Seven people are to be seated at a round table - HSC - SSCE Mathematics Extension 1 - Question 3 - 2002 - Paper 1

To find the arrangements where Kevin and Jill do not sit next to each other, we can first calculate the total arrangements without restrictions and subtract the cases where they are together.

1. Total arrangements without restrictions: 720 (as calculated above).
2. Treat Kevin and Jill as a single unit or block. Now we have 6 units (the K&J block plus five other individuals), which can be arranged in:

$$(6-1)! = 5! = 120.$$

Since Kevin and Jill can switch places within their block, we multiply by 2:

$$120 \times 2 = 240.$$

3. Arrangements where Kevin and Jill do not sit next to each other is:

$$720 - 240 = 480.$$

To show that a root exists between 3.7 and 3.8, we evaluate the function at both points:

1. Calculate $f(3.7)$: $$f(3.7) = e^{-3.7} - 3(3.7)^2\ \approx 0.024 - 41.37 = -41.346.$$
2. Calculate $f(3.8)$: $$f(3.8) = e^{-3.8} - 3(3.8)^2\ \approx 0.020 - 43.32 = -43.300.$$

Since $f(3.7)$ and $f(3.8)$ are both negative, we need to inspect the function further or calculate intermediate values to confirm a sign change between these points.

Newton's method formula is given by:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}.$$

1. Differentiate $f(x)$:

$$f'(x) = -e^{-x} - 6x.$$

2. Evaluate $f(3.8)$ and $f'(3.8)$:
   • From previous calculations, we already have $f(3.8)$. Now calculate:
   $$f'(3.8) \approx -0.020 - 22.8 \approx -22.82.$$
3. Applying Newton's method: $$x_{1} = 3.8 - \frac{-43.300}{-22.82} \approx 3.8 - 1.897 \approx 3.467.$$

Thus, a better approximation to three significant figures is approximately 3.47.

To verify, substitute $T$ into the differential equation:

1. Differentiate $T$: $$\frac{dT}{dt} = -kAe^{-kt}.$$
2. Substitute into rac{dT}{dt} = -k(T - 22): $$-kAe^{-kt} = -k((22 + Ae^{-kt}) - 22)\ = -kAe^{-kt}.$$

This holds true, thus verifying the solution.

Using the initial condition, at $t = 0$, $T(0) = 80$:

$$80 = 22 + A \Rightarrow A = 58.$$

Using the condition $T(10) = 60$:

$$60 = 22 + 58e^{-10k} \Rightarrow 38 = 58e^{-10k} \Rightarrow e^{-10k} = \frac{38}{58} \Rightarrow -10k = \ln\left(\frac{38}{58}\right)\Rightarrow k = \frac{-1}{10} \ln\left(\frac{38}{58}\right).$$

We want to solve for t such that $T(t) = 30$:

$$30 = 22 + 58e^{-kt} \Rightarrow 8 = 58e^{-kt} \Rightarrow e^{-kt} = \frac{8}{58} \Rightarrow -kt = \ln\left(\frac{8}{58}\right) \Rightarrow t = \frac{-1}{k} \ln\left(\frac{8}{58}\right).$$

Using the previously calculated value of $k$, substitute to find t. The final answer should be rounded to the nearest minute.