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A crystal ball with a diameter of 6 inches is being packaged for shipment. If the crystal ball is placed inside a circular cylinder with radius 3 inches and height 6 inches, how much volume will need to be filled with padding? (The volume of a sphere with radius r is 4/3πr3, and the volume of a right circular cylinder with radius r and height is πr2h.) Use "pi" for π.

If you invest $8000 into an account with an annual interest rate of r compounded annually, the amount of money you have in the account after one year is:

A=8000+8000r

Write this formula again with the right side in factored form.

Find the amount of money in this account at the end of one year if the interest rate is 19%.

There will be $ in his count at the end of year one.

None · Answer

I have news for you. It will be operationally impossible to get the ball into the cylinder because there is no clearance. Other than that, you already have the answer, πr²h - (4/3)πr³, Since r is the same for both ball and cylinder, you have 54π - 36π = 18π, which I am sure you can evaluate.

A = 8000(1 + r) where r is the interest rate expressed as a decimal fraction.
$9520

A = 8000(1 + r) where r is the interest rate expressed as a decimal fraction.
$9520

Battleaxe · Answer

Volume of sphere = (4/3)(pi)(r^3) = 4/3(3.14)(27) = 113.04in^3
Volume of cylinder = (pi)(r^2)(h) = (3.14)(9)(6) = 169.56in^3
Difference = 169.56 - 113.04 = 56.52in^3 needs to be filled with padding

A = P(1 + r/n)^(nt)
P = 8000
n = 1
t = 1
A = 8000(1 + r/1)^(1)(1)
A = 8000 + 8000r
A = 8000(1 + r)

if r = 0.19:
A = 8000(1 + 0.19/1)^1
A = 8000 + 1520 = $9520 balance after 1 year.

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DWRead · Answer

volume of padding = volume of cylinder - volume of sphere

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