🧠:<think>First, I need to calculate the total energy consumption for a single tournament. There are 20 computers, each consuming 500 watts per hour. The tournament lasts for 3 hours, so the total energy used per computer is 500 watts multiplied by 3 hours, which equals 1500 watt-hours. Converting this to kilowatt-hours by dividing by 1000 gives 1.5 kWh per computer. Multiplying by the 20 computers results in a total energy consumption of 30 kWh for the tournament.Next, to find the electricity cost, I'll multiply the total energy consumption by the cost per kilowatt-hour. So, 30 kWh multiplied by 0.12 per kWh equals 3.60. This is the cost of electricity for one tournament.For the second part, I need to calculate the total monthly revenue from 10 tournaments. Each tournament has 40 participants, and 30% of them opt for the monthly membership, while the remaining 70% pay the entry fee per tournament. First, I'll determine the number of monthly members: 30% of 40 participants is 12 members. The revenue from memberships is 12 members multiplied by 100, which equals 1200.Next, I'll calculate the number of participants paying the entry fee: 70% of 40 participants is 28 participants. The revenue from entry fees for one tournament is 28 participants multiplied by 15, totaling 420. For 10 tournaments, this amounts to 4200.Finally, adding the membership revenue and the entry fee revenue gives the total monthly revenue: 1200 plus 4200 equals 5400.</think>

🧠:<think>Alright, so I have this problem about vectors representing themes in Romantic era poetry. It's divided into two parts. Let me start with the first one.1. Linear Independence of Vectors in Set SOkay, so I have four vectors in a 3-dimensional space. The vectors are:- v₁ = (3, 5, 7)- v₂ = (2, 4, 6)- v₃ = (5, 7, 11)- v₄ = (4, 6, 8)I need to check if these vectors are linearly independent. Hmm, in a 3-dimensional space, the maximum number of linearly independent vectors is 3. Since we have four vectors here, they must be linearly dependent. So, they are not linearly independent. But the question also asks to find a non-trivial linear combination that equals the zero vector. To find such a combination, I can set up a system of equations. Let me denote the coefficients as a, b, c, d for vectors v₁, v₂, v₃, v₄ respectively. So, the equation is:a*v₁ + b*v₂ + c*v₃ + d*v₄ = 0Which translates to:3a + 2b + 5c + 4d = 0 ...(1)5a + 4b + 7c + 6d = 0 ...(2)7a + 6b + 11c + 8d = 0 ...(3)And since we have four variables and three equations, there will be infinitely many solutions, but we need a non-trivial one (not all coefficients zero).Let me try to solve this system. Maybe I can express it as a matrix and perform row operations.The augmented matrix is:[3 2 5 4 | 0][5 4 7 6 | 0][7 6 11 8 | 0]Let me write this as:Row 1: 3 2 5 4Row 2: 5 4 7 6Row 3: 7 6 11 8I can try to eliminate variables. Let's start by making the first element of Row 2 and Row 3 zero.First, let's make Row 2 = Row 2 - (5/3)Row 1.Calculating:Row 2: 5 - (5/3)*3 = 0 4 - (5/3)*2 = 4 - 10/3 = 2/3 7 - (5/3)*5 = 7 - 25/3 = (21 -25)/3 = -4/3 6 - (5/3)*4 = 6 - 20/3 = (18 -20)/3 = -2/3So Row 2 becomes: 0 2/3 -4/3 -2/3Similarly, Row 3 = Row 3 - (7/3)Row 1.Calculating:Row 3: 7 - (7/3)*3 = 0 6 - (7/3)*2 = 6 - 14/3 = (18 -14)/3 = 4/3 11 - (7/3)*5 = 11 - 35/3 = (33 -35)/3 = -2/3 8 - (7/3)*4 = 8 - 28/3 = (24 -28)/3 = -4/3So Row 3 becomes: 0 4/3 -2/3 -4/3Now, the matrix looks like:Row 1: 3 2 5 4Row 2: 0 2/3 -4/3 -2/3Row 3: 0 4/3 -2/3 -4/3Next, I can eliminate the second element in Row 3 using Row 2.Let me make Row 3 = Row 3 - 2*Row 2.Calculating:Row 3: 0 - 0 = 0 4/3 - 2*(2/3) = 4/3 - 4/3 = 0 -2/3 - 2*(-4/3) = -2/3 + 8/3 = 6/3 = 2 -4/3 - 2*(-2/3) = -4/3 + 4/3 = 0So Row 3 becomes: 0 0 2 0So now, the matrix is:Row 1: 3 2 5 4Row 2: 0 2/3 -4/3 -2/3Row 3: 0 0 2 0This is an upper triangular matrix. Now, let's write the equations:From Row 3: 2c = 0 => c = 0From Row 2: (2/3)b - (4/3)c - (2/3)d = 0. Since c=0, this becomes (2/3)b - (2/3)d = 0 => 2b - 2d = 0 => b = dFrom Row 1: 3a + 2b + 5c + 4d = 0. Since c=0 and b=d, this becomes 3a + 2d + 4d = 0 => 3a + 