answer:The given quadratic equation is 2x^{2}-5x+2=0. To find its roots, we can employ the quadratic formula x = frac{-b pm sqrt{b^2 - 4ac}}{2a}, where a=2, b=-5, and c=2. Computing the determinant Delta=b^2 -4ac = (-5)^2 - 4cdot2cdot2 = 25 - 16 = 9. Now we can find the roots: x_1 = frac{-(-5) + sqrt{9}}{2cdot2} = frac{5 + 3}{4} = 2 x_2 = frac{-(-5) - sqrt{9}}{2cdot2} = frac{5 - 3}{4} = frac{1}{2} Thus, the roots are x_1=2 and x_2=frac{1}{2}. The eccentricity (e) for an ellipse is 0 < e < 1 and for a hyperbola, it is e > 1. For a parabola, the eccentricity is exactly 1. Since x_1=2>1, it can serve as the eccentricity of a hyperbola. And x_2=frac{1}{2} in (0,1), making it suitable as the eccentricity of an ellipse. Therefore, the correct answer is: [boxed{A: Ellipse and Hyperbola}] This problem essentially tests the understanding of the relation between the eccentricity of conic sections and the roots of a quadratic equation.

answer:1. **Calculate the length of AB**: Since ABCD is a parallelogram, AB = DC = 15. 2. **Determine the length of AE**: Since EB = 3 and AB = 15, we find AE by subtracting EB from AB: [ AE = AB - EB = 15 - 3 = 12. ] 3. **Calculate the area of parallelogram ABCD**: Using AB as the base and DE as the altitude, the area is: [ text{Area} = AB times DE = 15 times 5 = 75. ] 4. **Use the area to find DF using the base AD**: Since AD = BC = 15, the area can also be expressed as: [ text{Area} = AD times DF = 15 times DF. ] Solving for DF: [ 75 = 15 times DF implies DF = frac{75}{15} = 5. ] Conclusion: The calculation reveals that the length of the altitude DF is 5. Thus, the final answer is (5). The final answer is boxed{(B) 5}

answer:The salesman sold 120 kilograms in the morning and 240 kilograms in the afternoon. To find the total amount of pears he sold that day, we simply add the two amounts together: 120 kg (morning) + 240 kg (afternoon) = 360 kg So, the salesman sold a total of boxed{360} kilograms of pears that day.

answer:(1) Let x = y = 0, we have f(0) + f(0) = f(0). Therefore, f(0) = 0. [ therefore f(0) = 0. ] Next, let y = -x. We get f(x) + f(-x) = f(0) = 0, which implies f(-x) = -f(x). Therefore, f(x) is an odd function on (-1,1). [ therefore boxed{text{f(x) is an odd function on (-1,1).}} ] (2) From dfrac{1-x}{1+x} > 0, we derive that -1 < x < 1, which means the domain of f(x) = lgdfrac{1-x}{1+x} is (-1,1). Moreover, [ f(x) + f(y) = lgdfrac{1-x}{1+x} + lgdfrac{1-y}{1+y} = lgleft(dfrac{1-x}{1+x} cdot dfrac{1-y}{1+y}right) = lgdfrac{1-x-y+xy}{1+x+y+xy}, ] [ fleft(dfrac{x+y}{1+xy}right) = lgdfrac{1-dfrac{x+y}{1+xy}}{1+dfrac{x+y}{1+xy}} = lgdfrac{1-x-y+xy}{1+x+y+xy}. ] Therefore, f(x) + f(y) = fleft(dfrac{x+y}{1+xy}right). When x < 0, 1-x > 1+x > 0, thus dfrac{1-x}{1+x} > 1, and therefore lgdfrac{1-x}{1+x} > 0. Hence, the function f(x) = lgdfrac{1-x}{1+x} meets these conditions. [ boxed{text{The function f(x) = lgdfrac{1-x}{1+x} satisfies the given conditions.}} ] (3) Suppose -1 < x_1 < x_2 < 1, then [ f(x_1) - f(x_2) = f(x_1) + f(-x_2) = fleft(dfrac{x_1 - x_2}{1 - x_1 x_2}right), ] [ because x_1 - x_2 < 0 text{ and } -1 < x_1 x_2 < 1, ] [ therefore dfrac{x_1 - x_2}{1 - x_1 x_2} < 0. ] By condition (2) we know fleft(dfrac{x_1 - x_2}{1 - x_1 x_2}right) > 0, thus f(x_1) - f(x_2) > 0, meaning f(x_1) > f(x_2). Therefore, f(x) is monotonically decreasing on (-1,1). Since fleft(-dfrac{1}{2}right) = 1, we have fleft(dfrac{1}{2}right) = -1. Setting F(x) = f(x) + dfrac{1}{2} = 0 yields 2f(x) = -1. And 2f(x) = f(x) + f(x) = fleft(dfrac{2x}{1+x^2}right). Because f(x) is monotonically decreasing on (-1,1), [ therefore dfrac{2x}{1+x^2} = dfrac{1}{2}, ] solving for x gives x = 2 pm sqrt{3}. However, since x in (-1,1), we have [ x = 2 - sqrt{3}. ] Thus, the solution to the original equation is x = 2 - sqrt{3}. [ boxed{x = 2 - sqrt{3}}. ]