数学试卷参考答案和评分标准
- 格式:doc
- 大小:192.50 KB
- 文档页数:4
第四次数学模拟试卷参考答案和评分标准
一.
二、填空题(本题有6小题,每小题5分,共30分)
11.x(x-9) 12.外切13.21 14.3 15.(1,4)或(3,4)16.(3,1)三、解答题(本题有8题,共80分)
17.(本题10
分)
(1)原式= = (前4后1共5分)
(2) 去分母,得x
x8
2
)3
3-
=
-
(. ·································································· 2分
解得,1
=
x.············································································· 4分经检验,1
=
x是原方程的根.
∴原方程的根是1
=
x.································································· 5分18.(本题9分)解:作图3
(1)如右图
(2)点A1、B1、C1的坐标
A1( 4 , -3 )
B1( 5 , 0 )
C1( 1 , -1 )
19.(本题8分)
解:(1)略(3分)
(2)根据题意得: 30
A
∠=︒,60
PBC
∠=︒
所以6030
APB
∠=︒-︒,所以APB A
∠=∠,所以AB=PB··························2分在Rt BCP
∆中,90,60
C PBC
∠=︒∠=︒,PC=450,
所以PB =
450
sin60
==
︒
······························2分答:略. ····················································1分
1
3
3
3
2
3
2-
+
-
⨯2
3
3
2
+
70
20. (本题10分)
(1)两个条形各1分,阴影1分,数字1分 (4分)
(2)今年销量的极差是 36 ,去年销量的中位数是 30 。(4分) (3)
(2分) 21.(本题9分) 解:(1)可以得到36个不同形式的二次函数 (图表略)(5分). (2)点在直线上的有(0,0),(1,1),(2,2),(3,3),(4,4),(5,5)六个,这样概率
为 (4分)
22.(本题10分) (1)x =1,y =2;x =2,y =6,代人y =ax 2+bx ,得⎩⎨
⎧=+=+6
242
b a b a ,解得⎩⎨⎧==11b a
∴y=x 2+x ;h=41x -165-(x 2+x)=-x 2+40x -165 (4分)
(2)h =-x 2+40x -165=-(x -20)2+235,
当x =20时h 最大,即开放20个月,纯收益达到最大 (3分) (3)∵x =4时h<0,x =5时h>0,
∴这套大型游乐设施开放5个月后,就能收回投资. (3分)
23.(本题10分)
解:(1)∵∠ABC 与∠ADC 互补,
∴∠ABC +∠ADC =180°. ∵∠A =90°,
∴∠C =360°-90°-180°=90°.(2分)
145
9.144100037
7073503939≈≈⨯++++
B D
C
61366=÷
(2)过点A 作AE ⊥BC,垂足为E.则线段AE 把四边形ABCD 分成△ABE 和四边形AECD 两部分,把△ABE 以A 点为旋转中心,逆时针旋转90°,则被分成的两部分重新拼成一个正方形. 过点A 作AF ∥BC 交CD 的延长线于F ,
∵∠ABC +∠ADC =180°,又∠ADF +∠ADC =180°, ∴∠ABC =∠ADF.
∵AD =AB ,∠AEC =∠AFD =90°,∴△ABE ≌△ADF.
∴AE =AF.∴四边形AECF 是正方形. (3分) (3)解法1:连结BD ,
∵∠C =90°,CD =6,BC =8,∆Rt BCD 中, 106822=+=BD . 又∵S 四边形ABCD =49,∴S △ABD =49-24=25. 过点A 作AM ⊥BD 垂足为M ,
∴S △ABD =2
1
×BD ×AM =25.∴AM =5.
又∵∠BAD =90°,∴△ABM ∽△ABD.
∴AM
MD BM AM =. 设BM =x ,则MD =10-x , ∴
5
105x
x -=
.解得x =5. ∴AB =25. (5分) 解法2:连结BD ,∠A =90°. 设AB =x ,BD =y ,则x 2
+y 2
=102
,① ∵
2
1
xy =25,∴xy =50.② 由①,②得:(x –y )2
=0. ∴x =y. 2x 2
=100. ∴x =25.
B
D
C