
it just disappeared
pls @ me
south inlet
ok so they integrated sec(u) tan(u)
dont you have to do inverse product rule or something?
whats the steps to integrating that
high lily
derivative of sec(u) is sec(u)tan(u)
south inlet
oh
high lily
they're just considering a common derivative
graceful quail
So I've tried to solve this by hand and with my calculator and for some reason I can't get the right derivative
obtuse pebbleBOT
So I've tried to solve this by hand and with my calculator and for some reason I...
graceful quail
random ocean
in this expression, this dy/dx +1 is in bracket
quiet ingot
is this a typo? and if not how would i solve (f g)(2)
obtuse pebbleBOT
is this a typo? and if not how would i solve (f g)(2)
heavy thicket
is this a typo? and if not how would i solve (f g)(2)
It’s f*g
reef cedar
obtuse pebbleBOT
reef cedar
im not sure how to begin answering this
reef cedar
Which means it’s in the direct center
ok
heavy thicket
Now the radius is between 12 and 78
And the degree restriction is shown
For better understanding. 7pi/18 is 70 degrees
reef cedar
so its between 290 and 70
reef cedar
I think it is best to draw it out like so
am I finding 4 coordinates
heavy thicket
Yes
reef cedar
Yes
so the right seats would be (12,-7π/18) and (78,-7π/18)
and left seats (12,7π/18) and (78,7π/18)
heavy thicket
As you face the stage
heavy thicket
and left seats (12,7π/18) and (78,7π/18)
Those coordinates are the performers left point of view
You are facing the stage
Not facing the audience
reef cedar
Those coordinates are the performers left point of view
so I just make them opposite
so the right seats would be (12,7π/18) and (78,7π/18) and left seats (12,-7π/18) and (78,-7π/18)
heavy thicket
Yes
obtuse pebbleBOT
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graceful quail
I am trying to find the critical points for this equation
obtuse pebbleBOT
I am trying to find the critical points for this equation
heavy thicket
,w differentiate 7x^9 + 7x^7 + 7x
warm shaleBOT
heavy thicket
I’m not sure if it’s the same
heavy thicket
Derivative either equals 0 or undefined
random ocean
cause is always grt than 0
random ocean
grt than 7 to be precise
heavy thicket
grt than 7 to be precise
Doesn’t matter
random ocean
Doesn’t matter
ye
graceful quail
Got it, so when it evaluates to something crazy like that it means there are no critical points?
warm shaleBOT
dusky ledge
Hello for lim as $ x→0$ of $(e^{7/x}-8x)^{x/2}$ I get lim does not exist
but choices are \
a) $1$\
b) $e^{3.5}$\
c) $3.5$\
d) $0$\
e) $infinity$
And in my attempts I chose and d, but both were wrong. I though I would make it something ^ 0 = 1, or 0 ^ 0, but the e should go zero from left and infinity from right. SO it is different hence limit does not exists. Where did I go wrong?
obtuse pebbleBOT
Hello for lim as $ x→0$ of $(e^{7/x}-8x)^{x/2}$ I get lim does not exist
but ch...
warm shaleBOT
Noby707
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
timid silo
How did you get it as not existing
tardy epoch
L'hopital or something?
dusky ledge
the lim of $e^{7/x}$ as x approches 0 is different from both sides
warm shaleBOT
Noby707
dusky ledge
L'hopital or something?
I will try that, but I could not get an undetermined form anywhere, I will try a few stuff then get back. Thank you!
tardy epoch
the lim of $e^{7/x}$ as x approches 0 is different from both sides
Yea but that's not the question. You have to treat the whole function entirely when taking the limit
dusky ledge
Yea, I can not get rid of the $e^{7/x}$, every time I derive it i get $-7/{x^{2}} * e^{7/x}$ the rest could be delt with somehow, but This is what is bothering me. I got to a long iteration of deriving and it is never ending.
warm shaleBOT
Noby707
tardy epoch
Show your derivative
upbeat island
does the original question have the limit to zero from the left/right?
agreeing with legloas here the graph doesn't seem to have a two-sided limit
dusky ledge
I can't simplify to get rid of the e, but if we disregard the -8 and -8x, we can get the limit to $e^{3.5}$ which is the answer given in the answer sheet. But I want to know how exactly we got there?
warm shaleBOT
Noby707
upbeat island
if thats the answer in the back, then the og prob should be
$\lim_{x \to 0^+} \left( e^{7/x} - 8x \right) ^{\frac{x}{2}}$
citrusmunch
upbeat island
i'll play with it
dusky ledge
i'll play with it
Thank you, What I know is the expression after L'H rule, should simplify to 7/2. But my algebra is not on point to get there, so right now I will review algebraic manipulation.
upbeat island
gotta go soon, but i'll type out some to get you started
warm shaleBOT
upbeat island
\begin{align*}
\lim_{x \to 0^+} \left( e^{7/x} - 8x \right) ^{\frac{x}{2}}
&= e^{\lim_{x \to 0^+} \frac{\ln \left( e^{\frac{7}{x}} - 8x \right) }{\frac{2}{x}}} \\
&\overset{\text{L'H}}{=} e^{\lim_{x \to 0^+} \frac{7 e^{\frac{7}{x}} + 8 x^2}{2 e^{\frac{7}{x}} - 16 x}}
\end{align*}
citrusmunch
upbeat island
couple more L'H needed, or you can make an argument with exponentials outpacing the polynomials and divide those out.
hopefully i didn't make an error somewhere but i gtg, this should get you started!
random ocean
is the ans for x>0+ inf, and x>0- is 1?
obtuse pebbleBOT
@dusky ledge Has your question been resolved?
timid silo
Dang
dusky ledge
couple more L'H needed, or you can make an argument with exponentials outpacing ...
