answer:Let's solve the given math problem step-by-step. Problem 1: Determine the value of ( n )Given:- The number of chapters released each month forms an arithmetic sequence.- The first month saw 3 chapters released.- By the ( n )-th month, a total of ( 2n ) chapters have been released.- The total number of chapters released over ( n ) months is 210.Let's denote the common difference of the arithmetic sequence by ( d ).The number of chapters released in the ( k )-th month can be expressed as:[ a_k = 3 + (k - 1)d ]The total number of chapters released in ( n ) months is the sum of the first ( n ) terms of the arithmetic sequence:[ S_n = frac{n}{2} (2a + (n-1)d) ]Here, ( a = 3 ), and the sum ( S_n ) is given to be 210:[ 210 = frac{n}{2} (2 cdot 3 + (n-1)d) ]We also know that the total number of chapters by the ( n )-th month is ( 2n ):[ S_n = 2n ]Thus, we have:[ frac{n}{2} (6 + (n-1)d) = 210 ]Multiplying both sides by 2:[ n (6 + (n-1)d) = 420 ]Since the total number of chapters by the ( n )-th month is ( 2n ):[ S_n = 2n ][ frac{n}{2} (2 cdot 3 + (n-1)d) = 2n ][ n (3 + frac{(n-1)d}{2}) = 2n ][ 3 + frac{(n-1)d}{2} = 2 ][ frac{(n-1)d}{2} = 2 - 3 ][ frac{(n-1)d}{2} = -1 ][ (n-1)d = -2 ][ d = frac{-2}{n-1} ]Substitute ( d ) back into the equation ( n (6 + (n-1)d) = 420 ):[ n left(6 + (n-1) left(frac{-2}{n-1}right)right) = 420 ][ n left(6 - 2right) = 420 ][ n cdot 4 = 420 ][ n = frac{420}{4} ][ n = 105 ]So, the value of ( n ) is:[ n = 15 ] Problem 2: Calculate the total number of weeks required to complete the anime adaptationGiven:- Each episode covers half a chapter.- The anime adaptation begins one month after the release of the ( n )-th chapter.The total number of chapters is ( 2n ):[ text{Total chapters} = 2 times 15 = 30 ]Each episode covers half a chapter, so the total number of episodes required is:[ text{Total episodes} = 30 times 2 = 60 ]Since one episode is released every week, the total number of weeks required is:[ text{Total weeks} = 60 ]Final Answer: The final answer is 60 weeks. I hope it is correct.

answer:Let's solve the given problem step-by-step. Part 1: Finding ( R_A(t) ) and ( R_B(t) )First, let's solve for ( R_A(t) ).The differential equation for Protocol A is:[ frac{dR_A(t)}{dt} = -k_A R_A(t) + alpha_A e^{-beta_A t} ]This is a linear non-homogeneous differential equation. The standard method to solve this is to use an integrating factor.The integrating factor ( mu(t) ) is:[ mu(t) = e^{int k_A dt} = e^{k_A t} ]Multiplying both sides of the differential equation by ( mu(t) ):[ e^{k_A t} frac{dR_A(t)}{dt} + k_A e^{k_A t} R_A(t) = alpha_A e^{(k_A - beta_A)t} ]The left side can be written as a derivative of a product:[ frac{d}{dt} left( e^{k_A t} R_A(t) right) = alpha_A e^{(k_A - beta_A)t} ]Integrate both sides with respect to ( t ):[ e^{k_A t} R_A(t) = alpha_A int e^{(k_A - beta_A)t} dt ][ e^{k_A t} R_A(t) = alpha_A cdot frac{1}{k_A - beta_A} e^{(k_A - beta_A)t} + C ]To find the constant ( C ), use the initial condition ( R_A(0) = 0 ):[ e^{0} R_A(0) = alpha_A cdot frac{1}{k_A - beta_A} e^{0} + C ][ C = 0 - alpha_A cdot frac{1}{k_A - beta_A} ]So, the solution for ( R_A(t) ) is:[ e^{k_A t} R_A(t) = alpha_A cdot frac{1}{k_A - beta_A} e^{(k_A - beta_A)t} - alpha_A cdot frac{1}{k_A - beta_A} ]Or:[ R_A(t) = alpha_A cdot frac{e^{(k_A - beta_A)t} - e^{-k_A t}}{k_A - beta_A} ]Similarly, for Protocol B, we have:[ frac{dR_B(t)}{dt} = -k_B R_B(t) + alpha_B e^{-beta_B t} ]Using the same method:[ mu(t) = e^{k_B t} ]Multiplying through by ( mu(t) ):[ e^{k_B t} frac{dR_B(t)}{dt} + k_B e^{k_B t} R_B(t) = alpha_B e^{(k_B - beta_B)t} ]Rewrite the left side as a derivative of a product:[ frac{d}{dt} left( e^{k_B t} R_B(t) right) = alpha_B e^{(k_B - beta_B)t} ]Integrate both sides:[ e^{k_B t} R_B(t) = alpha_B int e^{(k_B - beta_B)t} dt ][ e^{k_B t} R_B(t) = alpha_B cdot frac{1}{k_B - beta_B} e^{(k_B - beta_B)t} + C ]Using the initial condition ( R_B(0) = 0 ):[ C = - alpha_B cdot frac{1}{k_B - beta_B} ]So, the solution for ( R_B(t) ) is:[ e^{k_B t} R_B(t) = alpha_B cdot frac{e^{(k_B - beta_B)t} - e^{-k_B t}}{k_B - beta_B} ]Or:[ R_B(t) = alpha_B cdot frac{e^{(k_B - beta_B)t} - e^{-k_B t}}{k_B - beta_B} ] Part 2: Calculating ( t_A ) and ( t_B ) for 90% recoveryGiven:[ R_A(t_A) = 90 ][ alpha_A = 50, k_A = 0.1, beta_A = 0.05 ][ 90 = 50 cdot frac{e^{(0.1 - 0.05)t_A} - e^{-0.1 t_A}}{0.1 - 0.05} ][ 90 = 50 cdot 20 left(e^{0.05 t_A} - e^{-0.1 t_A}right) ][ 0.09 = e^{0.05 t_A} - e^{-0.1 t_A} ]Similarly for ( R_B(t_B) = 90 ):[ alpha_B = 60, k_B = 0.15, beta_B = 0.07 ][ 90 = 60 cdot frac{e^{(0.15 - 0.07)t_B} - e^{-0.15 t_B}}{0.15 - 0.07} ][ 90 = 60 cdot 12.5 left(e^{0.08 t_B} - e^{-0.15 t_B}right) ][ 0.12 = e^{0.08 t_B} - e^{-0.15 t_B} ]To solve these equations, numerical methods or iterative methods (such as Newton-Raphson) can be used to find the values of ( t_A ) and ( t_B ).Final Answer: The final answer is ( t_A ) and ( t_B ) need to be calculated numerically. I hope it is correct.

answer:To solve the given problem, we need to determine the constant (k) and then use it to find the time since the artifact was buried. Step 1: Determine the constant (k)We are given the following information:- The initial depth (D_0 = 10) meters.- The depth after 5 centuries (D(5) = 4) meters.The depth function is given by:[ D(t) = D_0 e^{-kt} ]Plugging the given values into the equation:[ 4 = 10 e^{-5k} ]To solve for (k), we first isolate the exponential term:[ e^{-5k} = frac{4}{10} ][ e^{-5k} = 0.4 ]Taking the natural logarithm of both sides:[ -5k = ln(0.4) ][ k = -frac{ln(0.4)}{5} ]Using a calculator to find the natural logarithm:[ ln(0.4) approx -0.91629 ]Therefore:[ k = -frac{-0.91629}{5} ][ k approx 0.18326 ] Step 2: Calculate the time (t) since the artifact was buriedWe are given:- The current depth (D(t) = 2) meters.- The initial depth (D_0 = 10) meters.- The constant (k approx 0.18326).The depth function is:[ D(t) = 10 e^{-0.18326t} ]We need to solve for (t) when (D(t) = 2):[ 2 = 10 e^{-0.18326t} ]Isolating the exponential term:[ e^{-0.18326t} = frac{2}{10} ][ e^{-0.18326t} = 0.2 ]Taking the natural logarithm of both sides:[ -0.18326t = ln(0.2) ][ t = frac{ln(0.2)}{-0.18326} ]Using a calculator to find the natural logarithm:[ ln(0.2) approx -1.60944 ]Therefore:[ t = frac{-1.60944}{-0.18326} ][ t approx 8.78 ]Final Answer: The time since the artifact was first buried is approximately 8.78 centuries. I hope it is correct.

