Solve the following quantitative comparison problem by plugging in the number 0,1,2, -2, and `1//2` in that order - when possible.
`{:("Column A",m>0,"Column B"),(m^(10),,m^(100)):}`

Options

A. If column A is larger
B. If column B is larger
C. If the columns are equal
D. If there is not enough information to decide

Answer

To solve the quantitative comparison problem, we will evaluate the expressions in Column A and Column B by plugging in the values \( m = 0, 1, 2, -2, \) and \( \frac{1}{2} \) in that order, while ensuring that \( m > 0 \).

### Step-by-Step Solution:

1. **Plug in \( m = 1 \)**:
- Column A: \( m^{10} = 1^{10} = 1 \)
- Column B: \( m^{100} = 1^{100} = 1 \)
- Comparison: Column A = Column B (both are 1)

2. **Plug in \( m = 2 \)**:
- Column A: \( m^{10} = 2^{10} = 1024 \)
- Column B: \( m^{100} = 2^{100} \) (which is much larger than 1024)
- Comparison: Column B > Column A

3. **Plug in \( m = 0 \)**:
- Since \( m \) must be greater than 0, we skip this value.

4. **Plug in \( m = -2 \)**:
- Since \( m \) must be greater than 0, we skip this value.

5. **Plug in \( m = \frac{1}{2} \)**:
- Column A: \( m^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \)
- Column B: \( m^{100} = \left(\frac{1}{2}\right)^{100} = \frac{1}{2^{100}} \) (which is much smaller than \( \frac{1}{1024} \))
- Comparison: Column A > Column B

### Summary of Results:
- For \( m = 1 \): Column A = Column B
- For \( m = 2 \): Column B > Column A
- For \( m = \frac{1}{2} \): Column A > Column B

### Conclusion:
Since we have different results for different values of \( m \) (equal for \( m = 1 \), Column B greater for \( m = 2 \), and Column A greater for \( m = \frac{1}{2} \)), we cannot definitively conclude which column is larger overall. Therefore, the answer is:

**D: There is not enough information to decide.**