Thank you a lot?
It is a weird question. But the idea was of how certain polynomials behave vs one another? which is faster which are not.
obtuse pebbleBOT
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obtuse pebbleBOT
shadow lava
Need help with these 3 questions
hardy gale
shadow lava
For the first question I put C
Last question D
Just need help when second question
obtuse pebbleBOT
shadow lava
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obtuse pebbleBOT
@dark cargo Has your question been resolved?
obtuse pebbleBOT
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solemn crater
obtuse pebbleBOT
solemn crater
Am I doing something wrong, or is this answer provided in the textbook just dead wrong?
Shouldn't that vertical asymptote still be at x=0, and the horizontal asymptote be at y=2?
bleak maple
Shouldn't that vertical asymptote still be at x=0, and the horizontal asymptote ...
if by reciprocal function they mean 1/x then yeah
they give the right answer on top but the graph doesnt match
solemn crater
Ok sweet. That's exactly what I thought, but I just wanted to make sure since that's the only answer I have, and mine looked way different lol.
Thank you @bleak maple!
.close
obtuse pebbleBOT
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wraith abyss
obtuse pebbleBOT
wraith abyss
how do i solve b
obtuse pebbleBOT
@wraith abyss Has your question been resolved?
cold cloak
If for this question instead of 3 we hve 4..wud there be a genral form for me to use and solve?
obtuse pebbleBOT
If for this question instead of 3 we hve 4..wud there be a genral form for me to...
cold cloak
Yes
cold cloak
How do u get tht by using permutations?
fickle swan
i did it mentally
cold cloak
I got it by giving each a letter a value like A,B ,c
cold cloak
Yes
cold cloak
Yes
fickle swan
has to be in the proper envelope
cold cloak
It means tht its 1- the proability tht no letters are in correct envelope
fickle swan
taht means that at least one letter is in the wrong envelope
thus giving us 2/3
yeah
cold cloak
OK.
cold cloak
Can we derive a general form for these type of questions?
fickle swan
now u have 24 different combinations
fickle swan
yeah
cold cloak
How do we do tht tho?
fickle swan
but thatll take longer the more envelopes and letters u have
cold cloak
I think 10
fickle swan
we have 4! number of permutations
cold cloak
are in wrong envelope
cold cloak
we have 4! number of permutations
Yes
mighty geyser
derangement
cold cloak
derangement
whts tht?
fickle swan
D(4)=4!(1−1/1!+1/2!−1/3!+1/4!)
mighty geyser
cold cloak
Noice
mighty geyser
there's a weird closed form in terms of rounding
fickle swan
9/24
cold cloak
thatll be the amount of permutations without any correct letters
oh..
fickle swan
ook
fickle swan
^thats the probability
obtuse pebbleBOT
cold cloak
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hot hazel
errant night
elo
errant night
just wanted to kno if my answer is right
soz for bad quality
DE = 10
EM = 9.5
DM = 8.5
just wanted to know if this is the wanted angle
is angle EMD 61.2?
or did i do it wrong'
obtuse pebbleBOT
@errant night Has your question been resolved?
obtuse pebbleBOT
@errant night Has your question been resolved?
errant night
<@&286206848099549185>
kindred oasis
Without approximating I got 65.2°, so there's probably something slightly off. I'm not sure where $cos^{-1}(\frac{b^2+c^2-a^2}{2bc})$ comes from.
I found EM in the same way you did, but then since M is the midpoint of BC, I used the triangle with vertexes M, E and the center of the base of the pyramid, which I'll call O. You found EM before, to find MO you can just divide in half the length of BC (since O is the center of the base of the pyramid). With those 2 I found the angle by doing:
$$cos^{-1}(\frac{4}{9.5}) = 65°$$
warm shaleBOT
Andrea276
kindred oasis
@errant night
cold cloak
If a hand of 7 cards is dealt what is the probability of it having 3 kings
cold cloak
Coz
the 7 card hand will have 3 kings and 4 normal cards
And those normal cards are chosen from the reaming cards right
sterile wing
well yes
cold cloak
so its 52-4 kings
sterile wing
Should be 48
cold cloak
so it shud be 52-3 right?
cold cloak
Yes
sterile wing
If it said at least 3 kings
cold cloak
OH..
sterile wing
Then we would have taken 49
sterile wing
Yes
cold cloak
.close
clever nacelle
obtuse pebbleBOT
clever nacelle
I need help with this
timid silo
,rccw
warm shaleBOT
timid silo
What do you need help in
warm canopy
Show the original question, I don't think this is true?
