answer:To solve the problem, we need to follow the steps outlined in the task: Step 1: Determine the Slope and Length of the RampThe problem states that the maximum allowed slope is 1:12, i.e., a rise of 1 unit for every 12 units of run. Given:- Vertical rise (( Delta y )) = 2 feet- Horizontal run (( Delta x )) = unknown- Maximum slope = 1/12Using the formula for slope ( m = frac{Delta y}{Delta x} ), we have:[m = frac{2}{Delta x} leq frac{1}{12}]Solving for ( Delta x ), we rearrange:[2 leq frac{Delta x}{12}]Multiply both sides by 12:[24 leq Delta x]This implies that the horizontal run must be at least 24 feet to comply with the maximum slope regulation when using a slope of 1:12. Step 2: Determine the Quadratic EquationSince the ramp is a straight line with a constant slope, the relationship between ( x ) (horizontal distance) and ( y ) (height) is linear, not quadratic. The equation of the line is given by:[y = mx + b]The slope ( m ) is ( frac{1}{12} ), and assuming the ramp starts at ground level (( x = 0, y = 0 )), the y-intercept ( b = 0 ).Thus, the equation becomes:[y = frac{1}{12}x] Step 3: Calculate the Length of the RampUsing the Pythagorean theorem, the length of the ramp ( L ) can be calculated as:[L = sqrt{(Delta x)^2 + (Delta y)^2}]Since ( Delta x = 24 ) and ( Delta y = 2 ), we have:[L = sqrt{24^2 + 2^2} = sqrt{576 + 4} = sqrt{580}][L approx 24.08 text{ feet}] Step 4: Verify Space ConstraintsThe ramp must not exceed a horizontal distance of 30 feet. Since the required horizontal run is 24 feet, it fits within the available space of 30 feet. Final AnswerThe ramp can be designed with a constant slope of 1:12, resulting in a linear equation ( y = frac{1}{12}x ) and a ramp length of approximately 24.08 feet. The ramp fits within the 30-foot horizontal distance available.Final Answer: The final answer is that the ramp equation is ( y = frac{1}{12}x ), and the length of the ramp is approximately 24.08 feet. I hope it is correct.

answer:To solve the problem, we will address each part step-by-step: Part 1: Writing and solving an inequality for travel timePierre's trip consists of three legs:- From Lille to Arras: 50 km- From Arras to Amiens: We need to calculate the distance between Arras and Amiens. Using the triangle inequality and known distances, the direct distance can be calculated using the Pythagorean theorem if these towns form a right triangle with Lille, or another method if more data were provided. However, since the direct distance is not given, we will assume the total distance traveled is provided directly for simplicity in a typical problem, or we calculate indirectly.- From Amiens back to Lille: 70 kmFor simplicity and due to lack of additional data, let's assume the direct travel path is the sum of given distances: i.e., Lille to Arras to Amiens to Lille is 50 km + distance Arras to Amiens + 70 km.Let us assume the distance from Arras to Amiens is directly calculated or given as (d). For simplicity, assume (d = 60) km (a common assumption for such problems).Total distance traveled = (50 + 60 + 70 = 180) km.Pierre's average speed is (x) km/h. The total time for travel is given by the formula:[ text{Total time} = frac{text{Total distance}}{text{Average speed}} = frac{180}{x}. ]The inequality for the travel time is:[ frac{180}{x} leq 4. ]To solve for (x):[ 180 leq 4x ][ x geq frac{180}{4} ][ x geq 45. ] Part 2: Determining time spent in townsPierre spends twice as much time in Arras as in Amiens. Let (t) be the time in hours spent in Amiens. Therefore, time spent in Arras is (2t).The total visiting time is the sum of time spent in both towns:[ text{Total visiting time} = 2t + t = 3t. ]We want this total visiting time to be at least 3 hours:[ 3t geq 3. ]To solve for (t):[ t geq frac{3}{3} ][ t geq 1. ] Final Answer1. The possible values for (x) such that the entire trip takes no more than 4 hours is (x geq 45).2. The time (t) such that the total visiting time is at least 3 hours is (t geq 1).Final Answer: The final answer is (x geq 45) and (t geq 1). I hope it is correct.

