answer:Answer: A, Chromosomal variation includes variations in chromosomal structure and the number of chromosomes, so A is correct; B, The exchange of segments between non-homologous chromosomes is a chromosomal structural variation (translocation), while the exchange of segments between non-sister chromatids of homologous chromosomes is a part of genetic recombination, making B incorrect; C, Cri-du-chat syndrome is caused by a partial deletion of human chromosome 5, so C is correct; D, Variations in chromosomal structure can change the number or arrangement of genes on the chromosome, so D is correct. Therefore, the answer is: boxed{B}. Analysis: 1. Variations in chromosomal structure refer to changes such as deletion, addition, inversion, or translocation of segments in one or several chromosomes within a cell. 2. Variations in the number of chromosomes refer to changes in the number of chromosomes within a cell, either an increase or a decrease.

answer:Triangle ABC is isosceles and right-angled, with its hypotenuse AB equal to 4, so AC = BC = 2sqrt{2} andS_{triangle ABC} = frac{1}{2} cdot (2sqrt{2})^2 = 4Let H be the midpoint of AC, and K be the projection of point H onto the hypotenuse AB. Then DH = 2, and HK = 1 as half the height CL of triangle ABC. From the right triangles DKH and ADH, we find thatDK = sqrt{HK^2 + DH^2} = sqrt{1 + 4} = sqrt{5}, quad AD = sqrt{AH^2 + DH^2} = sqrt{2 + 4} = sqrt{6}ThenS_{triangle ACD} = frac{1}{2} AC cdot DH = frac{1}{2} cdot 2sqrt{2} cdot 2 = 2sqrt{2}From the theorem of three perpendiculars, it follows that DK perp AB, soS_{triangle ABD} = frac{1}{2} AB cdot DK = frac{1}{2} cdot 4 cdot sqrt{5} = 2sqrt{5}We will calculate the volume of the pyramid in two ways. On one hand,V_{ABCD} = frac{1}{3} S_{triangle ABC} cdot DH = frac{1}{3} cdot 4 cdot 2 = frac{8}{3}On the other hand, if alpha is the angle between the faces ABD and ACD, thenV_{ABCD} = frac{2}{3} cdot frac{S_{triangle ABD} cdot S_{triangle ACD} sin alpha}{AD} = frac{2}{3} cdot frac{2sqrt{5} cdot 2sqrt{2}}{sqrt{6}} = frac{8sqrt{10}}{3sqrt{6}} sin alphaFrom the equationfrac{8sqrt{10}}{3sqrt{6}} sin alpha = frac{8}{3}we find that sin alpha = sqrt{frac{3}{5}}.## Answerarcsin sqrt{frac{3}{5}}

answer:3959

answer: Part (a)We need to show that for ( n = 0, 1, 2, ldots ),[ a_n = cos(theta_n) text{ for some } 0 < theta_n < frac{pi}{2}, ]and determine (theta_n).1. Base Case: For ( n = 0 ): [ a_0 = frac{1}{2} = cosleft(frac{pi}{3}right). ] Thus, (theta_0 = frac{pi}{3}), which satisfies (0 < theta_0 < frac{pi}{2}).2. Inductive Step: Assume that for some ( k geq 0 ), ( a_k = cos(theta_k) ) for some ( 0 < theta_k < frac{pi}{2} ). We need to show that ( a_{k+1} = cos(theta_{k+1}) ) for some ( 0 < theta_{k+1} < frac{pi}{2} ). By the given recurrence relation: [ a_{k+1} = sqrt{frac{1 + a_k}{2}}. ] Using the inductive hypothesis ( a_k = cos(theta_k) ): [ a_{k+1} = sqrt{frac{1 + cos(theta_k)}{2}}. ] Using the double-angle identity for cosine: [ cosleft(frac{theta_k}{2}right) = sqrt{frac{1 + cos(theta_k)}{2}}. ] Therefore: [ a_{k+1} = cosleft(frac{theta_k}{2}right). ] Let (theta_{k+1} = frac{theta_k}{2}). Since (0 < theta_k < frac{pi}{2}), it follows that (0 < theta_{k+1} < frac{pi}{2}).3. Conclusion: By induction, for all ( n geq 0 ), there exists (theta_n) such that ( a_n = cos(theta_n) ) and ( theta_n = frac{theta_0}{2^n} = frac{pi}{3 cdot 2^n} ). Part (b)We need to calculate:[ lim_{n to infty} 4^n (1 - a_n). ]1. From part (a), we have: [ a_n = cosleft(frac{pi}{3 cdot 2^n}right). ]2. Using the small-angle approximation for cosine, (cos(x) approx 1 - frac{x^2}{2}) for ( x ) close to 0: [ a_n approx 1 - frac{left(frac{pi}{3 cdot 2^n}right)^2}{2}. ]3. Therefore: [ 1 - a_n approx frac{left(frac{pi}{3 cdot 2^n}right)^2}{2} = frac{pi^2}{18 cdot 4^n}. ]4. Multiplying both sides by ( 4^n ): [ 4^n (1 - a_n) approx 4^n cdot frac{pi^2}{18 cdot 4^n} = frac{pi^2}{18}. ]5. Taking the limit as ( n to infty ): [ lim_{n to infty} 4^n (1 - a_n) = frac{pi^2}{18}. ]The final answer is (boxed{frac{pi^2}{18}}).

answer:Proof: frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}begin{array}{l}=frac{a^{2}}{a(b+c)}+frac{b^{2}}{b(c+a)}+frac{c^{2}}{c(a+b)} geqslant frac{(a+b+c)^{2}}{2(a b+b c+c a)} geqslant frac{3(a b+b c+c a)}{2(a b+b c+c a)}=frac{3}{2} .end{array}

answer:14. A circle can be circumscribed around an isosceles trapezoid. Considering the angles of the trapezoid, we conclude that its upper base is equal to the radius of the circumscribed circle, ( r = 1 partial ). The lower base is equal to the side of an equilateral triangle inscribed in this circle, i.e., it is equal to ( sqrt{3} partial mu ). The angle at the lower base of the trapezoid is ( 45^{circ} ), and the height is equal to the projection of the lateral side onto the lower base, ( h = frac{1}{2}(sqrt{3} - 1) partial ). It is then easy to calculate the area of the trapezoid. ( S = 0.5 ) sq. dm.