Oh you've just changed it from what you have written
warm canopy
(X ∩ Y ) ∪ (X ∩ Z) is a subset of X ∩ (Y ∪ Z)
This is what you want
tranquil quiver
reopen #help-2 message
obtuse pebbleBOT
reopen https://discord.com/channels/268882317391429632/359052604149465088/100360...
tranquil quiver
oh cool, in that case you could explore negative feedback loops in nature or something
@atomic hornet yeah maybe but what could i do with that
just create another model for it?
because that doesnt really mathematically extend on the topic
since its just another example of the same question
atomic hornet
Oh i thought it was just general research into the topic, not mathematically
tranquil quiver
yeah sorry if i misled you
atomic hornet
Its cool
tranquil quiver
but yeah its gotta be mathematical at least as well
after all it is a math class
atomic hornet
Fair lol
This looks promising https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2440438/
Not really related to temperature though
Its just like 'this question gives an example of a negative feedback loop, so I looked more into that, and found this' kinda thing
tranquil quiver
Not really related to temperature though
well rather than relate to the knowledge part of it it can relate to the mathematical part
atomic hornet
Yeah
tranquil quiver
this is honestly probs too complicated for the class
would of been good when we were doing logic maybe
but boolean network isnt close enough to trig or that i think
do you know of any good example of 3d trig models?
<@&286206848099549185>
obtuse pebbleBOT
@tranquil quiver Has your question been resolved?
tranquil quiver
well im really lost on what to do
obtuse pebbleBOT
@tranquil quiver Has your question been resolved?
nocturne cloak
Could do something basic on differential equations
Like dT/dx = -T would be a simple example
Heat flow is complicated in its full
@tranquil quiver
tranquil quiver
Like dT/dx = -T would be a simple example
not a bad idea but honestly its so hard to decide if something or not is good to do
the teacher isnt very helpful and they havent given an easy to understand expectation of what they want
nocturne cloak
I mean heat flow equations are doable
tranquil quiver
yes doable of course
but its just a question
of whether my teacher would accept that as a valid/good extension
we havent even done calculus and im unsure if he wants the extension to be purely trigonometry relted
related
nocturne cloak
Bruh how is that supposed to be only trig related
Distant relation could also be convergent/divergent series
obtuse pebbleBOT
@tranquil quiver Has your question been resolved?
obtuse pebbleBOT
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Available help channel!
junior kiln
$hi$
warm shaleBOT
TimOn
junior kiln
Just wanted to add the bot to my server
void pelican
junior kiln
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obtuse pebbleBOT
junior kiln
.reopen
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warm canopy
obtuse pebbleBOT
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timid silo
hi
sage iron
how can i approach 3
create a variable
grand oxide
so i was wondering, should i first multiply by sqrt3 - i to lose the i in the denominator[idk whats the name of the numbers below the fraction]
obtuse pebbleBOT
so i was wondering, should i first multiply by sqrt3 - i to lose the i in the de...
grand oxide
mine :D
warm canopy
blocked twice
?
warm canopy
so i was wondering, should i first multiply by sqrt3 - i to lose the i in the de...
Seems like a good start yes
grand oxide
Seems like a good start yes
then I was thinking something like, using the Moavro identity, again idk the english name
De Moivre's
close enough
ye thinking imma go with that
.close
radiant hornet
I dont get what the question is asking
obtuse pebbleBOT
I dont get what the question is asking
radiant hornet
Can someone help me with this
tardy epoch
the question is asking to find the sum of the red arcs in the figure on the left
obtuse pebbleBOT
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fervent cairn
Can anyone VC with me to help me understand what is the line of symmetry for y= -2(x+3)squared by 2 +2
obtuse pebbleBOT
Can anyone VC with me to help me understand what is the line of symmetry for y= ...
warm shaleBOT
KN
fervent cairn
? What u mean transformations
noble kindle
transformations done to the parent function y = x^2
fervent cairn
The question I sent is all the problem says nothing of transforming
noble kindle
well, if you can know how the function is graphed it would help
and knowing how it is graphed would require you to know the transformations
random ocean
Can anyone VC with me to help me understand what is the line of symmetry for y= ...
line of symmetry is x=a, where f(a-x)=f(a+x)
fervent cairn
Should I say that this is a quadratic equation
fervent cairn
I don’t know
random ocean
line of symmetry is x=a, where f(a-x)=f(a+x)
@fervent cairn use this
or its a simple quadratic, drawing rough graph by transformations would give u ur ans
noble kindle
well first of all do you know what a line of symmetry is
fervent cairn
It’s a symmetrical line up
noble kindle
right. its a line of the form x = n such that this line divides your porabola into mirror images
for the graph y = x^2, the line of symmetry would be where?
x = ... ?
fervent cairn
for the graph y = x^2, the line of symmetry would be where?
2?
noble kindle
type in x^2
for what value of x does the function y =x^2 cut into two mirror images of eachother
fervent cairn
0
noble kindle
right
fervent cairn
K so what
noble kindle
Your function is just a transformation of y=x^2
specifically, it moves how many units horizontally (left or right)?
fervent cairn
Left?
fervent cairn
None im guessing since I got 0 for x2
noble kindle
$y = -2(x+3)^2+2$
warm shaleBOT
KN
noble kindle
this is your function
this is a quadratic function in its vertex form
so $y = a(x+b)^2+k$
warm shaleBOT
KN
noble kindle
the b term tells you how many units it moves left/right
it is unintuitive, in the sense that, if your b is positive, you actually move to the left. If your b is negative, you actually move to the right
since your b = 3, it will move how many units in which direction
and that is your line of symmetry
meager granite
obtuse pebbleBOT
meager granite
before the “therefore” why is that equation that passes through the two points of tangency written like that?
where does -alpha^2 + beta^2 over alpha - beta come from
where does it all come from
tardy epoch
this?
meager granite
yup
tardy epoch
slope = rise over run
meager granite
I know that
tardy epoch
Two points with x values alpha and beta. what are their y values?
it's already given above
meager granite
how did they make it equal to that
tardy epoch
how did they make it equal to that
point slope form
tardy epoch
how did they make it equal to that
or explain what you mean with "it equal to that"
tardy epoch
the only thing it’s equal to is what I mean
what's "it"
meager granite
there's like 4 equations there. be specific
the middle one
tardy epoch
factor
meager granite
h how do they turn that big slope into a simple alpha plus beta
tardy epoch
(a^2 - b^2) = (a-b)(a+b)
obtuse pebbleBOT
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tranquil quiver
does anyone know of any examples of patterns in nature that can me modelled using the reciprocal trig functions (cot, sec, cosec)
obtuse pebbleBOT
does anyone know of any examples of patterns in nature that can me modelled usin...
