answer：Okay, so I have this problem about constructing an open-top rectangular pool. The volume needs to be 18 cubic meters, and the depth is 2 meters. The cost for the bottom is 200 yuan per square meter, and the walls are 150 yuan per square meter. I need to find the minimum cost to build this pool.Alright, let me start by understanding the problem. It's an open-top pool, so it doesn't have a lid or cover. It's rectangular, so it has a length, width, and depth. The volume is given, which is length × width × depth. The depth is fixed at 2 meters, so that should help in figuring out the area of the base.First, let me calculate the area of the bottom. Since the volume is 18 m³ and the depth is 2 m, the area of the base (length × width) should be 18 divided by 2, which is 9 m². So, the bottom area is 9 m².Now, the cost for the bottom is 200 yuan per square meter. So, the total cost for the bottom would be 9 m² × 200 yuan/m² = 1800 yuan. That's straightforward.Next, I need to figure out the cost for the walls. The pool is open-top, so there are four walls: two with length and two with width. Each of these walls has a height equal to the depth of the pool, which is 2 meters.So, the area of each wall can be calculated as follows:- The two longer walls: each has an area of length × depth.- The two shorter walls: each has an area of width × depth.Therefore, the total area of the walls is 2 × (length × depth) + 2 × (width × depth). Since the depth is 2 meters, this simplifies to 2 × (length × 2) + 2 × (width × 2) = 4 × length + 4 × width = 4 × (length + width).So, the total area of the walls is 4(l + w), where l is length and w is width.Now, the cost for the walls is 150 yuan per square meter. So, the total cost for the walls would be 4(l + w) × 150 yuan/m².But wait, I don't know the values of l and w yet, except that their product is 9 m² (since l × w = 9). So, I need to express the total cost in terms of a single variable and then find the minimum cost.Let me denote length as l and width as w. So, l × w = 9. Therefore, w = 9 / l.Substituting this into the total area of the walls, we get:Total wall area = 4(l + w) = 4(l + 9/l).So, the cost for the walls is 4(l + 9/l) × 150 yuan.Let me compute that:Cost for walls = 4(l + 9/l) × 150 = 600(l + 9/l).Now, the total cost of the pool is the sum of the cost for the bottom and the cost for the walls:Total cost = 1800 + 600(l + 9/l).So, I need to minimize this total cost with respect to l. That is, find the value of l that minimizes 1800 + 600(l + 9/l).To find the minimum, I can take the derivative of the cost function with respect to l, set it equal to zero, and solve for l.Let me denote the cost function as C(l):C(l) = 1800 + 600(l + 9/l).Taking the derivative of C with respect to l:C'(l) = d/dl [1800] + d/dl [600(l + 9/l)] = 0 + 600(1 - 9/l²).Set C'(l) = 0:600(1 - 9/l²) = 0.Divide both sides by 600:1 - 9/l² = 0.So, 1 = 9/l².Multiply both sides by l²:l² = 9.Take square root:l = 3 or l = -3.But since length can't be negative, l = 3 meters.So, the length is 3 meters. Then, the width w = 9 / l = 9 / 3 = 3 meters.Wait, so both length and width are 3 meters? That means the pool is actually a cube in terms of its base, but since the depth is 2 meters, it's not a cube but a square prism.But does that make sense? If both length and width are equal, the pool is square in shape, which might minimize the perimeter, hence the wall area.Let me verify that this is indeed a minimum. I can check the second derivative or analyze the behavior.Compute the second derivative:C''(l) = d/dl [600(1 - 9/l²)] = 600(0 + 18/l³) = 10800 / l³.At l = 3, C''(3) = 10800 / 27 = 400, which is positive. So, the function is concave upward at this point, meaning it's a minimum.Therefore, the minimum cost occurs when l = 3 meters and w = 3 meters.Now, let's compute the total cost:Total cost = 1800 + 600(l + 9/l).Substitute l = 3:Total cost = 1800 + 600(3 + 9/3) = 1800 + 600(3 + 3) = 1800 + 600 × 6 = 1800 + 3600 = 5400 yuan.So, the minimum cost is 5400 yuan.But let me double-check my calculations to make sure I didn't make any mistakes.First, volume is 18 m³, depth 2 m, so base area is 9 m². Correct.Cost for the bottom: 9 × 200 = 1800 yuan. Correct.For the walls, total area is 4(l + w). Since l × w = 9, so w = 9/l. Therefore, total wall area is 4(l + 9/l). Correct.Cost for walls: 4(l + 9/l) × 150 = 600(l + 9/l). Correct.Total cost: 1800 + 600(l + 9/l). Correct.Taking derivative: 600(1 - 9/l²). Setting to zero: 1 - 9/l² = 0, so l² = 9, l = 3. Correct.Second derivative: 10800 / l³, which is positive at l = 3, so minimum. Correct.Calculating total cost: 1800 + 600(3 + 3) = 1800 + 3600 = 5400. Correct.So, everything seems to check out. Therefore, the minimum cost is indeed 5400 yuan.Alternatively, another way to think about it is that for a given area, the shape that minimizes the perimeter is a square. Since the walls' cost depends on the perimeter (since each wall's area is length × depth, and there are two lengths and two widths), minimizing the perimeter would minimize the wall area, hence the cost.So, by making the base a square, we minimize the perimeter, which in turn minimizes the wall area and thus the total cost. That makes sense.Therefore, I'm confident that the minimum cost is 5400 yuan.**Final Answer**The minimum cost of constructing the pool is boxed{5400} yuan.

