answer：First, I need to determine the speed of the stream based on the given information about the man's swimming speeds upstream and downstream.I'll start by defining the variables:- Let ( v ) represent the speed of the stream in km/h.- The man's swimming speed in still water is 1.5 km/h.When swimming downstream, the stream's speed adds to the man's swimming speed, so his effective downstream speed is ( 1.5 + v ) km/h. Conversely, when swimming upstream, the stream's speed subtracts from his swimming speed, resulting in an effective upstream speed of ( 1.5 - v ) km/h.The problem states that it takes twice as long to swim upstream as it does to swim downstream. This implies that the distance covered upstream in a certain time is half the distance covered downstream in the same time, since speed is directly proportional to distance when time is constant.Let's denote the time it takes to swim downstream as ( t ) hours. Therefore, the time to swim upstream is ( 2t ) hours.The distances covered downstream and upstream can be expressed as:- Downstream distance: ( (1.5 + v) times t )- Upstream distance: ( (1.5 - v) times 2t )Since the distances are the same, I can set these two expressions equal to each other:[ (1.5 + v) times t = (1.5 - v) times 2t ]Next, I'll solve for ( v ):[ 1.5t + vt = 3t - 2vt ][ 3vt = 1.5t ][ v = frac{1.5t}{3t} ][ v = frac{1.5}{3} ][ v = 0.5 ]Therefore, the speed of the stream is 0.5 km/h.

answer：Okay, so I have this problem: I need to find all real numbers ( a ) such that the inequality ( |2 - x| + |x + 1| leq a ) holds true for every ( x ) in the interval [1, 5]. Hmm, let me try to figure this out step by step.First, I remember that absolute value expressions can represent distances on the number line. So, ( |2 - x| ) is the distance between ( x ) and 2, and ( |x + 1| ) is the distance between ( x ) and -1. So, the expression ( |2 - x| + |x + 1| ) is like adding these two distances together. Wait, so if I think of it that way, it's the sum of the distances from ( x ) to 2 and from ( x ) to -1. That makes me think of the triangle inequality or something related to geometry. Maybe I can visualize this on the number line.Let me sketch this out mentally. On the number line, we have points at -1, 2, and the variable point ( x ) somewhere between 1 and 5. The expression ( |2 - x| + |x + 1| ) is the sum of the distances from ( x ) to 2 and from ( x ) to -1. So, if I move ( x ) around between 1 and 5, this sum will change.I need to find the maximum value of this sum because the inequality has to hold for all ( x ) in [1, 5]. So, if I can find the maximum value of ( |2 - x| + |x + 1| ) in that interval, then ( a ) has to be at least that maximum value. That way, the inequality will hold for all ( x ) in [1, 5].Alright, so how do I find the maximum of this function? Maybe I can consider the function ( f(x) = |2 - x| + |x + 1| ) and analyze it over the interval [1, 5]. Since absolute value functions are piecewise linear, their sum will also be piecewise linear, and the maximum will occur either at a critical point or at the endpoints of the interval.Critical points occur where the expressions inside the absolute values change their sign, right? So, for ( |2 - x| ), the critical point is at ( x = 2 ), and for ( |x + 1| ), the critical point is at ( x = -1 ). But since our interval is [1, 5], ( x = -1 ) is outside of this interval, so the only critical point inside [1, 5] is at ( x = 2 ).So, I should evaluate ( f(x) ) at the endpoints ( x = 1 ) and ( x = 5 ), as well as at the critical point ( x = 2 ). Let's compute these:1. At ( x = 1 ): ( f(1) = |2 - 1| + |1 + 1| = |1| + |2| = 1 + 2 = 3 ).2. At ( x = 2 ): ( f(2) = |2 - 2| + |2 + 1| = |0| + |3| = 0 + 3 = 3 ).3. At ( x = 5 ): ( f(5) = |2 - 5| + |5 + 1| = | -3 | + |6| = 3 + 6 = 9 ).Hmm, interesting. So, at ( x = 1 ) and ( x = 2 ), the function ( f(x) ) is 3, and at ( x = 5 ), it's 9. So, it seems like the function reaches its maximum at ( x = 5 ).But wait, is that the only maximum? Let me check if the function is increasing or decreasing in the intervals [1, 2] and [2, 5]. Maybe I can take the derivative or analyze the slope.Since ( f(x) ) is piecewise linear, the slope will change at the critical points. Let's consider the intervals:1. For ( x < 2 ), both ( |2 - x| ) and ( |x + 1| ) are linear functions. Specifically, ( |2 - x| = 2 - x ) because ( x < 2 ), and ( |x + 1| = x + 1 ) because ( x + 1 > 0 ) for ( x