answer:To solve the problem, we need to address both parts of the question step-by-step. Part 1: Finding the number of weeks until the conference when both countries send the same number of delegates.We are given two polynomials:- The number of delegates from Country A: ( n^2 - 5n + 6 )- The number of delegates from Country B: ( 2n^2 - 7n + 3 )To find when both countries send the same number of delegates, we set the two polynomials equal to each other:[n^2 - 5n + 6 = 2n^2 - 7n + 3]Rearrange the equation to bring all terms to one side:[n^2 - 5n + 6 - 2n^2 + 7n - 3 = 0]Simplify:[-n^2 + 2n + 3 = 0]Multiply the entire equation by -1 to make the leading coefficient positive:[n^2 - 2n - 3 = 0]Now, factor the quadratic equation:[(n - 3)(n + 1) = 0]Set each factor to zero and solve for ( n ):1. ( n - 3 = 0 ) gives ( n = 3 )2. ( n + 1 = 0 ) gives ( n = -1 )Since ( n ) must be a positive integer (as it represents weeks), we have:[n = 3] Part 2: Calculate the difference in the number of delegates if the conference is scheduled in 3 weeks.We will evaluate both polynomials at ( n = 3 ).For Country A:[n^2 - 5n + 6 = 3^2 - 5 times 3 + 6 = 9 - 15 + 6 = 0]For Country B:[2n^2 - 7n + 3 = 2 times 3^2 - 7 times 3 + 3 = 2 times 9 - 21 + 3 = 18 - 21 + 3 = 0]Thus, in 3 weeks, both countries send the same number of delegates — 0 delegates each. Hence, Country A will send ( 0 - 0 = 0 ) more delegates than Country B.Final Answer: The final answer is 3 weeks for part 1 and 0 more delegates for part 2. I hope it is correct.

answer:To solve this problem, we need to set up a linear equation and an inequality based on the costs and savings associated with the filtration system. Step 1: Set up the linear equationThe initial cost of installing the water filtration system is 1,200. The ongoing cost for maintaining the system is 30 per month. The savings from not purchasing bottled water is 50 per month.To find the break-even point, we need to determine when the total savings equals the total costs. The savings per month is 50, and the cost per month is 30. Therefore, the net savings per month is:[ text{Net savings per month} = 50 - 30 = 20 text{ dollars.} ]Now, let ( m ) represent the number of months. We need to find when the total net savings equals the initial cost:[ 20m = 1200. ] Step 2: Solve the linear equationTo solve for ( m ), divide both sides of the equation by 20:[ m = frac{1200}{20} = 60. ]This means it will take 60 months for the restaurant to break even. Step 3: Set up and solve the inequalityTo find out when the restaurateur will begin to see a net financial gain, we need to determine when the total net savings exceeds the initial cost. This is expressed by the inequality:[ 20m > 1200. ]Divide both sides by 20 to solve for ( m ):[ m > frac{1200}{20} = 60. ]This means that after 60 months, the restaurateur will start to see a net financial gain.Final Answer: The final answer is 60 months. I hope it is correct.

answer:To solve this problem, we start by analyzing the quadratic function given:[ N(x) = -2x^2 + 40x + 100. ]This is a quadratic function of the form ( ax^2 + bx + c ), where ( a = -2 ), ( b = 40 ), and ( c = 100 ). Part 1: Finding the Maximum Number of New UsersThe vertex form of a quadratic function ( ax^2 + bx + c ) gives the maximum or minimum point of the parabola. Since the coefficient of ( x^2 ) is negative (( a = -2 )), the parabola opens downward, indicating a maximum point at the vertex.The ( x )-coordinate of the vertex for a quadratic function ( ax^2 + bx + c ) is given by:[ x = -frac{b}{2a}. ]Substituting the values of ( b ) and ( a ):[ x = -frac{40}{2(-2)} = -frac{40}{-4} = 10. ]So, the platform reaches the maximum number of new users 10 days after the campaign begins.To find the maximum number of new users, we substitute ( x = 10 ) back into the function ( N(x) ):[N(10) = -2(10)^2 + 40(10) + 100 = -2(100) + 400 + 100 = -200 + 400 + 100 = 300.]Thus, the maximum number of new users is 300. Part 2: Determining the Range of Days for at least 500 New UsersWe need to find the values of ( x ) for which ( N(x) geq 500 ).Set up the inequality:[-2x^2 + 40x + 100 geq 500.]Subtract 500 from both sides to set the equation to zero:[-2x^2 + 40x + 100 - 500 geq 0,]which simplifies to:[-2x^2 + 40x - 400 geq 0.]Divide every term by -2 to make the quadratic term positive, remembering to reverse the inequality sign:[x^2 - 20x + 200 leq 0.]Now, solve the quadratic equation ( x^2 - 20x + 200 = 0 ) using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -20 ), and ( c = 200 ):[x = frac{-(-20) pm sqrt{(-20)^2 - 4(1)(200)}}{2(1)},][x = frac{20 pm sqrt{400 - 800}}{2},][x = frac{20 pm sqrt{-400}}{2}.]Since the discriminant ((-400)) is negative, there are no real solutions, meaning the quadratic ( x^2 - 20x + 200 leq 0 ) has no real roots.Since the parabola opens upward and does not intersect the x-axis, it is always positive, meaning there are no real ( x ) values for which ( N(x) geq 500 ). Therefore, the campaign will never achieve at least 500 new users based on this model.Final Answer: The maximum number of new users is 300, reached 10 days after the campaign starts. The campaign never achieves 500 new users. I hope it is correct.

