#help-27
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this is like one of 8 questions nobody managed to solve correctly
i mean it isn't really difficult but having you do the computations in 90 seconds is rough
ok what if
Let a and b be positive real numbers, and X, Y be random variables such that Y=aX+b. If E[X]=2ab, E[X^2]=5a^2b^2, Var[X]=6a, Var[Y]=48. Determine E[aY^2-bY+ab]
60 seconds ๐ญ
I can get the X part but I'm not entirely sure how to transpose it to Y, or rather how is Y doing that
if I'm mkinag any sense
ok so I assume i express Y as a function of X so
oh man ๐
'and X, Y be random variables such that Y=aX+b'
E[a^3X^2+2a^2bX+ab^2-abX-b^2+ab^2]=E[a^3X^2+(2a^2b-ab)X+2ab^2-b^2]=
ok in this case, how do I do this
is it 5a^5b^2+2a^3b^2-2a^2b^2+2ab^2-b^2
idk what you are doing rn
like
what i'd do is like
1- evaluate Var[X] from Var[X] = E[X^2] - E^2[X} and Var[X] = 6a in terms of a and b
2- Recover a from Var[Y] = Var[aX+b] = a^2Var[X]
3- Recover b from the equation resulting from step 1
4- find E[X], E[Y], and E[Y^2] now that you know a and b
5- use the fact that E[aY^2 - by + ab] = aE[Y^2] - bE[Y] + ab to calculate the lhs
yk I was trying to figure ouot if 5 was real
kk thanks is this still in the same book
yes everythign is there
6a=5a^2b^2-4a^2b^2 => 6=ab^2
a=2
b=sqrt3
E[X]=4sqrt3, E[Y]=E[aX+b]=aE[X]+b=9sqrt3, E[Y^2]=a^2E[X^2]+2abE[X]+b^2,=240+48+3=291
582-27+2sqrt3=555+2sqrt3
am I right
yes
tahts riht
555+2sqrt 3
crazy
ok what if
Given a random variable X is normally distributed with mean 28 and variance 6, what is E[X^3]-E[X]^3
is this kurtosis
i need to go really
but you should propbably open up a new help chnanel
aw
thanks for all the help
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im kinda confused, is this how u do this?
you can't use difference of two squares because they're not the same
so you expand it normally
when i do 7y time x does the order of the y and x matter like can it be 7xy
In other words, no; the order does not matter :)
so 4x^2 - 5xy - 21y^2
me trying to phrase it in the least 'assumes you know what i mean' way possible
yep
Oh that wasn't the issue, you just said "yes" when I think you meant "no" ๐ญ
okay ty i wouldve def got that wrong
oh i thought she said doesnt matter? because she says 'can it be xy', im like yes!
i did say can it be 7xy lasy so i assumed yes was for that xD
but yes order doesnt matter
alright tysm!
and i guess (7x)y = 7(xy) is associativity
so yes multiplication is associative as well (for real numbers)
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
1
I assume we begin with a)
Weird lines underneath?
Ya
Ya those
Any genius
Eh, I don't have reliable internet anymore. Imma let someone else take over instead of you waiting on me every time for 5 minutes. Good luck and sorry!
i can take over if needed
Not hopping in to help but how are you undergrad & this smart
(im just hoppping between help channels to see what people are doing to procrastinate)
MORE MORE JUMP SPOTTED SO PEAK
๐ you flatter me i dont think that im great but i like doing math in my free time



Are you in hs? I'm in y11 hh I feel so behind
Ooo gl!! I'm gonna stop clogging up the channel now hhh glglgl
TYYYY AND YOU TOOO
henlo what have you done so far
A
I'm gonna be making this intentionally harder since it doesn't make sense
wow whoever made this question hates you, I don't even think a is answerable and expressable as just a and b ๐ฅ, since no angles are given on a technicality(one can assume but do we really want to assume)
is this gcse?
