answer:Let x = ki, where k is a real number. The given equation becomes [(ki)^4 + 2(ki)^3 + 6(ki)^2 + 34(ki) + 49 = 0,] which simplifies to [k^4 - 2ik^3 - 6k^2 + 34ik + 49 = 0.] To satisfy this equation for x=ki (pure imaginary), its real part and imaginary part must separately equal zero: - Real part: k^4 - 6k^2 + 49 = 0 - Imaginary part: -2ik^3 + 34ik = -2ik(k^2 - 17) = 0. From the imaginary part, -2ik(k^2 - 17) = 0. Since kneq 0 in pure imaginary numbers, we have k^2 = 17, thus k = pm sqrt{17}. Therefore, the pure imaginary solutions are: [boxed{isqrt{17}, -isqrt{17}}.]

answer:To find the number of smaller cuboids that a diagonal line passes through in a larger cuboid, we will use the principle of inclusion-exclusion. First, denote the edge lengths of the smaller cuboids as 2, 7, and 13. We are asked to fit these into a larger cuboid of length 2002. Next, we find the least common multiple (LCM) of 2, 7, and 13: [ text{LCM}(2, 7, 13) = 2 times 7 times 13 = 182 ] So, the length of the cuboid should be an integer multiple of 182: [ frac{2002}{182} = 11 ] Now, consider a cuboid with dimensions 2, 7, and 13 whose length is extended to 182. We need to calculate the smallest number of lengths that intersect a diagonal of a cuboid with length 2002. The intersecting smaller cuboids along this diagonal can be calculated using the inclusion-exclusion principle. The formula for the number of cuboids a diagonal passes through in L units long cuboid is: [ frac{L}{a} + frac{L}{b} + frac{L}{c} - frac{L}{ab} - frac{L}{ac} - frac{L}{bc} + frac{L}{abc} ] where a=2, b=7, c=13, and ab, ac and bc are the products of pairs, and abc includes all three modular factors. Given ( L = 182 ), [ frac{182}{2} = 91, quad frac{182}{7} = 26, quad frac{182}{13} = 14 ] [ frac{182}{2 times 7} = frac{182}{14} = 13, quad frac{182}{2 times 13} = frac{182}{26} = 7, quad frac{182}{7 times 13} = frac{182}{91} = 2 ] [ frac{182}{2 times 7 times 13} = frac{182}{182} = 1 ] Applying inclusion-exclusion principle: [ frac{182}{2} + frac{182}{7} + frac{182}{13} - frac{182}{2 times 7} - frac{182}{2 times 13} - frac{182}{7 times 13} + frac{182}{2 times 7 times 13} ] [ = 91 + 26 + 14 - 13 - 7 - 2 + 1 = 110 ] Since the original cuboid of length 2002 is an 11 times larger cuboid (as calculated before), it follows: [ 110 times 11 = 1210 ] Thus, the total number of smaller cuboids the diagonal line passes through is ( boxed{1210} ) which corresponds to choice D.

answer:1. Given that ABCD is a rectangle and G, F, E are points such that GR parallel FH parallel EO parallel AB. 2. The length AD = 12 and we are to find the length of AG. 3. Since ABCD is a rectangle and E is the midpoint of AD, we have AE = frac{AD}{2} = frac{12}{2} = 6. 4. Considering the similarity of triangles involving the parallel lines, we have: [ triangle EOH sim triangle BAH ] Therefore, the side ratios must be equal. 5. The triangles' sides provide the ratio: [ frac{OH}{AH} = frac{EO}{AB} ] Since E is the midpoint of AD, EO : CD = 1 : 2. 6. Considering point O is the midpoint of both AC and BD, these lines divide each other into equal parts. [ AC = BD = 2 times OC = 2 times OB = 2 times OD ] As E is the midpoint of AD, and [ frac{OH}{AH} = 1 : 2 ] 7. Utilizing the similarity of triangles and the intervals, we derive: [ frac{AF}{AE} = frac{FH}{EO} ] Consequently, Thus, [ frac{AH}{AO} = 2 : 3 ] 8. Given, [ AD = 12, quad AE = 6, quad AF = 4 ] 9. Knowing the parallel segments, we have: [ FH : AB = 1 : 3 ] 10. Since GR parallel FH, we find the ratio to be: [ frac{AG}{AF} = frac{AR}{AH} = frac{3}{4} ] Consequently, summing these lengths: [ AG = frac{3}{4} times AF = frac{3}{4} times 4 = 3 ] # Conclusion: [ boxed{3} ]

answer:(1) By definition, the domain set A consists of values where the logarithm argument is greater than zero, thus A={x|x^2 - x - 2 > 0}={x|x < -1 text{ or } x > 2}. The domain set B consists of values where the expression under the square root is non-negative, so B={x|3 - |x| geq 0}={x|-3 leq x leq 3}. Therefore, the intersection A cap B is the set of values where x satisfies both conditions, which yields A cap B={x|-3 leq x < -1 text{ or } 2 < x leq 3}. (2) For C to be a subset of B, it must be true that the endpoints of the interval defined by C also satisfy the domain conditions of B. Since m - 1 < m + 2 is always true, we have the following system of inequalities to satisfy [C subseteq B]: begin{cases} m - 1 geq -3 m + 2 leq 3 end{cases} Solving these inequalities we find -2 leq m leq 1. The range of possible values for m is thus boxed{[-2, 1]}.