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

i think c. but don't trust it.

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!