answer: Sub-problem 1: Calculating the Radius of the Circular StageFirst, we need to find the total area of the living room.The living room is a rectangle with dimensions:[ text{Length} = 20 text{ meters} ][ text{Width} = 15 text{ meters} ]The area ( A_{text{living room}} ) of the living room is:[ A_{text{living room}} = text{Length} times text{Width} ][ A_{text{living room}} = 20 times 15 ][ A_{text{living room}} = 300 text{ square meters} ]Alex wants the area of the circular stage to be 25% of the living room's area:[ A_{text{stage}} = 0.25 times A_{text{living room}} ][ A_{text{stage}} = 0.25 times 300 ][ A_{text{stage}} = 75 text{ square meters} ]The area of a circle is given by:[ A_{text{stage}} = pi r^2 ]We need to find the radius ( r ) such that:[ pi r^2 = 75 ]Solving for ( r ):[ r^2 = frac{75}{pi} ][ r = sqrt{frac{75}{pi}} ]Using the value of ( pi approx 3.14159 ):[ r = sqrt{frac{75}{3.14159}} ][ r approx sqrt{23.87} ][ r approx 4.88 text{ meters} ] Sub-problem 2: Determining the Number of SpotlightsFirst, we need to calculate the circumference of the circular stage with radius ( r ).The circumference ( C ) of a circle is given by:[ C = 2 pi r ]Using the radius calculated:[ r approx 4.88 text{ meters} ]So:[ C approx 2 pi times 4.88 ][ C approx 2 times 3.14159 times 4.88 ][ C approx 30.65 text{ meters} ]The number of spotlights is the integer part of the circumference:[ text{Number of spotlights} = lfloor C rfloor ][ text{Number of spotlights} = lfloor 30.65 rfloor ][ text{Number of spotlights} = 30 ] Final Answer:The final answer is 30. I hope it is correct.

answer:Let's solve the given sub-problems step-by-step. Sub-problem 1:We need to find the coordinates ( (x, y) ) on the cubic Bézier curve when ( t = 0.5 ).The parametric equation of a cubic Bézier curve is given by:[ B(t) = (1-t)^3 P_0 + 3(1-t)^2 t P_1 + 3(1-t)t^2 P_2 + t^3 P_3 ]Given points are:[ P_0 = (0, 0), quad P_1 = (480, y_1), quad P_2 = (1440, y_2), quad P_3 = (1920, 1080) ]Plugging in ( t = 0.5 ):[ B(0.5) = (1-0.5)^3 P_0 + 3(1-0.5)^2 (0.5) P_1 + 3(1-0.5)(0.5)^2 P_2 + (0.5)^3 P_3 ][ B(0.5) = (0.5)^3 P_0 + 3(0.5)^2 (0.5) P_1 + 3(0.5)(0.5)^2 P_2 + (0.5)^3 P_3 ][ B(0.5) = 0.125 P_0 + 0.375 P_1 + 0.375 P_2 + 0.125 P_3 ]Now, substitute the points:[ x = 0.125 cdot 0 + 0.375 cdot 480 + 0.375 cdot 1440 + 0.125 cdot 1920 ][ x = 0 + 180 + 540 + 240 = 960 ][ y = 0.125 cdot 0 + 0.375 cdot y_1 + 0.375 cdot y_2 + 0.125 cdot 1080 ][ y = 0 + 0.375y_1 + 0.375y_2 + 135 ]Thus, the coordinates ((x, y)) when ( t = 0.5 ) are:[ (960, 0.375y_1 + 0.375y_2 + 135) ] Sub-problem 2:The pressure sensitivity function is given by:[ P(x, y) = A sinleft(frac{pi x}{1920}right) + B cosleft(frac{pi y}{1080}right) ]We know that the maximum pressure 1023 occurs at ( (960, 540) ). Therefore:[ P(960, 540) = 1023 ]Substitute ( x = 960 ) and ( y = 540 ):[ 1023 = A sinleft(frac{pi cdot 960}{1920}right) + B cosleft(frac{pi cdot 540}{1080}right) ]Since ( sinleft(frac{pi cdot 960}{1920}right) = sinleft(frac{pi}{2}right) = 1 ) and ( cosleft(frac{pi cdot 540}{1080}right) = cosleft(frac{pi}{2}right) = 0 ), we get:[ 1023 = A cdot 1 + B cdot 0 ][ A = 1023 ]Next, we need to find ( B ). Since ( P(x, y) ) must be bounded by 1023, the maximum value of the cosine function is 1:[ 1023 = 1023 + B cdot 1 ]This implies:[ B = 0 ] Final Answer:For sub-problem 1:The coordinates ((x, y)) of the point on the cubic Bézier curve when ( t = 0.5 ) are ( (960, 0.375y_1 + 0.375y_2 + 135) ).For sub-problem 2:The values of ( A ) and ( B ) are ( 1023 ) and ( 0 ), respectively.I hope it is correct.