6d = 0 => a = -2dSo, we can express a, b, c in terms of d. Let me set d = t (a parameter). Then:a = -2tb = tc = 0d = tSo, the general solution is:a = -2tb = tc = 0d = tWe can choose t = 1 for simplicity. So, the coefficients are:a = -2, b = 1, c = 0, d = 1Therefore, the linear combination is:-2*v₁ + 1*v₂ + 0*v₃ + 1*v₄ = 0Let me verify this:Compute -2*v₁ + v₂ + v₄:-2*(3,5,7) + (2,4,6) + (4,6,8)= (-6, -10, -14) + (2,4,6) + (4,6,8)= (-6 + 2 + 4, -10 + 4 + 6, -14 + 6 + 8)= (0, 0, 0)Yes, that works. So, a non-trivial combination is -2v₁ + v₂ + v₄ = 0.2. Calculating Historical Influence on "Nostalgia"Now, the second part. We have a vector p = (1,1,2) representing "Nostalgia." We need to compute the dot product of p with each vector in S, then sum these dot products weighted by H(t) for t=18 and t=19.First, let's compute the dot products.Dot product formula: p · v = p₁*v₁ + p₂*v₂ + p₃*v₃So, for each vector in S:- p · v₁ = 1*3 + 1*5 + 2*7 = 3 + 5 +14 = 22- p · v₂ = 1*2 + 1*4 + 2*6 = 2 + 4 +12 = 18- p · v₃ = 1*5 + 1*7 + 2*11 = 5 +7 +22 = 34- p · v₄ = 1*4 + 1*6 + 2*8 = 4 +6 +16 = 26So, the dot products are 22, 18, 34, 26.Next, we need to compute the sum of these dot products weighted by H(t) for t=18 and t=19.H(t) = t² - 3t + 2First, compute H(18):H(18) = 18² - 3*18 + 2 = 324 -54 +2 = 272Then, H(19):H(19) = 19² - 3*19 + 2 = 361 -57 +2 = 306Now, the total historical influence is the sum of the dot products multiplied by H(t) for each t. Wait, actually, the problem says:"the sum of these dot products weighted by H(t) for t = 18 and t = 19."So, I think it means for each t, compute the sum of dot products multiplied by H(t). Or, maybe compute the sum of (dot product * H(t)) for each t.Wait, let me read again:"Calculate the dot product of p with each vector in S and then find the total historical influence on 'Nostalgia' by computing the sum of these dot products weighted by H(t) for t = 18 and t = 19."Hmm, so maybe for each t, compute the sum of (dot product * H(t)), but since H(t) is the same for all dot products at a given t, it's equivalent to H(t) multiplied by the sum of dot products.Wait, let me parse the sentence:"sum of these dot products weighted by H(t) for t = 18 and t = 19"So, for each t, compute the sum of (dot product * H(t)), but since H(t) is a scalar, it's H(t) multiplied by the sum of dot products.But actually, the wording is a bit ambiguous. It could mean either:1. For each t, compute the sum over vectors of (dot product * H(t)), which would be H(t) * (sum of dot products).Or,2. For each vector, compute (dot product * H(t)), then sum over vectors and t.But since t is given as 18 and 19, it's likely that for each t, compute the sum of dot products multiplied by H(t). So, two separate totals, one for t=18 and one for t=19.But the problem says "the total historical influence," which might imply a single total. Hmm.Wait, let me read again:"Calculate the dot product of p with each vector in S and then find the total historical influence on 'Nostalgia' by computing the sum of these dot products weighted by H(t) for t = 18 and t = 19."So, it's the sum of (dot products) weighted by H(t) for t=18 and t=19. So, perhaps for each vector, compute (dot product) * H(t) for t=18 and t=19, then sum all these.But that would be sum_{v in S} (p·v) * H(18) + sum_{v in S} (p·v) * H(19) = H(18)*sum(p·v) + H(19)*sum(p·v) = (H(18) + H(19)) * sum(p·v)Alternatively, if it's for each t, compute the sum of (p·v) * H(t), then sum over t.Wait, maybe it's:Total influence = sum_{v in S} (p·v) * H(t) for t=18 and t=19.But t is two different values, so perhaps it's sum_{v in S} (p·v) * H(18) + sum_{v in S} (p·v) * H(19) = (H(18) + H(19)) * sum(p·v)Alternatively, maybe it's sum_{v in S} (p·v) * [H(18) + H(19)]But let me think again.The problem says: "the sum of these dot products weighted by H(t) for t = 18 and t = 19."So, "these dot products" refers to the four dot products computed earlier (22,18,34,26). So, for each of these four dot products, weight them by H(t) for t=18 and t=19, then sum them all.But that would be ambiguous because each dot product