answer:According to the problem, taking two of these as conditions and the other as a conclusion, there are three propositions that can be constructed; they are as follows: ① If l perp alpha and l parallel beta, then alpha perp beta. ② If l perp alpha, alpha perp beta, and l nsubseteq beta, then l parallel beta. ③ If l parallel beta, alpha perp beta, and l nsubseteq alpha, then l perp alpha; Analysis shows: ① If l perp alpha and l parallel beta, then alpha perp beta. This proposition is correct. ② If l perp alpha, alpha perp beta, and l nsubseteq beta, then l parallel beta. This proposition is correct. ③ If l parallel beta, alpha perp beta, and l nsubseteq alpha, then l perp alpha. This proposition is incorrect because there could also be a parallel or intersecting positional relationship. Therefore, there are boxed{2} correct propositions. Hence, the answer is C.

answer:The sum of the angle measures of a decagon is (180^circ times (10-2) = 1440^circ), so each angle of a regular decagon measures (1440^circ / 10 = 144^circ). Therefore, (angle BCD = 144^circ), which means (angle BCQ = 180^circ - angle BCD = 36^circ). Similarly, (angle QAF = 36^circ). Since (angle ABC = 144^circ), the reflex angle at (B) that is an interior angle of (ABCQ) has measure (360^circ - 144^circ = 216^circ). The interior angles of quadrilateral (ABCQ) must sum to (360^circ), so we have: [ angle Q = 360^circ - angle QAF - (text{reflex } angle B) - angle BCQ = 360^circ - 36^circ - 216^circ - 36^circ = boxed{72^circ}. ]

answer:To solve this problem, we should verify the truth of each proposition P and q. First, let's analyze proposition P. To prove that P is true, we need to find at least one x_0 in (0, +infty) that satisfies the inequality x_0 + frac{1}{x_0} > 3. Take x_0 = 2, we have: 2 + frac{1}{2} = 2.5 > 3, This statement is false for this particular value of x_0, but this does not disprove the existence of other possible x_0 values that could satisfy the inequality. We can actually prove the initial inequality by analyzing the function f(x) = x + frac{1}{x} for x > 0. Calculating the minimum of f(x) within the domain given will show whether the proposition is true. To do this, we set the derivative of f(x): f'(x) = 1 - frac{1}{x^2}. Setting f'(x) = 0 to find the critical points, 1 - frac{1}{x^2} = 0 x^2 = 1 x = 1. Now, we evaluate f(x) at x = 1: f(1) = 1 + frac{1}{1} = 2 < 3. However, since f(1) yields the minimum value for this function when x > 0, and the value is less than 3, any other x > 1 will give a value of f(x) larger than 2, hence greater than 3. Thus, for x > 1, f(x) > 3. Hence, proposition P is true. Next, we examine proposition q: We need to show whether the inequality x^2 > 2^x holds for all x in the interval (2, +infty). To disprove q, we only need to find a counterexample. Let's take x = 4: 4^2 = 16, quad 2^4 = 16. The inequality does not hold since 16 leq 16 (it should be strictly greater). Therefore, proposition q is false. Since p is true and q is false, the correct answer is the one that contains p and the negation of q, which is p land (neg q). So the correct answer is: boxed{A}.

answer:To determine the line of symmetry for the function g(x), given that g(x) = g(4-x) for all x, let's follow similar steps as in the original problem: 1. For every point (x,y) that lies on the graph of y = g(x), the point (4-x, y) must also lie on this graph due to the given functional equation. 2. Rewriting the expression for reflection, we define x as k + (x-k), where k is the midpoint and reflection axis. For 4-x, we can write this as k - (x-k). 3. To find the value of k that satisfies both expressions as a reflection, equate k + (x-k) = 4-x and solve for k. Setting up the equation: [ 4-x = k - (x-k) implies 4-x = k - x + k implies 4 = 2k implies k = 2. ] Conclusion: The line of symmetry for the graph of y = g(x), when g(x) = g(4-x), is given by the vertical line boxed{x=2}.