answer：First, I need to simplify the expression for ( z ). The given complex number is ( z = |(sqrt{3} - i)i| + i^{2018} ).I'll start by calculating ( |(sqrt{3} - i)i| ). Multiplying ( (sqrt{3} - i) ) by ( i ) gives:[(sqrt{3} - i)i = sqrt{3}i - i^2 = sqrt{3}i + 1]Now, I'll find the modulus of ( sqrt{3}i + 1 ):[| sqrt{3}i + 1 | = sqrt{1^2 + (sqrt{3})^2} = sqrt{1 + 3} = sqrt{4} = 2]Next, I'll determine ( i^{2018} ). Since ( i ) cycles every 4 powers:[2018 div 4 = 504 text{ with a remainder of } 2]Thus:[i^{2018} = i^2 = -1]Now, I'll substitute these values back into the expression for ( z ):[z = |(sqrt{3} - i)i| + i^{2018} = 2 + (-1) = 1]Finally, I'll find the modulus of ( z ):[|z| = |1| = 1]

answer：Okay, so I have this problem here about functions and their inverses. Let me try to understand it step by step. The problem says:"Given that the function ( y = f(x) ) has an inverse function ( y = f^{-1}(x) ), if the graph of the function ( y = f(x) - 1 ) passes through the point ( (1, 2) ), then the graph of the function ( y = f^{-1}(x) + 1 ) passes through the point ______."Alright, so first, let me break this down. I know that if a function has an inverse, then the graph of the function and its inverse are symmetric with respect to the line ( y = x ). That means if the point ( (a, b) ) is on the graph of ( f(x) ), then the point ( (b, a) ) is on the graph of ( f^{-1}(x) ).Now, the problem mentions the function ( y = f(x) - 1 ). It says that this function passes through the point ( (1, 2) ). So, when ( x = 1 ), ( y = 2 ). Let me write that down:( 2 = f(1) - 1 )Hmm, so if I solve for ( f(1) ), I can find the value of the original function ( f(x) ) at ( x = 1 ). Let's do that:( 2 = f(1) - 1 )Adding 1 to both sides:( 2 + 1 = f(1) )( f(1) = 3 )Okay, so ( f(1) = 3 ). That means the point ( (1, 3) ) is on the graph of ( f(x) ).Since ( f(x) ) has an inverse function ( f^{-1}(x) ), the graph of ( f^{-1}(x) ) must pass through the point ( (3, 1) ). That's because, as I remembered earlier, the inverse function reflects the original function over the line ( y = x ), so the coordinates are swapped.Now, the problem asks about the graph of ( y = f^{-1}(x) + 1 ). So, this is the inverse function shifted up by 1 unit. I need to find a point that lies on this new graph.Since ( f^{-1}(x) ) passes through ( (3, 1) ), let's see what happens when we add 1 to the function. If I plug in ( x = 3 ) into ( f^{-1}(x) + 1 ), I get:( y = f^{-1}(3) + 1 )But we know that ( f^{-1}(3) = 1 ), because ( f(1) = 3 ). So substituting that in:( y = 1 + 1 = 2 )Therefore, when ( x = 3 ), ( y = 2 ). So the point ( (3, 2) ) is on the graph of ( y = f^{-1}(x) + 1 ).Let me just recap to make sure I didn't make a mistake. The function ( y = f(x) - 1 ) passes through ( (1, 2) ), which means ( f(1) = 3 ). Therefore, ( f^{-1}(3) = 1 ). Then, shifting the inverse function up by 1 gives ( y = f^{-1}(x) + 1 ), so plugging in ( x = 3 ) gives ( y = 2 ). That seems right.I think I've got it. The point ( (3, 2) ) is on the graph of ( y = f^{-1}(x) + 1 ).**Final Answer**The graph of the function ( y = f^{-1}(x) + 1 ) passes through the point boxed{(3, 2)}.