is a scalar, and we have two weights (H(18) and H(19)). So, perhaps it's:Total influence = sum_{v in S} (p·v) * H(t) for t=18 and t=19.But that would mean for each vector, compute (p·v) * H(18) + (p·v) * H(19), then sum over vectors.Which is equivalent to sum_{v in S} (p·v) * (H(18) + H(19))Alternatively, it could be interpreted as for each t, compute the sum of (p·v) * H(t), then add those two sums together.Either way, the result would be the same:Total influence = (H(18) + H(19)) * (sum of p·v)Because sum_{v} (p·v) * H(t) for t=18 and t=19 is H(18)*sum(p·v) + H(19)*sum(p·v) = (H(18) + H(19)) * sum(p·v)So, let's compute sum(p·v):Sum = 22 + 18 + 34 + 26 = 22+18=40, 34+26=60, total 100.H(18) = 272, H(19)=306So, total influence = (272 + 306) * 100 = 578 * 100 = 57,800Wait, but that seems very large. Alternatively, maybe it's H(t) applied to each dot product individually, then summed.But if it's for each vector, compute p·v * H(t) for t=18 and t=19, then sum all of them.So, for each vector, compute p·v * H(18) + p·v * H(19), then sum over vectors.Which is the same as sum(p·v) * (H(18) + H(19)) = 100 * 578 = 57,800Alternatively, if it's for each t, compute sum(p·v) * H(t), then sum over t.Which would be sum(p·v) * H(18) + sum(p·v) * H(19) = 100*272 + 100*306 = 27,200 + 30,600 = 57,800So, either way, the total influence is 57,800.But let me double-check if that's the correct interpretation.The problem says: "the sum of these dot products weighted by H(t) for t = 18 and t = 19."So, "weighted by H(t)" for t=18 and t=19. So, for each dot product, multiply by H(18) and H(19), then sum all.But that would mean for each vector, compute p·v * H(18) + p·v * H(19), then sum over vectors.Which is the same as sum(p·v) * (H(18) + H(19)) = 100 * 578 = 57,800Yes, that seems correct.So, the total historical influence is 57,800.But let me make sure I didn't misinterpret. If it's for each t, compute the sum of (p·v) * H(t), then sum over t, that's the same as 100*(272 + 306) = 57,800.Alternatively, if it's for each vector, compute p·v * H(18) + p·v * H(19), then sum over vectors, same result.So, I think that's the answer.Final Answer1. The vectors are linearly dependent, and a non-trivial combination is boxed{-2mathbf{v}_1 + mathbf{v}_2 + mathbf{v}_4 = mathbf{0}}.2. The total historical influence on "Nostalgia" is boxed{57800}.</think>

🧠:<think>Okay, so I need to figure out how many games the Houston Baptist Huskies won and lost in 1969. They played a total of 50 games, and the ratio of games won to games lost is the golden ratio, φ, which is (1 + √5)/2. Hmm, that sounds a bit tricky, but let's break it down step by step.First, let me recall what the golden ratio is. φ is approximately 1.618, but since we need an exact number, I should keep it as (1 + √5)/2. The ratio of games won to games lost is φ, so if I let the number of games won be W and the number of games lost be L, then W/L = φ.Also, we know that the total number of games played is 50, so W + L = 50. That gives me two equations:1. W/L = φ2. W + L = 50I can use these two equations to solve for W and L. Let me express W from the first equation. If W/L = φ, then W = φ * L.Now, substitute W in the second equation:φ * L + L = 50Factor out L:L * (φ + 1) = 50So, L = 50 / (φ + 1)But φ is (1 + √5)/2, so let's compute φ + 1:φ + 1 = (1 + √5)/2 + 1 = (1 + √5 + 2)/2 = (3 + √5)/2Therefore, L = 50 / [(3 + √5)/2] = 50 * [2 / (3 + √5)] = 100 / (3 + √5)Hmm, I have a denominator with a radical. I should rationalize it. Multiply numerator and denominator by (3 - √5):L = [100 * (3 - √5)] / [(3 + √5)(3 - √5)] = [100 * (3 - √5)] / (9 - 5) = [100 * (3 - √5)] / 4Simplify that:L = 25 * (3 - √5) ≈ 25 * (3 - 2.236) ≈ 25 * 0.764 ≈ 19.1Wait, that can't be right because the number of games lost should be an integer. Hmm, maybe I made a mistake in my calculations.Let me double-check. So, starting from L = 100 / (3 + √5). Rationalizing:Multiply numerator and denominator by (3 - √5):L = [100*(3 - √5)] / [(3)^2 - (√5)^2] = [100*(3 - √5)] / (9 - 5) = [100*(3 - √5)] / 4 = 25*(3 - √5)Calculating 25*(3 - √5):3 - √5 ≈ 3 - 2.236 ≈ 0.76425 * 0.764 ≈ 19.1So, L ≈ 19.1, but since the number of games lost must be an integer, maybe it's 19 or 20. Hmm, but we need an exact number, not an approximate. So perhaps the exact value is 25*(3 - √5), which is approximately 19.1. But since we can't have a fraction of a game, maybe the problem expects an exact fractional value? Or perhaps I need to represent it differently.Wait, maybe I should express W and L in terms of φ. Let me think.We have W = φ * L, and W + L = 50.So, substituting W:φ * L + L = 50 => L*(φ + 1) = 50But φ + 1 is equal to φ squared, because φ satisfies the equation φ^2 = φ + 1. So, φ + 1 = φ^2.Therefore, L = 50 / φ^2Similarly, W = φ * L = φ * (50 / φ^2) = 50 / φSo, W = 50 / φ and L = 50 / φ^2Since φ = (1 + √5)/2, let's compute 1/φ:1/φ = 2 / (1 + √5). Rationalizing the denominator:1/φ = [2*(1 - √5)] / [(1 + √5)(1 - √5)] = [2*(1 - √5)] / (1 - 5) = [2*(1 - √5)] / (-4) = [ -2*(1 - √5) ] / 4 = [2*(√5 - 1)] / 4 = (√5 - 1)/2So, 1/φ = (√5 - 1)/2 ≈ (2.236 - 1)/2 ≈ 1.236/2 ≈ 0.618Similarly, 1/φ^2 = (1/φ)^2 = [(√5 - 1)/2]^2 = (5 - 2√5 + 1)/4 = (6 - 2√5)/4 = (3 - √5)/2 ≈ (3 - 2.236)/2 ≈ 0.764/2 ≈ 0.382Therefore, W = 50 / φ = 50 * (√5 - 1)/2 = 25*(√5 - 1) ≈ 25*(2.236 - 1) ≈ 25*1.236 ≈ 30.9Similarly, L = 50 / φ^2 = 50*(3 - √5)/2 = 25*(3 - √5) ≈ 25*(3 - 2.236) ≈ 25*0.764 ≈ 19.1So, W ≈ 30.9 and L ≈ 19.1. But since the number of games must be integers, this is a problem. Maybe the problem expects us to use the exact values, even if they are not integers? But that doesn't make much sense because you can't win a fraction of a game.Wait, perhaps I made a wrong assumption. Maybe the ratio is games won to games lost, which is φ, but φ is approximately 1.618, so for every game lost, they won about 1.618 games. But since the number of games must be integers, maybe we need to find integers W and L such that W/L ≈ φ and W + L = 50.Alternatively, maybe the problem is designed such that W and L are exact multiples in the golden ratio, even if they are not integers. But that seems unlikely because you can't win a fraction of a game.Wait, maybe the problem is expecting an exact fractional answer, even though in reality, it's not possible. Let me check the problem statement again.It says, "the ratio of games won to games lost by the team forms a golden ratio (φ), where φ = (1 + √5)/2." So, it's a ratio, so it's possible that W and L are not integers, but in reality, they must be. Hmm, this is confusing.Wait, perhaps the problem is designed in such a way that W and L are integers, and their ratio is exactly φ. But since φ is irrational, that's impossible. So, maybe the problem is expecting us to express W and L in terms of φ, even if they are not integers. Or perhaps it's a trick question, and the numbers are meant to be exact, even though in reality, they can't be.Alternatively, maybe the problem is using the golden ratio in a different way. Let me think again.Wait, maybe the ratio is W:L = φ:1, so W = φ * L, and W + L = 50. So, substituting, φ * L + L = 50 => L*(φ + 1) = 50.But φ + 1 = φ^2, so L = 50 / φ^2, and W = 50 / φ.So, as I calculated earlier, W = 25*(√5 - 1) ≈ 30.9 and L ≈ 19.1.But since these are not integers, maybe the problem expects us to round them? Or perhaps it's a theoretical problem where fractional games are allowed? That seems odd.Wait, maybe I made a mistake in the ratio. The ratio of games won to games lost is φ, so W/L = φ. So, W = φ * L.But maybe the ratio is the other way around? Like, games lost to games won is φ? Wait, no, the problem says the ratio of games won to games lost is φ, so W/L = φ.Wait, let me check the problem again: "the ratio of games won to games lost by the team forms a golden ratio (φ), where φ = (1 + √5)/2." So, yes, W/L = φ.Hmm, maybe the problem is expecting an exact answer in terms of φ, even if it's not an integer. So, W = 50 / (1 + 1/φ) = 50 / (1 + (√5 - 1)/2) = 50 / [(2 + √5 - 1)/2] = 50 / [(1 + √5)/2] = 100 / (1 + √5) = same as before.Wait, but that's the same as L = 25*(3 - √5). So, perhaps the answer is W = 25*(√5 - 1) and L = 25*(3 - √5). Even though they are not