The water bill (R) is partly constant and partly varies as the cube of the volume (V). If R=4.25 when V=50 and R=5.16 when V=60, find the expression for R in terms of V.

Step 1: Formulate the relationship between the water bill (R), the constant part (C), and the volume (V). The problem states that the water bill R is partly constant and partly varies as the cube of the volume V. This can be written as: R = C + kV^3 where C is the constant part and k is the constant of proportionality. Step 2: Use the given scenarios to set up a system of linear equations for C and k. Scenario 1: When V = 50 litres, R = 4.25 cedis. Substitute these values into the equation: 4.25 = C + k(50)^3 4.25 = C + 125000k (1) Scenario 2: When V = 60 litres, R = 5.16 cedis. Substitute these values into the equation: 5.16 = C + k(60)^3 5.16 = C + 216000k (2) Step 3: Solve the system of equations for C and k. Subtract equation (1) from equation (2): (C + 216000k) - (C + 125000k) = 5.16 - 4.25 91000k = 0.91 Divide by 91000 to find k: k = (0.91)/(91000) k = 0.00001 Now substitute the value of k into equation (1) to find C: 4.25 = C + 125000(0.00001) 4.25 = C + 1.25 Subtract 1.25 from both sides to find C: C = 4.25 - 1.25 C = 3.00 Step 4: Write the expression for R in terms of V. Substitute the values of C and k back into the general equation R = C + kV^3: R = 3 + 0.00001V^3 The expression for R in terms of V is R = 3 + 0.00001V^3. 3 done, 2 left today. You're making progress.