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question:A retired UK parliament member from the Labour party who represented the area of Cleveland decided to use his knowledge and experience to analyze the voting patterns in his constituency over the past three election cycles. He noted the following observations:1. The number of votes received by the Labour party in each election can be modeled by the function ( L(n) = 2000 + 300n + 50n^2 ), where ( n ) is the number of election cycles since he first took office.2. The total number of votes cast in each election cycle can be modeled by the function ( T(n) = 10000 + 500n ).Sub-problems:1. Determine the rate of change of the number of votes received by the Labour party with respect to the election cycle number ( n ) after the third election (i.e., when ( n = 3 )). What does this rate of change signify in the context of the elections?2. Calculate the percentage of total votes received by the Labour party in the third election cycle. Based on this calculation and the given functions, discuss whether the Labour party's share of the total votes is increasing or decreasing over time.
answer:Let's solve each sub-problem step-by-step. Sub-problem 1: Rate of Change of Votes Received by the Labour PartyWe need to find the rate of change of the number of votes received by the Labour party with respect to the election cycle number ( n ) after the third election (i.e., when ( n = 3 )).The function for the number of votes received by the Labour party is:[ L(n) = 2000 + 300n + 50n^2 ]To find the rate of change, we need to compute the derivative of ( L(n) ) with respect to ( n ):[ L'(n) = frac{d}{dt}(2000 + 300n + 50n^2) ]Using the power rule for differentiation:[ L'(n) = 0 + 300 + 100n ][ L'(n) = 300 + 100n ]Now, evaluate ( L'(n) ) at ( n = 3 ):[ L'(3) = 300 + 100 cdot 3 ][ L'(3) = 300 + 300 ][ L'(3) = 600 ]This means the rate of change of the number of votes received by the Labour party at the third election cycle is 600 votes per election cycle. Interpretation:The rate of change signifies that, at the third election cycle, the number of votes received by the Labour party is increasing by 600 votes per election cycle. Sub-problem 2: Percentage of Total Votes Received by the Labour PartyFirst, we need to find the number of votes received by the Labour party and the total number of votes cast in the third election cycle.# Votes received by the Labour party in the third election:[ L(3) = 2000 + 300 cdot 3 + 50 cdot 3^2 ][ L(3) = 2000 + 900 + 450 ][ L(3) = 3350 ]# Total number of votes cast in the third election:[ T(3) = 10000 + 500 cdot 3 ][ T(3) = 10000 + 1500 ][ T(3) = 11500 ]Now, we calculate the percentage of total votes received by the Labour party:[ text{Percentage} = left( frac{L(3)}{T(3)} right) times 100 ][ text{Percentage} = left( frac{3350}{11500} right) times 100 ][ text{Percentage} approx 29.13% ] Determine if Labour Party's Share is Increasing or DecreasingTo determine if the Labour party's share of the total votes is increasing or decreasing over time, we can compute the percentage of total votes received in different election cycles and observe the trend.For the first election cycle (( n = 0 )):[ L(0) = 2000 + 300 cdot 0 + 50 cdot 0^2 = 2000 ][ T(0) = 10000 + 500 cdot 0 = 10000 ][ text{Percentage} = left( frac{2000}{10000} right) times 100 = 20% ]For the second election cycle (( n = 1 )):[ L(1) = 2000 + 300 cdot 1 + 50 cdot 1^2 = 2350 ][ T(1) = 10000 + 500 cdot 1 = 10500 ][ text{Percentage} = left( frac{2350}{10500} right) times 100 approx 22.38% ]We observe that the percentage increased from 20% in the first cycle to approximately 22.38% in the second cycle, and then to approximately 29.13% in the third cycle. This indicates that the Labour party's share of the total votes is increasing over time.Final Answer: The final answer is 600 and approximately 29.13%. I hope it is correct.
question:Brentwood is planning to install a new sustainable urban garden that aligns with local environmental goals and enhances the community's engagement with fitness and arts. The garden will feature a unique artistic layout and include sustainable energy sources such as solar panels.1. The area designated for the garden forms a region bounded by the parametric equations: [ x(t) = 3 cos(t) + cos(3t), quad y(t) = 3 sin(t) - sin(3t) ] where ( t ) ranges from ( 0 ) to ( 2pi ). Calculate the enclosed area of this region.2. The garden will also include a set of solar panels that are arranged in a spiral pattern described by the polar equation ( r(theta) = e^{0.05theta} ), where ( theta ) ranges from ( 0 ) to ( 4pi ). Determine the total length of the spiral path taken by the solar panels.
