Question

Please help if you can i keep getting stuck

Answer 1

Answer:

Part A) The cone couldn't contain all the ice cream if it melted.

Part B) The height of the cone would be $8\ cm$

Part C) The height of the cylinder would be $3\ cm$

Step-by-step explanation:

Part A) Determine whether the cone could contain all of the ice cream if it melted

step 1

Find the volume of the ice cream (sphere)

The volume is equal to

$V=\frac{4}{3}\pi r^{3}$

we have

$r=4/2=2\ cm$ -----> the radius is half the diameter

substitute

$V=\frac{4}{3}\pi (2)^{3}=\frac{32}{3}\pi\ cm^{3}$

step 2

Find the volume of the cone

The volume is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=4/2=2\ cm$ -----> the radius is half the diameter

$h=7.5\ cm$

substitute

$V=\frac{1}{3}\pi (2)^{2}(7.5)=\frac{30}{3}\pi\ cm^{3}$

step 3

Compare the volume of the sphere and the volume of the cone

$\frac{30}{3}\pi\ cm^{3} < \frac{32}{3}\pi\ cm^{3}$

The volume of the cone is less than the volume of the sphere

therefore

The 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 unchanged

The volume of the cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$V=\frac{32}{3}\pi\ cm^{3}$

$r=2\ cm$

substitute in the formula and solve for h

$\frac{32}{3}\pi=\frac{1}{3}\pi (2)^{2}h$

simplify

$32=(2)^{2}h$

$32=4h$

$h=32/4=8\ cm$

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

The volume of the cylinder is equal to

$V=\pi r^{2}h$

we have

$V=\frac{32}{3}\pi\ cm^{3}$

$r=2\ cm$

substitute in the formula and solve for h

$\frac{32}{3}\pi=\pi (2)^{2}h$

simplify

$\frac{32}{3}=(2)^{2}h$

$\frac{32}{3}=4h$

$h=\frac{32}{12}=2.67\ cm$

Round to the nearest whole centimeter

$2.67=3\ cm$