七年级数学期末模拟卷01(参考答案)【苏科版】七年级数学期末模拟卷(江苏专用)(苏科版七上第1~6章)-学易金卷:2023-2024学年初中上学期期末模拟考试.docx

七年级数学期末模拟卷01(参考答案)【苏科版】七年级数学期末模拟卷(江苏专用)(苏科版七上第1~6章)-学易金卷:2023-2024学年初中上学期期末模拟考试.docx

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2023-2024学年上学期期末模拟考试

七年级数学

1

2

3

4

5

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7

8

B

D

B

B

C

B

B

B

9.2.305×109 10.5 11.3

12. 13.4 14.158°42′

15.x2﹣4xy 16.32

17.(6分)

解:(1)原式=(﹣24)×12-(﹣24)×5

=﹣12+40+9

=37;··············································································3分

(2)原式=﹣1+6﹣9=﹣4.··························································6分

18.(6分)

解:(1)4-x3=

5(4﹣x)=3(x﹣3)﹣15,

20﹣5x=3x﹣9﹣15,

﹣5x﹣3x=﹣9﹣15﹣20,

﹣8x=﹣44,

x=112;

(2)5(x﹣3)﹣2(x﹣3)=0,

5x﹣15﹣2x+6=0,

5x﹣2x=15﹣6,

3x=9,

x=3.···········································································6分

19.(8分

解:∵x2+x=0,∴x2+x+1186=0+1186=1186,

故答案为:1186;·······································································2分

(1)∵x2+x﹣1=0,

∴x2+x=1,

∴x2+x+2016=1+2016=2017,

故答案为:2017;······································································4分

(2)∵a+b=5,

∴2(a+b)﹣4a﹣4b+21=2(a+b)﹣4(a+b)+21=﹣2(a+b)+21=﹣10+21=11;··········6分

(3)∵a2+2ab=20,b2+2ab=8,

∴2a2+4ab=40,3b2+6ab=24,

∴2a2+4ab﹣3b2﹣6ab=2a2﹣3b2﹣2ab=40﹣24=16.······································8分

20.(6分)

解:(1)主视图,俯视图如图所示:

······················································2分

(2)这个几何体的表面积=4+4+4+4+3+3=22(cm2);··································4分

(3)如果在这个几何体上再添加一些小立方块,并保持主视图和左视图不变,最多可以再添加2个小立方块.

故答案为:2.······································································6分

21.(8分)

解:(1)如图,直线CD即为所求;···················································2分

(2)如图,直线CE即为所求;······················································4分

(3)如图,在线段CA、CB、CE中,线段CE最短,理由是垂线段最短;

故答案为:CE,垂线段最短;························································6分

(4)S△ABC=3×7-12×1×7-12×

故答案为:8.·········································································8分

22.(8分)

解(1)∵直线AB、CD相交于点O,

∴∠AOC+∠AOD=180°,∠BOE+∠AOE=180°,

∵∠AOC=80°,

∴∠AOD=180°﹣80°=100°,

∵OE平分∠AOD,

∴∠AOE=12∠AOD=

∴∠BOE=180°﹣∠AOE=180°﹣50°=130°;········································4分

(2)由(1)得∠AOE=50°.

∵OF⊥OE,

∴∠EOF=90°,

即∠AOE+∠AOF=90°.

∴∠AOF=90°﹣∠AOE=90°﹣50°=40°.

∵∠AOC=80°,

∴∠C

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