answer：Okay, so I have this problem here where I need to find the intersection of two sets, A and B. Let me read it again carefully.Set A is defined as A = {x ∈ N* | -3 ≤ x ≤ 3}. Hmm, okay. So N* usually refers to the set of natural numbers, which are the positive integers starting from 1, right? So N* = {1, 2, 3, ...}. But here, x is in N*, and it's also between -3 and 3. Wait, that seems a bit confusing because natural numbers are positive, so how can x be between -3 and 3? Maybe it's just considering the natural numbers within that range. So, if x has to be a natural number and also between -3 and 3, then x can only be 1, 2, or 3. So, A = {1, 2, 3}. That makes sense.Now, set B is defined as B = {x | 0 < x < 4}. Okay, so this is the set of all real numbers x such that x is greater than 0 and less than 4. So, B is the interval (0, 4). But wait, the problem doesn't specify what kind of numbers x is. Is it natural numbers or real numbers? Hmm, in set A, x is a natural number, but in set B, it's just x. Maybe I should assume that x is a real number since it's not specified otherwise. So, B is all real numbers between 0 and 4, not including 0 and 4 themselves.Now, I need to find A ∩ B, which is the intersection of sets A and B. The intersection of two sets consists of all elements that are common to both sets. So, I need to find the elements that are in both A and B.Set A is {1, 2, 3}, and set B is (0, 4). So, let's see which elements of A are also in B. The elements of A are 1, 2, and 3. Are these numbers in B? Well, B includes all real numbers greater than 0 and less than 4. So, 1 is greater than 0 and less than 4, so 1 is in B. Similarly, 2 and 3 are also greater than 0 and less than 4, so they are in B as well. Therefore, all elements of A are in B, which means A ∩ B = {1, 2, 3}.Wait, but let me double-check. If B is (0, 4), which is an interval of real numbers, and A is {1, 2, 3}, which are integers, then technically, the intersection would still be {1, 2, 3} because those are the points where both sets overlap. So, I think my answer is correct.Let me look at the options given:A: {1, 2, 3}B: (0, 3]C: {-1, 1, 2, 3}D: {0, 1, 2}So, option A is exactly what I got: {1, 2, 3}. Option B is an interval from 0 to 3, including 3 but not 0. But since A only has integers, the intersection should also be integers, not an interval. So, B is not correct. Option C includes -1, which isn't in A or B, so that's wrong. Option D includes 0, which isn't in A or B, so that's also incorrect.Therefore, the correct answer should be A: {1, 2, 3}.

answer：Alright, let's tackle this problem step by step. So, we have the function f(x) = sin2x + √3cos2x, and we're supposed to translate its graph to the right by π/6 units to get g(x). Then, we need to determine which of the options A, B, C, or D is correct regarding the properties of g(x).First, I remember that when you translate a function to the right by a certain amount, you replace x with (x - h), where h is the amount you're shifting. In this case, h is π/6. So, g(x) should be f(x - π/6). Let me write that down:g(x) = f(x - π/6) = sin[2(x - π/6)] + √3cos[2(x - π/6)]Hmm, that looks a bit complicated. Maybe I can simplify it using some trigonometric identities. I recall that sin(A - B) and cos(A - B) can be expanded. Let me try expanding both terms.Starting with sin[2(x - π/6)]:sin[2x - 2*(π/6)] = sin(2x - π/3)Similarly, cos[2(x - π/6)]:cos[2x - 2*(π/6)] = cos(2x - π/3)So, now g(x) becomes:g(x) = sin(2x - π/3) + √3cos(2x - π/3)This still looks a bit messy. Maybe I can combine these terms into a single sine or cosine function using the amplitude-phase form. I remember that any function of the form A sinθ + B cosθ can be written as C sin(θ + φ), where C = √(A² + B²) and φ = arctan(B/A) or something like that.Let me apply that here. So, A is 1 (coefficient of sin) and B is √3 (coefficient of cos). Therefore, C should be √(1² + (√3)²) = √(1 + 3) = √4 = 2.Now, to find φ, which is the phase shift. φ = arctan(B/A) = arctan(√3/1) = arctan(√3). I remember that arctan(√3) is π/3 because tan(π/3) = √3.So, putting it all together, g(x) can be written as:g(x) = 2 sin(2x - π/3 + π/3) = 2 sin(2x)Wait, that simplifies nicely! The -π/3 and +π/3 cancel each other out, leaving just 2 sin(2x). That makes things much easier.Now, let's analyze the options given:A: The graph is symmetric about the point (π/6, 0)B: The graph is symmetric about the point (π/6, 0)C: The graph is strictly increasing on (0, π/4)D: The graph is strictly decreasing on (0, π/4)First, I notice that both A and B are the same, so they must be duplicates. Maybe that was a typo, but regardless, let's check them.Symmetry about a point usually means that if you rotate the graph 180 degrees around that point, it looks the same. For the sine function, which is symmetric about the origin, but after shifting, it might have a different center of symmetry.But in our