🧠:<think>Alright, so I have this problem about reconstructing an ancient Avar geometric artifact. It's a three-dimensional polyhedron with a symmetry group related to the dihedral group D_n, where n is significant in Avar culture, specifically related to constellations. They mention n=12, so I think that refers to the 12 constellations of the zodiac or something similar.First, I need to figure out the number of possible rotations and reflections in the dihedral group D_12. I remember that dihedral groups are the symmetry groups of regular polygons, consisting of rotations and reflections. The order of the dihedral group D_n is 2n, which means there are n rotations and n reflections. So for n=12, the order should be 24. That means there are 12 rotations and 12 reflections. Wait, let me make sure. The dihedral group D_n has n rotational symmetries (including the identity rotation) and n reflection symmetries. So yes, for n=12, that would be 12 rotations and 12 reflections, totaling 24 elements. So the symmetry group has 24 elements in total, reflecting both the rotational and reflectional symmetries of the artifact.Now, moving on to the second part. The surface of the artifact is decorated with a tessellation pattern made up of regular polygons, and each vertex corresponds to the vertices of a truncated icosahedron. I need to calculate the number of faces, vertices, and edges of this truncated icosahedron and verify Euler's characteristic.I recall that a truncated icosahedron is one of the Archimedean solids. It's the shape of a soccer ball, right? It has pentagons and hexagons. Let me try to remember the counts.An icosahedron has 12 faces, 20 vertices, and 30 edges. When you truncate it, each original vertex is replaced by a new face. Since an icosahedron has 12 faces, truncating each vertex (which was originally 20) would add 20 new faces. But wait, no, truncating a polyhedron replaces each vertex with a new face, and each original face becomes a new face as well, but modified.Wait, maybe I should think about the process. Truncating a polyhedron means cutting off each vertex, turning each original vertex into a new face. For an icosahedron, which has 12 faces, 20 vertices, and 30 edges, truncating it would result in:- Each original face (which is a triangle) becomes a hexagon because each corner is cut off.- Each original vertex becomes a new pentagonal face because the icosahedron has five edges meeting at each vertex.So, the number of faces after truncation would be the original 12 faces, each becoming a hexagon, plus 20 new pentagonal faces, one for each original vertex. So total faces = 12 + 20 = 32.Wait, but I think I might be mixing up the counts. Let me check again. An icosahedron has 12 triangular faces, 20 vertices, and 30 edges. When truncated, each original triangular face becomes a hexagon, and each original vertex becomes a pentagon. So yes, 12 hexagons and 20 pentagons, totaling 32 faces.Now, for the number of vertices. Each original vertex is replaced by a new face, and each original edge is replaced by a new edge. The number of vertices in the truncated icosahedron can be calculated by considering that each original edge is connected to two vertices, and truncating each vertex adds new edges.Wait, maybe it's better to use Euler's formula. Euler's formula states that for any convex polyhedron, V - E + F = 2.We know F = 32. Let's find V and E.But I need to find V and E. Alternatively, I can use the fact that each original edge is truncated, and each original vertex is replaced by a new face.Wait, another approach: Each original edge of the icosahedron is adjacent to two faces. When truncated, each original edge is replaced by a new edge connecting a hexagon and a pentagon. The number of edges in the truncated icosahedron can be calculated as follows:Each original edge is split into three edges? Wait, no. When you truncate a polyhedron, each original edge is replaced by a new edge connecting the new faces. Specifically, each original edge is adjacent to two original faces, and after truncation, each original edge will correspond to a new edge between a hexagon and a pentagon.Wait, actually, each original edge will be replaced by a new edge, and