1. **Solve the equation** $(7p + 9) \tan 53^\circ = 143.8$. 2. Calculate $\tan 53^\circ$ using a calculator: $\tan 53^\circ \approx 1.3270$. 3. Substitute into the equation: $(7p + 9) \times 1.3270 = 143.8$. 4. Divide both sides by 1.3270: $7p + 9 = \frac{143.8}{1.3270} \approx 108.34$. 5. Subtract 9 from both sides: $7p = 108.34 - 9 = 99.34$. 6. Divide both sides by 7: $p = \frac{99.34}{7} \approx 14.2$. --- 7. **Work out the area of the quadrilateral EFGH.** 8. The quadrilateral has right angles at E and H, with vertical sides EH = 8 cm and FG = 4 cm, and horizontal side HG = 11 cm. 9. The shape can be split into a rectangle and a right triangle by the diagonal EG. 10. The rectangle has dimensions 4 cm (FG) by 11 cm (HG), so area = $4 \times 11 = 44$ cm$^2$. 11. The triangle formed by EH and the difference in vertical sides is with height $8 - 4 = 4$ cm and base 11 cm. 12. Area of triangle = $\frac{1}{2} \times 11 \times 4 = 22$ cm$^2$. 13. Total area = rectangle area + triangle area = $44 + 22 = 66$ cm$^2$. --- 14. **Find the bearing from C to D on the map.** 15. Coordinates: C at (2,4), D at (8,7) in grid units. 16. Calculate horizontal difference: $\Delta x = 8 - 2 = 6$. 17. Calculate vertical difference: $\Delta y = 7 - 4 = 3$. 18. Bearing is measured clockwise from North. 19. Calculate angle $\theta = \arctan\left(\frac{\Delta x}{\Delta y}\right) = \arctan\left(\frac{6}{3}\right) = \arctan(2) \approx 63.4^\circ$. 20. Bearing from C to D = $0^\circ + 63.4^\circ = 63^\circ$ (nearest degree). --- 21. **Determine the best scale value A for the height axis graph.** 22. The height values range from 4 m to 58 m. 23. To choose a good scale, pick a value that divides the range into manageable intervals. 24. The range is $58 - 4 = 54$ m. 25. A suitable scale could be 10 m per interval, so $A = 10$. --- 26. **Determine the best scale value A for the distance from home axis.** 27. Maximum distance from home is Joe's 62 km at 3 hours. 28. The axis should cover at least 62 km. 29. Choose a scale dividing 62 into equal intervals, e.g., 10 km per interval. 30. So, $A = 10$ is a good scale value. **Final answers:** - $p \approx 14.2$ - Area of quadrilateral = $66.0$ cm$^2$ - Bearing from C to D = $63^\circ$ - Best scale for height axis $A = 10$ - Best scale for distance from home axis $A = 10$