2020-2021南京市联合体七年级数学上册期末试卷及答案

0a（第5题）－11一、选择题（本大题共8小题，每小题2分，共16分．在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡．．．相应位置．．．．上）1．－13的倒数是A．13B．－3C．－13D．32．马拉松(Marathon)是国际上非常普及的长跑比赛项目，全程距离26英里385码，折合约为42000米，用科学记数法表示42000为A．42×103B．4.2×104C．4.2×105D．42000×1053．对于代数式－1＋m的值，下列说法正确的是A．比－1大B．比－1小C．比m大D．比m小4．如图，点P是直线a外的一点，点A、B、C在直线a上，且PB⊥a，垂足为点B，PA⊥PC，则下列说法不正确的是A．线段PB的长是点P到直线a的距离B．PA、PB、PC三条线段中，PB最短C．线段AC的长是点A到直线PC的距离D．线段PC的长是点C到直线PA的距离5.有理数a在数轴上的位置如图所示，下列各数中，在0到1之间的是①－a－1，②│a＋1│，③2－│a│，④12│a│.A．②③④B．①③④C．①②③D．①②③④6．整式mx＋n的值随x的取值不同而不同，下表是当x取不同值时对应的整式的值：则关于x的方程－mx＋n＝8的解为A．x＝－3B．x＝0C．x＝1D．x＝2x－2－1012mx＋n－12－8－404PCAB（第4题）a【联合体数学】2020七上期末考试（试卷& 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 ）EBFCAOD（第15题）二、填空题（本大题共10小题，每小题2分，共20分．不需写出解答过程，请把答案直接填写在答题卡相应位置．．．．．．．上）7．黄山主峰一天早晨气温为－1℃，中午上升了8℃，夜间又下降了10℃，那么这天夜间黄山主峰的气温是▲℃．8．－2的绝对值是▲；12的相反数是▲．9.写出一个“系数是－2，次数是3”的单项式▲．10．若3a－2b＝4，则7＋9a－6b＝▲．11．已知x＝2是关于x的方程a(x＋1)＝2a＋x的解，则a的值是▲．12．一个角的余角比这个角小40°，则这个角的度数为▲．13．如图，数轴上有A、B、C三点，C为AB的中点，点A表示的数为－3.2，点B表示的数为2，则点C表示的数为▲．14．如图，要使图中平面展开图按虚线折叠成正方体后，相对面上两个数之积为12，则x－y＝▲．15．如图，直线AB与直线CD相交于点O，OE平分∠BOD，OF平分∠COE,若∠FOB的度数为30°，则∠AOC的度数为▲°．16.直线AB⊥CD，垂足为点O，直线EF经过点O，若锐角∠COE＝m°，则∠AOF＝▲°（用含m的代数式表示）．三、解答题（本大题共10小题，共68分．请在答题卡指定区域．．．．．．．内作答，解答时应写出文字说明、 证明 住所证明下载场所使用证明下载诊断证明下载住所证明下载爱问住所证明下载爱问 过程或演算步骤）17．（8分）计算（1）－32－9×－13（2）[（－2）3＋43]÷4＋－23ABC0（第13题）（第14题）18．（6分）先化简，再求值：(3a2b－ab2)－2(ab2－3a2b)，其中a＝13，b＝－3．19．（6分）解方程：（1）5x－2(3x－1)＝4，（2）x2－x－13＝1.20．（6分）如图，正方形网格线的交点叫格点，格点P是∠AOB的边OB上的一点，用网格画图，保留作图痕迹．（1）过点P画OB的垂线，交OA于点C；（2）线段▲的长度是点O到PC的距离；（3）PO＜OC的理由是▲.21（6分）如图是由一些棱长都为1cm的小正方体组合成的简单几何体．（1）画出该几何体的主视图、左视图和俯视图；（2）如果在这个几何体上再添加一些小正方体，并保持主视图和左视图不变，最多可以再添加▲块小正方体．POAB主视图俯视图左视图22．（7分）如图，直线AB、CD相交于点O，OE⊥AB，OE平分∠COF．（1）若∠AOF＝140°，求∠EOF的度数；（2）OB是∠DOF的角平分线吗？为什么？．23．（6分）某制造工厂计划若干天完成一批玩具的订货任务，如果每天生产玩具20个，那么就比订货任务少生成100个；如果每天生产玩具23个，那么就可超过订货任务20个，求原计划几天完成任务？24.（6分）已知点C在直线AB上，AC＝10cm，CB＝6cm，点M、N分别是AC、BC的中点．求线段AB、MN的长；25．（8分）如图，直线AB、CD相交于点O，∠AOD为锐角，OE⊥CD，OF平分∠BOD.（1）图中与∠AOE互余的角为▲；（2）若∠EOB＝∠DOB，求∠AOE的度数；（3）图中与锐角∠AOE互补角的个数随∠AOE的度数变化而变化，直接写出与∠AOE互补的角的个数及对应的∠AOE的度数.CFADBEOABCDEOF26．（9分）几何知识可以解决生活中许多距离最短的问题.让我们从书本一道习题入手进行探索.【回顾】（1）如图①，A、B是公路l两侧的两个村庄.现要在公路l上修建一个垃圾站C，使它到A、B两村庄的路程之和最小，请在图中画出点C的位置，并说明理由.【探索】（2）如图②，在B村庄附件有一个生态保护区，现要在公路l上修建一个垃圾站C，使它到A、B两村庄的路程之和最小，从B村庄到公路不能穿过生态保护区，请在图中画出点C的位置.（3）如图③，A、B是河两侧的两个村庄．现要在河上修建一座桥，使得桥与河岸垂直，且A村到B村的总路程最短，请在图中画出桥的位置.（保留画图痕迹）lB①AlB②生态保护区AB一条河A③说明：本评分标准每题给出了一种或几种解法供参考．如果考生的解法与本解答不同，参照本评分标准的精神给分．一、选择题（本大题共6小题，每小题2分，共12分）二、填空题（本大题共10小题，每小题2分，共20分）7．－3；8．2；－12．9．答案不唯一．10．19；11．2．