answer：First, I need to find the general formula for the sequence. The given condition is ( frac{a_n}{n} = n - 2 ). By multiplying both sides by ( n ), I can express ( a_n ) as ( a_n = n^2 - 2n ).Next, I want to determine which term in the sequence equals 15. So, I set up the equation ( n^2 - 2n = 15 ).To solve for ( n ), I rearrange the equation to form a quadratic equation: ( n^2 - 2n - 15 = 0 ).Using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -2 ), and ( c = -15 ), I calculate the discriminant: ( (-2)^2 - 4(1)(-15) = 4 + 60 = 64 ).Taking the square root of the discriminant gives ( sqrt{64} = 8 ). Plugging this back into the quadratic formula, I find ( n = frac{2 pm 8}{2} ).This results in two solutions: ( n = frac{10}{2} = 5 ) and ( n = frac{-6}{2} = -3 ). Since the term number cannot be negative, I discard ( n = -3 ).Therefore, the 15th term of the sequence corresponds to ( n = 5 ).

answer：First, I need to understand the scale provided in the blueprint, which states that one inch represents 50 meters.Next, I identify the length of the line segment on the blueprint, which is 7.5 inches.To find the actual length in meters, I multiply the length of the line segment by the scale factor:7.5 inches × 50 meters/inch.Calculating this gives:7.5 × 50 = 375 meters.Therefore, the line segment represents 375 meters.

answer：First, I need to round the number -3657.7421 to the nearest ten. To do this, I'll look at the digit in the tens place and the digit immediately to its right.The number is -3657.7421. The digit in the tens place is 5, and the digit to its right is 7.Since the digit to the right of the tens place is 7, which is greater than 5, I need to round the tens digit up by one. However, because the number is negative, rounding up actually means moving to a more negative number.So, rounding 5 up by one gives me 6. Therefore, the number becomes -3660.Thus, the rounded value of -3657.7421 to the nearest ten is -3660.

answer：Okay, so I have this problem about buying books for the school library. It's divided into two parts. Let me try to understand and solve each part step by step.**Part (1): Find the price of each literary classic and comic book.**Alright, the problem says that 20 literary classics and 40 comic books together cost 1520 yuan. Also, literary classics cost 440 yuan more than comic books for the same quantity. Hmm, I need to find the price per book for both types.Let me denote:- Let ( x ) be the price of one literary classic.- Let ( y ) be the price of one comic book.From the first statement: 20 literary classics and 40 comic books cost 1520 yuan. So, the equation would be:[ 20x + 40y = 1520 ]From the second statement: Literary classics cost 440 yuan more than comic books for the same quantity. I think this means that if I buy the same number of both, the total cost for literary classics is 440 yuan more than for comic books. So, for the same quantity, say 20 books each, the cost difference is 440 yuan. Wait, but the first statement already mentions 20 literary and 40 comic books. Maybe I need to interpret this differently.Wait, maybe it means that for the same number of books, the cost of literary classics is 440 yuan more than comic books. So, if I buy, say, 1 literary classic and 1 comic book, the literary classic is 440 yuan more expensive? That doesn't make sense because 440 yuan is a lot for a single book difference. Maybe it's per unit?Wait, let me read it again: "The literary classics cost 440 yuan more than the comic books for the same quantity." So, for the same number of books, the total cost for literary classics is 440 yuan more than for comic books. So, if I buy 20 literary classics and 20 comic books, the cost difference is 440 yuan. That makes more sense.So, the cost for 20 literary classics is 440 yuan more than the cost for 20 comic books. So, the equation would be:[ 20x = 20y + 440 ]Simplify this:[ x = y + 22 ]Okay, so now I have two equations:1. ( 20x + 40y = 1520 )2. ( x = y + 22 )I can substitute the second equation into the first one to find the values of ( x ) and ( y ).Substituting ( x = y + 22 ) into the first equation:[ 20(y + 22) + 40y = 1520 ]Let me expand this:[ 20y + 440 + 40y = 1520 ]Combine like terms:[ 60y + 440 = 1520 ]Subtract 440 from both sides:[ 60y = 1080 ]Divide both sides by 60:[ y = 18 ]So, the price of one comic book is 18 yuan. Now, substitute ( y = 18 ) back into ( x = y + 22 ):[ x = 18 + 22 = 40 ]Therefore, the price of one literary classic is 40 yuan, and the price of one comic book is 18 yuan.**Part (2): Find all the book purchasing plans that meet the conditions.**The school requires purchasing 20 more comic books than literary classics. So, if I buy ( x ) literary classics, I need to buy ( x + 20 ) comic books. The total number of books should be at least 72, and the total cost should not exceed 2000 yuan.Let me denote:- ( x ) = number of literary classics- ( x + 20 ) = number of comic booksTotal number of books:[ x + (x + 20) = 2x + 20 ]This should be at least 72:[ 2x + 20 geq 72 ]Subtract 20 from both sides:[ 2x geq 52 ]Divide by 2:[ x geq 26 ]So, the number of literary classics should be at least 26.Now, the total cost should not exceed 2000 yuan. The cost for literary classics is 40 yuan each, and for comic books, it's 18 yuan each.Total cost:[ 40x + 18(x + 20) leq 2000 ]Let me expand this:[ 40x + 18x + 360 leq 2000 ]Combine like terms:[ 58x + 360 leq 2000 ]Subtract 360 from both sides:[ 58x leq 1640 ]Divide both sides by 58:[ x leq frac{1640}{58} ]Let me calculate this:[ 58 times 28 = 1624 ][ 58 times 29 = 1682 ]Since 1640 is between 1624 and 1682, ( x ) must be less than or equal to 28 (since 28.62... is not an integer, and ( x ) must be an integer).So, ( x leq 28 )Combining both conditions:[ 26 leq x leq 28 ]Therefore, the possible values for ( x ) are 26, 27, and 28.Let me list the purchasing plans:1. If ( x = 26 ): - Literary classics: 26 - Comic books: 26 + 20 = 46 - Total books: 26 + 46 = 72 - Total cost: ( 40 times 26 + 18 times 46 = 1040 + 828 = 1868 ) yuan2. If ( x = 27 ): - Literary classics: 27 - Comic books: 27 + 20 = 47 - Total books: 27 + 47 = 74 - Total cost: ( 40 times 27 + 18 times 47 = 1080 + 846 = 1926 ) yuan3. If ( x = 28 ): - Literary classics: 28 - Comic books: 28 + 20 = 48 - Total books: 28 + 48 = 76 - Total cost: ( 40 times 28 + 18 times 48 = 1120 + 864 = 1984 ) yuanAll these plans satisfy the conditions of at least 72 books and a total cost not exceeding 2000 yuan.