answer：Okay, so I have this problem where I'm given three positive numbers a, b, and c, all greater than zero. I need to figure out something about the values of three expressions: a + 1/b, b + 1/c, and c + 1/a. The options are about whether all of them are greater than 2, all less than 2, at least one is not greater than 2, or at least one is not less than 2.Hmm, let me think. I remember that for any positive number x, the expression x + 1/x is always greater than or equal to 2. This is because of the AM-GM inequality, which says that the arithmetic mean is greater than or equal to the geometric mean. So, for x and 1/x, their arithmetic mean is (x + 1/x)/2, and their geometric mean is sqrt(x * 1/x) = 1. Therefore, (x + 1/x)/2 >= 1, which means x + 1/x >= 2.But in this problem, the expressions are a bit different. Instead of x + 1/x, we have a + 1/b, b + 1/c, and c + 1/a. So, each term is a different variable plus the reciprocal of another variable. That complicates things a bit because they're not symmetric like x + 1/x.Let me consider the options one by one.Option A: All greater than 2. So, is it possible that a + 1/b > 2, b + 1/c > 2, and c + 1/a > 2 all at the same time? Well, if I set a = b = c = 1, then each expression becomes 1 + 1 = 2, which is exactly 2. So, in this case, they're not greater than 2. But maybe if I choose different values?Suppose I set a = 2, b = 1/2, c = 2. Then, a + 1/b = 2 + 2 = 4, which is greater than 2. b + 1/c = 1/2 + 1/2 = 1, which is less than 2. c + 1/a = 2 + 1/2 = 2.5, which is greater than 2. So, in this case, two expressions are greater than 2, and one is less than 2. So, not all are greater than 2.Wait, but maybe if I choose a, b, c such that each is greater than 1? Let's say a = b = c = 2. Then, a + 1/b = 2 + 1/2 = 2.5 > 2, same for the others. So, in this case, all three expressions are greater than 2. Hmm, so sometimes they can all be greater than 2. But does that mean option A is always true? No, because in the first example, when a = b = c = 1, they are equal to 2, and in another case, one was less than 2. So, A is not necessarily always true.Option B: All less than 2. Is it possible for all three expressions to be less than 2? Let's test with a = b = c = 1. Then, each expression is 2, so not less than 2. If I set a = b = c = 1/2, then a + 1/b = 1/2 + 2 = 2.5 > 2, which is greater than 2. So, that doesn't work. Maybe another set? If I set a = 1/2, b = 2, c = 1/2. Then, a + 1/b = 1/2 + 1/2 = 1 < 2, b + 1/c = 2 + 2 = 4 > 2, c + 1/a = 1/2 + 2 = 2.5 > 2. So, only one is less than 2. It seems challenging to get all three expressions less than 2. Maybe it's impossible?Wait, let's try to see if it's possible. Suppose a + 1/b < 2, b + 1/c < 2, and c + 1/a < 2. Then, adding all three inequalities: a + 1/b + b + 1/c + c + 1/a < 6. But from the AM-GM inequality, we know that a + 1/a >= 2, b + 1/b >= 2, c + 1/c >= 2. So, adding these, a + 1/a + b + 1/b + c + 1/c >= 6. But in our case, we have a + 1/b + b + 1/c + c + 1/a < 6. This is a contradiction because the left side is equal to (a + 1/a) + (b + 1/b) + (c + 1/c) - (1/a + 1/b + 1/c) + (1/b + 1/c + 1/a). Wait, no, actually, it's just a rearrangement. Let me think again.Wait, no, actually, the sum a + 1/b + b + 1/c + c + 1/a is equal to (a + 1/a) + (b + 1/b) + (c + 1/c) - (1/a + 1/b + 1/c) + (1/b + 1/c + 1/a). Hmm, that seems messy. Maybe I should think differently.Alternatively, if I consider that a + 1/b < 2, then 1/b < 2 - a. Similarly, 1/c < 2 - b, and 1/a < 2 - c. Then, multiplying these inequalities: (1/b)(1/c)(1/a) < (2 - a)(2 - b)(2 - c). So, 1/(abc) < (2 - a)(2 - b)(2 - c).But I don't know if this helps. Maybe I can assume that a, b, c are all less than 2. Then, 2 - a, 2 - b, 2 - c are positive. But I'm not sure.Alternatively, maybe I can use the AM-GM inequality on the sum a + 1/b + b + 1/c + c + 1/a. Let's see:a + 1/b + b + 1/c + c + 1/a = (a + 1/a) + (b + 1/b) + (c + 1/c) >= 6.But from our assumption, a + 1/b + b + 1/c + c + 1/a < 6. So, this is a contradiction. Therefore, it's impossible for all three expressions to be less than 2. So, option B is false.Option C: At least one is not greater than 2. So, this means that at least one of the expressions is <= 2. Is this necessarily true?Wait, earlier when I set a = b = c = 2, all expressions were greater than 2. So, in that case, none of them were <= 2. Therefore, option C is not necessarily true.Option D: At least one is not less than 2. So, this means that at least one of the expressions is >= 2. Is this necessarily true?From the earlier analysis, when I tried to assume that all three expressions are less than 2, it led to a contradiction. Therefore, it's impossible for all three expressions to be less than 2. Hence, at least one of them must be >= 2. So, option D is true.Wait, let me double-check. If I assume that all three expressions are less than 2, then their sum is less than 6. But the sum is also equal to (a + 1/a) + (b + 1/b) + (c + 1/c), which is >= 6. Therefore, the assumption that all three expressions are less than 2 leads to a contradiction. Hence, at least one expression must be >= 2. So, D is correct.Therefore, the answer is D.

