answer:(1) When k = frac{1}{2}, we can solve the system of equations: begin{cases} y = frac{1}{2}x + frac{p}{4} y^2 = 2px end{cases} Eliminating y, we get 4x^2 + 28px + p^2 = 0. Thus, begin{align} x_1 + x_2 &= 7p, x_1x_2 &= frac{p^2}{4}. end{align} Hence, (x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = 49p^2 - p^2 = 48p^2. Since the chord MN has a length of 4sqrt{15}, we have: |MN|^2 = left(1 + frac{1}{4}right)(x_1 - x_2)^2 = frac{5}{4} times 48p^2 = 16 times 15, which implies p^2 = 4, and so p = 2. Therefore, the standard equation of the parabola C is y^2 = 4x. (2) From part (1), we know that the equation of the line l is y = k(x + 1). Substituting this into the equation of the parabola, we get ky^2 - 4y + 4k = 0. Let M(x_1, y_1), N(x_2, y_2), and Q(x_3, y_3) be the coordinates of points M, N, and Q, respectively. Since M and N lie on the parabola, we have y_1y_2 = 4. The slope of line MQ is given by: k_{MQ} = frac{y_1 - y_3}{x_1 - x_3} = frac{y_1 - y_3}{frac{y_1^2}{4} - frac{y_3^2}{4}} = frac{4}{y_1 + y_3}. The equation of line MB is y + 1 = frac{4}{y_1 + y_3}(x - 1). Thus, y_1 + 1 = frac{4}{y_1 + y_3}(x_1 - 1), and we can solve for y_1 to get: y_1 = -frac{4 + y_3}{1 + y_3}. Since y_1y_2 = 4, we have: frac{4}{y_2} = -frac{4 + y_3}{1 + y_3}, which simplifies to y_2y_3 + 4(y_2 + y_3) + 4 = 0. The equation of line QN is y - y_2 = frac{4}{y_2 + y_3}(x - x_2). Solving for x, we find that x = 1, and substituting this in the equation, we find that y = -4. Thus, the line QN passes through the fixed point boxed{(1, -4)}.

answer:Since f(1) < 2, and since f(x) is an odd function defined on mathbb{R}, therefore f(-1) = -f(1), thus f(-1) > -2. Since the smallest positive period of f(x) is 3, therefore f(2) = f(3-1) = f(-1), thus f(2) > -2. Hence, the answer is boxed{(-2, +infty)}.

answer:To find the total pounds of snacks Kelly bought, we need to add the weights of the peanuts, raisins, and almonds together: Peanuts: 0.1 pounds Raisins: 0.4 pounds Almonds: 0.3 pounds Total weight = 0.1 + 0.4 + 0.3 Total weight = 0.8 pounds Kelly bought a total of boxed{0.8} pounds of snacks.

answer:1. **Right Triangle Formation**: Given QA = 6, QB = 8, and QC = 10, we check if triangle QAB is a right triangle. [ QA^2 + QB^2 = 6^2 + 8^2 = 36 + 64 = 100 = 10^2 = QC^2. ] This shows that triangle QAB is a right triangle with Q as the right angle vertex. 2. **Rotate triangle QAC about A**: Rotate triangle QAC by 60^circ about A. Let Q' be the image of Q. We have Q'A = QA = 6, Q'C = QC = 10, and AC = AB = 10. 3. **Analyze triangle AQQ'**: Since angle Q'QA = 60^circ and QA = Q'A = 6, triangle AQQ' is equilateral and QQ' = 6. 4. **Determine angle Q'QB**: triangle QAB being a right triangle and triangle AQQ' equilateral implies angle Q'QB = 90^circ. 5. **Use trigonometric ratios**: Let angle BQ'Q = beta. Then, cos beta = frac{6}{10} = frac{3}{5} and sin beta = frac{4}{5}. 6. **Apply the Law of Cosines in triangle QBC**: [ BC^2 = QC^2 + QB^2 - 2 cdot QC cdot QB cdot cos(angle BQC), ] where angle BQC = 60^circ + beta. Using the cosine addition formula: [ cos(60^circ + beta) = cos 60^circ cos beta - sin 60^circ sin beta = frac{1}{2} cdot frac{3}{5} - frac{sqrt{3}}{2} cdot frac{4}{5} = frac{3}{10} - frac{2sqrt{3}}{5}. ] Plugging in the values: [ BC^2 = 100 + 64 - 2 cdot 10 cdot 8 cdot left(frac{3}{10} - frac{2sqrt{3}}{5}right) = 164 - 160 left(frac{3}{10} - frac{2sqrt{3}}{5}right) = 100 + 32sqrt{3}. ] 7. **Calculate the area of triangle ABC**: [ text{Area} = frac{sqrt{3}}{4} times (BC)^2 = frac{sqrt{3}}{4} times (100 + 32sqrt{3}) = 25sqrt{3} + 24. ] Approximating sqrt{3} approx 1.732, the area calculates to: [ 25 times 1.732 + 24 approx 43.3 + 24 = 67.3. ] Thus, the area of triangle ABC to the nearest integer is 67. The final answer is boxed{B}