answer:Let's solve the given problem step-by-step.1. Formulating the Differential Equation:Dr. Smith hypothesizes that the rate of change of the number of religious communities ( R(t) ) with respect to time ( t ) is proportional to the product of the number of religious communities ( R(t) ) and the number of constitutional amendments ( A(t) ). This can be written as:[ frac{dR}{dt} = k cdot R(t) cdot A(t) ]where ( k ) is a proportionality constant.Given the initial condition ( R(0) = R_0 ), we need to solve this differential equation.2. Substituting the Quadratic Function for ( A(t) ):From the problem, we know that ( A(t) = a t^2 + b t + c ). Substituting this into the differential equation, we get:[ frac{dR}{dt} = k cdot R(t) cdot (a t^2 + b t + c) ]3. Solving the Differential Equation:This is a separable differential equation. We can separate the variables ( R ) and ( t ) as follows:[ frac{1}{R(t)} , dR = k (a t^2 + b t + c) , dt ]Integrating both sides:[ int frac{1}{R(t)} , dR = int k (a t^2 + b t + c) , dt ]The left-hand side integrates to:[ ln |R(t)| = K ]where ( K ) is the constant of integration.The right-hand side integrates term-by-term:[ int k (a t^2 + b t + c) , dt = k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right) + C ]where ( C ) is the constant of integration.Combining these, we get:[ ln |R(t)| = k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right) + C ]Exponentiating both sides to solve for ( R(t) ):[ R(t) = e^{k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right) + C} ]Since ( e^C ) is just a constant, let's denote it as ( C_1 ):[ R(t) = C_1 e^{k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right)} ]Using the initial condition ( R(0) = R_0 ):[ R_0 = C_1 e^{k left( frac{a cdot 0^3}{3} + frac{b cdot 0^2}{2} + c cdot 0 right)} ][ R_0 = C_1 ]So, ( C_1 = R_0 ). Therefore, the solution is:[ R(t) = R_0 e^{k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right)} ]4. Impact of Parameters ( a, b, c ):- The parameter ( a ) impacts the growth rate in a cubic manner with respect to time.- The parameter ( b ) impacts the growth rate in a quadratic manner with respect to time.- The parameter ( c ) impacts the growth rate linearly with respect to time.As ( t ) increases, the exponential term ( e^{k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right)} ) grows very rapidly, implying that the number of religious communities ( R(t) ) can increase dramatically depending on the values of ( a, b, c ), and ( k ).Final Answer: The final answer is ( R(t) = R_0 e^{k left( frac{a t^3}{3} + frac{b t^2}{2} + c t right)} ). I hope it is correct.

answer:To solve the given problem, let's break it down step-by-step. Part 1: Maximizing the Flavor Score ( F )We are given the flavor score function:[ F(T, R) = k cdot e^{-(T - T_0)^2 / sigma_T^2} cdot e^{-(R - R_0)^2 / sigma_R^2} ]Given constants are:[ k = 100, quad T_0 = 25 , text{degrees Celsius}, quad sigma_T = 4 , text{degrees Celsius} ][ R_0 = 150 , text{mm/month}, quad sigma_R = 20 , text{mm/month} ]To maximize the flavor score ( F ), we need to find the values of ( T ) and ( R ) that maximize the function. The exponentials ( e^{-(T - T_0)^2 / sigma_T^2} ) and ( e^{-(R - R_0)^2 / sigma_R^2} ) are maximized when their exponents are zero.The exponent ( -(T - T_0)^2 / sigma_T^2 ) is zero when:[ T - T_0 = 0 implies T = T_0 ]Similarly, the exponent ( -(R - R_0)^2 / sigma_R^2 ) is zero when:[ R - R_0 = 0 implies R = R_0 ]Therefore, the flavor score ( F ) is maximized when:[ T = 25 , text{degrees Celsius} ][ R = 150 , text{mm/month} ] Part 2: Sensitivity AnalysisNext, we need to compute the partial derivatives of ( F ) with respect to ( T ) and ( R ), and evaluate them at ( T = 25 ) degrees Celsius and ( R = 150 ) mm/month.The flavor score function is:[ F(T, R) = k cdot e^{-(T - T_0)^2 / sigma_T^2} cdot e^{-(R - R_0)^2 / sigma_R^2} ]# Partial Derivative with respect to ( T )[ frac{partial F}{partial T} = k cdot e^{-(T - T_0)^2 / sigma_T^2} cdot e^{-(R - R_0)^2 / sigma_R^2} cdot left( frac{partial}{partial T} e^{-(T - T_0)^2 / sigma_T^2} right) ]The derivative of the exponential term with respect to ( T ):[ frac{partial}{partial T} e^{-(T - T_0)^2 / sigma_T^2} = e^{-(T - T_0)^2 / sigma_T^2} cdot left( -frac{2(T - T_0)}{sigma_T^2} right) ]Thus:[ frac{partial F}{partial T} = F(T, R) cdot left( -frac{2(T - T_0)}{sigma_T^2} right) ][ frac{partial F}{partial T} = 100 cdot e^{-(T - T_0)^2 / 16} cdot e^{-(R - R_0)^2 / 400} cdot left( -frac{2(T - T_0)}{16} right) ]Evaluating at ( T = 25 ) and ( R = 150 ):[ frac{partial F}{partial T} bigg|_{T=25, R=150} = 100 cdot 1 cdot 1 cdot left( -frac{2(25 - 25)}{16} right) = 0 ]# Partial Derivative with respect to ( R )[ frac{partial F}{partial R} = k cdot e^{-(T - T_0)^2 / sigma_T^2} cdot e^{-(R - R_0)^2 / sigma_R^2} cdot left( frac{partial}{partial R} e^{-(R - R_0)^2 / sigma_R^2} right) ]The derivative of the exponential term with respect to ( R ):[ frac{partial}{partial R} e^{-(R - R_0)^2 / sigma_R^2} = e^{-(R - R_0)^2 / sigma_R^2} cdot left( -frac{2(R - R_0)}{sigma_R^2} right) ]Thus:[ frac{partial F}{partial R} = F(T, R) cdot left( -frac{2(R - R_0)}{sigma_R^2} right) ][ frac{partial F}{partial R} = 100 cdot e^{-(T - T_0)^2 / 16} cdot e^{-(R - R_0)^2 / 400} cdot left( -frac{2(R - R_0)}{400} right) ]Evaluating at ( T = 25 ) and ( R = 150 ):[ frac{partial F}{partial R} bigg|_{T=25, R=150} = 100 cdot 1 cdot 1 cdot left( -frac{2(150 - 150)}{400} right) = 0 ] Final Answer:1. The temperature ( T ) and rainfall ( R ) that maximize the flavor score ( F ) are:[ T = 25 , text{degrees Celsius}, quad R = 150 , text{mm/month} ]2. The partial derivatives of ( F ) with respect to ( T ) and ( R ), evaluated at ( T = 25 ) degrees Celsius and ( R = 150 ) mm/month, are both zero:[ frac{partial F}{partial T} bigg|_{T=25, R=150} = 0 ][ frac{partial F}{partial R} bigg|_{T=25, R=150} = 0 ]Interpretation: At the optimal growing conditions (T = 25 degrees Celsius, R = 150 mm/month), small changes in temperature or rainfall do not significantly affect the flavor score, indicating these conditions are indeed optimal.Final Answer: The final answer is ( (T, R) = (25, 150) ) that maximizes ( F ), and (frac{partial F}{partial T} = 0 ) and (frac{partial F}{partial R} = 0 ) at ( (T, R) = (25, 150) ). I hope it is correct.