case, g(x) = 2 sin(2x). Let's check if it's symmetric about (π/6, 0). To test this, we can see if for any point (x, y) on the graph, the point (2π/6 - x, -y) is also on the graph. Let's pick x = π/6. Then, g(π/6) = 2 sin(2*(π/6)) = 2 sin(π/3) = 2*(√3/2) = √3. So, the point (π/6, √3) is on the graph. For symmetry about (π/6, 0), the point (2π/6 - π/6, -√3) = (π/6, -√3) should also be on the graph. But g(π/6) is √3, not -√3. So, it's not symmetric about (π/6, 0). Therefore, both A and B are incorrect.Now, moving on to options C and D. We need to determine if g(x) is strictly increasing or decreasing on the interval (0, π/4). To do this, let's find the derivative of g(x).g(x) = 2 sin(2x)g'(x) = 2 * 2 cos(2x) = 4 cos(2x)Now, let's analyze the sign of g'(x) on (0, π/4).The interval (0, π/4) for x translates to 2x in (0, π/2). Cosine is positive in the first quadrant (0 to π/2), so cos(2x) is positive for all x in (0, π/4). Therefore, g'(x) = 4 cos(2x) is positive on (0, π/4), which means g(x) is strictly increasing on that interval.So, option C is correct, and option D is incorrect.Wait, but let me double-check. Is there any chance that the derivative could be negative in that interval? Since 2x goes from 0 to π/2, and cosine is positive throughout that interval, the derivative remains positive. So, yes, g(x) is indeed strictly increasing on (0, π/4).Therefore, the correct answer is option C.

answer：First, I need to determine the speed of Wild Bill's horse. I'll denote the horse's speed as ( v_h ) feet per second.When Wild Bill fires the gun in the same direction the horse is running, the bullet's speed relative to the ground is the sum of the bullet's speed and the horse's speed. This gives me the equation:[400 + v_h]When he fires the gun in the opposite direction, the bullet's speed relative to the ground is the difference between the bullet's speed and the horse's speed. This gives me the equation:[400 - v_h]According to the problem, the bullet fired in the same direction is 40 feet per second faster than the bullet fired in the opposite direction. This leads to the equation:[400 + v_h = (400 - v_h) + 40]Simplifying this equation:[400 + v_h = 440 - v_h][2v_h = 40][v_h = 20]Therefore, Wild Bill's horse runs at a speed of 20 feet per second.

answer：First, calculate the cost of medication, which is 50% of the total bill:[5000 times 0.50 = 2500]Subtract the medication cost from the total bill to find the remaining amount:[5000 - 2500 = 2500]Next, determine the cost of overnight stays, which is 25% of the remaining bill:[2500 times 0.25 = 625]Subtract the cost of overnight stays from the remaining bill:[2500 - 625 = 1875]Subtract the cost of food (175) from the remaining amount:[1875 - 175 = 1700]The remaining amount is the cost of the ambulance ride:[boxed{1700}]

answer：Okay, so I have this problem about Lisa sending orders and using packing peanuts. Let me try to understand what's being asked here. First, the problem says that Lisa has sent 3 large orders and 4 small orders. Each large order requires 200 grams of packing peanuts, and each small order requires 50 grams. The question is asking how many grams of packing peanuts Lisa has used in total.Alright, so I need to find the total grams of packing peanuts used for all the orders she sent. That means I need to calculate the packing peanuts used for the large orders and the small orders separately and then add them together.Let me break it down step by step. First, for the large orders: Lisa sent 3 large orders, and each large order needs 200 grams of packing peanuts. So, to find the total grams for large orders, I should multiply the number of large orders by the grams per large order. So, 3 large orders multiplied by 200 grams per large order. Let me write that out: 3 * 200g. Hmm, 3 times 200 is 600. So, Lisa used 600 grams of packing peanuts for the large orders.Now, moving on to the small orders. Lisa sent 4 small orders, and each small order needs 50 grams of packing peanuts. So, similarly, I need to multiply the number of small orders by the grams per small order.That would be 4 small orders multiplied by 50 grams per small order. So, 4 * 50g.4 times 50 is 200. So, Lisa used 200 grams of packing peanuts for the small orders.Now, to find the total grams of packing peanuts used, I need to add the grams used for large orders and the grams used for small orders together.So, 600 grams (from large orders) plus 200 grams (from small orders). Let's add those up: 600 + 200 equals 800.Therefore, Lisa has used a total of 800 grams of packing peanuts for all her orders.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For the large orders: 3 * 200g = 600g. That seems right. For the small orders: 4 * 50g = 200g. That also looks correct.Adding them together: 600g + 200g = 800g. Yep, that adds up.So, I think I've got it right. Lisa used a total of 800 grams of packing peanuts for her orders.

