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20 10 answers to the second set of mathematical model test papers for the senior high school entrance examination in Yangpu district
First, multiple-choice questions (4 points for each small question, out of 24 points)

1.a; 2.b; 3.a; 4.c; 5.d; 6.B

Fill in the blanks (4 points for each small question in this big question, out of 48 points)

7.; 8.; 9.; 10.; 1 1..; 12.

13.; 14.; 15.36; 16.; 17.2 or 8; 18. 1

Iii. Answer questions (19~22 is 10, 23~24 is 12, and 25 is 14, out of 78).

19. solution: the original formula =-3 points.

= - .

= - .

When, the original formula =-.

= - .

20. Solution: Multiply both sides of the equation to get:-2 points.

Namely:-3 points.

So,-4 points.

After inspection, I gave up in order to increase my roots. -1 min.

So the solution of the original equation is

2 1. solution: a is AD⊥BC in D-.

Tanb =, ∴ If AD= k, BD=4k,-2 points.

∵∠C=600,∴DC=,AC = -。

∫ BC = 12, ∴∴-.

∴ AC = -。

22.( 1) 100 - .

(2) four-.

(3) 36 - .

(4) 1900 - .

(5) no. Because the height of some seventh-grade students does not represent the height of first-grade students. -2 points

23. It is proved that (1)∵M is the midpoint of AB, and AD⊥BC, BE⊥AC,

∴2 points, 2 points.

∴ me = MD,-。

△ med is an isosceles triangle-66.

(2) ∵∴∠ Mei = ∠ MEA,-1 min.

∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873

Similarly, ∴∠ MAD = ∠ MDA,-1 min.

∴∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠∠873

EMD =∣EMD

= 2 ∠ Mae-2 ∠ MAD = 2 ∠ DAC-65438。

24. solution: (1)∫ straight line passes through ∴∴ k = point c-1 point.

∴A(-2,0),-。

The parabola intersects with point A and point C of∴.

That is, the analytical formula of ∴ parabola is-2 points.

(2) Vertex D-can be found.

Let DH⊥y axis intersect with Y axis at h-point.

∴ - 6544

= - .

25. (The full mark of this question is 14, item (1) is 4, item (2) is 6, and item (3) is 4).

Solution: (1) △ ABF ∽△ GBC, △ FDE ∽△ CGE ∽ △ BCE-4 points.

(2) Method 1:∵ divided into equal parts ∠B, ∴∠ABE=∠EBC,

∵ad//bc,∴∠afb=∠ebc,∴∠abe=∠afb,∴ab=af,

∴AF=4,DF= 1,-。

∵ AD//BC, ∴DF:BC=DE:EC, ∴ DE =, CE =-2 points.

∵AD//BC,AB=CD,∴∠BCD=∠ABC

* CG shares ∠BCD, BE shares ∠ABC, ∴ CBG = ∠ BCG, ∴BG=CG.

Let BG = CG = x, then it is given by △FDE∽△CGE, DF:CG=DE:GE, ∴ Ge =-1.

And from △CGE∽△BCE, EC2=EG? EB is

∴, which means BG =-2 points.

Method 2: Find DF= 1,-65438+.

Get DE =, CE =-2 points.

Let DF:BC= 1:5 let EF=x, BE=5x, from △FDE∽△CGE, score-1.

And from △CGE∽△BCE, EC2=EG? EB, that is, get- 1

Get-1 minute.

(3)① When the time is right, point A is within ⊙ P ..-2 points.

(2) When point A is within ⊙P and point E is outside ⊙P-2.