💡To find the slope of line ( m ), we start by determining the midpoint of the line segment with endpoints ( (2, 8) ) and ( (8, -2) ). The midpoint formula is:[text{Midpoint} = left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right)]Substituting the given points:[text{Midpoint} = left( frac{2 + 8}{2}, frac{8 + (-2)}{2} right) = left( frac{10}{2}, frac{6}{2} right) = (5, 3)]Next, we know that line ( m ) passes through the midpoint ( (5, 3) ) and has a y-intercept of ( -2 ), meaning it also passes through ( (0, -2) ). The slope ( m ) of a line passing through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is calculated as:[text{Slope} = frac{y_2 - y_1}{x_2 - x_1}]Using the points ( (0, -2) ) and ( (5, 3) ):[text{Slope} = frac{3 - (-2)}{5 - 0} = frac{3 + 2}{5} = frac{5}{5} = 1]Therefore, the slope of line ( m ) is ( 1 ).

💡To determine the value of ( c ) such that ( x^2 + 14x + c ) is a perfect square, we can express it in the form ( (x + a)^2 ), where ( a ) is a constant.Expanding ( (x + a)^2 ) gives:[x^2 + 2ax + a^2]Comparing this with the given quadratic expression ( x^2 + 14x + c ), we can set up the following equations:[2a = 14 quad text{and} quad a^2 = c]Solving for ( a ) from the first equation:[a = frac{14}{2} = 7]Substituting ( a = 7 ) into the second equation to find ( c ):[c = 7^2 = 49]Therefore, the value of ( c ) is ( 49 ).

💡Okay, so Larry, Julius, and Nina are playing this game where they take turns throwing a ball at a bottle. Larry goes first, then Julius, then Nina, and then it repeats. The first person to knock the bottle off wins. Each time a player throws, there's a 1/3 chance they'll knock it off. I need to find the probability that Larry wins the game.Hmm, let me think about how to approach this. It seems like a probability problem involving turns and repeated attempts. Maybe I can model this as an infinite series where Larry has multiple chances to win on his turns.First, let's break down the possible scenarios where Larry can win. He can win on his first turn, or if everyone misses in the first round and then he wins on his second turn, and so on. So, the probability that Larry wins can be thought of as the sum of probabilities of him winning on each of his turns.Let me denote the probability that Larry wins as P. The probability that he wins on his first turn is straightforward: it's 1/3. If he doesn't win on his first turn, then Julius and Nina also have to miss for Larry to get another chance. The probability that Larry misses is 2/3, and similarly, Julius and Nina each have a 2/3 chance of missing. So, the probability that all three miss in the first round is (2/3)^3.If all three miss, then it's effectively the same situation as the beginning, just one round later. So, the probability that Larry wins on his second turn is (2/3)^3 * (1/3). Similarly, the probability that Larry wins on his third turn would be (2/3)^6 * (1/3), and so on.So, the total probability P that Larry wins is the sum of these probabilities:P = (1/3) + (2/3)^3 * (1/3) + (2/3)^6 * (1/3) + (2/3)^9 * (1/3) + ...This looks like an infinite geometric series where each term is (2/3)^{3n} * (1/3) for n = 0, 1, 2, ...The first term a of this series is 1/3, and the common ratio r is (2/3)^3 = 8/27.The formula for the sum of an infinite geometric series is S = a / (1 - r), provided that |r| < 1, which is true here since 8/27 is less than 1.Plugging in the values:S = (1/3) / (1 - 8/27) = (1/3) / (19/27) = (1/3) * (27/19) = 9/19.So, the probability that Larry wins the game is 9/19.Wait, let me double-check my steps to make sure I didn't make a mistake. I considered all the scenarios where Larry can win on his first, second, third, etc., turns. Each time, the probability of getting to that turn is (2/3)^{3n}, and then Larry has a 1/3 chance to win. Summing this up as a geometric series with a = 1/3 and r = 8/27 gives me 9/19. That seems correct.I think I've got it. Larry's probability of winning is 9/19.

