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2018-2019年北京市平谷区七年级上数学期末试卷+答案

平谷区2018—2019学年度第一学期期末质量监控试卷初 一 数 学 2019年1月一、选择题(本题共16分,每小题2分)1.2018年我国在人工智能领域取得显著成就,自主研发的人工智能“绝艺”获得全球最前沿的人工智能赛事冠军,这得益于所建立的大数据中心的规模和数据存储量,它们决定着人工智能深度学习的质量和速度,其中的一个大数据中心能存储58 000 000 000本书籍.将58 000 000 000用科学记数法表示应为( ) A. 95810⨯ B.105.810⨯ C. 115.810⨯ D. 110.5810⨯2.实数a ,b 在数轴上的对应点的位置如图所示,则错误的结论是( )A .0<aB .b a >C .0>+b aD .0<ab 3. 下列运算结果为负数的是( )A .π-3B .3-C .2)3(-D .)3(--4.如果x=-5是关于x 的方程135x m +=-的解,那么m 的值是( ) A .-40 B .-2 C .-4 D .45.若右图是某几何体的三视图,则这个几何体是( ) A 圆锥 B 圆柱 C 球D 三棱柱6.如果23x y -=,那么代数式y x +-24的值为( ) A .-1 B .4 C .-4 D .17.已知点A ,B ,C 是一条直线上的三点,若AB =5,BC =3则AC 长为( )A .8B .2C .8或2D .无法确定8.如图,用小石子按一定规律摆出以下图形:依照此规律,第n 个图形中小石子的个数是(n 为正整数) ( )A .nB. 13+nC. 3+nD. 23-n二、填空题(本题共16分,每小题2分)9.“a 的3倍与b 的一半的和”用代数式表示为 10. 单项式234x y -的系数是 ,次数是 .11.若94m -与m 互为相反数,则m = .12.如图, BD 平分∠ABC ,过点B 作BE 垂直BD ,若∠ABC =40°,则∠ABE= °13. 如果0)2019(12=-++n m ,那么n m 的值为 . 14.如图,直线AB 表示某天然气的主管道,现在要从主管道引一条分管道到某村庄P ,则沿图中线段 修建可使用料最省.理由是 15.我国元代数学家朱世杰所撰写的《算学启蒙》中有这样一道题:“良马日行二百四十里,驽马日行一百五十里,驽马先行一十二日,问良马几何追及之.” 译文:良马平均每天能跑240里,驽马平均每天能跑150里.现驽马出发12天后,良马从同一地点出发沿同一路线追它,问良马多少天能够追上驽马?设良马x 天能够追上驽马,根据题意可列一元一次方程 16. 将正方体骰子(相对面上的点数分别为1和6、2和5、3和4)放置于水平桌面上,如图1.将骰子向右翻滚90°,然后在桌面上按逆时针方向旋转90°,则完成一次变换.如图2.若骰子的初始位置为图1所示的状态,那么按上述规则连续完成2次变换后,骰子朝上一面的点数是________;连续完成2019次变换后,骰子朝上一面的点数是________.三、解答题(本题共50分,共10个小题,每小题5分) 17.计算:)20(23)6(17--+-+-18.计算:3)23(1655.2-⨯÷-19.计算:24)433281(⨯-+.20.解方程: x x x -=-+7)52(34.21.解方程:12271243xx -=-+ .22.化简()()2327322+---a a a a23.先化简,再求值: )45()2(32222ab b a ab b a ---,其中2=a ,1-=b .24. 列方程解应用题:甲班有45人,乙班有39人. 现在需要从甲、乙两班各抽调一些同学去参加歌咏比赛. 如果从乙班抽调的人数比甲班抽调的人数多4人,那么甲班剩余人数恰好是乙班剩余人数的1.5倍. 请问从甲、乙两班各抽调了多少参加歌咏比赛.25.阅读材料:对于任意有理数a ,b ,规定一种新的运算:a ⊙b =()1a a b +-,例如,2⊙5=2×(2+5)-1=13; (1)计算)2(⊙3-;(2)若5⊙)2(=-x ,求x 的值.26.金秋十月小鹏家的苹果园喜获丰收,共采摘苹果20筐,经过称重这20筐苹果的质量如下:(单位:千克)48,46,53,50,60,49,51,36,45,47,56,50,57,48,44,52,49,53,49,54在没带计算器的情况下,小鹏想帮父亲快速算出苹果的总质量.(1)小鹏通过观察发现,如果以千克为标准,把超出的质量记为正,可以得到上表中各数之和为;(2)因此,这20筐苹果的总质量为 .四、解答题(本题共18分,共3小题,其中第27题6分,28题6分,29题6分)27.已知直线AB上一点O,以O为端点画射线OC,作∠AOC的角平分线OD,作∠BOC的角平分线OE;(1)按要求完成画图;(2)通过观察、测量你发现∠DOE= °;(3)补全以下证明过程:证明:∵OD平分∠AOC(已知)∴∠DOC= ∠AOC()∵OE平分∠BOC(已知)∴∠EOC= ∠BOC()∵∠AOC+∠BOC= °∴∠DOE=∠DOC+∠EOC= (∠AOC+∠BOC)= °.28.暑假里某班同学相约一起去某公园划船,在售票处了解到该公园划船项目收费标准如下:(1)其中,两人船项目和八人船项目单价模糊不清,通过询问,了解到以下信息:①一只八人船每小时的租金比一只两人船每小时的租金的2倍少30元;②租2只两人船,3只八人船,游玩一个小时,共需花费630元.请根据以上信息,求出两人船项目和八人船项目每小时的租金;(2)若该班本次共有18名同学一起来游玩,每人乘船的时间均为 1小时,且每只船均坐满,试列举出可行的方案(至少四种),通过观察和比较,找到所有方案中最省钱的方案.29.阅读完成问题:数轴上,已知点A、B、C.其中,C为线段AB的中点:(1)如图,点A表示的数为-1,点B表示的数为3,则线段AB的长为,C点表示的数为 ;(2)若点A表示的数为-1,C点表示的数为2,则点B表示的数为 ;(3)若点A表示的数为t,点B表示的为t+2,则线段AB的长为 ,若C点表示的数为2,则t= ;(4)点A表示的数为x,点B表示的为2x,C点位置在-2至3之间(包括边界点),1若C点表示的数为x,则1x+2x+3x的最小值为 ,1x+2x+3x的最大值3为 .平谷区2018—2019学年度第一学期期末质量监控初一数学参考答案及评分标准 2019.1一、选择题(本题共16分,每小题2分)三、解答题(本题共50分,每小题5分) 17.解:)20(23)6(17--+-+-=2023617++-- ····································································································· 2分=202323++- ·········································································································· 4分 =20 ·························································································································· 5分18.