What are the answers to the following questions: 12. m9 / m2=?, 13. Find the gradient of the line passing through points (0,2) and (1,5). 14. If Q = [50, 30, 60]^T and P = [60, 20, 70]^T, what is Q+P?

The simplified expression is $\boxed{\text{B. } m^7}$.

13. Step 1: Identify the coordinates of the two points. Point B: $(x_1, y_1) = (0, 2)$ Point D: $(x_2, y_2) = (1, 5)$

$m = \frac{y_2 - y_1}{x_2 - x_1}$.

The gradient of the line is $\boxed{\text{D. } 3}$.

14. Step 1: Identify the given matrices P and Q.

The result of $Q+P$ is $\boxed{\text{B. } \begin{bmatrix} 110 \\ 50 \\ 130 \end{bmatrix}}$.

29. Step 1: State the given map distance and actual distance. Map distance = $16 \text{ cm}$ Actual distance = $640 \text{ km}$

Scale = $16 \text{ cm} : 640 \text{ km}$

Divide both sides by 16.

The scale of the map in statement form is $\boxed{1 \text{ cm represents } 40 \text{ km}}$.

30. Step 1: State the given map distance and actual distance. Map distance = $4 \text{ cm}$ Actual distance = $1.2 \text{ km}$

$1 \text{ km} = 1000 \text{ m}$
$1 \text{ m} = 100 \text{ cm}$
So, $1 \text{ km} = 1000 \times 100 \text{ cm} = 100,000 \text{ cm}$.
Actual distance = $1.2 \text{ km} \times 100,000 \text{ cm/km} = 120,000 \text{ cm}$.

Scale = $4 \text{ cm} : 120,000 \text{ cm}$

Divide both sides by 4.

The scale of the map is $\boxed{1:30,000}$.

31. Step 1: State the given map scale and map distance. Scale = $1:50000$ (This means $1 \text{ cm}$ on the map represents $50,000 \text{ cm}$ in actual distance). Map distance = $4 \text{ cm}$

Actual distance = Map distance $\times$ Scale factor
Actual distance = $4 \text{ cm} \times 50,000$
Actual distance = $200,000 \text{ cm}$

$1 \text{ km} = 100,000 \text{ cm}$
Actual distance in km = $\frac{200,000 \text{ cm}}{100,000 \text{ cm/km}}$
Actual distance = $2 \text{ km}$
The actual length of the road is $\boxed{2 \text{ km}}$.