answer：Alright, so I've got this problem about transporting apples using cars, and I need to figure out some relationships and maximize profit. Let me try to break it down step by step.First, the problem mentions three types of apples: A, B, and C. Each type has a different tonnage per car and a different profit per ton. There are 40 cars in total, and they need to transport exactly 200 tons of apples. Each car can only carry one type of apple and must be fully loaded. For part (1), I need to find the functional relationship between y and x, where x is the number of cars transporting type A apples, and y is the number for type B. Since there are 40 cars in total, the number of cars transporting type C would be 40 minus x minus y. Okay, so let's write that down: number of cars for C = 40 - x - y.Next, the total tons transported should be 200 tons. Each car carrying type A can carry 6 tons, type B 5 tons, and type C 4 tons. So, the total tons equation would be:6x + 5y + 4*(40 - x - y) = 200.Let me simplify that equation. First, distribute the 4:6x + 5y + 160 - 4x - 4y = 200.Combine like terms:(6x - 4x) + (5y - 4y) + 160 = 200.That simplifies to:2x + y + 160 = 200.Subtract 160 from both sides:2x + y = 40.So, solving for y gives:y = -2x + 40.Alright, so that's the functional relationship between y and x. That seems straightforward.Moving on to part (2), the problem states that the number of cars transporting each type of apple must be at least 4. So, x ≥ 4, y ≥ 4, and the number of cars for type C, which is 40 - x - y, must also be ≥ 4.Given that y = -2x + 40, let's substitute that into the condition for type C:40 - x - y ≥ 4.Substitute y:40 - x - (-2x + 40) ≥ 4.Simplify:40 - x + 2x - 40 ≥ 4.Which simplifies to:x ≥ 4.Wait, that's interesting. So, the condition for type C just gives us x ≥ 4, which we already have. So, our constraints are:x ≥ 4,y = -2x + 40 ≥ 4,and since y must be ≥ 4, let's solve for x:-2x + 40 ≥ 4,-2x ≥ -36,Divide both sides by -2 (remembering to reverse the inequality):x ≤ 18.So, x must be between 4 and 18, inclusive. Since x and y must be integers (you can't have a fraction of a car), x can take integer values from 4 to 18.Let's count how many possible values x can take:From 4 to 18 inclusive, that's 18 - 4 + 1 = 15 possible values.So, there are 15 possible arrangements for the cars.Now, part (3) asks to maximize the profit under these conditions. The profit is calculated based on the tons transported and the profit per ton for each type.The profit function would be:Profit = (Tons of A * Profit per ton A) + (Tons of B * Profit per ton B) + (Tons of C * Profit per ton C).Given that each car carries a certain tonnage, we can express the tons as:Tons of A = 6x,Tons of B = 5y,Tons of C = 4*(40 - x - y).So, substituting y = -2x + 40 into the tons of B and C:Tons of B = 5*(-2x + 40) = -10x + 200,Tons of C = 4*(40 - x - (-2x + 40)) = 4*(40 - x + 2x - 40) = 4x.So, Tons of C = 4x.Now, the profit function becomes:Profit = 6x*5 + (-10x + 200)*7 + 4x*8.Let's compute each term:6x*5 = 30x,(-10x + 200)*7 = -70x + 1400,4x*8 = 32x.Adding them all together:Profit = 30x - 70x + 1400 + 32x,Combine like terms:(30x - 70x + 32x) + 1400 = (-8x) + 1400.So, Profit = -8x + 1400.To maximize the profit, since the coefficient of x is negative (-8), the profit decreases as x increases. Therefore, to maximize profit, we need to minimize x.Given that x must be at least 4, the minimum value of x is 4.So, substituting x = 4 into the profit function:Profit = -8*4 + 1400 = -32 + 1400 = 1368.Therefore, the maximum profit is 1368 hundred dollars.Just to double-check, let's find the corresponding y and number of cars for C:y = -2x + 40 = -2*4 + 40 = -8 + 40 = 32,Number of cars for C = 40 - x - y = 40 - 4 - 32 = 4.So, 4 cars for A, 32 for B, and 4 for C. That satisfies all the constraints: each type has at least 4 cars, total cars are 40, and total tons are:6*4 + 5*32 + 4*4 = 24 + 160 + 16 = 200 tons.Everything checks out. So, the maximum profit is indeed 1368 hundred dollars.

