🧠:<think>Alright, so I have this problem where Officer Elena is analyzing graffiti incidents in five neighborhoods. The data given is in terms of x and y for each neighborhood. The first sub-problem is to find the values of x and y given that the total number of incidents is 200. The second part is about reducing the total incidents by 20% and figuring out the new numbers for each neighborhood.Starting with the first part. I need to set up a system of equations. Since there are two variables, x and y, I should have two equations. But wait, the problem only mentions the total being 200. That gives me one equation. Hmm, maybe I need another piece of information? Let me check the problem again.Wait, the problem says "formulate and solve a system of linear equations." It doesn't specify another equation, so perhaps I'm supposed to use the total as the only equation? But that would mean I have one equation with two variables, which usually means infinitely many solutions. That doesn't make sense because the problem is asking for specific values of x and y. Maybe I missed something.Looking back, the data is given for each neighborhood in terms of x and y. So each neighborhood's incidents are expressed as a linear combination of x and y. So, the total incidents would be the sum of all these expressions. Let me write that out.Neighborhood A: x + 2yNeighborhood B: 3x - yNeighborhood C: 2x + yNeighborhood D: x - 3yNeighborhood E: 4x + 5yTotal = A + B + C + D + E = (x + 2y) + (3x - y) + (2x + y) + (x - 3y) + (4x + 5y)Let me compute that step by step.First, combine the x terms:x + 3x + 2x + x + 4x = (1 + 3 + 2 + 1 + 4)x = 11xNow the y terms:2y - y + y - 3y + 5y = (2 - 1 + 1 - 3 + 5)y = (2 -1 is 1, 1 +1 is 2, 2 -3 is -1, -1 +5 is 4)y = 4ySo total incidents = 11x + 4y = 200So that's one equation: 11x + 4y = 200But we need another equation to solve for x and y. Wait, the problem doesn't provide another equation. Maybe I need to assume something else? Or perhaps I misread the problem.Wait, the problem says "formulate and solve a system of linear equations." It doesn't specify another equation, so maybe I'm supposed to use another condition? Or perhaps the problem expects me to realize that without another equation, it's impossible to solve, but that seems unlikely because the problem is asking for specific values.Wait, maybe the problem is only giving one equation because it's a system with two variables, but perhaps I need to find x and y such that all the neighborhood incidents are non-negative? Because you can't have negative graffiti incidents. So maybe I need to set up inequalities as well, but the problem says "formulate and solve a system of linear equations," so maybe it's just the one equation.But then, with one equation, we can't solve for two variables. Maybe the problem is expecting me to realize that and say that there's insufficient information? But that doesn't seem right because the problem is asking to find x and y.Wait, perhaps I made a mistake in adding up the terms. Let me double-check.Total x terms:Neighborhood A: xNeighborhood B: 3xNeighborhood C: 2xNeighborhood D: xNeighborhood E: 4xAdding them up: 1 + 3 + 2 + 1 + 4 = 11x. That seems correct.Total y terms:Neighborhood A: 2yNeighborhood B: -yNeighborhood C: yNeighborhood D: -3yNeighborhood E: 5yAdding them up: 2 -1 +1 -3 +5 = 4y. That also seems correct.So total is 11x + 4y = 200.So that's the only equation. Hmm. Maybe the problem is expecting me to express y in terms of x or vice versa, but the problem says "find the values of x and y," implying specific numerical values.Wait, maybe I need to consider that the number of incidents in each neighborhood must be non-negative. So, for each neighborhood, the expression must be greater than or equal to zero.So, let's write those inequalities:A: x + 2y ≥ 0B: 3x - y ≥ 0C: 2x + y ≥ 0D: x - 3y ≥ 0E: 4x + 5y ≥ 0These are all constraints. But without another equation, I can't solve for x and y uniquely. So, perhaps the problem is expecting me to express x and y in terms of each other? Or maybe I'm supposed to assume that x and y are integers? Or perhaps there's a typo in the problem.Wait, maybe I misread the problem. Let me check again.The problem says: "formulate and solve a system of linear equations to find the values of x and y." So, it's expecting a system, which usually has as many equations as variables. Since we have two variables, x and y, we need two equations. But the problem only gives one equation from the total.Hmm, maybe the problem is expecting me to realize that and say that there's insufficient information, but that seems unlikely because the problem is asking to find x and y.Wait, perhaps the problem is expecting me to use another condition, like the number of incidents in each neighborhood being positive, but that would be inequalities, not equations.Alternatively, maybe the problem is expecting me to realize that the system is underdetermined and express x and y in terms of each other, but the problem says "find the values," implying specific numbers.Wait, maybe I made a mistake in adding up the total. Let me check again.Total x terms: 1 + 3 + 2 + 1 + 4 = 11xTotal y terms: 2 -1 +1 -3 +5 = 4yYes, that's correct. So 11x + 4y = 200.So, unless there's another equation, I can't solve for x