integers, maybe that's the exact answer expected.Alternatively, perhaps the problem is expecting us to use the approximate value of φ, which is about 1.618, and then solve for W and L as integers. Let's try that.If W/L ≈ 1.618, then W ≈ 1.618 * L.And W + L = 50.So, substituting, 1.618*L + L = 50 => 2.618*L ≈ 50 => L ≈ 50 / 2.618 ≈ 19.098, which is approximately 19.1, so L ≈ 19 games, and W ≈ 31 games.But 19 + 31 = 50, so that works. So, perhaps the answer is 31 wins and 19 losses.But wait, if we take L = 19, then W = 50 - 19 = 31.Then, the ratio W/L = 31/19 ≈ 1.6316, which is close to φ ≈ 1.618, but not exact. So, it's an approximation.Alternatively, if we take L = 20, then W = 30, and W/L = 30/20 = 1.5, which is less than φ.So, 31/19 ≈ 1.6316 is closer to φ than 30/20 = 1.5.So, maybe the answer is 31 wins and 19 losses.But the problem says "the ratio forms a golden ratio," which is exact, not approximate. So, perhaps the answer is supposed to be in exact terms, even if they are not integers.So, W = 25*(√5 - 1) and L = 25*(3 - √5). Let me compute these exactly:√5 ≈ 2.236, so:W ≈ 25*(2.236 - 1) = 25*(1.236) ≈ 30.9L ≈ 25*(3 - 2.236) = 25*(0.764) ≈ 19.1So, approximately 30.9 wins and 19.1 losses. But since we can't have fractions, maybe the problem expects us to express it in terms of φ.Alternatively, perhaps the problem is designed such that W and L are integers, and the ratio is as close as possible to φ. In that case, 31/19 ≈ 1.6316 is the closest we can get with integers adding up to 50.So, maybe the answer is 31 wins and 19 losses.But the problem says "the ratio of games won to games lost by the team forms a golden ratio (φ)", which is exact. So, perhaps the answer is supposed to be in exact terms, even if they are not integers. So, W = 25*(√5 - 1) and L = 25*(3 - √5).Alternatively, maybe the problem is expecting us to write the exact values in terms of φ.Wait, let me think again.We have W/L = φ, so W = φ * L.And W + L = 50.So, substituting, φ * L + L = 50 => L*(φ + 1) = 50.But φ + 1 = φ^2, so L = 50 / φ^2.Similarly, W = φ * L = φ * (50 / φ^2) = 50 / φ.So, W = 50 / φ and L = 50 / φ^2.Since φ = (1 + √5)/2, let's compute 1/φ and 1/φ^2.1/φ = (√5 - 1)/2, as I calculated earlier.1/φ^2 = (3 - √5)/2.So, W = 50 * (√5 - 1)/2 = 25*(√5 - 1).Similarly, L = 50 * (3 - √5)/2 = 25*(3 - √5).So, the exact number of games won is 25*(√5 - 1) and lost is 25*(3 - √5).But these are not integers, so perhaps the problem expects us to leave it in this form.Alternatively, maybe the problem is expecting us to express it as fractions. Let me compute 25*(√5 - 1):√5 ≈ 2.236, so √5 - 1 ≈ 1.236, so 25*1.236 ≈ 30.9.Similarly, 3 - √5 ≈ 0.764, so 25*0.764 ≈ 19.1.So, approximately 30.9 wins and 19.1 losses, but since we can't have fractions, maybe the problem is designed to accept these exact forms.Alternatively, perhaps the problem is expecting us to use the continued fraction approximation of φ to find integers W and L such that W/L is as close as possible to φ.But in that case, the closest integers would be 31 and 19, as I thought earlier.But the problem says "the ratio forms a golden ratio," which is exact, so perhaps the answer is supposed to be in exact terms, even if they are not integers.So, perhaps the answer is W = 25*(√5 - 1) and L = 25*(3 - √5).Alternatively, maybe the problem is expecting us to express it as fractions, but since √5 is irrational, it can't be expressed as a fraction.Wait, maybe I should write the exact values as they are, even if they are not integers.So, for part 1, the exact number of games won is 25*(√5 - 1) and lost is 25*(3 - √5).For part 2, the total score is calculated by 3 points for each win and -1 point for each loss. So, total score = 3*W - 1*L.Substituting W and L:Total score = 3*(25*(√5 - 1)) - (25*(3 - √5)) = 75*(√5 - 1) - 25*(3 - √5)Let me expand this:= 75√5 - 75 - 75 + 25√5Wait, no, let me do it step by step.First, expand 3*(25*(√5 - 1)):= 75*(√5 - 1) = 75√5 - 75Then, expand (25*(3 - √5)):= 75 - 25√5So, total score = (75√5 - 75) - (75 - 25√5) = 75√5 - 75 - 75 + 25√5Combine like terms:75√5 + 25√5 = 100√5-75 -75 = -150So, total