answer：First, I need to understand the cost structure for developing the inventory management application. The developer charges a base fee of 500 plus an additional 3 per item in the inventory.For the first part, the total cost is given as 2,000. I'll set up a linear equation where the total cost equals the base fee plus the variable cost per item multiplied by the number of items. By plugging in the known total cost, I can solve for the number of items.In the second part, the owner wants to save at least 400 annually through optimization. The savings are calculated as 50 for every 100 items. I'll set up an inequality where the total savings are greater than or equal to 400 and solve for the minimum number of items required to meet this savings goal.

answer：First, I need to add the two decimal numbers, 4.3 and 3.88.To make the addition easier, I'll align the decimal points by rewriting 4.3 as 4.30.Now, I'll add the numbers column by column, starting from the rightmost digit.Adding the hundredths place: 0 + 8 equals 8.Next, adding the tenths place: 3 + 8 equals 11. I'll write down 1 and carry over 1 to the units place.Finally, adding the units place: 4 + 3 plus the carried-over 1 equals 8.Combining these results, the sum is 8.18.

answer：To determine the day when the maximum number of hymns is sung, I start by analyzing the quadratic equation ( h(x) = -2x^2 + 8x + 4 ). Since the coefficient of ( x^2 ) is negative, the parabola opens downward, indicating that the vertex will give the maximum value.The x-coordinate of the vertex of a parabola defined by ( ax^2 + bx + c ) is found using the formula ( x = -frac{b}{2a} ). Here, ( a = -2 ) and ( b = 8 ), so:[x = -frac{8}{2 times (-2)} = -frac{8}{-4} = 2]This means the maximum number of hymns is sung on the 2nd day. To find the number of hymns, I substitute ( x = 2 ) back into the equation:[h(2) = -2(2)^2 + 8(2) + 4 = -8 + 16 + 4 = 12]Therefore, 12 hymns are sung on the day when the maximum number is performed.Next, to find the number of hymn books distributed on this day, I use the given pattern ( b(x) = x^2 + 3x ). Substituting ( x = 2 ):[b(2) = (2)^2 + 3(2) = 4 + 6 = 10]So, 10 hymn books are distributed on the day when the maximum number of hymns is sung.