and y uniquely. Maybe the problem is expecting me to express y in terms of x or vice versa.Let me try that.From 11x + 4y = 200, we can solve for y:4y = 200 - 11xy = (200 - 11x)/4But that's just expressing y in terms of x. Unless there's another condition, we can't find specific values.Wait, maybe the problem is expecting me to assume that x and y are integers? Let me see.If x and y are integers, then (200 - 11x) must be divisible by 4.So, 200 is divisible by 4, 200/4=50. So, 11x must also leave a remainder that makes (200 -11x) divisible by 4.11 mod 4 is 3, so 11x mod 4 is 3x mod 4.So, 200 -11x ≡ 0 mod 4Which is equivalent to 0 - 3x ≡ 0 mod 4So, -3x ≡ 0 mod 4 => 3x ≡ 0 mod 4Since 3 and 4 are coprime, this implies x ≡ 0 mod 4.So, x must be a multiple of 4.Let me let x = 4k, where k is an integer.Then y = (200 -11*(4k))/4 = (200 -44k)/4 = 50 -11kSo, x =4k, y=50-11kNow, we also have the constraints that each neighborhood's incidents must be non-negative.So, let's write the expressions for each neighborhood:A: x + 2y =4k + 2*(50 -11k)=4k +100 -22k=100 -18k ≥0So, 100 -18k ≥0 => 18k ≤100 => k ≤100/18≈5.555. So, k ≤5B:3x - y=3*(4k) - (50 -11k)=12k -50 +11k=23k -50 ≥0So, 23k -50 ≥0 =>23k ≥50 =>k≥50/23≈2.173. So, k≥3 (since k must be integer)C:2x + y=2*(4k) + (50 -11k)=8k +50 -11k=50 -3k ≥0So, 50 -3k ≥0 =>3k ≤50 =>k ≤16.666. So, k ≤16, but from A, k≤5D:x -3y=4k -3*(50 -11k)=4k -150 +33k=37k -150 ≥0So, 37k -150 ≥0 =>37k ≥150 =>k≥150/37≈4.054. So, k≥5E:4x +5y=4*(4k) +5*(50 -11k)=16k +250 -55k=250 -39k ≥0So, 250 -39k ≥0 =>39k ≤250 =>k ≤250/39≈6.41. So, k≤6But from A, k≤5, and from D, k≥5. So, combining these, k must be 5.So, k=5Therefore, x=4*5=20y=50 -11*5=50 -55= -5Wait, y=-5? That can't be, because y represents something in the number of incidents, which can't be negative.Wait, that's a problem. So, y=-5, which is negative. That violates the constraint for neighborhood A: x +2y=20 +2*(-5)=20-10=10, which is non-negative. But y itself is negative, which might not be acceptable because y is a variable, not necessarily an incident count. Wait, but in the expressions for each neighborhood, y is multiplied by coefficients, so even if y is negative, the total could still be positive.But let's check all neighborhoods:A:20 +2*(-5)=10 ≥0B:3*20 - (-5)=60 +5=65 ≥0C:2*20 + (-5)=40 -5=35 ≥0D:20 -3*(-5)=20 +15=35 ≥0E:4*20 +5*(-5)=80 -25=55 ≥0So, all neighborhoods have non-negative incidents, even though y is negative. So, maybe y can be negative? Or perhaps y represents something else, like a factor that can be negative.But in the context of graffiti incidents, x and y might represent different factors, perhaps x is a positive factor and y is a negative factor, or vice versa. So, maybe y being negative is acceptable as long as the total incidents are positive.But let's see, if k=5, y=-5, which is acceptable because all neighborhoods have non-negative incidents.But let's check if k=4 would work.If k=4, then x=16, y=50 -44=6Check the constraints:A:16 +2*6=28 ≥0B:3*16 -6=48 -6=42 ≥0C:2*16 +6=32 +6=38 ≥0D:16 -3*6=16 -18=-2 <0. Not acceptable.So, k=4 gives D as negative, which is invalid.Similarly, k=5 gives D=35, which is okay.k=6: x=24, y=50 -66=-16Check D:24 -3*(-16)=24 +48=72 ≥0But y=-16, let's check all neighborhoods:A:24 +2*(-16)=24 -32=-8 <0. Not acceptable.So, k=6 invalidates A.So, the only valid k is 5, which gives y=-5, but all neighborhoods have non-negative incidents.So, x=20, y=-5.But y=-5 seems odd because it's negative, but mathematically, it works because the coefficients in the neighborhoods' expressions make the total non-negative.So, perhaps that's the solution.Now, moving on to the second sub-problem.Officer Elena wants to reduce the total number of incidents by 20%. So, the new total would be 200 - 0.2*200=160.She plans to allocate resources proportionally to the current number of incidents in each neighborhood. So, each neighborhood's incidents will be reduced by 20% as well.Wait, but the problem says "allocate resources proportionally to the current number of incidents in each neighborhood." So, does that mean each neighborhood's reduction is proportional to their current incidents? Or does it mean that the reduction is 20% overall, and each neighborhood's incidents are reduced by 20%?I think it's the latter. Because if she reduces the total by 20%, and she allocates resources proportionally, it's likely that each neighborhood's incidents are reduced by 20% as well.So, the new number of incidents in each neighborhood would be 80% of the original.So, for each neighborhood, new incidents = original incidents * 0.8But let's confirm.Total original incidents:200Total after reduction:160So, each neighborhood's incidents are reduced proportionally, meaning each is multiplied by 0.8.So, for example, Neighborhood A:10 incidents (from x=20, y=-5:20 +2*(-5)=10). So, new A=10*0.8=8Similarly, B:65*0.8=52C:35*0.8=28D:35*0.8=28E:55*0.8=44Let me check the total:8+52+28+28+44=160, which is correct.Alternatively, since the total is reduced by 20%, each neighborhood's incidents are reduced by 20% as well.So, the new incidents for each neighborhood would be 80% of the original.So, the new numbers are:A:8B:52C:28D:28E:44But let me make sure that this is the correct interpretation. The problem says "allocate resources proportionally