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初中学业水平考试数学试题参考答案第PAGE8页(共5页)
数学试题参考答案及评分标准
一选择题:本大题共12个小题,每小题5分,共60分。
题目
1
2
3
4
5
6
7
8
9
10
11
12
题号
A
D
C
D
B
A
C
C
D
B
A
B
二填空题:本大题共5个小题,每小题4分,共20分。
13a≥514x(x+3)(x—3)15(1,3)16—217(—2024,2024)
三解答题:本大题共7个小题,共70分。
18(本题满分8分)
解:整理方程组得,·········································2分
①×2—②得—7y=7,
y=1,····································································4分
把y=1代入①得x—2=3,
解得x=5,································································6分
∴方程组的解为··················································8分
19(本题满分8分)
证明:∵△ABC是等腰三角形,
∴∠EBC=∠DCB,·······················································2分
在△EBC与△DCB中,
∴BE=CD,BC=CB
∴△EBC≌△DCB(SAS),··················································6分
∴BD=CE······························································8分
20(本题满分10分)
解:(1)将点A(1,2)代入y=,得m=2,
∴双曲线的表达式为:y=,···············································1分
把A(1,2)和B(4,0)代入y=kx+b得:
y=,解得:,·········································3分
∴直线的表达式为:y=x+;·········································4分
(2)联立,解得,或,·························5分
∵点A的坐标为(1,2),
∴点B的坐标为(3,),···············································6分
∵S△AOB=S△AOB—S△AOB=OC·—OC·
=×4×2—×4×
=,
∴△AOB的面积为;···················································8分
(3)1m3····························································10分
21(本题满分10分)
解:(1)12099························································4分
(2)如图:
··············8分
(3)把“礼仪”“陶艺”“园艺”“厨艺”及“编程”等五门校本课程分别记为ABCDE,画树状图如下:
共有25种等可能的结果,其中小刚和小强两人恰好选到同一门课程的结果有5种,
∴小刚和小强两人恰好选到同—门课程的概率P=
············································10分
22(本题满分10分)
解:小明能运用以上数据,得到综合楼的高度,理由如下:
作EG⊥AB,垂足为G,作AH⊥CD,垂足为H,如图:
22题
22题答案图
····················2分
由题意知,EG=BF=40米,EF=BG=1288米,∠HAE=16°=∠AEG=16°,∠CAH=9°,
在Rt△AEG中,
tan∠AEG=,
∴tan16°=,即0287≈,············································4分
∴AG=40×0287=1148(米),
∴AB=AG+BG=1148+1288=2436(米),····································6分
∴HD=AB=2436米,
在Rt△ACH中,AH=BD=BF+FD=80米,
tan∠CAH=,
∴tan
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