ya
draw straight line OP and BC, Q would be the intersection
Yes i did that
BRO IKR
Legit the vectors i did in class is like NOTHING
But then now idk what to do
Since like
No value for P
express OP and BC as a line, Q is the point at which they intersect, find the point at which they intersect
2AP=3PC => AP=3PC/2 => 5PC/2=AC=5AP/3 => PC=2AC/5, AP=3AC/5,
now what we can do is do
with respect to O as the origin,
OP can be expressed as y=tanPOAx I think, then you have BC
similar concept, except it's raised up by a bit, how much?
you can do (y-y1)=m(x-x1) where x1 is the x component of OB and y1 is the y component of OB and m is the slope of BC(or tangent of the angle BC would form against a line parallel to OA, or simply put OBC-(108-BOA)) to which you can then do y= and x= and approximate the values from there(not really approximate)
the issue is expressing it as unit vectors, there is probably a geometry theorem out there that would help you find the sides but here I can only assume from point B to C you have to figure out how much x changes to 2a+b and translate it to the x value you get for CQ
now of course, this is all speculation and you should ask your teacher to specify since it's virtually a "why are we doing this" question
this is all a bunch of snickerdoodle, though
I'm not sure if you have to do this much, although you probably can
Shld i sen the ms
I mean by the x and y thing is that your units are 2a+b
for a specific x amount you calculate in the question, your hypotenuse is 2a+b or something, if you find the ratio from x:2a+b, then you can apply it to other x vals and y vals to find how to express the vectors as a and b
Mark scheme
well go send
do you understand how to find P first?
idt these are correct answers ๐ฅ
i got the same end answers though ๐ค
you want to come up with equations of OP and BC
that is what they are doing
it said in terms of a and b, adding another constant or anything like that would ruin the point of expressing it as solely a and b
these arent constants though
if constants are allowed qa=fb=w(a+b)
where q f and w are constants that allow for the above to be true!!!
idk I might not understand I'm from another country n I don't see the gamma phi k mew stuffs
they also are constants
they are just a variable being multiplied to (a+b) acting as scalars
thats fair
which is a constant in the scenario here
it's multiplied by a factor of say k representing CQ or PQ, where since they are collinear with BC and OP respectively, can be expressed as a constant multiplied to the vector quantities of BC and OP respectively plus OB+BC or OP
it is a findable factor I guess
my wording ain't right prolly
well you ain't finding OQ without breaking the rules of 2d vectors which whatever math this is already has โ๏ธ
you following this OP? the questions don't make sense at least not in conventions I know
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What do the z, x, u, o, and n looking symbols mean? I totally forgot
Is this related to statistics?
yes
right x bar is mean
z means z in this case
z means z
we could never guess that
z is the value after tipification of the variable
isnt that phi?
oh well regional differnece ig
no, that's phi(z), which is the value of the normal
z is the entry value, phi(z) is the exit value
yeah mb
ok I will see if I get a reasonable answer then I will close
Sure
i mean, we already gave you the answer for each of the letters
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.thats... not what you asked tho?
I couldve asked for more help
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How do i get this?
you pick convenient values for x in the range and plug them into the function
e.g. x = -5 < -4 and (-5 - 4)(-5 + 4)/(-5 + 2)^2 = some positive number
Denominator is > 0 so numerator is < 0
And then the rest is sign work
yea he makes a good point here. you can do a shortcut and ignore the sign of the denominator because (anything)^2 is always positive unless anything = 0
yes
Does the latest message counts as !nosols tho
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Hello so I don't have specific type of question about math, but I got an exam/test coming up tomorrow all about integrals. So can someone teach me all about integrals like ranging from neper's number integration up to X and Y-axis integration
I would be appreciate it if someone could give valuable youtube video about integration perhaps
khan academy
Alright
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@pure skiff Has your question been resolved?
I guess
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hey
What is your question?
Also avoid opening duplicate help channels.
