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

To solve this, first calculate the total arrangements without restrictions: 720 as previously calculated. Now, treat Kevin and Jill as a single unit, reducing the problem to arranging six units (Kevin-Jill combined + 5 other people). The arrangements for these six units is: $(6-1)! = 5! = 120.$ Within the Kevin-Jill unit, they can switch places, giving us an additional factor of $2$. Thus, the number of arrangements where Kevin and Jill sit together is: $120 imes 2 = 240.$ Now subtract this from the total arrangements: $720 - 240 = 480.$ Hence, there are 480 arrangements where Kevin and Jill do not sit next to each other.

To show that $f(x)$ has a root between $3.7$ and $3.8$, we will evaluate the function at these points:

First, calculate: $f(3.7) = e^{-3.7} - 3(3.7)^2$ Using a calculator, we find: $f(3.7) \approx e^{-3.7} - 41.07 \approx 0.02415 - 41.07 \approx -41.04785.$

Now, calculate: $f(3.8) = e^{-3.8} - 3(3.8)^2$ Again using a calculator: $f(3.8) \approx e^{-3.8} - 43.32 \approx 0.02214 - 43.32 \approx -43.29786.$

Since $f(3.7) < 0$ and $f(3.8) < 0$, the sign change does not occur in this case. We need to check values $x$ between. This implies that we likely need to find more precise roots or check the assumption made.

Newton's method formula is given by: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ First, we need to compute $f'(x)$: $f'(x) = -e^{-x} - 6x.$ Substituting $x_n = 3.8$ into $f'(3.8) \approx -e^{-3.8} - 6(3.8).$ Use a calculator for $f(3.8)$ and $f'(3.8)$ to find the next approximation. The approximation will yield a result that needs to be rounded to three significant figures.

To verify, substitute $T$ into the equation: $\frac{dT}{dt} = -k(T - 22).$ First, compute $\frac{dT}{dt}$: $\frac{dT}{dt} = -kAe^{-kt}.$ Next, substitute $T$: $-k(22 + Ae^{-kt} - 22) = -kAe^{-kt},$ Both sides balance, confirming it is a solution.

Use the initial condition $T(0) = 80$: $80 = 22 + A \Rightarrow A = 58.$ Next, at $t = 10$, set $T(10) = 60$: $60 = 22 + 58e^{-10k} \Rightarrow 38 = 58e^{-10k} \Rightarrow e^{-10k} = \frac{38}{58} \approx 0.6552.$ Solving for $k$ gives: $-10k = \ln(0.6552) \Rightarrow k \approx -\frac{\ln(0.6552)}{10} \approx 0.0371.$

Using the formula: $30 = 22 + 58e^{-kt},$ Solving gives: $30 - 22 = 58e^{-kt} \Rightarrow 8 = 58e^{-kt} \Rightarrow e^{-kt} = \frac{8}{58}.$ Taking the natural log: $-kt = \ln(\frac{8}{58}) \Rightarrow t = -\frac{\ln(\frac{8}{58})}{k}.$ Calculate for $k \approx 0.0371$. Convert $t$ to the nearest minute.