First, multiple-choice questions (3 points for each small question, 30 points for * * *):
1. The following deformation is correct ()
A. if x2=y2, then x = Yb. If x2 = y2, then x = y.
C. if x(x-2)=5(2-x), then x =-5d. If (m+n)x=(m+n)y, then x = y.
2. By May 9th, 20 10, there were 1600 Chinese and foreign journalists registered for the Shanghai World Expo, and 2 1600 was expressed as () by scientific counting method.
a . 0.2 16× 105 b . 2 1.6× 103 c . 2. 16× 103d . 2. 16× 104
3. The following calculation is correct ()
A.3a-2a= 1 B.x2y-2xy2= -xy2
C.3a2+5a2=8a4 D.3ax-2xa=ax
4. Rational numbers A and B are shown on the number axis as shown in Figure 3, and the following conclusion is wrong ().
A.b< a CD.
5. Given that the solution of equation 4x-3m=2 about x is x=m, then the value of m is ().
A2 B- 2c. 2 or 7d-2 or 7
6. The following statement is true ()
The coefficient of a is -2b. The number of 32ab3 is 6.
The constant term of c polynomial d.x2+x- 1 is 1.
7. The effective number to round 0.06097 to the approximate value of one thousandth is ().
0，6，0 B.0，6， 1，0 C.6，0，9 D.6， 1
8. A workshop planned to produce a batch of parts, and then produced 10 pieces per hour. It took 12 hour, which not only completed the task, but also produced 60 more pieces. Suppose the original plan is to produce x parts per hour, and the equation is ().
a . 13x = 12(x+ 10)+60 b . 12(x+ 10)= 13x+60
C.D.
9. As shown in the figure, C, O and B are on the same straight line, ∠ AOB = 90,
∠AOE=∠DOB, the following conclusions are drawn: ①∠ EOD = 90; ②∠COE =∠AOD； ③∠COE =∠DOB； ④ Coe+BOD = 90。 The correct number is ().
A. 1
10. As shown in the figure, fold a rectangular piece of paper along EF, with points C and D at points M and N respectively, ∠MFB= ∠MFE. Then ∠MFB= ().
A.30 B.36 C.45 D.72
Two. Fill in the blanks (3 points for each small question, *** 18 points):
The difference between 2 times of 1 1.x and 3 can be expressed as.
12. If the value of algebraic expression x+2y is 3, the value of algebraic expression 2x+4y+5 is.
13. It takes A yuan to buy a pen and B yuan to buy a notebook, so it takes Yuan to buy M pens and N notebooks.
14. If 5a2bm and 2anb are similar terms, then m+n=.
15.900-46027/= , 1800-42035/29"= .
16. If the ratio of an angle to its complementary angle is 1∶2, then this angle is a degree, and the ratio of this angle to its complementary angle is.
Third, answer questions (***8 small questions, 72 points):
17. (* *10) Calculation:
( 1)-0.52+ ;
(2) .
18. (* *10 minute) Solve the equation:
( 1)3(20-y)= 6y-4(y- 1 1)；
(2) .
19.(6 points) As shown in the figure, find the area of the shaded part below.
20.(7 points) Given that A=3x2+3y2-5xy and B=2xy-3y2+4x2, find:
( 1)2A-B； (2) When x=3 and y=, the value of 2A-B 。
2 1.(7 points) As shown in the figure, ∠BOC=2∠AOB, OD bisection ∠AOC, ∠BOD=
14, find ∠AOB degree.
22.( 10) The figure below shows the T-shaped pattern made of chess pieces.
As can be seen from the patterns, the 1 t pattern needs 5 pieces, the second t pattern needs 8 pieces, and the third t pattern needs 1 1.
(1) According to this rule, how many pieces do you need to make the eighth pattern?
(2) How many pieces do you need to make the nth pattern?
(3) How many pieces are needed to form the 20th10 pattern?
23.( 10) A middle school in our city always opens its gate at the specified time every noon. Xiaoming, a seventh-grade classmate, rides his bike from home to school at the same time every day. He arrived at school at noon on Monday at a speed of15km/h. As a result, he waited for 6 minutes at the school gate before opening the door. At noon on Tuesday, he arrived at school at a speed of 9 kilometers per hour. As a result, the school gate has been open for six minutes. At noon on Wednesday, Xiaoming wants to arrive at the school gate on time.
Please follow the following ideas to complete the solution process of this question:
Solution: If it takes T hours for Xiaoming to arrive at the school gate on time from home at noon on Wednesday, it will take several hours for Xiaoming to arrive at the school gate from home at noon on Monday and several hours for Xiaoming to arrive at the school gate from home at noon on Tuesday. According to the meaning of the problem, the equation is obtained:
24.( 12 point) As shown in the figure, there are three points A, B and C on the ray OM, which satisfy OA=20cm, AB=60cm and BC= 10cm (as shown in the figure). Point P starts from point O and moves at a constant speed along the om direction 1cm/ s, and point Q starts from.
(1) when PA=2PB, point q moves to.
The position is exactly one third of the line segment AB.
Point, find the movement speed of point Q;
(2) If the moving speed of point Q is 3cm/ s, how long will it take for P and Q to be 70cm apart?
(3) When the point P moves to the line segment AB, take the midpoint E and F of OP and AB respectively.
Reference answer:
First, multiple-choice questions: BDDCA, CDBCB.
Second, fill in the blanks:
1 1.2x-3； 12. 1 1 13 . am+bn
14.3 15.43033/, 137024/3 1" 16.300.
Third, answer questions:
17.( 1)-6.5; (2) .
18.( 1)y = 3.2； (2)x=- 1。
19.。
20.( 1)2 x2+9 y2- 12xy； (2)3 1.
2 1.280.
22.( 1) 26 pieces;
(2) Because the [1] th pattern has five pieces, the [2] th pattern has (5+3× 1) pieces, and the [3] th pattern has (5+3×2) pieces, the [n] th pattern can be obtained by a rule.
(3)3×20 10+2=6032 (pieces).
23.； ; From equation:, solution: t=0.4,
So the distance from home to school by bike for Xiao Ming is:15 (0.4-0.1) = 4.5 (km).
That is, the speed at which Xiao Ming arrives at the school gate on time by bike from home at noon on Wednesday is:
4.5÷0.4= 1 1.25 (km/h).
24.( 1) ① when p is on the line segment AB, from PA=2PB and AB=60, we can get:
PA=40, OP=60, so the movement time of point P is 60 seconds.
If AQ=, BQ=40, CQ=50, the moving speed of point Q is:
50÷60= (cm/s);
If BQ=, BQ=20, CQ=30, the moving speed of point Q is:
30÷60= (cm/s).
(2) When P is on the extension line of the line segment, from PA=2PB and AB=60, we can get:
PA= 120 and OP= 140, so the movement time of point p is 140 seconds.
If AQ=, BQ=40, CQ=50, the moving speed of point Q is:
50140 = (cm/s);
If BQ=, BQ=20, CQ=30, the moving speed of point Q is:
30÷ 140= (cm/s).
(2) Let the exercise time be t seconds, then:
① Before P meets Q: 90-(t+3t)=70, and the solution is t=5 seconds;
② After P meets Q: When Q moves to O and stops moving, Q moves for 30 seconds at most, while when P continues to move for 40 seconds, the distance between P and Q is 70cm, so t=70 seconds.
∴ After 5 seconds or 70 seconds, P and Q are 70 cm apart.
(3) let OP=xcm, point p be on line AB, 20≤x≤80, OB-AP=80-(x-20)= 100-x, ef = of-OE = (OA+)-OE = (20+).
∴。