to the current number of incidents in each neighborhood." So, if she allocates resources proportionally, it might mean that the reduction is proportional, not the incidents themselves. So, perhaps the reduction amount is proportional, not the incidents.Wait, that's a different interpretation. Let me think.If the total reduction is 40 (from 200 to 160), and she allocates the reduction proportionally to each neighborhood's current incidents, then each neighborhood's reduction would be (incidents / total) * 40.So, for example, Neighborhood A has 10 incidents, which is 10/200=0.05 of the total. So, reduction for A is 0.05*40=2. So, new incidents for A=10-2=8Similarly, B:65/200=0.325, reduction=0.325*40=13, new B=65-13=52C:35/200=0.175, reduction=0.175*40=7, new C=35-7=28D:35/200=0.175, reduction=7, new D=35-7=28E:55/200=0.275, reduction=0.275*40=11, new E=55-11=44So, same result as before:8,52,28,28,44.So, whether you reduce each neighborhood's incidents by 20% or allocate the total reduction proportionally, you get the same result because 20% reduction is equivalent to proportional allocation of the total reduction.So, the new incidents are:A:8B:52C:28D:28E:44So, that's the answer for the second part.But let me just make sure that the first part is correct. We found x=20, y=-5. Even though y is negative, all neighborhoods have non-negative incidents. So, that's acceptable.So, summarizing:1. x=20, y=-52. New incidents:A:8B:52C:28D:28E:44</think>

🧠:<think>Alright, so I'm trying to help Jane solve these differential equations for her carbon capture technology. Let me take this step by step.Starting with the first problem: the rate of CO₂ extraction, R(t), is given by the differential equation dR/dt = -kR + α sin(βt). Hmm, okay, this looks like a linear nonhomogeneous differential equation. I remember that to solve such equations, we can use an integrating factor or find the homogeneous solution and then a particular solution.First, let's write down the equation again:dR/dt + kR = α sin(βt)Yes, that's the standard form for a linear DE: dy/dt + P(t)y = Q(t). In this case, P(t) is k, a constant, and Q(t) is α sin(βt).So, the integrating factor, μ(t), is e^(∫P(t)dt) = e^(∫k dt) = e^(kt). Multiplying both sides of the equation by μ(t):e^(kt) dR/dt + k e^(kt) R = α e^(kt) sin(βt)The left side is the derivative of (e^(kt) R) with respect to t. So, we can write:d/dt [e^(kt) R] = α e^(kt) sin(βt)Now, integrate both sides with respect to t:∫ d/dt [e^(kt) R] dt = ∫ α e^(kt) sin(βt) dtSo, e^(kt) R = α ∫ e^(kt) sin(βt) dt + CNow, I need to compute the integral ∫ e^(kt) sin(βt) dt. I think integration by parts is the way to go here, or maybe use a formula for integrating e^(at) sin(bt) dt.I recall that ∫ e^(at) sin(bt) dt = e^(at)/(a² + b²) [a sin(bt) - b cos(bt)] + CLet me verify that. Let’s set u = e^(at), dv = sin(bt) dt. Then du = a e^(at) dt, v = -cos(bt)/b.So, ∫ e^(at) sin(bt) dt = -e^(at) cos(bt)/b + (a/b) ∫ e^(at) cos(bt) dtNow, integrate the second integral by parts again. Let u = e^(at), dv = cos(bt) dt. Then du = a e^(at) dt, v = sin(bt)/b.So, ∫ e^(at) cos(bt) dt = e^(at) sin(bt)/b - (a/b) ∫ e^(at) sin(bt) dtPutting it back into the previous equation:∫ e^(at) sin(bt) dt = -e^(at) cos(bt)/b + (a/b)[e^(at) sin(bt)/b - (a/b) ∫ e^(at) sin(bt) dt]Let me denote I = ∫ e^(at) sin(bt) dt. Then,I = -e^(at) cos(bt)/b + (a/b)(e^(at) sin(bt)/b - (a/b) I)Simplify:I = -e^(at) cos(bt)/b + (a e^(at) sin(bt))/b² - (a² / b²) IBring the (a² / b²) I term to the left:I + (a² / b²) I = -e^(at) cos(bt)/b + (a e^(at) sin(bt))/b²Factor I:I (1 + a² / b²) = e^(at) [ -cos(bt)/b + a sin(bt)/b² ]Multiply both sides by b² to eliminate denominators:I (b² + a²) = e^(at) [ -b cos(bt) + a sin(bt) ]Thus,I = e^(at) [ -b cos(bt) + a sin(bt) ] / (a² + b²) + CSo, the integral is:∫ e^(kt) sin(βt) dt = e^(kt) [ -β cos(βt) + k sin(βt) ] / (k² + β²) + CTherefore, going back to our equation:e^(kt) R = α [ e^(kt) ( -β cos(βt) + k sin(βt) ) / (k² + β²) ] + CDivide both sides by e^(kt):R(t) = α [ (-β cos(βt) + k sin(βt) ) / (k² + β²) ] + C e^(-kt)Now, apply the initial condition R(0) = R0.At t=0,R(0) = α [ (-β cos(0) + k sin(0) ) / (k² + β²) ] + C e^(0) = R0Simplify:R0 = α [ (-β * 1 + k * 0 ) / (k² + β²) ] + CSo,R0 = - α β / (k² + β²) + CTherefore, C = R0 + α β / (k² + β²)So, the general solution is:R(t) = α [ (-β cos(βt) + k sin(βt) ) / (k² + β²) ] + [ R0 + α β / (k² + β²) ] e^(-kt)We can write this as:R(t) = (α / (k² + β²)) ( -β cos(βt) + k sin(βt) ) + ( R0 + α β / (k² + β²) ) e^(-kt)Alternatively, factor out α / (k² + β²):R(t) = (α / (k² + β²)) ( -β cos(βt) + k sin(βt) ) + R0 e^(-kt) + ( α β / (k² + β²) ) e^(-kt)Combine the last two terms:R(t) = R0 e^(-kt) + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) + β e^(-kt) )Wait, let me check that again. The last term is ( α β / (k² + β²) ) e^(-kt), so when combined with R0 e^(-kt), it's:R(t) = R0 e^(-kt) + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) ) + ( α β / (k² + β²) ) e^(-kt)So, factor e^(-kt) from the last two terms:R(t) = R0 e^(-kt) + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) ) + ( α β / (k² + β²) ) e^(-kt)= [ R0 + ( α β / (k² + β²) ) ] e^(-kt) + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) )Alternatively, we can write it as:R(t) = e^(-kt) [ R0 + ( α β / (k² + β²) ) ] + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) )This seems correct. Let me check the dimensions. The homogeneous solution is R0 e^(-kt), which makes sense because as t increases, the extraction rate decays exponentially. The particular solution is a combination of sine and cosine terms, which oscillate over time, modulated by the constants.So, that's the solution for R(t).Now, moving on to the second part: the storage capacity S(t) is the integral of R(t) from 0 to t.So,S(t) = ∫₀ᵗ R(τ) dτWe need to compute this integral. Let's substitute R(τ) from the solution we found.So,S(t) = ∫₀ᵗ [ e^(-kτ) ( R0 + α β / (k² + β²) ) + (α / (k² + β²)) ( -β cos(βτ) + k sin(βτ) ) ] dτWe can split this integral into two parts:S(t) = ∫₀ᵗ e^(-kτ) ( R0 + α β / (k² + β²) ) dτ + ∫₀ᵗ (α / (k² + β²)) ( -β cos(βτ) + k sin(βτ) ) dτLet me compute each integral separately.First integral:I1 = ∫₀ᵗ e^(-kτ) ( R0 + α β / (k² + β²) ) dτLet me denote A = R0 + α β / (k² + β²), so:I1 = A ∫₀ᵗ e^(-kτ) dτ = A [ (-1/k) e^(-kτ) ]₀ᵗ = A ( (-1/k) e^(-kt) + 1/k ) = A (1 - e^(-kt))/kSubstituting back A:I1 = ( R0 + α β / (k² + β²) ) (1 - e^(-kt))/kSecond integral:I2 = (α / (k² + β²)) ∫₀ᵗ ( -β cos(βτ) + k sin(βτ) ) dτLet's compute the integral inside:∫ ( -β cos(βτ) + k sin(βτ) ) dτIntegrate term by term:∫ -β cos(βτ) dτ = -β [ sin(βτ)/β ] = - sin(βτ)∫ k sin(βτ) dτ = k [ -cos(βτ)/β ] = - (k / β) cos(βτ)So, the integral from 0 to t is:[ - sin(βτ) - (k / β) cos(βτ) ]₀ᵗ = [ - sin(βt) - (k / β) cos(βt) ] - [ - sin(0) - (k / β) cos(0) ]Simplify:= [ - sin(βt) - (k / β) cos(βt) ] - [ 0 - (k / β)(1) ]= - sin(βt) - (k / β) cos(βt) + k / βSo, putting it back into I2:I2 = (α / (k² + β²)) [ - sin(βt) - (k / β) cos(βt) + k / β ]Therefore, combining I1 and I2:S(t) = I1 + I2= ( R0 + α β / (k² + β²) ) (1 - e^(-kt))/k + (α / (k² + β²)) [ - sin(βt) - (k / β) cos(βt) + k / β ]Simplify this expression.First, let's handle the first term:Term1 = ( R0 + α β / (k² + β²) ) (1 - e^(-kt))/k= R0 (1 - e^(-kt))/k + ( α β / (k² + β²) ) (1 - e^(-kt))/k= R0 (1 - e^(-kt))/k + α β (1 - e^(-kt)) / [k(k² + β²)]Now, the second term:Term2 = (α / (k² + β²)) [ - sin(βt) - (k / β) cos(βt) + k / β ]= - α sin(βt) / (k² + β²) - α k cos(βt) / [β(k² + β²)] + α k / [β(k² + β²)]So, combining Term1 and Term2:S(t) = R0 (1 - e^(-kt))/k + α β (1 - e^(-kt)) / [k(k² + β²)] - α sin(βt)/(k² + β²) - α k cos(βt)/(β(k² + β²)) + α k/(β(k² + β²))Let me see if we can combine some terms. Notice that the last term is a constant, so it will contribute to the steady-state part of S(t).Looking at the terms involving e^(-kt):They are R0 (1 - e^(-kt))/k and α β (1 - e^(-kt)) / [k(k² + β²)]So, combining these:= [ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kt))Similarly, the other terms are:- α sin(βt)/(k² + β²) - α k cos(βt)/(β(k² + β²)) + α k/(β(k² + β²))So, putting it all together:S(t) = [ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kt)) - α sin(βt)/(k² + β²) - α k cos(βt)/(β(k² + β²)) + α k/(β(k² + β²))This expression can be further simplified by combining constants.Let me factor out 1/(k² + β²) where possible.First, the term [ R0 / k + α β / (k(k² + β²)) ] can be written as:= ( R0 (k² + β²) + α β ) / [k(k² + β²) ]Similarly, the other terms:- α sin(βt)/(k² + β²) - α k cos(βt)/(β(k² + β²)) + α k/(β(k² + β²))= [ - α sin(βt) - (α k / β) cos(βt) + α k / β ] / (k² + β²)So, combining everything:S(t) = [ ( R0 (k² + β²) + α β ) / (k(k² + β²)) ] (1 - e^(-kt)) + [ - α sin(βt) - (α k / β) cos(βt) + α k / β ] / (k² + β²)This is a bit messy, but it's the expression for S(t).Now, the storage capacity S(t) will reach its maximum capacity S_max at time T when S(T) = S_max.So, we need to solve for T in:S(T) = S_maxWhich is:[ ( R0 (k² + β²) + α β ) / (k(k² + β²)) ] (1 - e^(-kT)) + [ - α sin(βT) - (α k / β) cos(βT) + α k / β ] / (k² + β²) = S_maxThis equation is transcendental and likely cannot be solved analytically for T. Therefore, we would need to solve it numerically. However, perhaps we can express T in terms of the parameters, but it might not be straightforward.Alternatively, if the oscillatory terms average out over time, especially if β is small compared to k, the storage capacity might approach a steady-state value as t increases. But since the problem asks for the time T when S(T) = S_max, we need to consider the exact expression.Given the complexity, I think the answer would involve setting up the equation as above and noting that T must be found numerically. However, perhaps there's a way to express it more neatly.Alternatively, if we consider that the integral of R(t) can be expressed in terms of the homogeneous and particular solutions, maybe we can write S(t) as a combination of exponential and oscillatory terms, but I think we've already done that.So, in conclusion, S(t) is given by the expression we derived, and T is the solution to S(T) = S_max, which would require numerical methods unless specific values for the constants are given.But since the problem doesn't specify particular values, we can leave the answer in terms of the integral expression or the derived formula.Wait, perhaps I can write S(t) in a more compact form.Let me try:S(t) = ( R0 / k )(1 - e^(-kt)) + ( α β / (k(k² + β²)) )(1 - e^(-kt)) - ( α / (k² + β²) ) sin(βt) - ( α k / (β(k² + β²)) ) cos(βt) + ( α k / (β(k² + β²)) )So, combining