answer：Okay, so I have this problem where I need to find all non-empty subsets A, B, and C of the positive integers N* such that:1. A, B, and C are pairwise disjoint. That means A ∩ B = B ∩ C = C ∩ A = ∅.2. The union of A, B, and C is all of N*. So every positive integer is in exactly one of A, B, or C.3. For any a in A, b in B, and c in C, the sums a + c must be in A, b + c must be in B, and a + b must be in C.Hmm, this seems a bit abstract, but let me try to break it down.First, since A, B, and C are pairwise disjoint and their union is N*, every positive integer is in exactly one of these sets. So, they form a partition of N*.Now, the third condition is interesting. It tells me that adding an element from A and an element from C gives me another element in A. Similarly, adding an element from B and C gives me another element in B, and adding an element from A and B gives me an element in C.This seems to suggest some kind of structure where the sets are closed under addition with elements from C, or something like that.Let me think about the smallest elements in these sets. Maybe starting with the smallest positive integer, which is 1.Suppose 1 is in A. Then, for any c in C, 1 + c must be in A. Similarly, if 1 is in B, then 1 + c must be in B. But since 1 is in either A or B or C, and C is non-empty, let's see.Wait, if 1 is in C, then adding 1 to any a in A would give a + 1 in A, which might cause problems because 1 is in C, so a + 1 would have to be in A, but 1 is in C, so a + 1 is in A. Hmm, not sure if that's a problem yet.Alternatively, maybe 1 is in A. Then, 1 + c is in A for any c in C. So, if c is in C, then 1 + c is in A. Similarly, if c is in C, then 1 + c is in A, which suggests that A contains numbers that are 1 more than elements of C.But since A, B, and C are disjoint, this might create some kind of periodic structure.Wait, maybe I should consider the minimal element in C. Let's say the smallest element in C is t. Then, since A and B must cover all numbers less than t, because A ∪ B ∪ C = N*, and C starts at t.So, the numbers 1, 2, ..., t-1 are all in A or B.Now, for any a in A, a + t must be in A. Similarly, for any b in B, b + t must be in B. So, A and B are periodic with period t.But since A and B are disjoint, their periods can't overlap. So, A and B must partition the numbers modulo t.Wait, that might be the key. If A and B partition the residues modulo t, then adding t to any element in A or B keeps them in their respective sets.But what is t? It's the minimal element in C. So, t must be such that adding t to any element in A or B keeps them in A or B, respectively.Also, from condition 3, a + b must be in C. So, if a is in A and b is in B, their sum is in C. Since C is the set containing t and above, but actually, C can have smaller elements as well, but t is the minimal.Wait, no. If t is the minimal element in C, then all elements in C are at least t. But a and b could be less than t, so their sum could be less than 2t. But since C contains t, which is the smallest, the sum a + b must be in C, which is at least t.But a and b are at least 1, so a + b is at least 2. If t is 1, then C contains 1, but then a + b would have to be in C, which is 1, but a + b is at least 2, so t can't be 1.If t is 2, then C contains 2 and above. Then, a and b are in A or B, which are 1. So, a + b would be 2, which is in C. That works for the sum. But let's see if the other conditions hold.If t = 2, then A and B must be periodic with period 2. So, A could be all odd numbers, and B could be all even numbers, but wait, C is supposed to contain 2 and above, but if B is all even numbers, then B would include 2, which is in C, so that's a problem because B and C are disjoint.So, t can't be 2 because if t = 2, then C contains 2, but B would have to contain numbers like 4, 6, etc., which are even, but 2 is already in C, so B can't contain 2. So, maybe A is all odd numbers, and B is empty? But B has to be non-empty.Wait, no, B has to be non-empty. So, maybe t = 3.If t = 3, then C contains 3 and above. Then, A and B must cover 1 and 2. Let's say A = {1}, B = {2}. Then, let's check the conditions.For any a in A (which is 1), and c in C (which is 3,4,5,...), a + c = 4,5,6,... which should be in A. But A only contains 1, so 4 isn't in A. That's a problem.Wait, so maybe A needs to be more than just {1}. Maybe A is all numbers congruent to 1 mod 3. Similarly, B is all numbers congruent to 2 mod 3, and C is all multiples of 3.Let me check that.If A = {1,4,7,10,...