@tranquil quiver Has your question been resolved?
timid silo
My only question is
Whoops sorry wrong quesiton
this is the quesiton
*question
−m3≥−2
-m/3≥−2
would m be less than or equal to 6
or greater than or equal to 6
obtuse pebbleBOT
cerulean pine
would m be less than or equal to 6
yes
obtuse pebbleBOT
Available help channel!
meager frost
The ruler of a fiefdom is busy apportioning rations to her troops;
Times are tough, so she wants to give out as few rations as possible;
The apportionment is subject of these constraints:
- The troops are lined up single file;
- Some troops already have rations;
- Whenever a troop is awarded a ration, one troop immediately next to them on either side must also receive a ration (this effect does not stack);
- By the time apportionment is 1s complete, all troops must have even numbers of rations each, without regard to equal disbursement;
Troop lineup:
(represented as loaves of bread)
[ 2, 3, 4, 5, 6 ]
Solution:
[ 2, 4, 5, 5, 6 ]
^ ^
[ 2, 4, 6, 6, 6 ]
^ ^
Four rations are apportioned total, which is the minimum to answer this riddle, submit an algorithm for computing the least number 0f rations possible to hand out according to these constraints, in addition to determining if such apportionment is is possible in all circumstances or if some are in exception;
obtuse pebbleBOT
The ruler of a fiefdom is busy apportioning rations to her troops;
Times are tou...
meager frost
I've been stuck on this problem for 2 hours.
obtuse pebbleBOT
@meager frost Has your question been resolved?
charred plume
What progress have you made? Do you know when the apportionment is possible?
meager frost
Haven't made much headway. And no I don't know when the apportionment is possible
charred plume
Ok. I may not have time to get you to the end, but here's a few helpful ideas, hopefully:
Adding one to a number swaps it from even to odd or vice versa, much like toggling a light switch. So if it helps, you could simplify the problem a little. Instead of people with integer numbers of rations, just think of a row of light switches. Initially some are on (corresponding to a person with even number of rations), some are off. You have to turn all the switches on. But you can only flip pairs of adjacent switches at a time.
The only reason to suggest that is it's a bit more visual.
meager frost
Ohh. That makes it easier. Thank you
charred plume
At any rate... How many light switches are on at the beginning, and how many need to be on at the end? Is it possible, while only flipping two at once?
For the algorithm, try by hand, in the simplest cases. For example, only two switches off, but with a bunch of on switches between.
meager frost
So I wasn't how many troops their are total but I think I figured it out
My proposed algorithm is
- find the first troop in the line with an odd number of rations
- count the number of troops between this troop and the next troop with an odd number of rations, then add 1; keep track of this number
- disregard these two troops from here on out and repeat steps 1 and 2 for each extra pair of troops with an odd number of rations, adding the results of step 2 to your running total
- multiply your total by two for the final number of rations once no troops with an odd number of rations remain
charred plume
Yeah! Sounds solid to me.
obtuse pebbleBOT
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dense otter
obtuse pebbleBOT
dense otter
this is where i’m at and i’m not sure where to go
i’m stuck at the very bottom part, i know i’m supposed to get a Bessel equation for R but i’m not sure how to find the order of the bessel function
tardy epoch
i forget how to do these, but i can go over this with you if you want
https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Partial_Differential_Equations_(Walet)/10%3A_Bessel_Functions_and_Two-Dimensional_Problems/10.02%3A_Bessel’s_Equation
dense otter
i’m good on bessel function im just lost on this part
but i think i sort of get it now
ima try a couple of things and see what happens
.close
drowsy stump
obtuse pebbleBOT
drowsy stump
how do I explain why its approaching infinity and negative infinity ?
tardy epoch
compare powers of the leading terms to the top and bottom
true sphinx
Okay, so I wanted to calculate how much money I would have to spend in order to get my desired item from a loot box in a video game.
These are the 3 options:
First type has a chance of 17% to contain a pet and a 5% chance to contain a mount.
It costs $0.99.
Second type has a chance of 3% to contain a pet and a 25% chance to contain a mount.
It costs $1.20.
Third type has a chance of 6% to contain both a pet and a mount bundled together.
It costs $0.70.
What would be the most efficient way of getting both the pet and the mount?
My thought process (most likely very incorrect) goes something like this:
For the first box, it would be a total of 17%+5% chance of getting anything, so 22%.
That means around 4-5 attempts to get anything, so around 4-5$.
In the case that I get a pet, I can directly go to the 2nd loot box which has a higher chance of getting a mount, there I would need to do an average of 4 attempts to get the mount, so 4*1.20 = 4.80$
So best case I would get the pet and mount through loot boxes 1 and 2 with a total of 8.80$ - 9.80$ spent.