each original vertex will introduce new edges around the new pentagonal face.Let me think differently. The truncated icosahedron has two types of faces: hexagons and pentagons. Each hexagon is surrounded by hexagons and pentagons, and each pentagon is surrounded by hexagons.Each original edge of the icosahedron is adjacent to two triangles. After truncation, each original edge becomes a new edge between a hexagon and a pentagon. So the number of edges can be calculated as follows:Each original edge contributes one new edge, and each original vertex contributes edges around the new pentagonal face. Each original vertex had five edges, so truncating it would create a pentagon with five new edges. However, each new edge is shared between two vertices, so the total number of new edges from truncation is (20 vertices * 5 edges)/2 = 50 edges.But wait, the original edges were 30. Each original edge is split into a new edge, so we have 30 new edges from the original edges. Plus the 50 new edges from the truncated vertices. So total edges E = 30 + 50 = 80.Wait, but that might be double-counting. Let me check another way. Each face contributes edges, but each edge is shared by two faces.The truncated icosahedron has 12 hexagonal faces and 20 pentagonal faces. Each hexagon has 6 edges, and each pentagon has 5 edges. So total edges counted per face would be (12*6 + 20*5) = 72 + 100 = 172. But since each edge is shared by two faces, the actual number of edges is 172 / 2 = 86.Wait, that contradicts my previous calculation. Hmm. Let me see.Wait, maybe I made a mistake in the first approach. Let me try to use the formula for truncated polyhedrons.When you truncate a polyhedron, the number of vertices, edges, and faces changes as follows:- The original V vertices become V new faces (each vertex becomes a new face).- The original F faces become F new faces, but each original face is transformed into a new face with more edges.- The original E edges become E new edges, but each original edge is now connected to two new faces.Wait, perhaps it's better to use the formula for truncation. The truncated icosahedron can be represented as tI, where I is the icosahedron.The formula for the number of vertices, edges, and faces after truncation can be given by:- V' = V + 2E- E' = 3E- F' = F + VWait, no, that doesn't seem right. Let me recall the correct formula.Actually, when truncating a polyhedron, the number of new vertices is equal to the original number of edges, because each original edge is split into a new vertex. Wait, no, that's for dual polyhedrons.Wait, perhaps I should look up the properties of a truncated icosahedron. But since I can't do that, I'll try to derive it.An icosahedron has V=20, E=30, F=12.After truncation, each original vertex is replaced by a new face (a pentagon), and each original face is replaced by a new face (a hexagon). So the number of faces becomes F' = F + V = 12 + 20 = 32.Each original edge is adjacent to two original faces. After truncation, each original edge is replaced by a new edge connecting a hexagon and a pentagon. Additionally, each original vertex, which had degree 5, becomes a pentagon with 5 new edges. So the total number of edges E' can be calculated as:Each original edge contributes one edge, and each original vertex contributes 5 edges, but each new edge is shared by two vertices. So E' = E + (V * 5)/2 = 30 + (20 * 5)/2 = 30 + 50 = 80.Wait, but earlier when I counted edges by faces, I got 86. There's a discrepancy here. Let me check again.If F' = 32, with 12 hexagons and 20 pentagons, then the total number of edges is (12*6 + 20*5)/2 = (72 + 100)/2 = 172/2 = 86.But according to the truncation approach, I got E' = 80. So which one is correct?I think I made a mistake in the truncation approach. Let me think again.When you truncate a polyhedron, each original vertex is replaced by a new face, and each original face is replaced by a new face with more edges. The number of edges added per original vertex is equal to the degree of the vertex. For an icosahedron, each vertex has degree 5, so truncating each vertex adds 5 new edges. However, each new edge is shared between two vertices, so the total number of new edges from truncation