score = 100√5 - 150Alternatively, factor out 50:= 50*(2√5 - 3)But maybe it's better to leave it as 100√5 - 150.Alternatively, factor out 25:= 25*(4√5 - 6)But 100√5 - 150 is fine.Alternatively, factor out 50:= 50*(2√5 - 3)But I think 100√5 - 150 is the simplest form.But let me check my calculations again.Total score = 3W - LW = 25*(√5 - 1)L = 25*(3 - √5)So, 3W = 3*25*(√5 - 1) = 75*(√5 - 1) = 75√5 - 75Then, subtract L: 75√5 - 75 - 25*(3 - √5) = 75√5 - 75 -75 +25√5Wait, no, 25*(3 - √5) is 75 -25√5.So, subtracting that: 75√5 -75 -75 +25√5Combine like terms:75√5 +25√5 = 100√5-75 -75 = -150So, total score = 100√5 -150.Yes, that's correct.Alternatively, we can factor out 50:= 50*(2√5 - 3)But both forms are acceptable.So, summarizing:1. The exact number of games won is 25*(√5 - 1) and lost is 25*(3 - √5).2. The total score is 100√5 - 150 points.But let me check if there's another way to express this.Alternatively, since W = 50 / φ and L = 50 / φ^2, maybe we can express the total score in terms of φ.But I think the way I did it is fine.Alternatively, maybe we can compute the numerical value:√5 ≈ 2.236So, 100√5 ≈ 100*2.236 ≈ 223.6223.6 - 150 ≈ 73.6So, approximately 73.6 points.But since the problem asks for the exact number, we should keep it in terms of √5.So, the total score is 100√5 - 150.Alternatively, factor out 50: 50*(2√5 - 3), but both are correct.So, to answer the questions:1. Games won: 25*(√5 - 1), Games lost: 25*(3 - √5)2. Total score: 100√5 - 150But let me check if I can simplify 100√5 - 150 further.Alternatively, factor out 50: 50*(2√5 - 3), but that's just another form.Alternatively, factor out 25: 25*(4√5 - 6), but again, same thing.I think 100√5 - 150 is the simplest exact form.So, I think that's the answer.</think>

🧠:<think>Alright, so I have this problem where I need to figure out the total amount of money in an investment portfolio after 10 years, given that the principal is 500,000 and it's compounded continuously with a variable interest rate function r(t) = 0.03 + 0.02sin(t). Then, I also need to find the average interest rate over those 10 years. Hmm, okay, let me break this down step by step.First, I remember that when dealing with continuous compounding, the formula for the amount of money after time t is given by A = P * e^(∫r(t) dt), where P is the principal, and the integral is from 0 to t. So in this case, P is 500,000, and t is 10 years. So I need to compute the integral of r(t) from 0 to 10 and then exponentiate that result, multiply by the principal to get the total amount.Alright, so let's write that down. The amount A is:A = 500,000 * e^(∫₀¹⁰ r(t) dt)Given that r(t) = 0.03 + 0.02sin(t), so the integral becomes:∫₀¹⁰ (0.03 + 0.02sin(t)) dtI can split this integral into two parts:∫₀¹⁰ 0.03 dt + ∫₀¹⁰ 0.02sin(t) dtLet's compute each integral separately.First integral: ∫₀¹⁰ 0.03 dtThat's straightforward. The integral of a constant is just the constant multiplied by the variable, so:0.03 * t evaluated from 0 to 10.So that's 0.03*(10 - 0) = 0.3.Second integral: ∫₀¹⁰ 0.02sin(t) dtThe integral of sin(t) is -cos(t), so:0.02 * [-cos(t)] from 0 to 10.Calculating that:0.02 * [ -cos(10) + cos(0) ]I know that cos(0) is 1, and cos(10) is... hmm, I need to compute cos(10). Since 10 is in radians, right? Because in calculus, we usually work in radians. So cos(10 radians) is approximately... let me recall, cos(π) is -1, and 10 radians is about 1.59π, which is a bit more than π. So cos(10) is negative, but I need the exact value. Maybe I can compute it using a calculator.Wait, but since I don't have a calculator here, maybe I can just keep it symbolic for now. So:0.02 * [ -cos(10) + 1 ] = 0.02*(1 - cos(10))So putting it all together, the total integral is:0.3 + 0.02*(1 - cos(10))Therefore, the exponent in the formula for A is 0.3 + 0.02*(1 - cos(10)).So now, the total amount A is:500,000 * e^(0.3 + 0.02*(1 - cos(10)))Hmm, okay, so I need to compute this exponent. Let me compute each part step by step.First, 0.3 is straightforward.Then, 0.02*(1 - cos(10)). Let me compute 1 - cos(10) first.As I said, cos(10 radians). Let me think, 10 radians is approximately 572.958 degrees (since 1 radian is about 57.2958 degrees). So 10 radians is 572.958 degrees. That's a bit more than 360 degrees, so it's like 572.958 - 360 = 212.958 degrees. So cos(212.958 degrees). 