Mathematics, 02.12.2019 03:31 Mah1937

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. provide explanations so i can see where i messed up?

If you can i keep getting stuck ice cream in the shape of a sphere sits atop a cone as

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Answer from: hannahbear3747

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

$2.67=3\ cm$

Answer from: Quest

Its equal to d=7.07in

Answer from: Quest

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

For the figures below, assume they are made of semicircles, quarter circles and squares. for each sh

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