Answer:
The volume of the new can is [tex]V_2=160x^3\pi\ cm^3[/tex]
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube
Let
z ----> the scale factor
In this problem
[tex]z=x[/tex]
The volume of the original can is
[tex]V_1=\pi r^{2}h[/tex]
The volume of the new can is
[tex]V_2=z^{3}V_1[/tex]
[tex]V_2=x^3(\pi r^{2}h)[/tex]
we have
[tex]r=4\ cm\\h=10\ cm[/tex]
substitute
[tex]V_2=x^3(\pi (4)^{2}10)[/tex]
[tex]V_2=160x^3\pi\ cm^3[/tex]