Mr. Smith offers students two paying options for training. The first option costs R 1 025 per hour plus an additional of R4 800. The second option cost R1 345 per hour with no annual amount.

Option 1: Cost ($C_1$) = R 1 025 per hour + R 4 800. So, $C_1 = 1025H + 4800$.
Option 2: Cost ($C_2$) = R 1 345 per hour. So, $C_2 = 1345H$.

• For A (Option 1, $H=0$): $$A = 1025(0) + 4800 = 4800$$
• For B (Option 1, $H=20$): $$B = 1025(20) + 4800 = 20500 + 4800 = 25300$$
• For C (Option 2, $H=10$): $$C = 1345(10) = 13450$$
• For D (Option 1, $C_1 = 56050$): $$56050 = 1025D + 4800$$ $$56050 - 4800 = 1025D$$ $$51250 = 1025D$$ $$D = \frac{51250}{1025} = 50$$
• For E (Option 1, $H=40$): $$E = 1025(40) + 4800 = 41000 + 4800 = 45800$$
• For F (Option 2, $H=40$): $$F = 1345(40) = 53800$$

The missing values are:

• A: $\boxed{\text{R } 4800}$
• B: $\boxed{\text{R } 25300}$
• C: $\boxed{\text{R } 13450}$
• D: $\boxed{50 \text{ hours}}$
• E: $\boxed{\text{R } 45800}$
• F: $\boxed{\text{R } 53800}$

3.1.2
The formula for Option 1 is:

where $C_1$ is the cost and $H$ is the number of hours.

3.1.3
The dependent variable is the Cost (R).
Reason: The cost depends on the number of hours worked; as the number of hours changes, the total cost changes.

3.1.4
To draw the graphs:

• Heading: Cost of Training Options vs. Number of Hours
• X-axis: Label "Number of Hours (H)". Scale from 0 to 50, perhaps in increments of 5 or 10.
• Y-axis: Label "Cost (R)". Scale from 0 to 70000, perhaps in increments of 5000 or 10000.
• Plot points for Option 1: (0, 4800), (5, 9925), (10, 15050), (20, 25300), (30, 35550), (40, 45800), (50, 56050). Connect these points with a straight line and label it "Option 1".
• Plot points for Option 2: (0, 0), (5, 6725), (10, 13450), (20, 26900), (30, 40350), (40, 53800), (50, 67250). Connect these points with a straight line and label it "Option 2".
• Break-even point: Mark the intersection of the two lines.

3.1.5

Using Option 1:

Using Option 2:

The break-even point is $\boxed{\text{15 hours and R } 20175}$.