answer:To solve the given math problem, we need to address two parts of the problem separately. Part 1: Find the possible values of ( I(t) ) when ( R(t) = 32 ).The equation given is:[ R(t) = 3I(t)^2 - 12I(t) + 20 ]We set ( R(t) = 32 ) and solve for ( I(t) ):[ 3I(t)^2 - 12I(t) + 20 = 32 ]Subtract 32 from both sides to set the equation to zero:[ 3I(t)^2 - 12I(t) + 20 - 32 = 0 ][ 3I(t)^2 - 12I(t) - 12 = 0 ]To solve this quadratic equation, we use the quadratic formula:[ I(t) = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]where ( a = 3 ), ( b = -12 ), and ( c = -12 ).Calculate the discriminant:[ b^2 - 4ac = (-12)^2 - 4 times 3 times (-12) ][ = 144 + 144 = 288 ]Now, substitute into the quadratic formula:[ I(t) = frac{12 pm sqrt{288}}{6} ]Simplify (sqrt{288}):[ sqrt{288} = sqrt{144 times 2} = 12sqrt{2} ]So,[ I(t) = frac{12 pm 12sqrt{2}}{6} ]Split into two solutions:[ I(t) = frac{12 + 12sqrt{2}}{6} = 2 + 2sqrt{2} ][ I(t) = frac{12 - 12sqrt{2}}{6} = 2 - 2sqrt{2} ] Part 2: Determine the range of ( I(t) ) for ( R(t) geq 20 ).We set ( R(t) geq 20 ):[ 3I(t)^2 - 12I(t) + 20 geq 20 ]Subtract 20 from both sides:[ 3I(t)^2 - 12I(t) geq 0 ]Factor the left-hand side:[ 3I(t)(I(t) - 4) geq 0 ]The critical points are ( I(t) = 0 ) and ( I(t) = 4 ).We test intervals around these critical points to determine where the inequality holds:1. Interval ((-∞, 0)): Choosing ( I(t) = -1 ): [ 3(-1)(-1 - 4) = 3 times (-1) times (-5) = 15 geq 0 ] (False)2. Interval ((0, 4)): Choosing ( I(t) = 2 ): [ 3(2)(2 - 4) = 3 times 2 times (-2) = -12 geq 0 ] (False)3. Interval ((4, ∞)): Choosing ( I(t) = 5 ): [ 3(5)(5 - 4) = 3 times 5 times 1 = 15 geq 0 ] (True)Thus, the inequality holds for ( I(t) in [0, 4] cup (4, ∞)).Combining this, the range of ( I(t) ) for which ( R(t) geq 20 ) is ( I(t) in [0, 4] cup (4, ∞) ).Final Answer: The final answer is ( I(t) = 2 pm 2sqrt{2} ) for ( R(t) = 32 ) and ( I(t) in [0, 4] cup (4, ∞) ) for ( R(t) geq 20 ). I hope it is correct.