answer：Alright, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. First, I know that in an arithmetic sequence, each term is obtained by adding a common difference to the previous term. So, the general form of an arithmetic sequence is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.The problem gives me two equations:1. a_1 + a_3 + a_5 = 1052. a_2 + a_4 + a_6 = 99And I need to find the value of n that maximizes the sum of the first n terms, S_n.Okay, let me write down what each of these terms is in terms of a_1 and d.Starting with the first equation:a_1 is just a_1.a_3 is the third term, so that's a_1 + 2d.a_5 is the fifth term, which is a_1 + 4d.So, adding them together: a_1 + (a_1 + 2d) + (a_1 + 4d) = 105.Let me simplify that:a_1 + a_1 + 2d + a_1 + 4d = 105Combine like terms:3a_1 + 6d = 105I can factor out a 3:3(a_1 + 2d) = 105Divide both sides by 3:a_1 + 2d = 35Okay, so that's one equation: a_1 + 2d = 35.Now, moving on to the second equation:a_2 + a_4 + a_6 = 99Similarly, let's express each term:a_2 is a_1 + d.a_4 is a_1 + 3d.a_6 is a_1 + 5d.Adding them together: (a_1 + d) + (a_1 + 3d) + (a_1 + 5d) = 99Simplify:a_1 + d + a_1 + 3d + a_1 + 5d = 99Combine like terms:3a_1 + 9d = 99Factor out a 3:3(a_1 + 3d) = 99Divide both sides by 3:a_1 + 3d = 33Now, I have two equations:1. a_1 + 2d = 352. a_1 + 3d = 33Hmm, so I can set up a system of equations here. Let me subtract the first equation from the second to eliminate a_1.(a_1 + 3d) - (a_1 + 2d) = 33 - 35Simplify:a_1 + 3d - a_1 - 2d = -2Which gives:d = -2So, the common difference d is -2.Now, plug this back into one of the equations to find a_1. Let's use the first equation:a_1 + 2d = 35Substitute d = -2:a_1 + 2(-2) = 35Simplify:a_1 - 4 = 35Add 4 to both sides:a_1 = 39So, the first term is 39, and the common difference is -2.Now, I need to find the value of n that maximizes S_n, the sum of the first n terms.I remember that the sum of the first n terms of an arithmetic sequence is given by:S_n = n/2 * [2a_1 + (n - 1)d]Alternatively, it can also be written as:S_n = n(a_1 + a_n)/2Where a_n is the nth term, which is a_1 + (n - 1)d.Since we have a_1 and d, we can write S_n in terms of n.Let me use the first formula:S_n = n/2 * [2a_1 + (n - 1)d]Plugging in a_1 = 39 and d = -2:S_n = n/2 * [2*39 + (n - 1)*(-2)]Simplify inside the brackets:2*39 = 78(n - 1)*(-2) = -2n + 2So, adding them together:78 - 2n + 2 = 80 - 2nTherefore, S_n = n/2 * (80 - 2n)Simplify further:S_n = n/2 * (80 - 2n) = n*(40 - n) = 40n - n^2So, S_n = -n^2 + 40nHmm, this is a quadratic function in terms of n, and since the coefficient of n^2 is negative (-1), the parabola opens downward, meaning the vertex is the maximum point.The vertex of a parabola given by f(n) = an^2 + bn + c is at n = -b/(2a). In this case, a = -1 and b = 40.So, n = -40/(2*(-1)) = -40/(-2) = 20Therefore, the value of n that maximizes S_n is 20.Wait, but let me double-check. Since n has to be a positive integer, and in the context of the problem, n can't be a fraction or negative. So, n = 20 is the point where the sum is maximized.But just to be thorough, let me think about the sequence itself. Since the common difference d is negative (-2), the terms are decreasing. So, the sequence starts at 39 and each term is 2 less than the previous one.Therefore, the terms will eventually become negative, and the sum will start decreasing once the negative terms outweigh the positive ones.So, the maximum sum occurs just before the terms become negative or when the terms are still positive but adding a negative term would decrease the total sum.Let me find when the terms become negative.The nth term is a_n = a_1 + (n - 1)d = 39 + (n - 1)*(-2) = 39 - 2(n - 1) = 41 - 2nSet a_n > 0:41 - 2n > 041 > 2nn < 41/2 = 20.5So, the 20th term is a_20 = 41 - 2*20 = 41 - 40 = 1The 21st term is a_21 = 41 - 2*21 = 41 - 42 = -1So, the 21st term is negative. Therefore, adding the 21st term would decrease the sum.Hence, the maximum sum occurs at n = 20.Therefore, the value of n that maximizes S_n is 20.**Final Answer**The value of ( n ) that maximizes ( S_n ) is boxed{20}.

answer：First, I need to determine the quadratic equation that represents the number of videos collected each year. The equation is given as ( n^2 + bn + c ), where ( a = 1 ).In the first year (( n = 1 )), she collected 5 videos. Plugging this into the equation:[1^2 + b(1) + c = 5 1 + b + c = 5 b + c = 4 quad text{(Equation 1)}]By the fifth year (( n = 5 )), her total collection had grown to 85 videos. Plugging this into the equation:[5^2 + b(5) + c = 85 25 + 5b + c = 85 5b + c = 60 quad text{(Equation 2)}]Now, I can solve the system of equations:[begin{cases}b + c = 4 5b + c = 60end{cases}]Subtract Equation 1 from Equation 2:[(5b + c) - (b + c) = 60 - 4 4b = 56 b = 14]Substitute ( b = 14 ) back into Equation 1:[14 + c = 4 c = -10]So, the quadratic equation is ( n^2 + 14n - 10 ).Next, to find the total number of videos by the end of the 10th year, I'll calculate the sum of the videos collected each year from ( n = 1 ) to ( n = 10 ).Using the formula for the sum of a quadratic sequence:[S = sum_{n=1}^{10} (n^2 + 14n - 10)]Breaking it down:[S = sum_{n=1}^{10} n^2 + 14sum_{n=1}^{10} n - 10sum_{n=1}^{10} 1]Calculating each part:[sum_{n=1}^{10} n^2 = frac{10(10 + 1)(2 times 10 + 1)}{6} = 385 sum_{n=1}^{10} n = frac{10(10 + 1)}{2} = 55 sum_{n=1}^{10} 1 = 10]Plugging these back into the sum:[S = 385 + 14(55) - 10(10) S = 385 + 770 - 100 S = 1055]Therefore, by the end of the 10th year, she will have a total of 1,055 videos.

