1. The problem involves understanding and converting between different volume units and their reciprocals related to length measurements. 2. Recall the basic volume conversions: - $1\,m^3 = 1000\,dm^3$ - $1\,dm^3 = 1000\,cm^3$ - $1\,L = 1\,dm^3$ - $1\,dL = 0.1\,L$ 3. For the given lengths and volumes, we can calculate the reciprocal of volume per unit length or vice versa, depending on the context. 4. Example: For length $1\,dm$ and volume $0.91\,dm^3$, the reciprocal volume per length is calculated as: $$\text{Reciprocal} = \frac{1}{0.91\,dm^3} \approx 1.10\,dm^{-3}$$ 5. Similarly, for length $2\,cm$ and volume $1.41\,cm^3$: $$\text{Reciprocal} = \frac{1}{1.41\,cm^3} \approx 0.709\,cm^{-3}$$ 6. This process applies to all given pairs, converting volumes to consistent units if necessary before calculating reciprocals. 7. For volumes given in liters and deciliters, convert to cubic decimeters or cubic centimeters as needed: - $9.5\,L = 9.5\,dm^3$ - $0.356\,L = 0.356\,dm^3$ - $10.25\,dL = 1.025\,L = 1.025\,dm^3$ - $0.48\,dL = 0.048\,L = 0.048\,dm^3$ 8. Then calculate reciprocals similarly: $$\frac{1}{9.5} \approx 0.105\,dm^{-3}$$ $$\frac{1}{0.356} \approx 2.81\,dm^{-3}$$ $$\frac{1}{1.025} \approx 0.976\,dm^{-3}$$ $$\frac{1}{0.048} \approx 20.83\,dm^{-3}$$ 9. These reciprocals help understand the inverse relationship between volume and length in these compound units. Final answers are the reciprocals calculated above for each pair.