geq 1 ). So, ( f(x) = (2 - x) + (x + 1) = 2 - x + x + 1 = 3 ). So, in the interval [1, 2), the function is constant at 3.2. For ( x geq 2 ), ( |2 - x| = x - 2 ) because ( x geq 2 ), and ( |x + 1| = x + 1 ) as before. So, ( f(x) = (x - 2) + (x + 1) = x - 2 + x + 1 = 2x - 1 ). This is a linear function with a slope of 2, so it's increasing for ( x geq 2 ).Therefore, in the interval [2, 5], the function ( f(x) ) is increasing from 3 up to ( f(5) = 9 ). So, the maximum value of ( f(x) ) on [1, 5] is indeed 9 at ( x = 5 ).Therefore, for the inequality ( |2 - x| + |x + 1| leq a ) to hold for all ( x in [1, 5] ), ( a ) must be at least 9. So, the range of ( a ) is all real numbers greater than or equal to 9.Wait, let me double-check. If ( a = 9 ), then the inequality becomes ( |2 - x| + |x + 1| leq 9 ). At ( x = 5 ), this is exactly 9, so it holds. For all other ( x ) in [1, 5], since the function is either constant or increasing up to 9, the inequality will hold as well because the left side is less than or equal to 9.If ( a ) were less than 9, say ( a = 8 ), then at ( x = 5 ), the left side is 9, which is greater than 8, so the inequality would not hold. Therefore, ( a ) must be at least 9.So, putting it all together, the range of ( a ) is all real numbers ( a ) such that ( a geq 9 ). In interval notation, that's ( [9, infty) ).**Final Answer**The range of ( a ) is boxed{[9, +infty)}.

answer：Alright, so I have this problem where I need to show that the function φ(x) = 1/(1 + x²)^(3/2) is a solution to a certain Volterra integral equation. The equation is given as φ(x) = 1/(1 + x²) - ∫₀ˣ [t/(1 + x²)] φ(t) dt. Okay, let's break this down step by step.First, I need to understand what it means for φ(x) to be a solution to this integral equation. Essentially, I have to plug φ(x) into the right-hand side of the equation and verify that it equals the left-hand side, which is φ(x) itself. So, my goal is to substitute φ(x) into the integral and see if everything simplifies correctly.Let me write down the integral equation again for clarity:φ(x) = 1/(1 + x²) - ∫₀ˣ [t/(1 + x²)] φ(t) dt.Given that φ(x) = 1/(1 + x²)^(3/2), I'll substitute this into the integral:φ(x) = 1/(1 + x²) - ∫₀ˣ [t/(1 + x²)] * [1/(1 + t²)^(3/2)] dt.Okay, so now I have an expression on the right-hand side that involves an integral. My next step is to evaluate this integral and see if it simplifies to φ(x).Let's look at the integral more closely:∫₀ˣ [t/(1 + x²)] * [1/(1 + t²)^(3/2)] dt.I notice that 1/(1 + x²) is a constant with respect to t, so I can factor that out of the integral:1/(1 + x²) * ∫₀ˣ [t / (1 + t²)^(3/2)] dt.Now, the integral simplifies to:∫₀ˣ [t / (1 + t²)^(3/2)] dt.This looks like a standard integral that might have a known antiderivative. Let me think about substitution methods. If I let u = 1 + t², then du/dt = 2t, which means t dt = du/2. That seems promising because the numerator is t dt, and the denominator is (1 + t²)^(3/2), which would become u^(3/2).Let's perform the substitution:Let u = 1 + t² ⇒ du = 2t dt ⇒ t dt = du/2.When t = 0, u = 1 + 0 = 1.When t = x, u = 1 + x².So, substituting into the integral:∫₀ˣ [t / (1 + t²)^(3/2)] dt = ∫₁^{1+x²} [1 / u^(3/2)] * (du/2).Simplifying this:(1/2) ∫₁^{1+x²} u^(-3/2) du.Now, integrating u^(-3/2):The antiderivative of u^(-3/2) is u^(-1/2) / (-1/2) = -2 u^(-1/2).So, evaluating from 1 to 1 + x²:(1/2) [ -2 u^(-1/2) ] from 1 to 1 + x².Simplify the constants:(1/2) * (-2) [ u^(-1/2) ] from 1 to 1 + x² = - [ u^(-1/2) ] from 1 to 1 + x².Now, plug in the limits:- [ (1 + x²)^(-1/2) - 1^(-1/2) ] = - [ 1/√(1 + x²) - 1 ].Simplify this expression:-1/√(1 + x²) + 1 = 1 - 1/√(1 + x²).So, the integral ∫₀ˣ [t / (1 + t²)^(3/2)] dt equals 1 - 1/√(1 + x²).Now, going back to the expression we had earlier:1/(1 + x²) * [1 - 1/√(1 + x²)].Let's compute this:1/(1 + x²) * [1 - 1/√(1 + x²)] = [1/(1 + x²)] - [1/(1 + x²)^(3/2)].So, putting it all together, the right-hand side of the original equation becomes:1/(1 + x²) - [1/(1 + x²)] * [1 - 1/√(1 + x²)] = 1/(1 + x²) - [1/(1 + x²) - 1/(1 + x²)^(3/2)].Simplify this expression:1/(1 + x²) - 1/(1 + x²) + 1/(1 + x²)^(3/2) = 1/(1 + x²)^(3/2).Which is exactly φ(x). Therefore, φ(x) satisfies the integral equation.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the substitution, everything seems correct. The integral was evaluated properly, and the algebra steps seem fine. So, yes, φ(x) does satisfy the given Volterra integral equation.