answer：Okay, so I have this problem here: A bike travels a certain distance in 7 seconds and its speed is 50 m/s. How far did the bike travel? Hmm, let me try to figure this out step by step.First, I know that speed is basically how fast something is moving. In this case, the bike's speed is given as 50 meters per second (m/s). That means every second, the bike covers 50 meters. Okay, so if it's going at 50 meters every second, then in 7 seconds, it should cover more than 50 meters, right?I think the formula I need here is the basic distance formula. I remember it's something like distance equals speed multiplied by time. Let me write that down: Distance = Speed × Time. Yeah, that sounds right. So, if I plug in the numbers I have, it should give me the distance.So, the speed is 50 m/s, and the time is 7 seconds. Let me do the multiplication: 50 times 7. Hmm, 50 times 7 is 350. So, does that mean the bike traveled 350 meters? That seems straightforward, but let me double-check to make sure I didn't make any mistakes.Wait, does the unit make sense? Speed is in meters per second, and time is in seconds, so when I multiply them, the seconds cancel out, leaving me with meters. Yeah, that makes sense because distance should be in meters. So, 50 m/s times 7 s equals 350 meters. Okay, that checks out.Is there another way to think about this to confirm? Maybe breaking it down into smaller parts. If the bike goes 50 meters every second, then in one second, it's at 50 meters. In two seconds, it's at 100 meters, and so on. So, in 7 seconds, it would be 50 times 7, which is 350 meters. Yep, that matches what I got earlier.I guess I could also visualize it. Imagine the bike starting at a point and moving straight for 7 seconds. Each second, it moves 50 meters away from the starting point. After 7 seconds, it would have moved 50 meters seven times, totaling 350 meters. That makes sense.Is there anything else I need to consider? Maybe units conversion? Well, the units are already in meters and seconds, so I don't think I need to convert anything here. If the speed was given in a different unit, like kilometers per hour, I might have to convert it to meters per second first, but that's not the case here.What if the time wasn't a whole number? Like, if it was 7.5 seconds, would I still multiply the same way? I think so. The formula should work regardless of whether the time is a whole number or a decimal. So, in this case, since it's exactly 7 seconds, multiplying 50 by 7 gives me the exact distance.Let me also think about real-life scenarios. If I'm riding a bike at 50 m/s, that's actually pretty fast. To put it into perspective, 50 m/s is equivalent to 180 kilometers per hour or about 112 miles per hour. That's faster than most bikes can go; usually, bikes are around 15-30 m/s. So, maybe this is a very fast bike or perhaps it's a motorcycle. But regardless, the calculation remains the same.Another thing to consider is whether the bike is moving at a constant speed or if it's accelerating. The problem doesn't specify, but since it gives a constant speed of 50 m/s, I assume it's moving at that constant speed throughout the 7 seconds. If it were accelerating, we would need more information, like the acceleration rate, to calculate the distance.Also, if I wanted to find the time instead, I could rearrange the formula. Time equals distance divided by speed. But in this case, we're given both speed and time, so we just need to find the distance.Let me recap to make sure I haven't missed anything. The problem gives me speed (50 m/s) and time (7 s), and asks for distance. I used the formula Distance = Speed × Time, plugged in the numbers, and got 350 meters. I checked the units, thought about real-life applications, and considered whether there might be other factors like acceleration, but it seems like everything lines up.I think I'm confident with this answer. The bike traveled 350 meters in 7 seconds at a speed of 50 m/s.

