A cylindrical container with a radius of 6 cm and a height of 10 cm is filled with water to a depth of 6 cm. A sphere with a radius of 3 cm is placed at the bottom of the container. a) What is the volume of the water to the nearest tenth?
b) What is the volume of the sphere to the nearest tenth?
c) How much higher does the water level rise in the container, to the nearest tenth, after the
sphere is placed inside?

a. 452.4 cm³ to the nearest tenth b. 113.1 cm³ to the nearest tenth

c. 5.0 cm to the nearest tenth

Step-by-step explanation:

a. The volume of the water V₁ = πr²h since it is in a cylindrical container. r = 6 cm and h = 10 cm - 6 cm = 4 cm (since it is filled to a depth of 6 cm measuring from the top, so we have a height from the bottom of 10 cm - 6 cm = 4 cm).

So, V₁ = πr²h

= π × (6 cm)² × 4 cm

= 452.39 cm³

= 452.4 cm³ to the nearest tenth

b. The volume of the sphere V₂ = 4πr³/3 where r = radius of sphere = 3 cm.

V₂ = 4πr³/3

= 4π(3 cm)³/3

= 113.1 cm³ to the nearest tenth

c. When the sphere is placed in the container, the new volume of water in the container would be V = V₁ + V₂

So V = 452.4 cm³ + 113.1 cm³ = 565.5 cm³

Since the volume of water is still the volume of a cylinder, its height h is

h = V/πr²

= 565.5 cm³/π(6 cm)²

= 5.0 cm to the nearest tenth

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]