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: ( p q) Step 1: Apply De Morgan's Law: ( p q) ( p) ( q). Step 2: Simplify the double negations: ( p) p and ( q) q. Step 3: Combine the simplified terms: p q. Translation: Ndum plays handball and Che plays football. b) Translate into ordinary English: ( p q) Step 1: Apply De Morgan's Law: ( p q) ( p) q. Step 2: Simplify the double negation: ( p) p. Step 3: Combine the simplified terms: p q. Translation: Ndum plays handball or Che does not play football. --- 1. (i) a) Differentiate ((3x^3)/( x)) with respect to x. Step 1: Use logarithm properties to simplify the expression. y = ((3x^3)/( x)) = (3x^3) - ( x) y = 3 + (x^3) - ( x) y = 3 + 3 x - ( x) Step 2: Differentiate each term with respect to x. Recall that (d)/(dx)( k) = 0 for a constant k, and (d)/(dx)( u) = (1)/(u) (du)/(dx). (dy)/(dx) = (d)/(dx)( 3) + (d)/(dx)(3 x) - (d)/(dx)(( x)) (dy)/(dx) = 0 + 3((1)/(x)) - ((1)/( x) · x) (dy)/(dx) = (3)/(x) - ( x)/( x) (dy)/(dx) = (3)/(x) - x The derivative is (3)/(x) - x. 1. (i) b) A curve is defined parametrically by x = t^3, y = t(1 - (1)/(4)t^3). 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)(1 - (1)/(4)(-1)^3) = (-1)(1 - (1)/(4)(-1)) = (-1)(1 + (1)/(4)) = (-1)((5)/(4)) = -(5)/(4) The point is (-1, -(5)/(4)). Step 2: Find (dx)/(dt) and (dy)/(dt). x = t^3 (dx)/(dt) = 3t^2 y = t - (1)/(4)t^4 (dy)/(dt) = 1 - (1)/(4)(4t^3) = 1 - t^3 Step 3: Find the gradient (dy)/(dx) using the chain rule (dy)/(dx) = (dy/dt)/(dx/dt). (dy)/(dx) = (1 - t^3)/(3t^2) Step 4: Evaluate the gradient at t = -1. m = .(dy)/(dx)|_t=-1 = (1 - (-1)^3)/(3(-1)^2) = (1 - (-1))/(3(1)) = (1+1)/(3) = (2)/(3) Step 5: Use the point-slope form of a line y - y_0 = m(x - x_0). y - (-(5)/(4)) = (2)/(3)(x - (-1)) y + (5)/(4) = (2)/(3)(x + 1) Step 6: Clear denominators and rearrange into general form. Multiply by 12 (the LCM of 4 and 3). 12(y + (5)/(4)) = 12((2)/(3)(x + 1)) 12y + 15 = 8(x + 1) 12y + 15 = 8x