answer：Okay, so I have this problem where the king has 100 hunting dogs, and he wants to select three for hunting. He gathered 100 courtiers and asked each of them to pick the three fastest dogs. It turned out that for any two courtiers, there are two dogs that they both selected. I need to prove that the selections of some two courtiers completely coincide. Hmm, interesting.Alright, let me try to break this down. So, there are 100 courtiers, each selecting three dogs. The key condition here is that for any two courtiers, there are two dogs that they both selected. So, their selections overlap by at least two dogs. I need to show that under this condition, there must be at least two courtiers who have exactly the same three dogs selected.Let me think about how to approach this. Maybe I can model this as a problem in combinatorics or set theory. Each courtier's selection is a set of three dogs, and the condition is that the intersection of any two sets is at least two dogs. So, in other words, for any two sets (courtiers' selections), |A ∩ B| ≥ 2.I remember something about intersecting families in combinatorics. Maybe this is related to the Erdos-Ko-Rado theorem, which deals with intersecting families of sets. The classic EKR theorem states that for a family of k-element subsets of an n-element set, if every pair of subsets intersects, then the maximum size of such a family is C(n-1, k-1), provided n ≥ 2k.But in our case, the condition is stronger: not only do the subsets intersect, but their intersection is at least two elements. So, maybe we can use a similar approach or some combinatorial argument.Let me consider the number of possible triples. The total number of ways to choose three dogs out of 100 is C(100, 3), which is a huge number. But we only have 100 courtiers, each selecting a unique triple (assuming no two courtiers have the same selection, which is what we need to disprove).Wait, actually, the problem doesn't state that all selections are unique. It just says that each courtier picks three dogs, and any two courtiers share at least two dogs. So, maybe some selections are the same, but we need to show that at least two are.So, perhaps I can use the pigeonhole principle. If the number of possible triples is less than the number of courtiers, then by pigeonhole, at least two courtiers must have the same selection. But C(100, 3) is way larger than 100, so that approach won't work directly.Hmm, maybe I need to consider the structure of the selections. Since every pair of selections shares at least two dogs, the family of triples is highly intersecting. Maybe this restricts the number of possible triples significantly.Let me try to model this. Suppose each selection is a triple {a, b, c}. For any two triples, they share at least two elements. So, if I have two triples, say {a, b, c} and {a, b, d}, they share two elements a and b. Similarly, {a, b, c} and {a, c, d} share two elements a and c, and so on.So, perhaps all triples must share a common pair. Wait, but if all triples share a common pair, say {a, b}, then every triple is {a, b, x} for some x. In that case, the number of such triples is 98, since x can be any of the remaining 98 dogs. But we have 100 courtiers, so even if all triples share a common pair, we would need at least two courtiers to have the same triple, because 98 < 100. So, by pigeonhole, at least two courtiers would have the same triple.But wait, is it necessarily true that all triples share a common pair? Or could they share different pairs?Let me think. Suppose that not all triples share the same pair. Then, there are at least two different pairs that are shared among different triples. But if two triples share different pairs, say {a, b} and {a, c}, then their intersection is only {a}, which is only one dog, contradicting the condition that any two triples share at least two dogs.Ah, that's a key point. If two triples share different pairs, their intersection would only be one dog, which violates the given condition. Therefore, all triples must share the same pair. That is, there exists a specific pair {a, b} such that every triple is of the form {a, b, x}, where x is