answer:1. First, we need to find the area (a_n) between the curves (y = e^{-x}) and (y = e^{-x}|cos x|) over the interval ((n-1)pi leq x leq npi). This can be expressed as: [ a_n = int_{(n-1)pi}^{npi} left(e^{-x} - e^{-x}|cos x|right) , dx = int_{(n-1)pi}^{npi} e^{-x}(1 - |cos x|) , dx ] 2. To simplify the integral, we perform a change of variables. Let (x = (n-1)pi + y), where (0 leq y leq pi). Then (dx = dy), and the integral becomes: [ a_n = e^{-(n-1)pi} int_0^pi e^{-y}(1 - |cos y|) , dy ] 3. Next, we need to evaluate the integral: [ int_0^pi e^{-y}(1 - |cos y|) , dy ] This can be split into two integrals: [ int_0^pi e^{-y}(1 - |cos y|) , dy = int_0^pi e^{-y} , dy - int_0^pi e^{-y}|cos y| , dy ] 4. The first integral is straightforward: [ int_0^pi e^{-y} , dy = left[-e^{-y}right]_0^pi = 1 - e^{-pi} ] 5. For the second integral, we split it into two parts due to the absolute value: [ int_0^pi e^{-y}|cos y| , dy = int_0^{pi/2} e^{-y}cos y , dy + int_{pi/2}^pi e^{-y}(-cos y) , dy ] 6. Using integration by parts, we find: [ int e^{-y}cos y , dy = e^{-y}sin y + int e^{-y}sin y , dy ] [ int e^{-y}sin y , dy = -e^{-y}cos y - int e^{-y}cos y , dy ] Solving these, we get: [ int e^{-y}cos y , dy = frac{1}{2} e^{-y}(sin y - cos y) ] 7. Evaluating the definite integrals: [ int_0^{pi/2} e^{-y}cos y , dy = left[frac{1}{2} e^{-y}(sin y - cos y)right]_0^{pi/2} = frac{1}{2}(e^{-pi/2} - 1) ] [ int_{pi/2}^pi e^{-y}(-cos y) , dy = left[-frac{1}{2} e^{-y}(sin y - cos y)right]_{pi/2}^pi = frac{1}{2}(e^{-pi} - e^{-pi/2}) ] 8. Combining these results: [ int_0^pi e^{-y}|cos y| , dy = frac{1}{2}(e^{-pi/2} - 1) + frac{1}{2}(e^{-pi} - e^{-pi/2}) = frac{1}{2}(1 - e^{-pi}) ] 9. Therefore: [ int_0^pi e^{-y}(1 - |cos y|) , dy = (1 - e^{-pi}) - frac{1}{2}(1 - e^{-pi}) = frac{1}{2}(1 - e^{-pi}) ] 10. Substituting back, we get: [ a_n = e^{-(n-1)pi} cdot frac{1}{2}(1 - e^{-pi}) = frac{1}{2} e^{-(n-1)pi} (1 - e^{-pi}) ] 11. Summing (a_n) from (n = 1) to (infty): [ sum_{n=1}^infty a_n = frac{1}{2} (1 - e^{-pi}) sum_{n=1}^infty e^{-(n-1)pi} ] 12. The series (sum_{n=1}^infty e^{-(n-1)pi}) is a geometric series with the first term (a = 1) and common ratio (r = e^{-pi}): [ sum_{n=1}^infty e^{-(n-1)pi} = frac{1}{1 - e^{-pi}} ] 13. Therefore: [ sum_{n=1}^infty a_n = frac{1}{2} (1 - e^{-pi}) cdot frac{1}{1 - e^{-pi}} = frac{1}{2} ] The final answer is (boxed{frac{1}{2}})

answer:To find out how many minutes Latoya's call lasted, we need to calculate the total cost of the call and then divide that by the cost per minute. First, let's find out how much credit was used during the call: Initial credit - Remaining credit = Credit used 30.00 - 26.48 = 3.52 Now, we know that the call cost 3.52. To find out how many minutes the call lasted, we divide the cost of the call by the cost per minute: Minutes = Cost of call / Cost per minute Minutes = 3.52 / 0.16 per minute Now, let's do the math: Minutes = 3.52 / 0.16 Minutes = 22 So, Latoya's call lasted for boxed{22} minutes.

answer:**Analysis** This problem examines the application of the Sine Rule and the properties and applications of the sine function. By applying the Sine Rule, we can find the values of a and c, and then write the expression for the perimeter of triangle ADC. Using the graph and properties of the trigonometric functions, we can determine the range of the perimeter of triangle ADC. **Solution** In triangle ABC, B= frac{pi}{3}, AC=b= sqrt{3}, By the Sine Rule, we have frac {a}{sin A}= frac {c}{sin C}= frac {b}{sin B}= frac { sqrt {3}}{sin frac {pi}{3}}=2, therefore a=2sin A, c=2sin C, Since D is a point on segment BC and AB=AD, therefore triangle ABD is an equilateral triangle, AB=AD=BD=c, therefore The perimeter of triangle ADC is l=a-c+b+c=a+b=2sin A+ sqrt{3}, Given frac{pi}{3} < A < frac{2pi}{3}, therefore frac{ sqrt{3}}{2} < sin Aleqslant 1, therefore 2sin A+ sqrt{3}in(2 sqrt{3},2+ sqrt{3}]; Hence, the range of the perimeter of triangle ADC is (2 sqrt{3},2+ sqrt{3}]. Therefore, the correct answer is boxed{text{B}}.

answer:First, let's calculate the discount on the original price of the TV. Discount = 10% of 1700 = 0.10 * 1700 = 170 Now, let's subtract the discount from the original price to get the discounted price. Discounted price = Original price - Discount = 1700 - 170 = 1530 Next, we need to calculate the taxes on the discounted price. First, the 15% VAT: VAT = 15% of 1530 = 0.15 * 1530 = 229.50 Now, let's add the VAT to the discounted price. Price after VAT = Discounted price + VAT = 1530 + 229.50 = 1759.50 Next, we calculate the 8% sales tax on the discounted price. Sales Tax = 8% of 1530 = 0.08 * 1530 = 122.40 Finally, let's add the sales tax to the price after VAT to get the final price. Final price = Price after VAT + Sales Tax = 1759.50 + 122.40 = 1881.90 The final price of the television after applying the discount and accounting for both taxes is boxed{1881.90} .