北师大版2019-2020学年八年级上学期英语期末考试试卷B卷

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2019-2020学年北师大版八年级数学上册期末检测卷

2019-2020学年北师大版八年级数学上册期末检测卷

期末检测卷时间:120分钟一、选择题(每小题3分,共45分)满分:150分1.以下列哪组数为边长,可以得到直角三角形的是(A.9,16,25B.8,15,17)C.6,8,14D.10,12,132.在下列各数中是无理数的有()-0.333…,4,5,-π,2.0101001…(相邻两个1之间增加1个0) A.3个B.4个C.5个D.2个3.如图,直线AB∥C D,∠B=60°,∠C=40°,则∠E等于()A.70°B.80°C.90°D.100°4.一次函数y=x+2的图象与x轴交点的坐标是(A.(0,2)B.(0,-2))C.(2,0)D.(-2,0)5.下列结果错误的是()A.(-2)=2B.16的算术平方根是421 47 2C.12的算术平方根是D.(-π) 的算术平方根是 π26.下列不属于二元一次方程组的是()+y=3,-y=1x=3,xA. B.x-y=1 x+y=3,=1xy=3,x-y=1xC. D.y7.点P(3,-5)关于y轴对称的点的坐标为(A.(-3,-5)B.(5,3))C.(-3,5)D.(3,5)8.设a=19,a在两个相邻整数之间,则这两个整数是(A.1和2B.2和3) C.3和4D.4和59.已知k>0,则函数y=-kx+k的图象经过第________象限( A.一、二、三B.二、三、四)C .一、二、四D .一、三、四10.红领巾公园健走步道环湖而建,以红军长征路为主题,如图是利用平面直角坐标系 画出的健走步道路线上主要地点的大致分布图,这个坐标系分别以正东、正北方向为x 轴、 y 轴的正方向,如果表示遵义的点的坐标为(-5,7),表示腊子口的点的坐标为(4,-1),那 么这个平面直角坐标系原点所在位置是( A .泸定桥 B .瑞金 )C .包座D .湘江11.如图,直线 a ∥b ,∠1=85°,∠2=35°,则∠3 的度数为( A .85° B .60° C .50° D .35°)12.一个直角三角形的三边长分别是 6cm 、8cm 、xcm ,则 x 的值为( A .100 B .10 C .10 或 2 7 D .100 或 28)13.若 2a b 与 5a b 是同类项,则( ) 3x y 5 + 2 4y 2x - =1, =2 x =2,x A. B.y =-1 y =0, =2 x =3, y =1 xC. D. y14.如图,在边长为2 的正方形 AB C D 中剪去一个边长为 1 的小正方形 CEF G ,动点P 从点 A 出发,沿 A →D →E →F →G →B 的路线绕多边形的边匀速运动到点 B 时停止(不含点 A 和点 B),则△ABP 的面积 S 随着时间 t 变化的函数图象大致是( )15.如图,在长方形纸片AB C D 中,已知 A D =8,折叠纸片,使AB 边与对角线 AC 重 合,点 B 落在点 F 处,折痕为 AE ,且 EF =3,则 AB 的长为( A .3 B .4 C .5 D .6)二、填空题(每小题 5 分,共 25 分)16.直线 y =x +2 与 y 轴的交点坐标为________. 17 . 命 题 “ 同 位 角 相等 , 两 直 线 平 行 ” 中, 条 件 是 ______________, 结 论 是 ______________.18.已知点 P(5,-2),点 Q(3,a +1),且直线 P Q 平行于 x 轴,则 a =________.19.如图,已知 y =ax +b 和 y =kx 的图象交于点 P ,根据图象可得关于 x ,y 的二元一 -y +b =0,-y =0a x 次方程组 的解是________.k x20.已知点(-6,y )、(8,y )都在直线 y =-2x -6 上,则 y ,y 的大小关系是____________. 1 2 1 2三、解答题(共 80 分) 21.(8 分)计算: 1(1) 12+ -2313; (2)(1- 5)(1+ 5)+(1+ 5) .2 22.(8 分)解方程组:+y =1, 3x +2y =4, 2x -y =5.x (1) (2)4x +y =-8;23.(10 分)如图,在△AB C 中,C D 平分∠ACB ,D E ∥A C ,∠B =50°,∠ED C =30°. 求∠A D C 的度数.24.(12分)解古算题:“今有甲、乙二人持钱不知其数.甲得乙半而钱四十八,乙得甲太半而亦钱四十八.甲、乙持钱各几何?”题目大意是:甲、乙两人各带了若干钱.如果甲得到乙所有钱的一半,那么甲共有钱48文,如果乙得到甲所有钱的错误!,那么乙也共有钱48文.问甲、乙两人各带了多少钱?25.(12分)如图,直线PA经过点A(-1,0),P(1,2),直线PB是一次函数y=-x+3的图象.(1)求直线PA的表达式及Q点的坐标;(2)求四边形P Q O B的面积.26.(14分)我市某中学举行“中国梦·校园好声音”歌手大赛,初、高中部根据初赛成绩,各选出5名选手组成初中代表队和高中代表队参加学校决赛.两个队各选出的5名选手的决赛成绩如图所示.(1)根据图示填写下表;,平均数(分),中位数(分),众数(分)初中部,,85,高中部,85,,100(2)结合两队成绩的平均数和中位数,分析哪个队的决赛成绩较好;(3)计算两队决赛成绩的方差,并判断哪一个代表队选手成绩较为稳定.27.(16分)小明和爸爸从家步行去公园,爸爸先出发,一直匀速前行,小明后出发.家到公园的距离为2500m,如图是小明和爸爸所走的路程s(m)与步行时间t(min)的函数图象.(1)直接写出小明所走路程s与时间t的函数解析式;(2)小明出发多少时间与爸爸第三次相遇?(3)在速度都不变的情况下,小明希望比爸爸早20min到达公园,则小明在步行过程中停留的时间需作怎样的调整?参考答案与解析1.D 2.C 3.C 4.D 5.A 6.C 7.D 8.D 9.C 10.B 11.C 12.C 13.B14.B 解析:当点 P 在 A D 上时,△ABP 的底 AB 不变,高增大,∴△ABP 的面积 S 随着时间 t 的增大而增大;当点 P 在 DE 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的 面积 S 不变;当点 P 在 EF 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时 间 t 的增大而减小;当点 P 在 F G 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的面积 S 不变;当点 P 在 G B 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时间 t 的 增大而减小.故选 B.15.D 解析:∵四边形AB C D 是长方形,A D =8,∴BC =8.∵△AEF 是△AEB 翻折而 成,∴BE =EF =3,AB =AF ,∠AFE =∠B =90°,∴CE =BC -BE =8-3=5,∠CFE =90°. 在 Rt △CEF 中,C F = CE 2-EF 2= 52-32=4.设 AB =AF =x ,则 A C =AF +CF =x +4.在Rt △ABC 中,AC =AB +BC ,即(x +4) =x +8 ,解得 x =6,即 AB =6.故选 D.2 2 2 2 2 2 16.(0,2) 17.同位角相等 两直线平行=-4, =-2 x18.-3 19. 20.y >y 1 2y3 33 5 321.解:(1)原式=2 3+ -2× = .(4 分)3 3 (2)原式=1-( 5) +1+2 5+( 5) =1-5+1+2 5+5=2+2 5.(8 分)2 2 =-3, =4. x =2, y =-1. x22.解:(1) (4 分) (2) (8 分)y23.解:∵DE ∥A C ,∠E D C =30°,∴∠AC D =∠ED C =30°.(3 分)∵C D 平分∠ACB , ∴∠BC D =∠AC D =30°.(6 分)∵∠B =50°,∴∠A D C =∠B +∠BC D =50°+30°=80°.(10 分)1 x + y =48,2 =36, x24.解:设甲带了 x 文钱,乙带了 y 文钱,根据题意得 (6 分)解得 2=24.y3x +y =48, (11 分)答:甲、乙两人分别带了 36 文钱和 24 文钱.(12 分)25.解:(1)设直线 PA 的表达式为 y =kx +b .由直线 PA 经过点 A(-1,0),点 P(1,2),0=-k+b,2=k+b,k=1,b=1.可得解得∴直线PA的表达式为y=x+1.(4分)当x=0时,y=1,∴点Q的坐标为(0,1).(6分)(2)在y=-x+3中,令y=0,则x=3,∴点B的坐标为(3,0).∵点A的坐标为(-1,1 212720),∴OA=1,AB=4,(8分)∴S26.解:(1)858580(6分)=S-S=×4×2-×1×1=.(12分)四边形P Q O B△PA B△QA O(2)因为两个队的平均数都相同,初中部的中位数高,所以在平均数相同的情况下中位数高的初中部代表队的决赛成绩较好.(10分)1 51 5(3)s2=[(75-85)+(80-85)+(85-85)+(85-85)+(100-85)]=70,s=[(70222222高中部初中部-85)2+(100-85)2+(100-85)2+(75-85)2+(80-85)2]=160.因为s2<s2,所以初中代初中部高中部表队选手成绩较为稳定.(14分)50t(0≤t≤20),27.解:(1)s=1000(20<≤30),(6分)t50t-500(30<t≤60).(2)设小明的爸爸所走的路程s与步行时间t的函数解析式为s=k t+b,则25k+b=1000,k=30,解得则小明的爸爸所走的路程与步行时间的解析式为s=30t+=250,b=250,b250.(9分)当小明与爸爸第三次相遇时,50t-500=30t+250,解得t=37.5.(11分)答:小明出发37.5min后与爸爸第三次相遇.(12分)(3)令30t+250=2500,解得t=75,则小明的爸爸到达公园需要75min.(14分)∵小明到达公园需要的时间是60min,∴小明希望比爸爸早20min到达公园,则小明在步行过程中停留的时间需减少5min.(16分)参考答案与解析1.D 2.C 3.C 4.D 5.A 6.C 7.D 8.D 9.C 10.B 11.C 12.C 13.B14.B 解析:当点 P 在 A D 上时,△ABP 的底 AB 不变,高增大,∴△ABP 的面积 S 随着时间 t 的增大而增大;当点 P 在 DE 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的 面积 S 不变;当点 P 在 EF 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时 间 t 的增大而减小;当点 P 在 F G 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的面积 S 不变;当点 P 在 G B 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时间 t 的 增大而减小.故选 B.15.D 解析:∵四边形AB C D 是长方形,A D =8,∴BC =8.∵△AEF 是△AEB 翻折而 成,∴BE =EF =3,AB =AF ,∠AFE =∠B =90°,∴CE =BC -BE =8-3=5,∠CFE =90°. 在 Rt △CEF 中,C F = CE 2-EF 2= 52-32=4.设 AB =AF =x ,则 A C =AF +CF =x +4.在Rt △ABC 中,AC =AB +BC ,即(x +4) =x +8 ,解得 x =6,即 AB =6.故选 D.2 2 2 2 2 2 16.(0,2) 17.同位角相等 两直线平行=-4, =-2 x18.-3 19. 20.y >y 1 2y3 33 5 321.解:(1)原式=2 3+ -2× = .(4 分)3 3 (2)原式=1-( 5) +1+2 5+( 5) =1-5+1+2 5+5=2+2 5.(8 分)2 2 =-3, =4. x =2, y =-1. x22.解:(1) (4 分) (2) (8 分)y23.解:∵DE ∥A C ,∠E D C =30°,∴∠AC D =∠ED C =30°.(3 分)∵C D 平分∠ACB , ∴∠BC D =∠AC D =30°.(6 分)∵∠B =50°,∴∠A D C =∠B +∠BC D =50°+30°=80°.(10 分)1 x + y =48,2 =36, x24.解:设甲带了 x 文钱,乙带了 y 文钱,根据题意得 (6 分)解得 2=24.y3x +y =48, (11 分)答:甲、乙两人分别带了 36 文钱和 24 文钱.(12 分)25.解:(1)设直线 PA 的表达式为 y =kx +b .由直线 PA 经过点 A(-1,0),点 P(1,2),0=-k+b,2=k+b,k=1,b=1.可得解得∴直线PA的表达式为y=x+1.(4分)当x=0时,y=1,∴点Q的坐标为(0,1).(6分)(2)在y=-x+3中,令y=0,则x=3,∴点B的坐标为(3,0).∵点A的坐标为(-1,1 212720),∴OA=1,AB=4,(8分)∴S26.解:(1)858580(6分)=S-S=×4×2-×1×1=.(12分)四边形P Q O B△PA B△QA O(2)因为两个队的平均数都相同,初中部的中位数高,所以在平均数相同的情况下中位数高的初中部代表队的决赛成绩较好.(10分)1 51 5(3)s2=[(75-85)+(80-85)+(85-85)+(85-85)+(100-85)]=70,s=[(70222222高中部初中部-85)2+(100-85)2+(100-85)2+(75-85)2+(80-85)2]=160.因为s2<s2,所以初中代初中部高中部表队选手成绩较为稳定.(14分)50t(0≤t≤20),27.解:(1)s=1000(20<≤30),(6分)t50t-500(30<t≤60).(2)设小明的爸爸所走的路程s与步行时间t的函数解析式为s=k t+b,则25k+b=1000,k=30,解得则小明的爸爸所走的路程与步行时间的解析式为s=30t+=250,b=250,b250.(9分)当小明与爸爸第三次相遇时,50t-500=30t+250,解得t=37.5.(11分)答:小明出发37.5min后与爸爸第三次相遇.(12分)(3)令30t+250=2500,解得t=75,则小明的爸爸到达公园需要75min.(14分)∵小明到达公园需要的时间是60min,∴小明希望比爸爸早20min到达公园,则小明在步行过程中停留的时间需减少5min.(16分)参考答案与解析1.D 2.C 3.C 4.D 5.A 6.C 7.D 8.D 9.C 10.B 11.C 12.C 13.B14.B 解析:当点 P 在 A D 上时,△ABP 的底 AB 不变,高增大,∴△ABP 的面积 S 随着时间 t 的增大而增大;当点 P 在 DE 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的 面积 S 不变;当点 P 在 EF 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时 间 t 的增大而减小;当点 P 在 F G 上时,△ABP 的底 AB 不变,高不变,∴△ABP 的面积 S 不变;当点 P 在 G B 上时,△ABP 的底 AB 不变,高减小,∴△ABP 的面积 S 随着时间 t 的 增大而减小.故选 B.15.D 解析:∵四边形AB C D 是长方形,A D =8,∴BC =8.∵△AEF 是△AEB 翻折而 成,∴BE =EF =3,AB =AF ,∠AFE =∠B =90°,∴CE =BC -BE =8-3=5,∠CFE =90°. 在 Rt △CEF 中,C F = CE 2-EF 2= 52-32=4.设 AB =AF =x ,则 A C =AF +CF =x +4.在Rt △ABC 中,AC =AB +BC ,即(x +4) =x +8 ,解得 x =6,即 AB =6.故选 D.2 2 2 2 2 2 16.(0,2) 17.同位角相等 两直线平行=-4, =-2 x18.-3 19. 20.y >y 1 2y3 33 5 321.解:(1)原式=2 3+ -2× = .(4 分)3 3 (2)原式=1-( 5) +1+2 5+( 5) =1-5+1+2 5+5=2+2 5.(8 分)2 2 =-3, =4. x =2, y =-1. x22.解:(1) (4 分) (2) (8 分)y23.解:∵DE ∥A C ,∠E D C =30°,∴∠AC D =∠ED C =30°.(3 分)∵C D 平分∠ACB , ∴∠BC D =∠AC D =30°.(6 分)∵∠B =50°,∴∠A D C =∠B +∠BC D =50°+30°=80°.(10 分)1 x + y =48,2 =36, x24.解:设甲带了 x 文钱,乙带了 y 文钱,根据题意得 (6 分)解得 2=24.y3x +y =48, (11 分)答:甲、乙两人分别带了 36 文钱和 24 文钱.(12 分)25.解:(1)设直线 PA 的表达式为 y =kx +b .由直线 PA 经过点 A(-1,0),点 P(1,2),0=-k+b,2=k+b,k=1,b=1.可得解得∴直线PA的表达式为y=x+1.(4分)当x=0时,y=1,∴点Q的坐标为(0,1).(6分)(2)在y=-x+3中,令y=0,则x=3,∴点B的坐标为(3,0).∵点A的坐标为(-1,1 212720),∴OA=1,AB=4,(8分)∴S26.解:(1)858580(6分)=S-S=×4×2-×1×1=.(12分)四边形P Q O B△PA B△QA O(2)因为两个队的平均数都相同,初中部的中位数高,所以在平均数相同的情况下中位数高的初中部代表队的决赛成绩较好.(10分)1 51 5(3)s2=[(75-85)+(80-85)+(85-85)+(85-85)+(100-85)]=70,s=[(70222222高中部初中部-85)2+(100-85)2+(100-85)2+(75-85)2+(80-85)2]=160.因为s2<s2,所以初中代初中部高中部表队选手成绩较为稳定.(14分)50t(0≤t≤20),27.解:(1)s=1000(20<≤30),(6分)t50t-500(30<t≤60).(2)设小明的爸爸所走的路程s与步行时间t的函数解析式为s=k t+b,则25k+b=1000,k=30,解得则小明的爸爸所走的路程与步行时间的解析式为s=30t+=250,b=250,b250.(9分)当小明与爸爸第三次相遇时,50t-500=30t+250,解得t=37.5.(11分)答:小明出发37.5min后与爸爸第三次相遇.(12分)(3)令30t+250=2500,解得t=75,则小明的爸爸到达公园需要75min.(14分)∵小明到达公园需要的时间是60min,∴小明希望比爸爸早20min到达公园,则小明在步行过程中停留的时间需减少5min.(16分)。

