answer:To find the surface area of solid QVWX, calculate the area of each of its triangular faces and add them up. Areas of triangle QVX and triangle QWX: Both triangles are right-angled with legs of lengths 5 (half the side of the equilateral triangle) and 10 (half the prism's height). Therefore, the area of each triangle is: text{Area} = frac{1}{2} times 5 times 10 = 25. Area of triangle QVW: This triangle is equilateral with side length 5. The altitude (height) of an equilateral triangle can be found by h = frac{sqrt{3}}{2}a where a is the side length, so: h = frac{sqrt{3}}{2} times 5 = frac{5sqrt{3}}{2}. The area of triangle QVW is: text{Area} = frac{1}{2} times 5 times frac{5sqrt{3}}{2} = frac{25sqrt{3}}{4}. Area of triangle VWX: With VW = 5 and VX = WX = sqrt{125} (since these are the hypotenuses of the right triangles with legs 5 and 10), set up an altitude from X to VW at midpoint M, where VM = MW = 2.5. Using the Pythagorean Theorem: XM = sqrt{VX^2 - VM^2} = sqrt{125 - 6.25} = sqrt{118.75}. The area of triangle VWX is: text{Area} = frac{1}{2} times 5 times sqrt{118.75}. The total surface area of solid QVWX is: 50 + frac{25sqrt{3}}{4} + frac{5sqrt{118.75}}{2} = boxed{50 + frac{25sqrt{3}}{4} + frac{5sqrt{118.75}}{2}}.

answer:We follow similar steps as in the original problem: - Extend AD to E, where it intersects the perpendicular to BC from C. - BD^2 = BA^2 - AD^2 = 15^2 - 9^2 = 144, so BD = 12text{ cm}. - BC^2 = DC^2 - BD^2 = 26^2 - 12^2 = 484 - 144 = 340, so BC = sqrt{340} approx 18.44text{ cm}. Taking approximately 18 cm for ease of calculation in the next step. - Assume BCED is a rectangle. Then DE = BC = 18text{ cm}, CE = BD = 12text{ cm}. - AE = 18 + 9 = 27text{ cm}. - By the Pythagorean Theorem in triangle AEC, AC^2 = AE^2 + CE^2 = 27^2 + 12^2 = 729 + 144 = 873, thus AC = sqrt{873} approx 29.5text{ cm}. - So, AC approx boxed{29.5} to the nearest tenth of a centimeter.

answer:1. Solution: f'(x) = ln x + 1 (x > 0). Let f'(x) = 0, we get x = frac{1}{e} ldots (2 text{ points}) Since f'(x) < 0 when x in (0, frac{1}{e}); and f'(x) > 0 when x in (frac{1}{e}, +infty), Therefore, when x = frac{1}{e}, f(x)_{min} = -frac{1}{e} ldots (3 text{ points}) 2. Solution: F'(x) = 2x - (a - 2) - frac{a}{x} = frac{(2x - a)(x + 1)}{x} (x > 0) ldots When a leqslant 0, F'(x) > 0, the function F(x) is monotonically increasing on (0, +infty), and the monotonically increasing interval of the function F(x) is (0, +infty) ldots When a > 0, from F'(x) > 0, we get x > frac{a}{2}; from F'(x) < 0, we get 0 < x < frac{a}{2}. Therefore, the monotonically increasing interval of the function F(x) is (frac{a}{2}, +infty), and the monotonically decreasing interval is (0, frac{a}{2}) ldots (7 text{ points}) 3. Proof: Since x_1 and x_2 are two distinct real roots of the equation F(x) = m, from (1) we know that a > 0. Without loss of generality, let 0 < x_1 < x_2, then x_1^2 - (a - 2)x_1 - a ln x_1 = c, x_2^2 - (a - 2)x_2 - a ln x_2 = c. Subtracting the two equations, we get x_1^2 - (a - 2)x_1 - a ln x_1 - x_2^2 + (a - 2) cdot x_2 + a ln x_2 = 0, That is, x_1^2 + 2x_1 - x_2^2 - 2x_2 = ax_1 + a ln x_1 - ax_2 - a ln x_2 = a(x_1 + ln x_1 - x_2 - ln x_2). Therefore, a = frac{x_1^2 + 2x_1 - x_2^2 - 2x_2}{x_1 + ln x_1 - x_2 - ln x_2}. Since F'(frac{a}{2}) = 0, We need to prove that x_1 + x_2 > frac{x_1^2 + 2x_1 - x_2^2 - 2x_2}{x_1 + ln x_1 - x_2 - ln x_2}, That is, ln frac{x_1}{x_2} < frac{2x_1 - 2x_2}{x_1 + x_2}. Let t = frac{x_1}{x_2} (0 < t < 1). Define g(t) = ln t - frac{2t - 2}{t + 1}, then g'(t) = frac{1}{t} - frac{4}{(t + 1)^2} = frac{(t - 1)^2}{t(t + 1)^2}. Since t > 0, g'(t) geqslant 0 if and only if t = 1, so g(t) is an increasing function on (0, +infty). Also, g(1) = 0, so when t in (0, 1), g(t) < 0 always holds. Therefore, the original statement is proved ldots (12 text{ points}) The final answers are: 1. boxed{f(x)_{min} = -frac{1}{e}} 2. When a leqslant 0, boxed{text{Monotonically increasing interval of } F(x) text{ is } (0, +infty)}; When a > 0, boxed{text{Monotonically increasing interval of } F(x) text{ is } (frac{a}{2}, +infty) text{ and monotonically decreasing interval is } (0, frac{a}{2})} 3. boxed{x_1 + x_2 > a}

answer:1. Denote beta = angle DBO = angle ABD. Since overline{BD} bisects angle ABO, beta is also half of angle ABO. 2. By the Law of Sines on triangle DBO, we find: [ frac{DB}{sin phi} = frac{OD}{sin beta} Rightarrow OD = frac{DB sin beta}{sin phi}. ] 3. In right triangle ABD, sin beta can be expressed as: [ sin beta = frac{AD}{DB} = frac{2 - OD}{DB}. ] We substitute back in to get: [ OD = frac{(2 - OD)DB}{DB sin phi} Rightarrow OD = frac{2 - OD}{sin phi}. ] Solving for ( OD ), on simplifying we get, [ OD sin phi + OD = 2 Rightarrow OD = frac{2}{1 + sin phi} = frac{2}{1 + sin phi}. ] Thus, the solution is: (boxed{frac{2}{1 + sin phi}}).