answer:To solve this problem, we will break it down into two parts as described in the problem statement. Part 1: Determine the number of cubes and rectangular prisms in the set.1. Set up the system of equations: Let ( c ) be the number of cubes and ( r ) be the number of rectangular prisms. We know: - The total number of blocks is 30. - There are twice as many cubes as rectangular prisms. From these, we can create the following equations: [ c + r = 30 ] [ c = 2r ]2. Solve the system of equations: Substitute ( c = 2r ) into the first equation: [ 2r + r = 30 ] [ 3r = 30 ] [ r = 10 ] Substitute ( r = 10 ) back into ( c = 2r ): [ c = 2 times 10 = 20 ] Therefore, there are 20 cubes and 10 rectangular prisms. Part 2: Determine ( x ) using the total surface area.1. Write the equation for the total surface area: - The surface area of one cube with side length ( x ) is: [ 6x^2 ] - The surface area of one rectangular prism with dimensions ( x ), ( 2x ), and ( x + 1 ) is: [ 2(x cdot 2x + 2x cdot (x + 1) + (x + 1) cdot x) = 2(2x^2 + 2x(x + 1) + x(x + 1)) ] [ = 2(2x^2 + 2x^2 + 2x + x^2 + x) = 2(5x^2 + 3x) = 10x^2 + 6x ] - Total surface area for 20 cubes and 10 rectangular prisms: [ 20 times 6x^2 + 10 times (10x^2 + 6x) = 120x^2 + 100x^2 + 60x = 220x^2 + 60x ] Set this equal to the total surface area given (1500 cm²): [ 220x^2 + 60x = 1500 ]2. Solve for ( x ): First, simplify the equation by dividing all terms by 20: [ 11x^2 + 3x = 75 ] Rearrange it into a standard quadratic equation: [ 11x^2 + 3x - 75 = 0 ] Use the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 11 ), ( b = 3 ), and ( c = -75 ): [ x = frac{-3 pm sqrt{3^2 - 4 times 11 times (-75)}}{2 times 11} ] [ x = frac{-3 pm sqrt{9 + 3300}}{22} ] [ x = frac{-3 pm sqrt{3309}}{22} ] Calculate ( sqrt{3309} ), which is approximately 57.52: [ x = frac{-3 pm 57.52}{22} ] Since ( x ) must be positive, we take the positive root: [ x = frac{-3 + 57.52}{22} approx frac{54.52}{22} approx 2.478 ] Therefore, the side length ( x ) is approximately 2.478 cm.Final Answer: The final answer is ( x approx 2.478 ) cm. I hope it is correct.

