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20 14 ―― Mathematical answers of the joint entrance examination for grade one in Jiaozuo city in the 20 15 school year.
20 14――20 15 school year, the mathematics answer of the first grade entrance examination in Jiaozuo city is the reference answer of mathematics science.

1.BDCAA? BBABC? Civil band

Second, 13. ( 1)(2) 14. 15.5 16.(2, 19/8)

Third, answer questions.

17.( 1) The scatter diagram is as follows.

What is the linear regression equation of ∴? ............................, eight.

(3) For the question: y= 1 15, 6.5x+17.5 =115, x =15 ...

18.( 1)f(x)? =

=sin(2x+π/6)+3/2,

∴ f (the smallest positive period ∴f(x) is π...6 points.

(2) The maximum value of ∵ y = sin (2x+π/6) is 1, and the minimum value is-1.

∴f(x)==sin(2x+π/6)+3/2 The maximum value is 5/2, and the minimum value is 1/2.

19.( 1) The frequencies with a score in the range of [120, 130] are

1-(0.01+0.015+0.015+0.025+0.005) ×10 =1-0.7 = 0.3.

The frequencies with scores between [135, 150] are,

The frequency is 0. 175? ×? 60= 10.5≈ 1 1,

Therefore, the number of students with excellent grades is 1 1. ..........., eight.

The average score is 95× 0.1+105× 0.15× 0.15+125× 0.3+135× 0.25.

So it is estimated that the average score of mathematics in the whole grade of this exam is 12 1. ............ is 12.

20.2 1. certificate: (1) link a? C, straight triangular prism a? b? c? -in -ABC, quadrilateral AA? c? C is a rectangle,

∴ Point F is in A? C, and f is a? The midpoint of C.

At △A? In BC, ∫E and f are a respectively? b,A? The midpoint of c, so ef∨BC.

∵BC and plane ABC, EF? Aircraft ABC? EF∨ Plane ABC...........4 points

(II) in triangular prism a? b? c? -Yes. -ABC, b? B⊥ Airplane ABC, ∴? b? B⊥EF?

If (i) knows ef∑BC and AB⊥BC, then EF⊥AB,

∵? b? B∩AB=B,∴? EF⊥ Aircraft ABB? One,

∵EF and AEF, ∴ AEF⊥ ABB? A..........? 8 points

㈢? ..... 12 point

22. solution: (1)│CR│=2, and the equation of the circle is x? +(y-3)? =4..........3 points.

(2) Method 1: ① When k does not exist,

X=- 1, then P (- 1, 3-√ 3), Q (- 1, 3+√ 3), M (- 1, 3),

Obviously there is AC am = 9,

② when k exists, let y=k(x+ 1),

The equation of ∴l is y=kx+k,

P(x? ,y? ),Q(x? ,y? ),M(x? ,y? ),

∴,

Do you have an X? + 1+3y? =9, that is, simultaneous y=kx+k, x? +(y-3)? =4,

Substitute into the equation: x? + 1+3y? =9, so:

Solution: k=4/3,

To sum up, the equation of L is x=- 1 or 4x-3y+4=0. .............. 12 point

Method 2: ∫M is the midpoint of PQ line,

According to the vertical diameter theorem, that is, CM⊥PQ, that is, am MC = 0,

(1) if k exists, let the straight line l be y-0=k(x+ 1), that is, y=kx+k,?

Distance from the center of the circle c (0 0,3) to the straight line l

The solution is k=4/3,

The equation of the straight line is 4x-3y+4=0,

② If k does not exist, the straight line through a (- 1, 0) is x=- 1,

It is also satisfied that the distance from c (0 0,3) to the straight line x=- 1 is 1.

To sum up, the equation of a straight line is 4x-3y+4=0 or x=- 1.

Method 3: A (- 1, 0), C (0 0,3), set point M(x? ,y? ), then:

From the meaning of the question: AC am = x? + 1+3y? =9,x? = 8-3 years old ①

Because m is the midpoint of the chord PQ, AM⊥CM,

AM CM=x? (x? + 1)+y? (y? -3)=0, and substitute it into Formula ① to get:

(8-3y? )(9-3y? )+y? (y? -3)=0, sorting:

(y? -3)( 10y? -24)=0, and the solution is: y? =3 or y? = 12/5

The coordinate of m is (-1, 3) or (4/5, 15/5), so the equation of straight line L is 4x-3y+4=0 or x=- 1.