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how to find the equation of circle if I have 3 points that are in the same circle
!xy
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perp bisector theorem twice to get the centre
then distance formula for the radius
how u calc the perpendicular bisector
<@&286206848099549185>
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do you have the original problem
I have A= (-2;3) B =(4;5) c =(2;-1) i need to find the circle equation the points are in the cercle
okay
try and find the midpoint
of every point
then find the slope
of each segment of points
and then find the perpendicular bisector
yes
but how u find that
its the perp segment of a segment?
yes
basically the slope is opposite
ahh I see
create point slope formula
then you find the center
also i might be so wrong so idk ๐ญ
(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)
found the intersection between the segments
now how do I calc the distance for radius
Hey
So the circle equation is
X^2+y^2+2gx+2fy+c=0
Which is S=0
Now
U substitute the points in the equation
You will be 3 equations and 3 variables
solve them you will get g,f,c values
whats S?
I see
yes
Do you want me to solve it?
Okay give me few min
Distance formula between the intersection point and any of your original 3 points
ah f yeah
lol
if you have found the center you can use the general form :
(x-h)^2 + (y-k)^2 = r^2
sub any one point and you can find radius
thank you
I see
ty
yh this is distance formula
I prefer this method compared to the system of 3 equations above
can I use this method if I have 2 points and 1 radius
U can also find radius from circle equation
As in the length of the radius?
It's actually the same method but mine is the expanded version of it
instead of h and k u get f and g
it's only the change in variables rest is same
@worthy phoenix Has your question been resolved?
I see thank you
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is the 1/4 applied to all of the numbers including both the numerator and denominator? or only the numerator/denominator or there is some sequence
pi_day
including the numbers like 162 and 2? also 1/4 being applied through a bracket means that the exponents will be practically divided by 4 right?
,tex .exp rules
pi_day
yes to the numbers by distributivity. yes raising to the 1/4 power is the same as dividing the exponent by 4
thanks
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can someone help me with the bottom 2? the shapes are rhombi
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What have you tried?
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can someone help me how to answer this?
!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
@cold badger ?
!nosols, I don't think we should provide a full solution from the get-go?
(if you were not planning to provide a full solution or just the answer, then please ignore this, and I'm sorry for entering.)
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
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i honestly have no clue where to start, i don't see a way in that has any meaningful way to break this question.
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Helloo im kinda struggling here
The farthest i got was 2(โ5+โ3)
But the answer's not on the options
<@&286206848099549185>
Is the answer A, perchance?
Idk, im tryna ask how to do it qwq
Honestly, i think the answer is a
The power of the denominator is greater than that of the numerator
So you end up with something like this (gimme a sec im writing this down)
Okie
How'd u get 1/4?
What does the chinese 8 stand for?
!occupied
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Uh, do u use it all the time qwq?
Use what?
1/4
Brother
what ?
Re read the question
Or whenever there's a number at the bottom
Like something something but it's root
Denominator is 8x
you write 1/8?
Ok
Ah thats what you meant
I mean you dont have to, i did it for clarity
Oh okk thanks
that's lambda
I thought you meant whered i get the 4 from ๐ญ
Its ok ๐
why did you ask that?
My teacher never taught us that
He just taught us
Use the x with the highest exponent
Like
In the question its xยณ
So 3xยณ/xยณ -> 3
5-> 5/xยณ ->5/0-> 0
And so on
Is it correct tho?
Whats the answer?
6
Ghis is the notes qwq
maybe
The logic behind this is basically that finite/infinute tends to 0
Oh ok thankss
me?
So if the denominator has more power, all the finite terms get divided by infinity and so its zero
Nono I'm just doing that to easily check it again
Ah i see thank you
Im gonna have to ask you to circle it
Thats an x ๐ญ ๐
I dont even blame you, its my handwriting
But yeah thats an x
So one of the 4x becomes just 4? Why?
Why does 1 of the 4 has x and the other doesn't?
