Question

A plane intersects the center of a sphere with a volume of about 113.1 m3. What is the area of the cross section? Round to the nearest tenth.

Please help however answers will get brainiest pls help.

Answer 1

Answer:

Step-by-step explanation:

$if ~radius=r \\volume =\frac{4}{3} \pi r^3\\113.1=\frac{4}{3} \pi r^3\\113.1 \times \frac{3}{4 \pi } =r^3\\\frac{339.3 }{4 \times 3.14} =r^3\\r^3=\frac{339.3}{12.56} \approx 27\\r=3\\area =\pi r^2=\pi *3^2=9\pi =9*3.14=28.26 \approx 28.3 ~m^2$