In the case of the third loot box it would be 100/6 = 16.6 (16-17) attempts at getting both items together, so, on average, 160.70 or 170.70 dollars, which is 11.20$ or 11.90$.
So in my (most likely terrible) solutions, it would make sense to try and directly go for one or the other.
obtuse pebbleBOT
Okay, so I wanted to calculate how much money I would have to spend in order to ...
royal basin
hang on
do i understand correctly that the rewards from box types 1 and 2 are mutually exclusive?
true sphinx
Do you mean if they are the same items?
royal basin
no
what i mean is, do i understand correctly that box type 1 gives you only a pet, or only a mount, or nothing?
true sphinx
Oh yeah, in that case you're absolutely correct. It has the 17% chance to give only a pet, 5% chance to give only mount, or nothing at all
The third box has the 6% chance to give both at the same time or nothing at all
royal basin
right
let me see
i think i know what to do
okay so.
box type 3 is actually the easiest to calculate
true sphinx
I got neck deep into trying something with Geometric distribution, but yeah.. didn't get far 😛
royal basin
there is a 94% chance of each particular box failing
so if you get n boxes of the third type the probability you don't get anything is 0.94^n
and the chance of success is 1 minus that
for box types 1 and 2, the chance of success in n boxes is:
1 - P(only get nothing or pets) - P(only get nothing or mounts) + P(only get nothing)
i've graphed these chances on the y axis against your budget on the x-axis
red for type 1, green for type 2, blue for type 3
so it looks like type 3 boxes give you the most bang for your buck
true sphinx
Okay, so let's say I open 10 (type 3) loot boxes, that would mean:
0.94^10 and the success of getting what I want is 1-0.94^10 = 0.46 = 46%?
royal basin
,calc 1 - 0.94^10
warm shaleBOT
Result:
0.4613848859051
yeah
46.14% more like
just as a comparison, here are the costs to reach 95% certainty you'll get what you want for each box type:
type 1: $58.41
type 2: $118.80
type 3: $34.30
true sphinx
Okay, so let's take the same 10 box example for the type 1 boxes, it would how then?
1 - 0.83^10 - 0.95*10 + 0.78^10 // This would be my chance of getting anything?
true sphinx
oh yeah shii
royal basin
,calc 1-0.83^10-0.95^10+0.78^10
warm shaleBOT
Result:
0.32946040720193
1 - 0.83^10 - 0.95^10 + 0.78^10 // Chance of getting anything on type 1 boxes
royal basin
33% chance you'll get both a pet and a mount
true sphinx
Okay, but then I have the odd case that let's say I take type 2 boxes, which are obviously more costly, but I quickly get a mount (since it has the 25% chance), then proceed to open type 1 boxes to get the pet. Is this possible to account for somehow?
true sphinx
for box types 1 and 2, the chance of success in n boxes is:
1 - P(only get noth...
Also, this quite helped explain the logic behind the calculations, thanks.
royal basin
hm. this feels tricky
lots of casework to be done
but i have a feeling type 3 boxes may still outclass even that strat
wooden warren
show this
royal basin
first instinct is to raise both sides to the power of 2n in your ineq
to get $n^n \overset?< n!^2$
warm shaleBOT
Ann
royal basin
and then $n!^2 = \prod_{k=1}^n k(n-k+1)$ by interleaving the product of $1:n$ in ascending order with the same product in descending order
warm shaleBOT
Ann
obtuse pebbleBOT
@wooden warren Has your question been resolved?
wooden warren
and then?
wooden warren
and then $n!^2 = \prod_{k=1}^n k(n-k+1)$ by interleaving the product of $1:n$ in...
if we prove that (n-k+1)k>n we've won right?
royal basin
yes
royal basin
there's a hint that we should start by showing this
this should say k goes up to n-1 btw
wooden warren
yeah it does
Okay thanks a lot this was easier than i d imagined
you're freakin brilliant at maths
obtuse pebbleBOT
wooden warren
.close
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Available help channel!
plucky shale
Flip a coin and arrange outcomes in a sequence. If there are consecutive heads, they merge into one head.
obtuse pebbleBOT
Flip a coin and arrange outcomes in a sequence. If there are consecutive heads, ...
plucky shale
I have come up with a method to replicate the entire sample space, but it seems a bit ugly.
wooden cipher
whats the question/problem part
plucky shale
Hang on a sec
wooden cipher
oh so youre looking for the probability for length?
plucky shale
basically, you make a sequence of HTHTH, where you put tails between heads and then insert tails and heads into the slots
plucky shale
oh so youre looking for the probability for length?
yep.
The question: If you add up all the possibilities for a starting sequence length n, what length do you get?
wooden cipher
did you make this up or is there a picture you can provide of the source
wooden cipher
thats 6 lol
plucky shale
....whoops
if I add in length
and without the length
Basically, it iterates through the possible merged lengths first (1 to n), and for each of the merged lengths, it goes through the possible number of heads that are shown
I uh, plugged it into desmos, being at wit's end
This ^ gives a linear result
It computes the average length for the sequence length n
I would like to prove it somehow for all n if it is true
wooden cipher
hmm im not sure, this is a hard problem
obtuse pebbleBOT
@plucky shale Has your question been resolved?
plucky shale
Flip a coin and arrange outcomes in a sequence. If there are consecutive heads, ...
need to prove that the average length of a sequence length n, as n increases, corresponds to a linear relation
<@&286206848099549185>
obtuse pebbleBOT
@plucky shale Has your question been resolved?
plucky shale
<@&286206848099549185>
plucky shale
.close
lean pilot
would someone be able to help me with part 2 of this?
obtuse pebbleBOT
would someone be able to help me with part 2 of this?
lean pilot
i got this result for k^
obtuse pebbleBOT
@lean pilot Has your question been resolved?
novel knoll
gonna use Y instead of hat{k} for the min of n samples
so we want to find dist of Y
\begin{align*} F_Y(y)&=P(Y \leq y)=1-P(y\geq y)\ &=1-P(X_1\geq y,X_2\geq y,\dots,X_n\geq y)\end{align*}
now use they are independent and plug in
warm shaleBOT
ScapeProf
novel knoll
would someone be able to help me with part 2 of this?