2019-2020学年浙江省嘉兴市北京师范大学南湖附属学校八年级第一学期期中英语试卷.(含解析)

2019-2020学年浙江省嘉兴市北京师范大学南湖附属学校八年级第一学期期中英语试卷.(含解析)

2019-2020学年浙江省嘉兴市南湖区北师大南湖附校八年级第一学期期中考试英语试卷二、完形填空。

阅读下面短文,掌握其大意,然后在各题所给的四个选项中选出一个最佳答案。

Alex was a giraffe. He lived in a zoo with his mum, dad and other giraffe friends. As Alex grew, he___16__his neck was much shorter than the others’. He began to get__17__. But his mother told him that he was __18__. She said it did not matter how short or long his __19__ was.Alex tried not to __20__ his short neck. But as his friends grew taller and taller, his neck _21__ short. Sometimes the other giraffes__22___ him, but his parents told him not to worry. They said that he should be happy with _23__ the way he was. He tried, but he could not even play in some giraffe sports. He was just too _24___.One day, one of the zoo workers came and took Alex away. Alex was afraid. Would he have to leave the zoo just __25__ he didn’t have a long neck? The zoo worker calmed (使镇静) Alex down and gave him some food. He felt__26__. Then he led Alex into another area of the zoo. There was a small __27___ between the two areas. Alex was the only giraffe that could go through it.Soon Alex’s __28__ came to see him. “I told you that you were special.” his mother said. “The zoo has made a place just for __29__ because so many people want to see how special you are!”She was right. Tourists pointed to him and felt___30___. They looked happy to see something so special at the zoo.16 A.believed B.found C.hoped D.agreed17 A.careful B.great C.interested D.worried18 A.fat B.thin C.special mon19 A.leg B.neck C.foot D.hair20 A.go over B.find out C.think about D.look for21 A.stayed B.became C.turned D.felt22 A.looked after B.fell behind ughed at D.heard of23 A.herself B.himself C.themselves D.ourselves24 A.tall B.short C.fat D.thin25 A.after B.when C.because D.before26 A.better B.worse C.sadder D.angrier27 A.box B.hole C.window D.door28 A.grandparents B.parents C.teachers D.friends29 A.him B.you D.me30 A.excited B.angry C.sad D.heavy【答案】BDCBC ACBBC ADBBA【解答】16.考察动词。

北师大版八年级上册数学期中考试试题含答案

北师大版八年级上册数学期中考试试题含答案

北师大版八年级上册数学期中考试试卷一、选择题。

(每小题只有一个正确答案,每小题3分)1.下列哪个点在函数112y x =+的图象上()A .(2,1)B .(2,1)-C .(2,0)-D .(2,0)2.如图,两个较大正方形的面积分别为225、289,且中间夹的三角形是直角三角形,则字母A 所代表的正方形的面积为()A .4B .8C .16D .643.已知点P (m+3,2m+4)在x 轴上,那么点P 的坐标为()A .(﹣1,0)B .(1,0)C .(﹣2,0)D .(2,0)4.△ABC 的三条边分别为a ,b ,c ,下列条件不能判断△ABC 是直角三角形的是()A .a 2+b 2=c 2B .a=5,b=12,c=13C .∠A=∠B+∠CD .∠A :∠B :∠C=3:4:55.下列各式的计算中,正确的是()A =B =C =D=-6.在函数y =1x -中,自变量x 的取值范围是()A .x≥1B .x≤1且x≠0C .x≥0且x≠1D .x≠0且x≠17.已知直角三角形两边的长为3和4,则此三角形的周长为()A .12B .C .12或D .以上都不对8.如图,长为8cm 的橡皮筋放置在x 轴上,固定两端A 和B ,然后把中点C 向上拉升3cm 至D 点,则橡皮筋被拉长了()A .2cmB .3cmC .4cmD .5cm9.化简二次根式)AB C D10.如图,在正方形ABCD 纸片上有一点P ,PA =1,PD =2,PC =3,现将△PCD 剪下,并将它拼到如图所示位置(C 与A 重合,P 与G 重合,D 与D 重合),则∠APD 的度数为A .150°B .135°C .120°D .108°11|1|0-=b ,那么()2017a b +的值为()A .-1B .1C .20173D .20173-12.如图1,点G 为BC 边的中点,点H 在AF 上,动点P 以每秒2cm 的速度沿图1的边运动,运动路径为G→C→D→E→F→H ,相应的△ABP 的面积y (cm 2)关于运动时间t (s )的函数图象如图2,若AB =6cm ,则下列结论正确的个数有()①图1中BC 长4cm ;②图1中DE 的长是6cm ;③图2中点M 表示4秒时的y 值为24cm 2;④图2中的点N 表示12秒时y 值为15cm 2.A .4个B .3个C .2个D .1个二、填空题13.-27的立方根为________________,________.14.已知函数y =(a+1)x+a 2﹣1,当a_____时,它是一次函数;当a_____时,它是正比例函数.15.如图,△ABC 的边BC 在数轴上,AB ⊥BC ,且BC =3,AB =1,以C 为圆心,AC 长为半径画圆分别交数轴于点A′、点A″,那么数轴上点A′、点A″所表示的数分别是_____、_____.16.如图,在平面直角坐标系中,点A 1,A 2,A 3…都在x 轴上,点B 1,B 2,B 3…都在直线y =x 上,OA 1=1,且△B 1A 1A 2,△B 2A 2A 3,△B 3A 3A 4,…△B n A n A n +1…分别是以A 1,A 2,A 3,…A n …为直角顶点的等腰直角三角形,则△B 10A 10A 11的面积是________.三、解答题17.计算:|13|+(2019﹣20﹣(12)﹣2182818(263)(263)32)2--19.如图,在平面直角坐标系中,正方形ABCD 和正方形EFGC 面积分别为64和16.(1)请写出点A ,E ,F 的坐标;(2)求S △BDF .204792737272,请你观察上述式子规律后解决下面问题.(1)规定用符号[m]表示实数m 的整数部分,例如:[45]=0,[π]=3,填空:10+2]=;[5=.(2)如果a ,5b ,求a 2﹣b 2的值.21.如图,在长方形ABCD 中,AB =8,AD =10,点E 为BC 上一点,将△ABE 沿AE 折叠,使点B 落在长方形内点F 处,且DF =6.(1)试说明:△ADF 是直角三角形;(2)求BE 的长.22.先阅读下面的解题过程,然后再解答.我们只要找到两个数a ,b ,使a b m +=,ab n =,即22m +==0)b => .这里7m =,12n =,由于437+=,4312⨯=,所以227,+=,2+..23.(1)如图1,长方体的长为4cm,宽为3cm,高为12cm.求该长方体中能放入木棒的最大长度;(2)如图2,长方体的长为4cm,宽为3cm,高为12cm.现有一只蚂蚁从点A处沿长方体的表面爬到点G处,求它爬行的最短路程.(3)若将题中的长方体换成透明圆柱形容器(容器厚度忽略不计)的高为12cm,底面周长为10cm,在容器内壁离底部3cm的点B处有一饭粒,此时一只蚂蚁正好在容器外壁且离容器上沿3cm的点A处.求蚂蚁吃到饭粒需要爬行的最短路程是多少?24.在平面直角坐标系中,已知点A(-3,-1),B(-1,0),C(-2,3),请在图中画出△ABC,并画出与△ABC关于y轴对称的图形.25.如图(1),是两个全等的直角三角形(直角边分别为a,b,斜边为c)(1)用这样的两个三角形构造成如图(2)的图形,利用这个图形,证明:a2+b2=c2;(2)用这样的两个三角形构造图3的图形,你能利用这个图形证明出题(1)的结论吗?如果能,请写出证明过程;(3)当a=3,b=4时,将其中一个直角三角形放入平面直角坐标系中,使直角顶点与原点重合,两直角边a,b分别与x轴、y轴重合(如图4中Rt△AOB的位置).点C为线段OA 上一点,将△ABC沿着直线BC翻折,点A恰好落在x轴上的D处.①请写出C、D两点的坐标;②若△CMD为等腰三角形,点M在x轴上,请直接写出符合条件的所有点M的坐标.参考答案1.C【分析】分别把x=2和x=−2代入解析式求出对应的y值来判断点是否在函数图象上.【详解】解:(1)当x=2时,y=2,所以(2,1)不在函数112y x=+的图象上,(2,0)也不在函数112y x=+的图象上;(2)当x=−2时,y=0,所以(−2,1)不在函数112y x=+的图象上,(−2,0)在函数112y x=+的图象上.故选C.【点睛】本题考查的知识点是一次函数图象上点的坐标特征,即直线上的点的坐标一定适合这条直线的解析式.2.D【分析】根据正方形的面积等于边长的平方,由正方形PQED的面积和正方形PRQF的面积分别表示出PR2及PQ2,又三角形PQR为直角三角形,根据勾股定理求出QR2,即为所求正方形的面积.【详解】解:∵正方形PQED的面积等于225,∴即PQ2=225,∵正方形PRGF的面积为289,∴PR2=289,又∵△PQR为直角三角形,根据勾股定理得:PR2=PQ2+QR2,∴QR2=PR2﹣PQ2=289﹣225=64,则正方形QMNR的面积为64.故选:D.【点睛】此题考查了勾股定理,以及正方形的面积公式.勾股定理最大的贡献就是沟通“数”与“形”的关系,它的验证和利用都体现了数形结合的思想,即把图形的性质问题转化为数量关系的问题来解决.能否由实际的问题,联想到用勾股定理的知识来求解是本题的关键.3.B【分析】根据x轴上点的纵坐标为0列方程求出m的值,再求解即可.【详解】∵点P(m+3,2m+4)在x轴上,∴2m+4=0,解得m=−2,∴m+3=−2+3=1,∴点P的坐标为(1,0).故选B.【点睛】本题考查的知识点是点的坐标,解题关键是熟记x轴上的点纵坐标为0.4.D【分析】根据勾股定理的逆定理及三角形内角和定理对各选项进行逐一判断即可.【详解】解:A、a2+b2=c2,是直角三角形,故本选项不符合题意;B、∵52+122=132,∴此三角形是直角三角形,故本选项不符合题意;C、∵∠A+∠B+∠C=180°,∠A=∠B+∠C∴∠A=90°,∴此三角形是直角三角形,故本选项不符合题意;D、设∠A=3x,则∠B=4x,∠C=5x,∵∠A+∠B+∠C=180°,∴3x+4x+5x=180°,解得x=15°∴∠C=5×15°=75°,∴此三角形不是直角三角形,故本选项符号要求;故选D.【点睛】本题考查勾股定理及三角形内角和定理,熟知以上知识是解答此题的关键.5.D【分析】根据二次根式的乘法法则对A进行判断;根据二次根式的除法法则对B进行判断;根据二次根式的加减法对C、D进行判断.【详解】解:A、原式=A选项错误;B、原式==B选项错误;CC选项错误;D=-,所以D选项正确.故选:D.【点睛】本题考查了二次根式的混合运算:先把各二次根式化简为最简二次根式,然后进行二次根式的乘除运算,再合并即可.在二次根式的混合运算中,如能结合题目特点,灵活运用二次根式的性质,选择恰当的解题途径,往往能事半功倍.6.C【分析】根据分式和二次根式有意义的条件进行计算即可.【详解】由题意得:x≥0且x﹣1≠0.解得:x≥0且x≠1.故x的取值范围是x≥0且x≠1.故选C.【点睛】本题考查了函数自变量的取值范围问题,掌握分式和二次根式有意义的条件是解题的关键.7.C【详解】设Rt△ABC的第三边长为x,①当4为直角三角形的直角边时,x为斜边,由勾股定理得,,此时这个三角形的周长=3+4+5=12;②当4为直角三角形的斜边时,x为直角边,由勾股定理得,=,此时这个三角形的周长.故选C8.A 【分析】根据勾股定理可以得到AD 和BD 的长度,然后用AD+BD-AB 的长度即为所求.【详解】根据题意可得BC=4cm ,CD=3cm ,根据Rt △BCD 的勾股定理可得BD=5cm ,则AD=BD=5cm ,所以橡皮筋被拉长了(5+5)-8=2cm .【点睛】主要考查了勾股定理解直角三角形.9.B 【分析】首先根据二次根式有意义的条件求得a 、b 的取值范围,然后再利用二次根式的性质进行化简即可【详解】202a a ∴+<∴<-a a a ∴∙=--故选B【点睛】本题考查了二次根式的性质及化简,解题的关键是根据二次根式有意义的条件判断字母的取值范围.本题需要重点注意字母和式子的符号.10.B 【分析】连接PG ,由题意得出PD =GD =2,∠CDP =∠ADG ,得出∠PDG =∠ADC =90°,得出△PDG 是等腰直角三角形,由等腰直角三角形的性质得出∠GPD =45°,PGPD =,得出AP 2+PG 2=AG 2,由勾股定理的逆定理得出∠GPA =90°,即可得出答案.【详解】解:连接PG ,如图所示:∵四边形ABCD 是正方形,∴AD =CD ,∠ADC =90°,AG =PC =3,∵PA =1,PD =2,PC =3,将△PCD 剪下,并将它拼到如图所示位置(C 与A 重合,P 与G 重合,D 与D 重合),∴PD =GD =2,∠CDP =∠ADG ,∴∠PDG =∠ADC =90°,∴△PDG 是等腰直角三角形,∴∠GPD =45°,PG PD =,∵AG =PC =3,AP =1,PG =,∴AP 2+PG 2=AG 2,∴∠GPA =90°,∴∠APD =90°+45°=135°;故选:B .【点睛】本题考查了勾股定理、勾股定理的逆定理、正方形的性质、等腰直角三角形的判定与性质等知识,熟练掌握正方形的性质和勾股定理的逆定理是解题的关键.11.A【分析】根据算术平方根和绝对值的非负性,确定a 、b 的值,再代入代数式求值即可.【详解】解:由题意得:a+2=0,b-1=0,即a=-2,b=1所以,()()()201720172017==211=1a b +-+--故答案为A.【点睛】本题主要考查了非负数的性质,利用非负数的性质确定待定的字母的值是解答的关键12.C【分析】理解问题的过程,能够通过图象得到函数是随自变量的增大,知道函数值是增大还是减小.【详解】解:由图象可得:0~2秒,点P在GC上运动,则GC=2×2=4cm,∵点G是BC中点,∴BC=2GC=8cm,故①不合题意;由图象可得:2﹣4秒,点P在CD上运动,则第4秒时,y=S△ABP =12×6×8=24cm2,故③符合题意;由图象可得:4﹣7秒,点P在DE上运动,则DE=2×3=6cm,故②符合题意;由图象可得:当第12秒时,点P在H处,∵EF=AB﹣CD=6﹣4=2cm,∴t=22=1s,∴AH=8+6﹣2×(12﹣5﹣1)=6,∴y=S△ABP =12×6×6=18cm2,故④不合题意,∴正确的是②③,故选:C.【点睛】本题考查了动点问题的函数图象,关键是能根据函数图象的性质和图象上的数据分析得出函数的类型和所需要的条件,结合实际意义得到正确的结论.13.-3;2 ;【分析】根据立方根、平方根的定义和倒数乘积等于1即可解题.【详解】解:(1)∵(-3)×(-3)×(-3)=-27,∴-27的立方根为-3;(24=±2;(3)∵(1⎛⨯= ⎝⎭,∴5的倒数为故答案为:-3;±2;14.≠1,=1【分析】根据一次函数的定义、正比例函数的定义,可得答案.【详解】解:已知函数y =(a+1)x+a 2﹣1,当a=-1时,a+1=0,y=a 2﹣1,∴当a≠﹣1时,它是一次函数;当a =1时,a 2﹣1=0,它是正比例函数,故答案为:≠1,=1.【点睛】本题主要考查了一次函数和正比例函数的定义,一次函数y kx b =+的定义条件是:k 、b 为常数,0k ≠,自变量次数为1,0b =是一次函数是正比例函数.15.1、1【解析】【分析】根据勾股定理求出AC ,得到OA′和OA′′的长,根据数轴的概念解答即可.【详解】由勾股定理得,AC ,则CA′=CA′′,∴OA′﹣1,OA′′+1,∴A′、点A″所表示的数分别是1故答案为:1【点睛】本题考查的是勾股定理、实数与数轴,如果直角三角形的两条直角边长分别是a ,b ,斜边长为c,那么a2+b2=c2.16.217【解析】【分析】根据OA1=1,可得点A1的坐标为(1,0),然后根据△OA1B1,△B1A1A2,△B2B1A2,△B2A2A3,△B3B2A3…都是等腰直角三角形,求出A1A2,B1A2,A2A3,B2A3…的长度,然后找出规律,求出点B10的坐标.结合等腰直角三角形的面积公式解答.【详解】∵OA1=1,∴点A1的坐标为(1,0).∵△OA1B1是等腰直角三角形,∴A1B1=1,∴B1(1,1).∵△B1A1A2是等腰直角三角形,∴A1A2=1,B1A2∵△B2B1A2为等腰直角三角形,∴A2A3=2,∴B2(2,2),同理可得:B3(22,22),B4(23,23),…B n(2n﹣1,2n﹣1),∴点B10的坐标是(29,29),∴△B10A10A11的面积是:12×29×29=217.故答案为:217.【点睛】本题考查了一次函数图象上点的坐标特征:一次函数y=kx+b,(k≠0,且k,b为常数)的图象是一条直线,直线上任意一点的坐标都满足函数关系式y=kx+b.也考查了等腰直角三角形的性质.17【分析】首先计算乘方,然后从左向右依次计算,求出算式的值是多少即可.【详解】解::|1(2019﹣)0﹣(1 2)﹣21+1﹣44【点睛】此题主要考查实数的运算,解题的关键是熟知实数的性质.18.﹣3【分析】根据二次根式的混合运算顺序,先对各项利用二次根式的乘除化简,再用加减法进行计算即可.【详解】((22222⎡⎤⎡--+-⨯⎢⎥⎢⎣⎦⎣5(243)(29=+---3=.【点睛】本题考查了二次根式的混合运算、平方差公式、完全平方公式,解决本题的关键是熟练运用公式.19.(1)A (0,8),E (8,4),F (12,4);(2)S △BDF =32【分析】(1)根据正方形的面积求出两个正方形的边长,再求出OG ,然后写出各点的坐标即可;(2)根据S △BDF =S △BDC +S 梯形BCGF ﹣S △DGF 列式计算即可得解.【详解】解:(1)∵正方形ABCD 和正方形EFGC 面积分别为64和16,∴正方形ABCD 和正方形EFGC 的边长分别为8和4,∴OG =8+4=12,∴A (0,8),E (8,4),F (12,4);(2)S △BDF =S △BDC +S 梯形BCGF ﹣S △DGF ,=12×8×8+12×(4+8)×4﹣12×(8+4)×4,=32+24﹣24,=32.【点睛】本题考查了坐标与图形性质,三角形的面积,难点在于(2)列出BDF ∆的面积的表达式.20.(1)5,1;(2)a 2﹣b 2的值为7【分析】(1)根据题目中所给规律即可得结果;(2)把无理数的整数部分和小数部分分别表示出来,再代入计算即可.【详解】解:(1的整数部分为33,∴2]5+=;[51=.故答案为5、1.(2)根据题意,得34<< ,859∴<+<,583a ∴=-.152<514b ∴==-1a b ∴+=,7a b -=.22()()a b a b a b ∴-=+-7=-.∴22a b -的值为7.【点睛】本题考查了估算无理数的大小,解决本题的关键是根据无理数的整数部分确定小数部分.21.(1)见解析;(2)BE =4.【分析】(1)由折叠的性质可知AF=AB=8,然后再依据勾股定理的逆定理可证明△ADF 为直角三角形;(2)由题意可证点E 、D 、F 在一条直线上,设BE=x ,则EF=x ,DE=6+x ,EC=10-x ,在Rt △CED 中,依据勾股定理列方程求解即可.【详解】(1)将△ABE 沿AE 折叠,使点B 落在长方形内点F 处,∴AF =AB =8,∵AF 2+DF 2=62+82=100=102=AD 2,∴∠AFD =90°∴△ADF 是直角三角形(2)∵折叠∴BE =EF ,∠B =∠AFE =90°又∵∠AFD =90°∴点D ,F ,E 在一条直线上.设BE =x ,则EF =x ,DE =6+x ,EC =10-x ,在Rt △DCE 中,∠C =90°,∴CE 2+CD 2=DE 2,即(10-x )2+82=(6+x )2.∴x =4.∴BE =4.【点睛】本题主要考查的是翻折的性质、勾股定理的逆定理、勾股定理的定理,依据勾股定理列出关于x 的方程是解题的关键.22.见解析【分析】应先找到哪两个数的和为13,积为42.再判断是选择加法,还是减法.【详解】根据题意,可知13m =,42n =,由于7613+=,7642⨯=,所以2213+==【点睛】此题考查二次根式的性质与化简,解题关键在于求得13m =,42n =.23.(1)13cm ;(2;(3)13(cm )【分析】(1)利用勾股定理直接求出木棒的最大长度即可.(2)将长方体展开,利用勾股定理解答即可;(3)将容器侧面展开,建立A 关于EF 的对称点A′,根据两点之间线段最短可知A′B 的长度即为所求.【详解】解:(1)由题意得:如图,该长方体中能放入木棒的最大长度是:=;cm13()(2)①如图,AG,②如图,AG=,③如图,AG ,;(3) 高为12cm ,底面周长为10cm ,在容器内壁离容器底部3cm 的点B 处有一饭粒,此时蚂蚁正好在容器外壁,离容器上沿3cm 与饭粒相对的点A 处,5A D cm ∴'=,12312BD AE cm =-+=,∴将容器侧面展开,作A 关于EF 的对称点A ',连接A B ',则A B '即为最短距离,13()A B cm '=.【点睛】本题考查了平面展开—最短路径问题,将图形展开,利用轴对称的性质和勾股定理进行计算是解题的关键.同时也考查了同学们的创造性思维能力.24.画图见解析.【解析】分析:首先在平面直角坐标系中描出各点,然后顺次连接得到△ABC ,找出三个顶点关于y 轴对称的点坐标,然后顺次连接,得出对称后的图形.详解:如图所示:点睛:本题主要考查的是图形的轴对称,属于基础题型.关于y 轴对称的两个点,他们的横坐标互为相反数,纵坐标相等.25.(1)见解析;(2)能,见解析;(3)①C 、D 两点的坐标为C (0,32),D (2,0);②符合条件的所有点M 的坐标为:(716,0)、(92,0);、(﹣2,0)、(﹣12,0)【分析】(1)根据梯形的面积的两种表示方法即可证明;(2)根据四边形ABCD 的面积的两种表示方法即可证明;(3)①根据翻折的性质和勾股定理即可求解;②根据等腰三角形的性质分四种情况求解即可.【详解】解:(1)∵S 梯形ABCD =211222ab c ⨯+S 梯形ABCD =()()12a b a b ++21112()()222ab c a b a b ∴⨯+=++22222ab c a ab b ∴+=++222c a b ∴=+.(2)连接BD ,如图:S 四边形ABCD =()21122c a b a +-,S 四边形ABCD =21122ab b +,∴221111()2222c a b a ab b +-=+,222c a b ∴=+.(3)①设OC a =,则4AC a =-,又5AB =,根据翻折可知:5BD AB ==,4CD AC a ==-,532OD BD OB =-=-=.在Rt COD ∆中,根据勾股定理,得22(4)4a a -=+,解得32a =.3(0,)2C ∴,(2,0)D .答:C 、D 两点的坐标为3(0,)2C ,(2,0)D .②如图:当点M 在x 轴正半轴上时,CM DM =,设CM DM x ==,则2223(2)()2x x =-+,解得2516x =,7216x ∴-=,7(16M ∴,0);CD MD =,35422=-=,59222+=,9(2M ∴,0);当点M 在x 轴负半轴上时,CM CD =,2OM OD == ,(2,0)M ∴-;DC DM =,35422=-=,51222OM ∴=-=,1(2M ∴-,0).∴符合条件的所有点M 的坐标为:7(16,0)、9(2,0)、(2,0)-、1(2-,0).【点睛】本题考查了等腰三角形的判定和性质,勾股定理,折叠的性质,是三角形的综合题,解决本题的关键是分情况讨论思想的运用.。

