Translate the logical statements ~("~p or ~q") and ~("~p and q") into English, and differentiate ln(3x3/sin x).

Here are the solutions to the questions:

9. (ii) Given the statements:
p: Ndum plays handball
q: Che plays football

a) Translate into ordinary English: $\sim(\sim p \lor \sim q)$

Step 1: Apply De Morgan's Law: $\sim(\sim p \lor \sim q) \equiv \sim(\sim p) \land \sim(\sim q)$.

Step 2: Simplify the double negations: $\sim(\sim p) \equiv p$ and $\sim(\sim q) \equiv q$.

Step 3: Combine the simplified terms: $p \land q$.
Translation: Ndum plays handball and Che plays football.

b) Translate into ordinary English: $\sim(\sim p \land q)$

Step 1: Apply De Morgan's Law: $\sim(\sim p \land q) \equiv \sim(\sim p) \lor \sim q$.

Step 2: Simplify the double negation: $\sim(\sim p) \equiv p$.

Step 3: Combine the simplified terms: $p \lor \sim q$.
Translation: Ndum plays handball or Che does not play football.

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1. (i) a) Differentiate $\ln\left(\frac{3x^3}{\sin x}\right)$ with respect to $x$.

Step 1: Use logarithm properties to simplify the expression.

$$y = \ln\left(\frac{3x^3}{\sin x}\right) = \ln(3x^3) - \ln(\sin x)$$ $$y = \ln 3 + \ln(x^3) - \ln(\sin x)$$ $$y = \ln 3 + 3\ln x - \ln(\sin x)$$

Step 2: Differentiate each term with respect to $x$.
Recall that $\frac{d}{dx}(\ln k) = 0$ for a constant $k$, and $\frac{d}{dx}(\ln u) = \frac{1}{u} \frac{du}{dx}$.

$$\frac{dy}{dx} = \frac{d}{dx}(\ln 3) + \frac{d}{dx}(3\ln x) - \frac{d}{dx}(\ln(\sin x))$$ $$\frac{dy}{dx} = 0 + 3\left(\frac{1}{x}\right) - \left(\frac{1}{\sin x} \cdot \cos x\right)$$ $$\frac{dy}{dx} = \frac{3}{x} - \frac{\cos x}{\sin x}$$ $$\frac{dy}{dx} = \frac{3}{x} - \cot x$$

The derivative is $\boxed{\frac{3}{x} - \cot x}$.

1. (i) b) A curve is defined parametrically by $x = t^3$, $y = t\left(1 - \frac{1}{4}t^3\right)$. Find the equation of the tangent to the curve at the point where $t = -1$.

Step 1: Find the coordinates $(x_0, y_0)$ of the point of tangency when $t = -1$.

$$x_0 = (-1)^3 = -1$$ $$y_0 = (-1)\left(1 - \frac{1}{4}(-1)^3\right) = (-1)\left(1 - \frac{1}{4}(-1)\right) = (-1)\left(1 + \frac{1}{4}\right) = (-1)\left(\frac{5}{4}\right) = -\frac{5}{4}$$

The point is $\left(-1, -\frac{5}{4}\right)$.

Step 2: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

$$x = t^3 \implies \frac{dx}{dt} = 3t^2$$ $$y = t - \frac{1}{4}t^4 \implies \frac{dy}{dt} = 1 - \frac{1}{4}(4t^3) = 1 - t^3$$

Step 3: Find the gradient $\frac{dy}{dx}$ using the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

$$\frac{dy}{dx} = \frac{1 - t^3}{3t^2}$$

Step 4: Evaluate the gradient at $t = -1$.

$$m = \left.\frac{dy}{dx}\right|_{t=-1} = \frac{1 - (-1)^3}{3(-1)^2} = \frac{1 - (-1)}{3(1)} = \frac{1+1}{3} = \frac{2}{3}$$

Step 5: Use the point-slope form of a line $y - y_0 = m(x - x_0)$.

$$y - \left(-\frac{5}{4}\right) = \frac{2}{3}(x - (-1))$$ $$y + \frac{5}{4} = \frac{2}{3}(x + 1)$$

Step 6: Clear denominators and rearrange into general form. Multiply by 12 (the LCM of 4 and 3).

$$12\left(y + \frac{5}{4}\right) = 12\left(\frac{2}{3}(x + 1)\right)$$ $$12y + 15 = 8(x + 1)$$ $$$$