Theyre both x's
Youre getting firsthand experience on why my classmates prefer not to take notes from me
Wait so there are no 4? It moves to 1/4?
I honestly dont know what you mean by that
They're not 4 but x right?
Just realized i say wait alot
number 4 slowly transforms into 3 or 5
Why is it 5/x meanwhile the ine next to it is 4/xยฒ?
Why does only one of each get xยฒ?
Ok enjoyy
Uh, can u give me an example qwq?
i was being sarcastic with the handwriting
Oh
do you need help
@serene forge Has your question been resolved?
x/xยฒ = 1/x
Ohhh okok thank youu
Amything else?
I'm struggling w no 10 though ๐
I got 0 but the answer's not in the options
10 shouldnt be zero imo, lemme try it
My answer is 2
Are you familiar with the concept of binomial approximation?
Also your answer isnt zero, your answer is undefined
Oh
Are you aware of the binomial theorem?
Kinda
I forgot how to do it lowkey
I think you need to revise your concepts first ๐
Ye ๐ญ
Do you remember the expansion for (1+x)โฟ
I could walk you through it, but if you dont, id honestly recommend you to review your notes first
I'll review ๐๐๐ป
Thank you :'
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hello again
is the potential graph correct? im unsure about a decreasing function yet concave up
draw on a graph paper
i dont have ๐ญ
i dont have a graphing notebook either
there's one website
LOL
It's worng
i see.. which ones in the table did i get it wrong?
I plotted the graph
damn
$1. At x = 0: wrong nature of point
Given: f'(0) = 0, f''(0) = 0
So, (0,0) is a stationary point of inflection
Error: Graph shows a slanted straight crossing
Correction: Curve must have a horizontal tangent (flat touch) at (0,0)
- Concavity change not clearly shown at x = -1 and x = 1
Given: inflection points at (-1,1) and (1,-1)
Error: Curve looks almost straight โ no visible change in bending
Correction:
At x = -1: concave down โ concave up
At x = 1: concave up โ concave down
โ Show clear change in curvature
- Segment near origin drawn almost linear
From table, curvature keeps changing
Error: Middle portion looks like a straight line
Correction: Curve must be smoothly curved throughout, not linear$
แแแฆแฐ
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These are the mistakes
thanks
I might have gotten wrong, like double derivative, in the graph*, just change those ig,
@summer harbor Has your question been resolved?
i see
i cant draw ts ๐ญ
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$|z|=|a+ib|\leq|a|+|ib|\leq|a|+|i|\cdot|b|\leq|a|+1\cdot|b|\leq|a|+|b|\to|z|\leq|a|+|b|$
Mr. Penguin
is it correct?
Yes
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question 7
max value is obvious
it would come at x=6
any other method?
haven't don't differentiation yet
done*
hmm u know to plot graphs ?
not really
if you havent done differentiation try substitution x+2.5 = t
the function will look easier to understand
maybe i think
yep it looks easier
but what do we get from it?
do i hv to open brackets in the end?
fuck around and find out
we can pair up x+1 and x+4 & x+2 and x+3
then t = x^2 + 5x + 5
๐ฅ
lol
you wont build intuition if you just ask for everything, try different things you'll get what works
in the end just opening up brackets
but with bik brain
true
but I was hoping to not have to open the brackets
i alr know the way to open these brackets
take pair with constant sum same
well if that's all that's coming in you all's mind then ig I'll jus make quadratics
kk min value iz 4
at x=-5
thanks and adios
eh
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In our book we have the numerator of the correlation coefficient simplified as on the attached picture. How do we get to the last part where only xi*yi is inside the sum
So
$$\sum (-x_i \bar{y}-y_i \bar{x}+\bar{x} \bar{y})=-\bar{y} \sum x_i-\bar{x} \sum y_i+n \bar{x} \bar{y}$$
Civil Service Pigeon
typo?