北师大版八年级上英语期末测试题及答案

北师大版八年级上英语期末测试题及答案

北师大版八年级上英语期末测试题及答案第一部分选择题(共25个小题,每题1分,满分25分)1. - How do you like the movie?- __B__. It's very interesting.A. Never mindB. I like it very muchC. It doesn't matterD. I hope so2. - What does your father do?- __C__. He is a doctor.A. SorryB. Excuse meC. He is a doctorD. Don't mention it...第二部分非选择题(共30个小题,每题2分,满分60分)1. 根据短文内容,回答问题。

My name is Tom. I'm from the United States. I am twelve years old.I live in New York with my parents and my sister. I go to school from Monday to Friday. I like reading books and playing basketball. My favorite subject is English. I can speak a little Chinese. I want to visit China one day.Question: Where does Tom live?Answer: __He lives in New York.__...第三部分写作(满分15分)请根据下列要点写一篇介绍你的家乡的短文:1.所在省市;2.地理位置和面积;3.人口和民族分布;4.风景和名胜;5.特色食物;6.你对家乡的感受。

答案略以上是北师大版八年级上英语期末测试题及答案。

北师大版2019-2020学年七年级上学期英语期中十校联考试卷D卷

北师大版2019-2020学年七年级上学期英语期中十校联考试卷D卷

北师大版2019-2020学年七年级上学期英语期中十校联考试卷D卷姓名:________ 班级:________ 成绩:________一、单项填空。

请从A、B、C、D四个选项中选出可以填入空白处 (共5题;共10分)1. (2分)—Hi, Daniel. How was your trip to Qingdao?—________. The beaches there are very beautiful and we also enjoyed some fresh seafood.A . Not badlyB . Very muchC . Very goodD . So well2. (2分)What time ______ your best friend go to school?A . isB . doC . does3. (2分)—When did Neil Armstrong walk on the moon?—____July 20th,1969.A . InB . AtC . ForD . On4. (2分)The girl is Linda Black. Linda is her ________ name.A . firstB . lastC . familyD . middle5. (2分)- Bad luck! I lost _______ new pen yesterday.- Don't worry. I'll lend _____ to you.A . my; meB . mine; myC . mine; mineD . my; mine二、交际配对。

(共1题;共5分)6. (5分)根据短文内容,从短文后的方框ABCDEFG选项中,选出适当的选项B: Certainly. Go along this street. Turn left into Xingfu Street, and the museum is on your right.A:________B: It's about thirty minutes.A: I see.________B: Yes, you can. A No.103 bus will take you there.A:________B: Over there. Look! The bus is coming.A: Thank you very much.B:________三、完形填空。

2024-2025学年北师大版(2019)八年级科学上册月考试卷507

2024-2025学年北师大版(2019)八年级科学上册月考试卷507

2024-2025学年北师大版(2019)八年级科学上册月考试卷507考试试卷考试范围:全部知识点;考试时间:120分钟学校:______ 姓名:______ 班级:______ 考号:______总分栏题号一二三四五六总分得分评卷人得分一、单选题(共8题,共16分)1、为了缓解我国北方地区水资源不足的状况而实施的南水北调工程,将主要改变水循环环节中的( )A. 蒸发B. 水汽输送C. 降水D. 地表径流2、14.如图所示的电路中,闭合开关S,当滑动变阻器的滑片P向右移动时,电压表、电流表示数的变化情况是 ( )15.A. 电压表、电流表示数均变大B. 电压表、电流表示数均变小C. 电压表示数变大,电流表示数变小D. 电压表示数变小,电流表示数变大3、下列操作中只发生物理变化的是()A. 将二氧化碳通入水中B. 将二氧化碳通入澄清石灰水中C. 将二氧化碳倾倒入盛有燃着蜡烛的烧杯中D. 将二氧化碳加压降温压缩成干冰4、含有硫的煤,在火炉中如果不完全燃烧,则排放出能污染空气的有毒气体是( )A. 和B.C. COD. 和CO5、通过语言功能建立的条件反射应该是()A. 看到老师走上讲台,同学们马上起立向老师敬礼B. 听到主人唤它的“名字”,小狗就向主人跑过来C. 看到有人突然横穿马路,司机立即刹车停了下来D. 听了老师讲的故事,同学们都忍不住笑了6、如图所示,在探究并联电路中的电流关系时,小明同学用电流表测出A、B、C三处的电流分别为I A=0.4A、I B=0.2A、I C=0.2A,在表格中记录数据后,下一步应该做的是()A. 整理器材,结束实验B. 分析数据,得出结论C. 换用不同规格的小灯泡,再测出几组电流值D. 换用电流表的另一量程,再测出一组电流值7、关于人和绿色植物呼吸作用的描述,正确的是()A. 都要吸入二氧化碳B. 人吸入氧气,绿色植物吸入二氧化碳C. 都要吸入氧气D. 人吸入二氧化碳,绿色植物吸入氧气8、如图所示,将原来置于黑暗环境中的绿色植物移至光下后,CO2的吸取量发生改变,下列各项叙述正确的是()A. 曲线AB段表示绿色植物没有进行光合作用B. 曲线BC段表示绿色植物只进行光合作用C. 在B点,光合作用和呼吸作用的速率相等D. 整段曲线表明,随着光照强度增强,光合作用增强,呼吸作用减弱评卷人得分二、填空题(共8题,共16分)9、信息的磁记录:通过 ____ 的方法来记录信息。

2019-2020学年北师大版八年级数学第一学期期末测试题(含答案)

2019-2020学年北师大版八年级数学第一学期期末测试题(含答案)