answer:Let's solve the given problem step-by-step. Problem 1: Enclosed Area of the RegionThe parametric equations for the region are given by:[x(t) = 3 cos(t) + cos(3t), quad y(t) = 3 sin(t) - sin(3t)]where ( t ) ranges from ( 0 ) to ( 2pi ).The formula for the area enclosed by a parametric curve ( (x(t), y(t)) ) is:[A = frac{1}{2} int_{a}^{b} left( x(t) frac{dy}{dt} - y(t) frac{dx}{dt} right) dt]First, we need to find the derivatives (frac{dx}{dt}) and (frac{dy}{dt}):[frac{dx}{dt} = -3 sin(t) - 3 sin(3t)][frac{dy}{dt} = 3 cos(t) - 3 cos(3t)]Next, substitute ( x(t) ), ( y(t) ), (frac{dx}{dt}), and (frac{dy}{dt}) into the area formula:[A = frac{1}{2} int_{0}^{2pi} left[ (3 cos(t) + cos(3t))(3 cos(t) - 3 cos(3t)) - (3 sin(t) - sin(3t))(-3 sin(t) - 3 sin(3t)) right] dt]Simplify the integrand:[A = frac{1}{2} int_{0}^{2pi} left[ (3 cos(t) + cos(3t))(3 cos(t) - 3 cos(3t)) + (3 sin(t) - sin(3t))(3 sin(t) + 3 sin(3t)) right] dt][A = frac{1}{2} int_{0}^{2pi} left[ 9 cos^2(t) - 9 cos(t) cos(3t) + 3 cos(t) cos(3t) - 3 cos^2(3t) + 9 sin^2(t) - 9 sin(t) sin(3t) - sin(3t) sin(3t) right] dt][A = frac{1}{2} int_{0}^{2pi} left[ 9 cos^2(t) + 9 sin^2(t) - 3 cos^2(3t) - sin^2(3t) right] dt]Using the Pythagorean identities ( cos^2(t) + sin^2(t) = 1 ):[A = frac{1}{2} int_{0}^{2pi} left[ 9 - 3 cos^2(3t) - sin^2(3t) right] dt]Given the symmetry and periodicity, we know that the integral of (cos^2(3t)) and (sin^2(3t)) over one period is (pi). Therefore:[A = frac{1}{2} int_{0}^{2pi} 6 dt = frac{1}{2} cdot 6 cdot 2pi = 6pi]Final Answer: The final answer is 6pi. I hope it is correct. Problem 2: Total Length of the Spiral PathThe polar equation for the spiral is:[r(theta) = e^{0.05theta}]where ( theta ) ranges from ( 0 ) to ( 4pi ).The formula for the length ( L ) of a curve given in polar coordinates ( r(theta) ) is:[L = int_{a}^{b} sqrt{ left( frac{dr}{dtheta} right)^2 + r^2 } , dtheta]First, we calculate (frac{dr}{dtheta}):[frac{dr}{dtheta} = 0.05 e^{0.05theta}]Substitute (frac{dr}{dtheta}) and ( r(theta) ) into the length formula:[L = int_{0}^{4pi} sqrt{ left( 0.05 e^{0.05theta} right)^2 + left( e^{0.05theta} right)^2 } , dtheta][L = int_{0}^{4pi} sqrt{ 0.0025 e^{0.1theta} + e^{0.1theta} } , dtheta][L = int_{0}^{4pi} sqrt{ e^{0.1theta} (0.0025 + 1) } , dtheta][L = int_{0}^{4pi} sqrt{ 1.0025 e^{0.1theta} } , dtheta][L = int_{0}^{4pi} sqrt{ 1.0025 } cdot e^{0.05theta} , dtheta][L = sqrt{ 1.0025 } int_{0}^{4pi} e^{0.05theta} , dtheta][sqrt{ 1.0025 } approx 1.00125][L approx 1.00125 int_{0}^{4pi} e^{0.05theta} , dtheta]The integral of ( e^{0.05theta} ) is:[int e^{0.05theta} , dtheta = frac{1}{0.05} e^{0.05theta} = 20 e^{0.05theta}]Evaluating this from ( 0 ) to ( 4pi ):[L approx 1.00125 left[ 20 e^{0.05 cdot 4pi} - 20 e^{0.05 cdot 0} right]][L approx 1.00125 left[ 20 e^{0.2pi} - 20 right]][L approx 1.00125 cdot 20 left( e^{0.2pi} - 1 right)]Using ( e^{0.2pi} approx 1.718 ):[L approx 1.00125 cdot 20 left( 1.718 - 1 right)][L approx 1.00125 cdot 20 cdot 0.718][L approx 1.00125 cdot 14.36][L approx 14.38]Final Answer: The final answer is 14.38. I hope it is correct.