💡Okay, so I have this problem here where Alex has 8 different kinds of lunch meat and 12 different kinds of cheese. He wants to make a sandwich using one kind of meat and any three kinds of cheese. The question is asking how many different sandwiches he can make. Hmm, let's break this down step by step.First, I need to understand what's being asked. Alex is making sandwiches, and for each sandwich, he needs to choose one type of lunch meat and three types of cheese. So, the total number of different sandwiches would depend on how many ways he can combine these meats and cheeses.Let me start with the lunch meat. There are 8 different kinds, and he wants to use one kind per sandwich. So, how many ways can he choose one kind of meat from 8? That seems straightforward. It's just 8 choices, right? So, for the meat part, there are 8 possible options.Now, moving on to the cheese. There are 12 different kinds of cheese, and he wants to use any three kinds for each sandwich. This part is a bit more complex because it involves combinations. When the order doesn't matter, like choosing three cheeses out of twelve, we use combinations, not permutations. So, I need to calculate how many ways there are to choose 3 cheeses out of 12.The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items we want to choose. In this case, n is 12 and k is 3. Plugging those numbers into the formula, I get C(12, 3) = 12! / (3!(12 - 3)!).Calculating that, 12! is 12 × 11 × 10 × 9! and 3! is 6, so it simplifies to (12 × 11 × 10) / (3 × 2 × 1) = 220. So, there are 220 ways to choose three cheeses out of twelve.Now, to find the total number of different sandwiches, I need to multiply the number of meat choices by the number of cheese combinations. That's because for each meat choice, there are 220 different cheese combinations possible. So, 8 (meat choices) multiplied by 220 (cheese combinations) equals 1760.Wait, let me double-check that. If I have 8 meats and for each meat, I can pair it with 220 different sets of cheeses, then yes, 8 × 220 should give me the total number of unique sandwiches. 8 × 200 is 1600, and 8 × 20 is 160, so adding those together gives 1760. That seems correct.I don't think I've missed anything here. The key was recognizing that the cheese selection is a combination problem since the order of the cheeses doesn't matter in a sandwich. If the order did matter, like if he was layering the cheeses in a specific sequence, then we would have used permutations instead. But since it's just any three kinds, combinations are the way to go.Also, it's important to note that Alex is using one kind of meat per sandwich, not multiple meats. So, we're only choosing one out of the eight, which is a simple selection without any combination involved there.In summary, by calculating the number of ways to choose the meat and the number of ways to choose the cheeses separately and then multiplying them together, we get the total number of different sandwiches Alex can make. That total is 1760.

💡First, I recognize that the set ( A ) is defined as ( A = {1, -7} ).Next, I need to determine the relationship between the element ( -7 ) and the set ( A ).Since ( -7 ) is explicitly listed within the set ( A ), it is clear that ( -7 ) is an element of ( A ).Therefore, the correct symbol to use is ( in ), indicating that ( -7 ) is an element of ( A ).