2019-2020学年八年级数学第一学期期末测试卷一、选择题(本大题10小题,每小题3分,共30分.)在每小题列出的四个选项中,只有个正确选项,请将正确答案写在答题卷的相应位置1.下列实数中,不是无理数的是()A.B.﹣C.2π(π表示圆周率)D.22.下列各点中,位于第二象限的是()A.(8,﹣1)B.(8,0)C.(﹣,3)D.(0,﹣4)3.下列各组数据中,不是勾股数的是()A.3,4,5B.7,24,25C.8,15,17D.5,7,94.如图,在△ABC中,∠A=80°,点D在BC的延长线上,∠ACD=145°,则∠B是()A.45°B.55°C.65°D.75°5.某小组长统计组内5人一天在课堂上的发言次数分別为3,3,0,4,5.关于这组数据,下列说法错误的是()A.众数是3B.中位数是0C.平均数3D.方差是2.86.一次函数y=﹣2x﹣1的图象大致是()A.B.C.D.7.如图所示,下列推理及括号中所注明的推理依据错误的是()A.∵∠1=∠3,∴AB∥CD(内错角相等,两直线平行)B.∵AB∥CD,∴∠1=∠3(两直线平行,内错角相等)C.∵AD∥BC,∴∠BAD+∠ABC=180°(两直线平行,同旁内角互补)D.∵∠DAM=∠CBM,∴AB∥CD(两直线平行,同位角相等)8.下列说法正确的是()A.1的平方根是1B.﹣8的立方根是﹣2C.=±2D.=﹣29.小明中午放学回家自己煮面条吃,有下面几道工序:(1)洗锅盛水2分钟;(2)洗菜3分钟;(3)准备面条及佐料2分钟;(4)用锅把水烧开7分钟;(5)用烧开的水煮面条和菜要3分钟.以上各工序除(4)外,一次只能进行一道工序,小明要将面条煮好,最少用()A.14分钟B.13分钟C.12分钟D.11分钟10.体育课上,20人一组进行足球比赛,每人射点球5次,已知某一组的进球总数为49个,进球情况记录如下表,其中进2个球的有x人,进3个球的有y人,由题意列出关于x与y的方程组为()A.B.C.D.二、填空题(本大题6小题,每小题4分,共24分)请将下列各题的正确答案写在答题卷的相应位置11.计算:=;|﹣|=.12.命题“若a2>b2,则a>b”的逆命题是,该逆命题是(填“真”或“假”)命题.13.计算:(3+)()=.14.小明某学期的数学平时成绩70分,期中考试80分,期末考试85分,若计算学期总评成绩的方法如下:平时:期中:期末=3:3:4,则小明总评成绩是分.15.有大小两种货车,2辆大货车与1辆小货车一次可以运货7吨,1辆大货车与2辆小货车一次可以运货5吨.则1辆大货车与1辆小货车一次可以运货吨.16.在平面直角坐标系xOy中,点A1,A2,A3,…和B1,B2,B3,…分别在直线y=kx+b和x轴上.△OA1B1,△B1A2B2,△B2A3B3,…都是等腰直角三角形,如果A1(1,1),A2(,).那么点A3的纵坐标是,点A2013的纵坐标是.三、解答题(一)(本大题共3小题,每小题6分,共18分)17.计算:(2﹣1)2﹣()÷.18.解方程组:19.如图,在平面直角坐标系中,Rt△ABC的三个顶点坐标为A(﹣3,0),B(﹣3,﹣3),C (﹣1,﹣3)(1)填空:AC=;(2)在图中作出△ABC关于x轴对称的图形△DEF.四、解答题(二)(本大题共3小题,每小题7分,共21分)20.据市旅游局发布信息,今年春节假期期间,我市外来与外出旅游的总人数为226万人,分别比去年同期增长30%和20%,去年同期外来旅游比外出旅游的人数多20万人.求我市去年外来和外出旅游的人数.21.我区某中学开展“社会主义核心价值观”演讲比赛活动,九(1)、九(2)班根据初赛成绩各选出5名选手参加复赛,两个班各选出的5名选手的复赛成绩(满分为100分)如图所示.根据图中数据解决下列问题:(1)九(1)班复赛成绩的中位数是分,九(2)班复赛成绩的众数是分;(2)小明同学已经算出了九(1)班复赛的平均成绩=85分;方差S2=[(85﹣85)2+(75﹣85)2+(80﹣85)2+(85﹣85)2+(100﹣85)2]=70(分2),请你求出九(2)班复赛的平均成绩x2和方差S22;(3)根据(2)中计算结果,分析哪个班级的复赛成绩较好?22.已知,直线PQ∥MN,△ABC的顶点A与B分别在直线MN与PQ上,点C在直线AB的右侧,且∠C=45°,设∠CBQ=∠α,∠CAN=∠β.(1)如图1,当点C落在PQ的上方时,AC与PQ相交于点D,求证:∠β=∠α+45°.请将下列推理过程补充完整:证明:∵∠CDQ是△CBD的一个外角(三角形外角的定义),∴∠CDQ=∠α+∠C(三角形的一个外角等于和它不相邻的两个内角的和)∵PQ∥MN(),∴∠CDQ=∠β().∴∠β=(等量代换).∵∠C=45°(已知),∴∠β=∠α+45°(等量代换)(2)如图2,当点C落在直线MN的下方时,BC与MN交于点F,请判断∠α与∠β的数量关系,并说明理由.五、解答题(本大题共3小题,每小题9分,共27分)请将正确答案写在答题卷的相应位置23.如图1所示,小亮家与学校之间有一超市,小亮骑车由家匀速行驶去学校,然后在校学习8小时.最后放学骑车匀速回家(上学与放学均不在超市停留).图2中的折线OABC表示小亮离家的距离y(km)与离家的时间x(h)之间的函数关系.根据已上信息,解答下列问题:(1)小亮上学的速度为km/h,放学回家的速度为km/h;(2)求线段BC所表示的y与x之间的函数关系;(3)如果小亮两次经过超市的时间间隔为8.48小时,那么超市离小亮家多远?24.如图,在△ABC中,∠C=90°,将△ACE沿着AE折叠以后C点正好落在AB边上的点D处.(1)当∠B=28°时,求∠AEC的度数;(2)当AC=6,AB=10时,①求线段BC的长;②求线段DE的长.25.已知:如图,在平面直角坐标系中,点O是坐标系原点,在△AOC中,OA=OC,点A坐标为(﹣3,4),点C在x轴的正半轴上,直线AC交y轴于点M,将△AOC沿AC折叠得到△ABC,请解答下列问题:(1)点C的坐标为;(2)求线段OM的长;(3)求点B的坐标.2019-2020学年八年级数学第一学期期末测试卷参考答案与试题解析一、选择题(本大题10小题,每小题3分,共30分.)在每小题列出的四个选项中,只有个正确选项,请将正确答案写在答题卷的相应位置1.下列实数中,不是无理数的是()A.B.﹣C.2π(π表示圆周率)D.2【分析】根据无理数、有理数的定义逐一对每个选择支进行判断.【解答】解:是分数,属于有理数,故选项A正确;﹣,2π,2是无理数.故选:A.【点评】此题主要考查了无理数的定义,注意:带根号的开不尽方的数是无理数,无限不循环小数为无理数,含π的数是无理数.如2π,,0.8080080008…(每两个8之间依次多1个0)等形式.2.下列各点中,位于第二象限的是()A.(8,﹣1)B.(8,0)C.(﹣,3)D.(0,﹣4)【分析】依据位于第二象限的点的横坐标为负,纵坐标为正,即可得到结论.【解答】解:∵位于第二象限的点的横坐标为负,纵坐标为正,∴位于第二象限的是(﹣,3)故选:C.【点评】本题主要考查了点的坐标,解题时注意:位于第二象限的点的横坐标为负,纵坐标为正.3.下列各组数据中,不是勾股数的是()A.3,4,5B.7,24,25C.8,15,17D.5,7,9【分析】欲判断是否为勾股数,必须根据勾股数是正整数,同时还需验证两小边的平方和是否等于最长边的平方.【解答】解:A、32+42=52,能构成直角三角形,是整数,故错误;B、72+242=252,能构成直角三角形,是整数,故错误;C、82+152=172,构成直角三角形,是正整数,故错误;D、52+72≠92,不能构成直角三角形,故正确;故选:D.【点评】此题主要考查了勾股数的定义,熟记勾股数的定义是解题的关键.4.如图,在△ABC中,∠A=80°,点D在BC的延长线上,∠ACD=145°,则∠B是()A.45°B.55°C.65°D.75°【分析】利用三角形的外角的性质即可解决问题;【解答】解:在△ABC中,∵∠ACD=∠A+∠B,∠A=80°,∠ACD=145°,∴∠B=145°﹣80°=65°,故选:C.【点评】本题考查三角形的外角,解题的关键是熟练掌握基本知识,属于中考常考题型.5.某小组长统计组内5人一天在课堂上的发言次数分別为3,3,0,4,5.关于这组数据,下列说法错误的是()A.众数是3B.中位数是0C.平均数3D.方差是2.8【分析】根据方差、众数、平均数、中位数的含义和求法,逐一判断即可.【解答】解:将数据重新排列为0,3,3,4,5,则这组数的众数为3,中位数为3,平均数为=3,方差为×[(0﹣3)2+2×(3﹣3)2+(4﹣3)2+(5﹣3)2]=2.8,故选:B.【点评】本题考查了众数、中位数、平均数以及方差,解题的关键是牢记概念及公式.6.一次函数y=﹣2x﹣1的图象大致是()A.B.C.D.【分析】先根据一次函数的系数判断出函数图象所经过的象限,由此即可得出结论.【解答】解:在y=﹣2x﹣1中,∵﹣2<0,﹣1<0,∴此函数的图象经过二、三、四象限,故选:D.【点评】本题考查的是一次函数的图象,熟知当k<0,b>0时,一次函数y=kx+b的图象在一、二、四象限是解答此题的关键.7.如图所示,下列推理及括号中所注明的推理依据错误的是()A.∵∠1=∠3,∴AB∥CD(内错角相等,两直线平行)B.∵AB∥CD,∴∠1=∠3(两直线平行,内错角相等)C.∵AD∥BC,∴∠BAD+∠ABC=180°(两直线平行,同旁内角互补)D.∵∠DAM=∠CBM,∴AB∥CD(两直线平行,同位角相等)【分析】依据内错角相等,两直线平行;两直线平行,内错角相等;两直线平行,同旁内角互补;同位角相等,两直线平行进行判断即可.【解答】解:A.∵∠1=∠3,∴AB∥CD(内错角相等,两直线平行),正确;B.∵AB∥CD,∴∠1=∠3(两直线平行,内错角相等),正确;C.∵AD∥BC,∴∠BAD+∠ABC=180°(两直线平行,同旁内角互补),正确;D.∵∠DAM=∠CBM,∴AD∥BC(同位角相等,两直线平行),错误;故选:D.【点评】本题主要考查了平行线的性质与判定,平行线的判定是由角的数量关系判断两直线的位置关系.平行线的性质是由平行关系来寻找角的数量关系.8.下列说法正确的是()A.1的平方根是1B.﹣8的立方根是﹣2C.=±2D.=﹣2【分析】根据平方根、算术平方根的定义逐一判别可得.【解答】解:A.1的平方根是±1,此选项错误;B.﹣8的立方根是﹣2,此选项正确;C.=2,此选项错误;D.=2,此选项错误;故选:B.【点评】本题主要考查平方根与立方根,解题的关键是掌握平方根和算术平方根及立方根的定义.9.小明中午放学回家自己煮面条吃,有下面几道工序:(1)洗锅盛水2分钟;(2)洗菜3分钟;(3)准备面条及佐料2分钟;(4)用锅把水烧开7分钟;(5)用烧开的水煮面条和菜要3分钟.以上各工序除(4)外,一次只能进行一道工序,小明要将面条煮好,最少用()A.14分钟B.13分钟C.12分钟D.11分钟【分析】根据统筹方法,烧开水时可洗菜和准备面条及佐料,这样可以节省时间,所以小明所用时间最少为(1)、(4)、(5)步时间之和.【解答】解:第一步,洗锅盛水花2分钟;第二步,用锅把水烧开7分钟,同时洗菜3分钟,准备面条及佐料2分钟,总计7分钟;第三步,用烧开的水煮面条和菜要3分钟.总计共用2+7+3=12分钟.故选:C.【点评】解决问题的关键是读懂题意,采用统筹方法是生活中常用的有效节省时间的方法,本题将数学知识与生活相结合,是一道好题.10.体育课上,20人一组进行足球比赛,每人射点球5次,已知某一组的进球总数为49个,进球情况记录如下表,其中进2个球的有x人,进3个球的有y人,由题意列出关于x与y的方程组为()A.B.C.D.【分析】设进2个球的有x人,进3个球的有y人,根据20人共进49个球,即可得出关于x,y的二元一次方程组,此题得解.【解答】解:设进2个球的有x人,进3个球的有y人,根据题意得:,即.故选:A.【点评】本题考查了由实际问题抽象出二元一次方程组,找准等量关系,正确列出二元一次方程组是解题的关键.二、填空题(本大题6小题,每小题4分,共24分)请将下列各题的正确答案写在答题卷的相应位置11.计算:=;|﹣|=2.【分析】根据二次根式的分母有理化和二次根式的性质分别计算可得.【解答】解:==,|﹣|==2,故答案为:,2.【点评】本题主要考查二次根式的分母有理化,解题的关键是掌握二次根式的有理化方法和二次根式的性质.12.命题“若a2>b2,则a>b”的逆命题是如a>b,则a2>b2,,该逆命题是(填“真”或“假”)假命题.【分析】先写出命题的逆命题,然后在判断逆命题的真假.【解答】解:如a2>b2,则a>b”的逆命题是:如a>b,则a2>b2,假设a=1,b=﹣2,此时a>b,但a2<b2,即此命题为假命题.故答案为:如a>b,则a2>b2,假.【点评】此题考查了命题与定理的知识,写出一个命题的逆命题的关键是分清它的题设和结论,然后将题设和结论交换.在写逆命题时要用词准确,语句通顺.13.计算:(3+)()=+1.【分析】利用多项式乘法展开,然后合并即可.【解答】解:原式=3﹣6+7﹣2=+1.故答案为+1.【点评】本题考查了二次根式的混合运算:先把各二次根式化简为最简二次根式,然后进行二次根式的乘除运算,再合并即可.在二次根式的混合运算中,如能结合题目特点,灵活运用二次根式的性质,选择恰当的解题途径,往往能事半功倍.14.小明某学期的数学平时成绩70分,期中考试80分,期末考试85分,若计算学期总评成绩的方法如下:平时:期中:期末=3:3:4,则小明总评成绩是79分.【分析】按3:3:4的比例算出本学期数学总评分即可.【解答】解:本学期数学总评分=70×30%+80×30%+85×40%=79(分).故答案为:79.【点评】本题考查了加权成绩的计算,平时成绩:期中考试成绩:期末考试成绩=3:3:4的含义就是分别占总数的30%、30%、40%.15.有大小两种货车,2辆大货车与1辆小货车一次可以运货7吨,1辆大货车与2辆小货车一次可以运货5吨.则1辆大货车与1辆小货车一次可以运货4吨.【分析】设1辆大货车一次可以运货x吨,1辆小货车一次可以运货y吨,由“2辆大货车与1辆小货车一次可以运货7吨,1辆大货车与2辆小货车一次可以运货5吨”,即可得出关于x,y的二元一次方程组,将方程组的两方程相加再除以3,即可求出结论.【解答】解:设1辆大货车一次可以运货x吨,1辆小货车一次可以运货y吨,根据题意得:,(①+②)÷3,得:x+y=4.故答案为:4.【点评】本题考查了二元一次方程组的应用,找准等量关系,正确列出二元一次方程组是解题的关键.16.在平面直角坐标系xOy中,点A1,A2,A3,…和B1,B2,B3,…分别在直线y=kx+b和x轴上.△OA1B1,△B1A2B2,△B2A3B3,…都是等腰直角三角形,如果A1(1,1),A2(,).那么点A3的纵坐标是,点A2013的纵坐标是()2012.【分析】先求出直线y =kx +b 的解析式,求出直线与x 轴、y 轴的交点坐标,求出直线与x 轴的夹角的正切值,分别过等腰直角三角形的直角顶点向x 轴作垂线,然后根据等腰直角三角形斜边上的高线与中线重合并且等于斜边的一半,利用正切值列式依次求出三角形的斜边上的高线,即可得到A 3的坐标,进而得出各点的坐标的规律.【解答】解:∵A 1(1,1),A 2(,)在直线y =kx +b 上,∴,解得,∴直线解析式为y =x +;设直线与x 轴、y 轴的交点坐标分别为N 、M ,当x =0时,y =,当y =0时, x +=0,解得x =﹣4,∴点M 、N 的坐标分别为M (0,),N (﹣4,0),∴tan ∠MNO ===,作A 1C 1⊥x 轴与点C 1,A 2C 2⊥x 轴与点C 2,A 3C 3⊥x 轴与点C 3,∵A 1(1,1),A 2(,),∴OB 2=OB 1+B 1B 2=2×1+2×=2+3=5,tan ∠MNO ===,∵△B 2A 3B 3是等腰直角三角形,∴A 3C 3=B 2C 3,∴A 3C 3==()2,同理可求,第四个等腰直角三角形A 4C 4==()3,依此类推,点A n 的纵坐标是()n ﹣1.∴A2013=()2012故答案为:,()2012.【点评】本题考查的是一次函数图象上点的坐标特点,熟知一次函数图象上各点的坐标一定适合此函数的解析式是解答此题的关键.三、解答题(一)(本大题共3小题,每小题6分,共18分)17.计算:(2﹣1)2﹣()÷.【分析】先利用二次根式的除法法则和完全平方公式运算,然后把各二次根式化简为最简二次根式后合并即可.【解答】解:原式=8﹣4+1﹣(﹣)=9﹣4﹣2+=9﹣5.【点评】本题考查了二次根式的混合运算:先把各二次根式化简为最简二次根式,然后进行二次根式的乘除运算,再合并即可.在二次根式的混合运算中,如能结合题目特点,灵活运用二次根式的性质,选择恰当的解题途径,往往能事半功倍.18.解方程组:【分析】方程组利用代入消元法求出解即可.【解答】解:,把①代入②得:3x﹣2x+3=8,解得:x=5,把x=5代入①得y=7,则原方程组的解为.【点评】此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.19.如图,在平面直角坐标系中,Rt△ABC的三个顶点坐标为A(﹣3,0),B(﹣3,﹣3),C(﹣1,﹣3)(1)填空:AC=;(2)在图中作出△ABC关于x轴对称的图形△DEF.【分析】(1)利用勾股定理求解可得;(2)分别作出点B与点C关于x轴的对称图形,再与点A首尾顺次连接即可得.【解答】解:(1)AC==,故答案为:;(2)所画图形如下所示,其中△DEF即为所求,【点评】本题主要考查作图﹣轴对称变换,解题的关键是熟练掌握轴对称变换的定义和性质,并据此得出变换后的对应点及勾股定理.四、解答题(二)(本大题共3小题,每小题7分,共21分)20.据市旅游局发布信息,今年春节假期期间,我市外来与外出旅游的总人数为226万人,分别比去年同期增长30%和20%,去年同期外来旅游比外出旅游的人数多20万人.求我市去年外来和外出旅游的人数.【分析】设我市去年外来旅游的有x万人,外出旅游的有y万人,根据去年同期外来旅游比外出旅游的人数多20万人及今年外来与外出旅游的人数与去年人数之间的关系,即可得出关于x,y的二元一次方程组,解之即可得出结论.【解答】解:设我市去年外来旅游的有x万人,外出旅游的有y万人,根据题意得:,解得:.答:我市去年外来旅游的有100万人,外出旅游的有80万人,【点评】本题考查了二元一次方程组的应用,找准等量关系,正确列出二元一次方程组是解题的关键.21.我区某中学开展“社会主义核心价值观”演讲比赛活动,九(1)、九(2)班根据初赛成绩各选出5名选手参加复赛,两个班各选出的5名选手的复赛成绩(满分为100分)如图所示.根据图中数据解决下列问题:(1)九(1)班复赛成绩的中位数是85分,九(2)班复赛成绩的众数是100分;(2)小明同学已经算出了九(1)班复赛的平均成绩=85分;方差S2=[(85﹣85)2+(75﹣85)2+(80﹣85)2+(85﹣85)2+(100﹣85)2]=70(分2),请你求出九(2)班复赛的平均成绩x2和方差S22;(3)根据(2)中计算结果,分析哪个班级的复赛成绩较好?【分析】(1)利用众数、中位数的定义分别计算即可;(2)利用平均数和方差的公式计算即可;(3)利用方差的意义进行判断.【解答】解:(1)九(1)班复赛成绩的中位数是85分,九(2)班复赛成绩的众数是100分;故答案为:85,100;(2)九(2)班的选手的得分分别为70,100,100,75,80,所以九(2)班成绩的平均数=(70+100+100+75+80)=85,九(2)班的方差S22=[(70﹣85)2+(100﹣85)2+(100﹣85)2+(75﹣85)2+(80﹣85)2]=160;(3)平均数一样的情况下,九(1)班方差小,所以九(1)班的成绩比较稳定.【点评】本题考查了方差:方差是反映一组数据的波动大小的一个量.方差越大,则平均值的离散程度越大,稳定性也越小;反之,则它与其平均值的离散程度越小,稳定性越好.也考查了统计图.22.已知,直线PQ∥MN,△ABC的顶点A与B分别在直线MN与PQ上,点C在直线AB的右侧,且∠C=45°,设∠CBQ=∠α,∠CAN=∠β.(1)如图1,当点C落在PQ的上方时,AC与PQ相交于点D,求证:∠β=∠α+45°.请将下列推理过程补充完整:证明:∵∠CDQ是△CBD的一个外角(三角形外角的定义),∴∠CDQ=∠α+∠C(三角形的一个外角等于和它不相邻的两个内角的和)∵PQ∥MN(已知),∴∠CDQ=∠β(两直线平行,同位角相等).∴∠β=∠α+∠C(等量代换).∵∠C=45°(已知),∴∠β=∠α+45°(等量代换)(2)如图2,当点C落在直线MN的下方时,BC与MN交于点F,请判断∠α与∠β的数量关系,并说明理由.【分析】(1)根据题意可以写出推理过程,从而可以解答本题;(2)根据三角形外角的性质和三角形的内角和即可得到结论..【解答】解:(1)证明:∵∠CDQ是△CBD的一个外角(三角形外角的定义),∴∠CDQ=∠α+∠C(三角形的一个外角等于和它不相邻的两个内角的和)∵PQ∥MN(已知),∴∠CDQ=∠β(两直线平行,同位角相等).∴∠β=∠α+∠C(等量代换).∵∠C=45°(已知),∴∠β=∠α+45°(等量代换);故答案为:已知,两直线平行,同位角相等,∠α+∠C,(2)证明:∵∠CFN是△ACF的一个外角(三角形外角的定义),∴∠CFN=∠β+∠C(三角形的一个外角等于和它不相邻的两个内角的和),∵PQ∥MN(已知),∴∠CFN=∠α(两直线平行,同位角相等)∴∠α=∠β+∠C(等量代换).∵∠C=45°(已知),∴∠α=∠β+45°(等量代换).【点评】本题考查了三角形外角的性质,平行线的性质,解题的关键是明确题意,找出所求问题需要的条件,利用数形结合的思想解答.五、解答题(本大题共3小题,每小题9分,共27分)请将正确答案写在答题卷的相应位置23.如图1所示,小亮家与学校之间有一超市,小亮骑车由家匀速行驶去学校,然后在校学习8小时.最后放学骑车匀速回家(上学与放学均不在超市停留).图2中的折线OABC表示小亮离家的距离y(km)与离家的时间x(h)之间的函数关系.根据已上信息,解答下列问题:(1)小亮上学的速度为5km/h,放学回家的速度为3km/h;(2)求线段BC所表示的y与x之间的函数关系;(3)如果小亮两次经过超市的时间间隔为8.48小时,那么超市离小亮家多远?【分析】(1)根据题意和图象中的数据可以求得小亮上学的速度和放学回家的速度;(2)根据图象中的数据和题意可以求得线段BC所表示的y与x之间的函数关系;(3)由题意可知,小明从家到超市和从超市到家的时间之和是总的时间减去两次经过超市的时间间隔,从而可以解答本题.【解答】解:(1)由题意可得,小明上学的速度为:3÷0.6=5km/h,放学回家的速度为:3÷(9.6﹣0.6﹣8)=3km/h,故答案为:5,3;(2)设线段BC所表示的y与x之间的函数关系式为y=kx+b,将B(8.6,3)、C(9.6,0)代入y=kx+b,得,得,∴线段BC所表示的y与x之间的函数关系式为y=﹣3x+28.8(8.6≤x≤9.6);(3)设超市离家skm,=9.6﹣8.48,解得:s=2.1.答:超市离家2.1km.【点评】本题考查一次函数的应用,解答本题的关键是明确题意,利用一次函数的性质和数形结合的思想解答.24.如图,在△ABC中,∠C=90°,将△ACE沿着AE折叠以后C点正好落在AB边上的点D处.(1)当∠B=28°时,求∠AEC的度数;(2)当AC=6,AB=10时,①求线段BC的长;②求线段DE的长.【分析】(1)在Rt△ABC中,利用互余得到∠BAC=62°,再根据折叠的性质得∠CAE=∠CAB =31°,然后根据互余可计算出∠AEC=59°;(2)①在Rt△ABC中,利用勾股定理即可得到BC的长;②设DE=x,则EB=BC﹣CE=8﹣x,依据勾股定理可得,Rt△BDE中DE2+BD2=BE2,再解方程即可得到DE的长.【解答】解:(1)在Rt△ABC中,∠ABC=90°,∠B=28°,∴∠BAC=90°﹣28°=62°,∵△ACE沿着AE折叠以后C点正好落在点D处,∴∠CAE=∠CAB=×62°=31°,Rt△ACE中,∠ACE=90°∴∠AEC=90°﹣31°=59°.(2)①在Rt△ABC中,AC=6,AB=10,∴BC===8.②∵△ACE沿着AE折叠以后C点正好落在点D处,∴AD=AC=6,CE=DE,∴BD=AB﹣AD=4,设DE=x,则EB=BC﹣CE=8﹣x,∵Rt△BDE中,DE2+BD2=BE2,∴x2+42=(8﹣x)2,解得x=3.即DE的长为3.【点评】本题考查了折叠问题,折叠是一种对称变换,它属于轴对称,解题时常设要求的线段长为x,然后根据折叠和轴对称的性质用含x的代数式表示其他线段的长度,选择适当的直角三角形,运用勾股定理列出方程求出答案.25.已知:如图,在平面直角坐标系中,点O是坐标系原点,在△AOC中,OA=OC,点A坐标为(﹣3,4),点C 在x 轴的正半轴上,直线AC 交y 轴于点M ,将△AOC 沿AC 折叠得到△ABC ,请解答下列问题:(1)点C 的坐标为 (5,0) ;(2)求线段OM 的长;(3)求点B 的坐标.【分析】(1)利用勾股定理求出OA 的长即可解决问题;(2)求出直线AC 的解析式,利用待定系数法即可解决问题;(3)只要证明AB =AC =5,AB ∥x 轴,即可解决问题;【解答】解:(1)∵A (﹣3,4),∴OA ==5,∴OA =OC =5,∴C (5,0),故答案为(5,0);(2)设直线AC 的解析式y =kx +b ,函数图象过点A 、C ,得,解得,∴直线AC 的解析式y =﹣x +,当x =0时,y =,即M (0,),∴OM =.(3)∵△AOC沿着AC折叠得到△ABC,∴OA=BA,OC=BC,且∠ACO=∠ACB,又∵OA=OC,∴AB=AC=OC,∴∠BAC=∠ACB,∴∠ACO=∠BAC,∴AB∥x轴,由(1)知,C(5,0),∴OC=5.∵AB=AC=OC,∴AB=5.∵A坐标为(﹣3,4),AB∥x轴,∴B坐标为(2,4).【点评】本题属于三角形综合题,考查了翻折变换,等腰三角形的性质,一次函数的应用等知识,解题的关键是熟练掌握基本知识,属于中考常考题型.。