some other dog.Therefore, all 100 triples must be of the form {a, b, x}, with x ranging over the remaining 98 dogs. But since we have 100 courtiers, and only 98 possible x's, by the pigeonhole principle, at least two courtiers must have the same x, meaning their triples are identical.So, that would mean that at least two courtiers have the same selection of three dogs, which is exactly what we needed to prove.Wait, let me double-check this reasoning. Suppose that all triples share a common pair {a, b}. Then, each triple is {a, b, x_i}, where x_i is one of the remaining 98 dogs. Since there are only 98 different x's, but 100 courtiers, two courtiers must share the same x, hence the same triple. That makes sense.But what if not all triples share the same pair? Suppose there are two different pairs, say {a, b} and {a, c}. Then, the triple {a, b, x} and {a, c, y} would only share the dog a, which is only one dog, violating the condition that any two triples share at least two dogs. Therefore, all triples must share the same pair.Therefore, the only way for all triples to satisfy the condition is if they all share a common pair, and hence, by the pigeonhole principle, two triples must coincide.So, I think this reasoning holds. It shows that under the given condition, the selections must all share a common pair, and with more courtiers than the number of possible extensions of that pair, two courtiers must have the same triple.Yeah, that seems solid. I don't see any flaws in this logic. So, I can conclude that there must be at least two courtiers whose selections completely coincide.

answer：Okay, so I have this function f(x) that's defined for all real numbers. It has a property where f(x + 1) equals twice f(x). That seems like a recursive relationship, right? So if I know the value of f at some point x, I can find it at x + 1 by just doubling it. Cool, but how does that help me find f(21/2)?Well, the problem also tells me that when x is between -1 and 0, inclusive, f(x) is equal to x cubed. So for x in (-1, 0], f(x) = x³. That gives me a starting point because I can use that to find f at some other points by applying the recursive property.I need to find f(21/2). Let me write that down: f(21/2). Hmm, 21 divided by 2 is 10.5. So I need to find f(10.5). But I don't know f(10.5) directly. However, I know that f(x + 1) = 2f(x). Maybe I can work backwards from 10.5 to get into the interval where I know f(x).Let me think. If I subtract 1 from 10.5, I get 9.5. So f(10.5) = 2f(9.5). But I still don't know f(9.5). Maybe I can keep subtracting 1 until I get into the interval (-1, 0]. Let's see how many times I need to subtract 1 from 10.5 to get into that interval.Starting from 10.5:10.5 - 1 = 9.59.5 - 1 = 8.58.5 - 1 = 7.57.5 - 1 = 6.56.5 - 1 = 5.55.5 - 1 = 4.54.5 - 1 = 3.53.5 - 1 = 2.52.5 - 1 = 1.51.5 - 1 = 0.50.5 - 1 = -0.5Okay, so I subtracted 1 eleven times to get from 10.5 to -0.5. That means f(10.5) = 2^11 * f(-0.5). Because each time I subtract 1, I multiply by 2, and I did that 11 times.Now, f(-0.5) is given because -0.5 is in the interval (-1, 0]. So f(-0.5) = (-0.5)^3. Let me calculate that: (-0.5)^3 is -0.125, which is -1/8.So now, f(10.5) = 2^11 * (-1/8). Let me compute 2^11. 2^10 is 1024, so 2^11 is 2048. Therefore, f(10.5) = 2048 * (-1/8).Calculating that, 2048 divided by 8 is 256. So 2048 * (-1/8) is -256. Therefore, f(10.5) is -256.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from f(10.5), I subtracted 1 eleven times to get to -0.5, which is correct because 10.5 - 11 = -0.5. Then, f(-0.5) is indeed (-0.5)^3 = -1/8. Multiplying 2^11 by -1/8 gives -256. Yep, that seems right.So, after working through it step by step, I'm confident that f(21/2) is -256.

answer：First, calculate the total number of dimes by multiplying the number of dime piles by the number of coins per pile: 3 piles × 4 coins/pile = 12 dimes.Next, subtract the number of dimes from the total number of coins to find the number of quarters: 20 coins - 12 dimes = 8 quarters.Finally, determine the number of quarter piles by dividing the number of quarters by the number of coins per pile: 8 quarters ÷ 4 coins/pile = 2 piles of quarters.