Answer:
The volume of the new can is
![V_2=160x^3\pi\ cm^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/onxfd8ceoa6km3e4ha52rjrfphazaz1ow8.png)
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](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ya8ygeudr1990e9yssmllaxm1tfhn0wvrx.png)
The volume of the original can is
![V_1=\pi r^(2)h](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dfz4l955ecqlx7mdlkzcvrsqrp4pe0suoe.png)
The volume of the new can is
![V_2=z^(3)V_1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/c3ggbils9mtt0vqta90glrwfjrlzec2dck.png)
![V_2=x^3(\pi r^(2)h)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t1t6177khn5achd8qnq93l50zjlkikee2d.png)
we have
![r=4\ cm\\h=10\ cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/aafb1iugegsbn5rtsv4d44hg9gbvevbgol.png)
substitute
![V_2=x^3(\pi (4)^(2)10)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/898obz2ykv1bttlpi5nae5uusvrzqm8yj6.png)
![V_2=160x^3\pi\ cm^3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/onxfd8ceoa6km3e4ha52rjrfphazaz1ow8.png)