question:The charismatic frontman of a renowned Britpop band is planning a special concert tour that features unique vocal performances and energetic stage setups. For the tour, the frontman has decided to experiment with acoustics and lighting to enhance the experience for the audience. The stage is designed as an elliptical area with semi-major axis (a) meters and semi-minor axis (b) meters, while the lighting setup involves a complex pattern of moving spotlights.1. Suppose the frontman wants the path of each spotlight to trace a Lissajous curve on the stage, given by the parametric equations ( x(t) = A sin(at + delta) ) and ( y(t) = B sin(bt) ), where ( A = frac{a}{2} ), ( B = frac{b}{2} ), ( a = 3 ), and ( b = 2 ). If the phase shift (delta) is chosen such that the curve intersects itself at the origin, find the possible values of (delta).2. During one of the performances, the frontman sings a note that causes the air pressure in the concert hall to oscillate with a frequency given by the function ( P(t) = P_0 cos(omega t + phi) ), where ( P_0 ) is the amplitude, (omega) is the angular frequency, and (phi) is the phase shift. If the note creates a standing wave with nodes at ( x = 0 ) and ( x = L ), where ( L = 10 ) meters, determine the possible wavelengths and frequencies of the note, given that the speed of sound in air is ( v = 343 ) m/s.
answer:To solve the given problem, we will address each part separately. Part 1: Lissajous Curve Intersection at the OriginThe parametric equations for the Lissajous curve are:[ x(t) = A sin(at + delta) ][ y(t) = B sin(bt) ]with ( A = frac{a}{2} = frac{3}{2} ) and ( B = frac{b}{2} = 1 ).For the curve to intersect at the origin, we need:[ x(t) = 0 quad text{and} quad y(t) = 0. ]1. Condition for ( x(t) = 0 ): [ frac{3}{2} sin(3t + delta) = 0 ] [ Rightarrow sin(3t + delta) = 0 ] [ Rightarrow 3t + delta = npi quad text{for} quad n in mathbb{Z}. ]2. Condition for ( y(t) = 0 ): [ 1 cdot sin(2t) = 0 ] [ Rightarrow sin(2t) = 0 ] [ Rightarrow 2t = mpi quad text{for} quad m in mathbb{Z}. ]From the second condition, we have:[ t = frac{mpi}{2}. ]Substitute ( t = frac{mpi}{2} ) into the first condition:[ 3left(frac{mpi}{2}right) + delta = npi ][ Rightarrow frac{3mpi}{2} + delta = npi ][ Rightarrow delta = npi - frac{3mpi}{2} ][ Rightarrow delta = pi left(n - frac{3m}{2}right). ]For the intersection at the origin, ( n - frac{3m}{2} ) must be an integer. This is possible when ( m ) is even (say ( m = 2k )), which gives:[ delta = pi left(n - frac{3(2k)}{2}right) = pi (n - 3k). ]Thus, the possible values of (delta) are integer multiples of (pi). Part 2: Standing Wave Frequencies and WavelengthsThe standing wave has nodes at ( x = 0 ) and ( x = L ), with ( L = 10 ) meters. The general form of standing wave nodes for wavelengths is:[ n frac{lambda}{2} = L ][ Rightarrow lambda = frac{2L}{n} = frac{20}{n} quad text{for} quad n = 1, 2, 3, ldots ]The frequency ( f ) is related to the wavelength (lambda) by the speed of sound:[ v = f lambda ][ Rightarrow f = frac{v}{lambda} = frac{343}{frac{20}{n}} = frac{343n}{20}. ]Therefore, the possible wavelengths are (lambda = frac{20}{n}) meters, and the corresponding frequencies are ( f = frac{343n}{20} ) Hz, where ( n = 1, 2, 3, ldots ).Final Answer: The possible values of (delta) are integer multiples of (pi), (lambda = frac{20}{n}) meters, and ( f = frac{343n}{20} ) Hz for ( n = 1, 2, 3, ldots ). I hope it is correct.
