Kala’s company makes solid metal balls for various industrial uses. A customer wants aluminum balls that have a diameter of 6 in. If Kala must make 70 of those balls, how much aluminum will she need?
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7,912.8 inch³ of aluminum is needed to make 70 metal balls.
Step-by-step explanation:
Step 1:
The area of a sphere is calculated by multiplying [tex]\frac{4}{3}[/tex] with π and the cube of the radius (r³).
In the given problem, the diameter is 6 inches. So the radius of the spheres is 3 inches.
The volume of a sphere,[tex]V = \frac{4}{3} \pi r^{3} .[/tex]
So if r = 3 inches and π = 3.14,
[tex]V = \frac{4}{3} (3.14) (3^{3}) = 113.04[/tex] cubic inches.
So for a single aluminum ball, 113.04 inch³ of aluminum is needed.
Step 2:
If 113.04 inch³ of aluminum is used for 1 ball, then we multiply 70 with this value to determine the aluminum needed for 70 such balls.
The aluminum needed for 70 metal balls [tex]= 70(113.04) = 7,912.8[/tex] inch³.
So 7,912.8 inch³ of aluminum is needed to make 70 metal balls.