Question

PLEASE HELP!

Answer 1

Answer:

The volume of the new can is $V_2=160x^3\pi\ cm^3$

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

$z=x$

The volume of the original can is

$V_1=\pi r^{2}h$

The volume of the new can is

$V_2=z^{3}V_1$

$V_2=x^3(\pi r^{2}h)$

we have

$r=4\ cm\\h=10\ cm$

substitute

$V_2=x^3(\pi (4)^{2}10)$

$V_2=160x^3\pi\ cm^3$