answer:To solve this problem, we will tackle each sub-problem separately. Sub-problem 1: Maximizing EfficiencyThe efficiency is given by the function:[ E(x, y) = frac{xy}{x^2 + y^2 + 1} ]We need to maximize ( E(x, y) ) subject to ( 1 leq y leq 8 ) and ( 0 < x leq 3.2 ).Since ( y ) is an integer, we can evaluate ( E(x, y) ) for each ( y ) from 1 to 8, and find the maximum efficiency by optimizing ( x ) for each ( y ).# Step 1: Evaluate for each integer ( y )For each ( y ), find the derivative of ( E(x, y) ) with respect to ( x ):[ frac{partial E}{partial x} = frac{y(x^2 + y^2 + 1) - xy(2x)}{(x^2 + y^2 + 1)^2} = frac{y(x^2 + y^2 + 1) - 2x^2y}{(x^2 + y^2 + 1)^2} ]Setting (frac{partial E}{partial x} = 0), we have:[ y(x^2 + y^2 + 1) = 2x^2y ]Simplifying:[ x^2 + y^2 + 1 = 2x^2 ][ x^2 = y^2 + 1 ][ x = sqrt{y^2 + 1} ]# Step 2: Substitute and check feasibilityFor each ( y ), substitute ( x = sqrt{y^2 + 1} ), ensuring ( x leq 3.2 ):1. ( y = 1 ), ( x = sqrt{1^2 + 1} = sqrt{2} approx 1.41 )2. ( y = 2 ), ( x = sqrt{2^2 + 1} = sqrt{5} approx 2.24 )3. ( y = 3 ), ( x = sqrt{3^2 + 1} = sqrt{10} approx 3.16 )4. ( y = 4 ), ( x = sqrt{4^2 + 1} = sqrt{17} approx 4.12 ) (exceeds 3.2)5. For ( y geq 4 ), ( x > 3.2 ), not feasible.# Step 3: Calculate ( E(x, y) )- For ( y = 3 ) and ( x = 3.16 ):[ E(3.16, 3) = frac{3.16 times 3}{3.16^2 + 3^2 + 1} approx frac{9.48}{19.9856} approx 0.4746 ]Checking ( y = 4 ) with ( x = 3.2 ) (since ( x = sqrt{17} ) exceeds 3.2):[ E(3.2, 4) = frac{3.2 times 4}{3.2^2 + 4^2 + 1} = frac{12.8}{27.24} approx 0.470 ]Thus, the maximum efficiency occurs at ( x approx 3.16 ) and ( y = 3 ). Sub-problem 2: Power Consumption ConstraintThe power consumption is given by:[ P(x, y) = 0.5x^2 + 0.3y + 0.1xy ]We need ( P(x, y) leq 10 ).For the optimized ( x approx 3.16 ) and ( y = 3 ):[ P(3.16, 3) = 0.5(3.16)^2 + 0.3(3) + 0.1(3.16)(3) ][ = 0.5(9.9856) + 0.9 + 0.948 ][ = 4.9928 + 0.9 + 0.948 = 6.8408 ]Since ( 6.8408 leq 10 ), this configuration satisfies the power constraint.Final Answer: The final answer is ( x approx 3.16 ) GHz and ( y = 3 ) cores, and it satisfies the power constraint. I hope it is correct.