北师大版2019-2020学年八年级下学期语文期末考试试卷B卷精版

北师大版2019-2020学年八年级下学期语文期末考试试卷B卷精版

北师大版2019-2020学年八年级下学期语文期末考试试卷B卷姓名:________ 班级:________ 成绩:________一、选择题 (共2题;共4分)1. (2分)下列划线成语使用不正确的一项是()A . 当改革的浪潮以摧枯拉朽之势席卷旧的司法鉴定制度时,我国司法鉴定的一个新的时代拉开了序幕。

B . 那些对自己的事业有探索精神并乐此不疲的人,最终都走向了成功。

C . 侦探小说中眼花缭乱的情节让我一头雾水,完全忘记了如何思考。

D . 家风是一种“软约束”,通过潜移默化的影响,实现对家庭成员行为、作风、操守的有效约束。

2. (2分)下列句子中没有语病的一项是()A . 近几年,国产奶粉的质量问题频发,从客观上推进了我国消费者对“洋奶粉”的依赖心理,使得“洋奶粉”格外受宠。

B . 望着白云缭绕的巍巍香炉峰和飞流直下、势不可挡的庐山瀑布,无不使游览者感受到大自然的壮美雄奇和神功伟力。

C . 国家有关部门组织核安全方面的专家,用9个月时间对在建核电机组、待建核电机组及核燃料循环设施等进行了安全检查。

D . 中学生之所以喜欢网络小说的原因,在于这些作品大多思想情感丰富细腻,人物形象栩栩如生,而且叙述方式自由活泼。

二、句子默写 (共1题;共5分)3. (5分) (2019七上·阳江月考) 根据课文默写古诗文。

(1)生活中表示既善于从正面学习,也善于从反面借鉴的意思时,我们常引用《论语》中的话:________,________。

(2)儒家经典让我们获益匪浅。

《论语》中的“________,________?”常用来表达当别人不了解甚至误解自己时应当采取的正确态度。

(3)孔子在《论语·述而》中论述君子对富贵的正确态度是:________,________。

(4)每年一度的亚洲博鳌论坛,华夏儿女喜迎各国嘉宾,我们可以引用《论语》中的“________,________?”来诠释这份情怀。

2019-2020新北师大版八年级上册数学期末测试卷含答案最新版

2019-2020新北师大版八年级上册数学期末测试卷含答案最新版

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2022-2023学年英语北师大版(2019)必修三单元测试卷 Unit 7 Art