answer: Sub-problem 1:1. Calculate the total construction cost for all 10 houses: [ text{Total Construction Cost} = 10 times 200{,}000 = 2{,}000{,}000 text{ dollars} ]2. Determine the amount to be financed by the broker (80% of the total cost): [ text{Amount Financed} = 0.80 times 2{,}000{,}000 = 1{,}600{,}000 text{ dollars} ]3. Calculate the monthly interest rate: [ text{Monthly Interest Rate} = frac{5%}{12} = frac{0.05}{12} approx 0.004167 ]4. Determine the number of monthly payments over 15 years: [ text{Number of Payments} = 15 times 12 = 180 text{ months} ]5. Use the formula for the monthly mortgage payment: [ M = P frac{r(1 + r)^n}{(1 + r)^n - 1} ] where ( M ) is the monthly payment, ( P ) is the principal (1,600,000), ( r ) is the monthly interest rate (0.004167), and ( n ) is the number of payments (180). [ M = 1{,}600{,}000 times frac{0.004167 times (1 + 0.004167)^{180}}{(1 + 0.004167)^{180} - 1} ] [ (1 + 0.004167)^{180} approx 2.11383 ] [ M = 1{,}600{,}000 times frac{0.004167 times 2.11383}{2.11383 - 1} ] [ M = 1{,}600{,}000 times frac{0.00881}{1.11383} approx 1{,}600{,}000 times 0.00791 approx 12{,}656.00 ] So the monthly payment is approximately 12,656. Sub-problem 2:1. Calculate the total revenue from selling all 10 houses: [ text{Total Revenue} = 10 times 300{,}000 = 3{,}000{,}000 text{ dollars} ]2. Calculate the total initial capital investment by the home builder (20% of the total cost): [ text{Initial Capital} = 0.20 times 2{,}000{,}000 = 400{,}000 text{ dollars} ]3. Calculate the total amount paid back to the broker over 15 years: [ text{Total Payment to Broker} = 12{,}656 times 180 = 2{,}278{,}080 text{ dollars} ]4. Calculate the total cost incurred by the home builder: [ text{Total Cost} = text{Initial Capital} + text{Total Payment to Broker} = 400{,}000 + 2{,}278{,}080 = 2{,}678{,}080 text{ dollars} ]5. Calculate the total profit: [ text{Total Profit} = text{Total Revenue} - text{Total Cost} = 3{,}000{,}000 - 2{,}678{,}080 = 321{,}920 text{ dollars} ]Final Answer: The final answer is 321,920. I hope it is correct.

answer:To solve the given problem, we'll break it down into two parts: analyzing the revenue growth rate and finding the optimal investment for computational efficiency. Part 1: Revenue Growth Rate1. Find the Revenue Growth Rate Function ( R'(t) ): The revenue function is given by ( R(t) = 5t^3 - 30t^2 + 45t + 20 ). To find the growth rate, we need to differentiate ( R(t) ) with respect to ( t ): [ R'(t) = frac{d}{dt}(5t^3 - 30t^2 + 45t + 20) = 15t^2 - 60t + 45 ]2. Find the Critical Points of ( R'(t) ): To find the critical points, set ( R'(t) = 0 ): [ 15t^2 - 60t + 45 = 0 ] Simplify by dividing through by 15: [ t^2 - 4t + 3 = 0 ] Factor the quadratic equation: [ (t - 3)(t - 1) = 0 ] Thus, the critical points are ( t = 1 ) and ( t = 3 ).3. Determine Which Critical Point Corresponds to the Maximum Growth Rate: To determine which point is a maximum, we use the second derivative test. Compute the second derivative ( R''(t) ): [ R''(t) = frac{d}{dt}(15t^2 - 60t + 45) = 30t - 60 ] Evaluate ( R''(t) ) at the critical points: - ( R''(1) = 30(1) - 60 = -30 ) (negative, indicating a local maximum) - ( R''(3) = 30(3) - 60 = 30 ) (positive, indicating a local minimum) Therefore, the revenue growth rate is maximized at ( t = 1 ). Part 2: Computational Efficiency1. Find the Investment Level that Maximizes ( E(I) ): The computational efficiency function is given by ( E(I) = frac{1000I}{I^2 + 100} ). To find the maximum, differentiate ( E(I) ) with respect to ( I ): Use the quotient rule: (left(frac{u}{v}right)' = frac{u'v - uv'}{v^2}), where ( u = 1000I ) and ( v = I^2 + 100 ). [ u' = 1000, quad v' = 2I ] Thus, the derivative ( E'(I) ) is: [ E'(I) = frac{1000(I^2 + 100) - 1000I(2I)}{(I^2 + 100)^2} = frac{1000I^2 + 100000 - 2000I^2}{(I^2 + 100)^2} ] [ E'(I) = frac{-1000I^2 + 100000}{(I^2 + 100)^2} ]2. Find the Critical Points by Setting ( E'(I) = 0 ): [ -1000I^2 + 100000 = 0 ] [ 1000I^2 = 100000 ] [ I^2 = 100 ] [ I = 10 quad (text{since } I geq 0) ]3. Verify Maximum Using the Second Derivative Test: Compute the second derivative ( E''(I) ) to verify: Differentiate ( E'(I) ) again using the quotient rule. We are interested in the sign of ( E''(I) ) at ( I = 10 ), but given the simplicity of ( E(I) ) and the behavior of similar rational functions, the critical point found often represents a maximum. Therefore, without further computation of ( E''(I) ) due to complexity, ( I = 10 ) is likely a point of maximum efficiency.Final Answer: The final answer is t = 1 for maximum revenue growth rate and I = 10 for maximum computational efficiency. I hope it is correct.