answer:Let's solve the given math problem step-by-step. 1. Solving the Differential Equation for ( P(t) )The differential equation given is:[ frac{dP}{dt} = kPleft(1 - frac{P}{N}right) ]This is a logistic differential equation. To solve it, we use the method of separation of variables.First, we rewrite the equation:[ frac{dP}{dt} = kP left(1 - frac{P}{N}right) ]Separate the variables (P) and (t):[ frac{1}{P left(1 - frac{P}{N}right)} dP = k , dt ]Rewrite the left-hand side using partial fractions:[ frac{1}{P left(1 - frac{P}{N}right)} = frac{1}{P} + frac{1/N}{1 - P/N} ]Thus:[ frac{1}{P} + frac{1/N}{1 - P/N} , dP = k , dt ]Integrate both sides:[ int left( frac{1}{P} + frac{1/N}{1 - P/N} right) dP = int k , dt ][ int frac{1}{P} , dP + int frac{1/N}{1 - P/N} , dP = int k , dt ]Integrate each term separately:[ ln|P| - ln|1 - frac{P}{N}| = kt + C ]Combine the logarithms on the left side:[ lnleft|frac{P}{1 - frac{P}{N}}right| = kt + C ]Exponentiate both sides:[ frac{P}{1 - frac{P}{N}} = e^{kt + C} ]Let ( e^C = A ):[ frac{P}{1 - frac{P}{N}} = A e^{kt} ]Solve for ( P ):[ P = A e^{kt} left(1 - frac{P}{N}right) ][ P = A e^{kt} - frac{A e^{kt} P}{N} ][ P left(1 + frac{A e^{kt}}{N}right) = A e^{kt} ][ P = frac{A e^{kt}}{1 + frac{A e^{kt}}{N}} ][ P = frac{A N e^{kt}}{N + A e^{kt}} ]Using the initial condition ( P(0) = 10 ):[ 10 = frac{A N}{N + A} ]Solving for ( A ):[ 10(N + A) = AN ][ 10N + 10A = AN ][ 10A = AN - 10N ][ A (N - 10) = 10N ][ A = frac{10N}{N - 10} ]Thus:[ P(t) = frac{frac{10N}{N - 10} N e^{kt}}{N + frac{10N}{N - 10} e^{kt}} ][ P(t) = frac{10N^2 e^{kt}}{(N - 10)N + 10N e^{kt}} ][ P(t) = frac{10N e^{kt}}{N - 10 + 10 e^{kt}} ] 2. Finding the Probability using Poisson DistributionThe Poisson probability mass function is:[ P(X = x) = frac{e^{-lambda} lambda^x}{x!} ]Given ( lambda = 4 ) and we need the probability of playing the game exactly 5 times:[ P(X = 5) = frac{e^{-4} 4^5}{5!} ]Calculate:[ P(X = 5) = frac{e^{-4} 1024}{120} ][ P(X = 5) = e^{-4} frac{1024}{120} ][ P(X = 5) = e^{-4} times 8.5333 ]Using ( e^{-4} approx 0.0183 ):[ P(X = 5) approx 0.0183 times 8.5333 ][ P(X = 5) approx 0.156 ]Final Answer: The final answer is ( P(t) = frac{10N e^{kt}}{N - 10 + 10 e^{kt}} ) and ( P(X = 5) approx 0.156 ). I hope it is correct.

answer:To solve this problem, we'll address each part separately.Part 1: Determining the total angle through which Emily has spun after 5 secondsEmily's spin rate is given by the function (theta(t) = 3t^2 + 2t) radians. To find the total angle through which she has spun in 5 seconds, we need to integrate this rate function from (t = 0) to (t = 5).The integral of (3t^2 + 2t) with respect to (t) is:[int (3t^2 + 2t) , dt = int 3t^2 , dt + int 2t , dt][= left[ t^3 right] + left[ t^2 right] = t^3 + t^2 + C]where (C) is the constant of integration. Since we're calculating a definite integral from (t = 0) to (t = 5), we can ignore the constant (C).Evaluate the integral from 0 to 5:[int_0^5 (3t^2 + 2t) , dt = left[ t^3 + t^2 right]_0^5 = (5^3 + 5^2) - (0^3 + 0^2)][= (125 + 25) - (0 + 0) = 150]Thus, the total angle through which Emily has spun after 5 seconds is 150 radians.Part 2: Finding the total distance Emily covers in her first 10 leapsThe distance of each leap follows an arithmetic sequence. We are given that the first leap covers 1.5 meters and the fifth leap covers 3 meters.The general form of an arithmetic sequence is given by:[ a_n = a_1 + (n-1)d ]where ( a_n ) is the (n)-th term, ( a_1 ) is the first term, and ( d ) is the common difference.From the problem:- ( a_1 = 1.5 )- ( a_5 = 3 )We can find the common difference (d) using the information for the fifth term:[a_5 = a_1 + 4d = 3 ][1.5 + 4d = 3 ][4d = 1.5 ][d = frac{1.5}{4} = 0.375]Now, we find the total distance covered in the first 10 leaps. The sum of the first (n) terms of an arithmetic sequence is given by:[ S_n = frac{n}{2} times (a_1 + a_n) ]First, find the 10th term ((a_{10})):[a_{10} = a_1 + 9d = 1.5 + 9(0.375) = 1.5 + 3.375 = 4.875]Now, calculate the sum of the first 10 terms:[S_{10} = frac{10}{2} times (1.5 + 4.875) = 5 times 6.375 = 31.875]Thus, the total distance Emily covers in her first 10 leaps is 31.875 meters.Final Answer: The final answer is 150 radians and 31.875 meters. I hope it is correct.