解:)827(1655.2原式-⨯÷-= =82751625⨯⨯ ····························································································· 3分 =27 ···················································································································· 5分19.解:24)433281(⨯-+=244324322481⨯-⨯+⨯ ······················································································· 3分 =18163-+ ··············································································································· 4分 =1 ····························································································································· 5分20.解:去括号,得 x x x -=-+71564. ······································································ 2分移项,得 15764+=++x x x . ········································································ 3分 合并同类项,得 2211=x . ··············································································· 4分 系数化为1,得 2=x . ····················································································· 5分21.解:去分母(方程两边同乘以12),得)27(12)43(6x x -=-+. ····················································································· 1分去括号,得x x 27122418-=-+. ··································································· 2分 移项,得 12247218+-=+x x . ··································································· 3分 合并同类项,得 520-=x . ············································································· 4分 系数化为1,得 41-=x . ·················································································· 5分 22.原式:446273222--=-+--=a a a a a a ················································································ 3分······························································································································· 5分23.解:)45()2(32222ab b a ab b a ---=b a 2623ab -b a 25-24ab + ············································································ 2分 =22ab b a +. ········································································································ 3分当2=a ,1-=b 时,原式=22)1(2)1(2-⨯+-⨯=-2. ·········································································· 5分24.解:解:设从甲班抽调了x 人参加歌咏比赛…………… 1分根据题意列方程,得 [])4(392345+-=-x x . ……….……… 3分解得:x=15.…………………………….……… 4分∴x+4=19答:从甲班抽调了15人参加歌咏比赛,从乙班抽调了19人参加歌咏比赛………5分25.解:(1)21)23(3)2(⊙3=--⨯=- ……………………….……… 2分12251-2451-)2(25⊙)2(由题意,)2(-==-=-=+--=-x x x x x26.(1)50 (不唯一) …………………1分................................3分可以得到上表中各数之和为-3 ; .........................................4分(2)因此,这20筐苹果的总质量为 997 .分四、27.(1)……………………………………………2分(2) 90° …………………………………………………3分(3)∵OD 平分∠AOC(已知)∴∠DOC=21∠AOC ( 角平分线定义 )∵OE 平分∠BOC (已知)11 / 11∴∠EOC=21∠BOC (角平分线定义 )……………………………………………4分 ∵∠AOC+∠BOC= 180 °; …………………………………………5分∴∠DOE=∠DOC+∠EOC=21(∠AOC+∠BOC )= 90 °………………………………6分 28.(1)设:两人船每艘x 元,则八人船每艘(2x-30)元 ……………………1 由题意,可列方程630)302(3x 2=-+x (2)解得:x=90∴2x-30=150答:两人船每艘90元,则八人船每艘150元 (3)29.(1)4;1 (对一个即给1分,两个都对也给1分) (1)(2)5 (2)(3)2;1 (4)(4)-6;9 (6)。

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