212.958 degrees is in the third quadrant, where cosine is negative. 212.958 - 180 = 32.958 degrees. So cos(212.958) = -cos(32.958). Cos(32.958 degrees) is approximately 0.8443. So cos(212.958) is approximately -0.8443.But wait, in radians, 10 radians is 10, so I need to compute cos(10). Let me use the Taylor series or something? Or maybe use a calculator approximation.Wait, maybe I can recall that cos(10) is approximately -0.83907. Let me verify that. Yes, cos(10) is approximately -0.83907. So 1 - cos(10) is 1 - (-0.83907) = 1 + 0.83907 = 1.83907.Therefore, 0.02*(1 - cos(10)) = 0.02*1.83907 ≈ 0.0367814.So the exponent is 0.3 + 0.0367814 ≈ 0.3367814.Therefore, A ≈ 500,000 * e^(0.3367814)Now, e^0.3367814. Let me compute that. I know that e^0.3 ≈ 1.349858, and e^0.3367814 is a bit higher.Alternatively, I can use the Taylor series for e^x around x=0.3.But maybe it's easier to use a calculator approximation.Alternatively, since 0.3367814 is approximately 0.3368.So e^0.3368 ≈ ?I know that ln(1.4) ≈ 0.3365, so e^0.3365 ≈ 1.4. Therefore, e^0.3368 is approximately 1.4001 or something close.Wait, let me compute it more accurately.We know that ln(1.4) ≈ 0.33647, so e^0.33647 = 1.4.So 0.3367814 is 0.33647 + 0.0003114.So e^(0.33647 + 0.0003114) = e^0.33647 * e^0.0003114 ≈ 1.4 * (1 + 0.0003114) ≈ 1.4 * 1.0003114 ≈ 1.400436.So e^0.3367814 ≈ approximately 1.4004.Therefore, A ≈ 500,000 * 1.4004 ≈ 500,000 * 1.4004.Calculating that: 500,000 * 1.4 = 700,000, and 500,000 * 0.0004 = 200. So total is 700,000 + 200 = 700,200.Wait, but that seems a bit off because 0.3367814 is a bit more than 0.33647, so the exponent is a bit more, so the result is a bit more than 1.4, so 500,000 * 1.4004 is 700,200.But let me check with a calculator if possible.Alternatively, maybe I can compute e^0.3367814 more accurately.We can use the Taylor series expansion of e^x around x=0.33647.Let me denote x = 0.33647 + 0.0003114.So e^x = e^(0.33647 + 0.0003114) = e^0.33647 * e^0.0003114.We know e^0.33647 = 1.4.e^0.0003114 ≈ 1 + 0.0003114 + (0.0003114)^2/2 ≈ 1 + 0.0003114 + 0.000000048 ≈ 1.000311448.Therefore, e^x ≈ 1.4 * 1.000311448 ≈ 1.400436.So, yes, approximately 1.400436.Therefore, A ≈ 500,000 * 1.400436 ≈ 500,000 * 1.400436.Calculating that:500,000 * 1.4 = 700,000500,000 * 0.000436 = 500,000 * 0.0004 = 200, and 500,000 * 0.000036 = 18. So total is 200 + 18 = 218.Therefore, total A ≈ 700,000 + 218 = 700,218.So approximately 700,218.Wait, but let me check if I did the multiplication correctly.Wait, 500,000 * 1.400436 = 500,000 * 1 + 500,000 * 0.4 + 500,000 * 0.000436Which is 500,000 + 200,000 + 218 = 700,218.Yes, that's correct.So the total amount is approximately 700,218.But wait, let me think again. Is this correct? Because the integral of r(t) from 0 to 10 is 0.3 + 0.02*(1 - cos(10)).Wait, cos(10) is approximately -0.83907, so 1 - cos(10) is 1 - (-0.83907) = 1.83907.So 0.02*1.83907 ≈ 0.0367814.So the exponent is 0.3 + 0.0367814 ≈ 0.3367814.So e^0.3367814 ≈ 1.400436.Therefore, 500,000 * 1.400436 ≈ 700,218.Yes, that seems consistent.Alternatively, maybe I can use a calculator to compute e^0.3367814 more accurately.But since I don't have a calculator here, I think 1.4004 is a reasonable approximation.So, to summarize, the total amount after 10 years is approximately 700,218.Now, moving on to the second part: calculating the average interest rate over the 10-year period.The average value of a function r(t) over the interval [a, b] is given by (1/(b - a)) * ∫ₐᵇ r(t) dt.In this case, a = 0, b = 10, so the average rate r_avg is (1/10) * ∫₀¹⁰ r(t) dt.But wait, we already computed ∫₀¹⁰ r(t) dt earlier, which was 0.3 + 0.02*(1 - cos(10)) ≈ 0.3367814.Therefore, r_avg = (1/10) * 0.3367814 ≈ 0.03367814.So, converting that to a percentage, it's approximately 3.367814%.Therefore, the average interest rate is approximately 3.3678%.But let me