1. **State the problem:** We have a torpedo depth modeled by a quadratic pattern with depths at 1, 2, and 3 seconds given. We need to find the depth at 5 seconds, verify the formula for depth at n seconds, find the maximum depth, and find when the torpedo is at 104 m depth for the second time. 2. **Calculate depth at 5 seconds (3.1):** Given depths: $T_1=36$, $T_2=71$, $T_3=104$. We assume the depth follows $T_n = an^2 + bn + c$. Using the points: $$a(1)^2 + b(1) + c = 36$$ $$a(2)^2 + b(2) + c = 71$$ $$a(3)^2 + b(3) + c = 104$$ Which gives: $$a + b + c = 36$$ $$4a + 2b + c = 71$$ $$9a + 3b + c = 104$$ Subtract first from second: $$3a + b = 35$$ Subtract second from third: $$5a + b = 33$$ Subtract these two: $$(5a + b) - (3a + b) = 33 - 35 \\ 2a = -2 \\ a = -1$$ Plug $a=-1$ into $3a + b = 35$: $$3(-1) + b = 35 \\ -3 + b = 35 \\ b = 38$$ Plug $a$ and $b$ into $a + b + c = 36$: $$-1 + 38 + c = 36 \\ 37 + c = 36 \\ c = -1$$ So, $T_n = -n^2 + 38n - 1$. Calculate $T_5$: $$T_5 = -(5)^2 + 38(5) - 1 = -25 + 190 - 1 = 164$$ 3. **Show formula for $T_n$ (3.2):** From above, we derived $T_n = -n^2 + 38n - 1$ using the given data points. 4. **Calculate maximum depth (3.3):** Since $T_n$ is quadratic with $a = -1 < 0$, it opens downward, so maximum at vertex. Vertex at: $$n = -\frac{b}{2a} = -\frac{38}{2(-1)} = 19$$ Calculate max depth: $$T_{19} = -(19)^2 + 38(19) - 1 = -361 + 722 - 1 = 360$$ Maximum depth is 360 metres. 5. **Find when depth is 104 m for second time (3.4):** Solve $T_n = 104$: $$-n^2 + 38n - 1 = 104 \\ -n^2 + 38n - 105 = 0 \\ n^2 - 38n + 105 = 0$$ Use quadratic formula: $$n = \frac{38 \pm \sqrt{38^2 - 4(1)(105)}}{2} = \frac{38 \pm \sqrt{1444 - 420}}{2} = \frac{38 \pm \sqrt{1024}}{2} = \frac{38 \pm 32}{2}$$ Two solutions: $$n = \frac{38 - 32}{2} = 3$$ $$n = \frac{38 + 32}{2} = 35$$ The second time depth is 104 m is at $n=35$ seconds. **Final answers:** - Depth at 5 seconds: 164 m - Depth formula: $T_n = -n^2 + 38n - 1$ - Maximum depth: 360 m at 19 seconds - Second time at 104 m depth: 35 seconds