answer:First, let's calculate the cost of the grapes before tax: Cost of grapes = Quantity of grapes * Price per kg = 8 kg * 70/kg = 560 Now, let's calculate the sales tax on the grapes: Sales tax on grapes = Cost of grapes * Sales tax rate = 560 * 8% = 560 * 0.08 = 44.80 So, the total cost of the grapes including tax is: Total cost of grapes = Cost of grapes + Sales tax on grapes = 560 + 44.80 = 604.80 Next, let's calculate the cost of the mangoes before tax: Cost of mangoes = Quantity of mangoes * Price per kg = 9 kg * 55/kg = 495 Now, let's calculate the sales tax on the mangoes: Sales tax on mangoes = Cost of mangoes * Sales tax rate = 495 * 11% = 495 * 0.11 = 54.45 So, the total cost of the mangoes including tax is: Total cost of mangoes = Cost of mangoes + Sales tax on mangoes = 495 + 54.45 = 549.45 Finally, let's calculate the total amount Andrew paid to the shopkeeper including the tax for both grapes and mangoes: Total amount paid = Total cost of grapes + Total cost of mangoes = 604.80 + 549.45 = 1154.25 Therefore, Andrew paid boxed{1154.25} to the shopkeeper including the tax.

answer:To calculate the present worth of 3600 due in 2 years at a 20% per annum compound interest rate, we can use the formula for the present value (PV) of a future sum of money: PV = FV / (1 + r)^n Where: - PV is the present value (what we're trying to find) - FV is the future value (3600 in this case) - r is the annual interest rate (20% or 0.20) - n is the number of years (2 years in this case) Plugging in the values: PV = 3600 / (1 + 0.20)^2 PV = 3600 / (1.20)^2 PV = 3600 / 1.44 PV = 2500 So, the present worth of 3600 due in 2 years at a 20% per annum compound interest rate is boxed{2500} .

answer:Solution: Given the set P=(-infty,0] cup (3, +infty) and Q={0,1,2,3}, then complement_{mathbb{R}}P=(0,3], so (complement_{mathbb{R}}P) cap Q={1,2,3}. Therefore, the correct choice is: boxed{C}. This is based on the definition of complement and intersection sets. The problem examines the definition and application of sets, which is a fundamental question.

answer:1. **Define the Geometry and Notation**: Let's consider a regular n-gon with center O and its inscribed circle having radius r. We denote the point X by the vector boldsymbol{x} = overrightarrow{OX} and the unit vectors boldsymbol{e}_1, ldots, boldsymbol{e}_n as vectors pointing from the center O to the midpoints of the sides of the n-gon. 2. **Distance Calculation**: The distance from point X to the i-th side of the n-gon can be expressed as: [ left| (boldsymbol{x}, boldsymbol{e}_i) - r right| ] where (boldsymbol{x}, boldsymbol{e}_i) is the dot product of vectors boldsymbol{x} and boldsymbol{e}_i. 3. **Sum of Squared Distances**: The sum of the squares of these distances is: [ sum_{i=1}^n left| (boldsymbol{x}, boldsymbol{e}_i) - r right|^2 ] 4. **Expand the Expression**: Expanding the squared absolute value, we obtain: [ sum_{i=1}^n left( (boldsymbol{x}, boldsymbol{e}_i)^2 - 2r (boldsymbol{x}, boldsymbol{e}_i) + r^2 right) ] This can be rewritten as: [ sum_{i=1}^n (boldsymbol{x}, boldsymbol{e}_i)^2 - 2r sum_{i=1}^n (boldsymbol{x}, boldsymbol{e}_i) + sum_{i=1}^n r^2 ] 5. **Summation of Constants**: Since r is a constant, the terms involving r can be factored out: [ sum_{i=1}^n r^2 = n r^2 ] 6. **Vector Properties**: Because boldsymbol{e}_i are unit vectors, their dot products with boldsymbol{x} can be treated individually. Specifically, the sum of dot products for a regular polygon is symmetrical around the center. 7. **Given Problem Relation**: According to problem 6.73, the sum sum_{i=1}^n (boldsymbol{x}, boldsymbol{e}_i)^2 is known to be equal to frac{n d^2}{2} where d is the distance from X to the center O of the n-gon: [ sum_{i=1}^n (boldsymbol{x}, boldsymbol{e}_i)^2 = frac{n d^2}{2} ] 8. **Conclusion**: Combining all these pieces, we obtain: [ sum_{i=1}^n left|(boldsymbol{x}, boldsymbol{e}_i) - rright|^2 = frac{n d^2}{2} + n r^2 = n left( frac{d^2}{2} + r^2 right) ] Thus, the proof is complete: [ boxed{n left( r^2 + frac{d^2}{2} right)} ]