Mathematics examination paper in Pudong New Area, two days before the end of the second semester of the 96 academic year.

One choice: (each question in this big question 12 points)

1。 The ordinate of the intersection of the straight line and the Y axis is ................................................... ().

(A)(B)(C)3(D)？ 3。

2。 You can set the original equation to instead of solving the equation ... ()

(A)(B)；

Items (c) and (d).

3。 The following formula, the real root of the system of equations ............................................................ ().

Items (a), (b), (c) and (d).

4。 It is known that the two diagonals AC and BD of the parallelogram ABCD intersect at the point o to 8cm and the length of the long side of 12cm is equal to BC and 6cm, then the circumference of △BOC is equal to ..............., ............ and .................. ().

(A) 14(B) 15(C) 16(D) 17 .

5。 The following proposition is a false proposition ........................................................................ ()

The trapezoid of (a) is equal to two diagonal lines, and the two diagonal lines of the rectangle of (b) are equally divided;

The diamonds of (c) are perpendicular to each other, and each diagonal of the square of (d) bisects the diagonal.

6。 Afterwards, determine the event ........................................................................ ().

(a) Equation X has a real number solution; On the (b real number solution) equation of x

(c) Equation X has a real number solution; (d) The equation about x has a real number solution.

Fill in the blanks (this big question 12 questions, 36 points out of 3 points for each question)

. Equation.

8。 If the function is a function, then.

9。 If the function images of point A (2, m) and point B (4, n) then have a size relationship of m, n is: ten thousand yuan. (">," = "or"

10。 If the equation of x is an unrelated root and x = 2, then the value of k.

1 1。 Please write the solution of the quadratic equation of two variables.

12。 The sum of the internal angles of a heptagon is equal to degrees.

13。 Given that the square ABCD is 8cm, the diagonal BD distance of the midpoint AB M on one side is equal to

The centimeter side length is equal to.

14。 The hypotenuse of an isosceles right-angled triangle is 5 cm high, so the midpoint of the length of two right-angled sides of this triangle is equal to cm.

15。 Two elements of a vector: size.

16。 Known parallelogram ABCD, set, vector, vector

=。

17。 Equipped with three red balls, five yellow balls and six black balls, these balls are all the same except the color, so the probability of taking out a black ball in this bag.

18。 The probability of is composed of two randomly selected numbers 2, 4 and 6. All randomly selected two-digit words have a number, which is divisible by 3.

Answer: (52 points for the big question)

19。 (The full mark of this question is 6)

Solution of the equation

20。 (The full mark of this question is 6)

Figure, known carrier. Find: Vector ( 1)，(2)。

2 1。 Title (7 points)

As we all know, because in the parallelogram ABCD, the AP of BC and CD2 cm both bisect the ∠BAD AC side BC at point P. 。

Find the difference of PC length.

22。 Title (7 points)

Two walks from B to A, B, A to 35km, first turn left, then B bike, as shown in the figure, and answer under the information icon provided by the relationship between travel time and distance:

(1)(B) Leave later than A;

(2)B, catch up with a after the departure time.

(3) How many hours before A arrives at B?

23。

Students (Question 8) learned that after the Sichuan earthquake, they took out pocket money and took part in fund-raising activities. Raise funds for students from Class A and Class B in 840 yuan * * * to donate to Class B. Class B donated $5 per capita, with a per capita donation of 1 ,000 yuan, two categories less than that of Class A, so as to increase the number of students in Class A and Class B.

24。 (topic 8 points)

It is known that: AM△ABC midline, D is the midpoint of line segment AM, AM = AC, AE‖BC.

Proof: quadrilateral EBCA isosceles trapezoid.

/a & gt；

25。 (Question of full mark 10)

It is known that, as in the rhombic ABCD, the transverse ray CD with AB = 4 ∠ B = 60 ∠ PAQ = 60 is at the fixed point of point P, and the distance of point P is set, at point B, X and PQ = Y.

(1) proves that △APQ is an equilateral triangle;

(2) Find the resolution function y about x and write its definition domain;

(3) If PD⊥ AQ, the value of BP.

/a & gt；

Pudong New Area, the second semester, the first two days of the school year ended.

