answer:**Analysis:** To solve this problem, we need to consider the equivalence relationships between propositions. True (or false) propositions appear in pairs. Among the original proposition and its converse, inverse, and contrapositive, the number of true propositions must be an even number. Therefore, the correct answer is boxed{text{B}}. **Key Point:** This question mainly examines the relationships among the four types of propositions.

answer:To find the angle of inclination (theta) of the line, we first need to find the slope (m). The slope of a line in the form ax + by + c = 0 is given by -a/b. For the given line x+sqrt{3}y-3=0, we have a = 1 and b = sqrt{3}. Therefore, the slope is: m = -dfrac{a}{b} = -dfrac{1}{sqrt{3}} The tangent of the angle of inclination is equal to the slope of the line, so we have: tan(theta) = -dfrac{1}{sqrt{3}} The angle theta can be found by taking the arctangent of both sides: theta = arctanleft(-dfrac{1}{sqrt{3}}right) Since we are looking for the angle in the counterclockwise direction from the positive x-axis, we need to add pi to the result because the slope is negative: theta = pi + arctanleft(-dfrac{1}{sqrt{3}}right) = dfrac{5pi}{6} Thus, the angle of inclination is boxed{dfrac{5pi}{6}}.

answer:Given the cone's cross-section along its vertex, which is a right triangle with a vertex angle of 90 degrees, the ratio of the base to the height of a right triangle in this context equals to 1 (base = height). Thus, if we denote the radius as ( r ), the base of the triangle becomes ( 2r ) and the height is ( r ) itself. Using the volume formula ( V = frac{1}{3} pi r^2 h ), and substituting ( h = r ) and ( V = 16384 pi ): [ frac{1}{3} pi r^2 times r = 16384pi ] Dividing both sides by (pi): [ frac{1}{3} r^3 = 16384 ] Multiplying both sides by 3: [ r^3 = 49152 ] Taking the cube root of both sides, we find ( r ): [ r = sqrt[3]{49152} = 36.8 ] Hence, the height ( h ), which equals ( r ), is: [ h = boxed{36.8} ]

answer:Given the recursion (a_{n + 2} = frac{a_n + 2017}{1 + a_{n + 1}}) for (n geq 1), we obtain the following from the setting: a_3(a_2 + 1) = a_1 + 2017, ;; a_4(a_3 + 1) = a_2 + 2017, ;; a_5(a_4 + 1) = a_3 + 2017. By subtracting consecutive equations, we find: a_3 - a_1 = (a_3 + 1)(a_4 - a_2) ;; text{and} ;; a_4 - a_2 = (a_4 + 1)(a_5 - a_3). Suppose (a_3 - a_1 neq 0). This leads to (a_4 - a_2 neq 0), (a_5 - a_3 neq 0), and so forth. The inequality |a_3 - a_1| > |a_4 - a_2| > |a_5 - a_3| > dots leads to a contradiction as these differences cannot indefinitely decrease while remaining positive integers. Thus, (a_{n+2} - a_n = 0) for all (n geq 1), indicating that all terms with odd indices are equal, and all terms with even indices are equal. Therefore, (a_1 = frac{a_1 + 2017}{a_2 + 1}), leading to (a_1a_2 = 2017 = 2017 cdot 1). The minimum value of (a_1 + a_2) occurs when ({a_1, a_2} = {1, 2017}), providing a sum of (boxed{2018}).