the first two terms:= [ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kt)) - ( α / (k² + β²) ) sin(βt) - ( α k / (β(k² + β²)) ) cos(βt) + ( α k / (β(k² + β²)) )Notice that the last term is a constant, so it can be combined with the other constants if any. But in this case, it's just a constant term.So, S(t) = [ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kt)) - ( α / (k² + β²) ) sin(βt) - ( α k / (β(k² + β²)) ) cos(βt) + ( α k / (β(k² + β²)) )This is as simplified as it gets.Therefore, to find T such that S(T) = S_max, we set up the equation:[ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kT)) - ( α / (k² + β²) ) sin(βT) - ( α k / (β(k² + β²)) ) cos(βT) + ( α k / (β(k² + β²)) ) = S_maxThis equation would need to be solved numerically for T, as it's unlikely to have an analytical solution due to the combination of exponential and trigonometric functions.So, summarizing:1. The solution for R(t) is:R(t) = R0 e^(-kt) + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) ) + ( α β / (k² + β²) ) e^(-kt)Which can be written as:R(t) = e^(-kt) [ R0 + α β / (k² + β²) ] + (α / (k² + β²)) ( -β cos(βt) + k sin(βt) )2. The storage capacity S(t) is:S(t) = [ R0 / k + α β / (k(k² + β²)) ] (1 - e^(-kt)) - ( α / (k² + β²) ) sin(βt) - ( α k / (β(k² + β²)) ) cos(βt) + ( α k / (β(k² + β²)) )And the time T when S(T) = S_max is found by solving the above equation for T, which typically requires numerical methods.I think that's the extent of the solution without specific values for the constants. Let me double-check the integration steps to ensure I didn't make a mistake.In the integral of e^(kt) sin(βt) dt, I used the standard formula and verified it via integration by parts, so that seems correct. Then, applying the integrating factor correctly led to the solution for R(t). The integral for S(t) was split into two parts, each integrated correctly, so that should be fine too.Yes, I believe this is correct.</think>

🧠:<think>First, I need to determine if the number of jokes follows a quadratic sequence. A quadratic sequence has a second difference that is constant. Let me calculate the first and second differences between the given terms.The given number of jokes are: 12, 15, 21, 30, and 45.Calculating the first differences:15 - 12 = 321 - 15 = 630 - 21 = 945 - 30 = 15Now, the second differences:6 - 3 = 39 - 6 = 315 - 9 = 6Hmm, the second differences are not constant. This suggests that the sequence might not be purely quadratic. However, since the problem states that it follows a quadratic sequence, I'll proceed by assuming a quadratic formula of the form ( J(n) = an^2 + bn + c ).Using the given data points, I can set up a system of equations:For n = 1: ( a(1)^2 + b(1) + c = 12 ) → ( a + b + c = 12 )For n = 2: ( a(2)^2 + b(2) + c = 15 ) → ( 4a + 2b + c = 15 )For n = 3: ( a(3)^2 + b(3) + c = 21 ) → ( 9a + 3b + c = 21 )Now, I'll solve this system of equations step by step.Subtracting the first equation from the second:( (4a + 2b + c) - (a + b + c) = 15 - 12 )( 3a + b = 3 ) → Equation (4)Subtracting the second equation from the third:( (9a + 3b + c) - (4a + 2b + c) = 21 - 15 )( 5a + b = 6 ) → Equation (5)Subtracting Equation (4) from Equation (5):( (5a + b) - (3a + b) = 6 - 3 )( 2a = 3 ) → ( a = frac{3}{2} )Substituting ( a = frac{3}{2} ) into Equation (4):( 3(frac{3}{2}) + b = 3 )( frac{9}{2} + b = 3 )( b = 3 - frac{9}{2} = -frac{3}{2} )Finally, substituting ( a ) and ( b ) into the first equation:( frac{3}{2} - frac{3}{2} + c = 12 )( 0 + c = 12 )( c = 12 )So, the general formula is ( J(n) = frac{3}{2}n^2 - frac{3}{2}n + 12 ).Next, to find the total number of jokes by the end of the 10th episode, I need to calculate the sum of jokes from episode 1 to episode 10. The sum of a quadratic sequence can be found using the formula for the sum of the first n terms of a quadratic sequence.The sum ( S(n) ) is given by:( S(n) = frac{n}{6}(2a(1)^2 + (2a + b)(n - 1) + a(n - 1)^2 + b(n - 1) + c) )Plugging in the values:( a = frac{3}{2} ), ( b = -frac{3}{2} ), ( c = 12 ), and ( n = 10 ):( S(10) = frac{10}{6}(2(frac{3}{2})(1)^2 + (2(frac{3}{2}) - frac{3}{2})(10 - 1) + frac{3}{2}(10 - 1)^2 - frac{3}{2}(10 - 1) + 12) )Simplifying step by step:( 2(frac{3}{2}) = 3 )( (3 - frac{3}{2}) = frac{3}{2} )( frac{3}{2}(81) = frac{243}{2} )( -frac{3}{2}(9) = -frac{27}{2} )Combining all terms inside the parentheses:( 3 + frac{3}{2}(9) + frac{243}{2} - frac{27}{2} + 12 )( = 3 + frac{27}{2} + frac{216}{2} + 12 )( = 3 + 13.5 + 108 + 12 )( = 136.5 )Finally, multiplying by ( frac{10}{6} ):( frac{10}{6} times 136.5 = frac{1365}{6} = 227.5 )Since the number of jokes should be an integer, the total number of jokes Alex would have memorized by the end of the 10th episode is 228.</think>

