answer:1. Consider the case when a=0. The function becomes f(x)=4x-3, which is an increasing function on the interval [2,+infty) due to the properties of linear functions. This does not satisfy the given condition. 2. If a < 0, the graph of the function f(x) is a downward-opening parabola. The axis of symmetry is given by x=- frac {2a+2}{a}leqslant 2, which implies aleqslant - frac {1}{2}. 3. When a > 0, the graph of the function f(x) is an upward-opening parabola, which does not meet the requirement of the problem. Therefore, the range of values for a is aleqslant - frac {1}{2}. The answer is boxed{A}. This problem requires an understanding of the monotonicity of linear and quadratic functions and the ability to analyze cases based on the coefficient of the x^2 term, which is the letter a. The problem involves classifying and discussing the function based on whether a is 0, positive, or negative, and then solving each case using the properties of the function's graph.

answer:Step 1: Rotate the coordinate axes by angle alpha The given equation is: [ 8 x^{2}+4 x y+5 y^{2}-56 x-32 y+80=0. ] We perform a rotation of the coordinate axes by an angle alpha using the following transformations: [ x = x_1 cos alpha - y_1 sin alpha, ] [ y = x_1 sin alpha + y_1 cos alpha. ] Substituting these into the equation, we get: [ begin{aligned} 8 (x_1 cos alpha - y_1 sin alpha)^2 + 4 (x_1 cos alpha - y_1 sin alpha)(x_1 sin alpha + y_1 cos alpha) + 5 (x_1 sin alpha + y_1 cos alpha)^2 - 56 (x_1 cos alpha - y_1 sin alpha) - 32 (x_1 sin alpha + y_1 cos alpha) + 80 = 0. end{aligned} ] Step 2: Expand and collect terms involving ( x_1 y_1 ) Expanding and simplifying, we collect the terms involving (x_1 y_1): [ begin{aligned} & 8(x_1^2 cos^2 alpha - 2 x_1 y_1 cos alpha sin alpha + y_1^2 sin^2 alpha) + & 4(x_1^2 cos alpha sin alpha + x_1 y_1 (cos^2 alpha - sin^2 alpha) - y_1^2 sin alpha cos alpha) + & 5(x_1^2 sin^2 alpha + 2 x_1 y_1 sin alpha cos alpha + y_1^2 cos^2 alpha) - & 56(x_1 cos alpha - y_1 sin alpha) - & 32(x_1 sin alpha + y_1 cos alpha) + 80 = 0. end{aligned} ] Step 3: Set the condition to nullify the coefficient of x_1 y_1 We want the coefficient of (x_1 y_1) to be zero. After combining terms involving (x_1 y_1), the coefficient is: [ -16 cos alpha sin alpha + 4 (cos^2 alpha - sin^2 alpha) + 10 sin alpha cos alpha = 0. ] This simplifies to: [ -6 cos alpha sin alpha + 4 cos^2 alpha - 4 sin^2 alpha = 0, ] or equivalently: [ 2 tan^2 alpha + 3 tan alpha - 2 = 0. ] Step 4: Solve for tan alpha Solving the quadratic equation: [ 2 t^2 + 3 t - 2 = 0, ] we find: [ t = frac{-3 pm sqrt{9 + 16}}{4} = frac{-3 pm 5}{4}. ] Thus: [ t = frac{1}{2} quad text{or} quad t = -2. ] We choose tan alpha = frac{1}{2}. Therefore: [ cos alpha = frac{2}{sqrt{5}}, quad sin alpha = frac{1}{sqrt{5}}, ] [ cos^2 alpha = frac{4}{5}, quad sin^2 alpha = frac{1}{5}, quad cos alpha sin alpha = frac{2}{5}. ] Step 5: Substitute back into the equation Substituting these values: [ 8 (x_1^2 frac{4}{5} - 2 x_1 y_1 frac{2}{5} + y_1^2 frac{1}{5}) + 4 (x_1^2 frac{2}{5} + x_1 y_1 frac{3}{5} - y_1^2 frac{2}{5}) + 5 (x_1^2 frac{1}{5} + 2 x_1 y_1 frac{2}{5} + y_1^2 frac{4}{5}) - ] [ 56 (x_1 frac{2}{sqrt{5}} - y_1 frac{1}{sqrt{5}}) - 32 (x_1 frac{1}{sqrt{5}} + y_1 frac{2}{sqrt{5}}) + 80 = 0. ] Simplifying, we get: [ 9 x_1^2 + 4 y_1^2 - frac{144}{sqrt{5}} x_1 - frac{64}{sqrt{5}} y_1 + 80 = 0. ] Step 6: Complete the square Completing the square for (x_1) and (y_1): [ 9 left(x_1^2 - frac{16}{sqrt{5}} x_1 right) + 4 left(y_1^2 - frac{2}{sqrt{5}} y_1 right) + 80 = 0, ] [ 9 left(x_1 - frac{8}{sqrt{5}}right)^2 + 4 left(y_1 - frac{1}{sqrt{5}}right)^2 = 36. ] Step 7: Normalize to canonical form Normalize the equation: [ frac{(x_1 - frac{8}{sqrt{5}})^2}{4} + frac{(y_1 - frac{1}{sqrt{5}})^2}{9} = 1. ] This is the equation of an ellipse with: [ boxed{frac{(x_2)^2}{4} + frac{(y_2)^2}{9} = 1}. ]

answer:To calculate the total distance Alice walks in a week from Monday to Friday, we break down the calculation into daily and weekly totals. 1. **Daily Distance Calculation:** - In the morning, Alice walks through a large grass field to get to school, covering a distance of 10 miles. - In the afternoon, to avoid the hot sun, she takes a longer path through a forest on her way home, covering 12 miles. - Therefore, the total distance Alice walks in one day is the sum of the morning and afternoon distances: 10 , text{miles} + 12 , text{miles} = 22 , text{miles}. 2. **Weekly Distance Calculation:** - The calculation above is for Monday. For the remaining four days, from Tuesday to Friday, she repeats the same distance each day. - Thus, for these four days, she walks a total of 4 times 22 , text{miles} = 88 , text{miles}. - Adding Monday's distance to the total for Tuesday to Friday gives us the total distance for the week: 22 , text{miles} + 88 , text{miles} = 110 , text{miles}. Therefore, by Friday, Alice will have walked a total of boxed{110 , text{miles}} for the week.

answer:Since S_n = n + (n-1) times 2 + (n-2) times 2^2 + ldots + 2 times 2^{n-2} + 2^{n-1}, 2S_n = 2n + (n-1) times 2^2 + (n-2) times 2^3 + ldots + 2 times 2^{n-1} + 2^n, Subtracting the two equations, we get -S_n = n - (2 + 2^2 + ldots + 2^{n-1}) - 2^n, Simplifying, we find the answer is option boxed{text{D}}.