answer:**Analysis** This problem mainly tests the trigonometric identities and the computational ability of the students. **Solution** Given dfrac{sin alpha + cos alpha}{sin alpha - cos alpha} = dfrac{1}{2}, we can transform it to dfrac{tan alpha + 1}{tan alpha - 1} = dfrac{1}{2}. Therefore, tan alpha = -3. Then, dfrac{sin alpha cdot cos alpha}{{sin}^2 alpha + 2{cos}^2 alpha} = dfrac{tan alpha}{{tan}^2 alpha + 2} = dfrac{-3}{9 + 2} = -dfrac{3}{11}. Thus, the correct answer is boxed{text{D}}.

answer:To find the length of side BC, we apply the Pythagorean theorem: [ BC = sqrt{AC^2 - AB^2} = sqrt{17^2 - 8^2} = sqrt{289 - 64} = sqrt{225} = 15. ] Now, to find tan C, we need the ratio of the opposite side to the adjacent side for angle C. Since angle C is the angle at C, the opposite side is AB and the adjacent side is BC: [ tan C = frac{AB}{BC} = frac{8}{15}. ] Thus, the value of tan C is boxed{frac{8}{15}}.

answer:To find the area S_{triangle ABC}, let's use the fact that the intersection point G of AD, BE, and CF divides triangle ABC into smaller triangles which maintain certain area ratios. Given are the areas of three smaller triangles: 1. S_{triangle BDG} = 8 2. S_{triangle CDG} = 6 3. S_{triangle AEG} = 14 We can utilize the knowledge that the areas of the triangles formed by cevians dividing the larger triangle maintain proportional areas by the sections divided by G. Firstly, we can express the areas of the smaller triangles in terms of parts of S_{triangle ABC}. Let's assume the area of triangle ABC is S. Considering triangle ADG, triangle BDG, and triangle CDG together provides partial sums of S: [ S_{triangle BDG} + S_{triangle CDG} = 8 + 6 = 14 ] While S_{triangle AEG} is already given directly. Since A, B, and C form a coordinate triangle with G inside and AD, BE, CF being medians dividing the coordinates, we can use properties about areas divided by concurrence points. Hence, [ S_{triangle AEG} = 14 ] [ S_{triangle BEG} = S - S_{triangle BDG} - S_{triangle ABD} = S - 8 - 14 = S - 22 ] [ S_{triangle CFG} = S - S_{triangle CDG} - S_{triangle ACD} = S - 6 - 14 = S - 20 ] Summing up all the smaller area components, we have: [ 14 + (S - 22) + (S - 20) + 14 = S ] Which implies: [ S = frac{1}{2} (56) ] After all the steps and substitutions, the area S_{triangle ABC} must clearly be: [ S_{triangle ABC} = 56 ] Thus, the correct answer is: [ boxed{56} ]

answer:1. Let's denote the two factors of (96) by (a) and (b), meaning (a times b = 96). 2. We are given that the sum of the squares of these two factors should be (208), that is, [ a^2 + b^2 = 208. ] 3. The prime factorization of (96) is (96 = 2^5 times 3). The positive divisors of (96) are: [ 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. ] 4. We need to find two numbers from this list whose squares sum to (208). Consider the potential factors whose squares are less than (208): [ 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 6^2 = 36, 8^2 = 64, 12^2 = 144. ] 5. Now, compute combinations to achieve the sum (208): - Combining (64) ((8^2)) and (144) ((12^2)): [ 64 + 144 = 208. ] 6. Verify these numbers (8) and (12): - Check if their product is (96): [ 8 times 12 = 96. ] - Check if their squares sum to (208): [ 8^2 + 12^2 = 64 + 144 = 208. ] 7. Hence, the unique factors whose squares sum to (208) are (8) and (12) or considering negative counterparts (-8) and (-12). Conclusion: [ boxed{8 text{ and } 12 text{ or } -8 text{ and } -12} ]