🧠:<think>Alright, so I have this problem about modeling the political influence and voter turnout in Victorian England. It's divided into two parts: the Voting Influence Model and the Voter Turnout Analysis. Let me tackle them one by one.Starting with the first part: the Voting Influence Model. The function given is ( I(t) = frac{a}{1 + b e^{-ct}} ). I need to find the constants ( a ), ( b ), and ( c ). The information provided is that in 1850 (( t = 0 )), the influence was 20 units, and in 1870 (( t = 20 )), it was 50 units.Okay, so let's plug in ( t = 0 ) first. When ( t = 0 ), ( I(0) = 20 ). Substituting into the equation:( 20 = frac{a}{1 + b e^{0}} )Since ( e^{0} = 1 ), this simplifies to:( 20 = frac{a}{1 + b} )So, ( a = 20(1 + b) ). I'll keep this as equation (1).Next, when ( t = 20 ), ( I(20) = 50 ). Plugging into the function:( 50 = frac{a}{1 + b e^{-20c}} )From equation (1), I can substitute ( a ) here:( 50 = frac{20(1 + b)}{1 + b e^{-20c}} )Let me simplify this equation. Multiply both sides by the denominator:( 50(1 + b e^{-20c}) = 20(1 + b) )Divide both sides by 10 to make it simpler:( 5(1 + b e^{-20c}) = 2(1 + b) )Expanding both sides:( 5 + 5b e^{-20c} = 2 + 2b )Subtract 2 from both sides:( 3 + 5b e^{-20c} = 2b )Bring the ( 2b ) to the left:( 3 + 5b e^{-20c} - 2b = 0 )Factor out ( b ):( 3 + b(5 e^{-20c} - 2) = 0 )Hmm, so we have:( b(5 e^{-20c} - 2) = -3 )Let me write this as:( b(2 - 5 e^{-20c}) = 3 )So, ( b = frac{3}{2 - 5 e^{-20c}} ). Let's call this equation (2).Now, we have two equations: equation (1) and equation (2). But there are three variables: ( a ), ( b ), and ( c ). So, I need another equation or a way to relate these variables.Wait, maybe I can assume something about the function's behavior as ( t ) approaches infinity. In the long run, the influence ( I(t) ) should approach ( a ), because ( e^{-ct} ) goes to zero. So, the maximum influence is ( a ). But we don't have information about the maximum influence. Hmm, maybe I need another condition.Wait, perhaps the function is a logistic growth curve, which typically has an inflection point where the growth rate is maximum. But without more data points, it's hard to determine. Alternatively, maybe I can express ( a ) in terms of ( b ) from equation (1) and substitute into equation (2), but that might not help directly.Alternatively, maybe I can express ( e^{-20c} ) in terms of ( b ) from equation (2):From equation (2):( 2 - 5 e^{-20c} = frac{3}{b} )So,( 5 e^{-20c} = 2 - frac{3}{b} )Thus,( e^{-20c} = frac{2}{5} - frac{3}{5b} )Let me denote this as equation (3).Now, going back to equation (1):( a = 20(1 + b) )So, if I can find ( b ) and ( c ), I can find ( a ).But I still have two equations (equation (2) and equation (3)) with two variables ( b ) and ( c ). Let me see if I can solve for ( c ) in terms of ( b ) from equation (3):From equation (3):( e^{-20c} = frac{2}{5} - frac{3}{5b} )Take natural logarithm on both sides:( -20c = lnleft( frac{2}{5} - frac{3}{5b} right) )Thus,( c = -frac{1}{20} lnleft( frac{2}{5} - frac{3}{5b} right) )Let me denote this as equation (4).Now, substitute equation (4) into equation (2):( b = frac{3}{2 - 5 e^{-20c}} )But from equation (3), ( 2 - 5 e^{-20c} = frac{3}{b} ), so:( b = frac{3}{frac{3}{b}} )Wait, that simplifies to ( b = frac{3}{frac{3}{b}} = b ). Hmm, that's just an identity, which doesn't help.Maybe I need another approach. Let me consider that the function ( I(t) ) is a logistic function, which typically has the form ( frac{a}{1 + b e^{-ct}} ). The parameters ( a ), ( b ), and ( c ) can be determined if we have two points and perhaps an inflection point. But since we don't have the inflection point, maybe we can assume that the function is symmetric or something else.Alternatively, let me try to express everything in terms of ( b ). From equation (1), ( a = 20(1 + b) ). From equation (2), ( b = frac{3}{2 - 5 e^{-20c}} ). From equation (3), ( e^{-20c} = frac{2}{5} - frac{3}{5b} ).Let me substitute ( e^{-20c} ) from equation (3) into equation (2):( b = frac{3}{2 - 5 left( frac{2}{5} - frac{3}{5b} right)} )Simplify the denominator:( 2 - 5 left( frac{2}{5} - frac{3}{5b} right) = 2 - 2 + frac{3}{b} = frac{3}{b} )So,( b = frac{3}{frac{3}{b}} = b )Again, this is just an identity, so it doesn't help. Hmm, seems like I'm going in circles.Maybe I need to make an assumption or use another method. Let's consider that the function ( I(t) ) is a logistic curve, which has a characteristic S-shape. The point where the curve is steepest is the inflection point, which occurs at ( t = frac{ln(b)}{c} ). But without knowing the inflection point, it's hard to determine ( c ).Alternatively, maybe I can set up a system of equations. Let me write down the two equations we have:1. ( a = 20(1 + b) )2. ( 50 = frac{a}{1 + b e^{-20c}} )Substitute ( a ) from equation 1 into equation 2:( 50 = frac{20(1 + b)}{1 + b e^{-20c}} )Simplify:( 50(1 + b e^{-20c}) = 20(1 + b) )Divide both sides by 10:( 5(1 + b e^{-20c}) = 2(1 + b) )Expand:( 5 + 5b e^{-20c} = 2 + 2b )Subtract 2:( 3 + 5b e^{-20c} = 2b )Rearrange:( 5b e^{-20c} = 2b - 3 )Divide both sides by ( b ) (assuming ( b neq 0 )):( 5 e^{-20c} = 2 - frac{3}{b} )Let me denote this as equation (5):( 5 e^{-20c} = 2 - frac{3}{b} )Now, from equation (1):( a = 20(1 + b) )We have two equations (equation 5 and equation 1) with three variables. It seems we need another equation or a way to relate ( c ) and ( b ).Wait, perhaps I can express ( e^{-20c} ) in terms of ( b ) from equation (5):( e^{-20c} = frac{2}{5} - frac{3}{5b} )Let me denote this as equation (6).Now, if I can find another relationship, perhaps by considering the derivative at a certain point, but since we don't have information about the rate of change, that might not be feasible.Alternatively, maybe I can assume that the function is symmetric or has a certain property. But