A: 21,111.55 dm³
- • ~ • - _ ☆ _ - • ~ • -
Solution (if needed to be shown):
Using the formula:
[tex]\pi \: r {}^{2} h[/tex]
For tank A, the volume would be:
[tex] = \: 3.1416 \: (8 \: dm) {}^{2} \:(15 \: dm)[/tex]
[tex] = 3.1416 \: (64 \: dm {}^{2} )(15 \: dm)[/tex]
[tex] = 3.1416 \: (960 \: dm {}^{3} )[/tex]
[tex] = 3015.936 \: d {m}^{3} \: or \: 3015.93 \: dm {}^{3} [/tex]
As for tank B, the volume would be:
[tex] = 3.1416 \: (16 \: dm) {}^{2} \: (30 \: dm)[/tex]
[tex] = 3.1416 \: (256 \: dm {}^{2} )(30 \: dm)[/tex]
[tex] = 3.1416 \: (7680 \: dm {}^{3} )[/tex]
[tex] = 24127.48 \: dm {}^{3} [/tex]
To find out how much greater tank B is than tank A, proceed to subtraction:
[tex] = 24127.48 \: dm {}^{3} \: - \: 3015.93 \: dm {}^{3} [/tex]
[tex] = 21111.55 \: dm {}^{3} [/tex]
