湖南省益阳市2020年七年级期末学业水平考试试卷(扫描版)

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25. (本题 12 分)
解:(1)因为∠ABC 和∠ACB 的平分线 BO 与 CO 相交于点 O,
所以∠EBO=∠OBC = 1 ABC ,∠FCO=∠OCB = 1 ACB
2
2
又∠ABC=50°,∠ACB=60°,
所以∠OBC=25°,∠OCB=30°
所以∠BOC=180°-∠OBC -∠OCB=125° ·············································6 分
(2)设第一天购买了 x 瓶,第二天购买 y 瓶, (x>y 且 x、y 均为正整数) ①第一、二天均超过 30 瓶但不超过 50 瓶,享受八折优惠,
依题意得
x + y = 70 2.4x + 2.4y
=
183
,无解;·················································8
解:原式 = 6x4 − 3x3 + 3x2 − 3x4 + 4x3 − 2x2 = 3x4 + x3 + x2 ····························6 分 当 x = −1 时,原式 = 3 (−1)4 + (−1)3 + (−1)2 = 3 −1+1 = 3 . ···················8 分
9 10
答案
D
B
C
D
B
D
C
A
C
A
一、(本题共 10 个小题,每小题 4 分,共 40 分) 二、(本题共 8 个小题,每小题 4 分,共 32 分)
11. a(m + 3)(m − 3) ;12.2;13. 4 ;14.0;15.ab;16.52 ;17. 90°;18.6. 9
三、解答题(本题 8 个小题,共 78 分) 19.(本题 8 分)
21. (本题 8 分) 解:利用图形平移的性质及连接两点的线中,线段最短,可知: AC+CD+DB=(ED+DB)+CD=EB+CD. 而 CD 的长度又是平行线 PQ 与 MN 之间的距离,所以 AC+CD+DB 最短. ······8 分
22. (本题 10 分) 解:(1)因为 37.3 + 36.9 + 37.2 + a + 37 + 37.1 + 36.8 + 36.7 = 37 ···························4 分 8 所以 a = 37 (2)中 位 数 和 众 数 均 为 37 ······························································10 分
7/9Байду номын сангаас
解得:
x
y
= 1000 = 11000
············································································ 8

答:N95 口罩购进 1000 副和一次性口罩购进 11000 副. ········································10 分
26. (本题 12 分) 解:(1)因为(一)班一次性购买了纯净水 70 瓶,
8/9
所以享受六折优惠,即一班付出:70×3×60%=126 元, 因为两班共付出了 309 元, 所以(二)班付出了:309-126=183 元, 所以(一)班比(二)班少付多:183-126=57 元, 答:(一)班比(二)班少付 57 元.···························································6 分
(2)因为∠BOC=130°,
A
所以∠1+∠2=50°
因为∠1:∠2=3:2
所以 1 = 3 50 = 30 , 2 = 2 50 = 20 B
5
5
E
1
O 2F
C
因为 EF∥BC
所以∠OBC=∠1=30°,∠OCB=∠2=20°
因为∠ABC 和∠ACB 的平分线 BO 与 CO 相交于点 O,
所以∠ABC=60°,∠ACB=40°.························································12 分

②第一天超过 50 瓶,享受六折优惠,第二天不超过 30 瓶,不享受优惠,
依题意得
x + y = 70 1.8x + 3y =
183
,无整数解;·············································10

③第一天超过 30 瓶但不超过 50 瓶,享受八折优惠,第二天不超过 30 瓶,不 享受优惠,
24.(本题 10 分) 解:设 N95 口罩 x 副和一次性口罩购进 y 副, ·····················································2 分
依题意得:
x + 15x
y +
= 12000 1.5y = 31500

···················································5 分
23. (本题 10 分) 解:如图,过点 D 作 DE∥AB 交 BC 于点 E. ················································2 分 ∴ ∠A+∠2=180°,∠B+∠3=180°(两直线平行,同旁内角互补). ·········6 分 又∵ ∠3=∠1+∠C, ∴ ∠A+∠B+∠C+∠1+∠2=360°, ···················································8 分 即∠A+∠B+∠C+∠ADC=360°.························································10 分
益阳市 2020 年期末学业水平考试试卷(七年级)
数学
时量:90 分钟 满分:150 分
题一 次 1~8
二 9~14
三 15 16 17


18 19 20 21

总分
22
得 分
试题卷
1/9
2/9
3/9
4/9
5/9
2020 年上学期考试七年级数学试卷 参考答案及评分标准
题号
1
2
3
4
5
6
7
8
依题意得
x + y = 70 2.4x + 3y =
183
,解得
x y
= =
45 25

答:第一天购买 45 瓶,第二天购买 25 瓶. ·····································12 分
9/9
20.(本题 8 分) 解:因为 x﹣y=1, 所以(x﹣y)2=1, 即 x2+y2﹣2xy=1; ··············································································4 分 因为 x2+y2=9,
6/9
所以 2xy=9﹣1, 解得 xy=4. ······················································································8 分
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