is (V * degree)/2 = (20 * 5)/2 = 50.Additionally, the original edges are still present but now connect the new faces. Each original edge is now connected to a hexagon and a pentagon, so they remain as edges. So the total number of edges is original edges (30) plus new edges from truncation (50), totaling 80.But this contradicts the face-based calculation which gave 86. Hmm.Wait, perhaps the face-based calculation is incorrect because the truncated icosahedron has a different structure. Let me think about the actual structure.A truncated icosahedron has 12 regular pentagonal faces and 20 regular hexagonal faces. Each pentagon is surrounded by hexagons, and each hexagon is surrounded by pentagons and hexagons. Each pentagon has 5 edges, and each hexagon has 6 edges. So total edges counted per face: 12*5 + 20*6 = 60 + 120 = 180. Since each edge is shared by two faces, total edges E = 180 / 2 = 90.Wait, that's even more. Now I'm confused.Wait, no, that can't be right. Because a truncated icosahedron is a well-known shape, and I think its edge count is 90. Let me check:Wait, actually, the truncated icosahedron has 90 edges. Because each of the 12 pentagons contributes 5 edges, and each of the 20 hexagons contributes 6 edges, totaling 12*5 + 20*6 = 60 + 120 = 180, but each edge is shared by two faces, so 180/2 = 90.So E = 90.Now, using Euler's formula V - E + F = 2.We have F = 32, E = 90, so V = E - F + 2 = 90 - 32 + 2 = 60.Wait, so V = 60.But let's verify this with another method. The truncated icosahedron has V = 60 vertices. Each vertex is where a pentagon and two hexagons meet. So each vertex is of degree 3.The number of edges can also be calculated as (V * degree)/2 = (60 * 3)/2 = 90, which matches.So, to summarize:- Faces (F): 32 (12 pentagons + 20 hexagons)- Edges (E): 90- Vertices (V): 60Now, let's verify Euler's formula: V - E + F = 60 - 90 + 32 = 2, which holds true.So, the truncated icosahedron has 60 vertices, 90 edges, and 32 faces.Wait, but earlier I thought the number of faces was 32, which is correct, but the edge count was conflicting. It seems the correct edge count is 90, not 80 or 86. So I must have made a mistake in the truncation approach earlier.I think the confusion came from how the edges are counted after truncation. The correct way is to consider the face-based calculation, which gives E = 90, and that aligns with Euler's formula when V=60 and F=32.So, to conclude:1. The dihedral group D_12 has 24 elements: 12 rotations and 12 reflections. The order of the group is 24, reflecting both rotational and reflectional symmetries.2. The truncated icosahedron has 32 faces (12 pentagons and 20 hexagons), 60 vertices, and 90 edges. Euler's characteristic is verified as 60 - 90 + 32 = 2.</think>

🧠:<think>Alright, so I'm trying to figure out this problem about Crewe Alexandra F.C.'s performance analysis. It's divided into two parts, and I need to tackle them step by step. Let me start with the first part.Problem 1: They mention a complex function ( f(z) = a + bi ), where ( a ) is the average number of wins per season, and ( b ) is the average goal difference per season. The values of ( a ) and ( b ) are given as functions of ( n ), the number of key players injured each season. Specifically, ( a = 12 + frac{1}{n} ) and ( b = -5 + frac{2}{n} ). We need to find the modulus of ( f(z) ) when ( n = 3 ).Okay, modulus of a complex number ( a + bi ) is calculated as ( sqrt{a^2 + b^2} ). So, first, I need to compute ( a ) and ( b ) when ( n = 3 ).Let me compute ( a ) first:( a = 12 + frac{1}{3} )( 1/3 ) is approximately 0.3333, so adding that to 12 gives:( a = 12.3333 )Now, computing ( b ):( b = -5 + frac{2}{3} )( 2/3 ) is approximately 0.6667, so subtracting that from -5 gives:( b = -5 + 0.6667 = -4.3333 )So, the complex function ( f(z) ) when ( n = 3 ) is:( f(z) = 12.3333 - 4.3333i )Now, to find the modulus:( |f(z)| = sqrt{(12.3333)^2 + (-4.3333)^2} )Calculating each term:( (12.3333)^2 ) is approximately ( 12.3333 * 12.3333 ). Let me compute that:12 * 12 = 14412 * 0.3333 ≈ 40.3333 * 12 ≈ 40.3333 * 0.3333 ≈ 0.1111So, adding all together:144 + 4 + 4 + 0.1111 ≈ 152.1111Wait, that seems a bit rough. Maybe I should compute it more accurately.12.3333 squared:Let