answer：Alright, so I've got this multiple-choice question here about functions. It's asking which one is both an even function and decreases on the interval (0, +∞). The options are A: y = x², B: y = x³, C: y = x⁻², and D: y = x⁻³. Hmm, okay, let me break this down step by step.First, I need to remember what an even function is. From what I recall, an even function satisfies the condition f(-x) = f(x) for all x in its domain. So, if I plug in -x into the function, it should give me the same result as when I plug in x. That means the graph of the function is symmetric with respect to the y-axis. Got it.Next, the function needs to decrease on the interval (0, +∞). So, for x values greater than zero, as x increases, the value of the function should decrease. That means the function is monotonically decreasing in that interval. Okay, so I need to check both the evenness and the monotonicity on the positive side.Let's go through each option one by one.Option A: y = x². I know that y = x² is a parabola opening upwards. Is it even? Let's test f(-x). If I plug in -x, I get (-x)² = x², which is the same as f(x). So, yes, it's even. Now, does it decrease on (0, +∞)? Well, the derivative of y = x² is 2x. On the interval (0, +∞), 2x is positive, which means the function is increasing, not decreasing. So, option A doesn't satisfy both conditions.Option B: y = x³. This is a cubic function. Is it even? Let's check f(-x). Plugging in -x, we get (-x)³ = -x³, which is -f(x). That means it's an odd function, not even. So, already, it doesn't meet the first condition. But just to be thorough, does it decrease on (0, +∞)? The derivative is 3x², which is always positive for x ≠ 0. So, it's increasing on (0, +∞) as well. So, option B is out.Option C: y = x⁻². That's the same as y = 1/x². Let's check if it's even. f(-x) = (-x)⁻² = (-1)⁻² * x⁻² = (1/(-1)²) * x⁻² = 1 * x⁻² = x⁻², which is equal to f(x). So, yes, it's even. Now, does it decrease on (0, +∞)? Let's find the derivative. The derivative of y = x⁻² is dy/dx = -2x⁻³. On the interval (0, +∞), x is positive, so x⁻³ is positive, and multiplying by -2 makes it negative. A negative derivative means the function is decreasing. So, option C satisfies both conditions.Option D: y = x⁻³. That's y = 1/x³. Is it even? Let's check f(-x). f(-x) = (-x)⁻³ = (-1)⁻³ * x⁻³ = -1/x³, which is -f(x). So, it's an odd function, not even. Therefore, it doesn't meet the first condition. But just to confirm, does it decrease on (0, +∞)? The derivative is dy/dx = -3x⁻⁴. On (0, +∞), x⁻⁴ is positive, so the derivative is negative, meaning the function is decreasing. However, since it's not even, it doesn't satisfy both requirements.So, after going through each option, only option C meets both criteria: it's an even function and decreases on (0, +∞).**Final Answer**boxed{C}

answer：Alright, so I've got this math problem here, and I need to figure out whether "x > 0" is a sufficient, necessary, both, or neither condition for "x^{2020} > 0." Hmm, okay, let's break this down step by step.First, I remember that when dealing with exponents, especially even exponents, the result is always non-negative because any number multiplied by itself an even number of times can't be negative. So, x^{2020} is definitely going to be positive or zero. But wait, the problem says "x^{2020} > 0," which means it's strictly greater than zero. So, that tells me that x can't be zero because zero raised to any power is still zero, which doesn't satisfy the inequality.So, if x^{2020} > 0, then x must not be zero. That means x can be any positive number or any negative number, right? Because both positive and negative numbers raised to an even power become positive. So, x just needs to be anything except zero for x^{2020} to be greater than zero.Now, the question is about the condition "x > 0." Is this a sufficient condition, a necessary condition, both, or neither for "x^{2020} > 0"?Let me recall what these terms mean. A sufficient condition means that if the condition is met, then the statement is true. But it doesn't necessarily mean that the statement can't be true without the condition. On the other hand, a necessary