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A solid metal ball, 6 inches in diameter, weight 32 lbs., along with another solid ball of the same metal, 4 inches in diameter, (weight not given), are melted together to form another ball.  

Determine diameter and weight of the new ball.


Radius of first ball = 6/2 = 3 inches
volume =   (4/3)πr₁³   
    = (4/3)π(3³)   
    ≅ 113.097 in³

Density of metal:
32 lbs/113.097 in³ = (32/113.097) lb/in³

Radius of smaller ball = 4/2 = 2 inches
volume =   (4/3)πr₂³
    = (4/3)π(2³)   
    ≅ 33.510 in³

33.510 in³ ×  (32/113.097) lb/in³ ≅ 9.481 lbs
The second ball weighs 9.481 lbs.

Third ball weighs 32 + 9.481 = 41.481 lbs.
volume = 41.481 lbs × 113.097 in³/(32 lbs) = 146.608 in³
(4/3)πr₃³ = 146.609 in³
r₃³ = 146.609×3/(4π)
r₃ = ∛(146.609×3/(4π)) ≅  3.271 inches
Diameter of third ball = 6.542 inches

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Janet Yang


I can answer questions in Algebra, Basic Math, Calculus, Differential Equations, Geometry, Number Theory, and Word Problems. I would not feel comfortable answering questions in Probability and Statistics or Topology because I have not studied these in depth.


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