me write 12.3333 as 12 + 1/3, so:( (12 + 1/3)^2 = 12^2 + 2*12*(1/3) + (1/3)^2 = 144 + 8 + 1/9 = 152 + 1/9 ≈ 152.1111 )Similarly, ( (-4.3333)^2 ):Again, 4.3333 is 4 + 1/3, so:( (4 + 1/3)^2 = 16 + 2*(4)*(1/3) + (1/3)^2 = 16 + 8/3 + 1/9 ≈ 16 + 2.6667 + 0.1111 ≈ 18.7778 )So, modulus squared is ( 152.1111 + 18.7778 ≈ 170.8889 )Therefore, modulus is the square root of 170.8889.Calculating square root of 170.8889:I know that 13^2 = 169 and 14^2 = 196, so it's between 13 and 14.Compute 13.0^2 = 169.013.1^2 = 171.61Wait, 13.1^2 is 171.61, which is higher than 170.8889.So, the square root is between 13.0 and 13.1.Let me compute 13.0^2 = 169.013.05^2: Let's compute 13 + 0.05( (13 + 0.05)^2 = 13^2 + 2*13*0.05 + 0.05^2 = 169 + 1.3 + 0.0025 = 170.3025 )Hmm, 170.3025 is still less than 170.8889.Next, 13.07^2:Compute 13.07^2:= (13 + 0.07)^2 = 13^2 + 2*13*0.07 + 0.07^2= 169 + 1.82 + 0.0049= 170.8249That's very close to 170.8889.Difference: 170.8889 - 170.8249 = 0.064So, each additional 0.01 in the square adds approximately 2*13.07*0.01 + (0.01)^2 ≈ 0.2614 + 0.0001 ≈ 0.2615 per 0.01 increase.Wait, actually, the derivative of x^2 is 2x, so the approximate change in x needed is delta_x ≈ delta_y / (2x)Here, delta_y = 0.064, x = 13.07So, delta_x ≈ 0.064 / (2*13.07) ≈ 0.064 / 26.14 ≈ 0.00245So, adding 0.00245 to 13.07 gives approximately 13.07245So, sqrt(170.8889) ≈ 13.07245Therefore, modulus is approximately 13.0725But let me verify:13.07245^2:Compute 13 + 0.07245= (13)^2 + 2*13*0.07245 + (0.07245)^2= 169 + 1.8837 + 0.00525≈ 169 + 1.8837 = 170.8837 + 0.00525 ≈ 170.88895Which is very close to 170.8889, so that's accurate.Therefore, modulus is approximately 13.0725But since the original values were given to four decimal places, maybe we can keep it to four decimal places as well.So, modulus ≈ 13.0725Alternatively, since the problem might expect an exact fractional form.Wait, let's see:We had ( a = 12 + 1/3 ) and ( b = -5 + 2/3 )So, ( a = 37/3 ) and ( b = -13/3 )Therefore, modulus squared is ( (37/3)^2 + (-13/3)^2 = (1369/9) + (169/9) = (1369 + 169)/9 = 1538/9 )So, modulus is sqrt(1538/9) = sqrt(1538)/3Compute sqrt(1538):What's sqrt(1538)?Well, 39^2 = 1521, 40^2=1600, so sqrt(1538) is between 39 and 40.Compute 39.2^2 = 39^2 + 2*39*0.2 + 0.2^2 = 1521 + 15.6 + 0.04 = 1536.6439.2^2 = 1536.641538 - 1536.64 = 1.36So, sqrt(1538) ≈ 39.2 + 1.36/(2*39.2) ≈ 39.2 + 1.36/78.4 ≈ 39.2 + 0.0173 ≈ 39.2173Therefore, sqrt(1538)/3 ≈ 39.2173 / 3 ≈ 13.0724Which matches our earlier decimal approximation.So, modulus is approximately 13.0724, which is about 13.0724.So, I can write that as approximately 13.0724, or as an exact fraction sqrt(1538)/3.But since the problem didn't specify, probably decimal is fine.So, modulus is approximately 13.0724.But let me see if I can write it as a fraction.Wait, 1538 divided by 9 is 170.888..., so sqrt(170.888...). Hmm, not sure if that simplifies.Alternatively, maybe we can write it as sqrt(1538)/3, but that might not be necessary.So, for the first part, modulus is approximately 13.0724.Problem 2: Now, using this modulus ( x ), which is approximately 13.0724, we need to compute ( g(x) = e^{c x} ) where ( c = 0.1 ).So, ( g(x) = e^{0.1 * 13.0724} )Compute the exponent first:0.1 * 13.0724 = 1.30724So, ( g(x) = e^{1.30724} )Now, I need to compute ( e^{1.30724} ). Let me recall that ( e^1 = 2.71828 ), ( e^{1.3} ) is approximately 3.6693, and ( e^{1.30724} ) is slightly higher.Let me compute it step by step.First, let's compute ( e^{1.3} ):We know that ( e^{1.3} ≈ 3.6693 )Now, 1.30724 is 1.3 + 0.00724So, ( e^{1.30724} = e^{1.3} * e^{0.00724} )Compute ( e^{0.00724} ):We can approximate this using the Taylor series expansion around 0:( e^x ≈ 1 + x + x^2/2 + x^3/6 )Here, x = 0.00724Compute:1 + 0.00724 + (0.00724)^2 / 2 + (0.00724)^3 / 6First, 0.00724 squared:0.00724 * 0.00724 ≈ 0.0000524Divide by 2: ≈ 0.0000262Third term: (0.00724)^3 ≈ 0.000000379Divide by 6: ≈ 0.000000063So, adding all together:1 + 0.00724 = 1.00724Plus 0.0000262: 1.0072662Plus 0.000000063: ≈ 1.007266263So, ( e^{0.00724} ≈ 1.007266 )Therefore, ( e^{1.30724} ≈ e^{1.3} * 1.007266 ≈ 3.6693 * 1.007266 )Compute 3.6693 * 1.007266:First, 3.6693 * 1 = 3.66933.6693 * 0.007266 ≈ ?Compute 3.6693 * 0.007 = 0.02568513.6693 * 0.000266 ≈ 0.000975So, total ≈ 0.0256851 + 0.000975 ≈ 0.02666Therefore, total ( e^{1.30724} ≈ 3.6693 + 0.02666 ≈ 3.69596 )So, approximately 3.696Alternatively, using a calculator for better precision:But since I don't have a calculator, let me see if I can compute it more accurately.Alternatively, use the fact that ( e^{1.30724} ) can be computed as ( e^{1 + 0.30724} = e * e^{0.30724} )We know ( e ≈ 2.71828 )Compute ( e^{0.30724} ):Again, using Taylor series around 0:( e^{0.30724} ≈ 1 + 0.30724 + (0.30724)^2 / 2 + (0.30724)^3 / 6 + (0.30724)^4 / 24 )Compute each term:1) 12) 0.307243) (0.30724)^2 / 2 ≈ 0.09438 / 2 ≈ 0.047194) (0.30724)^3 / 6 ≈ 0.02905 / 6 ≈ 0.004845) (0.30724)^4 / 24 ≈ 0.00893 / 24 ≈ 0.000372Adding them up:1 + 0.30724 = 1.30724+ 0.04719 = 1.35443+ 0.00484 = 1.35927+ 0.000372 ≈ 1.359642So, ( e^{0.30724} ≈ 1.359642 )Therefore, ( e^{1.30724} = e * e^{0.30724} ≈ 2.71828 * 1.359642 )Compute 2.71828 * 1.359642:First, compute 2 * 1.359642 = 2.719284Then, 0.71828 * 1.359642Compute 0.7 * 1.359642 = 0.95175Compute 0.01828 * 1.359642 ≈ 0.02486So, total ≈ 0.95175 + 0.02486 ≈ 0.97661Therefore, total ( e^{1.30724} ≈ 2.719284 + 0.97661 ≈ 3.69589 )Which is approximately 3.6959, which matches our earlier approximation.So, ( g(x) ≈ 3.6959 )Therefore, the potential increase in player attraction is approximately 3.6959.But let me see if I can express this more accurately or if there's a better way.Alternatively, since I know that ( e^{1.30724} ) is approximately 3.696, I can write that as the answer.So, summarizing:1. The modulus of ( f(z) ) when ( n = 3 ) is approximately 13.0724.2. The exponential function ( g(x) = e^{0.1 * 13.0724} ≈ e^{1.30724} ≈ 3.696 )Therefore, the potential increase in player attraction is approximately 3.696.But let me check if I made any miscalculations.Wait, in the first part, I had ( a = 12 + 1/3 ≈ 12.3333 ) and ( b = -5 + 2/3 ≈ -4.3333 ). Then, modulus squared is ( (12.3333)^2 + (-4.3333)^2 ≈ 152.1111 + 18.7778 ≈ 170.8889 ). So, modulus is sqrt(170.8889) ≈ 13.0724. That seems correct.Then, for the second part, exponent is 0.1 * 13.0724 ≈ 1.30724, and ( e^{1.30724} ≈ 3.696 ). That seems correct.Alternatively, using a calculator, if I compute 1.30724, e^1.30724 is approximately 3.696. So, that seems accurate.Therefore, I think my calculations are correct.Final Answer1. The modulus of ( f(z) ) when ( n = 3 ) is boxed{13.07}.2. The potential increase in player attraction for the next season is boxed{3.696}.</think>

🧠:<think>Okay, so I have this problem where I need to calculate the area of an island that's shaped like an irregular polygon. The vertices are given as A(0, 0), B(5, 0), C(7, 4), D(3, 8), and E(-2, 5). Then, based on that area, I need to figure out how much power the island can generate if 20% of it is covered with solar panels. Each square kilometer can generate 1.5 megawatts, so I have to find the total potential power.First, I remember that there's a formula for calculating the area of a polygon when you know the coordinates of its vertices. I think it's called the shoelace formula. Let me recall how that works. I believe it involves multiplying coordinates in a specific way and then taking half the absolute difference between the sums.So, the formula is something like:Area = (1/2) * |sum over i (x_i * y_{i+1} - x_{i+1} * y_i)|Where the vertices are listed in order, either clockwise or counterclockwise, and the last vertex connects back to the first one.Let me write down the coordinates again to make sure I have them right:A(0, 0)B(5, 0)C(7, 4)D(3, 8)E(-2, 5)And then back to A(0, 0) to complete the polygon.So, I need to set up a table or something where I can compute each term x_i * y_{i+1} and x_{i+1} * y_i for each pair of consecutive vertices.Let me list the vertices in order and pair them:1. A to B: (0,0) to (5,0)2. B to C: (5,0) to (7,4)3. C to D: (7,4) to (3,8)4. D to E: (3,8) to (-2,5)5. E to A: (-2,5) to (0,0)Now, for each pair, I'll compute x_i * y_{i+1} and x_{i+1} * y_i.Starting with A to B:x_i = 0, y_i = 0x_{i+1} = 5, y_{i+1} = 0So, x_i * y_{i+1} = 0 * 0 = 0x_{i+1} * y_i = 5 * 0 = 0Next, B to C:x_i = 5, y_i = 0x_{i+1} = 7, y_{i+1} = 4x_i * y_{i+1} = 5 * 4 = 20x_{i+1} * y_i = 7 * 0 = 0C to D:x_i = 