Hollow sphere

The volume of the hollow ball is 3432 cm3. What is its internal radius when the wall thickness is 3 cm?

Correct result:

r =  8 cm

Solution:

V=3432 cm3 h=3 cm V=V1V2 V1=4/3π (r+h)3 V2=4/3π r3 3/4/π V=(r+h)3r3 k=3/4/π V=3/4/3.1416 3432819.3296 9r2+27r+27=819.3296  9r2+27r792.33=0  a=9;b=27;c=792.33 D=b24ac=27249(792.33)=29252.8656 D>0  r1,2=b±D2a=27±29252.8718 r1,2=1.5±9.50192728988 r1=8.00192728988 r2=11.0019272899   Factored form of the equation:  9(r8.00192728988)(r+11.0019272899)=0 r>0 r=[r1]=[8.0019]=8=8 cm V1=4/3π (r+h)3=4/3 3.1416 (8+3)35575.2798 V2=4/3π r3=4/3 3.1416 832144.6606 V3=V1V2=5575.27982144.66063430.6192

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