II. Fill in the blanks (the score for this question is *** 16, with 4 points for each small question)
9.k≥ 10。 27
1 1.2 points.
(2) 2 points
12. (2 points in each space) 2,2
Iii. Answer the question (30 points for this question, 5 points for each small question)
13. solution: the original formula = =4 (the first step is to give a number 1 minute, and the second step is to give a correct result 1 minute).
14. solution: original formula = (2 points for the first step and 3 points for the correct result for the second step)
=
15. Prove that quadrilateral is isosceles trapezoid,
.2 points
Yes, the midpoint,
.3 points
.4 points
Ae = de 5 points
16. Solution: Original formula =
3 points
4 points
5 points
17. Solution: If Type A robot carries kilograms of chemical raw materials per hour, then Type B robot carries (-20) kilograms per hour, 1 minute.
According to the meaning of the question: 3 points
The score for solving this equation is .4.
It is proved to be the solution of the equation, which accords with the practical significance, so -20=80. Five points.
A: Robots A and B respectively carry 100 kg and 80 kg of chemical raw materials every hour.
18. Solution: (1) 25 (1)
(2) 50 ( 1)
Draw a bar chart (1)
(3)5 people; (The score of this column for the equation is 1, and the answer given is 1).
Iv. Answer (this question ***2 1 point, 19, 6 points for 20 questions, 5 points for 2 1 question and 4 points for 22 questions)
19.
(1) Method 1: Connect as shown in the figure.
1 point
Say it again,
, 2 points
Namely.
Is the tangent of 0.3 point.
(2) Solution: Starting from (1), it is an isosceles right triangle. Four points.
, is the diameter,
.5 points
.
.6 points
20. Solution: (1)∵ Quadrilateral ABCD is a square.
∴∠BCF+∠FCD=900
BC=CD 1。
∵△ECF is an isosceles right triangle,
∴∠ECD+∠FCD=900.CF=CE 2。
∴∠BCF=∠ECD.
∴△BCF≌△DCE 3 points
(2) In △BFC, BC=5, CF=3, ∠ BFC = 900. ∴ BF = .4 points.
∵△bcf≌△dce,∴de=bf=4,∠bfc=∠dec=∠fce=900.∴DE∥FC
∴△DGE∽△CGF 5 points
∴ DG: GC = DE: CF = 4: 3.6。
2 1. Solution: (1) Point A( 1, 1) is on the image of the inverse proportional function.
K=2。 The analytical formula of the inverse proportional function is:. 1.
The analytical formula of linear function is:
Point A( 1, 1) is on the image of the linear function. Two points.
The analytical formula of linear function is
(2) Point A( 1, 1)
① when ∠OB 1A=90 o, that is, B 1A⊥OB 1.
∠AOB 1 = 45o∴b 1a = ob 1。 ∴ B 1 (1, 0) 4 o'clock.
② When ∠ o a b2 = 90 ∠ ao b2 = ∠ ab2o = 45,
B 1 is the midpoint of B2OB2 (2,0).
To sum up, the coordinates of point B are (1, 0) or (2,0) .5 minutes.
22.( 1)5 1.
(2) 15
V. Answer questions (this question is ***2 1 point, 23 questions are 6 points, 24 questions are 7 points, and 25 questions are 8 points)
23. Solution: (1)∫ The image of quadratic function passes through point C(0, -3),
∴ c =-3。 1.
Substituting points A (3 3,0) and B (2 2,3) gives the following results.
Solution: a= 1, b=-2.
-2 points.
The formula is:, so the symmetry axis is x= 1.
(2) According to the meaning of the question, BP = OQ = 0.1t.
Point b and point c have the same ordinate,
∴BC∥OA.
After passing through point B, point P is BD⊥OA, PE⊥OA, and the vertical feet are D and E respectively.
To make a quadrilateral ABPQ isosceles trapezoid, only PQ = AB, Pb ≠ AQ is needed.
That is QE = AD = 1.
QE = OE-OQ =(2-0. 1t)-0. 1t = 2-0.2t
∴2-0.2t= 1.
T = 5。 -Three points.
It is also known that Pb = 0.5 and AQ = 2.5.
That is, when t=5 seconds, the quadrilateral ABPQ is an isosceles trapezoid.
② Let the intersections of the symmetry axis with BC and X axis be f and g respectively.
The symmetry axis x= 1 is the perpendicular bisector of the line segment BC,
∴ BF = CF = OG = 1。 -Four points.
∫BP = OQ，
∴PF=QG.
And ≈PMF =∠QMG,
∴△MFP≌△MGQ.
∴MF=MG.
Point m is the midpoint of FG-5.
∴S=
=
By =
∴S=
BC=2，OA=3。
It takes 20 seconds to move from point P to point C and stop moving.
∴0<； t≤20
When t=20 seconds, the area s has a minimum of 3-6 points.
24. Solution: (1)
Set and intersect at this point, and then
∫ The midpoint is 1 point.
There are two more points.
∴ swept area = 3 points
(2)①, 4 points.
(2) When, take the midpoint, then
5 points
6 points
Yes,
At this moment,
To sum up, 7 o'clock
25.
(1) proves that ∵ is an equilateral triangle.
1 point
Is the midpoint
∴
∵
∴
2 points
∴
The trapezoid is an isosceles trapezoid. 3 points
(2) solution: in equilateral,
∴
∴
Four points.
∵ ∴
Five points.
(3) Solution: ① If yes, yes.
Then quadrangles and quadrangles are parallelograms.
6 points
When, yes.
Then quadrangles and quadrangles are parallelograms.
∴
When or, a quadrilateral with two vertices at p, m and a, b, c and d is a parallelogram.
At this time, the parallelogram has four .7 points.
② It is a right triangle.
∵
When the minimum value is taken,
∴ is the midpoint of the year, and
Eight points.
(Note: The solutions to many problems in this volume are not unique. Please give points according to the grading standard. )