Question

proposed multiplication or division | is this possible? | result
(3.0 mg)·(0.017 g) = ? | ∘ yes ∘ no |
\\(\frac{1.0\\ \text{dl}^2}{0.10\\ \text{l}}\\) = ? | ∘ yes ∘ no |
(5.0 kg)·(8.0 \\(\text{m}^2\\)) = ? | ∘ yes ∘ no |

Explanation:

Step1: Analyze first operation

To multiply \( (3.0 \, \text{mg}) \cdot (0.017 \, \text{g}) \), we need to convert units to be the same. Recall that \( 1 \, \text{g} = 1000 \, \text{mg} \), so \( 0.017 \, \text{g} = 0.017 \times 1000 \, \text{mg} = 17 \, \text{mg} \). Now we can multiply: \( 3.0 \, \text{mg} \times 17 \, \text{mg} = 51 \, \text{mg}^2 \)? Wait, no—wait, actually, the problem is about whether we can multiply with different units. Wait, no, the key is unit consistency for multiplication/division in terms of physical meaning, but actually, mathematically, we can multiply the numbers and units, but physically, does it make sense? Wait, no, the first operation: \( 3.0 \, \text{mg} \) (mass) and \( 0.017 \, \text{g} \) (mass) can be converted to same unit. Wait, maybe the question is about unit compatibility. Wait, no, the first row: \( (3.0 \, \text{mg}) \cdot (0.017 \, \text{g}) \). Let's check units: \( \text{mg} \times \text{g} = \text{mg} \times 1000 \, \text{mg} = 1000 \, \text{mg}^2 \), but physically, multiplying two masses gives a product with unit \( \text{mass}^2 \), which is not a standard physical quantity, but mathematically, can we do the multiplication? Wait, maybe the question is about whether the units are compatible for multiplication (i.e., same type, can be converted). Wait, no, the first step: convert \( 0.017 \, \text{g} \) to \( \text{mg} \): \( 0.017 \, \text{g} = 17 \, \text{mg} \). Then \( 3.0 \, \text{mg} \times 17 \, \text{mg} = 51 \, \text{mg}^2 \), but maybe the question is about whether we can perform the operation (mathematically, yes, but physically, it's a product of masses). Wait, maybe the problem is about unit consistency for meaningful operations. Wait, no, let's re-express:

First row: \( (3.0 \, \text{mg}) \times (0.017 \, \text{g}) \). Convert \( 0.017 \, \text{g} \) to \( \text{mg} \): \( 0.017 \times 1000 = 17 \, \text{mg} \). Then \( 3.0 \times 17 = 51 \), units \( \text{mg} \times \text{mg} = \text{mg}^2 \). But is this "possible"? Mathematically, yes, we can multiply the numbers and units. Wait, maybe the answer is yes? Wait, no, maybe I'm overcomplicating. Wait, the second row: \( \frac{1.0 \, \text{dL}^2}{0.10 \, \text{L}} \). Convert \( \text{dL} \) to \( \text{L} \): \( 1 \, \text{dL} = 0.1 \, \text{L} \), so \( 1 \, \text{dL}^2 = (0.1 \, \text{L})^2 = 0.01 \, \text{L}^2 \). Then \( \frac{1.0 \, \text{dL}^2}{0.10 \, \text{L}} = \frac{0.01 \, \text{L}^2}{0.10 \, \text{L}} = 0.1 \, \text{L} \). So that's possible. Third row: \( (5.0 \, \text{kg}) \times (8.0 \, \text{m}^2) \). Units: \( \text{kg} \times \text{m}^2 \), which is a unit of mass times area, not a standard physical quantity, but mathematically, we can multiply the numbers: \( 5.0 \times 8.0 = 40 \), units \( \text{kg} \cdot \text{m}^2 \). But the question is "Is this possible?"—maybe "possible" means whether we can perform the arithmetic (multiply/divide the numbers and units, even if the result is a non-standard unit). Wait, let's re-express each:

1. First row: \( 3.0 \, \text{mg} \times 0.017 \, \text{g} \). Convert \( 0.017 \, \text{g} \) to \( \text{mg} \): \( 0.017 \times 1000 = 17 \, \text{mg} \). Then \( 3.0 \times 17 = 51 \), units \( \text{mg} \times \text{mg} = \text{mg}^2 \). So mathematically, we can do it (yes), result \( 51 \, \text{mg} \cdot \text{g} \) (but converted to same unit: \( 3.0 \, \text{mg} = 0.003 \, \text{g} \), so \( 0.003 \, \text{g} \times 0.017 \, \text{g} = 0.000051 \, \text{g}^2 = 5.1 \times 10^{-5} \, \text{g}^2 \), or \( 51 \, \text{mg}^2 \)). Wait, maybe the first row: is it possible? Yes, because we can convert units and multiply. Wait, no—wait, the problem is maybe about whether the operation is valid in terms of unit types (e.g., mass times mass is okay, length squared over length is okay, mass times area is okay mathematically). Wait, let's check each:

- First row: \( (3.0 \, \text{mg}) \cdot (0.017 \, \text{g}) \): units are mass (mg) and mass (g), so same physical quantity, different units. So we can convert to same unit and multiply. So possible (yes). Result: convert \( 3.0 \, \text{mg} = 0.003 \, \text{g} \), then \( 0.003 \, \text{g} \times 0.017 \, \text{g} = 0.000051 \, \text{g}^2 = 5.1 \times 10^{-5} \, \text{g}^2 \), or \( 3.0 \, \text{mg} \times 17 \, \text{mg} = 51 \, \text{mg}^2 \).

- Second row: \( \frac{1.0 \, \text{dL}^2}{0.10 \, \text{L}} \): \( 1 \, \text{dL} = 0.1 \, \text{L} \), so \( \text{dL}^2 = (0.1 \, \text{L})^2 = 0.01 \, \text{L}^2 \). Then \( \frac{1.0 \times 0.01 \, \text{L}^2}{0.10 \, \text{L}} = \frac{0.01 \, \text{L}^2}{0.10 \, \text{L}} = 0.1 \, \text{L} \). So possible (yes), result \( 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \)).

- Third row: \( (5.0 \, \text{kg}) \cdot (8.0 \, \text{m}^2) \): units are mass (kg) and area (m²), so mathematically, we can multiply the numbers: \( 5.0 \times 8.0 = 40 \), units \( \text{kg} \cdot \text{m}^2 \). So possible (yes), result \( 40 \, \text{kg} \cdot \text{m}^2 \).

Wait, but maybe the first row: the user might think that different units (mg and g) can't be multiplied, but actually, we can convert to same unit. So:

First row:

Step1: Check unit compatibility

\( 3.0 \, \text{mg} \) and \( 0.017 \, \text{g} \) are both mass units. Convert \( 0.017 \, \text{g} \) to \( \text{mg} \): \( 0.017 \times 1000 = 17 \, \text{mg} \). Now we can multiply: \( 3.0 \, \text{mg} \times 17 \, \text{mg} = 51 \, \text{mg}^2 \) (or convert \( 3.0 \, \text{mg} \) to \( \text{g} \): \( 0.003 \, \text{g} \times 0.017 \, \text{g} = 0.000051 \, \text{g}^2 \)). So "is this possible?" → yes. Result: \( 5.1 \times 10^{-5} \, \text{g}^2 \) (or \( 51 \, \text{mg}^2 \)).

Second row:

Step2: Analyze second operation

\( \frac{1.0 \, \text{dL}^2}{0.10 \, \text{L}} \). Convert \( \text{dL} \) to \( \text{L} \): \( 1 \, \text{dL} = 0.1 \, \text{L} \), so \( \text{dL}^2 = (0.1 \, \text{L})^2 = 0.01 \, \text{L}^2 \). Now divide: \( \frac{1.0 \times 0.01 \, \text{L}^2}{0.10 \, \text{L}} = \frac{0.01 \, \text{L}^2}{0.10 \, \text{L}} = 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \)). So "is this possible?" → yes. Result: \( 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \)).

Third row:

Step3: Analyze third operation

\( (5.0 \, \text{kg}) \cdot (8.0 \, \text{m}^2) \). Units are mass (kg) and area (m²). Mathematically, we can multiply the numerical values and units: \( 5.0 \times 8.0 = 40 \), units \( \text{kg} \cdot \text{m}^2 \). So "is this possible?" → yes. Result: \( 40 \, \text{kg} \cdot \text{m}^2 \).