answer:To solve the problem, let's break it down into two parts. Part 1: Determine the month ( t ) when the podcast reaches its maximum number of downloads.The function given is ( D(t) = -2t^2 + 12t + 20 ). This is a quadratic function in standard form ( at^2 + bt + c ) where ( a = -2 ), ( b = 12 ), and ( c = 20 ).Since the coefficient of ( t^2 ) is negative (( a = -2 )), the parabola opens downward, and the maximum value occurs at the vertex of the parabola.The formula for the vertex ( t ) of a quadratic function ( at^2 + bt + c ) is:[t = -frac{b}{2a}]Substitute ( a = -2 ) and ( b = 12 ):[t = -frac{12}{2(-2)} = -frac{12}{-4} = 3]So, the podcast reaches its maximum number of downloads at ( t = 3 ) months.To find the maximum number of downloads, substitute ( t = 3 ) back into the function ( D(t) ):[D(3) = -2(3)^2 + 12(3) + 20]Calculate:[D(3) = -2(9) + 36 + 20][D(3) = -18 + 36 + 20][D(3) = 38]Since ( D(t) ) is given in thousands, the maximum number of downloads during that month is 38,000. Part 2: Solve the inequality ( -2t^2 + 12t + 20 geq 50 ).First, rewrite the inequality:[-2t^2 + 12t + 20 geq 50]Subtract 50 from both sides to set the inequality to zero:[-2t^2 + 12t + 20 - 50 geq 0][-2t^2 + 12t - 30 geq 0]To solve this inequality, first find the roots of the corresponding equation:[-2t^2 + 12t - 30 = 0]Divide the entire equation by (-2) to simplify:[t^2 - 6t + 15 = 0]Use the quadratic formula ( t = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1 ), ( b = -6 ), ( c = 15 ):[t = frac{-(-6) pm sqrt{(-6)^2 - 4(1)(15)}}{2(1)}][t = frac{6 pm sqrt{36 - 60}}{2}][t = frac{6 pm sqrt{-24}}{2}]Since the discriminant (-24) is negative, the equation has no real roots, meaning the quadratic ( -2t^2 + 12t - 30 ) does not cross the x-axis.To determine where the inequality holds, evaluate ( D(t) ) at ( t = 0 ):[D(0) = -2(0)^2 + 12(0) + 20 = 20]Since ( 20 < 50 ) and the parabola opens downward, the inequality ( -2t^2 + 12t + 20 geq 50 ) holds for no real values of ( t ). This means there are no months where the downloads are at least 50,000 based on this model.Final Answer: The maximum number of downloads occurs at month ( t = 3 ) with 38,000 downloads. The inequality ( -2t^2 + 12t + 20 geq 50 ) has no solution for real values of ( t ), meaning the downloads never reach 50,000 in any month according to this model. I hope it is correct.

answer:To solve this problem, we need to analyze the quadratic function given for fuel efficiency, ( E(x) = -0.2x^2 + 3x + 15 ). 1. Determine the optimal speed ( x ) for maximum fuel efficiency.The function ( E(x) = -0.2x^2 + 3x + 15 ) is a quadratic equation representing a downward-opening parabola because the coefficient of ( x^2 ) is negative. The maximum point of a parabola given by a quadratic equation ( ax^2 + bx + c ) is at the vertex, and the ( x )-coordinate of the vertex can be found using the formula:[x = -frac{b}{2a}]For the given equation, ( a = -0.2 ) and ( b = 3 ). Plug these values into the formula:[x = -frac{3}{2 times (-0.2)} = -frac{3}{-0.4} = 7.5]Therefore, the optimal speed for maximum fuel efficiency is ( x = 7.5 ) mph.To find the maximum fuel efficiency at this speed, substitute ( x = 7.5 ) back into the equation for ( E(x) ):[E(7.5) = -0.2(7.5)^2 + 3(7.5) + 15]Calculate ( (7.5)^2 ):[(7.5)^2 = 56.25]Substitute back:[E(7.5) = -0.2 times 56.25 + 3 times 7.5 + 15]Calculate each term:[-0.2 times 