answer：Okay, so I'm trying to solve this problem about Company X's sales. They have a list of unit sales for one day: 50, 50, 97, 97, 97, 120, 125, 155, 199, 199, 239. The question says there was one unit sale that was greater than the mean sale price but less than a certain sale price. I need to find that certain sale price.First, I think I need to find the mean sale price. To do that, I should add up all the sales and then divide by the number of sales. Let me list out the sales again to make sure I have them all: 50, 50, 97, 97, 97, 120, 125, 155, 199, 199, 239. That looks like 11 sales in total.Now, let's add them up. I'll do this step by step to avoid mistakes:50 + 50 = 100100 + 97 = 197197 + 97 = 294294 + 97 = 391391 + 120 = 511511 + 125 = 636636 + 155 = 791791 + 199 = 990990 + 199 = 11891189 + 239 = 1428Wait, that doesn't seem right. Earlier, I thought the total was 1318, but now I'm getting 1428. Maybe I made a mistake in adding. Let me try again.Starting over:50 + 50 = 100100 + 97 = 197197 + 97 = 294294 + 97 = 391391 + 120 = 511511 + 125 = 636636 + 155 = 791791 + 199 = 990990 + 199 = 11891189 + 239 = 1428Hmm, still 1428. Maybe the initial total was wrong. Let me check the original problem again. It says the total is 1318, but when I add them up, I get 1428. There must be a mistake in the initial problem statement or in my addition.Wait, let's count the number of sales again. There are 11 sales: two 50s, three 97s, one 120, one 125, one 155, two 199s, and one 239. That's 2+3+1+1+1+2+1 = 11 sales.Let me add them again carefully:50 + 50 = 100100 + 97 = 197197 + 97 = 294294 + 97 = 391391 + 120 = 511511 + 125 = 636636 + 155 = 791791 + 199 = 990990 + 199 = 11891189 + 239 = 1428Okay, I'm consistently getting 1428 as the total. Maybe the initial problem had a typo, or I misread the numbers. For now, I'll proceed with 1428 as the total.Now, to find the mean sale price, I divide the total by the number of sales, which is 11.Mean = 1428 / 11Let me calculate that. 11 times 129 is 1419, because 11*120=1320 and 11*9=99, so 1320+99=1419. Then, 1428 - 1419 = 9. So, 1428 / 11 = 129 + 9/11 ≈ 129.818.So, the mean sale price is approximately 129.82.Now, the problem says there was one unit sale that was greater than the mean sale price but less than a certain sale price. I need to find that certain sale price.Looking at the list of sales: 50, 50, 97, 97, 97, 120, 125, 155, 199, 199, 239.First, let's identify which sales are greater than the mean of approximately 129.82.Looking at the list:50, 50, 97, 97, 97, 120, 125, 155, 199, 199, 239.So, 120 is less than 129.82, 125 is also less than 129.82, 155 is greater, 199 is greater, 199 is greater, and 239 is greater.Wait, so 120 and 125 are below the mean, and 155, 199, 199, 239 are above.But the problem says there was one unit sale that was greater than the mean but less than a certain sale price. So, I think it's referring to the smallest sale that is greater than the mean, which would be 155.But the problem says "greater than the mean sale price but less than a certain sale price." So, the certain sale price would be the next higher sale after 155, which is 199.Wait, but there are two 199s. So, the certain sale price would be 199.But let me double-check. The mean is approximately 129.82. The sales above the mean are 155, 199, 199, 239.So, the smallest sale above the mean is 155, and the next one is 199. So, the certain sale price would be 199.But the problem says "greater than the mean sale price but less than a certain sale price." So, the sale that is greater than the mean but less than 199 would be 155.Wait, but there's only one sale between 129.82 and 199, which is 155.So, the certain sale price is 199.But let me make sure. If the certain sale price is 199, then the sale that is greater than the mean but less than 199 is 155.Yes, that makes sense.So, the certain sale price is 199.