Wait, but maybe the first row was intended to be "no" because the units are different? No, because we can convert units. Wait, maybe the problem is about whether the operation is a valid physical operation (e.g., multiplying two masses is not a standard physical quantity, but mathematically, it's allowed). So:

First row: yes, result \( 5.1 \times 10^{-5} \, \text{g}^2 \) (or \( 51 \, \text{mg}^2 \))

Second row: yes, result \( 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \))

Third row: yes, result \( 40 \, \text{kg} \cdot \text{m}^2 \)

But let's recheck:

First row: \( 3.0 \, \text{mg} = 0.003 \, \text{g} \), so \( 0.003 \, \text{g} \times 0.017 \, \text{g} = 0.000051 \, \text{g}^2 = 5.1 \times 10^{-5} \, \text{g}^2 \)

Second row: \( 1 \, \text{dL} = 0.1 \, \text{L} \), so \( 1 \, \text{dL}^2 = 0.01 \, \text{L}^2 \), then \( 0.01 \, \text{L}^2 / 0.10 \, \text{L} = 0.1 \, \text{L} \)

Third row: \( 5.0 \, \text{kg} \times 8.0 \, \text{m}^2 = 40 \, \text{kg} \cdot \text{m}^2 \)

So for each row:

1. Is this possible? Yes. Result: \( 5.1 \times 10^{-5} \, \text{g}^2 \) (or \( 51 \, \text{mg}^2 \))

2. Is this possible? Yes. Result: \( 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \))

3. Is this possible? Yes. Result: \( 40 \, \text{kg} \cdot \text{m}^2 \)

But maybe the first row: the original units are mg and g, so when multiplying, the units are mg·g, which is a valid product unit. So:

First row:

Step1: Convert units

\( 0.017 \, \text{g} = 0.017 \times 1000 \, \text{mg} = 17 \, \text{mg} \)

Step2: Multiply

\( 3.0 \, \text{mg} \times 17 \, \text{mg} = 51 \, \text{mg}^2 \) (or \( 3.0 \, \text{mg} = 0.003 \, \text{g} \), \( 0.003 \, \text{g} \times 0.017 \, \text{g} = 0.000051 \, \text{g}^2 \))

Second row:

Step1: Convert units

\( 1.0 \, \text{dL}^2 = (1.0 \times 0.1 \, \text{L})^2 = 0.01 \, \text{L}^2 \)

Step2: Divide

\( \frac{0.01 \, \text{L}^2}{0.10 \, \text{L}} = 0.1 \, \text{L} \)

Third row:

Step1: Multiply numbers

\( 5.0 \times 8.0 = 40 \)

Step2: Multiply units

\( \text{kg} \times \text{m}^2 = \text{kg} \cdot \text{m}^2 \)

So results:

1. Yes, \( 5.1 \times 10^{-5} \, \text{g}^2 \) (or \( 51 \, \text{mg}^2 \))

2. Yes, \( 0.1 \, \text{L} \) (or \( 1 \, \text{dL} \))

3. Yes, \( 40 \, \text{kg} \cdot \text{m}^2 \)

Answer:

First row: Is this possible? $\boldsymbol{\text{yes}}$, Result: $\boldsymbol{5.1 \times 10^{-5} \, \text{g}^2}$ (or $\boldsymbol{51 \, \text{mg}^2}$)
Second row: Is this possible? $\boldsymbol{\text{yes}}$, Result: $\boldsymbol{0.1 \, \text{L}}$ (or $\boldsymbol{1 \, \text{dL}}$)
Third row: Is this possible? $\boldsymbol{\text{yes}}$, Result: $\boldsymbol{40 \, \text{kg} \cdot \text{m}^2}$

(Note: If the question expects "no" for some, maybe I made a mistake. Wait, maybe the first row: multiplying mg and g (different units) without conversion—mathematically, you can multiply the numbers and units, so $3.0 \times 0.017 = 0.051$, units $\text{mg} \cdot \text{g}$, which is $0.051 \, \text{mg} \cdot \text{g}$. But converting to same unit is better. So maybe the first row is yes, second yes, third yes.)