question:A law student at the University of Botswana is conducting research on the economic impact of legal reforms on the local economy. To quantify this, the student is using a model that involves differential equations to predict the growth rate of businesses under different legal scenarios.Sub-problem 1: The student models the number of businesses ( B(t) ) at time ( t ) with the differential equation:[ frac{dB}{dt} = k B left( 1 - frac{B}{L} right) - frac{m B^2}{B + n} ]where ( k ) is the growth rate coefficient, ( L ) is the carrying capacity of the market, ( m ) is a legal penalty factor, and ( n ) is a stabilization constant. Given ( B(0) = B_0 ), find the general solution for ( B(t) ).Sub-problem 2: Suppose the student wants to analyze the impact of a new legal reform that modifies the penalty factor ( m ) to a new value ( m^prime ). Determine the new equilibrium points of the system and discuss their stability in terms of ( k ), ( L ), ( m^prime ), and ( n ).
answer:Let's tackle the sub-problems one by one. Sub-problem 1: General Solution for ( B(t) )We start with the differential equation:[ frac{dB}{dt} = k B left( 1 - frac{B}{L} right) - frac{m B^2}{B + n} ]This differential equation is non-linear and complicated due to the term (frac{m B^2}{B + n}). Finding an explicit analytical solution is quite challenging. However, we can proceed by analyzing the qualitative behavior of the system. Sub-problem 2: Equilibrium Points and Stability Analysis# Finding Equilibrium PointsEquilibrium points occur when (frac{dB}{dt} = 0). Setting the right-hand side of the differential equation to zero gives:[ k B left( 1 - frac{B}{L} right) - frac{m B^2}{B + n} = 0 ]Let's denote the equilibrium points by ( B = B_e ). Then,[ k B_e left( 1 - frac{B_e}{L} right) = frac{m B_e^2}{B_e + n} ]Dividing both sides by ( B_e ) (assuming ( B_e neq 0 )):[ k left( 1 - frac{B_e}{L} right) = frac{m B_e}{B_e + n} ]Rearranging terms:[ k left( 1 - frac{B_e}{L} right) (B_e + n) = m B_e ]Expanding and simplifying:[ k B_e + k n - frac{k B_e^2}{L} - frac{k n B_e}{L} = m B_e ]Collecting like terms:[ frac{k B_e^2}{L} + (m - k - frac{k n}{L}) B_e - k n = 0 ]This is a quadratic equation in ( B_e ):[ frac{k}{L} B_e^2 + left( m - k - frac{k n}{L} right) B_e - k n = 0 ]Using the quadratic formula ( B_e = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = frac{k}{L} ), ( b = m - k - frac{k n}{L} ), and ( c = -k n ), we get:[ B_e = frac{ -left( m - k - frac{k n}{L} right) pm sqrt{left( m - k - frac{k n}{L} right)^2 + frac{4k^2 n}{L}} }{2 frac{k}{L}} ]Simplifying:[ B_e = frac{ L left( -m + k + frac{k n}{L} right) pm sqrt{L^2 left( m - k - frac{k n}{L} right)^2 + 4k^2 n L}}{2k} ][ B_e = frac{L left( -m + k + frac{k n}{L} right) pm sqrt{L^2 left( m - k - frac{k n}{L} right)^2 + 4k^2 n L}}{2k} ][ B_e = frac{L left( -m + k + frac{k n}{L} right) pm sqrt{L^2 (m - k)^2 + k^2 n^2 + 2Lk n (m - k) + 4k^2 n L}}{2k} ]This can be further simplified, but it is already a complicated expression. # Stability AnalysisTo determine the stability of the equilibrium points, we need to analyze the behavior of the system around these points. This involves linearizing the system around the equilibrium points and examining the eigenvalues of the resulting Jacobian matrix. However, due to the complexity of the equation, a numerical approach or further simplification might be necessary.Final Answer: The equilibrium points are the roots of the quadratic equation (frac{k}{L} B_e^2 + left( m - k - frac{k n}{L} right) B_e - k n = 0). Their exact values and stability depend on the specific parameters ( k ), ( L ), ( m^prime ), and ( n ).
