answer:To solve this problem, we need to analyze the quadratic function ( E(t) = -2t^2 + 12t - 16 ). Part 1: Maximum Effectiveness1. Find the vertex of the parabola: The vertex of a quadratic function ( f(t) = at^2 + bt + c ) occurs at ( t = -frac{b}{2a} ). In our function: [ a = -2, quad b = 12, quad c = -16 ] [ t = -frac{12}{2 times -2} = -frac{12}{-4} = 3 ]2. Calculate the maximum effectiveness: Substitute ( t = 3 ) back into the equation to find ( E(3) ): [ E(3) = -2(3)^2 + 12(3) - 16 ] [ = -2 times 9 + 36 - 16 ] [ = -18 + 36 - 16 ] [ = 2 ] So, the maximum effectiveness of the treatment is ( 2 ) units at ( t = 3 ) hours. Part 2: Time Range for Effectiveness of at Least 4 Units1. Set up the inequality: We need to solve for ( t ) such that ( E(t) geq 4 ): [ -2t^2 + 12t - 16 geq 4 ]2. Rearrange the inequality: [ -2t^2 + 12t - 16 - 4 geq 0 ] [ -2t^2 + 12t - 20 geq 0 ]3. Solve the quadratic equation: First, solve the equation ( -2t^2 + 12t - 20 = 0 ). Divide the entire equation by (-2) to simplify: [ t^2 - 6t + 10 = 0 ] Use the quadratic formula: [ t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] Here, ( a = 1 ), ( b = -6 ), ( c = 10 ). Calculate the discriminant: [ b^2 - 4ac = (-6)^2 - 4 times 1 times 10 = 36 - 40 = -4 ] The discriminant is negative, which means there are no real roots. This implies that the quadratic equation ( t^2 - 6t + 10 = 0 ) has no real solutions, and thus ( E(t) = -2t^2 + 12t - 16 ) never reaches or exceeds 4. Therefore, there is no range of ( t ) for which the effectiveness is at least 4 units.Final Answer: The final answer is ( t = 3 ) hours for maximum effectiveness of 2 units; effectiveness never reaches 4 units. I hope it is correct.

answer:To solve this problem, we'll address each part step-by-step.1. Calculate the seating capacity 3 years after the last renovation. We need to find ( C(3) ) using the given polynomial function: [ C(x) = 5000x^2 + 25000x + 50000 ] Substitute ( x = 3 ): [ C(3) = 5000(3)^2 + 25000(3) + 50000 ] Calculate ( 5000 times 9 ): [ 5000 times 9 = 45000 ] Calculate ( 25000 times 3 ): [ 25000 times 3 = 75000 ] Add these results to the constant term: [ C(3) = 45000 + 75000 + 50000 = 170000 ]2. Determine the range of years ( x ) within which the seating capacity remains under 200,000 seats. We need to solve the inequality: [ 5000x^2 + 25000x + 50000 < 200000 ] Subtract 200,000 from both sides to set the inequality to less than zero: [ 5000x^2 + 25000x + 50000 - 200000 < 0 ] Simplify the inequality: [ 5000x^2 + 25000x - 150000 < 0 ] Divide the entire inequality by 5000 to simplify: [ x^2 + 5x - 30 < 0 ] To solve ( x^2 + 5x - 30 = 0 ), use the quadratic formula: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where ( a = 1 ), ( b = 5 ), and ( c = -30 ). Calculate the discriminant: [ b^2 - 4ac = 5^2 - 4(1)(-30) = 25 + 120 = 145 ] Now find the roots: [ x = frac{-5 pm sqrt{145}}{2} ] Calculate the approximate values: [ x_1 = frac{-5 + sqrt{145}}{2} approx 3.52 ] [ x_2 = frac{-5 - sqrt{145}}{2} approx -8.52 ] The quadratic ( x^2 + 5x - 30 ) is negative between the roots ( x_2 ) and ( x_1 ). Since ( x ) represents years since the last renovation, only non-negative values are meaningful. Thus, the valid range for ( x ) is: [ 0 leq x < 3.52 ]Final Answer: The final answer is ( C(3) = 170000 ) and the range of years ( x ) is ( 0 leq x < 3.52 ). I hope it is correct.

answer:To solve this problem, we will go through both parts step by step.Part 1: Finding the speed of the original train ( v ) and the new train1. Original Train: - The original train took 40 minutes to travel 30 km. - Convert the time to hours: ( 40 text{ minutes} = frac{40}{60} = frac{2}{3} text{ hours} ). - The speed of the original train ( v ) is given by the formula: [ v = frac{text{distance}}{text{time}} = frac{30 text{ km}}{frac{2}{3} text{ hours}} = 30 times frac{3}{2} = 45 text{ km/h} ]2. New Train: - The new train takes 10 minutes less, so it travels the same 30 km in ( 40 - 10 = 30 text{ minutes} ). - Convert 30 minutes to hours: ( 30 text{ minutes} = frac{30}{60} = frac{1}{2} text{ hours} ). - Let the speed of the new train be ( v_{text{new}} ). - The speed of the new train is given by: [ v_{text{new}} = frac{30 text{ km}}{frac{1}{2} text{ hours}} = 30 times 2 = 60 text{ km/h} ]Part 2: Speed of the new train through the tunnel- The new train takes 6 minutes to pass through an 8 km tunnel.- Convert 6 minutes to hours: ( 6 text{ minutes} = frac{6}{60} = frac{1}{10} text{ hours} ).- Let ( v_{text{new}} ) be the speed of the new train.- The speed of the new train is given by: [ v_{text{new}} = frac{text{distance}}{text{time}} = frac{8 text{ km}}{frac{1}{10} text{ hours}} = 8 times 10 = 80 text{ km/h} ]Final Answer: The final answer is ( v = 45 text{ km/h} ) for the original train, and the new train speed is either 60 km/h based on the trip time or 80 km/h based on the tunnel. I hope it is correct.

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