answer：Okay, so I have this problem here about arranging books on a shelf. There are two Arabic books, three German books, and four Spanish books, making a total of nine books. The conditions are that the two Arabic books must be placed together and the four Spanish books must be placed together. I need to figure out how many different ways the books can be arranged under these conditions. The answer choices are given, so that might help me check my work as I go.First, let me try to visualize the problem. There are three different categories of books: Arabic, German, and Spanish. Each category has a different number of books. The constraints are that all the Arabic books must be together and all the Spanish books must be together. The German books don't have any constraints, so they can be placed anywhere individually.So, if I think about this, the two Arabic books need to be treated as a single unit, and the four Spanish books also need to be treated as a single unit. That way, I can simplify the problem by considering these units instead of individual books.Let me break it down step by step.1. **Treating the Arabic and Spanish books as single units:** - Arabic books: 2 books → 1 unit - Spanish books: 4 books → 1 unit - German books: 3 books → 3 units So, in total, we have 1 (Arabic) + 3 (German) + 1 (Spanish) = 5 units to arrange.2. **Arranging these 5 units:** - The number of ways to arrange 5 distinct units is 5 factorial, which is 5! = 5 × 4 × 3 × 2 × 1 = 120 ways. Wait, but hold on. Are these units all distinct? The Arabic unit is one, the Spanish unit is another, and each German book is individual. So, yes, each unit is distinct because they belong to different categories or are individual books. So, 5! is correct here.3. **Considering the internal arrangements within the Arabic and Spanish units:** - For the Arabic unit: There are 2 books, and they can be arranged among themselves in 2! ways. So, 2! = 2 × 1 = 2 ways. - For the Spanish unit: There are 4 books, and they can be arranged among themselves in 4! ways. So, 4! = 4 × 3 × 2 × 1 = 24 ways.4. **Combining all these together:** - The total number of arrangements is the number of ways to arrange the units multiplied by the number of ways to arrange the books within each unit. - So, total arrangements = 5! × 2! × 4! = 120 × 2 × 24. Let me compute that step by step: - 120 × 2 = 240 - 240 × 24 = ? Hmm, let's calculate 240 × 24: - 240 × 20 = 4800 - 240 × 4 = 960 - So, 4800 + 960 = 5760 So, total arrangements would be 5760.Wait, but let me double-check my reasoning because sometimes when dealing with permutations, it's easy to make a mistake.First, I considered the Arabic and Spanish books as single units, which gives me 5 units in total. Then, I calculated the permutations of these 5 units, which is 5!. Then, I multiplied by the permutations within the Arabic unit (2!) and within the Spanish unit (4!). So, 5! × 2! × 4! = 120 × 2 × 24 = 5760. That seems correct.But let me think again if there's another way to approach this problem. Maybe by considering the positions of the Arabic and Spanish blocks.Alternatively, I can think of the problem as arranging the blocks and the individual German books. So, the two blocks (Arabic and Spanish) and the three German books. So, that's 5 items to arrange.Yes, that's the same as before. So, 5! ways to arrange them, and then multiply by the permutations within each block.Another way: If I didn't consider the blocks, the total number of arrangements without any constraints would be 9! = 362880. But since we have constraints, we have to adjust for that.But in this case, since the constraints are that certain books must be together, the method I used is the standard approach for such problems. So, I think 5760 is correct.Looking back at the answer choices: (A) 1440, (B) 2880, (C) 5760, (D) 182440, (E) 362880.5760 is option C, so that should be the answer.But just to make sure, let me think if I considered all the possibilities correctly.- We have two blocks: Arabic (A) and Spanish (S), and three German books (G1, G2, G3).- The number of ways to arrange these five items is 5!.- Then, within the Arabic block, the two books can be arranged in 2! ways.- Within the Spanish block, the four books can be arranged in 4! ways.So, total arrangements: 5! × 2! × 4! = 120 × 2 × 24 = 5760.Yes, that seems solid. I don't see any mistakes in this reasoning.Alternatively, if I think about the positions on the shelf, the Arabic block can start at position 1, 2, 3, 4, 5, or 6 (since it's two books). Similarly, the Spanish block can start at various positions, but this approach would complicate things because the blocks can overlap in different ways. The first method is more straightforward.Therefore, I'm confident that the correct answer is 5760, which is option C.**Final Answer**boxed{C}