#help-42
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@balmy saffron Has your question been resolved?
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Wow a new help room
So by definition, the compact subsets are the closed and bounded subsets
I guess B is unbound in R^3
Because there is not constrain for the z component
Oh yay I get it
But if I have to do a formal proof for “the statement D is not compact”, what can I do?
Basically I just do this question intuitively
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Closed by @scenic merlin
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âś…
So I tried to consider y=kx
The answer is not does not exist
May be rationalize can help you to simplify the denominator ,then it will be easy( i think)
Ok I will try
Or you can try approching (0,0) form both the axis like put y=0 and the. X=0 then check their limit
Yay I tried this but i guess I still require the rationalisation and doesn’t prove the limit exists
So the limit is 2
If you approach on the x-axis and y-axis separately
I think you will see the answer.
Ya I tried
But we will get sqrt(2x^2+1) or sqrt(-5y^2+1)
polar coordinates?
And this works
You mean fixing one of x or y?
Ok
idk if it helps ofc
I think it works too
Sqrt2-1 as the denominator
Then we shall simplify the expression by cancelling out each other’s
Thank you guys!

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Closed by @scenic merlin
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so uhh im trying to solve for x but im being left with a remainder
,rotate
i just based it off how my teacher solved it
?
this is not how you solve equations
do you see the difference in the operations now?
did i misinterpret it
In step 1 x is being added on the right side
ohh you mean the sign changed?
but to get to step 2 x was subtracted from both sides
9 was added to both sides
ahhh I see mb
sills
It does say combine similar terms, which is saying get everything with x on one side, and then everything else on the other
ahhh fair enough
guess I misinterpreted it
Solving basic linear equations with variables on both sides of the equation.
oooo thanks
gimme a sec
omg
dang thanks
I got the answer
is it right
thanks man
I'm very proud of you
Closed by @dark knot
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How could I simplify this further? I'm using chain rule to derive here, then substituting t for 5.
@surreal crypt Has your question been resolved?
<@&286206848099549185>
Can you use a calculator?
my main problem is that (-15 625)^(2/3) wont work
just figured it out thanks
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@bold charm Has your question been resolved?
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@brisk cipher Has your question been resolved?
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guys
i never know whether for the coefficient of restitution
say a is the speed of A after collision, and b is the speed of B after collision
is the speed of separation a - b or b - a
@fringe leaf Has your question been resolved?
,tex (\frac{|vb - va|}{|ua - ub|}) = CR?
heethar
@fringe leaf Has your question been resolved?
is this just a fact thats always the case?
its always b - a ???
It isn't always, seems like there are special cases. But this was provided for the case of a one-dimensional collision involving two objects.
what are the special cases?
i only need 1 dimensional with 2 objects
are there any special cases 1d with 2 objects?
There's one for an object bouncing off of a stationary object where CR is V/U where V is the speed of the object after impact and U is the speed of the object before impact.
Object dropped from rest is sqrt(h/H)
h = bounce height
H = drop height
stuff like that
ah alright, ok thanks
Was that the one you were looking for?
I find it very strange that coefficient of restitution is denoted as e
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@fringe leaf Has your question been resolved?
oh yes, thanks for the help forgot to close it.
Closed by @fringe leaf
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how would you do this
@tough granite Has your question been resolved?
,rotate
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whats the series of transformation for that ?
Don't open multiple channels
the guy left me
do you know
Close one of the channels you have open
ok
Hello
Well, I wasn't good with series, so you should wait patiently for someone to come by
oh ok
Hello
hi
I NEED HELP
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Why the caps?
If you read it, then you should know that you need to open your own channel
@small sierra Has your question been resolved?
@small sierra Has your question been resolved?
@small sierra Has your question been resolved?
rearrange the first equation so that its only x, and then work from inside out with stretches and translations
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