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Sagot :
The surface area of a sphere is [tex]S=4 \pi r^{2} [/tex] where r is the radius;
The circumference is [tex]2 \pi r=27 in[/tex] so that [tex]r= \frac{27}{2 \pi }=4.5 in [/tex] using this value we can evaluate the surface area of 1 ball as:
[tex]S=4 \pi r^{2} =4 \pi (4.5)^{2} =254.3 in^{2}[/tex]
For 100 ball you get: [tex]100*254.3=25430 in^{2} [/tex]
The circumference is [tex]2 \pi r=27 in[/tex] so that [tex]r= \frac{27}{2 \pi }=4.5 in [/tex] using this value we can evaluate the surface area of 1 ball as:
[tex]S=4 \pi r^{2} =4 \pi (4.5)^{2} =254.3 in^{2}[/tex]
For 100 ball you get: [tex]100*254.3=25430 in^{2} [/tex]
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