.
lean pilot
thanks for the ping
ill look thru that now
so is this equal to the pdf of the pareto distribution?
novel knoll
Plug in complement of CDF
lean pilot
we've done an example similar to this but the example was the normal distribution, so i'm a bit confused on how this works
do we do something like d/d F(y)
could you clarify a bit please i'm sorta lost
novel knoll
You are given the PDF of pareto dist
We need the CDF
So either find it in your book or integrate the PDF to find it
lean pilot
alrighty and how would i plug in the complement of what i find
novel knoll
P(X_1>=y)=1-P(X_1<=y)
lean pilot
lemme give it a go
tysm
im working in circles now
so i've gotten the CDF which is -k^(alpha)*x^(alpha)
lean pilot
P(X_1>=y)=1-P(X_1<=y)
and i'm trying to sub that complement in but i just reach a dead end
novel knoll
The CDF should be like 1-(k/x)^alpha
lean pilot
i just put it in wolfram
yeah it gives that but without the 1
sorry made a mistake typing but yeah that's what i used when b=k and a=alpha
novel knoll
,w CDF pareto
warm shaleBOT
lean pilot
i'll try to see what happened
but ok lets use 1-(k/x)^alpha
what would the next step be
novel knoll
Plug that in
novel knoll
P(X_1>=y)=1-P(X_1<=y)
.
lean pilot
yea im just a bit confused on what you mean by plug it in could you show me please
do we let P(X_1 <= y) = (k/x)^alpha
so we get P(X_1 >= y) = 1-(k/x)^alpha which is the cdf for the pareto dist?
lean pilot
it's not the opposite way around?
so its not P(X_1 <= y) = (k/y)^alpha cause that'll make
P(X_1 >= y) = 1-(k/y)^alpha, which is the CDF of the pareto dist
lean pilot
ah righto
lean pilot
So P(X_1>=y)=(k/y)^alpha
and then what do you do after this
lean pilot
so i went back to 1-P(X>=1)P(X>=2)...P(X>=n)
and i think simplifying gives you
1-n(k/y)^alpha
lean pilot
and we have a pareto dist for n=...
oh k
ok im confused how we got n independent paretos for k and alpha
from 1-n(k/y)^alpha
but this is what we want right?
novel knoll
1-(k/y)^alpha_1 * (k/y)^alpha_2 * …
lean pilot
ohhh each alpha is different
ah that makes sense since it was for X_1
so then would i just conclude by saying that it gives you n paretos for k, alpha_i
also k doesn't change the same way alpha does right?
novel knoll
1-(k/y)^alpha_1 * (k/y)^alpha_2 * …
And you need to simplify this
And notice the exponent is another real number
And hence fits definition of CDF pareto
random ocean
compact shadow
$(x\cos(B)+y\cos(A)-z)^{2}+(x\sin(B)-y\sin(A))^{2}=0$
warm shaleBOT
Cogwheels of the mind
obtuse pebbleBOT
Available help channel!
versed gale
obtuse pebbleBOT
versed gale
I'm doubting rules 1 and 2, I would appreciate some clarification
Here's my apparent 'counterexample' for rule 1:
Assume A = {1} and B = {5, 6, 7}, then AxB = {(1,5), (1,6), (1,7)} is a surjection from A to B, but |A| is less than |B|.
Similarly, for rule 2:
Assume A = {1, 2, 3} and B = {4}, then R: A -> B, where R = {(1,4)} is injective, but |A| > |B|.
Is there something wrong with these counterexamples?
novel knoll
What? Surjection is a function f that maps an element x to every element y (so for every y there is an x such that f(x)=y)
Since A={1} f can only map to 1 element
versed gale
Ah I see, it wasn't mentioned in the text that a surjection needs to be a function
There was only talk of surjective, total, injective, and bijective relations
Strange
Fair then, that obviously makes sense
novel knoll
yeah needs to be functions
spare forge
I'm kind of a noob at competition math and I was trying to solve this problem in a book and I can't quite get it:
Find the smallest m/n in which you have 0. ......251
So what I thought was that if you want the smallest m/n, you need to find a value in front of 251 like 1251 where the value isn't prime but divides 10 to the some power. Like 1251/10000 would end in 0.1251 and 1251 isn't prime. But here is where I got stuck: 10000 is always some 2^n times 5^n, but any number ending in 251 can never divide 2 and 5. Can you help me find my mistake?
obtuse pebbleBOT
I'm kind of a noob at competition math and I was trying to solve this problem in...
steep summit
I'm kind of a noob at competition math and I was trying to solve this problem in...
i think it'd have to be something like "find the smallest m" or "find the smallest n" or something?
otherwise it just doesn't make sense
spare forge
i think it'd have to be something like "find the smallest m" or "find the smalle...
