answer:Since sin 2alpha = -sin alpha, we have sin alpha (2cos alpha + 1) = 0, which gives sin alpha = 0, or cos alpha = -frac{1}{2}, If sin alpha = 0, then tan alpha = 0, If cos alpha = -frac{1}{2}, then sin alpha = pm frac{sqrt{3}}{2}, thus tan alpha = pm sqrt{3}. Therefore, the answer is: boxed{pm sqrt{3} text{ or } 0}. From sin 2alpha = -sin alpha, we get sin alpha (2cos alpha + 1) = 0, leading to sin alpha = 0, cos alpha = -frac{1}{2}, and subsequently. This problem tests the evaluation of trigonometric functions, double angle formulas, and basic trigonometric relationships, assessing reasoning and computational skills, and is considered a medium-level question.

answer:Given that the base edge length of the regular square pyramid is 2sqrt{2} cm, thus, the area of the base is 8text{cm}^2, since the volume is 8text{cm}^3, therefore, the height h=3, hence, the slant height h'=sqrt{2+9}=sqrt{11}, thus, its lateral surface area is 4 times frac{1}{2} times 2sqrt{2} times sqrt{11} = 4sqrt{22} So, the answer is: boxed{4sqrt{22}text{cm}^2} By using the volume formula, we find the height h=3, and by utilizing its properties, we find the slant height h'=sqrt{2+9}=sqrt{11}. Then, by applying the formula for the area of a triangle, we can solve it. This question examines the volume and surface area of a spatial geometric body, belonging to a calculation problem with moderate difficulty.

answer:Since 2 specific books must be included in every selection, we need to choose the remaining 3 books from the 6 books left (after excluding the 2 specific ones). The number of ways to choose 3 books from 6 is calculated using the combination formula: [ binom{6}{3} = frac{6!}{3!(6-3)!} = frac{6 times 5 times 4}{3 times 2 times 1} = 20 ] Thus, there are boxed{20} ways to choose 5 books from 8 such that the selection includes the 2 specific books.

answer:# Problem: [ 24.11 text{In sunlight, a sphere is placed on a flat surface. At a certain moment, the farthest point that the shadow of the sphere reaches is } 10 text{meters} text{from the point where the sphere touches the surface. At the same moment, a vertical meter stick casts a shadow of length } 2 text{meters. Then what is the radius of the sphere (in meters)? Assume the sunlight is parallel and the meter stick is a line segment.} text{(A) } frac{5}{2}| text{(B) } 9-4 sqrt{5} text{(C) } 8 sqrt{10}-23 text{(D) } 6-sqrt{15} text{(E) } 10 sqrt{5}-20 ] To determine the radius ( r ) of the sphere, we proceed as follows: 1. **Given Data**: - The length of the shadow of the meter stick is ( 2 text{meters} ). - The distance from where the shadow of the sphere touches the ground to the point where the sphere touches the ground is ( 10 text{meters} ). 2. **Observation**: - Let’s denote the height of the meter stick as ( A = 1 text{meter} ). - The shadow length ( AD ) for ( A ) is ( 2 text{meters} ). 3. **Triangle Proportions**: - By considering the similar triangles formed by the shadows and the sunlight direction, we have: [ triangle CEA sim triangle DBA ] - Let ( CE = r ). Then, the radius of the sphere ( r ) is the height in the triangle involving the sphere. - Using the proportion: [ frac{CE}{AC} = frac{BD}{AD} ] - Here, ( AC = r ) and ( AD = 5 sqrt{5} ). Hence, substituting the values: [ frac{r}{5-r} = frac{10}{5 sqrt{5}} ] 4. **Solve for ( r )**: Start by cross-multiplying and simplifying: [ sqrt{5} r = 10 - 2r ] [ sqrt{5} r + 2r = 10 ] [ r (sqrt{5} + 2) = 10 ] [ r = frac{10}{sqrt{5} + 2} ] We rationalize the denominator: [ r = frac{10 (sqrt{5} - 2)}{(sqrt{5} + 2)(sqrt{5} - 2)} ] [ r = frac{10 (sqrt{5} - 2)}{5 - 4} ] [ r = frac{10 (sqrt{5} - 2)}{1} ] [ r = 10 sqrt{5} - 20 ] # Conclusion: The radius of the sphere is ( 10 sqrt{5} - 20 ). [ boxed{E} ]