2022-2023学年英语北师大版(2019)必修三单元测试卷 Unit 7 Art

2022-2023学年英语北师大版(2019)必修三单元测试卷 Unit 7 Art 学校:___________姓名:___________班级:___________考号:___________ 一、阅读理解Top Public Sculpture Parks to Visit in AmericaKasmin Sculpture Garden (New York City)This quiet sculpture garden in Manhattan's Chelsea neighborhood is far from the crowds. Owned and operated by Kasmin Gallery, this exhibition space can be viewed from the nearby High Line. It is designed by Future Green, a Brooklyn, landscape architect studio, and it stands beside a famous building designed by Zaha Hadid. There's a current exhibition featuring bronze (铜) sculptures by Alma Allen, which shows the artist's regard for Utah.Tippet Rise Art Center (Fishtail)This sculpture garden is worth the trip to the Beartooth Mountains in Fishtail. It is a 12,500-acre ranch (牧场), which is peppered with public art, including sculptures by Mark di Suvero, among others. This summer, the ranch will be open to those who are hiking or traveling by bike.Storm King Art Center (New Windsor)By far the most popular sculpture park in upstate New York, it is a 500-acre sculpture park in Hudson Valley. Since opening in 1960, it has grown to include dozens of sculptures that change over time. In its collection, the park owns sculptures by famous artists including Carl Andre, Louise Bourgeois, and Daniel Buren.Olympic Sculpture Park (Seattle)This outdoor park was created by the nearby Seattle Art Museum and features a large red sculpture by Alexander Calder called Eagle, as well as Wake by Richard Serra. Since 2007, this waterfront park has brought creativity to Elliott Bay. The landscape design fits in with the local roads and skyline, facing the harbor in what's recognized as Seattle's largest downtown green space.1、What can be learned about the garden in Manhattan's Chelsea neighborhood?A. It is run by Kasmin Gallery.B. It is designed by Zaha Hadid.C. It has become a part of the High Line.D. It stands for Alma Allen's respect for Utah.2、Which of the following parks is located in Hudson Valley?A. Kasmin Sculpture Garden.B. Tippet Rise Art Center.C. Storm King Art Center.D. Olympic Sculpture Park.3、Where can you see the sculpture Wake?A. In New York City.B. In Seattle.C. In New Windsor.D. In Fishtail.Matt Doogue, a 34-year-old nature photographer, had been suffering from depression when he first found his passion for taking pictures of insects and his work is now featured inhouse," says Doogue. "When I attempted to end my life, I knew I needed to see someone. I went to the doctors and got treatment, but I know that I needed something more and that's when I started photography."Now a dad of two, Doogue found that looking at insects through a camera helped him in ways he could never have imagined and it proved to be the lifeline he needed. It had a calming effect that helped him to disconnect from stress. And his astonishing images, showing insects and spiders in amazing detail against brightly colored backgrounds, caught the eye of publishers at National Geographic. "I ended up as one of their featured photographers," recalled Doogue. "It was the peak of my career. It was incredible."Originally from Salford, Greater Manchester, he now lives in Armadale, West Lothian, Scotland. Though he fears that Scotland is in the middle of an epidemic of male suicide, he believes that sharing his love of nature photography can help others to cope with their mental health issues as well. "I think the problem is this man-up approach; the idea that men need to be strong puts so much pressure on young males to be fine all the time," says Doogue. "This is why I try and be so open about my own experience. Whenever I am out with my camera, I don't think about my other worries. It is just me and the environment around me. You can lose yourself in a spider making its web."4、What does the underlined phrase "hit rock bottom" in paragraph 1 mean?A. Be in the worst possible situation.B. Reach the bottom of a valley.C. Crash into the lowest part of a rock.D. Launch an attack on the rock bottom.5、How did photography benefit Doogue?A. It gave him a new way to express himself.B. It helped him to escape from pressure.C. It provided him with life-saving skills.D. It offered him an opportunity to explore nature.6、What caused Scottish men to develop mental health problems?A. The way men employ to solve problems.B. The lack of love for men's life and work.C. The worry that men get separated from people.D. The belief that men are expected to be strong.7、What is the main purpose of the author in writing the text?A. To warn the severity of mental problems.B. To show the benefits of nature photography.C. To advise readers to get close to nature.D. To introduce Doogue's fighting against depression.The sun is setting, brightening your kids' faces as they play in the waves. You reach for your phone for this perfect moment. But before you do, here's a bit of surprising science: Taking photos is not the perfect way to keep memory as you think.Taking too many pictures could actually harm the brain's ability to keep memories, says Elizabeth Loftus, a psychology professor at the University of California, Irvine. So we get the photo but kind of lose the memory.Photography "outsources" memories. It works in two ways: We either shake off the responsibility of remembering moments when taking pictures, or we're so distracted (分散注意力的) by the process that we miss the moment altogether.The first explanation is the loss of memory. People know that their camera is recording that moment, so they don't try to remember. Similarly, if you write down someone's phone number, you're less likely to remember it offhand because your brain tells you there's just no need. That's all well and good—until that piece of paper goes missing.The other is distraction. We're distracted by the process of taking a photo—how we hold our phone, composing the photo, such as smiling faces, the background to our liking and clear image, all of which uses up our attention that could otherwise help us memorize.However, taking photos can benefit memory—when done mindfully. While taking a photo may be distracting, the act of preparation by focusing on visual details around has some upsides. When people take the time to zoom in (拉近镜头) on specific things, memories become strengthened.Another benefit is that we recall moments more accurately with the photos. Memory hasbeen reshaped with the help of new information and new experiences. Thus, photos or videos help us recall moments as if they really happened.Memories die away without a visual record backing them up. Therefore, a photo is an excellent tool to help remember when done purposefully, which is worth exploring further.8、What is the purpose of the first paragraph?A. To introduce the topic.B. To call on readers not to take photos.C. To show the interest in taking photos.D. To make us think of similar experience.9、Why does photography "outsource" memories?A. Photos are more detailed than memories.B. Taking photos is helpful for us to memorize.C. People depend more on photos to remember than their brains.D. Many sources influence people's memories during photo-taking.10、What may likely be discussed next?A. Situations when taking photos is better.B. How to stay focused while taking photos.C. When distraction is most likely to happen.D. How to use photo-taking to memorize better.11、Which of the following could be the best title for the text?A. Photography Does Help to MemoriesB. Too Many Photos Taken Results in Poor MemoriesC. Remember the Moment and Take Photos ProperlyD. The Fewer Photos We Take, the Better We Will RememberThe first model of Apple's iPhone was launched in June 2007. Since then, many different smartphones have been introduced. The devices now influence our daily lives in many ways. One thing that has changed is that many people now use their phones to easily take pictures without the need for a camera. Not surprisingly, this change has caused major business problems for camera manufacturers.Of course, the camera built into the first iPhone 15 years ago did not include a high-quality camera able to compete with separate camera models. But over the years, smartphone makers have invested heavily in research and development to change that. Today, many smartphones have high-quality cameras designed to produce better pictures. And most phonedevices also offer powerful tools to improve the quality of the pictures we take.Japan's Camera &Tmaging Products Association (CIPA) said the digital camera market continually expanded starting in 1999. It experienced its first decrease in 2009 —and continued to fall thereafter. The biggest change appeared from 2010 to 2020, when worldwide camera shipments fell about 93 percent, CIPA reported. The decreases were mainly caused by drops in shipments of digital cameras that have built-in lenses.However, camera makers have had more success selling digital cameras with interchangeable lenses. This is because these cameras are generally targeted at professional photographers who demand higher quality. Such cameras can produce "high image quality that distinguishes them from smartphones," CIPA said.But this does not mean that professional photographers never use smartphones to capture pictures. Brynn Anderson is based with the AP in Atlanta, Georgia. She said: "Sometimesphotographed. Using a phone makes it easier for me to get comfortable moments that might not happen otherwise." Rodrigo Abd, an AP photographer in Buenos Aires, Argentina says using the iPhone makes it easier for him "to always be attentive" to everyday events when not covering a news story. Oded Balilty is based in Tel Aviv, Israel. "It is definitely an alternative tool," he said of the iPhone. But he added: "It's the photographer not the device, that determines the quality of a photo."12、What is the potential cause of the first decrease of digital cameras in 2009?A. Less money was invested to improve digital cameras.B. Cameras had been built into smartphones and improved.C. Fewer digital cameras with built-in lenses were producedD. The digital camera market stopped promoting new products.13、Why do digital cameras with interchangeable lenses enjoy good sales.A. They are more affordable.B. They have superb shooting quality.C. They offer the power to beautify photos.D. They are specially designed for professionals.14、What does the underlined word "intimidating " in the last paragraph probably pean?A. Amusing.B. Demanding.C. Rewarding.D. Scaring.15、What does Oded Balilty mean?A. The level of the photographer depends on the iPhone.B. The iPhone completely replaces his professional tool.C. The professional skills of the photographer is crucial.D. The iPhone enables him to work at any time and place.二、七选五16、Art in 21th Century LifeThe word "art" usually brings to mind images of white-walled galleries, abstract paintings costing millions of dollars, far removed from our everyday experience. ①________ The Internet has changed the idea that art appreciation is only for the noble. ②_______ You can even visit several museums around the world using virtual reality headsets, without leaving your home.③_______ Art has always been a vehicle for self-expression, but social media have made it much easier to share amateur work with the whole world. Where an amateur artist or musician might one have shared their work with a circle of friends and family, they can now sell their work to anyone in the world."A picture is worth a thousand words" is the motto of data visualization enthusiasts. The amount of information available today can be overwhelming, so some statisticians (数据分析师) made it their mission to present this mass of data using infographics that are easy for the public to understand. ④________ Of course, graphs can be abused to mislead the audience, so we need to take care to interpret them the right way.Another way that art facilitates education is by helping us to conceptualize things that are invisible to the naked eye. ⑤_______ Likewise, the mind-bending concepts in physics such as black holes can be better understood with the help of illustrators who have backgrounds in both art and science.A. Many platforms of social media help teaching the public art.B. Yet art is indeed closer than many would believe it to be nowadays.C. As most of our environment is man-made, everything in it contains art.D. Netizens are not only consumers of art but creators and participants, too.E. Biology students would find videos such as "The inner Life of the Cell" helpful.F. They present information in visually appealing ways instead of using dry numbers.G. It has enabled more people than ever to have access to visual art and music of all types.三、完形填空(15空)Let me tell you a secret. There are no wrong answers when you're talking about art.engagement with an artwork is this: Do you like it or not?the Emperor's people.you don't like it, that's OK! It doesn't matter what other people say or think.An art historian called Ernst Gombrich believes that a viewer "completed" the artwork,17、A. giving B. saying C. communicating D. connecting18、A. back away B. put away C. break off D. carry off19、A. Now that B. In that C. Even though D. As though20、A. pain B. fear C. surprise D. regret21、A. belief B. argument C. concern D. reason22、A. simply B. probably C. largely D. usually23、A. fresh B. equal C. unique D. different24、A. warned B. told C. reminded D. cheated25、A. supporters B. ministers C. soldiers D. subjects26、A. recognize B. describe C. explain D. wonder27、A. heart B. innocence C. audience D. truth28、A. Obviously B. Actually C. Similarly D. Accordingly29、A. appreciate B. create C. comment D. study30、A. surprising B. relaxing C. puzzling D. striking31、A. viewing B. collecting C. describing D. understanding四、语法填空32、I loved art from a young age. However, I grew up in the countryside where there were not many ①____ (opportunity). But I was fortunate since my father was teaching at a high school. I spent my childhood playing in the schoolyard, reading at my father's office and②_____ (draw)in the classrooms. There was nothing much ③______ (play)with, but he hada lot of chalks. I would draw on the blackboard and later, the playground ④_____ (become)my canvas.At my elementary school, art classes were very simple, so I taught myself. I would look at picture stories, posters, and sometimes advertisements and make copies of ⑤______ (they). The turning point came when I was about 10 years old. My father took me to visit ⑥_____ artist, who was my father's art teacher and told me, "You can't just be copying. You should observe real objects, real things and draw from life."As he was talking, he drew a profile(侧影)of my father, ⑦______ looked totally like my father and I was ⑧_____ (surprise). I learned my first lesson in art from that experience—to draw from ⑨_____ (observe). In the mid-1990s, my paintings ⑩______ (exhibit)in the Dallas Fort Worth area and I started gaining recognition and awards.五、书面表达33、为让学生体验中国绘画艺术,感受中国画的魅力,学生会打算本周六组织英语俱乐部成员和交换生去参观中国画画展。

北师大版2019-2020学年度第一学期八年级物理期末试题(含解析)

北师大版2019-2020学年度第一学期八年级物理期末试题(含解析)

第1页 共8页 ◎ 第2页 共8页绝密★启用前北师大版2019-2020学年度第一学期八年级物理期末试题试卷副标题考试时间:90分钟;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第I 卷(选择题)请点击修改第I 卷的文字说明一、单选题1.你认为最符合实际的是 ( ) A .人体的正常体温为35℃ B .冰箱冷冻室的温度约为5℃ C.阳山盛夏中午室外温度可达36℃D .阳山最低气温可达零下30℃2.图中四个图像,能正确描述非晶体凝固的是A .B .C . D.3.从冰箱内取出的冰棍周围会弥漫着“白气”;水烧开后水壶嘴会喷出“白气”。

下列分析正确的是 A .冰棍周围的“白气”是冰熔化成的小水珠 B .这两种情况的“白气”都是水蒸气C .壶嘴喷出的“白气”是壶嘴喷出的水蒸气液化成的小水珠D .这两种情况的“白气”都是空气中的水蒸气液化而成的小水珠 4.如图使用厚刻度尺测量木块的长度,其中正确的测量图是A .B .C .D .5.小米同学在用托盘太平测物质质量时,她将天平调好估计物体的质量30g ,将物体和砝码分别放在左右盘中,发现指针向右偏,她应A .减小砝码B .增加砝码C .把平衡螺母向右调D .把平衡螺母向左调 6.对密度定义式mVρ=的下列几种解释中,属于正确表达的是 A .质量越大的物体,密度越大 B .密度与物体的体积成反比C .当一个物体的体积减少一半,其密度减少一半.D .密度与物体的体积、质量无关7.2017年4月,中国自主研制的首艘货运飞船“天舟一号”在海南文昌航天发射场使用“长征七号”运载火箭发射,并与“天宫二号”顺利完成自动交会对接,如图所示,对接完成后,若认为“天舟一号”处于静止状态,则选取的参照物是A .地球B .太阳C .“长征七号”D .“天宫二号”8.甲乙两同学沿平直路面步行,他们运动的路程随时间变化的规律如图所示,下列说法中不正确的是A .甲同学比乙同学晚出发4sB .4s ﹣8s 内,甲乙同学都做匀速直线运动C .0﹣8s 内.甲乙两同学运动的路程相等D .8s 末甲乙两同学速度相等第3页共8页◎第4页共8页9.如图,将正在发声的音叉缓慢接触用细线挂起的乒乓球,乒乓球会反复被弹开,利用这一实验可以研究A.声音传播的快慢B.音调是否与频率有关C.声音产生的原因D.声音能否在真空中传播10.小明看到如图所示斜插入水中的筷子向上折,如图所示的四幅光路图中,能正确说明产生这一现象的原因的是( )A.B.C.D.第II卷(非选择题)请点击修改第II卷的文字说明二、填空题11.自然界中的云、雨、雪、霜等现象,都是_____的物态变化形成的.其中雾和露都是由空气中的水蒸气遇冷_______(填一物态变化名称,下同)形成的小水滴;雪和霜都是由空气中的水蒸气遇冷_____形成的小冰晶.这些小水滴和小冰晶_____(填“吸热”或“放热”)后又会发生_____和_____,形成水蒸气,从而构成自然界的水循环,如下图所示.12.一千多年前,唐朝的大诗人李白曾感叹长江的壮美景观:“天门中断楚江开,碧水东流至此回,两岸青山相对出,孤帆一片日边来。

北师大版初中英语八上Unit1单元测试试题试卷含答案

北师大版初中英语八上Unit1单元测试试题试卷含答案

Unit 1 Television 单元测试一、单项选择(共15小题;共45分)1.A __________ is something that people watch on TV or listen to on the radio.A. projectB. programmeC. quiz2. He was really worried that his coach might ________ /kɪk/ him off the team.A. kickB. keyC. killD. kid从下面选项中选出可以替换划线部分的最佳选项。

3. When a journalist(记者)________ someone such as a famous person, they ask them a series of questions.A. interviewsB. followsC. teaches4. — Stop, Mary! There is a(n) ________ which asks us not to swim in the river.— OK. Thanks.A. expressionB. warningC. permission5. My grandmother worked 12 hours a day in the past, but ________ a little money.A. spentB. madeC. wasted6. Helen is learning to use a computer ________ to draw pictures.A. programB. patternC. space7. — Mom, I’m thirteen now. Can I ride a bike to school?— Yes, you can. But you must follow the traffic ________ on the way.A. rulesB. plansC. safetyD. accident8. — What’s your favorite ________?— It’s Animal World.A. newsB. foodC. planD. program9. —What’s your favorite ________?—Happy Camp.A. foodB. sportC. programD. subject10. If you ________ someone or something, you hit them forcefully(强有力的) with your foot.A. kickB. teachC. take11. Cannes ________ the annual film festival(年度电影节).A. hostsB. haveC. hostess12. — I ________ my English book.— Look! Is that your English book in the lost and found case?A. lostB. foundC. lookD. thank13. I ________ my computer game. I must find it.A. foundB. haveC. lost14. — My favorite TV ________ is A Bite of China(《舌尖上的中国》).— We like it, too. My mother often cooks delicious food after watching it.A. instructionB. guidebookC. entertainmentD. program15. A bad storm ________ the city. For this reason, the sports meeting was put off.A. beatB. fellC. hitD. fought二、完形填空(共1小题;共30分)阅读下面的短文,掌握其大意,然后从短文后各题所给的选项中,选择最佳选项。

2019_2020学年高中英语Unit6DesignPeriodSeven练习(含解析)北师大版必修2

2019_2020学年高中英语Unit6DesignPeriodSeven练习(含解析)北师大版必修2

Period Seven Grammar—Prepositions (time,place &movement)&Relativeclauses(Ⅰ)感知以下课文原句,完成方框下的小题(一)1.表示时间的介词有:句1中的during,句2中的between,句6中的in。

2.表示地点的介词有:句5中的near,句7中的in,句8中的above。

3.表示动作的介词有:句3中的along,句4中的across。

(二)1.定语从句可以用来修饰和限定正在被谈论的人,如句1的expert,句4的farmer,句6的parents,句8的people;或修饰和限定正在被谈论的物如句2的something,句3的tombs,句5的papercuts,句7的papercuts。

2.在定语从句中要用下列关系代词:(1)用来指代人的有:who(在从句中作主语如句4或宾语),whom(在从句中作宾语如句1和句8),that(在从句中作主语或宾语)。

(2)用来指代物的有:which/that(在从句中作主语如句3、句7或宾语如句2和句5)。

(3)用来表示所属关系的有:whose(在从句中作定语如句6)。

3.若关系代词which,that或who/whom在从句中作宾语而不作主语时,可以省略(如句1、句2和句5)一、表示时间、地点和动作的介词(一)表示时间的介词的区别1.at,in和onat后接时间点,即“在”钟点、(做某事的)时刻等,如at8o’clock,atnoon等。

in后接时间段,即“在”较长的一段时间内,如inthemorning,inthefuture等。

on后接特定的日子、具体的日期、星期几、节日等。

如onMonday,onChristmasmorning等。

2.since和from“since+时间点”意为“自从(过去某时)以来”,表示从过去到现在的一段时间的过程,常与现在完成时连用。

from表示从时间的某一点开始,不涉及与现在的关系。

新版北师大版八年级上B卷训练题(共10套)

新版北师大版八年级上B卷训练题(共10套)

训练题一(50分)一、填空题(每小题4分,共20分)1、点P(a-1,-b+2)关于x轴的对称点与关于y轴对称的点的坐标相同,则a,b的值分别是。

2、点Q(3-a,5 -a)在第二象限,则= .3. 如图,沿矩形ABCD的对角线BD折叠,点C落在点E的位置,已知BC=8㎝,AB=6㎝,那么折叠后的重合部分的面积是 .4.在平面直角坐标系中,已知A(2,-2),在坐标轴上确定一点P,使△AOP为等腰三角形,则符合条件的点P的坐标为 .5.等腰梯形ABCD中,AD//BC,对角线AC和BD相交于E,已知,∠ADB=60 ,BD=12,且BE∶ED=5∶1,则这个梯形的周长是_____________.第5题第6题二(本题8分)6.在西湖公园的售票处贴有如下的海报:(1)如果八年级(8)班27名同学去西湖公园开展活动,那么他们至少要花多少钱买门票?(2)你能针对该班参加活动各种可能的人数,设计合理的买票方案吗?三. (本题10)7.如图,在正方形ABCD中,E为AD的中点,BF=DF+DC.求证:∠ABE=∠FBC.四、(本题12分)8.如图,矩形OABC在平面直角坐标系内(O为坐标原点),点A在x轴上,点C在y轴上,点B的坐标分别为,点E是BC的中点,点H在OA上,且AH=,过点H且平行于y轴的HG与EB交于点G,现将矩形折叠,使顶点C落在HG上,并与HG上的点D重合,折痕为EF,点F为折痕与y轴的交点。

(1)求∠CEF的度数和点D的坐标;(4分)(2)求折痕EF所在直线的函数表达式;(3分)(3)若点P在直线EF上,当⊿PFD为等腰三角形时,试问满足条件的点P有几个?请求出点P的坐标,并写出解答过程。