without more data points, it's challenging.Wait, perhaps I can express ( c ) in terms of ( b ) from equation (6):( -20c = lnleft( frac{2}{5} - frac{3}{5b} right) )Thus,( c = -frac{1}{20} lnleft( frac{2}{5} - frac{3}{5b} right) )Let me denote this as equation (7).Now, I can express ( a ) in terms of ( b ) from equation (1):( a = 20(1 + b) )So, if I can find ( b ), I can find ( a ) and ( c ).But I still have only one equation (equation 5) with two variables ( b ) and ( c ). It seems like I need another condition or perhaps make an assumption.Wait, maybe I can consider that the function ( I(t) ) approaches ( a ) as ( t ) approaches infinity. So, the maximum influence is ( a ). If I can estimate ( a ), perhaps from the trend.Looking at the data points: at ( t = 0 ), ( I = 20 ); at ( t = 20 ), ( I = 50 ). It's increasing, so ( a ) must be greater than 50. Maybe I can assume that the influence continues to grow and perhaps reaches a certain value, but without more data, it's hard.Alternatively, maybe I can consider the function's behavior at another point. For example, perhaps the midpoint of the logistic curve is at ( t = frac{ln(b)}{c} ). But without knowing the midpoint, it's difficult.Wait, another approach: let's consider that the logistic function can be written as ( I(t) = frac{a}{1 + b e^{-ct}} ). The derivative ( I'(t) ) is ( frac{a b c e^{-ct}}{(1 + b e^{-ct})^2} ). The maximum rate of change occurs at the inflection point, where ( I''(t) = 0 ). But without knowing the inflection point, this might not help.Alternatively, maybe I can use the fact that the function is sigmoidal and passes through the points (0,20) and (20,50). Let me try to solve for ( b ) and ( c ) numerically.From equation (5):( 5 e^{-20c} = 2 - frac{3}{b} )Let me denote ( x = e^{-20c} ). Then,( 5x = 2 - frac{3}{b} )From equation (1):( a = 20(1 + b) )But I need another equation. Wait, perhaps I can express ( x ) in terms of ( b ):( x = frac{2 - frac{3}{b}}{5} )Now, ( x = e^{-20c} ), so ( c = -frac{1}{20} ln x )But without another equation, it's still tricky.Wait, maybe I can assume that the function is symmetric around the midpoint. The midpoint of the logistic curve is at ( t = frac{ln(b)}{c} ). The midpoint value is ( frac{a}{2} ). But we don't know the midpoint value or the time.Alternatively, perhaps I can make an educated guess for ( b ) and solve for ( c ), then check if it makes sense.Let me try to assume a value for ( b ). Let's say ( b = 3 ). Then from equation (1):( a = 20(1 + 3) = 80 )From equation (5):( 5 e^{-20c} = 2 - frac{3}{3} = 2 - 1 = 1 )Thus,( e^{-20c} = frac{1}{5} )So,( -20c = ln(1/5) = -ln 5 )Thus,( c = frac{ln 5}{20} approx frac{1.609}{20} approx 0.08045 )Let me check if this works.So, ( a = 80 ), ( b = 3 ), ( c approx 0.08045 )Now, let's verify the second condition: at ( t = 20 ), ( I(20) = 50 )Compute ( I(20) = frac{80}{1 + 3 e^{-0.08045 * 20}} )Calculate the exponent:( 0.08045 * 20 = 1.609 )So, ( e^{-1.609} approx e^{-ln 5} = 1/5 = 0.2 )Thus,( I(20) = frac{80}{1 + 3 * 0.2} = frac{80}{1 + 0.6} = frac{80}{1.6} = 50 )Perfect! It works. So, the assumption ( b = 3 ) gives us consistent results.Therefore, the constants are:( a = 80 )( b = 3 )( c = frac{ln 5}{20} approx 0.08045 )But to be precise, ( c = frac{ln 5}{20} ), so we can leave it as ( frac{ln 5}{20} ) instead of approximating.So, summarizing:( a = 80 )( b = 3 )( c = frac{ln 5}{20} )Now, moving on to the second part: Voter Turnout Analysis. The function ( V(t) ) is a sinusoidal function with a period of 20 years. It peaks at 30,000 in 1860 (( t = 10 )) and hits a minimum of 10,000 in 1870 (( t = 20 )).A general sinusoidal function can be written as:( V(t) = A sin(Bt + C) + D )Or( V(t) = A cos(Bt + C) + D )Since the problem mentions a sinusoidal function, either sine or cosine can be used. Let's choose cosine for simplicity, as it might align better with the given peak.First, let's find the amplitude ( A ). The peak is 30,000 and the minimum is 10,000. The amplitude is half the difference between the maximum and minimum.So,( A = frac{30,000 - 10,000}{2} = 10,000 )The vertical shift ( D ) is the average of the maximum and minimum:( D = frac{30,000 + 10,000}{2} = 20,000 )So, the function is:( V(t) = 10,000 cos(Bt + C) + 20,000 )Next, the period is 20 years. The period ( T ) of a cosine function is given by ( T = frac{2pi}{B} ). So,( 20 = frac{2pi}{B} )Thus,( B = frac{2pi}{20} = frac{pi}{10} )Now, we need to find the phase shift ( C ). We know that the function peaks at ( t = 10 ). The cosine function reaches its maximum at ( 0 ), ( 2pi ), etc. So, we need to set the argument ( Bt + C ) equal to ( 0 ) when ( t = 10 ).Thus,( frac{pi}{10} * 10 + C = 0 )Simplify:( pi + C = 0 )So,( C = -pi )Therefore, the function is:( V(t) = 10,000 cosleft( frac{pi}{10} t - pi right) + 20,000 )Alternatively, using the identity ( cos(theta - pi) = -cos(theta) ), we can write:( V(t) = -10,000 cosleft( frac{pi}{10} t right) + 20,000 )But let me verify this. At ( t = 10 ):( V(10) = 10,000 cosleft( frac{pi}{10} * 10 - pi right) + 20,000 = 10,000 cos(pi - pi) + 20,000 = 10,000 cos(0) + 20,000 = 10,000 * 1 + 20,000 = 30,000 ). Correct.At ( t = 20 ):( V(20) = 10,000 cosleft( frac{pi}{10} * 20 - pi right) + 20,000 = 10,000 cos(2pi - pi) + 20,000 = 10,000 cos(pi) + 20,000 = 10,000 * (-1) + 20,000 = 10,000 ). Correct.Alternatively, if I had used a sine function, I would have needed to adjust the phase shift accordingly, but the cosine function works well here.So, the voter turnout function is:( V(t) = 10,000 cosleft( frac{pi}{10} t - pi right) + 20,000 )Or simplified as:( V(t) = -10,000 cosleft( frac{pi}{10} t right) + 20,000 )Either form is correct, but perhaps the first form is more explicit.Now, the problem asks to predict the influence and voter turnout in the year 1880. Since 1880 is 30 years after 1850, ( t = 30 ).First, let's compute the influence ( I(30) ):We have ( I(t) = frac{80}{1 + 3 e^{- (ln 5 / 20) t}} )Simplify the exponent:( - (ln 5 / 20) * 30 = - (30/20) ln 5 = - (3/2) ln 5 = ln(5^{-3/2}) = ln(1 / 5^{3/2}) )So,( e^{- (ln 5 / 20) * 30} = e^{ln(1 / 5^{3/2})} = 1 / 5^{3/2} = 1 / (sqrt{5^3}) = 1 / (sqrt{125}) = 1 / (5 sqrt{5}) approx 1 / 11.1803 approx 0.08944 )Thus,( I(30) = frac{80}{1 + 3 * 0.08944} = frac{80}{1 + 0.26832} = frac{80}{1.26832} approx 63.04 )So, the influence in 1880 is approximately 63.04 units.Next, the voter turnout ( V(30) ):Using the function ( V(t) = 10,000 cosleft( frac{pi}{10} * 30 - pi right) + 20,000 )Simplify the argument:( frac{pi}{10} * 30 = 3pi )So,( V(30) = 10,000 cos(3pi - pi) + 20,000 = 10,000 cos(2pi) + 20,000 = 10,000 * 1 + 20,000 = 30,000 )Wait, that's interesting. So, the voter turnout in 1880 is 30,000, same as in 1860. But let me double-check.Alternatively, using the other form:( V(t) = -10,000 cosleft( frac{pi}{10} * 30 right) + 20,000 = -10,000 cos(3pi) + 20,000 = -10,000 * (-1) + 20,000 = 10,000 + 20,000 = 30,000 )Yes, same result. So, the voter turnout peaks again in 1880.Wait, but the period is 20 years, so from 1860 (t=10) to 1880 (t=30) is 20 years, which is one full period. So, it makes sense that the function returns to the peak value.Therefore, the predictions are:Influence: approximately 63.04 unitsVoter Turnout: 30,000But let me express the influence more precisely. Earlier, I approximated ( e^{- (ln 5 / 20) * 30} approx 0.08944 ). Let me compute it more accurately.Compute ( 5^{3/2} = sqrt{125} approx 11.18033989 )So, ( 1 / 11.18033989 approx 0.0894427191 )Thus,( I(30) = frac{80}{1 + 3 * 0.0894427191} = frac{80}{1 + 0.268328157} = frac{80}{1.268328157} )Compute 80 / 1.268328157:1.268328157 * 63 = 80.000000001Wait, that's interesting. So, 1.268328157 * 63 ≈ 80Thus, ( I(30) ≈ 63 )Wait, let me check:1.268328157 * 63:1 * 63 = 630.268328157 * 63 ≈ 16.883Total ≈ 63 + 16.883 ≈ 79.883, which is close to 80. So, actually, ( I(30) = 63 ) exactly.Wait, let me compute 1.268328157 * 63:1.268328157 * 60 = 76.099689421.268328157 * 3 = 3.804984471Total ≈ 76.09968942 + 3.804984471 ≈ 79.90467389Which is approximately 79.905, very close to 80. So, ( I(30) ≈ 63 ) is a good approximation, but actually, it's slightly less than 63.Wait, let me compute 80 / 1.268328157:Using calculator:80 / 1.268328157 ≈ 63.04Yes, so approximately 63.04 units.But since the problem might expect an exact expression, let me see:( I(30) = frac{80}{1 + 3 e^{- (ln 5 / 20) * 30}} = frac{80}{1 + 3 e^{- (3/2) ln 5}} = frac{80}{1 + 3 * 5^{-3/2}} )Simplify ( 5^{-3/2} = 1 / 5^{3/2} = 1 / (5 sqrt{5}) )So,( I(30) = frac{80}{1 + 3 / (5 sqrt{5})} = frac{80}{1 + 3/(5 sqrt{5})} )To rationalize the denominator:Multiply numerator and denominator by ( 5 sqrt{5} ):( I(30) = frac{80 * 5 sqrt{5}}{5 sqrt{5} + 3} )Simplify numerator:( 80 * 5 sqrt{5} = 400 sqrt{5} )Denominator:( 5 sqrt{5} + 3 )So,( I(30) = frac{400 sqrt{5}}{5 sqrt{5} + 3} )We can factor out ( sqrt{5} ) from the denominator:( 5 sqrt{5} + 3 = sqrt{5}(5) + 3 )But it's not straightforward to simplify further. Alternatively, we can rationalize the denominator by multiplying numerator and denominator by ( 5 sqrt{5} - 3 ):( I(30) = frac{400 sqrt{5} (5 sqrt{5} - 3)}{(5 sqrt{5} + 3)(5 sqrt{5} - 3)} )Compute denominator:( (5 sqrt{5})^2 - (3)^2 = 25 * 5 - 9 = 125 - 9 = 116 )Numerator:( 400 sqrt{5} * 5 sqrt{5} = 400 * 5 * 5 = 400 * 25 = 10,000 )( 400 sqrt{5} * (-3) = -1200 sqrt{5} )So,( I(30) = frac{10,000 - 1200 sqrt{5}}{116} )Simplify:Divide numerator and denominator by 4:( frac{2500 - 300 sqrt{5}}{29} )So,( I(30) = frac{2500 - 300 sqrt{5}}{29} )Approximate value:Compute ( sqrt{5} ≈ 2.23607 )So,( 300 * 2.23607 ≈ 670.821 )Thus,( 2500 - 670.821 ≈ 1829.179 )Divide by 29:( 1829.179 / 29 ≈ 63.075 )So, approximately 63.08 units.Therefore, the influence in 1880 is approximately 63.08 units, and the voter turnout is 30,000.To summarize:1. Constants for the influence function: - ( a = 80 ) - ( b = 3 ) - ( c = frac{ln 5}{20} )2. Voter turnout function: - ( V(t) = 10,000 cosleft( frac{pi}{10} t - pi right) + 20,000 )Predictions for 1880 (( t = 30 )):- Influence: approximately 63.08 units- Voter Turnout: 30,000I think that's it. Let me just double-check the calculations to ensure there are no errors.For the influence function:At ( t = 0 ), ( I(0) = 80 / (1 + 3) = 20 ). Correct.At ( t = 20 ), ( I(20) = 80 / (1 + 3 e^{- (ln 5 / 20)*20}) = 80 / (1 + 3 e^{-ln 5}) = 80 / (1 + 3/5) = 80 / (8/5) = 50 ). Correct.At ( t = 30 ), as calculated, approximately 63.08. Correct.For the voter turnout:At ( t = 10 ), ( V(10) = 30,000 ). Correct.At ( t = 20 ), ( V(20) = 10,000 ). Correct.At ( t = 30 ), ( V(30) = 30,000 ). Correct.Everything checks out.</think>