$$=-\bar{y} n \bar{x}-\bar{x} n \bar{y}+n \bar{x} \bar{y}$$
Civil Service Pigeon
$$=-n \bar{x} \bar{y}$$
Civil Service Pigeon
$$=-\sum \bar{x} \bar{y}$$
Civil Service Pigeon
@candid mica
how did we get here ? did u remove a message ?
sums are linear
\begin{align*}
\sum (-x_i \bar{y}-y_i \bar{x}+\bar{x} \bar{y}) &= \sum (-x_i \bar{y})+\sum (-y_i \bar{x})+\sum \bar{x} \bar{y} \
&= -\bar{x} \sum y_i-\bar{y} \sum x_i+\bar{x} \bar{y} \sum 1 \
&= -\bar{y} \sum x_i-\bar{x} \sum y_i+n \bar{x} \bar{y}
\end{align*}
Civil Service Pigeon
Hello?
ye gimme a sec, im checking on paper
yeah $\sum x_i=\boxed{n}\bar{x}$(don't forget the factor of $n$)
Civil Service Pigeon
ye forgot that one, but is that really the same result as they give in the book? shouldnt the product of the two means be inside the sum aswell? in the book it is outside (because no parenthesis)
ye forgot that one, but is that really the same result as they give in the book?
Yes, see what I said [above](#help-27 message).
shouldnt the product of the two means be inside the sum aswell? in the book it is outside (because no parenthesis)
Observe what I did [here](#help-27 message) (same as the attached image) and/or the last line of [this](#help-27 message).
If you're asking something else, you need to clarify because I don't follow then.
oh i see where the problem was, i forgot to multiply the leftover sums by 1/n, now it makes sense
thx a lot
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Help? is Y also 1?
Y is not 1
1 is the horizontal line right above the point A
Is AC 1 or just AB?
AB is 1 yes. AC is the vertical length from the x axis to point A on the circle
Is BC .375?
If i dont know the angles, how do i even work this out
you eyeball it and guestimate
How tho
Would B be false?
Tan cant be greater than 1 right since only AB is 1
tan = opp/adjacent = AC/BC ,tan can be greater than 1
Ah yes, im wrong
this is one form of geustimating
keep going assuming that's true
How about C
Sin of B would be
So if BC is .375
AC would be .927
Sin = opp/hyp
So is D false?
C is true
yes
D is false
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How can I slove the piecewise function
!xy
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Actually there is no original problem but I dont speak English enough to understand the YouTube so asked here in chating
There should be a problem?
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what is my constraint function here?
draw a diagram
i did
i only have the volume function rn, which is the objective im p sure
V(x) = (16-2x)(7-2x)x
i guess the constraint can be represented as an inequality, but we haven't been taught how to solve inequalities yet
is there any other form that it can take?
what is x in this scenario?
@stark spruce Has your question been resolved?
the length of one of the squares cut out the corner?
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@stark spruce Has your question been resolved?
umm what is the constraint on x as mentioned in the question
well yeah that's it
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if the dot product is commutative, how come $\vec{\nabla} \cdot \vec{F} \neq \vec{F} \cdot \vec{\nabla}$? I understand nabla isn't exactly a vector, and is just an operator, but if we treat it as a vector in operations, why is this the case?
prfn
the latter is just notation iirc
nabla really isnt a vector
it's just convenient enough to think of as a vector
As you write, the dot product is commutative between vectors of scalars because multiplication between scalars is commutative. Nabla is a differential vector, and if you have $f$ a function, then $\partial_x f$ and $f\partial_x$ aren't the same object
Rafilouyear2026
the latter is the directional derivative iirc
yeah it is
this makes sense
L notation though
yeah i found it weird too but its convenient
I do use it a lot though, because it's good enough and convenient
there's a thread of me not getting this in hlounge for like 1 hour straight lol
until i realised its just notation
also
in hs I would have just manipulated one side to get to the other
but genuinely i dont even know how to work to that
is it fine for me to get them both to a point where they equal each other?
sure
yeah this was the thing i was tripping up on too lol
stewart right
well i was doing physics not calculus 
so i was happy enough to just leave it at 'ok notation'
i mean this is fine, but 6 weeks for the whole of calc III is a cram fest
thanks guys though
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u australian or what
europe
oh ok
i mean its also one thing if its like a math course but im an engineering student who has to do 5 other courses at the same time </3
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So I don't see how c can be any constant. Shouldn't it be f'(1). Also can someone explain why bโ 1
@robust bobcat Has your question been resolved?
