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    新疆和硕县第二中学2021-2022学年七年级上学期期末考试数学试题(word版 含答案)

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    新疆和硕县第二中学2021-2022学年七年级上学期期末考试数学试题(word版 含答案)

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    这是一份新疆和硕县第二中学2021-2022学年七年级上学期期末考试数学试题(word版 含答案),共5页。试卷主要包含了请将答案正确填写在答题卡上,下列运算中,正确的是,,其中x=﹣2,y=1等内容,欢迎下载使用。
    2021-2022学年第一学期期末考试试七年级数学(问卷)注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上   评卷人得分  一、选择题(本题8题,每4分,共32分)1.在﹣(﹣2),10,﹣1这四个数中,负数是(  )A.﹣(﹣2 B1 C0 D.﹣12.据统计,2021年和硕县人口总数约为5.8万人.将数据5.8万用科学记数法表示为(  )A58×104 B5.8×104 C0.58×103 D5.8×1053.某几何体从三个方向看到的平面图形都相同,这个几何体是(  )A       B        C     D4.下列运算中,正确的是(  )A2a2+3a35a5 B3x+3y6xy C3mn3nm0 D7x5x25.若x=﹣1是关于x的方程2x+m1的解,则m+1的值是(  )A4 B2 C.﹣2 D.﹣16.将“祝你考试成功”这六个字分别写在一个正方体的六个面上.若这个正方体的展开图如图所示,则在这个正方体中,与“你”字相对的字是(  )A.考            B.试            C.成            D.功    7一艘轮船从A港顺流行驶到B港,比它从B港逆流行驶到A港少用3小时,若船在静水中的速度为26千米/时,水流的速度为2千米/时,求A港和B港相距多少千米.设A港和B港相距x千米.根据题意,可列出的方程是(  )A BC D8.若21222423824162532…,则22022的末位数字是(  )A2 B4 C8 D6 评卷人得分  二、填空题(本题6题,每3分,共18分) 9.要把一根细木条固定在墙上,至少需要钉两个钉子,其中蕴含的数学道理是                  10.如图所示,一副三角板(直角顶点重合)摆放在桌面上,若∠AOC125°,则∠BOD                                                                11.如图所示是计算机某计算程序,若开始输入x=2,则最后输出的结果是                12.代数式3x83互为相反数,则x        13.若| b2 | +a+320,则(a + b2022的值为      14.某种商品的标价为120元,若以九折降价出售,仍可获利20%,设该商品的进货价为x元,可列方程为                        三、解答题(本题7,共50分)15.计算(共2小题,每小题4分,共8分)1)(﹣12)﹣(﹣20+(﹣8)﹣15         2)(﹣14  + 16 ÷(﹣23| 13 |   16.解方程(共2小题,每小题5分,共10分) 16x7 4x5                          2 1     17.(6分)先化简,再求值:4xy﹣(2x2+5xyy2+2x2+3xy),其中x=﹣2y1      186分)如图,线段AB8cmC是线段AB上一点,AC3.2cmMAB的中点,NAC的中点,求线段MN的长.        19.(6分)如图,已知∠AOB90°,∠BOC60°,OD平分∠AOC.求∠BOD的度数.        20.(7分)和硕县家庭用水收取水费规定如下:若每年每户用水不超过144立方米,每立方米水价按1.8元收费;若超过144立方米,其中没超过144立方米的部分仍按每立方米1.8元收费,超过144立方米的部分按每立方米2.7元收费.1)若小龙家2021年用水148立方米,应交水费多少元?2)若小龙家2020年的水费平均每立方米2.1元,那么他家这一年共用了多少立方米的水?     21.(7分)如图,M是线段AB上一点,且AB10cmCD两点分别从MB同时出发,1cm/s3cm/s的速度沿直线BA向左运动,运动方向如箭头所示(C在线段AM上,D在线段BM上).1)当点CD运动了2s,求这时AC + MD的值.2)若点CD运动时,总有MD 3AC,求AM的长.   
    2021-2022学年第一学期期末考试试七年级数学答案一、选择题(本题8题,每4分,共32分)题号12345678答案DBCCABAB填空题(本题6题,每3分,共18分)9 两点确定一条直线         10 55°         11 22   12                         13.  1             14 120×0.9 - x = 20%x三、解答题(本题7,共50分)15.计算(共2小题,每小题4分,共8分)1)解:原式= -12 + 20 - 8 -15··············································1                            =15·······················································42)解:原式= 1 + 16 ÷(-8-2···········································1             =1+(﹣2)﹣2···············································2             =3························································416.解方程(共2小题,每小题5分,共10分)1)解:移项,得  6x4x=5+7·······································2合并同类项,得  2x2·········································4系数化为1,得  x1··········································52)解:去分母,得  24x1)= 6﹣(3x1)························1去括号,得  8x263x+1··································2移项,得 8x+3x6+1+2······································3合并同类项,得 11x9·······································4系数化为1,得 x   ·······································517.先化简,再求值(6分)解:原式= 4xy - 2x2 - 5xy + y2 + 2x2 + 6xy·····································2=5xy + y2························································4x=﹣2y1原式 = 5×(﹣2)×1 + 12··········································5=9····························································618.解:∵MAB的中点,AB8cmAMBM4cm····················································2NAC的中点,AC3.2cmANCN1.6cm····················································4MNAMAN41.62.4cm························································619.解:∵∠AOB90°,∠BOC60°,∴∠AOC=∠AOB+BOC150°······································2OD平分∠AOC∴∠AOD   AOC75°···········································4∴∠BOD=∠AOB﹣∠AOD15°·····································620.(1)解:144×1.8 + 2.7×(148144)······································1=259.2+10.8=270(元)·····················································2答:小龙家2021年应交水费270.·······································32)解:设小龙家2020年共用了x立方米的水································4列方程得:  2.1x = 144×1.8+2.7x144       解得      x=216····················································6答:小龙家2020共用了216立方米的水.·································721.(1)解:当点CD运动了2s时,CM2 cmBD6 cm,···················1AB10cmCM2cmBD6cmAC+MDABCMBD10262 cm;························32)解:∵CD两点的速度分别为1cm/s3 cm/sBD3CM.····················································4又∵MD3ACBD+MD3CM+3AC,即BM3AM,······························6AM   AB2.5cm.············································7 
     

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