Write each decimal volume as a sum using whole-number amounts of the listed units.
a) \(4.007\,\text{m}^3\), using \(\text{m}^3\) and \(\text{dm}^3\)
b) \(12.65\,\text{dm}^3\), using \(\text{dm}^3\) and \(\text{cm}^3\)
c) \(2.3004\,\text{m}^3\), using \(\text{m}^3\), \(\text{dm}^3\), and \(\text{cm}^3\)
d) \(5.08\,\text{L}\), using \(\text{L}\) and \(\text{mL}\)
Hints
- Keep the whole-number part in the original unit.
- Convert the decimal part to the next smaller unit by multiplying by \(1000\).
- Repeat the process when the converted value still has a decimal part.
Solution
1. For part a, \(0.007\,\text{m}^3=7\,\text{dm}^3\), so \(4.007\,\text{m}^3=4\,\text{m}^3+7\,\text{dm}^3\).
2. For part b, \(0.65\,\text{dm}^3=650\,\text{cm}^3\), so \(12.65\,\text{dm}^3=12\,\text{dm}^3+650\,\text{cm}^3\).
3. For part c, \(0.3004\,\text{m}^3=300.4\,\text{dm}^3=300\,\text{dm}^3+400\,\text{cm}^3\). Therefore, \(2.3004\,\text{m}^3=2\,\text{m}^3+300\,\text{dm}^3+400\,\text{cm}^3\).
4. For part d, \(0.08\,\text{L}=80\,\text{mL}\), so \(5.08\,\text{L}=5\,\text{L}+80\,\text{mL}\).
Answer
a) \(4\,\text{m}^3+7\,\text{dm}^3\)
b) \(12\,\text{dm}^3+650\,\text{cm}^3\)
c) \(2\,\text{m}^3+300\,\text{dm}^3+400\,\text{cm}^3\)
d) \(5\,\text{L}+80\,\text{mL}\)