answer:To determine which geometric element can uniquely determine a plane, let's analyze each option step by step: - **Option A: Three points** Three points can determine a plane, but they must not be collinear. If the three points are collinear, they lie on the same line and hence infinite planes can pass through this line. Therefore, three points can uniquely determine a plane only if they are non-collinear. So, A cannot be universally correct because it doesn't specify that the points are non-collinear. - **Option B: Center and two points on a circle** A circle in a plane is determined by its center and any point on it. However, if we have the center and two points on the circle, these three points can uniquely determine the circle, but not necessarily a unique plane unless specified that they are non-collinear. If the two points and the center are collinear, this does not uniquely determine a plane because multiple planes can contain that same line. So, B is also incorrect. - **Option C: Two sides of a trapezoid** A trapezoid is a quadrilateral with at least one pair of parallel sides. The two sides of a trapezoid (whether they are the parallel sides or one parallel and one non-parallel side) lie in the same plane. Since the sides of a trapezoid must be part of the same figure, and a trapezoid is a planar figure, any two sides of a trapezoid uniquely determine the plane of the trapezoid. Therefore, C is correct because the given sides must coexist in the same plane by definition. - **Option D: A point and a line** A point and a line can determine a plane, but only if the point does not lie on the line. If the point lies on the line, then an infinite number of planes can pass through that line, and thus, the plane is not uniquely determined. So, D is incorrect because it does not specify the positional relationship between the point and the line. **Conclusion:** The only option that can always uniquely determine a plane, based on the given conditions, is option C: Two sides of a trapezoid. Therefore, the correct answer is boxed{text{C}}.

answer:To solve the given problem, let's follow the solution step by step: 1. We start with the given conditions x > 0, y > 0, and x + y = 1. We need to find the minimum value of the expression frac{1}{x} + frac{9}{y}. 2. Rewriting the expression using the condition x+y=1, we get: [ frac{1}{x} + frac{9}{y} = frac{x+y}{x} + frac{9(x+y)}{y} ] [ = frac{x}{x} + frac{y}{x} + frac{9x}{y} + frac{9y}{y} ] [ = 1 + frac{y}{x} + 9 + frac{9x}{y} ] [ = 10 + frac{y}{x} + frac{9x}{y} ] 3. Applying the AM-GM inequality (Arithmetic Mean - Geometric Mean inequality) to frac{y}{x} + frac{9x}{y}, we have: [ frac{y}{x} + frac{9x}{y} geq 2sqrt{frac{y}{x} cdot frac{9x}{y}} ] [ = 2sqrt{9} ] [ = 2cdot 3 ] [ = 6 ] 4. Adding 10 to both sides of the inequality derived from applying AM-GM, we get: [ 10 + frac{y}{x} + frac{9x}{y} geq 10 + 6 ] [ = 16 ] 5. Equality in the AM-GM inequality holds when frac{y}{x} = frac{9x}{y}. Given that x + y = 1, solving these equations simultaneously gives us x = frac{1}{4} and y = frac{3}{4}. 6. Therefore, the minimum value of frac{1}{x} + frac{9}{y}, given the conditions x > 0, y > 0, and x + y = 1, is boxed{16}.

answer:**Analysis** This problem mainly tests the knowledge of trigonometric identities and double-angle formulas. Pay attention to the signs. **Solution** Given: dfrac{2sin (pi-alpha)+sin 2alpha}{cos ^{2} dfrac{alpha}{2}} = dfrac{2sin alpha+2sin alphacos alpha}{ dfrac{1}{2}(1+cos alpha)} = dfrac{4sin alpha(1+cos alpha)}{1+cos alpha} =4sin alpha. Therefore, the answer is boxed{4sin alpha}.

answer:Since f(x) is an odd function that is monotonically decreasing on (0,+infty), we can infer that f(x) is also monotonically decreasing on (-infty,0). Given that f(4)=0, it follows that f(-4)=0 due to the odd function property. We have the inequality f(x-3) leqslant 0. There are two cases to consider for the value of x-3: 1. When x - 3 > 0, i.e., x > 3: For f(x) to be less than or equal to zero, we know x-3 must be greater than or equal to 4 since f(x) is monotonically decreasing on (0,+infty) and f(4)=0. Thus, we get: [ x geqslant 7 ] 2. When x - 3 < 0, i.e., x < 3: In this case, f(x-3) should be less than or equal to f(-4). Since f(-4)=0 and f(x) is monotonically decreasing on (-infty,0), it follows that: [ -1 leqslant x < 3 ] Combining these two cases, the range of x is: [ boxed{[-1,3) cup [7,+infty)} ] By following the properties of odd functions and the monotonicity on symmetric intervals, we deduce the range for the condition f(x-3) leqslant 0.