(5分)(备用图)训练题二(50分)一、填空题(每小题4分,共20分)x的图象上,则点Q(a,3a﹣5)位于第_________ 21.在平面直角坐标系xOy中,点P(2,a)在正比例函数y=12象限.22.若一次函数y=kx+b,当﹣2≤x≤6时,函数值的范围为﹣11≤y≤9,则此一次函数的解析式为_________ .23.已知y=√4x−1+√1−4x+9,则√36x+y= _________.24.如图,已知在△ABC中,AD、AE分别是边BC上的高线和中线,AB=9cm,AC=7cm,BC=8cm则DE的长为_________ .25.如图,已知菱形ABC1D1的边长AB=1cm,∠D1AB=60°,则菱形AC1C2D2的边长AC1= cm,四边形AC2C3D3也是菱形,如此下去,则菱形AC8C9D9的边长= _________ cm.(四边都相等的四边形称为菱形)第4题图第5题图二、(8分) 26.小颖和小亮上山游玩,小颖乘坐缆车,小亮步行,两人相约在山顶的缆车终点会合.已知小亮行走到缆车终点的路程是缆车到山顶的线路长的2倍.小颖在小亮出发后50min 才乘上缆车,缆车的平均速度为180m/min.设小亮出发x min后行走的路程为y m,图中的折线表示小亮在整个行走过程中y与x的函数关系.(1)小亮行走的总路程是m,他途中休息了min;(2分)(2)①当50<x<80时,求y与x的函数关系式;②当小颖到达缆车终点时,小亮离缆车终点的路程是多少?三、(10分)27.甲、乙两家体育用品商店出售同样的乒乓球拍和乒乓球,乒乓球拍每付定价20元,乒乓球每盒定价5元。

新教材北师大2019高一上英语期中考试卷

新教材北师大2019高一上英语期中考试卷

北师大2019新教材2020-2021学年第一学期高一英语期中考试英语试题第一部分听力(共两节,满分30分)第一节(共5小题;每小题1. 5分,满分7. 5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What did the woman buy for her husband?A. A stamp.B. A tie.C. A wallet.2. Where does the conversation most probably take place?A. In a hotel.B. In a bank.C. In a post office.3. How much should the man pay if he rents the car for two days?A. $ 19.B. $38.C. $ 129.4. What does the woman mean?A. She prefers to put on more clothes.B. She wants to leave the windows open.C. She doesn't mind closing the windows.5. Why does the woman like the black house better?A. Its yard is bigger.B. It's more beautiful.C. It has more rooms.第二节(共15小题;每小题1. 5分,满分22. 5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

北师大版2020届中考英语模拟试卷(十二)B卷

北师大版2020届中考英语模拟试卷(十二)B卷

北师大版2020届中考英语模拟试卷(十二)B卷姓名:________ 班级:________ 成绩:________一、语法选择(共15小题;每小题1分,满分15分) (共1题;共15分)1. (15分) I like my bike. It brings me a lot of fun. When I was an1boy, my father bought it for me. It is my “horse”. I often go to school2my bike. I like riding a bike 3 it is very convenient. If I go out by bike, it is 4 to find a place for parking it. Bikes are very cheap, so most people can buy one. There are three people in my family. 5of us has a bike. My parents are teachers in my school. In the morning, Mum, Dad and I each 6 to school. However on a rainy day we walk to school, because the road is bad for riding. I think riding bikes can make us7 . It not only helps us exercise more but also 8 the air clean. The bike is very important in Chi nese people’s life. So riding bikes can help you to 9 more about China and the Chinese people. Now people in the world say China is the country of bikes because there are 10 bikes in China. I hope people all round the world like them.(1)A . eight years oldB . eight year oldC . eight-year-oldD . eight-years-old(2)A . byB . onC . fromD . in(3)A . soB . butC . whileD . because(4)A . hardB . lateC . happyD . easy(5)A . EachB . BothC . NoneD . Some (6)A . walkB . runC . driveD . ride (7)A . strangeB . busyC . strongD . poor (8)A . feelsB . keepsC . seesD . likes (9)A . tellB . knowC . visitD . ask (10)A . so manyB . so muchC . too much二、完形填空(共10小题;每小题1.5分,满分15分) (共1题;共10分)2. (10分)完形填空On May Day holiday of this year, my parents and I went to the Great Wall. On the way I felt 1, because this was my first visit. I had dreamed 2 this trip many times. We took a train first, and then we took a bus to 3 there. When it appeared in front of me, I couldn't help 4. I said, "What a wonderful 5 it is!"When I walked on the Great Wall, I felt very proud because it is 6wall in the world. At the same time, I couldn't believe my eyes. I wondered how the ancient people built it by 7. I wondered how many people took part in building it. It was a mystery to me.I almost lost 8 in the beautiful scenery around it. "He 9 doesn't reach the Great Wall is nota 10 man. Now I'm on the top! I'm a true man," I said loudly.But, 11 there are many visitors every day, there are many problems for the Great Wall. For example, many visitors throw 12 about or spit on the Great Wall and some visitors 13 carved their names on it.As visitors, I think we 14 try our best to protect the Great Wall, not to destroy it. It's our duty. The past is not only for us to enjoy 15 for the people in the future.(1)A . worriedB . excitedC . lonelyD . proud(2)A . ofB . atC . onD . in(3)A . arrive atB . reachC . arrive inD . get(4)B . smilingC . jumpingD . reading (5)A . pictureB . sightC . holidayD . trip(6)A . the tallestB . the highestC . the farthestD . the longest (7)A . handB . machineC . toolD . foot(8)A . meB . myselfC . oneselfD . itself(9)A . whatB . whenC . whoD . how(10)A . falseB . strongC . realD . true (11)A . whenB . whileC . asD . why (12)A . rubbishB . stoneC . foodD . juice (13)A . everB . neverC . evenD . always (14)A . mayB . canC . wouldD . should (15)A . butB . but alsoC . alsoD . and三、阅读理解(共20小题;每小题2分,满分40分) (共4题;共40分)3. (10分)阅读理解With the development of economics(经济), the problem of left-behind children(留守儿童) has become a serious social problem. More and more people haveRealized we should do something to help them. One of the biggest problems is that the children are all hungry for the love from their parents. Many of them can just get a call or a letter from their parents half a year, a year or even several years, which makes some children lost their parent's faces. When they meet their own troubles they have no one to talk with. Nobody gives them enough care. Because of being too far away from their parents, the left-behind children's study is the second biggest problem. From a survey(调查)about their study, only two percent of the children get good results, while ten percent of them are common and eighty-eight percent of them are poor. What a pity!There are also many other problems of the left-behind children, such as having no ability to protect themselves, unhealthy lifestyles, spending too much pocket Money and being crazy about Internet and so on. In a word, to solve all the problems of the left -behind children needs the government, families and schools to try their best together.(1)How many problems are mentioned in the passage?A . Two.B . Three.C . Four.D . More than four.(2)Which problem is NOT mentioned in the passage?A . Having no place to live.B . Being crazy about the Internet.C . Unhealthy lifestyles.D . Not study well.(3)What's the meaning of "some children lost their parents faces"A . Some children lost their parents.B . Some children couldn't find their parents.C . Some children can't remember what their parents look like.D . Some children can't get any information about their parents.(4)Which of the following sentences is RIGHT?A . All of the left-behind children can see their parents often.B . The left-behind children can see their parents often.C . The left-behind children have much money.D . The left-behind children problem is a social problem.(5)What is the best title of the article?A . The Left-behind ChildrenB . What is Left-behind ChildrenC . The Problems of the Left-behind ChildrenD . How to help the Left-behind Children4. (10分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中选出最佳选项。

北师大版2019-2020学年八年级上册期末复习解答题常考题型综合练习题(解析版)

北师大版2019-2020学年八年级上册期末复习解答题常考题型综合练习题(解析版)

北师大版2019-2020学年八年级上册期末复习解答题常考题型综合练习题1.已知,在△ABC中,∠ACB=90°,CD⊥AB垂足为D,BC=6,AC=8,求AB与CD的长.2.解二元一次方程组3.解方程组4.解方程组5.解方程组:6.如图是一块地的平面图,AD=4 m,CD=3 m,AB=13 m,BC=12 m,∠ADC=90°,求这块地的面积.7.如图,甲、乙两艘轮船同时从港口O出发,甲轮船以20 海里/时的速度向南偏东45°方向航行,乙轮船向南偏西45°方向航行.已知它们离开港口O两小时后,两艘轮船相距50海里,则乙轮船平均每小时航行多少海里?8.如图,笔直的公路上A、B两点相距25km,C、D为两村庄,DA⊥AB于点A,CB⊥AB于点B,已知DA=15km,CB=10km,现在要在公路的AB段上建一个土特产品收购站E,使得C、D两村到收购站E的距离相等,则收购站E应建在离A点多远处?9.如图,某公路上A,B两点的正南方有D,C两村庄,现要在公路AB上建一个车站E,使C,D两村到E 站的距离相等,已知AB=50km,DA=20km,CB=10km,请你设计出E站的位置,并计算车站E距A点多远?10.如图,有一个直角三角形纸片,两直角边AC=18cm,BC=24cm,现将直角边AC沿直线AD折叠,使它落在斜边AB上,且与AE重合,你能求出BD的长吗?11.如图,四边形ABCD中,AB=BC=4,CD=6,DA=2,且∠B=90°,求:(1)AC的长;(2)∠DAB的度数.12.在一次消防演习中,消防员架起一架25米长的云梯AB;如图斜靠在一面墙上,梯子底端B离墙角C的距离为7米.(1)求这个梯子的顶端距地面AC有多高?(2)如果消防员接到命令,按要求将梯子底部在水平方向滑动后停在DE的位置上(云梯长度不变),测得BD长为8米,那么云梯的顶部在下滑了多少米?13.如图,某会展中心在会展期间准备将高5m,长13m,宽2m的楼梯上铺地毯,已知地毯每平方米18元,请你帮助计算一下,铺完这个楼道至少需要多少元钱?14.如图,某中学有一块四边形的空地ABCD,学校计划在空地上种植草皮,经测量∠A=90°,AB=3m,BC=12m,CD=13m,DA=4m,若每平方米草皮需要200元,问学校需要投入多少资金买草皮?15.李明准备租用一辆出租车搞个体营运,现有甲乙两家出租车公司可以和他签订合同,设汽车每月行驶x 千米,应付给甲公司的月租费y1元,应付给乙公司的月租费是y2元,y1、y2与x之间的函数关系的图象如图所示,请根据图象回答下列问题:(1)当汽车每月行驶的路程________时,甲乙两家公司的月租费一样;当汽车每月行驶的路程________时,甲公司的月租费比乙公司的月租费高.(2)分别求出y1、y2与x之间的函数关系式.16.已知一次函数y=kx+b的图象经过点(-1,-5),且与正比例函数y= x的图象相交于点(2,a),求:(1)a的值.(2)k,b的值.(3)这两个函数图象与x轴所围成的三角形的面积。

