Reference answers to math test questions

First, multiple choice questions

1.A

This topic analyzes and examines the rational number multiplication operation. Two numbers are multiplied, the same sign is positive, the different sign is negative, and the absolute value is multiplied. ∴3×(a4)= a3×4 = L2。

2.B

This topic analyzes the relationship between three-dimensional graphics and three views. The front view is a figure viewed from the front, with a rectangle on the left and a rectangle on the right.

3.D

Analyze this problem and investigate the scientific notation for expressing large numbers, and write it as a× l0n (1 ≤ a < 10, where n is a positive integer), where n is equal to the number of digits of the integer minus 1, ∴ n = 6- 1 = 5, ∴.

4.C

Analyze this problem and examine the meaning of median. First, arrange the data in descending order: 25, 25, 28, 30, 35, 37. The average of the middle two numbers =29, and the median is 29.

5.B

Check multiplication, division and exponentiation of algebraic expressions. Option a is the same as the base power, the base is the same, the index is added, ∴, A is incorrect. The power of option B with the same cardinality is multiplied. (a2)3-a2×3=a6, B is correct, C is divided by the same base power, the base is unchanged, and the exponent is subtracted. , C is incorrect, D option, ∴ D is incorrect.

6.C

Analyze this problem and investigate the solution of one-dimensional linear inequalities. If you solve inequality ①, you will get x; A 2 ∴ The solution set of this inequality group is A 2.

7.C

In this paper, the properties of rhombus are investigated: ∵ rhombus ABCD, ∴AB=BC=CD=AD= × 16=4, ∵∠ A = 60, ∴△ABD is an equilateral triangle, ∴ BD = AB = 4.

8.A

Analyze this problem and investigate the addition and subtraction of fractions. Add and subtract fractions with the same denominator, the denominator remains the same, and the numerator is added and subtracted. The result of fractional calculation must be the simplest fractional or algebraic expression.

9.B

This topic analyzes the data analysis in statistics. The sample size is as high as 280 people, 80 of whom want to hold cultural performances, accounting for. Using the sample to estimate the total population, there are about 1400×400 students in the whole school who want to hold a cultural performance.

10.B

By analyzing this problem, the properties of linear functions are studied. From the image, y decreases with the increase of x, ∴k-2.

1 1.D

By analyzing this problem, the properties of isosceles trapezoid are studied. The diagonal lines of isosceles trapezoid are equal, ∴A is correct, the waist of isosceles trapezoid is equal, ∴ AB = DC. And ∴BC=BC, ∴△ ABC △ DCB (SSS). ∴∠ OBC =

12.D

This question analyzes and examines the comprehensive application of acute trigonometric function, Pythagorean theorem and circle theorem, connecting AB, ∵BO⊥AO, ∴∠boa = 90°, ∴AB as the diameter, ∫ point D as the center, ∴D on AB, AB=, and.

13.C

By analyzing this problem, the symmetry of quadratic function like parabola is studied. The heights of the second and sixth seconds are equal, and the symmetry axis = 4.

14.C

This topic analyzes and investigates the problem of discovering the law. The original law is that the first and second numbers are odd or even. If the first number is n, then * * has (2n-l) numbers, and the result is equal to the square of the average of the first and second numbers. The first number in option A is odd, and the last number is even. A is not correct. The first number in option B is 1 000. ∴ The number of English * * is 2× 1 005— 1=2 009, while the number of B*** is 2 0 13, and B is incorrect. C conforms to the law in the topic. * The first number is 1 006. ∴ The number of English * * is 2×1006—1= 2011. The result is equal to () 2 = 20 1 12. D, the first number is 1 007.

15.A

In this paper, the properties of the square and the method of judging congruent triangles are investigated. ∫ square Abd, BCMN, ∴AB=AE=ED=BD, CB=BN, ∠ CBN = ∠ Abd = 90. If you pass N, NK⊥BD is in K, pass C, CQ ⊥. ∴ ABC = ∠ KBN. bc = bn，∠ CQB = ∠ bKN = 90，∴△bcq?△bnk。 ∴ GQ = NK，S△ABC= AB？ CQ，S△BDN= BD？ Knot ∵AB=BD,CQ=NK,∴S△ABC=S△BON. ∫△ABC?△CGM，∴ S2 = S2。 Similarly, S3=S 1, ∴ S 1 = S2 = S3.

Second, fill in the blanks

16. 19

17.(A-3)2

18.x 1=0，x2=2

19. 1 10

20.(3,6)

2 1.4

Third, answer questions.

22.( 1) solution: (a+b) (a-b)+2b2 = a 2-B2+2b2 ............................................................................................................ (2 (2

= A2+B2 ........................................... (3 points)

(2) Solution: 2 .............................. (5 points).

