Calculate the density of a mixture given the densities and volumes of its components.

Here are the solutions to the problems: Question 1: The expression to be evaluated using logarithm tables is not visible in the provided image. Therefore, this question cannot be answered. Question 2: Calculate the density of the mixture. Step 1: Calculate the mass of acid. Mass of acid = Density of acid × Volume of acid Mass of acid = 1.08 g/cm^3 × 200 cm^3 = 216 g Step 2: Calculate the mass of alcohol. Mass of alcohol = Density of alcohol × Volume of alcohol Mass of alcohol = 0.8 g/cm^3 × 300 cm^3 = 240 g Step 3: Calculate the total mass of the mixture. Total mass = Mass of acid + Mass of alcohol Total mass = 216 g + 240 g = 456 g Step 4: Calculate the total volume of the mixture. Total volume = Volume of acid + Volume of alcohol Total volume = 200 cm^3 + 300 cm^3 = 500 cm^3 Step 5: Calculate the density of the mixture. Density of mixture = Total massTotal volume Density of mixture = 456 g500 cm^3 = 0.912 g/cm^3 The density of the mixture is 0.912 g/cm^3. Question 3: Find the magnitude of vector OM. Step 1: Find the coordinates of point M. Point M divides line PQ in the ratio 2:1. Given P(5, 1) and Q(11, 4). Using the section formula M = ((nx_1 + mx_2)/(m+n), (ny_1 + my_2)/(m+n)), where (x_1, y_1) = (5, 1), (x_2, y_2) = (11, 4), m=2, n=1. M_x = (1 × 5 + 2 × 11)/(2+1) = (5 + 22)/(3) = (27)/(3) = 9 M_y = (1 × 1 + 2 × 4)/(2+1) = (1 + 8)/(3) = (9)/(3) = 3 So, the coordinates of M are (9, 3). Step 2: Determine the vector OM. The vector OM is the position vector of M, which is 9 \\ 3 . Step 3: Calculate the magnitude of vector OM. |OM| = sqrt(9^2 + 3^2) |OM| = sqrt(81 + 9) |OM| = sqrt(90) |OM| = sqrt(9 × 10) = 3sqrt(10) The magnitude of vector OM is 3sqrt(10). Question 4: The income tax rates table is not visible in the provided image. Therefore, this question cannot be answered. Question 5: The formula to make a subject of is not visible in the provided image. Therefore, this question cannot be answered. Question 6: A line passes through points (2, 5) and has a gradient of 2. a) Determine its equation in the form y=mx+c. Step 1: Use the given gradient m=2 and the point (x,y) = (2,5) in the equation y=mx+c. 5 = 2(2) + c 5 = 4 + c c = 5 - 4 c = 1 Step 2: Write the equation of the line. y = 2x + 1 The equation of the line is y = 2x + 1. b) Find the angle it makes with x-axis. Step 1: Use the relationship between gradient and angle. The gradient m = , where is the angle the line makes with the positive x-axis. = 2 = (2) ≈ 63.43^ The angle the line makes with the x-axis is 63.43^ (to two decimal places). Question 7: Find the value of P when Q=5. Step 1: Write the general equation for P. P is partly constant and partly varies as the cube of Q. P = k + cQ^3 where k and c are constants. Step 2: Use the given conditions to form a system of equations. When Q=1, P=23: 23 = k + c(1)^3 23 = k + c (Equation 1) When Q=2, P=44: 44 = k + c(2)^3 44 = k + 8c (Equation 2) Step 3: Solve the system of equations for k and c. Subtract Equation 1 from Equation 2: (44 - 23) = (k + 8c) - (k + c) 21 = 7c c = (21)/(7) = 3 Substitute c=3 into Equation 1: 23 = k + 3 k = 23 - 3 = 20 Step 4: Write the specific equation for P. P = 20 + 3Q^3 Step 5: Find P when Q=5. P = 20 + 3(5)^3 P = 20 + 3(125) P = 20 + 375 P = 395 The value of P when Q=5 is 395. Question 8: Determine the translation vector T and find the coordinates of A' and B'. Given vertices A(1, 2), B(3, 5), C(4, 1). Image of C is C'(6, -2). a) Determine the translation vector T. Step 1: Use the formula Image = Original + Translation Vector. C' = C + T 6 \\ -2 = 4 \\ 1 + T T = 6 \\ -2 - 4 \\ 1 T = 6-4 \\ -2-1 = 2 \\ -3 The translation vector is T = 2 \\ -3 . b) Find the co-ordinates of A' and B' under the translation vector T. Step 1: Find A'. A' = A + T A' = 1 \\ 2 + 2 \\ -3 = 1+2 \\ 2-3 = 3 \\ -1 So, A' is (3, -1). Step 2: Find B'. B' = B + T B' = 3 \\ 5 + 2 \\ -3 = 3+2 \\ 5-3 = 5 \\ 2 So, B' is (5, 2). The coordinates are A'(3, -1)