answer：Alright, so I have this equation to solve: (2^{y^2 - 1} = (2^{y - 1})^2). Hmm, okay, let me try to figure this out step by step. First, I know that when dealing with exponential equations, especially with the same base, I can set the exponents equal to each other. That might simplify things. Let me rewrite the right side to make it easier. The right side is ((2^{y - 1})^2). I remember that when you raise a power to another power, you multiply the exponents. So, that should be (2^{2(y - 1)}). Let me write that down:[2^{y^2 - 1} = 2^{2(y - 1)}]Okay, now both sides have the same base, which is 2. Since the bases are equal, I can set the exponents equal to each other. That gives me:[y^2 - 1 = 2(y - 1)]Alright, now I have a quadratic equation. Let me simplify this. First, I'll expand the right side:[y^2 - 1 = 2y - 2]Now, I'll bring all terms to one side to set the equation to zero. Subtract (2y) and add 2 to both sides:[y^2 - 2y + 1 = 0]Hmm, this looks like a quadratic equation. Let me see if I can factor it. The equation is (y^2 - 2y + 1 = 0). I notice that this is a perfect square trinomial because ((y - 1)^2 = y^2 - 2y + 1). So, factoring it gives:[(y - 1)^2 = 0]To solve for (y), I take the square root of both sides:[y - 1 = 0]Adding 1 to both sides gives:[y = 1]So, the solution is (y = 1). Since the problem asks for the sum of all positive real numbers (y) that satisfy the equation, and there's only one solution, which is 1, the sum (T) is just 1.Let me double-check my work to make sure I didn't make any mistakes. Starting from the original equation:[2^{y^2 - 1} = (2^{y - 1})^2]Simplify the right side:[2^{y^2 - 1} = 2^{2(y - 1)}]Set the exponents equal:[y^2 - 1 = 2y - 2]Bring all terms to one side:[y^2 - 2y + 1 = 0]Factor:[(y - 1)^2 = 0]So, (y = 1). Plugging this back into the original equation to verify:Left side: (2^{1^2 - 1} = 2^{1 - 1} = 2^0 = 1)Right side: ((2^{1 - 1})^2 = (2^0)^2 = 1^2 = 1)Both sides are equal, so (y = 1) is indeed a valid solution. Since it's the only solution, the sum (T) is 1.I think that's it. I don't see any other solutions, and my steps seem correct. So, the answer should be (T = 1).**Final Answer**boxed{B}