1.A 2。 B 3。 B 4。 D 5。 B 6。 C
7.B 8。 C 9。 A 10。 B 1 1。 B 12。 A
Fill in the blanks (this topic is entitled ***8 small questions, with 3 points for each small question and 24 points for * * *).
13. 14.7 15 (Ⅰ) 13.0; (Ⅱ) 16 270
17.6 18. 19.(6,0) 20.①③④
Three. (This big question is ***4 small questions, each with 4 points, *** 16 points)
2 1. solution: the original formula = 3 points.
= 4 points
22. Solution: Let the analytical expression of this straight line be as follows: Substitute two coordinates (1, 2) and (3,0) to get it.
2 points
3 points for the solution
So the analytical formula of this straight line is .4 points.
23. Solution: Multiply the two sides of the equation.
.2 points
Get 0.3 points.
Test: When ≠0.
Therefore, it is the solution of the original fractional equation. 4 points
24. Solution: (1)P (the number obtained is exactly 0) = .2 points.
(2) Method 1: Draw a tree diagram as follows:
3 points
There are 9 possible outcomes * * *, of which 5 meet the conditions.
So p (the absolute values of two numbers are equal) =.
Method 2: The list grid is as follows:
0 1
( , )
( ,0)
( , 1)
0 (0, )
(0,0) (0, 1)
1 ( 1, )
( 1,0) ( 1, 1)
3 points
There are 9 possible outcomes * * *, of which 5 meet the conditions.
So p (the absolute values of two numbers are equal) = 4 points.
Four, (this big topic ***4 small questions, each small question 5 points, ***20 points)
25. Solution: Suppose that manufacturer B sold a tool holder during this period.
According to the meaning of the question, get .3 points.
Get 0.4 points.
Number of blades sold: 50× 400 = 20,000 blades.
A: During this period, Manufacturer B sold 400 handles, 20,000 blades and 5 points.
Note: For the solution of binary linear equation, refer to the fraction.
26. Solution: (1) As shown in the figure:
2 points
(2) B.3 points
(3) According to the meaning of the question:
= 100 (person)
A: It is estimated that the number of all ninth-grade girls in this school who have improved their grades after training is 100. Five points.
27. Solution: (1) proves that ∵ is the tangent of big ∵ O, ∴∞= 90.
∫∫, ∴∠ Bad = 90. That's ...
Point a is on the small ⊙O, and ∴AD is the tangent of the small ⊙ O. 2 point.
(2)∫∨∴ The quadrilateral is a parallelogram.
Three points
∵ ∥ ,∴ .
∴ .
Say it again,
0.5 points
28. Solution: (1)√.
∴
The value range of ∴ is 0 ≤≤ 10. 1.
(2) equilateral triangle.
∴ .
That is, when, decimeter. Two points.
(3) When the umbrella is most open, the points are coincident.
Connect,. Obey.
∵ ,
The quadrangle is a diamond,
Yes, the bisector,
.
In Rt
.
Yes, the bisector,
∴ .
∴ ~ .
∴ .∴ 。
∴ .
∴ (square decimeter) .5 points
Verb (abbreviation of verb) (this big question * * 1 small question, *** 12 points)
Solution: (1) Order, get.
∴ The coordinate of point A is (2,0) .2 minutes.
This is an isosceles triangle. 3 points
(2) existence.
.5 points
(3) When 0 < < 2, as shown in figure 1, and the axis is H, suppose.
Figure 1
∫A(2，0)，C(，0)，
∴ .∴ .
∴
Replace, acquire
.
∵ ,
9 o'clock
When, does not exist.
When the axis is H as shown in Figure 2, set.
Figure 2
∫A(2，0)，C(，0)，
∴ ,∴ .
∴
Substitute,
Yes
∵ ,
12 point
Note: The solution is based on thinking. If it is not excluded, deduct 1 point.
Six, (this big topic * * 1 small topic, *** 12 points)
30. Solution: (1), 0.3 points.
Note: Give 1 point for each correct answer.
(2) existence. For the differences of the selected graphs, the following proofs are given:
Select the diagram 1. There is a straight bisector in Figure 1, which is proved as follows:
Figure 1
Method 1:
Prove that sum is congruent equilateral triangle,
∴ ,
∴ .
Here we go again.
∴ .
H point is on the middle vertical line of the line segment.
A little more: ∴ On the vertical line of the line segment.
∴ Divide 8 points vertically into a straight line.
Method 2:
Prove that sum is congruent equilateral triangle,
∴ ,
∴ .
Here we go again.
∴
∴ .
live in harmony with
∵ , ,
∴ ≌ .∴
∴ is the bisector of the vertex of an isosceles triangle.
Perpendicular bisector in a straight line. 8 points
Select Figure 2. There is a straight line bisected vertically in Figure 2, which is proved as follows:
Figure 2
∵
∴
Say it again,
∴ .
H point is on the middle vertical line of the line segment.
A little more: ∴ On the vertical line of the line segment.
Perpendicular bisector in a straight line. 8 points
Note: (1) If Method 2 is selected for proof in Figure 2, please refer to Method 2 above for scoring;
(2) If Figure 3 or Figure 4 is selected for authentication, score according to the above authentication process.
(3) In odd numbers,
Even numbers, 10 points.
(4) existence. In odd numbers, straight lines are divided vertically.
When it is an even number, the straight line is vertically divided. 12 points.
Note: In questions (3) and (4), you will get 1 point for each correct answer.