Two containers designed to hold water are side by side, both in the shape of a cylinder. Container A has a diameter of 10 feet and a height of 8 feet. Container B has a diameter of 12 feet and a height of 7 feet. Container A is full of water and the water is pumped into Container B until Container A is empty. To the nearest tenth, what is the percent of Container B that is full after the pumping is complete?

79.4 %

Step-by-step explanation:

volume of cylinder : pi×r²×h

attention ! the question gives us the diameters, so for the radius, we need to cut them in half.

Va = pi × (10/2)² × 8 = pi×25×8 = 628.31853... ft³

Vb = pi × (12/2)² × 7 = pi×36×7 = 791.68135... ft³

100% Vb = 791.68135... ft³

1% Vb = 7.9168135... ft³

now, 628.31853... ft³ have been pumped into B.

so, how often does 1% of the volume of B fit into that filled volume ?

628.31853... / 7.9168135... = 79.36507937...% = 79.4%

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!