The problem states that the cost C is partly constant and partly varies as the square root of the number n of students. This can be written as:

Step 1: Formulate the equation for the cost C in terms of the number of students n. The problem states that the cost C is partly constant and partly varies as the square root of the number n of students. This can be written as: C = k_1 + k_2 sqrt(n) where k_1 is the constant part and k_2 is the constant of proportionality. Step 2: Use the given data to find the values of k_1 and k_2. We are given two data points: 1. When n = 144, C = 640. 2. When n = 64, C = 560. Substitute these values into the equation: For n = 144: 640 = k_1 + k_2 sqrt(144) 640 = k_1 + 12k_2 (1) For n = 64: 560 = k_1 + k_2 sqrt(64) 560 = k_1 + 8k_2 (2) Subtract equation (2) from equation (1): (640 - 560) = (k_1 + 12k_2) - (k_1 + 8k_2) 80 = 4k_2 k_2 = (80)/(4) k_2 = 20 Substitute k_2 = 20 into equation (2): 560 = k_1 + 8(20) 560 = k_1 + 160 k_1 = 560 - 160 k_1 = 400 a) Equation connecting C and n: Substitute the values of k_1 and k_2 back into the general equation: C = 400 + 20sqrt(n) The equation connecting C and n is C = 400 + 20sqrt(n). b) Cost of feeding 196 students: Substitute n = 196 into the equation found in part (a): C = 400 + 20sqrt(196) C = 400 + 20(14) C = 400 + 280 C = 680 The cost of feeding 196 students is \680.00. c) Number of students that can be fed with 700.00: Substitute C = 700 into the equation found in part (a): 700 = 400 + 20sqrt(n) 700 - 400 = 20sqrt(n) 300 = 20sqrt(n) sqrt(n) = (300)/(20) sqrt(n) = 15 Square both sides to find n: n = 15^2 n = 225 The number of students that can be fed with 700.00 is 225 students. Send me the next one 📸