question:Two pastry students, Alex and Jamie, are working on perfecting their chocolate cake recipes. Alex claims that his cake density increases linearly with the amount of chocolate used, while Jamie asserts that her cake density follows a quadratic relationship with the amount of chocolate. For their friendly rivalry, they decide to mathematically determine whose cake achieves a higher density given a specific amount of chocolate.1. Given that Alex's cake density ( D_A ) (in grams per cubic centimeter) as a function of the amount of chocolate ( C ) (in grams) is modeled by the linear equation ( D_A(C) = aC + b ), and Jamie's cake density ( D_J ) is modeled by the quadratic equation ( D_J(C) = pC^2 + qC + r ), find the amount of chocolate ( C ) at which both cakes have the same density. Assume ( a, b, p, q, ) and ( r ) are real constants, and ( a neq 0 ), ( p neq 0 ).2. If Alex uses 50 grams of chocolate and the density of his cake is 10 grams per cubic centimeter, and Jamie uses 50 grams of chocolate and the density of her cake is 12 grams per cubic centimeter, find the value of ( C ) where Jamie's cake density exceeds Alex's cake density by 5 grams per cubic centimeter.
answer:Let's solve each part of the problem step by step. Part 1: Finding the amount of chocolate ( C ) at which both cakes have the same density.Given:[ D_A(C) = aC + b ][ D_J(C) = pC^2 + qC + r ]To find the value of ( C ) where both densities are equal:[ D_A(C) = D_J(C) ][ aC + b = pC^2 + qC + r ]Rearrange the equation to form a quadratic equation:[ pC^2 + qC + r - aC - b = 0 ][ pC^2 + (q - a)C + (r - b) = 0 ]This is a standard quadratic equation of the form ( pC^2 + (q - a)C + (r - b) = 0 ). Solving for ( C ) using the quadratic formula ( C = frac{-B pm sqrt{B^2 - 4AC}}{2A} ), where ( A = p ), ( B = q - a ), and ( C = r - b ):[ C = frac{-(q - a) pm sqrt{(q - a)^2 - 4p(r - b)}}{2p} ] Part 2: Finding the value of ( C ) where Jamie's cake density exceeds Alex's cake density by 5 grams per cubic centimeter.Given:- Alex's cake density at 50 grams of chocolate is 10 grams per cubic centimeter.- Jamie's cake density at 50 grams of chocolate is 12 grams per cubic centimeter.Using the given information to find constants ( a ) and ( b ) for Alex:[ D_A(50) = a(50) + b = 10 ][ 50a + b = 10 quad text{(1)} ]For Jamie:[ D_J(50) = p(50)^2 + q(50) + r = 12 ][ 2500p + 50q + r = 12 quad text{(2)} ]We need ( D_J(C) = D_A(C) + 5 ):[ pC^2 + qC + r = aC + b + 5 ]Rearrange to form a quadratic equation:[ pC^2 + qC + r - aC - b - 5 = 0 ][ pC^2 + (q - a)C + (r - b - 5) = 0 ]Substitute the known values for ( a ) and ( b ) from equation (1):[ b = 10 - 50a ]Substituting ( b ) back into the quadratic equation:[ pC^2 + (q - a)C + (r - (10 - 50a) - 5) = 0 ][ pC^2 + (q - a)C + (r - 10 + 50a - 5) = 0 ][ pC^2 + (q - a)C + (r + 50a - 15) = 0 ]We know that ( D_A(50) = 10 ) and ( D_J(50) = 12 ):[ 2500p + 50q + r = 12 ][ 50a + b = 10 ][ b = 10 - 50a ]Now we need to solve for ( C ) using the quadratic equation:[ C = frac{-(q - a) pm sqrt{(q - a)^2 - 4p(r + 50a - 15)}}{2p} ]Given the complexity and lack of specific numerical values for ( p ), ( q ), and ( r ), we can only provide a general form of the solution. Final Answer:The value of ( C ) where Jamie's cake density exceeds Alex's cake density by 5 grams per cubic centimeter can be found using the quadratic formula:[ C = frac{-(q - a) pm sqrt{(q - a)^2 - 4p(r + 50a - 15)}}{2p} ]I hope it is correct.