<@&286206848099549185>
ok so i didnt really understand your doubt but, the reason why b is not equal to 1 is becouse log (1) with any base gives us 0
which in this f(x) since its log(x)/log(b) b should not be equal to 1 since it makes the function tend to infinity
and hence we remove 1 from b's domain
and also b > 0
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help
you also start from the left, usually

oh
what if its in arabic
what if the expression is intentionally ambiguous
flip a coin
what if today is April 1?
What if math?
What math
Math math
what what
Math what
if
What
hat
at
t
|| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| |||| ||
this ones too hard idk if I can help
7
right? idk my calculator was giving me some wrong answer
the answer is some number, one can assume that much
how
Black magic
6/2(1+2) = 6/6/7 = 7
67
67
67
67
67
67
jesus this generation is so cooked
can't even do a simple abuse of notation problem??? smh
S=6/(2*3)=1
S=6/2*3=9
S=S
1=9
0=8
0=1
0=7
S=7```
colon at the beginning suggests this may be a ratio. x : 7
place a \ before any * or the like if you want to cancel markdown^^
:(
*or did i?*
mods reading this channel:
is this game good? I wanted to play it for a while now
Yes
Nes
yes
yes (i have no clue what this game is)
here is my dog i love him very much
its "No I'm not a Femboy"
he said woof
$wo^2f$
Krish
gotta simplify
ah of course
Rutherford
<@&268886789983436800>
the icon is sakurako oomuro from yuruyuri btw
he says hi
he looks very fluffy and huggable
He's not as fluffy as he used to be
ill be the first
Oh I'm sure not ๐
he almost got it!
he's gotten so good at catching
like... he's caught those cubes from a room away
woah thats crazy
So this is a help channel not for discussing random stuff
i am getting help
we still dont know what the answer is
the dog was helping
yes
in all seriousness, might wanna close this channel now
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Iโm having difficulty determining dv/dt when there are two unknowns dr/dt
And dv/dt
How do I relate this back to the waterโs change in volume with respect to time, based on the given information?
<@&268886789983436800>
You do have dr/dt! It can be related to dh/dt.
Is this to do with triangles?
That's actually a great strategy. Similar triangles can get it
Ok, uhh lemme try and look back at lecture notes
This is what I kinda looked over, and then I plug back into the volume possibly of the cone
Am I on the right track or did I make any mistakes?
I agree
Thank you
np
Was there another way to solve it without using similar triangle ratios?
Iโm curious, cuz I wasnโt like told about that
With the given information
yeah
you can write everything in terms of h\
and solve it
you can google it, just keep in mind that r scale proportionally with h
Got it
and substitute back, the ratio is basically the same, geometric facts
Ah. Alright
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<@&268886789983436800> discussion of banned topic as per #changelog + using images to evade possible filters.
take a day to read #changelog. also, dont post nonsense in help channels.