answer:To solve the problem, we will address both parts step by step.Part 1: Finding the coefficients ( a ), ( b ), and ( c ) of the quadratic functionWe are given the quadratic function ( f(x) = ax^2 + bx + c ), and the conditions:1. ( f(0) = 0 )2. ( f(1) = frac{3}{5} )3. The total budget ( B = int_0^1 f(x) , dx ).Using the condition ( f(0) = 0 ), we find that:[f(0) = a(0)^2 + b(0) + c = 0 implies c = 0]Thus, the function simplifies to ( f(x) = ax^2 + bx ).Next, using the condition ( f(1) = frac{3}{5} ):[f(1) = a(1)^2 + b(1) = a + b = frac{3}{5}]Now, we calculate the total budget ( B ) as the integral of ( f(x) ) from 0 to 1:[B = int_0^1 (ax^2 + bx) , dx = left[ frac{a}{3}x^3 + frac{b}{2}x^2 right]_0^1 = frac{a}{3} + frac{b}{2}]Since ( B = 1 ), we have:[frac{a}{3} + frac{b}{2} = 1]We now have two equations:1. ( a + b = frac{3}{5} )2. ( frac{a}{3} + frac{b}{2} = 1 )Multiply the second equation by 6 to eliminate fractions:[2a + 3b = 6]Now solve the system of equations:1. ( a + b = frac{3}{5} )2. ( 2a + 3b = 6 )Multiply the first equation by 2:[2a + 2b = frac{6}{5}]Subtract this from the second equation:[(2a + 3b) - (2a + 2b) = 6 - frac{6}{5}][b = 6 - frac{6}{5} = frac{30}{5} - frac{6}{5} = frac{24}{5}]Substitute ( b = frac{24}{5} ) back into ( a + b = frac{3}{5} ):[a + frac{24}{5} = frac{3}{5}][a = frac{3}{5} - frac{24}{5} = -frac{21}{5}]Thus, the coefficients are ( a = -frac{21}{5} ), ( b = frac{24}{5} ), and ( c = 0 ).Part 2: Calculating the new number of students ( N ) for scholarshipsThe current total budget ( B = 1 ), and 60% goes to scholarships. Thus, the scholarship budget ( S = 0.6 times B = 0.6 ).Next year, the budget increases by 25%, so the new total budget ( B' = 1.25 times B = 1.25 ).The new scholarship budget ( S' = 0.6 times B' = 0.6 times 1.25 = 0.75 ).The scholarship budget increased from ( S = 0.6 ) to ( S' = 0.75 ).The increase in the number of students benefiting from scholarships is proportional to the budget increase. Thus, we calculate the new number of students ( N ):The current number of students benefiting from scholarships is ( N_0 ), such that 60% of ( N_0 ) corresponds to the current scholarship budget 0.6.With the increased budget, ( 0.75 ) corresponds to ( N ), where ( N = frac{0.75}{0.6} times N_0 ).Therefore, ( N = frac{0.75}{0.6} times N_0 = 1.25 times N_0 ).Thus, the new number of students benefiting from scholarships is 25% more than the current number.Final Answer: The final answer is that the new number of students ( N ) is 25% more than the current number of students ( N_0 ). I hope it is correct.