answer:Let's solve the given problems step-by-step. Problem 1: Number of Shares ( P(t) ) at ( t = 10 )The logistic growth model is given by:[ P(t) = frac{L}{1 + e^{-k(t - t_0)}} ]Given:- ( L = 5000 )- ( k = 0.3 )- ( t_0 = 5 )We need to find ( P(t) ) at ( t = 10 ).Substitute the given values into the logistic growth model:[ P(10) = frac{5000}{1 + e^{-0.3(10 - 5)}} ]First, calculate the exponent:[ -0.3(10 - 5) = -0.3 times 5 = -1.5 ]Then, calculate ( e^{-1.5} ):[ e^{-1.5} approx 0.22313 ]Now, substitute this value into the logistic growth model:[ P(10) = frac{5000}{1 + 0.22313} = frac{5000}{1.22313} ]Finally, compute the number of shares:[ P(10) approx frac{5000}{1.22313} approx 4088.2 ] Problem 2: Engagement Rate ( E(t) ) at ( t = 10 )The engagement rate model is given by:[ E(t) = a ln(bt + 1) ]Given:- ( a = 0.8 )- ( b = 0.5 )We need to find ( E(t) ) at ( t = 10 ).Substitute the given values into the engagement rate model:[ E(10) = 0.8 ln(0.5 times 10 + 1) ]Simplify the argument of the logarithm:[ 0.5 times 10 + 1 = 5 + 1 = 6 ]Now, calculate the logarithm:[ ln(6) approx 1.79176 ]Finally, compute the engagement rate:[ E(10) = 0.8 times 1.79176 approx 1.4334 ] Final AnswersFinal Answer: The final answer for the number of shares ( P(t) ) at ( t = 10 ) is approximately 4088.2.Final Answer: The final answer for the engagement rate ( E(t) ) at ( t = 10 ) is approximately 1.4334.

answer:1. To solve the logistic growth differential equation: [ frac{dN}{dt} = rNleft(1 - frac{N}{K}right), ] we start by separating the variables: [ frac{1}{N left(1 - frac{N}{K}right)} frac{dN}{dt} = r. ] To simplify the left-hand side, we use partial fractions. We rewrite: [ frac{1}{N left(1 - frac{N}{K}right)} = frac{1}{N} + frac{1}{K - N}. ] Therefore, the equation becomes: [ left( frac{1}{N} + frac{1}{K - N} right) frac{dN}{dt} = r. ] Separating variables, we get: [ left( frac{1}{N} + frac{1}{K - N} right) dN = r , dt. ] Integrating both sides: [ int left( frac{1}{N} + frac{1}{K - N} right) dN = int r , dt. ] We can split the integral on the left-hand side: [ int frac{1}{N} , dN + int frac{1}{K - N} , dN = int r , dt. ] These integrals are straightforward: [ ln|N| - ln|K - N| = rt + C, ] where ( C ) is the constant of integration. Combining the logarithms: [ ln left| frac{N}{K - N} right| = rt + C. ] Exponentiating both sides to solve for ( N ): [ left| frac{N}{K - N} right| = e^{rt + C}. ] Let ( A = e^C ), then: [ frac{N}{K - N} = A e^{rt}. ] Solving for ( N ): [ N = frac{A e^{rt} (K - N)}{1 + A e^{rt}}. ] Thus: [ N = frac{A K e^{rt}}{1 + A e^{rt}}. ] To find ( A ), we use the initial condition ( N(0) = N_0 ): [ N(0) = frac{A K}{1 + A} = N_0. ] Solving for ( A ): [ A = frac{N_0}{K - N_0}. ] Substituting ( A ) back into the solution: [ N(t) = frac{frac{N_0}{K - N_0} K e^{rt}}{1 + frac{N_0}{K - N_0} e^{rt}}. ] Simplifying: [ N(t) = frac{N_0 K e^{rt}}{(K - N_0) + N_0 e^{rt}}. ]2. Now, we apply the given values: ( N_0 = 50 ), ( K = 5000 ), and ( r = 0.1 ). We need to determine the time ( t ) when the number of adopters is 90% of the carrying capacity: [ N(t) = 0.9K = 0.9 times 5000 = 4500. ] Thus: [ 4500 = frac{50 cdot 5000 e^{0.1t}}{(5000 - 50) + 50 e^{0.1t}}. ] Simplifying: [ 4500 = frac{250000 e^{0.1t}}{4950 + 50 e^{0.1t}}. ] Cross-multiplying: [ 4500 (4950 + 50 e^{0.1t}) = 250000 e^{0.1t}. ] Expanding: [ 4500 times 4950 + 4500 times 50 e^{0.1t} = 250000 e^{0.1t}. ] Simplifying: [ 22275000 + 225000 e^{0.1t} = 250000 e^{0.1t}. ] Isolating ( e^{0.1t} ): [ 22275000 = 250000 e^{0.1t} - 225000 e^{0.1t}. ] [ 22275000 = 25000 e^{0.1t}. ] Solving for ( e^{0.1t} ): [ e^{0.1t} = frac{22275000}{25000} = 891. ] Taking the natural logarithm: [ 0.1t = ln(891). ] Solving for ( t ): [ t = frac{ln(891)}{0.1} approx frac{6.791}{0.1} = 67.91. ]Final Answer: The final answer is ( t approx 67.91 ) years. I hope it is correct.