it said smallest m over n so like
steep summit
well then i don't know
spare forge
;-;
Im assuming its where m+n is the smallest
But I'm still confused because 251 is prime
steep summit
if it's asking for smallest m+n then yeah it'd be 1251
spare forge
if it's asking for smallest m+n then yeah it'd be 1251
201/799 is 0.25156 though
and 201+799 is 1000
steep summit
oh i thought 251 had to be the last digits
spare forge
Nah it can be anywhere
as long as it has 251
Sorry i phrased the question wrong
In that case do you still have any idea on how to solve it
Because plugging in a 2510 still makes this fraction huge
steep summit
,calc 40/159
warm shaleBOT
Result:
0.25157232704403
thing is 251 is really close to 1/4
in an actual math competition you'll probably wanna find some bounds on m and n to prove it or something
spare forge
thing is 251 is really close to 1/4
Yeah I thought that too
Anyways thanks!
.close
obtuse pebbleBOT
steep summit
32/127 works i think
obtuse pebbleBOT
Available help channel!
lament eagle
Write down the steps and process you follow to determine the total number of manhole covers in the county. This county would be San Joaquin County in California. Do I start by getting the population of the county? I'm not even sure how to start going about this. Thank you all in advance!
obtuse pebbleBOT
Write down the steps and process you follow to determine the total number of man...
random ocean
population density, yeah one of the factors
its more like an economical problem and less of a mathematical
high lily
number of streets would probs be more useful if you can find that
rough knoll
obtuse pebbleBOT
rough knoll
i can zoom in on the pdf
timid silo
Just make me able to read the relevant problem information
rough knoll
timid silo
And whar example did they get this from?
timid silo
Not an expert at this, but looks like a line in general form
As they are finding what C gives 3x+2y = 0
rough knoll
i'm supposed to find the value of 'k' at the end there but obviously 4h-(-9h) is not 0
timid silo
Are you solving a system of linear equations or what?
rough knoll
equation of a circle
timid silo
If your problem is that 4h-9h isnt 0 then scale one of the equations
2x+4y-10=0 is the same solutions as 4x+8y-20=0
That may work, or it may not work, but its the best I got
rough knoll
"2x+4y-10=0" where did this come from?
steep summit
3x+2y comes from the tangent to circle condition
timid silo
"2x+4y-10=0" where did this come from?
I made it as an example
rough knoll
3x+2y comes from the tangent to circle condition
what does that mean? that's the actual base formula?
timid silo
3x-2y can be writen as y=1.5x
Same thing in other form
It probably comes from it being a tangent line
steep summit
the equation of the given tangent line is 2x-3y+9=0
the slope of this line is 2/3
that means the slope of the line from that tangent point to the center of the circle would have to be -3/2
steep summit
that means the slope of the line from that tangent point to the center of the ci...
the question also gives you that this line passes through the tangent point as well
so the example was pretty much solving for the equation of thta line
rough knoll
isn't it solving for the value of h and k?
cuz it's then used in the example to find the center
then radius
steep summit
yep that's what they're doing
steep summit
i'd use the y=mx+c all the way
x-4y-15=0 has a slope of 1/4
implying we have y=-4x+c
sorry -4 since we took the negative of the reciprocal
and you also know that (3, -3) lies on this point so you can plug that into (x, y) and get -3 = -4(3) + c which yields c = 9 i believe
so the equation of the line would be y=-4x+9 or 4x+y-9=0
rough knoll
and you also know that (3, -3) lies on this point so you can plug that into (x, ...
u mean -9?
steep summit
im using y=mx+c so the c has a slightly different meaning in this context
but yeah the final equation is still 4x+y-9=0
rough knoll
cool, but how do i find h and k?
rough knoll
this is what i'm currently working on
on here i already did 4x+y-9=0 but couldnt find h and k
how do i find h and k?
timid silo
Picture them as linear equations, find the intersection
rough knoll
which would mean???
it's 1:31am where i'm at rn so u might need to explain it like i'm 5 years old. how do i find h and k from 4h+k-9=0?
timid silo
You have two equations, lines
Where theese intersect the solution is
So this gives
h = 2.435
k = -0.739
rough knoll
what website is that?
timid silo
Desmos app/website
Its free (really, no ads, and no premium!) and used by millions
obtuse pebbleBOT
@rough knoll Has your question been resolved?
marsh plaza
When proving statement p by contradiction, we use ~p -> c, where c is a false statement
To show that this implication is true, we assume that antecedent (~p) is true and that it implies consequent (c) is also true
But how can c be true if we know it is false?
obtuse pebbleBOT
When proving statement `p` by contradiction, we use `~p -> c`, where `c` is a fa...
drowsy cosmos
It is a known false statement which turns out to be true
like you come to the conclusion 0=1 then you know something went wrong
marsh plaza
It is a known false statement which turns out to be true
How can a false statement be true? That doesn't make sense
drowsy cosmos
if you remember the truth table for implication it says from a wrong assumption you can come to any conclusion
so you assume something wrong and with fruther implication you get to some statement which you know is wrong
that tells you that your assumptions were faulty
kind hawk
if ~p was true, then by our argument c would also have to be true. but c is false, so ~p cannot be true, so p has to be true
But how can
cbe true if we know it is false
that's the point of a proof by contradiction. it can't. so the assumption we did earlier that~pwas true had to be a wrong assumption
obtuse pebbleBOT
@marsh plaza Has your question been resolved?