Reference answers and scores

Multiple choice question:

1。 D 2 .d，3 .c； 4。 c； 5。 a； 6。 B.

Second, fill in the blanks:

7.8 8。 ≠ 1; 9。 & gt； 10.4; 1 1。 Wait; 12.900; 13。 ; 14.5; 15。 Direction 16. 17。 18。 .

Answer the question:

19。 Solution: ② Yes: y = 2 times. ..............................................................................( 1)

① Generation 5×2 = 20. ........................................................................( 1)

∴x = 2 .....................................................................................( 1)

When x = 2 and y = 4; When x =-2 and y = -4. ..................................................。 ( 1)

The solution of the equation is ..................................... and ........... (2).

20。 Mapping solution: every 2 points, 1 point ends.

2 1。 Solution: In parallelogram ABCD,

∵AD‖BC∴∠DAP =∠APB ..........................................................(2)

∠ DAP = ∠ BAP ∠ APB = ∠ DAP ................................................. (1)

∴AB = BP .....................................................................................(2)

∫ab = cd,∴pc = BC-BP = 2 .............................................................(2)

22。 Solution: (1) 2; ..................................................................( 1)

(2)2; .....................................................................( 1)

(3) The driving time of the resolution function is 5 t.. ..........................................( 1)

When S = 35 and T = 7. ........................................................................( 1)

Let b leave, and the analytic function S = KT+B changes with time.

According to these problems, the solution is a function of distance and time.

The analytical formula of ∴b is S = 10 ton -20. .................................... (1)

When S = 35 and T = 5.5. .....................................................................( 1)

∴7-5.5 = 1.5。

A: B arrived in .................................................... B 1.5 hours earlier than A. ( 1)

23。 Solution: The number of students in Class B is (x +2) of the tenth number of students in Class A. ..................( 1)

Also according to the meaning of the topic. ......................................................(3)

It's over. .........................................................( 1)

Solution. Tester ............................................................ (1)

: is the root of the original equation, but it does not meet the meaning of the question, so it is abandoned.

..............................................., .............................................. (1 min)

A: There are 40 students in Class A and 42 students in Class B.. ......................( 1)

24。 Proof: ∵AE‖BC∴∠AED =∠MCD∠EAD =∠CMD. ...........................( 1)

* AD = MD,∴△AED≌△MCD. ......................................................( 1)

∴AE = cm .................................................................................. (1)

∫BM = cm,∴ae = BM. Should

Quadrilateral parallelogram. ......................................................( 1)

∴EB = AM ....................................................................( 1)

AM = AC,∴EB = AC ...................................................................( 1)

∵AE‖BC, EB, AC's parallelogram EBCA is trapezoidal. ......................( 1)

∴ trapezoid isosceles trapezoid. ...............................................................( 1)

25。 Solution: (1) Link AC.

Ab = bc In the rhombic ABCD, ∠ B = 60 ∴△ ABC is an equilateral triangle. .................................( 1)

∴AC = AB，∠BAC =∠ BCA = 60。

∠∠paq = 60 ,∴∠bap =∠caq ...................................................。 ( 1)

AB‖CD∠B = 60 ∴∠BCD = 120 .

∴∠ACQ =∠B = 60 .

∴△ABP≌△ACQ。 ........................................................................( 1)

∴AP = AQ ....................................................................( 1)

∴△APQ is an equilateral triangle. ............................................................... AP = PQ = Y, that is, (1).

(2) Multiply △APQ is an equilateral triangle.

AH⊥BC is at point H, AB = 4, BH = 2, ∠ B = 60, AH =. .........( 1)

That's it. ....................................( 1)

For the domain defined by x≥0. ..................................................................( 1)

(3) In (i), at point P, ∵PD⊥AQ is on the side BC, and AP = PQ, ∴AQ PD is perpendicular to it.

∴AD = DQ .

∴CQ = 0 .....................................................................................( 1)

∫BP = cq,∴bp = 0 .

(ii) When point P is located on the extension line of BC side, point P is located at

Similarly, BP = 8. ..................................................................( 1)

In short, BP = 0 or BP = 8.

The first letter of the Greek alphabet.

The first letter of the Greek alphabet.