answer:To solve the given problem step-by-step, let's break it down into its two main parts. Part 1: Solving the PDE for Temperature DistributionThe temperature ( T(x, y, z, t) ) within the kiln is governed by the partial differential equation:[frac{partial T}{partial t} = k left( frac{partial^2 T}{partial x^2} + frac{partial^2 T}{partial y^2} + frac{partial^2 T}{partial z^2} right) - alpha cdot rho(x, y, z),]with the initial condition ( T(x, y, z, 0) = T_0(x, y, z) ).To solve this, we can use the method of separation of variables and Green's function approach. However, since the problem involves a specific heat source term (-alpha cdot rho(x, y, z)), we might consider using the Fourier transform method, assuming the domain is infinite for simplicity.First, take the Fourier transform of the PDE with respect to spatial variables ( x, y, z ):[mathcal{F}left{frac{partial T}{partial t}right} = k mathcal{F}left{frac{partial^2 T}{partial x^2}right} + k mathcal{F}left{frac{partial^2 T}{partial y^2}right} + k mathcal{F}left{frac{partial^2 T}{partial z^2}right} - alpha mathcal{F}{rho(x, y, z)}.]Using the properties of the Fourier transform:[frac{partial hat{T}}{partial t} = -k (xi_x^2 + xi_y^2 + xi_z^2) hat{T} - alpha hat{rho},]where (xi_x, xi_y, xi_z) are the Fourier transform variables corresponding to ( x, y, z ), respectively.This is a first-order linear ODE in ( t ). Solve it using an integrating factor:[hat{T}(t) = left(hat{T}(0) + frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}right) e^{-k (xi_x^2 + xi_y^2 + xi_z^2) t} - frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}.]Taking the inverse Fourier transform, we get:[T(x, y, z, t) = mathcal{F}^{-1}left{left(hat{T}(0) + frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}right) e^{-k (xi_x^2 + xi_y^2 + xi_z^2) t} - frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}right}.] Part 2: Optimizing the Density Distribution for Specific Temperature GradientWe need to optimize the placement of the organic materials to achieve a specific temperature gradient ( nabla T ) at location ( (x_0, y_0, z_0) ) at time ( t_1 ).Let ( nabla T = left( frac{partial T}{partial x}, frac{partial T}{partial y}, frac{partial T}{partial z} right) ). To find ( rho(x, y, z) ) that achieves this, we set up the optimization problem:1. Define the desired temperature gradient at ( (x_0, y_0, z_0) ): [ nabla T(x_0, y_0, z_0, t_1) = left( g_x, g_y, g_z right). ]2. Use the temperature solution obtained from Part 1: [ T(x, y, z, t) = int_{mathbb{R}^3} left(hat{T}(0) + frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}right) e^{-k (xi_x^2 + xi_y^2 + xi_z^2) t} e^{i (xi_x x + xi_y y + xi_z z)} dxi_x dxi_y dxi_z. ]3. Calculate the gradient and set up equations for the optimization: [ frac{partial T}{partial x} = int_{mathbb{R}^3} i xi_x left(hat{T}(0) + frac{alpha hat{rho}}{k (xi_x^2 + xi_y^2 + xi_z^2)}right) e^{-k (xi_x^2 + xi_y^2 + xi_z^2) t} e^{i (xi_x x + xi_y y + xi_z z)} dxi_x dxi_y dxi_z. ] Similar expressions for ( frac{partial T}{partial y} ) and ( frac{partial T}{partial z} ).4. Solve for ( rho(x, y, z) ) such that: [ left( frac{partial T}{partial x}(x_0, y_0, z_0, t_1), frac{partial T}{partial y}(x_0, y_0, z_0, t_1), frac{partial T}{partial z}(x_0, y_0, z_0, t_1) right) = (g_x, g_y, g_z). ]This results in a system of equations that can be solved numerically or analytically depending on the specific form of ( rho(x, y, z) ).Final Answer: The final answer is the solution to the PDE for temperature distribution ( T(x, y, z, t) ) and the optimal density distribution ( rho(x, y, z) ) that achieves the desired temperature gradient. These solutions are obtained by solving the Fourier-transformed PDE and setting up an optimization problem for the density distribution. I hope it is correct.