marsh plaza
Thanks, I think I understand now
lament garnet
claim
lament garnet
does anyone know how to name each side a, instead of a_1, a_2
on the geogebra app
pls ping me when answering, thanks!
obtuse pebbleBOT
@lament garnet Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
fading ravine
I'm reading Kolmogorov's Introductory Real Analysis, and limit points of a subset E of a topological space is defined such as a point $p$ such that $\lvert N_p \cap E \rvert = \infty$ for every neighborhood of $p$
obtuse pebbleBOT
I'm reading Kolmogorov's Introductory Real Analysis, and limit points of a subse...
warm shaleBOT
Andrew071
fading ravine
So I'm confused on this part
Suppose $T$ contains an infinite set with no limit point. Then $T$ contains a countable set $X \coloneqq {x_1,x_2,\dots}$ with no limit points. Hence the sets $X_n \coloneqq {x_n,x_{n+1},\dots }$ for $n\geq 1$ form a closed family.
warm shaleBOT
Andrew071
fading ravine
But I don't see why it should be the case that each $X_n$ is closed
warm shaleBOT
Andrew071
fading ravine
Like what's stopping us from having say a point $p \in T$ such that the only neighborhoods of $p$ are ${p,x_1}$ and $T$, and $p\neq x_j$ for any $j$?
warm shaleBOT
Andrew071
fading ravine
Because then $p \in \bar X_1$, the closure of $X_1$ but $p\notin X_1$ so that $X_1$ is not closed
warm shaleBOT
Andrew071
fading ravine
If it matters, the hypothesis is that $T$ is a compact space (no other restrictions like first countable or Hausdorff)
warm shaleBOT
Andrew071
fading ravine
<@&286206848099549185>
warm canopy
#point-set-topology might help you better
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
graceful bone
obtuse pebbleBOT
graceful bone
can't seem to figure out where I'm going wrong with this flux integral
<@&286206848099549185>
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languid tendon
Can someone help me better understand this problem?
I know it integrates to arcsin, and the function is 2(arcsin(1/2)-arcsin(0)). But I am having trouble getting the answer without the unit circle
obtuse pebbleBOT
Can someone help me better understand this problem?
I know it integrates to arcs...
languid tendon
I guess another way to ask is: I need help with inverse trig functions because I didn't really learn them when I took trig
cedar lichen
If arcsin(1/2) = x, then sinx = 1/2
You just need the value of x that spits out 1/2
languid tendon
so sin(pi/6), and sin(5pi/6) = 1/2 right
cedar lichen
Yes, but we only take the "simplest" value
In particular, for arcsin, the range is [-pi/2, pi/2]
So we take the value of x in that range
languid tendon
is there a rule for the restriction?
cedar lichen
We want arcsin to be a function, so we have to restrict it
languid tendon
right, so we just take 1 of them and you said the "simplest one"
cedar lichen
Which is the smallest, yeah
languid tendon
so -pi/2 is that going clockwise on the circle?
cedar lichen
As for why it's that interval, sin goes from y = -1 to 1 on that interval, so the range of arcsin will go through all the values
languid tendon
oh I see
cedar lichen
so -pi/2 is that going clockwise on the circle?
Yes. Negative angles are clockwise
languid tendon
then with the 2(pi/6) it simplifies to pi/3
cedar lichen
Yep
languid tendon
I have my next exam this Friday. I really hope problems like this are not on it lol. I can't seem to do them without the unit circle
cedar lichen
What's wrong with using the unit circle
languid tendon
I can't use it on the exam
I'm sure I could write it down from memory, but I don't think I remember all of it
cedar lichen
You only need to remember the first quadrant
The other quadrants come about from reflections and such
For example, quad II is the same as quad I, but the x values are all negative
languid tendon
but the radians
it goes from pi/# until pi/2 which then it goes 2pi/3, 3pi/4, etc
what are they increasing by?
cedar lichen
There's no set increase, it's just special values
The first quadrant is 0, π/6, π/4, π/3, and π/2
The other angles can be found by adding and subtracting π
For example, 3pi/4 is just π - π/4
languid tendon
so you subtract it when going around counter-clockwise?
cedar lichen
It depends. It really helps to draw a picture of the angle
languid tendon
so like 5pi/6, how do you get that one
cedar lichen
Notice it's only π/6 away from a 180 degree rotation
Thus the values are going to be the same as π/6, but because it's in quad II, the x value is negative
languid tendon
ok I see how those work now. when you said it's only pi/6 away I see how it's applied now
so 3pi/4 is pi/4 away
cedar lichen
Yep
5pi/4 is π/4 past 180, so it's in the third quadrant
So both x and y are negative
languid tendon
right. yeah this looks a lot simpler now lol. I remember doing a test when I took trig a few years ago and literally just writing the whole thing from memory. and I missed some because of that. but looking at it as reflections makes it so much easier
but yeah thanks for the help, it all makes sense now
.close
low cipher
this is more of a logic question. I have an 8 deck shoe of cards. for example. each deck has 52 cards, making it a total of 416 cards in the shoe.
lets say for example that i've played 32 cards. How can i figure out how many decks are left in the shoe, aka how many decks are left in play, including the one that is being played at the moment. So in context, that would still be 8 since an entire deck of cards hasnt been played yet.
obtuse pebbleBOT
this is more of a logic question. I have an 8 deck shoe of cards. for example. e...
low cipher
but when 60 cards have been played, it would be 7, since a full deck of cards has been now played and the next deck has been started