X = 3 .......................................... (6 points)

It is verified that x=3 is the solution ................................................................................. of the original equation (7 points).

23.( 1) solution: ∫∠A+∠b+∠C = 180, ∠ A = 60,

∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873

∠∠b:∠c = 1:5, ∴∠ B+5 ∠ B =120 ......................................... (2 points).

∴∠ B = 20 .................................. (3 points)

(2) Prove that ∵ quadrilateral ADCD is a square, ∴AB=CB, ∠ ABM = ∠ CBM ..................... (5 points).

∵BM is the edge of public * * *, ∴△ ABM△ CBM ....................................... (6 points).

∴ am = cm ..................................................... (7 points)

24. Solution 1: In this tour, there are X teachers and Y students ............... (1 minute).

........................................ (5 points)

Solution ................................................ (7 points)

A: On this tour, there are 10 teachers and 65,438+000 students ................................................................................ (8 points).

Solution 2: In this tour, there are X teachers ......................................... (1 minute).

40x+20 (110-x) = 2 400 ..................................... (5 points).

The solution is x =10 ............................................................. (6 points).

110-x =110 =100 (person) ................................................................................................................

A: On this tour, there are 10 teachers, 1000 students ........................................................................................................................ (8 points).

25. Solution: (1) The probability of flying to get a type A pen is (2 points).

(2) According to the meaning of the question, the list is as follows:

glad

Feifei a b c d

A(A，A)(A，B)(A，C)(A，D)

B(B，A)(B，B)(B，C)(B，D)

(? (C, A) (C, B) (C, C) (C, D)

D(D，A)(D，B)(D，C)(D，D)

Or tree:

............................................... (6 points)

As can be seen from the table (or tree diagram), there are 16 possible results, and 4 of them are qualified.

∴P (same model) = ................................... (8 points)

26. Solution: (1)①∫BD = ab, ∴ ∠ D = ∠ bad. ...............................................................................................................

∴∠ ABC = ∠ D+∠ Bad = 2 ∠ D = 30 ............................... (2 points).

∴∠ d =15 ............................................. (3 points)

② ∵∠ C = 90，∴∠CAD = 90∠∠d = 90∣∠d = 75 ................................

∫≈ABC = 30 ,ac=m,∴bd=ab=2m,bc= m,∴cd=cb+bd=(2+)m

............................................... (5 points)

∴∴ Tan ∠ CAD = .................... (6 points)

(2) The coordinates of point ∫m are (2,0), ∠ Omn = 75, ∠ Mon = 90,

∴ON=OM？ tan∠OMN=2×(2+ )=4+2。

The coordinate of point N is (0,4+) ...................................... (7 points).

Let the functional expression of the straight line MN be y=,

Then .......................... (8 points)

It is found that the function expression of line MN is

Y = ............................... (9 points)

27. solution: (1) ∵ parabola y= passing through points a (0 0,8) and c (6 6,0),

Get a solution

................................. (2 points)

(2)①∵OA=8,OC=6,∴AC=

So, QE⊥BC at point E,

∴

.................... (3 points)

=...(4 points)

When m=5, s takes the maximum .......................... (5 points).

(2) There is a point f on the parabola symmetry axis L, so that △FDQ is a right triangle, and four points F*** satisfy the conditions.

Coordinates are: F 1 (,8), F2 (,4), F3 (,), F4 (,).

................................................ (9 points)

28.( 1) Proof: ∫∠ACD =∠BCE, ∴∠ACD+∠DCE=∠BCE+∠DCE.

∴∠ ACE = ∠ DCB ........................... (1min)

∵CA=CD, CE=CB, ∴△ ace △ dcb ............................. (2 points)

(2) △ AMC ∽△ DMP .................... (3 points)

Reason: ∫△ace?△dcb, ∴ ∠ CAE = ∠ CDB ......................................... (4 points).

∠∠AMC =∠DMP, ∴△ AMC ∽△ DMP ................................. (5 points)

(3)∫△AMC∽△DMP, ∴...........................(6 points)

And ≈DMA =∠PMC,

∴△AMD∽△CMP.

∴∠ ADC = ∠ APC ........................................ (7 points)

Similarly, BEC = BPC.

CA = CD，CB=CE，

∴∠ADC=∠DAC= ( 180 -∠ACD)，

∠BEC=∠EBC= ( 180 -∠BCE)。

∫∠ACD =∠BCE，

∴∠ ADC = ∠ BEC ............................................ (8 points)

∴∠ APC = ∠ BPC ............................................ (9 points)

2+(-2)=0