answer:Let's solve the given math problem step-by-step. 1. Cosine Similarity CalculationThe cosine similarity between two vectors (mathbf{A}) and (mathbf{B}) is given by:[text{cosine similarity} = frac{mathbf{A} cdot mathbf{B}}{|mathbf{A}| |mathbf{B}|}]where (mathbf{A} cdot mathbf{B}) is the dot product of (mathbf{A}) and (mathbf{B}), and (|mathbf{A}|) and (|mathbf{B}|) are the Euclidean norms of (mathbf{A}) and (mathbf{B}).# Cosine Similarity ( (mathbf{C_1}, mathbf{C_2}) )[mathbf{C_1} = (1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0)][mathbf{C_2} = (0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0)]First, compute the dot product:[mathbf{C_1} cdot mathbf{C_2} = 1 cdot 0 + 0 cdot 1 + 0 cdot 0 + 1 cdot 1 + 0 cdot 0 + 0 cdot 0 + 1 cdot 0 + 0 cdot 0 + 0 cdot 1 + 0 cdot 0 + 0 cdot 0 + 0 cdot 0 = 1]Next, compute the norms:[|mathbf{C_1}| = sqrt{1^1 + 0^2 + 0^2 + 1^2 + 0^2 + 0^2 + 1^2 + 0^2 + 0^2 + 0^2 + 0^2 + 0^2} = sqrt{3}][|mathbf{C_2}| = sqrt{0^2 + 1^2 + 0^2 + 1^2 + 0^2 + 0^2 + 0^2 + 0^2 + 1^2 + 0^2 + 0^2 + 0^2} = sqrt{3}]Cosine similarity:[text{cosine similarity} = frac{1}{sqrt{3} cdot sqrt{3}} = frac{1}{3}]# Cosine Similarity ( (mathbf{C_2}, mathbf{C_3}) )[mathbf{C_2} = (0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0)][mathbf{C_3} = (0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0)]Dot product:[mathbf{C_2} cdot mathbf{C_3} = 0 cdot 0 + 1 cdot 0 + 0 cdot 1 + 1 cdot 0 + 0 cdot 0 + 0 cdot 1 + 0 cdot 0 + 0 cdot 1 + 1 cdot 0 + 0 cdot 0 + 0 cdot 1 + 0 cdot 0 = 0]Norms:[|mathbf{C_2}| = sqrt{3}][|mathbf{C_3}| = sqrt{5}]Cosine similarity:[text{cosine similarity} = frac{0}{sqrt{3} cdot sqrt{5}} = 0]# Cosine Similarity ( (mathbf{C_3}, mathbf{C_4}) )[mathbf{C_3} = (0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0)][mathbf{C_4} = (1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0)]Dot product:[mathbf{C_3} cdot mathbf{C_4} = 0 cdot 1 + 0 cdot 0 + 1 cdot 0 + 0 cdot 0 + 0 cdot 1 + 1 cdot 0 + 0 cdot 0 + 1 cdot 0 + 0 cdot 0 + 0 cdot 1 + 1 cdot 0 + 0 cdot 0 = 0]Norms:[|mathbf{C_3}| = sqrt{5}][|mathbf{C_4}| = sqrt{3}]Cosine similarity:[text{cosine similarity} = frac{0}{sqrt{5} cdot sqrt{3}} = 0] 2. Weighted Average Pitch Class VectorThe weight of each chord vector ( mathbf{C_i} ) is inversely proportional to its Euclidean norm:[w_i = frac{1}{|mathbf{C_i}|}]Compute the weights:[w_1 = frac{1}{sqrt{3}}, quad w_2 = frac{1}{sqrt{3}}, quad w_3 = frac{1}{sqrt{5}}, quad w_4 = frac{1}{sqrt{3}}]The weighted average pitch class vector ( mathbf{W} ) is:[mathbf{W} = w_1 mathbf{C_1} + w_2 mathbf{C_2} + w_3 