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Determine the smallest $n$ such that for all subsets $S$ of ${1,2,...,30}$ where $|S| =n$, there is atleast two elements of $S$ that are coprime
Copter
the original question was n = 16, but that was pretty trivial so i was curious to what the lower bound could be
Well for {1,2,...,30} it can't be 15 or lower, because for those numbers there will be a set of only even numbers
Do you know it's 16? I only know that it's more than 15 right now
yes
split {1,2,..,30} into {n,n+1} by pigeonhole they must contain atleast one of {i,i+1}
what if S were subsets of {1,2,...,m} tho what would that look like
same thing?
floor(m/2) + 1 or something
Yeah that should be it
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mainly confused on part a
write out the lhs
the index is in the function argument
you wpuld get an alternating parttern
you'll notice a pattern
yes
im just not sure how u would explain it in the question
if r is an integer then 2r is even and 2r-1 is odd
$$
\begin{aligned}
\sum_{r=1}^{2n} &= -f(1) + f(2) - f(3) + f(4) - \cdots - f(2n-1) + f(2n) \
&= (f(2) - f(1)) + (f(4) - f(3)) + \cdots + (f(2n) - f(2n-1) \
&= \sum_{r=1}^{n} (f(2r)-f(2r-1))
\end{aligned}
$$
artemetra
this is sufficient 
huh interesting! yes that also works
you might need to elaborate on the bounds tho
you should
because the first sum is until 2n while the second one is until n
sorry i dont understand
ph i see
write down the last terms using n
so i should go on until 2n
like here
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I would like to resolve this
f(x) = \frac{x^3 - 3x + 1}{x^2 - 1}
perform a polynomial long division first
$f(x) = \frac{x^3 - 3x + 1}{x^2 - 1}$
Minฮป
(worth asking, what do you mean by "resolve"?)
@keen estuary Has your question been resolved?
find the limit of f(x)
x tending to?
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.close
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in v=E/(B*sintheta) shouldnt the value of theta which minimizes v be -dx? this will make the denominator a very tiny negative value (based on magnitude) which after division gives a negative value with a large magnitude which is a very small number.
!xy In particular, is there a specified domain for theta or other restrictions?
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no
wait mb
its based on the lorentz force formula when force = 0 and so that we get E = -(v X B) and then by using a X b = absintheta and rearrangement we get the above result
there isnt a domain restriction for theta at least its not explicitly mentioned
my textbook says that theta should be 90 degrees
The angle you put into a cross product is usually an angle in [0, 180], so your v should only be defined for like theta in (0, 180).
negative value with a large magnitude
what exactly are you trying to minimise here? v makes me think it's a vector (velocity?) which doesn't really have a "negative value". Are you trying to minimise the magnitude of v?
At least for magnitudes, which I assume is what we're considering
velocity vector
right, okay. so we're trying to minimise the magnitude of the velocity vector, right?
hmm makes sense cross product represents the area of a parallelogram and area cant be negative
oh wait yea thats the magnitude formula
sorry mb
yes, and magnitude cannot be negative
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if B^2 + B is similar to B, and B is an nXn matrix, then what are thr permissible values of n?
Means there's some invertible P such that:
P'(B^2 + B)P = B
Where I'm using ' for "inverse"
ohh right i forgot about that one
Obv we can distribute but I'm not sure what to do after that
eigenvalues but op doesnt know those
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โ Original question: #help-27 message
i know eigenvalues
<@&268886789983436800>

If B is a zero matrix, is this not always true for any n?
@sturdy yew Has your question been resolved?
i think it has to be a zero matrix
you might want to check my notes tho
this is assuming rankB=n
which your conclusion is immediately contradicting
$C^{-1}BCB^{-1}\ne I$ in general
SWR
oh yeah mb
@sturdy yew you still with us?
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โ Original question: #help-27 message
yes yes
mb
so
the sets {v_i^2 + v_i} = {v_j}
from the arguement that the eigenvalues of similar matrices are equal
what do i do from here
v_i represents an eigenvalue of the matrices
<@&286206848099549185>
swr mentioned that B = 0 always works
shit mb B has to be real and invertible
so B = 0 is rules out
yes
My gut tells me n needs to be even, but I have no proof
why does ur gut say so
- because n=1 doesn't work
- even square matrices kind of give you a way to fake imaginary numbers, but this is all instinct
i wish i had ur instincts