Where can I find police auto or repo auctions?
I'm looking for police auctions for cars or repo car auctions online or in PA ... free would be a huge plus... but I've signed up and paid for one and it was a fake so I want to know its a 100 percent real! Thanks!
US $15,000.00
you should be able to find these in your local newspaper under auction
These should be avoided; you don’t know the condition of the vehicles, the maintenance history, or anything important when purchasing a car from either one of these places. Look at http://www.autotrader.com or another reliable ’site. Be sure to see the car in person 1st before any $$ changes hands. Be wary of some of the websites (won’t use names but one ends with “list”)…lots of ripoff people out there :}
Your band now has both a scary guitar, a snare drum and a really cool set of towers to play in, but will the Abstract Propeller and the Flummox Roller Coaster keep the repo men distracted long enough for you to move the Camaro and Sunbeam somewhere safe.
Take a cross section of the sphere and cylinder. Let R be the radius of the sphere as the problem states, and let r be the radius of the cylinder and h be the height of the cylinder.
Then (h/2)^2+r^2=R^2
For a cylinder, V=pi*r^2h
Thus V=pi*(R^2-(h^2/4))*h = pi(R^2h-h^3/4)
We wish to maximize the volume.
dV/dh = pi*(R^2-3h^2/4)
set dV/dh = 0
Then h=2R/sqrt(3)
We can justify that this value of h produces a maximum by using the second derivitive test. Note that the second derivitive of V with respect to h is
-3pi*h/2
and when h=2R/sqrt(3) we have
-3*pi*R/sqrt(3)
which is negative for all R>0 (and the radius of a sphere certainly is)
Thus the graph of V is concave down, and V ' =0, hence a maximum occurs there.
When h=2R/sqrt(3), we have that
V=4*pi*R^3/(3*sqrt(3))
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