💡Alright, so I'm trying to solve this problem about a project that converts carbon dioxide into usable chemical products. The problem has two parts, and I need to figure out both. Let me take it step by step.First, the problem gives me a piecewise function for the monthly processing cost ( y ) in terms of the monthly processing volume ( x ) (in tons). The function is:[y = begin{cases} frac{1}{3}x^{3} - 80x^{2} + 5040x, & text{if } x in [120, 144) frac{1}{2}x^{2} - 200x + 80000, & text{if } x in [144, 500)end{cases}]Additionally, it tells me that the value of the chemical products obtained from processing one ton of carbon dioxide is 200 yuan. If the project isn't profitable, the state will provide compensation.**Part (1):** When ( x in [200, 300] ), determine whether the project can be profitable. If it is profitable, calculate the maximum profit; if not, calculate the minimum monthly subsidy from the state required to prevent the project from incurring losses.Okay, so for ( x ) between 200 and 300, we need to check if the project makes a profit. To do this, I think I need to calculate the profit function, which is revenue minus cost.Revenue from processing ( x ) tons would be ( 200x ) yuan, since each ton gives 200 yuan. The cost is given by the function ( y ). So, the profit ( S ) should be:[S = 200x - y]Since ( x ) is in [200, 300], we'll use the second part of the piecewise function for ( y ):[y = frac{1}{2}x^{2} - 200x + 80000]So, substituting ( y ) into the profit equation:[S = 200x - left( frac{1}{2}x^{2} - 200x + 80000 right)]Let me simplify this:[S = 200x - frac{1}{2}x^{2} + 200x - 80000][S = -frac{1}{2}x^{2} + 400x - 80000]Hmm, that's a quadratic function in terms of ( x ). It looks like a downward-opening parabola because the coefficient of ( x^2 ) is negative. The maximum profit (if any) would occur at the vertex of this parabola.The vertex of a parabola given by ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Here, ( a = -frac{1}{2} ) and ( b = 400 ), so:[x = -frac{400}{2 times -frac{1}{2}} = -frac{400}{-1} = 400]Wait, but our interval is [200, 300], and 400 is outside this interval. That means the maximum profit within [200, 300] would occur at one of the endpoints.Let me calculate ( S ) at ( x = 200 ) and ( x = 300 ):At ( x = 200 ):[S = -frac{1}{2}(200)^2 + 400(200) - 80000][S = -frac{1}{2}(40000) + 80000 - 80000][S = -20000 + 80000 - 80000 = -20000]At ( x = 300 ):[S = -frac{1}{2}(300)^2 + 400(300) - 80000][S = -frac{1}{2}(90000) + 120000 - 80000][S = -45000 + 120000 - 80000 = -5000]So, at both ends, the profit is negative. That means the project isn't profitable in this interval. The maximum profit is actually the least negative value, which is at ( x = 300 ), giving a profit of -5000 yuan. Therefore, the project incurs a loss, and the state needs to provide a subsidy to cover this loss.The minimum subsidy required would be the amount needed to make the project break even, which is the absolute value of the maximum loss. So, the minimum monthly subsidy required is 5000 yuan.**Part (2):** At what monthly processing volume (in tons) does the project achieve the lowest average processing cost per ton?Alright, so now I need to find the value of ( x ) that minimizes the average processing cost per ton. The average processing cost per ton is given by ( frac{y}{x} ).Given the piecewise function for ( y ), the average cost will also be piecewise:[frac{y}{x} = begin{cases} frac{frac{1}{3}x^{3} - 80x^{2} + 5040x}{x} = frac{1}{3}x^{2} - 80x + 5040, & text{if } x in [120, 144) frac{frac{1}{2}x^{2} - 200x + 80000}{x} = frac{1}{2}x - 200 + frac{80000}{x}, & text{if } x in [144, 500)end{cases}]So, I need to minimize each piece separately and then compare the minimums.**First piece: ( x in [120, 144) )**The average cost function is:[frac{y}{x} = frac{1}{3}x^{2} - 80x + 5040]This is a quadratic function in terms of ( x ), opening upwards (since the coefficient of ( x^2 ) is positive). The minimum occurs at the vertex.The vertex is at ( x = -frac{b}{2a} ), where ( a = frac{1}{3} ) and ( b = -80 ):[x = -frac{-80}{2 times frac{1}{3}} = frac{80}{frac{2}{3}} = 80 times frac{3}{2} = 120]So, the minimum average cost in this interval occurs at ( x = 120 ). Let's compute the average cost here:[frac{y}{x} = frac{1}{3}(120)^2 - 80(120) + 5040][= frac{1}{3}(14400) - 9600 + 5040][= 4800 - 9600 + 5040][= (4800 + 5040) - 9600][= 9840 - 9600 = 240]So, the average cost is 240 yuan per ton at ( x = 120 ).**Second piece: ( x in [144, 500) )**The average cost function is:[frac{y}{x} = frac{1}{2}x - 200 + frac{80000}{x}]This function isn't a simple quadratic, so I'll need to find its minimum by taking the derivative and setting it equal to zero.Let me denote ( f(x) = frac{1}{2}x - 200 + frac{80000}{x} ).Compute the derivative ( f'(x) ):[f'(x) = frac{1}{2} - frac{80000}{x^2}]Set ( f'(x) = 0 ):[frac{1}{2} - frac{80000}{x^2} = 0][frac{1}{2} = frac{80000}{x^2}][x^2 = 80000 times 2 = 160000][x = sqrt{160000} = 400]So, the critical point is at ( x = 400 ). Since this is within the interval [144, 500), we can consider it.Now, let's compute the average cost at ( x = 400 ):[f(400) = frac{1}{2}(400) - 200 + frac{80000}{400}][= 200 - 200 + 200][= 200]So, the average cost is 200 yuan per ton at ( x = 400 ).Additionally, I should check the endpoints of this interval to ensure that 400 is indeed the minimum.At ( x = 144 ):[f(144) = frac{1}{2}(144) - 200 + frac{80000}{144}][= 72 - 200 + approx 555.56][= (72 + 555.56) - 200][= 627.56 - 200 = 427.56]At ( x = 500 ):[f(500) = frac{1}{2}(500) - 200 + frac{80000}{500}][= 250 - 200 + 160][= 250 + 160 - 200][= 410 - 200 = 210]So, at ( x = 500 ), the average cost is 210 yuan per ton, which is higher than at ( x = 400 ).Therefore, the minimum average cost in the second interval is 200 yuan per ton at ( x = 400 ).Comparing the two intervals:- First interval minimum: 240 yuan at ( x = 120 )- Second interval minimum: 200 yuan at ( x = 400 )Since 200 is less than 240, the overall minimum average processing cost per ton is achieved at ( x = 400 ).**Summary:**1. For ( x in [200, 300] ), the project isn't profitable. The maximum loss is 5000 yuan, so the state needs to provide a minimum subsidy of 5000 yuan.2. The project achieves the lowest average processing cost per ton at a monthly processing volume of 400 tons.**Final Answer**(1) The minimum monthly subsidy required is boxed{5000} yuan.(2) The lowest average processing cost per ton is achieved at a monthly processing volume of boxed{400} tons.