question:A professor of archaeology specializing in prehistoric civilizations is planning an expedition to explore a newly discovered area suspected to have relics from an ancient society. The site is located on a triangular piece of land bordered by a river on one side. The coordinates of the vertices of this triangular area are A(2, 3), B(8, 5), and C(5, 9). The river runs exactly parallel to the line segment AB. The professor wants to determine the best location for a base camp that maximizes the potential to reach all points within the triangular area efficiently.1. Calculate the coordinates of the centroid of the triangular piece of land. This point is considered as a potential location for the base camp.2. The professor also wants to ensure easy access to water from the river. Calculate the equation of the line parallel to AB that passes through the centroid. Determine the shortest distance from this line to the river, which is represented by line AB.Note: Use your knowledge of coordinate geometry, vectors, and distance calculations to solve the problem.
answer:To solve this problem, we need to find the centroid of the triangle and determine the shortest distance from the centroid to the line AB, which represents the river. Step 1: Find the Centroid of the TriangleThe coordinates of the vertices of the triangle are A(2, 3), B(8, 5), and C(5, 9). The formula for the centroid (G) of a triangle with vertices ((x_1, y_1)), ((x_2, y_2)), and ((x_3, y_3)) is given by:[Gleft(frac{x_1 + x_2 + x_3}{3}, frac{y_1 + y_2 + y_3}{3}right)]Substitute the coordinates of the vertices:[Gleft(frac{2 + 8 + 5}{3}, frac{3 + 5 + 9}{3}right) = Gleft(frac{15}{3}, frac{17}{3}right) = G(5, frac{17}{3})]So, the coordinates of the centroid are (G(5, frac{17}{3})). Step 2: Find the Equation of Line ABLine AB is defined by the points A(2, 3) and B(8, 5). First, find the slope of line AB:[text{slope of AB} = frac{5 - 3}{8 - 2} = frac{2}{6} = frac{1}{3}]The equation of a line in point-slope form is (y - y_1 = m(x - x_1)). Using point A(2, 3) and the slope (frac{1}{3}):[y - 3 = frac{1}{3}(x - 2)]Simplifying, we get:[y = frac{1}{3}x + frac{1}{3} times (-2) + 3 = frac{1}{3}x + frac{9}{3} - frac{2}{3} = frac{1}{3}x + frac{7}{3}]Thus, the equation of line AB is:[y = frac{1}{3}x + frac{7}{3}] Step 3: Find the Equation of the Line Parallel to AB through the CentroidA line parallel to AB will have the same slope. Thus, the slope is (frac{1}{3}). Now use the point-slope form with the centroid (G(5, frac{17}{3})):[y - frac{17}{3} = frac{1}{3}(x - 5)]Simplifying, we get:[y = frac{1}{3}x + frac{1}{3} times (-5) + frac{17}{3} = frac{1}{3}x - frac{5}{3} + frac{17}{3} = frac{1}{3}x + frac{12}{3} = frac{1}{3}x + 4]Thus, the equation of the line parallel to AB through the centroid is:[y = frac{1}{3}x + 4] Step 4: Calculate the Shortest Distance from the Centroid Line to Line ABThe shortest distance (d) between two parallel lines (y = mx + c_1) and (y = mx + c_2) is given by:[d = frac{|c_2 - c_1|}{sqrt{1 + m^2}}]Where (m = frac{1}{3}), (c_1 = frac{7}{3}), and (c_2 = 4).Substitute these values into the formula:[d = frac{left|4 - frac{7}{3}right|}{sqrt{1 + left(frac{1}{3}right)^2}} = frac{left|frac{12}{3} - frac{7}{3}right|}{sqrt{1 + frac{1}{9}}} = frac{left|frac{5}{3}right|}{sqrt{frac{10}{9}}}][d = frac{frac{5}{3}}{frac{sqrt{10}}{3}} = frac{5}{sqrt{10}} = frac{5sqrt{10}}{10} = frac{sqrt{10}}{2}]Final Answer: The final answer is (frac{sqrt{10}}{2}). I hope it is correct.