answer:To solve this problem, we need to find the dimensions of the larger rectangle and the smaller rectangle, then verify if the smaller rectangle can be cut out from the larger rectangle. Part 1: Dimensions of the Larger RectangleWe are given that the area of the larger rectangle is represented by the polynomial (x^2 + 4x - 60). We also know that the length of the rectangle is 4 inches longer than its width.1. Let's denote the width of the larger rectangle as (x).2. Therefore, the length of the larger rectangle is (x + 4).3. The area of the larger rectangle can be expressed as the product of its length and width: [ text{Area} = x(x + 4) = x^2 + 4x ]4. We are given that this area is (x^2 + 4x - 60). Therefore, we have: [ x^2 + 4x = x^2 + 4x - 60 + 60 ] Simplifying, we get: [ x^2 + 4x - 60 = x^2 + 4x ] This simplifies to: [ 0 = -60 ]5. However, we should equate the polynomial directly to zero to find the values of (x) that satisfy the equation: [ x^2 + 4x - 60 = 0 ]6. We can solve this quadratic equation by factoring. We need two numbers that multiply to (-60) and add to (4). These numbers are (10) and (-6).7. Therefore, the equation factors as: [ (x + 10)(x - 6) = 0 ]8. Solving for (x), we get: [ x + 10 = 0 quad text{or} quad x - 6 = 0 ] [ x = -10 quad text{or} quad x = 6 ]9. Since a dimension cannot be negative, (x = 6).Thus, the width of the larger rectangle is 6 inches, and the length is (6 + 4 = 10) inches. Part 2: Dimensions of the Smaller Rectangle1. We are given that the area of the smaller rectangle is 15 square inches and its width is 3 inches.2. Let the length of the smaller rectangle be (l). Then, the area is given by: [ 3 times l = 15 ]3. Solving for (l), we have: [ l = frac{15}{3} = 5 , text{inches} ] VerificationNow we verify if the smaller rectangle can be cut from the larger rectangle:- The dimensions of the larger rectangle are 10 inches (length) and 6 inches (width).- The dimensions of the smaller rectangle are 5 inches (length) and 3 inches (width).Since both the length and width of the smaller rectangle (5 inches and 3 inches, respectively) are within the dimensions of the larger rectangle (10 inches by 6 inches), it is indeed possible to cut out the smaller rectangle from the larger rectangle.Final Answer: The dimensions of the larger rectangle are 10 inches (length) and 6 inches (width), and the dimensions of the smaller rectangle are 5 inches (length) and 3 inches (width). I hope it is correct.

answer:To determine the values of the constants ( a ), ( b ), and ( c ) in the quadratic model ( G(P) = aP^2 + bP + c ), we need to solve the following system of equations based on the given data:1. When ( P = 2 ), ( G = 3 ): [ a(2)^2 + b(2) + c = 3 quad Rightarrow quad 4a + 2b + c = 3 ] 2. When ( P = 5 ), ( G = 7 ): [ a(5)^2 + b(5) + c = 7 quad Rightarrow quad 25a + 5b + c = 7 ] 3. When ( P = 6 ), ( G = 12 ): [ a(6)^2 + b(6) + c = 12 quad Rightarrow quad 36a + 6b + c = 12 ]This system of equations can be written as:[begin{align*}4a + 2b + c &= 3 quad quad quad (1) 25a + 5b + c &= 7 quad quad quad (2) 36a + 6b + c &= 12 quad quad quad (3)end{align*}]To solve this system, we will first eliminate ( c ) by subtracting equation (1) from equations (2) and (3):Subtract equation (1) from equation (2):[(25a + 5b + c) - (4a + 2b + c) = 7 - 3 21a + 3b = 4 quad quad quad (4)]Subtract equation (1) from equation (3):[(36a + 6b + c) - (4a + 2b + c) = 12 - 3 32a + 4b = 9 quad quad quad (5)]Now, we will solve equations (4) and (5) simultaneously. First, let's simplify equation (4):[7a + b = frac{4}{3} quad quad quad (6)]And equation (5):[8a + b = frac{9}{4} quad quad quad (7)]Subtract equation (6) from equation (7) to eliminate ( b ):[(8a + b) - (7a + b) = frac{9}{4} - frac{4}{3} a = frac{9}{4} - frac{4}{3} = frac{27}{12} - frac{16}{12} = frac{11}{12}]Now that we have ( a = frac{11}{12} ), substitute back into equation (6) to find ( b ):[7left(frac{11}{12}right) + b = frac{4}{3} frac{77}{12} + b = frac{4}{3} b = frac{4}{3} - frac{77}{12} = frac{16}{12} - frac{77}{12} = -frac{61}{12}]Finally, substitute ( a ) and ( b ) back into equation (1) to find ( c ):[4left(frac{11}{12}right) + 2left(-frac{61}{12}right) + c = 3 frac{44}{12} - frac{122}{12} + c = 3 -frac{78}{12} + c = 3 c = 3 + frac{78}{12} = 3 + frac{39}{6} = 3 + 6.5 = 9.5]Thus, the values of the constants are:[a = frac{11}{12}, quad b = -frac{61}{12}, quad c = 9.5]Now, to find the expected economic growth rate when ( P = 4 ):[ G(4) = a(4)^2 + b(4) + c = frac{11}{12}(16) - frac{61}{12}(4) + 9.5 = frac{176}{12} - frac{244}{12} + 9