7, y_i = 4x_{i+1} = 3, y_{i+1} = 8x_i * y_{i+1} = 7 * 8 = 56x_{i+1} * y_i = 3 * 4 = 12D to E:x_i = 3, y_i = 8x_{i+1} = -2, y_{i+1} = 5x_i * y_{i+1} = 3 * 5 = 15x_{i+1} * y_i = -2 * 8 = -16E to A:x_i = -2, y_i = 5x_{i+1} = 0, y_{i+1} = 0x_i * y_{i+1} = -2 * 0 = 0x_{i+1} * y_i = 0 * 5 = 0Now, let me sum up all the x_i * y_{i+1} terms:0 (from A-B) + 20 (B-C) + 56 (C-D) + 15 (D-E) + 0 (E-A) = 0 + 20 + 56 + 15 + 0 = 91Next, sum up all the x_{i+1} * y_i terms:0 (A-B) + 0 (B-C) + 12 (C-D) + (-16) (D-E) + 0 (E-A) = 0 + 0 + 12 -16 + 0 = -4Now, subtract the second sum from the first sum:91 - (-4) = 91 + 4 = 95Take the absolute value (which is still 95) and multiply by 1/2:Area = (1/2) * 95 = 47.5 square kilometers.Wait, let me double-check my calculations because sometimes it's easy to make a mistake with the signs or the multiplication.First, the x_i * y_{i+1} terms:A-B: 0*0=0B-C:5*4=20C-D:7*8=56D-E:3*5=15E-A:-2*0=0Total: 0+20=20, 20+56=76, 76+15=91, 91+0=91. That seems right.Now, the x_{i+1} * y_i terms:A-B:5*0=0B-C:7*0=0C-D:3*4=12D-E:-2*8=-16E-A:0*5=0Total: 0+0=0, 0+12=12, 12-16=-4, -4+0=-4. That also seems correct.So, 91 - (-4) = 95, half of that is 47.5. So, the area is 47.5 square kilometers.Okay, that seems solid.Now, moving on to the second part. The president wants to cover 20% of the island's area with solar panels. So, 20% of 47.5 km².Let me calculate 20% of 47.5:20% is 0.2, so 0.2 * 47.5 = ?0.2 * 40 = 80.2 * 7.5 = 1.5So, 8 + 1.5 = 9.5 km².So, 9.5 square kilometers will be covered with solar panels.Each square kilometer can generate 1.5 megawatts of power. So, total power generation is 9.5 * 1.5 MW.Calculating that:9 * 1.5 = 13.50.5 * 1.5 = 0.75So, 13.5 + 0.75 = 14.25 MW.Therefore, the total potential power generation is 14.25 megawatts.Wait, just to make sure I didn't make a mistake in the multiplication:9.5 * 1.5Break it down:9 * 1.5 = 13.50.5 * 1.5 = 0.75Adding them together: 13.5 + 0.75 = 14.25. Yep, that's correct.So, summarizing:1. The area of the island is 47.5 square kilometers.2. 20% of that is 9.5 square kilometers, which can generate 14.25 megawatts of power.I think that's all. I don't see any mistakes in my calculations, so I'm confident with these answers.Final Answer1. The area of the island is boxed{47.5} square kilometers.2. The total potential power generation capacity is boxed{14.25} megawatts.</think>

🧠:<think>Okay, so I have this problem where a diplomatic historian from Denmark is looking at trade relations with Brazil, Argentina, and Chile over the past 10 years. The trade amounts are modeled by these exponential functions plus linear terms. I need to figure out two things: first, when the total trade will exceed 1 billion USD, and second, the rate of change at 15 years. Hmm, let me break this down step by step.First, let me write down the given functions to make sure I have them right.Denmark-Brazil trade: ( T_B(t) = 50e^{0.05t} + 10t )Denmark-Argentina trade: ( T_A(t) = 40e^{0.06t} + 15t )Denmark-Chile trade: ( T_C(t) = 30e^{0.07t} + 20t )So, each of these is a function of time t, where t is the number of years since the study began. The units are in million USD.Problem 1: Determine the year t when the total trade first exceeds 1 billion USD. Since the functions are in million USD, 1 billion USD is 1000 million USD. So, I need to find t such that ( T_B(t) + T_A(t) + T_C(t) > 1000 ).Let me write the total trade function:( T(t) = T_B(t) + T_A(t) + T_C(t) )Substituting the given functions:( T(t) = 50e^{0.05t} + 10t + 40e^{0.06t} + 15t + 30e^{0.07t} + 20t )Let me combine like terms:The exponential terms are 50e^{0.05t}, 40e^{0.06t}, and 30e^{0.07t}.The linear terms are 10t, 15t, and 20t. Adding those together: 10 + 15 + 20 = 45t.So, ( T(t) = 50e^{0.05t} + 40e^{0.06t} + 30e^{0.07t} + 45t )We need to find t such that ( T(t) > 1000 ).Hmm, this is a transcendental equation, meaning it can't be solved algebraically easily. I might need to use numerical methods or trial and error to approximate t.Let me see if I can plug in some values of t and see when T(t) crosses 1000.First, let's note that t is in years, starting from when the study began. So t=0 is the starting point.Let me try t=20:Compute each term:50e^{0.05*20} = 50e^{1} ≈ 50*2.718 ≈ 135.940e^{0.06*20} = 40e^{1.2} ≈ 40*3.32 ≈ 132.830e^{0.07*20} = 30e^{1.4} ≈ 30*4.055 ≈ 121.6545t = 45*20 = 900Adding them up: 135.9 + 132.8 + 121.65 + 900 ≈ 135.9+132.8=268.7; 268.7+121.65=390.35; 390.35+900=1290.35 million USD. So at t=20, T(t)=1290.35, which is above 1000. So, the year is before 20.Wait, but the study is over the past 10 years, so t=0 is 10 years ago, and t=10 is now. So, t=20 is 10 years into the future. Hmm, but the question is about when the total trade first exceeds 1 billion. So, is t=20 the answer? Wait, but let's check t=15.Compute T(15):50e^{0.05*15}=50e^{0.75}≈50*2.117≈105.8540e^{0.06*15}=40e^{0.9}≈40*2.4596≈98.38430e^{0.07*15}=30e^{1.05}≈30*2.858≈85.7445t=45*15=675Adding up: 105.85 + 98.384 = 204.234; 204.234 +85.74=290; 290 +675=965 million USD. So, T(15)=965 <1000.So, between t=15 and t=20, the total trade crosses 1000 million USD.Let me try t=18:50e^{0.05*18}=50e^{0.9}≈50*2.4596≈122.9840e^{0.06*18}=40e^{1.08}≈40*2.944≈117.7630e^{0.07*18}=30e^{1.26}≈30*3.525≈105.7545t=45*18=810Total: 122.98 +117.76=240.74; 240.74 +105.75=346.49; 346.49 +810=1156.49 million. So, T(18)=1156.49 >1000.Wait, that's over 1000. So, between t=15 and t=18, it crosses 1000.Wait, but t=15 is 965, t=18 is 1156. Let me try t=16.T(16):50e^{0.05*16}=50e^{0.8}≈50*2.2255≈111.27540e^{0.06*16}=40e^{0.96}≈40*2.6117≈104.46830e^{0.07*16}=30e^{1.12}≈30*3.065≈91.9545t=45*16=720Total: 111.275 +104.468=215.743; 215.743 +91.95=307.693; 307.693 +720=1027.693 million. So, T(16)=1027.693 >1000.So, between t=15 and t=16, the total trade crosses 1000. Let's check t=15.5.Compute T(15.5):50e^{0.05*15.5}=50e^{0.775}≈50*2.171≈108.5540e^{0.06*15.5}=40e^{0.93}≈40*2.535≈101.430e^{0.07*15.5}=30e^{1.085}≈30*2.958≈88.7445t=45*15.5=697.5Total: 108.55 +101.4=209.95; 209.95 +88.74=298.69; 298.69 +697.5≈996.19 million. So, T(15.5)=996.19 <1000.So, between t=15.5 and t=16, the total trade crosses 1000. Let's try t=15.75.T(15.75):50e^{0.05*15.75}=50e^{0.7875}≈50*2.199≈109.9540e^{0.06*15.75}=40e^{0.945}≈40*2.574≈102.9630e^{0.07*15.75}=30e^{1.1025}≈30*3.012≈90.3645t=45*15.75=708.75Total: 109.95 +102.96=212.91; 212.91 +90.36=303.27; 303.27 +708.75≈1012.02 million. So, T(15.75)=1012.02 >1000.So, between t=15.5 and t=15.75, T(t) crosses 1000. Let's try t=15.6.T(15.6):50e^{0.05*15.6}=50e^{0.78}≈50*2.182≈109.140e^{0.06*15.6}=40e^{0.936}≈40*2.55≈10230e^{0.07*15.6}=30e^{1.092}≈30*2.98≈89.445t=45*15.6=702Total: 109.1 +102=211.1; 211.1 +89.4=300.5; 300.5 +702=1002.5 million. So, T(15.6)=1002.5 >1000.So, between t=15.5 and t=15.6, T(t) crosses 1000. Let's try t=15.55.T(15.55):50e^{0.05*15.55}=50e^{0.7775}≈50*2.176≈108.840e^{0.06*15.55}=40e^{0.933}≈40*2.542≈101.6830e^{0.07*15.55}=30e^{1.0885}≈30*2.97≈89.145t=45*15.55=700.25Total: 108.8 +101.68=210.48; 210.48 +89.1=300. (approx); 300 +700.25=1000.25 million. So, T(15.55)=1000.25 >1000.So, very close to t=15.55. Let's try t=15.54.T(15.54):50e^{0.05*15.54}=50e^{0.777}≈50*2.175≈108.7540e^{0.06*15.54}=40e^{0.9324}≈40*2.541≈101.6430e^{0.07*15.54}=30e^{1.0878}≈30*2.97≈89.145t=45*15.54≈700. (exact: 45*15=675, 45*0.54=24.3, total=699.3)Total: 108.75 +101.64=210.39; 210.39 +89.1=300. (approx); 300 +699.3=999.3 million. So, T(15.54)=999.3 <1000.So, between t=15.54 and t=15.55, T(t) crosses 1000. Let's approximate.At t=15.54, T=999.3At t=15.55, T=1000.25The difference is 1000.25 -999.3=0.95 over 0.01 years.We need to find t where T(t)=1000.So, the fractional part is (1000 -999.3)/0.95 ≈0.7/0.95≈0.7368 of the interval between 15.54 and 15.55.So, t≈15.54 +0.7368*0.01≈15.54 +0.007368≈15.5474.So, approximately t≈15.547 years. Since the question asks for the year to the nearest year, we need to round this to the nearest whole number.15.547 is approximately 15.55, which is closer to 16 than 15. But wait, let me think. Since t is the number of years since the study began, and the study is over the past 10 years, so t=0 is 10 years ago, t=10 is now, t=15 is 5 years into the future, and t=16 is 6 years into the future.But the question is asking for the year when the total trade first exceeds 1 billion. So, the exact time is approximately 15.55 years after the study began, which is about 5.55 years into the future from now (since t=10 is now). But the question is about