}, B = {2,5,8,11,...}, and C = {3,6,9,12,...}.Now, let's check condition 3.Take a in A, say 1, and c in C, say 3. a + c = 4, which is in A. Good.Take b in B, say 2, and c in C, say 3. b + c = 5, which is in B. Good.Take a in A, say 1, and b in B, say 2. a + b = 3, which is in C. Good.Similarly, take a = 4 and c = 6. 4 + 6 = 10, which is in A. Good.Take b = 5 and c = 6. 5 + 6 = 11, which is in B. Good.Take a = 4 and b = 5. 4 + 5 = 9, which is in C. Good.So, this seems to satisfy all conditions.But is this the only possibility? Let me see.Suppose instead that A is {2,5,8,...}, B is {1,4,7,...}, and C is {3,6,9,...}. Would that work?Let's check.Take a = 2, c = 3. 2 + 3 = 5, which is in A. Good.Take b = 1, c = 3. 1 + 3 = 4, which is in B. Good.Take a = 2, b = 1. 2 + 1 = 3, which is in C. Good.Similarly, a = 5, c = 6. 5 + 6 = 11, which is in A. Good.b = 4, c = 6. 4 + 6 = 10, which is in B. Good.a = 5, b = 4. 5 + 4 = 9, which is in C. Good.So, this also works. So, it seems that swapping A and B also works.Are there any other possibilities? Let's think.Suppose t is larger than 3, say t = 4.Then, C contains 4 and above. A and B must cover 1,2,3.Let's say A = {1,2}, B = {3}. Then, let's check.Take a = 1, c = 4. 1 + 4 = 5, which should be in A. But 5 is not in A, which only has 1,2. So, that's a problem.Alternatively, maybe A = {1,3}, B = {2}. Then, a = 1, c = 4. 1 + 4 = 5, which should be in A. But 5 isn't in A. So, that's also a problem.Alternatively, maybe A = {1}, B = {2,3}. Then, a = 1, c = 4. 1 + 4 = 5, which should be in A. But 5 isn't in A. So, again, a problem.It seems that if t = 4, it's hard to satisfy the condition that a + c is in A for all a in A and c in C, because A would have to include numbers like 1 + 4 = 5, which would require A to be infinite, but then A would have to include all numbers of the form 1 + 4k, which might conflict with B.Wait, maybe A is {1,5,9,...}, B is {2,6,10,...}, and C is {3,4,7,8,11,12,...}. But then, let's check.Take a = 1, c = 3. 1 + 3 = 4, which is in C. But it should be in A. So, that's a problem.Alternatively, maybe A is {1,5,9,...}, B is {2,6,10,...}, and C is {3,4,7,8,11,12,...}.But then, a + c = 1 + 3 = 4, which is in C, but it should be in A. So, that doesn't work.Alternatively, maybe A is {1,2,5,6,9,10,...}, B is {3,4,7,8,11,12,...}, and C is { something else }, but this seems too vague.I think the problem arises because if t is larger than 3, the conditions become too restrictive. For example, if t = 4, then A and B have to cover 1,2,3, but adding 4 to any element in A or B would require those sets to include numbers beyond their initial elements, which might not be possible without overlapping with C.So, maybe t can't be larger than 3. We saw that t = 3 works, and t = 2 doesn't because of the overlap issue, and t = 1 is impossible because a + b would have to be 1, which is impossible since a and b are at least 1.Therefore, the only possible value for t is 3, leading to the partition where A, B, and C are the sets of numbers congruent to 1, 2, and 0 modulo 3, respectively.But wait, in my earlier example, I had A as {1,4,7,...}, which is 1 mod 3, B as {2,5,8,...}, which is 2 mod 3, and C as {3,6,9,...}, which is 0 mod 3. That works.Alternatively, if I swap A and B, making A = {2,5,8,...} and B = {1,4,7,...}, that also works because the conditions are symmetric with respect to A and B.So, are these the only possible solutions? It seems so because any other partition would either fail the addition conditions or not cover all positive integers without overlap.Therefore, the only solutions are the partitions where A, B, and C are the residue classes modulo 3, either A = 1 mod 3, B = 2 mod 3, C = 0 mod 3, or A = 2 mod 3, B = 1 mod 3, C = 0 mod 3.I think that's it. So, the answer should be these two possible partitions.