12．65°；13．－0.6；14．3；15．80．16．(90±m)三、解答题（本大题共11小题，共88分）17．（本题8分）(1)解：－32－9×－13＝－9＋3······················································································································2分＝－6···························································································································4分(2)[（－2）3＋43]÷4＋－23＝[－8＋43]×14＋－23································································································2分＝－53＋－23···············································································································3分＝－73·····························································································································4分18．（本题6分）解：原式＝3a2b－ab2－2ab2＋6a2b·····················································································2分＝9a2b－3ab2························································································································4分当a＝13，b＝－3时，原式＝－12„„„„„„„„„„„„„„„„„„„„6分19．（6分）（1）5x－2(3x－1)＝4解：5x－6x＋2＝4„„„„„„„„„„„„„„„„„„„„„„„„1分－x＝2„„„„„„„„„„„„„„„„„„„„„„„„„2分x＝－2„„„„„„„„„„„„„„„„„„„„„„„„3分题号123456答案BBDCDA【联合体数学】2020七上期末考试答案（2）x2－x－13＝1.解：3x－2(x－1)＝6„„„„„„„„„„„„„„„„„„„„„„„„1分3x－2x＋2＝6„„„„„„„„„„„„„„„„„„„„„„„„„2分x＝4„„„„„„„„„„„„„„„„„„„„„„„„3分20．（6分）(1)如图所示，直线PC即为所求„„„„„„„„„2分(2)OP„„„„„„„„„„„„„4分(3)直线外一点与直线上各点连成的所有线段中，垂线段最短„„„„„„„„„„„„„6分21（6分）解：（1）„„„„„„„„„„„„„„„4分（2）6.„„„„„„„„„„„„„„„„„„„„6分22．（7分）(1)∵OE⊥AB，∴∠AOE＝90°．„„„„„„„„„„„„„„„1分又∵∠AOF＝140°，∴∠EOF＝∠AOF－∠AOE＝140°－90°＝50°．„„„„„„„„„„„„„„„3分(2)∵OE⊥AB，∴∠AOE＝∠BOE＝90°．∵OE平分∠COF，∴∠COE＝∠FOE．∴∠AOE－∠COE＝∠BOE－∠FOE，即∠AOC＝∠BOF．„„„„„„„„„„„„5分∵∠AOC＝∠DOB，∴∠DOB＝∠BOF，即OB平分∠DOF．„„„„„„„„„„„„„„7分POABC主视图俯视图左视图23．（6分）解：设原计划x天完成任务，由题意得20x＋100＝23x－20„„„„„„„„„„„„„„„„3分x＝40答：原计划40天完成任务．„„„„6分24.（6分）情况一、AB＝AC＋CB＝10＋6＝16„„„„„„„„„„„„„„„1分∵AC＝10cm，BC＝6cm，点M、N分别是AC、BC的中点．∴MC＝12AC＝5NC＝12CB＝3∴MN＝MC＋NC＝5＋3＝8„„„„„„„„„„„„„„„3分情况二、AB＝AC－CB＝10－6＝4„„„„„„„„„„„„„„„4分∵AC＝10cm，BC＝6cm，点M、N分别是AC、BC的中点．∴MC＝12AC＝5NC＝12CB＝3∴MN＝MC－NC＝5－3＝2„„„„„„„„„„„„„„„6分25．（8分）（1）∠AOD、∠BOC„„„„„„„„„„„„„„„„2分（2）设∠AOD＝x°．∵∠AOD＝x°，∴∠BOD＝180°－∠AOD＝180°－x°∠BOC＝∠AOD＝x°．∵OE⊥CD，∴∠EOC＝∠EOD＝90°．又∵∠EOB＝∠DOB，∴90°＋x°＝180°－x°，即x＝45．∴∠AOE＝∠EOD－∠AOD＝90°－45°＝45°．„„„„„„„„„„„„„5分（3）当锐角∠AOE＝30°时，互补角有2个，为∠EOB、∠EOF．当锐角∠AOE＝45°时，互补角有3个，为∠EOB、∠AOC、∠DOB．当锐角∠AOE不等于45°或30°时，互补角有1个，为∠EOB．„„„„„„„8分26．（9分）(1)画图正确„„„„„„„„„„„„„„„„1分理由：两点之间，线段最短．„„„„„„„„„„„„„„„„3分如图所示，点C即为所求。(2)如图所示，点C即为所求。画图正确„„„„„„„„„„„„„„„„6分(3)如图所示，线段CD即为所求。画图正确„„„„„„„„„„„„„„„„9分lB①AClB②生态保护区ACB一条河A③A′CD