condition means that for the statement to be true, the condition must be met, but meeting the condition doesn't guarantee the statement is true.So, if "x > 0" is a sufficient condition for "x^{2020} > 0", that would mean that whenever x is greater than zero, x^{2020} is definitely greater than zero. Is that true? Well, yes, because any positive number raised to any power remains positive. So, if x > 0, then x^{2020} > 0 is always true. Therefore, "x > 0" is a sufficient condition.But is it a necessary condition? That would mean that x^{2020} > 0 can only be true if x > 0. But wait, earlier I thought that x can also be negative, right? Because negative numbers raised to an even power are positive. So, if x is negative, say x = -1, then (-1)^{2020} = 1, which is still greater than zero. Therefore, "x > 0" isn't a necessary condition because x^{2020} > 0 can be true even if x isn't greater than zero—it could be less than zero.So, putting it all together, "x > 0" is a sufficient condition because it guarantees that x^{2020} > 0, but it's not a necessary condition because x^{2020} > 0 can be true without x > 0. Therefore, the correct answer should be that "x > 0" is a sufficient but not necessary condition for "x^{2020} > 0."I think that makes sense. Let me just double-check to make sure I didn't miss anything. If x > 0, then yes, x^{2020} is definitely positive. But if x is negative, it's still positive because of the even exponent. So, x doesn't have to be positive for x^{2020} to be positive, which confirms that "x > 0" isn't necessary. So, yeah, I feel confident with that answer.**Final Answer**boxed{A}

answer：Okay, so I have this problem about a geometric sequence with 10 terms. The product of the odd terms is 2, and the product of the even terms is 64. I need to find the common ratio. Hmm, let me think about how to approach this.First, I remember that in a geometric sequence, each term is the previous term multiplied by the common ratio, which is usually denoted as ( q ). So, the sequence looks like ( a_1, a_1 q, a_1 q^2, a_1 q^3, ldots ) up to 10 terms.The problem mentions the product of the odd terms and the product of the even terms. Let me clarify what that means. The odd terms would be the 1st, 3rd, 5th, 7th, and 9th terms. Similarly, the even terms are the 2nd, 4th, 6th, 8th, and 10th terms. So, there are 5 odd terms and 5 even terms.Let me write down the expressions for these terms:- Odd terms: ( a_1, a_3, a_5, a_7, a_9 )- Even terms: ( a_2, a_4, a_6, a_8, a_{10} )Since it's a geometric sequence, each term can be expressed in terms of ( a_1 ) and ( q ):- ( a_1 = a_1 )- ( a_3 = a_1 q^2 )- ( a_5 = a_1 q^4 )- ( a_7 = a_1 q^6 )- ( a_9 = a_1 q^8 )Similarly, for the even terms:- ( a_2 = a_1 q )- ( a_4 = a_1 q^3 )- ( a_6 = a_1 q^5 )- ( a_8 = a_1 q^7 )- ( a_{10} = a_1 q^9 )Now, the product of the odd terms is given as 2. Let me compute that:Product of odd terms:( a_1 times a_3 times a_5 times a_7 times a_9 = a_1 times (a_1 q^2) times (a_1 q^4) times (a_1 q^6) times (a_1 q^8) )Let me simplify this:Each term is ( a_1 ) multiplied by ( q ) raised to an even power. There are 5 terms, so:( (a_1)^5 times q^{2 + 4 + 6 + 8} )Wait, let me compute the exponents:2 + 4 + 6 + 8 = 20So, the product becomes:( (a_1)^5 times q^{20} = 2 )Similarly, the product of the even terms is 64. Let me compute that:Product of even terms:( a_2 times a_4 times a_6 times a_8 times a_{10} = (a_1 q) times (a_1 q^3) times (a_1 q^5) times (a_1 q^7) times (a_1 q^9) )Simplifying this:Each term is ( a_1 ) multiplied by ( q ) raised to an odd power. There are 5 terms, so:( (a_1)^5 times q^{1 + 3 + 5 + 7 + 9} )Let me compute the exponents:1 + 3 + 5 + 7 + 9 = 25So, the product becomes:( (a_1)^5 times q^{25} = 64 )Now, I have two equations:1. ( (a_1)^5 times q^{20} = 2 ) (Product of odd terms)2. ( (a_1)^5 times q^{25} = 64 ) (Product of even terms)Hmm, I can divide the second equation by the first to eliminate ( (a_1)^5 ). Let me try that.Dividing equation 2 by equation 1:( frac{(a_1)^5 times q^{25}}{(a_1)^5 times q^{20}} = frac{64}{2} )Simplify:( q^{5} = 32 )Okay, so ( q^5 = 32 ). I need to solve for ( q ). Let me recall that 32 is ( 2^5 ). So,( q^5 = 2^5 )Taking the fifth root of both sides:( q = 2 )So, the common ratio ( q ) is 2. Let me check if this makes sense.If ( q = 2 ), then let's see what ( (a_1)^5 ) would be from equation 1:( (a_1)^5 times 2^{20} = 2 )So,( (a_1)^5 = frac{2}{2^{20}} = 2^{-19} )Therefore, ( a_1 = (2^{-19})^{1/5} = 2^{-19/5} ). That seems a bit complicated, but it's a valid value for ( a_1 ).Let me check equation 2 with ( q = 2 ):( (a_1)^5 times 2^{25} = 64 )Substituting ( (a_1)^5 = 2^{-19} ):( 2^{-19} times 2^{25} = 2^{6} = 64 )Yes, that works out. So, the calculations are consistent.Therefore, the common ratio is 2, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this function f(x) = x² / (1 + x²), and I need to do two things with it. First, I need to find the values of f(2) + f(1/2) and f(3) + f(1/3). Then, I have to prove that f(x) + f(1/x) is a constant. Hmm, let me start with the first part.Alright, for f(2), I just plug in x = 2 into the function. So that would be (2)² / (1 + (2)²). Let me calculate that. 2 squared is 4, and 1 plus 4 is 5. So f(2) is 4/5. Got that.Now, f(1/2). That means I plug in x = 1/2. So, (1/2)² is 1/4, and 1 plus (1/2)² is 1 + 1/4, which is 5/4. So f(1/2) is (1/4) divided by (5/4). Dividing fractions is the same as multiplying by the reciprocal, so that's (1/4) * (4/5) = 1/5. Okay, so f(1/2) is 1/5.Now, adding f(2) and f(1/2) together: 4/5 + 1/5. That's straightforward; the denominators are the same, so I just add the numerators. 4 + 1 is 5, so 5/5 is 1. So f(2) + f(1/2) equals 1.Let me check that again to make sure I didn't make a mistake. f(2) is 4/5, f(1/2) is 1/5, adding them gives 1. Yeah, that seems right.Now, moving on to f(3) + f(1/3). Let's do f(3) first. Plugging in x = 3, so 3 squared is 9, and 1 + 9 is 10. So f(3) is 9/10.Next, f(1/3). Plugging in x = 1/3, so (1/3) squared is 1/9, and 1 + 1/9 is 10/9. Therefore, f(1/3) is (1/9) divided by (10/9). Again, dividing fractions is multiplying by the reciprocal, so (1/9) * (9/10) = 1/10. So f(1/3) is 1/10.Adding f(3) and f(1/3) together: 9/10 + 1/10. The denominators are the same, so adding the numerators: 9 + 1 = 10, so 10/10 is 1. Therefore, f(3) + f(1/3) is also 1.Hmm, interesting. Both f(2) + f(1/2) and f(3) + f(1/3) equal 1. I wonder if this is a coincidence or if there's a pattern here. Maybe the function has some symmetry when you plug in x and 1/x.Let me think about that. The function is f(x) = x² / (1 + x²). If I plug in 1/x instead of x, what happens? Let's see: f(1/x) = (1/x)² / (1 + (1/x)²). Simplifying that, (1/x²) / (1 + 1/x²). Let me write that as (1/x²) divided by ( (x² + 1)/x² ). Dividing by a fraction is the same as multiplying by its reciprocal, so it becomes (1/x²) * (x² / (x² + 1)) = 1 / (x² + 1).So f(1/x) simplifies to 1 / (x² + 1). Now, let's add f(x) and f(1/x) together. f(x) is x² / (1 + x²), and f(1/x) is 1 / (1 + x²). Adding them together: x² / (1 + x²) + 1 / (1 + x²). Since the denominators are the same, I can add the numerators: x² + 1, all over (1 + x²). So that's (x² + 1) / (x² + 1), which simplifies to 1.Wow, so f(x) + f(1/x) is always equal to 1, regardless of the value of x. That's why when I plugged in 2 and 1/2, and 3 and 1/3, I got 1 both times. So this isn't a coincidence; it's a general property of this function.Let me just recap to make sure I understand. When I take f(x) and f(1/x), their sum is always 1 because when you add x²/(1 + x²) and 1/(1 + x²), the numerators add up to (x² + 1), which cancels out the denominator. So it's a neat little identity.I think that makes sense. It's interesting how sometimes functions have these reciprocal properties. I guess it's because when you take the reciprocal of x, you're essentially flipping the fraction, and in this case, the function's structure allows the sum to simplify to a constant.So, in summary, for part (1), both f(2) + f(1/2) and f(3) + f(1/3) equal 1. For part (2), we've shown that f(x) + f(1/x) is always 1, which is a constant. That wraps up both parts of the problem.**Final Answer**(1) The values are boxed{1} and boxed{1}.(2) It is proven that ( f(x) + fleft(frac{1}{x}right) ) is a constant, specifically boxed{1}.