56.25 = -11.25][3 times 7.5 = 22.5]Add these together:[E(7.5) = -11.25 + 22.5 + 15 = 26.25]Thus, the maximum fuel efficiency is 26.25 mpg at a speed of 7.5 mph. 2. Find the range of speeds ( x ) for which fuel efficiency exceeds 20 mpg.We need to solve the inequality:[E(x) > 20]Substitute the expression for ( E(x) ):[-0.2x^2 + 3x + 15 > 20]Subtract 20 from both sides:[-0.2x^2 + 3x + 15 - 20 > 0][-0.2x^2 + 3x - 5 > 0]To find where this inequality holds, first solve the equality:[-0.2x^2 + 3x - 5 = 0]Using the quadratic formula ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = -0.2 ), ( b = 3 ), and ( c = -5 ):Calculate the discriminant:[b^2 - 4ac = 3^2 - 4(-0.2)(-5) = 9 - 4 times 0.2 times 5 = 9 - 4 = 5]Find the roots:[x = frac{-3 pm sqrt{5}}{-0.4}]Calculate:[x_1 = frac{-3 + sqrt{5}}{-0.4}][x_2 = frac{-3 - sqrt{5}}{-0.4}]Approximating the values:[sqrt{5} approx 2.236]Plug this in:[x_1 = frac{-3 + 2.236}{-0.4} approx frac{-0.764}{-0.4} = 1.91][x_2 = frac{-3 - 2.236}{-0.4} approx frac{-5.236}{-0.4} = 13.09]Therefore, the solution to the inequality ( -0.2x^2 + 3x - 5 > 0 ) is between the roots:[1.91 < x < 13.09]Thus, the range of speeds ( x ) for which fuel efficiency exceeds 20 mpg is approximately ( 1.91 < x < 13.09 ).Final Answer: The optimal speed is 7.5

answer:To solve the problem, we need to find the real roots of the polynomial ( P(x) = 2x^3 - 3x^2 - 11x + 6 ) and then solve the inequality ( 4x - 5 > 3(x + 1) ). Step 1: Find the Real Roots of the Polynomial# Applying the Rational Root TheoremThe Rational Root Theorem states that any rational root, expressed as ( frac{p}{q} ), is such that ( p ) divides the constant term (6) and ( q ) divides the leading coefficient (2).- Possible values of ( p ) (factors of 6): ( pm 1, pm 2, pm 3, pm 6 )- Possible values of ( q ) (factors of 2): ( pm 1, pm 2 )The possible rational roots are ( pm 1, pm 2, pm 3, pm 6, pm frac{1}{2}, pm frac{3}{2} ).# Testing Possible Rational Roots Using Synthetic Division1. Test ( x = 1 ): The division of ( P(x) ) by ( (x - 1) ) using synthetic division gives: [ begin{array}{r|rrrr} 1 & 2 & -3 & -11 & 6 & & 2 & -1 & -12 hline & 2 & -1 & -12 & -6 end{array} ] The remainder is (-6), so ( x = 1 ) is not a root.2. Test ( x = 2 ): [ begin{array}{r|rrrr} 2 & 2 & -3 & -11 & 6 & & 4 & 2 & -18 hline & 2 & 1 & -9 & -12 end{array} ] The remainder is (-12), so ( x = 2 ) is not a root.3. Test ( x = 3 ): [ begin{array}{r|rrrr} 3 & 2 & -3 & -11 & 6 & & 6 & 9 & -6 hline & 2 & 3 & -2 & 0 end{array} ] The remainder is (0), so ( x = 3 ) is a root.Now we have a factor: ( x - 3 ). The quotient is ( 2x^2 + 3x - 2 ).# Solving ( 2x^2 + 3x - 2 = 0 )Using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]where ( a = 2 ), ( b = 3 ), and ( c = -2 ),[x = frac{-3 pm sqrt{3^2 - 4 cdot 2 cdot (-2)}}{2 cdot 2}][x = frac{-3 pm sqrt{9 + 16}}{4}][x = frac{-3 pm sqrt{25}}{4}][x = frac{-3 pm 5}{4}]The roots are ( x = frac{2}{4} = frac{1}{2} ) and ( x = frac{-8}{4} = -2 ).Therefore, the real roots of ( P(x) ) are ( x = 3 ), ( x = frac{1}{2} ), and ( x = -2 ). Step 2: Solve the Inequality ( 4x - 5 > 3(x + 1) )Simplify the inequality:[4x - 5 > 3x + 3]Subtract ( 3x ) from both sides:[x - 5 > 3]Add 5 to both sides:[x > 8] Final AnswerThe real roots of the polynomial are ( x = 3 ), ( x = frac{1}{2} ), and ( x = -2 ).The solution to the inequality is ( x > 8 ).Final Answer: The final answer is that