Front view, left view and top view A.7 B.8 C.9 D. 10 7. In the convex quadrilateral ABCD, ∠ C = 1200, ∠ B = ∠ D = 900, AB = 6, BC = 2, then ad = () a.2b.6c.4d.68 Let n(n≥2) positive integers. Blood pressure must be even. C. when n is even, p is odd. D. when n is odd, p is even. 2. Fill in the blanks (***6 small questions, 5 points for each small question, out of 30 points) 9. If known, the value of the polynomial is. 10. Arrange five integers from the largest to the smallest, and the median is 4. If the only pattern in this example is 7, the maximum value of the sum of these five integers is. x 1 1. Fill in a number in each box in Figure 8, so that there is 1 in each row and column. The number of x in the small box in the upper right corner should be. 12. In △ABC, AB = 15cm, AC = 13cm, and the height of BC = 12cm. Then the area of triangle ABC is. Oxbacdy ... 1 21213. As shown in the figure, there is an animation program, and the square area ABCD at the top of the screen represents the black object A, where A (1, 1) b (2,1). 2) D (1, 2), use a signal gun to emit signals in a straight line. When the signal encounters area A, A turns from black to white, and when the value range of B is _ _ _ _ _ _, A can turn from black to white. 14. If the positive integer n has the following properties: one eighth of n is a square number and one ninth of n is a cubic number. Then the minimum hope number is. 3. Answer questions (***4 questions, score 12, 12, 12, 14, out of 50 points) and score reviewers 15. Four real numbers are known, and if four relationships are known, (2) the values obtained respectively.
Scoring reviewer 16. A middle school organized a spring outing for students, and the travel company provided several medium-sized buses. At first, each car took 28 people. After a period of departure, I found that a student was late and didn't get on the bus. Now I decided to pick him up in an empty car. When I got it back, I drove an empty car in time to redistribute everyone, so I just split the rest of the car equally. It is known that each car cannot carry more than 32 people. So how many students are there in this group of spring outing?
17. In △ABC, ∠C=, D is the midpoint of AB, E and F are on BC and AC respectively, ∠EDF=. As shown in figure 1, give the reviewer a score (1). If e is the midpoint of BC, EF and AF, BE. And explain the reasons; (2) As shown in Figure 2, when F moves on AC, point E moves on BC. What is the quantitative relationship BEtween EF and AF and be during exercise? And explain why.
Scoring reviewer 18. The known straight line (1) means that no matter whether k is any real number not equal to 1, this straight line passes through a certain point and the coordinates of this fixed point are obtained. (2) If point B is (5,0), point P is on the Y-axis, and point A is the fixed point determined in (1), let △PAB be an isosceles triangle, and find the analytical formula of straight line PA. Reference answers and grading suggestions for the examination questions of the second grade mathematics competition in 2008 I. Multiple choice questions (* * 8 small questions, 5 points for each small question, out of 40 points) Question 12345678 Answer CDC ABDBD II. Fill in the blanks (***6 small questions, 5 points for each small question, out of 30 points) 9.310.231.112.84cm2 or 24 cm2 (2 points for a correct answer)/kloc- -5 ≤ b ≤-114.20000000006 The score is 12, 12 and14, with a full score of 50)/ (12) solution: (1) From+= 4+8 = 60. ∴, ∴ ...2 points and -= 4-8 =-4, get, ... 2 points when appropriate,, ... 2 points when appropriate, ... 2 points 16. (12 score ≤ 32. It is easy to know that the number of tourists is equal to, when an empty car leaves, the number of students on this spring outing can be expressed as, thus listing the equation … 2 points, so … does not meet the condition of ≤32. Then, if it meets the meaning of the question ... 2 points, so the number of passengers is equal to = 29× 29 = 84 1 (people) ... 2 points A: there are 30 original cars, and there are 84 1 students in this spring outing. 17.( 12 points) solution: (1) ef2 = af2+be2 ...1point ∵ is the midpoint of AB and BC respectively. Df = CE = BE...3 points edf =, ∴ EF2 = DF2+DE2 = AF2+BE2 ...1point (2) EF2 = AF2+BE2... 1 point extends FD to. Df = DG, ∠gbd =∠a .∠EDF =, ∴ EF = EG... 1 point and ∠ GBD = ∠ A, ∴BG∨AC,. ... 1 min ∴ eg2 = be2+bg2 = be2+af2 ∴ ef2 = af2+be2 ... 2 min 18. (14 points) Solution: (1) From the meaning of the question, 4) ... 5 points (2) Discuss in three situations: ① Let P 1(0, m 1) satisfy p1b = It is easy to get AB=5, so the analytical formula of point P2 (0 0,0) and straight line P2A is: ... 2 points ③ Let P3 (0,m3) meet the requirement of P 1A=AB, which is obtained by Pythagorean theorem, that is, the analytical formula of P3 (0,P4 (0,0) and straight line P3A: ...