express this more precisely.We had ∫₀¹⁰ r(t) dt = 0.3 + 0.02*(1 - cos(10)).So r_avg = (0.3 + 0.02*(1 - cos(10)))/10.Which is 0.03 + 0.002*(1 - cos(10)).We already computed 1 - cos(10) ≈ 1.83907.So 0.002*1.83907 ≈ 0.00367814.Therefore, r_avg ≈ 0.03 + 0.00367814 ≈ 0.03367814, which is approximately 3.3678%.So, rounding to four decimal places, it's 3.3678%.Alternatively, if we want to express it more precisely, we can keep more decimal places.But for practical purposes, 3.3678% is a good approximation.Wait, let me double-check the integral computation.We had ∫₀¹⁰ r(t) dt = ∫₀¹⁰ 0.03 dt + ∫₀¹⁰ 0.02sin(t) dt.First integral: 0.03*10 = 0.3.Second integral: 0.02*(-cos(10) + cos(0)) = 0.02*(1 - cos(10)).Yes, that's correct.So, 0.3 + 0.02*(1 - cos(10)).We computed 1 - cos(10) ≈ 1.83907, so 0.02*1.83907 ≈ 0.0367814.So total integral ≈ 0.3 + 0.0367814 ≈ 0.3367814.Therefore, average rate ≈ 0.3367814 / 10 ≈ 0.03367814, which is 3.367814%.So, yes, that seems correct.Alternatively, if I use more precise value of cos(10), let's see.cos(10) is approximately -0.8390715290764524.So 1 - cos(10) = 1 - (-0.8390715290764524) = 1 + 0.8390715290764524 ≈ 1.8390715290764524.Therefore, 0.02*1.8390715290764524 ≈ 0.03678143058152905.So total integral ≈ 0.3 + 0.03678143058152905 ≈ 0.33678143058152905.Therefore, average rate ≈ 0.33678143058152905 / 10 ≈ 0.033678143058152905, which is approximately 3.3678143058152905%.So, rounding to, say, four decimal places, it's 3.3678%.Therefore, the average interest rate is approximately 3.3678%.So, to recap:1. The total amount after 10 years is approximately 700,218.2. The average interest rate over the 10-year period is approximately 3.3678%.I think that's it. Let me just make sure I didn't make any calculation errors.Wait, when I computed e^0.3367814, I approximated it as 1.4004, leading to A ≈ 700,218. Let me verify that exponent again.We have:A = 500,000 * e^(0.3367814)If I use a calculator, e^0.3367814 is approximately e^0.3367814 ≈ 1.400436.Yes, so 500,000 * 1.400436 ≈ 700,218.Yes, that seems correct.Alternatively, if I use more precise exponent:0.3367814.Let me compute e^0.3367814 step by step.We know that e^0.3 = 1.349858.e^0.3367814 = e^(0.3 + 0.0367814) = e^0.3 * e^0.0367814.We have e^0.3 ≈ 1.349858.Now, e^0.0367814.We can compute this using the Taylor series:e^x ≈ 1 + x + x²/2 + x³/6 + x⁴/24.Where x = 0.0367814.So:1 + 0.0367814 + (0.0367814)^2 / 2 + (0.0367814)^3 / 6 + (0.0367814)^4 / 24.Compute each term:First term: 1.Second term: 0.0367814.Third term: (0.0367814)^2 / 2 ≈ (0.0013525) / 2 ≈ 0.00067625.Fourth term: (0.0367814)^3 / 6 ≈ (0.000050) / 6 ≈ 0.00000833.Fifth term: (0.0367814)^4 / 24 ≈ (0.00000184) / 24 ≈ 0.0000000767.Adding them up:1 + 0.0367814 = 1.0367814+ 0.00067625 = 1.03745765+ 0.00000833 ≈ 1.037466+ 0.0000000767 ≈ 1.0374660767.So e^0.0367814 ≈ 1.037466.Therefore, e^0.3367814 ≈ e^0.3 * e^0.0367814 ≈ 1.349858 * 1.037466.Compute that:1.349858 * 1.037466.Let me compute 1.349858 * 1.037466.First, 1 * 1.349858 = 1.349858.0.03 * 1.349858 = 0.04049574.0.007 * 1.349858 = 0.009449.0.000466 * 1.349858 ≈ 0.000629.Adding them up:1.349858 + 0.04049574 = 1.39035374+ 0.009449 ≈ 1.39980274+ 0.000629 ≈ 1.40043174.So, e^0.3367814 ≈ 1.40043174.Therefore, A ≈ 500,000 * 1.40043174 ≈ 700,215.87.So approximately 700,215.87, which is about 700,216.So, rounding to the nearest dollar, it's 700,216.Wait, earlier I had 700,218, but with this more precise calculation, it's 700,216. The slight difference is due to the approximation in the Taylor series.But regardless, it's approximately 700,216.So, to be precise, I can say approximately 700,216.Alternatively, if I use a calculator, e^0.3367814 is approximately 1.400436, so 500,000 * 1.400436 = 700,218.So, depending on the precision, it's either 700,216 or 700,218. But since the integral was approximated, it's acceptable to say approximately 700,218.So, in conclusion:1. The total amount after 10 years is approximately 700,218.2. The average interest rate over the 10-year period is approximately 3.3678%.I think that's it. I don't see any mistakes in my calculations now.</think>