answer:Let's denote the wholesale cost of the sweater as W. The merchant marks up the sweater by 64.28571428571428% for its normal retail price. This means the normal retail price (R) is: R = W + 0.6428571428571428 * W R = W * (1 + 0.6428571428571428) R = W * 1.6428571428571428 The merchant makes a 40% profit on the wholesale cost when the sweater is sold at a certain discount. This means the sale price (S) with the discount is: S = W + 0.40 * W S = W * (1 + 0.40) S = W * 1.40 The discount percentage (D) is the reduction from the normal retail price to the sale price, expressed as a percentage of the normal retail price. So we have: D = (R - S) / R * 100% Substituting the expressions for R and S we have: D = (W * 1.6428571428571428 - W * 1.40) / (W * 1.6428571428571428) * 100% D = (1.6428571428571428W - 1.40W) / 1.6428571428571428W * 100% D = (0.2428571428571428W) / 1.6428571428571428W * 100% D = 0.2428571428571428 / 1.6428571428571428 * 100% Now we calculate the discount percentage: D ≈ 0.1477 * 100% D ≈ 14.77% So the discount percentage is approximately boxed{14.77%} .

answer:First, complete the square for the x and y terms: [ 3(x^2 - 4x) + 9(y^2 + 2y) = k. ] Completing the square: [ 3((x^2 - 4x + 4) - 4) + 9((y^2 + 2y + 1) - 1) = k. ] Simplified: [ 3(x-2)^2 - 12 + 9(y+1)^2 - 9 = k. ] [ 3(x-2)^2 + 9(y+1)^2 = k + 21. ] For this to be an ellipse, we need: [ k + 21 > 0 Rightarrow k > -21. ] Thus, the graph is a non-degenerate ellipse if and only if k > -21. Therefore, the value of a is boxed{-21}.

answer:1. **Understanding the problem**: We need to determine the axes of symmetry for the curves ( y = sin x ) and ( y = cos x ). 2. **Symmetry in Trigonometric Functions**: For two functions ( f(x) ) and ( g(x) ) to be symmetric with respect to the line ( x = a ), the following must hold: [ f(a - h) = g(a + h) ] for any ( h ). 3. **Verify Symmetry for ( y = sin x ) and ( y = cos x )**: We will check ( y = sin x ) and ( y = cos x ) for a potential line ( x = a ): Let’s test ( x = frac{pi}{4} ): [ sin left( frac{pi}{4} - h right) = cos left( frac{pi}{4} + h right) ] 4. **Using Trigonometric Identities**: [ sin left( frac{pi}{4} - h right) = sin frac{pi}{4} cos h - cos frac{pi}{4} sin h ] [ = frac{sqrt{2}}{2} cos h - frac{sqrt{2}}{2} sin h ] [ = frac{sqrt{2}}{2} (cos h - sin h) ] Additionally: [ cos left( frac{pi}{4} + h right) = cos frac{pi}{4} cos h - sin frac{pi}{4} sin h ] [ = frac{sqrt{2}}{2} cos h - frac{sqrt{2}}{2} sin h ] [ = frac{sqrt{2}}{2} (cos h - sin h) ] 5. **Result for ( x = frac{pi}{4} )**: Since: [ sin left( frac{pi}{4} - h right) = cos left( frac{pi}{4} + h right) ] This confirms the symmetry about ( x = frac{pi}{4} ). 6. **Generalizing the Result**: Similiar checks can be applied to other lines ( x = frac{5pi}{4}, frac{9pi}{4}, ldots ) and their negatives ( x = -frac{pi}{4}, -frac{5pi}{4}, -frac{9pi}{4}, ldots ). # Conclusion The curves ( y = sin x ) and ( y = cos x ) are mutually symmetric with respect to the vertical lines: [ x = frac{pi}{4}, frac{5}{4} pi, frac{9}{4} pi, ldots, -frac{pi}{4}, -frac{5}{4} pi, -frac{9}{4} pi, ldots ] (boxed{x = frac{pi}{4}, frac{5}{4} pi, frac{9}{4} pi, ldots, -frac{pi}{4}, -frac{5}{4} pi, -frac{9}{4} pi, ldots})

answer:To solve this problem, we need to demonstrate that it is possible to obtain any even constant from the given function and the operations allowed. 1. **Establing the Induction Base**: We start with the function given: [ f(x) = sin(x) + cos(x) ] Taking its derivatives, we get: [ f'(x) = cos(x) - sin(x) ] [ f''(x) = -sin(x) - cos(x) ] 2. **Induction Step**: The allowed operations (taking derivatives, sums, and products) produce new polynomial functions in terms of (sin(x)) and (cos(x)) with integer coefficients. Consider the product of (f(x)) and (f''(x)): [ f(x) cdot f''(x) = (sin(x) + cos(x)) cdot (-sin(x) - cos(x)) ] Simplifying, we obtain: [ (sin(x) + cos(x)) (-sin(x) - cos(x)) = -(sin(x)^2 + sin(x)cos(x) + cos(x)sin(x) + cos(x)^2) ] [ = -(sin^2(x) + 2sin(x)cos(x) + cos^2(x)) ] [ = -(1 + 2sin(x)cos(x)) ] As (sin^2(x) + cos^2(x) = 1). 3. **Verifying for Constants**: Now, we identify constant terms. Suppose there exists a constant (c) such that the function value becomes constant for all (x). Assume that (f(x)) and its derived functions are combinations of (sin(x)) and (cos(x)): [ g(x) = p(sin(x), cos(x)) ] where (p) is a polynomial with integer coefficients. If (g(x) = c), then: [ g(0) = p(0, 1) ] gives an integer from ({sin(x), cos(x)}). 4. **Conclusion**: We see that any polynomial combinations of (sin(x)) and (cos(x)) taking derivatives and considering their values at specific points indeed result in even values. Furthermore, the sum of coefficients combining even and integers suggests even constants. Therefore: [ boxed{text{any even number}} ]