answer:Let's solve the problem step-by-step: Step 1: Calculate the relative velocity of Car A with respect to Car B.First, we need to convert the velocities of Car A and Car B into vector form.# Car A's Velocity:- Speed: 50 km/h- Direction: 30 degrees north of eastWe can decompose this into its eastward (x) and northward (y) components:[ v_{A_x} = 50 cos(30^circ) = 50 times frac{sqrt{3}}{2} = 25sqrt{3} , text{km/h} ][ v_{A_y} = 50 sin(30^circ) = 50 times frac{1}{2} = 25 , text{km/h} ]So, the velocity vector of Car A is:[ vec{v}_A = (25sqrt{3}, 25) , text{km/h} ]# Car B's Velocity:- Speed: 40 km/h- Direction: 60 degrees south of eastWe can decompose this into its eastward (x) and southward (y) components:[ v_{B_x} = 40 cos(60^circ) = 40 times frac{1}{2} = 20 , text{km/h} ][ v_{B_y} = -40 sin(60^circ) = -40 times frac{sqrt{3}}{2} = -20sqrt{3} , text{km/h} ]So, the velocity vector of Car B is:[ vec{v}_B = (20, -20sqrt{3}) , text{km/h} ]# Relative Velocity of Car A with respect to Car B:[ vec{v}_{A/B} = vec{v}_A - vec{v}_B ][ vec{v}_{A/B} = (25sqrt{3}, 25) - (20, -20sqrt{3}) ][ vec{v}_{A/B} = (25sqrt{3} - 20, 25 + 20sqrt{3}) ]So, the relative velocity vector is:[ vec{v}_{A/B} = (25sqrt{3} - 20, 25 + 20sqrt{3}) , text{km/h} ] Step 2: Determine the resultant force experienced by Car C.First, we need to convert the relative velocities into forces. The force on Car C by Car A and Car B can be found using the formula ( vec{F} = m vec{a} ). Since Car C is initially at rest and struck simultaneously by Car A and Car B, the acceleration ( vec{a} ) can be considered the change in velocity over the collision time ( t ). For simplicity, we'll use the mass and velocity vectors directly.# Force by Car A on Car C:[ vec{F}_A = m_A vec{v}_A ][ m_A = 1500 , text{kg} ][ vec{v}_A = (25sqrt{3}, 25) , text{km/h} ]Converting km/h to m/s by multiplying by ( frac{1000}{3600} ):[ vec{v}_A = (25sqrt{3} times frac{1000}{3600}, 25 times frac{1000}{3600}) = (frac{25000sqrt{3}}{3600}, frac{25000}{3600}) , text{m/s} ][ vec{F}_A = 1500 times (frac{25000sqrt{3}}{3600}, frac{25000}{3600}) ][ vec{F}_A = (frac{37500000sqrt{3}}{3600}, frac{37500000}{3600}) ][ vec{F}_A = (10416.67sqrt{3}, 10416.67) , text{N} ]# Force by Car B on Car C:[ vec{F}_B = m_B vec{v}_B ][ m_B = 1200 , text{kg} ][ vec{v}_B = (20, -20sqrt{3}) , text{km/h} ]Converting km/h to m/s by multiplying by ( frac{1000}{3600} ):[ vec{v}_B = (20 times frac{1000}{3600}, -20sqrt{3} times frac{1000}{3600}) = (frac{20000}{3600}, frac{-20000sqrt{3}}{3600}) , text{m/s} ][ vec{F}_B = 1200 times (frac{20000}{3600}, frac{-20000sqrt{3}}{3600}) ][ vec{F}_B = (frac{24000000}{3600}, frac{-24000000sqrt{3}}{3600}) ][ vec{F}_B = (6666.67, -6666.67sqrt{3}) , text{N} ]# Resultant Force on Car C:[ vec{F}_{text{resultant}} = vec{F}_A + vec{F}_B ][ vec{F}_{text{resultant}} = (10416.67sqrt{3}, 10416.67) + (6666.67, -6666.67sqrt{3}) ][ vec{F}_{text{resultant}} = (10416.67sqrt{3} + 6666.67, 10416.67 - 6666.67sqrt{3}) ]The resultant force vector is:[ vec{F}_{text{resultant}} = (10416.67sqrt{3} + 6666.67, 10416.67 - 6666.67sqrt{3}) , text{N} ]Final Answer: The final answer is ( (10416.67sqrt{3} + 6666.67, 10416.67 - 6666.67sqrt{3}) , text{N} ). I hope it is correct.