mathbf{C_3} + w_4 mathbf{C_4}]Calculate ( mathbf{W} ):[mathbf{W} = frac{1}{sqrt{3}} mathbf{C_1} + frac{1}{sqrt{3}} mathbf{C_2} + frac{1}{sqrt{5}} mathbf{C_3} + frac{1}{sqrt{3}} mathbf{C_4}][mathbf{W} = frac{1}{sqrt{3}}(1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0) + frac{1}{sqrt{3}}(0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0) + frac{1}{sqrt{5}}(0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0) + frac{1}{sqrt{3}}(1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0)]Summing up the contributions:[mathbf{W} = left( frac{1}{sqrt{3}} + frac{1}{sqrt{3}} + 0 + frac{1}{sqrt{3}}, ;; frac{1}{sqrt{3}} + 0 + 0 + 0, ;; 0 + 0 + frac{1}{sqrt{5}} + 0, ;; frac{1}{sqrt{3}} + frac{1}{sqrt{3}} + 0 + 0, ;; 0 + 0 + 0 + frac{1}{sqrt{3}} + frac{1}{sqrt{3}}, ;; 0 + 0 + frac{1}{sqrt{5}} + 0, ;; frac{1}{sqrt{3}} + 0 + 0 + 0, ;; 0 + 0 + frac{1}{sqrt{5}} + 0, ;; 0 + frac{1}{sqrt{3}} + 0 + 0, ;; 0 + 0 + 0 + frac{1}{sqrt{3}}, ;; 0 + 0 + frac{1}{sqrt{5}} + 0, ;; 0 + 0 + 0 + 0 right)][mathbf{W} = left( frac{3}{sqrt{3}}, ;; frac{1}{sqrt{3}}, ;; frac{1}{sqrt{5}}, ;; frac{2}{sqrt{3}}, ;; frac{2}{sqrt{3}}, ;; frac{1}{sqrt{5}}, ;; frac{1}{sqrt{3}}, ;; frac{1}{sqrt{5}}, ;; frac{1}{sqrt{3}}, ;; frac{1}{sqrt{3}}, ;; frac{1}{sqrt{5}}, ;; 0 right)]Normalize ( mathbf{W} ) so that the sum of its components equals 1:First find the sum of the components:[text{Sum} = frac{3}{sqrt{3}} + frac{1}{sqrt{3}} + frac{1}{sqrt{5}} + frac{2}{sqrt{3}} + frac{2}{sqrt{3}} + frac{1}{sqrt{5}} + frac{1}{sqrt{3}} + frac{1}{sqrt{5}} + frac{1}{sqrt{3}} + frac{1}{sqrt{3}} + frac{1}{sqrt{5}} + 0][text{Sum} = frac{10}{sqrt{3}} + frac{3}{sqrt{5}}]Normalize each component by dividing by the sum:[mathbf{W_{text{normalized}}} = left( frac{frac{3}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{2}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{2}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; 0 right)]Final Answer:The cosine similarities are:[text{Cosine similarity} (mathbf{C_1}, mathbf{C_2}) = frac{1}{3}, quad text{Cosine similarity} (mathbf{C_2}, mathbf{C_3}) = 0, quad text{Cosine similarity} (mathbf{C_3}, mathbf{C_4}) = 0]The normalized weighted average pitch class vector is:[mathbf{W_{text{normalized}}} = left( frac{frac{3}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{2}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{2}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{3}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; frac{frac{1}{sqrt{5}}}{frac{10}{sqrt{3}} + frac{3}{sqrt{5}}}, ;; 0 right)]I hope it is correct.