answer：Alright, so I have this trigonometry problem here: Given that sin(α - 3π/2) = 3/5, I need to find the value of cos2α. Hmm, okay, let's break this down step by step.First off, I remember that trigonometric functions have these angle addition and subtraction formulas. Maybe I can use one of those to simplify sin(α - 3π/2). Let me recall the sine subtraction formula: sin(A - B) = sinA cosB - cosA sinB. So, applying that here, sin(α - 3π/2) would be sinα cos(3π/2) - cosα sin(3π/2).Now, I need to remember the values of cos(3π/2) and sin(3π/2). I think cos(3π/2) is 0 because at 3π/2 radians, which is 270 degrees, the cosine value is 0. Similarly, sin(3π/2) is -1 because sine at 270 degrees is -1. So plugging those in, we get sinα * 0 - cosα * (-1). Simplifying that, sinα * 0 is 0, and -cosα * (-1) is just cosα. So, sin(α - 3π/2) simplifies to cosα.Wait, so sin(α - 3π/2) = cosα. But the problem says sin(α - 3π/2) = 3/5. That means cosα = 3/5. Okay, so now I know that cosα is 3/5.Now, the question asks for cos2α. I remember there are double angle formulas for cosine. Let me recall them: cos2α can be expressed in a few ways, like cos²α - sin²α, or 2cos²α - 1, or 1 - 2sin²α. Since I already know cosα, maybe I can use the formula that only involves cosα. That would be 2cos²α - 1.So, plugging in cosα = 3/5, we get 2*(3/5)² - 1. Let's calculate that. First, square 3/5: (3/5)² = 9/25. Then multiply by 2: 2*(9/25) = 18/25. Now subtract 1: 18/25 - 1. Hmm, 1 is the same as 25/25, so 18/25 - 25/25 = -7/25.So, cos2α should be -7/25. Let me just double-check my steps to make sure I didn't make any mistakes. Starting from sin(α - 3π/2) = 3/5, I used the sine subtraction formula and correctly identified cos(3π/2) as 0 and sin(3π/2) as -1. That led me to cosα = 3/5. Then, using the double angle formula for cosine, I correctly substituted and calculated to get -7/25. Everything seems to check out.I guess another way to think about it is using the unit circle. If sin(α - 3π/2) = 3/5, then α - 3π/2 is an angle whose sine is 3/5. But since sine is periodic and has certain symmetries, maybe there's another way to interpret this angle. However, the method I used seems straightforward and reliable, so I think I'm confident with the answer.Just to be thorough, let me consider if there are other possible values for α that could satisfy sin(α - 3π/2) = 3/5. Sine is positive in the first and second quadrants, so α - 3π/2 could be in either of those. But regardless of the specific angle, since we're dealing with cosα, which is 3/5, and then using the double angle formula, the result should be consistent. So, I don't think there are multiple possible answers here; it should just be -7/25.Alright, I think I've covered all the bases here. I've used the sine subtraction formula correctly, identified the necessary trigonometric values, applied the double angle formula accurately, and double-checked my work. I feel pretty confident that cos2α is indeed -7/25.**Final Answer**The value of ( cos 2α ) is boxed{-dfrac{7}{25}}.

answer：First, calculate the total age of all 20 persons by multiplying the average age by the number of persons: 20 × 15 = 300 years.Next, determine the total age of the first group of 5 persons: 5 × 14 = 70 years.Then, calculate the total age of the second group of 9 persons: 9 × 16 = 144 years.Add the total ages of the two groups: 70 + 144 = 214 years.Finally, subtract this sum from the total age of all 20 persons to find the age of the 15th person: 300 - 214 = 86 years.