Please help if you can i keep getting stuck Ice cream in the shape of a sphere sits atop a cone as shown in the diagram below. Assume there is no ice cream inside the cone until after the ice cream melts. The diameter of the sphere and the diameter of the cone are both 4cm, and the height of the cone is 7.5 cm. Part A: Determine whether the cone could contain all of the ice cream if it melted. Part B: What would be the smallest cone in height in whole centimeters that would allow the cone to contain all of the melted ice cream if the diameter of the cone remains unchanged. Part C: If the container of the ice cream changed to a cylinder as shown in the diagram below, what would be the smallest height of the cylinder needed to the nearest whole centimeter to contain the melted ice cream. Assume there is no ice cream n the cylinder before the ice cream melts. Please provide explanations so i can see where i messed up?

Answer:Part A) The cone couldn't contain all the ice cream if it melted.Part B) The height of the cone would be [tex]8\ cm[/tex]Part C) The height of the cylinder would be [tex]3\ cm[/tex]Step-by-step explanation:Part A) Determine whether the cone could contain all of the ice cream if it meltedstep 1Find the volume of the ice cream (sphere)The volume is equal to[tex]V=\frac{4}{3}\pi r^{3}[/tex]we have[tex]r=4/2=2\ cm[/tex] -----> the radius is half the diametersubstitute[tex]V=\frac{4}{3}\pi (2)^{3}=\frac{32}{3}\pi\ cm^{3}[/tex]step 2Find the volume of the coneThe volume is equal to[tex]V=\frac{1}{3}\pi r^{2}h[/tex]we have[tex]r=4/2=2\ cm[/tex] -----> the radius is half the diameter[tex]h=7.5\ cm[/tex]substitute[tex]V=\frac{1}{3}\pi (2)^{2}(7.5)=\frac{30}{3}\pi\ cm^{3}[/tex]step 3Compare the volume of the sphere and the volume of the cone[tex]\frac{30}{3}\pi\ cm^{3} < \frac{32}{3}\pi\ cm^{3}[/tex]The volume of the cone is less than the volume of the spherethereforeThe cone couldn't contain all the ice cream if it melted.Part B) What would be the smallest cone in height in whole centimeters that would allow the cone to contain all of the melted ice cream if the diameter of the cone remains unchangedThe volume of the cone is equal to[tex]V=\frac{1}{3}\pi r^{2}h[/tex]we have[tex]V=\frac{32}{3}\pi\ cm^{3}[/tex][tex]r=2\ cm[/tex]substitute in the formula and solve for h[tex]\frac{32}{3}\pi=\frac{1}{3}\pi (2)^{2}h[/tex]simplify[tex]32=(2)^{2}h[/tex][tex]32=4h[/tex][tex]h=32/4=8\ cm[/tex]Part C) If the container of the ice cream changed to a cylinder as shown in the diagram below, what would be the smallest height of the cylinder needed to the nearest whole centimeter to contain the melted ice creamThe volume of the cylinder is equal to[tex]V=\pi r^{2}h[/tex]we have[tex]V=\frac{32}{3}\pi\ cm^{3}[/tex][tex]r=2\ cm[/tex]substitute in the formula and solve for h[tex]\frac{32}{3}\pi=\pi (2)^{2}h[/tex]simplify[tex]\frac{32}{3}=(2)^{2}h[/tex][tex]\frac{32}{3}=4h[/tex][tex]h=\frac{32}{12}=2.67\ cm[/tex]Round to the nearest whole centimeter[tex]2.67=3\ cm[/tex]