the year, not the exact time. So, if the study started 10 years ago, and we're looking for t≈15.55, that would be 15.55 years after the study began, which is 5.55 years into the future from the current year (t=10). So, if the current year is, say, 2023 (assuming the study is from 2013 to 2023), then t=15.55 would be around mid-2028.55, which is approximately 2029. But since the question is about the year t, not the calendar year, but the year since the study began, which is t=16.Wait, no, the question says "the year t (to the nearest year)", so t is the number of years since the study began. So, t=15.55 is approximately 16 years since the study began. So, the answer is t=16.But let me double-check. At t=15.55, it's just over 1000, so the first year when it exceeds is t=16.Wait, but actually, the trade is continuous, so the exact crossing point is at t≈15.55, which is between 15 and 16. Since the question asks for the year when it first exceeds, and since t is measured in whole years, we need to see if at t=15, it's below, and at t=16, it's above. So, the first year when it exceeds is t=16.Therefore, the answer to part 1 is t=16.Now, moving on to problem 2: Calculate the rate of change of the total trade amount at t=15 years.The rate of change is the derivative of T(t) with respect to t, evaluated at t=15.So, first, let's find T'(t).Given:( T(t) = 50e^{0.05t} + 40e^{0.06t} + 30e^{0.07t} + 45t )The derivative is:( T'(t) = 50*0.05e^{0.05t} + 40*0.06e^{0.06t} + 30*0.07e^{0.07t} + 45 )Simplify:( T'(t) = 2.5e^{0.05t} + 2.4e^{0.06t} + 2.1e^{0.07t} + 45 )Now, evaluate this at t=15.Compute each term:2.5e^{0.05*15}=2.5e^{0.75}≈2.5*2.117≈5.29252.4e^{0.06*15}=2.4e^{0.9}≈2.4*2.4596≈5.9032.1e^{0.07*15}=2.1e^{1.05}≈2.1*2.858≈6.001845 remains 45.Adding them up:5.2925 +5.903=11.1955; 11.1955 +6.0018≈17.1973; 17.1973 +45≈62.1973 million USD per year.So, the rate of change at t=15 is approximately 62.2 million USD per year.But let me compute more accurately.First, compute each exponential term precisely.Compute e^{0.05*15}=e^{0.75}=2.1170000166So, 2.5*2.1170000166≈5.2925e^{0.06*15}=e^{0.9}=2.4596031112.4*2.459603111≈5.903047466e^{0.07*15}=e^{1.05}=2.8583742472.1*2.858374247≈6.002585919Now, summing up:5.2925 +5.903047466=11.1955474711.19554747 +6.002585919=17.1981333917.19813339 +45=62.19813339So, approximately 62.198 million USD per year. Rounded to, say, two decimal places, 62.20 million USD per year.But since the question doesn't specify the precision, maybe we can leave it as is or round to the nearest whole number. 62.2 is approximately 62 million USD per year.Wait, but let me check the calculations again to ensure accuracy.Compute each term:2.5e^{0.75}=2.5*2.117=5.29252.4e^{0.9}=2.4*2.4596≈5.9032.1e^{1.05}=2.1*2.858≈6.002Sum: 5.2925 +5.903=11.1955; 11.1955 +6.002=17.1975; 17.1975 +45=62.1975≈62.2 million USD/year.Yes, that seems correct.So, the rate of change at t=15 is approximately 62.2 million USD per year.But let me express it more precisely. Since the functions are in million USD, the derivative is in million USD per year.So, the rate of change is approximately 62.2 million USD per year.Alternatively, if we want to express it as a whole number, 62 million USD per year.But since the decimal is .2, which is 20%, it's better to keep one decimal place, so 62.2 million USD/year.Wait, but let me check if I did the derivative correctly.Yes, the derivative of 50e^{0.05t} is 50*0.05e^{0.05t}=2.5e^{0.05t}Similarly for the others.Yes, that's correct.So, the final answers are:1. The year t is 16.2. The rate of change at t=15 is approximately 62.2 million USD per year.But let me make sure about the first part. Since t is the number of years since the study began, and the study is over the past 10 years, so t=0 is 10 years ago, t=10 is now, and t=16 is 6 years into the future. So, the total trade first exceeds 1 billion USD in the 16th year since the study began, which is 6 years from now. But the question is about the year t, not the calendar year, so the answer is t=16.Yes, that makes sense.Final Answer1. The total trade first exceeds 1 billion USD in year boxed{16}.2. The rate of change of the total trade amount at ( t = 15 ) years is boxed{62.2} million USD per year.</think>