1. The problem states: t is proportional to the square root of d, with t = 12 when d = 4. 2. To express t in terms of d, write $$t = k\sqrt{d}$$ for some constant k. 3. Substitute the known values (t=12, d=4) to find k: $$12 = k \sqrt{4} = 2k \implies k = \frac{12}{2} = 6$$ 4. Therefore, the formula is: $$t = 6\sqrt{d}$$ 5. To calculate t when d = 9: $$t = 6 \sqrt{9} = 6 \times 3 = 18$$ 6. The light intensity E is inversely proportional to the square of the distance r. Given E = 4 when r = 50, express E as: $$E = \frac{k}{r^2}$$ 7. Substitute to find k: $$4 = \frac{k}{50^2} = \frac{k}{2500} \implies k = 4 \times 2500 = 10000$$ 8. So, $$E = \frac{10000}{r^2}$$ 9. Calculate E when r = 20: $$E = \frac{10000}{20^2} = \frac{10000}{400} = 25$$ 10. Calculate r when E = 1600: $$1600 = \frac{10000}{r^2} \implies r^2 = \frac{10000}{1600} = 6.25 \implies r = \sqrt{6.25} = 2.5$$ 11. P is directly proportional to the cube of Q. When Q = 15, P = 1350, express P as: $$P = kQ^3$$ 12. Find k: $$1350 = k \times 15^3 = k \times 3375 \implies k = \frac{1350}{3375} = 0.4$$ 13. Formula: $$P = 0.4 Q^3$$ 14. Calculate P when Q = 20: $$P = 0.4 \times 20^3 = 0.4 \times 8000 = 3200$$ 15. F is inversely proportional to the square of the distance d, with F = 12 when d = 2: $$F = \frac{k}{d^2}$$ 16. Find k: $$12 = \frac{k}{2^2} = \frac{k}{4} \implies k = 48$$ 17. Formula: $$F = \frac{48}{d^2}$$ 18. Calculate F when d = 5: $$F = \frac{48}{5^2} = \frac{48}{25} = 1.92$$ 19. Calculate d when F = 3: $$3 = \frac{48}{d^2} \implies d^2 = \frac{48}{3} = 16 \implies d = 4$$ 20. D is directly proportional to $t^2$, and D = 8 when t = 4: $$D=k t^2$$ 21. Find k: $$8 = k \times 4^2 = 16k \implies k = \frac{8}{16} = 0.5$$ 22. Formula: $$D = 0.5 t^2$$ 23. Find t when D = 50: $$50 = 0.5 t^2 \implies t^2 = 100 \implies t = 10$$ (positive value) 24. P is directly proportional to $q^2$ with P = 270 when q = 7.5: $$P = k q^2$$ 25. Find k: $$270 = k \times (7.5)^2 = k \times 56.25 \implies k = \frac{270}{56.25} = 4.8$$ 26. Formula: $$P = 4.8 q^2$$ 27. Find q when P = 9: $$9 = 4.8 q^2 \implies q^2 = \frac{9}{4.8} = 1.875 \implies q = \sqrt{1.875} \approx 1.369$$ 28. R is inversely proportional to the square of c. Given R = 30 when c = 4: $$R = \frac{k}{c^2}$$ 29. Find k: $$30 = \frac{k}{4^2} = \frac{k}{16} \implies k = 480$$ 30. Formula: $$R = \frac{480}{c^2}$$ 31. Find c when R = 1920: $$1920 = \frac{480}{c^2} \implies c^2 = \frac{480}{1920} = 0.25 \implies c = 0.5$$ 32. A is proportional to $T^2$ and also proportional to $r^2$, given T = 47 when r = 0.25. We can write $$A = k T^2 r^2$$ but since A proportional to both, assume $$A = k (T r)^2$$. 33. Given T = 47 and r = 0.25, find k from known or consider ratio. Without A's value, rewrite problem considering $T \propto r^a$ or use given relation. 34. Problem asks: find r when T = 365. 35. Assuming the relation proportional to both squares, and given values imply constant k with $T = 47$ at $r=0.25$, then to find r when $T=365$, use ratio: $$\frac{T_2}{T_1} = \frac{r_2}{r_1}$$ assuming $T \propto r$. 36. So: $$\frac{365}{47} = \frac{r}{0.25} \implies r = 0.25 \times \frac{365}{47} \approx 1.94$$ 37. Final answer rounded to 3 significant figures: $$r = 1.94$$ 38. T is directly proportional to $\sqrt{x}$, with T = 400 when x = 625: $$T = k \sqrt{x}$$ 39. Find k: $$400 = k \sqrt{625} = k \times 25 \implies k = 16$$ 40. Formula: $$T = 16 \sqrt{x}$$ 41. Calculate T when x = 56.25: $$T = 16 \sqrt{56.25} = 16 \times 7.5 = 120$$ 42. P is directly proportional to $r^3$, with P = 343 when r = 3.5: $$P = k r^3$$ 43. Find k: $$343 = k \times (3.5)^3 = k \times 42.875 \implies k = \frac{343}{42.875} = 8$$ 44. Formula: $$P = 8 r^3$$