第一章 直线与圆 单元测试 2024-2025学年高二上学期数学北师大版(2019)选择性必修第一册

第一章 直线与圆 单元测试 2024-2025学年高二上学期数学北师大版(2019)选择性必修第一册

第一章 直线与圆 单元测试一、单选题1.若直线l 斜率为k ,向量在直线l 上,且向量在方向上的投影的模是其在方向上投影的模的2倍,则该直线的斜率k 的值为( )A .2B .C .D .2.已知圆:与圆:关于直线对称,则的方程为( )A .B .C .D .3.已知直线与圆:()交于A ,两点,且线段关于圆心对称,则( )A .1B .2C .4D .54.已知点,,,点是直线上的动点,若恒成立,则最小正整数( )A .1B .2C .3D .45.已知定点和直线,则点到直线的距离的最大值为( )A .BC .D .6.若点P 在直线上,点Q 在圆上,则线段PQ 长度的最小值为( )A .B .C .D .7.莱莫恩定理指出:过的三个顶点作它的外接圆的切线,分别和所在直线交于点,则三点在同一条直线上,这条直线被称为三角形的线.在平面直角坐标系中,若三角形的三个顶点坐标分别为,则该三角形的线的方程为( )A .B .C .D .8.直线l 过点,则直线l 的方程为( )A .B .C .D .二、多选题9.已知直线与圆交于,两点,点为线段的中点,且点的坐标为.当)A .B .的最小值为C .存在点,使D.存在,使10.下列说法正确的是( )A .已知直线过点,且在轴上截距等于轴上截距2倍,则直线的方程为B .直线没有倾斜角C .,,“直线与直线垂直”是“”的必要不充分条件D .已知直线的斜率满足,则它的倾斜角的取值范围是或11.已知直线l ∶x +y -2-a =0在x 轴和y 轴上的截距相等,则a 的值可以是( )A .0B .1C .-1D .-2.三、填空题12.已知斜率均为负的直线与直线平行,则两条直线之间的距离为 .13.已知圆和圆,M 、N 分别是圆C 、D 上的动点,P 为x 轴上的动点,则的最小值是 .14.过点,且与直线垂直的直线方程是.四、解答题15.圆内有一点,AB 为过点P 且倾斜角为的弦.(1)当时,求AB 的长;(2)当弦AB 被点P 平分时,写出直线AB 的方程.16.圆过、两点,且圆心在直线上.(1)求圆的方程;(2)若直线在轴上的截距是轴上的截距的2倍,且被圆截得的弦长为6,求直线的方程.17.已知动点与点的距离是它与原点的距离的2倍.m m ()1,0i =()0,1j = 122±12±M ()2211x y ++=N ()()22231x y -+-=l l 210x y --=210x y -+=230x y +-=230x y +-=20x y r -+=C ()()22213x y r ++-=0r >B AB r =()0,1A ()10B ,(),0C t M AC 2MA MB ≤t =()2,0P -()():34330l m x y m m ++-+=∈R P l d 34120x y +-=221x y +=12575175225()Lemoine ABC V ,,A B C BC,CA,AB ,,P Q R ,,P Q R Lemoine xOy ()()()0,1,2,0,0,4A B C -Lemoine 2320x y --=2380x y +-=32220x y +-=23320x y --=(1,1),(2,4)A B -2y x =-2y x =--2y x =-+2y x =+:0(R)l mx y m m --=∈222:()0O x y r r +=>A B Q AB T (3,0)1m =2r =AB A 45ATO ∠=︒m 54QO QT ⋅=-l ()2,1P x y l 240x y +-=10x +=R a ∈R b ∈210ax y +-=()1210a x ay +-+=3a =l k 11k -≤<α045α≤< 135180α≤< :0l bx ay +=:20m ax by a ++=()22:21C x y +-=22:610300D x y x y +--+=PM PN +()1,5-126x y+=228x y +=()1,2P -α3π4α=C ()0,3()4,5C 80-+=x y C l x y C l (,)M x y (3,0)P O(1)求动点的轨迹的方程;(2)求的最小值;(3)经过原点的两条互相垂直的直线分别与轨迹相交于,两点和,两点,求四边形ACBD 的面积的最大值.M E x y O E A B C D S参考答案1.D【分析】设出,求出向量在和方向上的投影的模,从而得到,求出直线斜率.【详解】设,则向量在方向上的投影的模为,向量在方向上的投影的模为,则,故该直线的斜率.故选:D 2.C【分析】根据两点的坐标,求其中点坐标以及斜率,根据对称轴与两对称点连接线段的关系,可得答案.【详解】由题意得,,则的中点的坐标为,直线的斜率.由圆与圆关于对称,得的斜率.因为的中点在上,所以,即.故选:C.3.D【分析】先求得圆心的坐标,进而列出关于的方程,解之即可求得的值.【详解】圆:的圆心,由圆心在直线上,可得,解之得.故选:D 4.D【分析】先设出,得到的方程为:,由得到圆的方程,结合点到直线的距离公式,求出的最小值即可.【详解】设,由在上,得:,即,由得:,化简得,依题意,线段与圆,至多有一个公共点,故(),m a b = m()1,0i =()0,1j = 2a b=(),m a b =m()1,0i =m ia i⋅=m ()0,1j =m jb j⋅= 2ab =12b k a ==±()0,1M -()2,3N MN ()1,1MN 31220MNk +==-M N l l 112l MN k k -==-MN l ()1112y x -=--230x y +-=C r r C ()()22213x y r ++-=(1,3)C -(1,3)C -20x y r -+=230r --+==5r (,)M x y AM 0x ty t +-=2MA MB ≤t (,)M x y M AC 1xy t+=0x ty t +-=2MA MB ≤()2222(1)41x y x y ⎡⎤+-≤-+⎣⎦22418((339x y -++≥AM 22418()()339x y -++=41,33⎛⎫- ⎪⎝⎭解得:,是使恒成立的最小正整数,由于,故选:D5.B【分析】先求得直线所过定点,然后根据两点间的距离公式求得正确答案.【详解】直线,即,由解得,所以直线过定点,所以的最大值为故选:B 6.B【分析】求出圆的圆心和半径,判断直线与圆的位置关系,则线段PQ 长度的最小值为圆心到直线的距离减去半径即可.【详解】圆的圆心为,半径,因为圆心到直线的距离为,所以线段PQ长度的最小值为.故选:B 7.B【分析】待定系数法求出外接圆方程,从而得到外接圆在处的切线方程,进而求出的坐标,得到答案.【详解】的外接圆设为,,解得,外接圆方程为,即,易知外接圆在处切线方程为,又,令得,,,在处切线方程为,又,令得,,则三角形的线的方程为,即故选:B.8.D2t ≥2t ≤t 2MA MB ≤324<<4t ∴=l ()():34330l m x y m m ++-+=∈R ()33430m x x y +++-=303430x x y +=⎧⎨+-=⎩33x y =-⎧⎨=⎩l ()3,3Q -d =221x y +=(0,0)O 1r =34120x y +-=1215d ==>127155-=,A C ,P R ABC V 220x y Dx Ey F ++++=104201640E F D F E F ++=⎧⎪∴++=⎨⎪-+=⎩034D E F =⎧⎪=⎨⎪=-⎩∴22340x y y ++-=2232524x y ⎛⎫++=⎪⎝⎭A 1y =:124x y BC +=-1y =52x =,152P ⎛⎫∴ ⎪⎝⎭()0,4C -4y =-:12xAB y +=4y =-10x =()10,4R ∴-Lemoine 410514102y x +-=+-2380x y +-=【分析】根据直线的两点式方程运算求解.【详解】因为,则线l 的方程为,整理得,所以直线l 的方程为.故选:D.9.AD【分析】利用圆的弦长公式判断A 、B ;假设存在点,求出直线方程,判断与圆的位置关系,判断C ,求出点的轨迹方程,可判断D.【详解】当时,直线,圆心到直线的距离,又,解得,A 正确;由上可知圆,圆心到直线的距离,则,B 错误;若,则直线斜率为,从而直线:,此时圆心到直线的距离,则直线与圆相离,即不存在点,使,C 错误;设点,因为直线过定点,则,即,化简为,为点的轨迹方程,若,则,即,得,故存在存在,使,D 正确.故选:AD.10.CD【分析】根据截距的概念可判定A ,根据倾斜角的定义可判定B ,利用两直线垂直的位置关系可判定C ,根据倾斜角与斜率的关系可判定D.【详解】对于A ,当直线在两个坐标轴的截距都是0时,显然直线方程为,故A 错误;B ,直线倾斜角是,故B错误;对于C ,若直线与直线垂直,则有或,所以不满足充分性,反之时,此时两直线垂直,满足必要性,故C 正确;对于D ,由直线的斜率与倾斜角的关系知:12,14-≠≠()()114121x y ---=---2y x =+2y x =+A AT Q 1m =:10l x y --=O d AB ===2r =22:4O x y +=O d ==AB ===>45ATO ∠=︒AT 1-AT 30x y +-=O 2d r >=AT O A 45ATO ∠=︒(),Q x y ():1(R)l y m x m =-∈()1,0C 222OQ QC OC +=()2222211x y x y ++-+=221124x y ⎛⎫-+= ⎪⎝⎭Q 54QO QT ⋅=- ()2534x x y -⋅-+=-()2534x x x x -⋅-+-=-[]50,18x =∈m 54QO QT ⋅=- 12y x =10x +=90 210ax y +-=()1210a x ay +-+=()1400a a a a +-=⇒=3a =3a =k满足的直线,则它的倾斜角的取值范围是或,故D 正确.故选:CD 11.ABCD【分析】求出两坐标轴上的截距,进而判断的可能取值.【详解】令y =0,得到直线在x 轴上的截距是,令x =0,得到直线在y 轴上的截距为2+a ,∴不论a 为何值,直线l 在x 轴和y 轴上的截距总相等.故选:ABCD.12.33/133【分析】利用斜率为负的两直线平行,找到,表示出直线,利用两平行线间的距离公式计算即可.【详解】因为斜率均为负的直线与直线平行,所以同号,且,解得:,所以直线与直线,所以这两条直线之间的距离为.13【分析】先得到,当且仅当三点共线,且三点共线时,等号成立,设C 关于x 轴的对称点,求出的最小值,进而得到的最小值.【详解】的圆心为,半径为1,,圆心为,半径为2,结合两圆位置可得,,当且仅当三点共线,且三点共线时,等号成立,设C 关于x 轴的对称点,连接,与轴交于点,此点即为所求,此时,即为的最小值,故的最小值为11k -≤<α045α≤< 135180α≤< a 2a +a =:0l bx ay +=:20m ax by a ++=,a b 02b a a b a=≠a =:0l x +=:10m x +=d ==3-3PM PN PC PD +≥+-,,P M C ,,P N D ()0,2C '-PC PD +PM PN +()22:21C x y +-=()0,2C ()()2222:610300354D x y x y x y +--+=⇒-+-=()3,5D 3PM PN PC CM PD DN PC PD +≥-+-=+-,,P M C ,,P N D ()0,2C '-CD'x P C D =='PC PD +PM PN +3314.【分析】根据垂直求出斜率,再由点斜式方程可得答案.【详解】直线的斜截式为,故斜率是,所以所求直线的斜率是,所以所求直线方程是,即.故答案为:.15.(2)【分析】(1)根据倾斜角以及求解出直线的方程,再根据半径、圆心到直线的距离、半弦长构成的直角三角形求解出;(2)根据条件判断出,结合和点坐标可求直线的方程.【详解】(1)圆的圆心,半径因为,所以直线的斜率,所以,即,所以圆心到的距离所以(2)因为弦被平分,所以,又因为,所以,所以,即.16.(1),,【分析】(1)先求得两点,的中垂线方程,再与联立,求得圆心即可;(2)先由直线且被圆截得的弦长为6,求得圆到直线的距离,再分截距为零和不为零求解.【详解】(1)解:两点,的中垂线方程为:,联立,解得圆心,则,故圆的方程为:;(2)由直线且被圆截得的弦长为6,故圆心到直线的距离为,3160x y -+=126x y+=36y x =-+3-13()1513y x -=+3160x y -+=3160x y -+=250x y -+=()1,2P -AB AB OP AB ⊥AB k P AB 228x y +=()0,0O r =3π4α=AB 3πtan14AB k ==-()()():211AB y x -=-⨯--:10AB x y +-=O AB d AB ===AB P OP AB ⊥20210OP k -==---12AB k =()()1:212AB y x -=--:250AB x y -+=()22825x y +-=0y ±=2160x y +--=2160x y +-+=()0,3()4,580-+=x y l C C l ()0,3()4,5280x y +-=80-+=x y ()0,8C =5r C ()22825x y +-=l C C l 4d =A .若直线过原点,可知直线的斜率存在,设直线为:,此时直线的方A .若直线不过原点,设直线为:,此时直线的方程为:,综上:直线,,.17.(1)(2)(3)7【分析】(1),根据两点间的距离公式化简可得方程;(2),法一:换元后与圆的方程联立,利用判别式法求解最小值;法二:几何法,利用直线与圆的位置关系列不等式求出最小值;法三:三角换元,结合辅助角公式利用余弦函数的性质求解最小值;(3),根据直线是否存在斜率进行分类讨论,当直线存在斜率时,利用点斜式写出两直线的方程,分别求出弦长,将四边形的面积用弦长表示,即可求出最大值.【详解】(1)由已知得化简得,即,所以动点的轨迹的方程为:;(2)法一:设,得,代入轨迹的方程消去并整理得,∴,即,解得故的最小值为;法二:设,即,由(1)的结论可知,轨迹是以点为圆心,半径长为2的圆,由题意可知,直线和圆有公共点,所以圆心到直线的距离不大于半径,即,解得故的最小值为;法三:由(1)可设,,则,因为,所以当时,y kx =4d k ==⇒=l 0y ±=12202x yx y a a a+=⇒+-=48d a ⇒=±l 2160x y +--=2160x y +-+=l 0y ±=2160x y +--=2160x y +-+=22(1)4x y ++=1--=22230x y x ++-=22(1)4x y ++=M E 22(1)4x y ++=x y t -=y x t =-E y ()2222(1)30x t x t +-+-=()22Δ4(1)830t t =---≥2270t t +-≤11t --≤≤-+x y -1--x y t -=0x y t --=E (1,0)-0x y t --=22(1)4x y ++=2≤11t --≤≤-+x y -1--12cos 2sin x y θθ=-+⎧⎨=⎩(02π)θ≤<π12cos 2sin 14x y θθθ⎛⎫-=-+-=-++ ⎪⎝⎭πcos 14θ⎛⎫+≥- ⎪⎝⎭3π4θ=的最小值为;(3)i )若两直线都有斜率,可设直线AB 的方程为,则直线CD 的方程为,由(1)的结论可知,轨迹是以点为圆心,半径长为2的圆.到直线AB 的距离同理,所以,ⅱ)若AB 、CD 两直线中有一条没有斜率,则另一条的斜率为0,此时线段AB 、CD 的长分别为4(或4、,所以.综上所述,当且仅当,即时,四边形ACBD 的面积取得最大值,最大值为7.x y -1--(0)y kx k =≠1=-y x kE 1(1,0)O -1O d =||AB ==CD ==11||||22S AB CD ==⨯==7=≤1||||72S AB CD ==<21112k =+1k =±S。

第三章 整式及其加减B卷压轴题考点训练(解析版)(北师大版,成都专用)

第三章 整式及其加减B卷压轴题考点训练(解析版)(北师大版,成都专用)

第三章整式及其加减B卷压轴题考点训练一、填空题=,=1212240a a an a=++-()216na a =-元,∴当20n >时,该户应缴纳的水费为()216na a -元;故答案为:()216na a -;(3)解:∵12224´=,∴12x >,当1220x <£时,甲用水量超过312m 但不超过320m ,乙用水量超过320m ,∴()()()12212 1.5212220122 1.5402022x x ´+-´´+´+-´´+--´´243362424804x x =+-+++-()116x =-元;当2028x <<时,甲的用水量超过320m ,乙的用水量超过312m 但不超过320m ,∴()()()1222012 1.522022122401223x x ´+-´´+-´´+´+--´´242448024843x x=++-++-()76x =+元,当2840x ££时,甲的用水量超过320m ,乙的用水量不超过312m ,∴()()()1222012 1.522022402x x ´+-´´+-´´+-´2424480802x x=++-+-()248x =+元;综上所述,当1220x <£时,甲,乙两户一个月共缴纳的水费()116x -元;当2028x <<时,甲,乙两户一个月共缴纳的水费()76x +元;当2840x ££时,甲,乙两户一个月共缴纳的水费()248x +元.【点睛】本题主要考查了有理数的四则混合计算的实际应用,整式加减计算的实际应用,正确理解题意利用分类讨论的思想求解是解题的关键.。

最新版2019-2020年北师大版九年级英语全册上学期11月份月考检测题及答案-精编试题

最新版2019-2020年北师大版九年级英语全册上学期11月份月考检测题及答案-精编试题

北师大版初三上学期11月份月考英语模拟试卷时间:90分钟分数:120分一、听力测试(本题共30分)听录音,每题读两遍Ⅰ、听句子,选出与所听句子相匹配的图画。

(本题共5分,每小题1分)(略)Ⅱ、听对话,根据问题选出正确答案。

(本题共10分,每小题2分)()6.A.At the train station B. At the bus stationC.At the airport()7.A.He likes it. B. He doesn’t like it. C. It’s too quiet.()8.A.Jesssica. B.Ben. C.Susan.()9.A.On foot. B.By bus C.In a car()10.A.Sue B.Bob C.Nobody.Ⅲ、听对话,根据对话内容选出最佳答案完成各句。

(本题共5分,每小题1分)()11.Jenny has been looking after ____girl these days.A. a six-year-oldB. a seven-year-oldC. aneight-year-old()12.The girl’s parents are____ now.A. in the UKB. in the USAC. in France()13.The little girl didn’t go with her parents because____.A. her parents were too busy to look after herB. she liked to stay with JennyC. she had to go to school here()14.Jenny and the girl would like to watch some____.A. cartoon filmsB. action filmsC. science fiction films ()15.Actually, it’s _____for Jenny to look after the girl.A. boringB. funC. badⅥ、听短文,根据所听内容选出短文中画线部分的同义词或同一短语。

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北师大版2019-2020学年八年级上学期英语期末考试试卷B卷姓名:________ 班级:________ 成绩:________一、单选题 (共10题;共20分)1. (2分)—kind of newspapers do you like?—English newspapers.A . WhatB . WhoC . HowD . When2. (2分)Tom is ________ building a new factory in the city. It will pollute the water.A . forB . againstC . disagreeD . agree3. (2分)Of the twin sisters, Tracy is _______.A . the tallerB . the tallC . tallerD . the tallest4. (2分)Which do you like _____, coffee, tea or milk?A . moreB . bestC . the betterD . better5. (2分)— She _______ herself in the accident.— Really? Do I need ________her?A . touched; visitB . hurt; visitC . touched; to visitD . hurt; to visit6. (2分)He did ________ for his children. What a good father!A . so muchB . manyC . so manyD . too many7. (2分)- There are some boys ______ basketball over there.- Let's go and join them.A . are playingB . playingC . to play8. (2分)— Let's to have lunch.— Sorry, I'm not .A . go; fullB . to go; fullC . go; hungryD . to go; hungry9. (2分)Grace _________ this game every time we play.A . winsB . wonC . will winD . has won10. (2分)—______ are you going with for vacation?—My parents.A . HowB . WhomC . WhatD . Why二、完形填空 (共1题;共10分)11. (10分)根据短文理解, 从A、B、C三个选项中选择最佳答案填空。

Billy is a boy of fifteen. His parents 1 three years ago. One day when he was walking in the street, he found a wallet. He 2 it to the owner, Mr. Baker. Mr. Baker gave his thanks to the boy. As the boy had no job, Mr. Baker made him 3 for him in his factory. Billy worked so 4 that Mr. and Mrs. Baker were happy with him.Mr. Baker loved planting trees. The week before last, he 5 a few trees home, planted them in the garden himself and watered them every day. Several days 6, he had to leave for another city. 7 he started, he said to Billy. "Take good care of the trees. Some boys near our house always want to steal 8." "Don't worry about them, sir." answered Billy, "I'll try my best 9 them." Six days passed and Mr. Baker came back. He asked, "Has anyone ever come to steal the trees?" "No, sir." said Billy, "To stop someone from 10 the trees, I pulled them up six days ago. I have hidden them for almost a week!"(1)A . deadB . was deadC . died(2)A . turnedB . returnedC . gave(3)A . workingB . workedC . work(4)A . goodB . badC . hard(5)A . was bringingB . broughtC . brings(6)A . agoB . laterC . before(7)A . As soon asB . WhenC . Before(8)A . itB . themC . one(9)A . watchingB . watchedC . to watch(10)A . stealingB . cuttingC . hurting三、阅读理解 (共4题;共51分)12. (6分)阅读下面短文,从各题所给的A、B、C、D四个选项中选出最佳选项。

Safe Travel AbroadPreparing to go traveling abroad can be really stressful, especially if you don't know what to expect. Here is a list of important things to remember, so you know exactly what you need to do before you take off.What vaccines(疫苗) do I need?Go to your doctor for advice as soon as you know you're going traveling. They will know what vaccines you will need and will be able to advise you on when you needA . work in insurance companiesB . need medical treatmentC . plan to travel abroadD . want to take vaccines(2)The text above may be from the part of in a newspaper.A . newsB . lifeC . sportsD . advertisement(3)The main purpose of the text is to__________.A . help you make a travel planB . tell you how to keep safe while travelingC . give you some advice on travel preparationsD . suggest you buy insurance for your family13. (10分)阅读理解(A)Jim went to school to study history, but when he finished his first year, he didn't pass the examand the teacher said to him, “You have to leave school.”When the boy's father heard this news, he was much worried. He came to the school to see the teacher. He found the teacher in the office and asked the teacher to let Jim go on with his studies the next year.“He is a good boy. He always does well in history,” said Jim's father, “and if you let him pass the exam this time, he will do better next year and pass the exam.”“No, no, that's impossible!” answered the teacher, “Last mo nth I asked him when Napoleon(拿破仑)died and he didn't know.”“Please sir, give him another chance.” said Jim's father, “You know we don't have any newspapers in our house, so nobody in our family knew that Mr. Napoleon was ill.”(1)The boy in this story was good at history.(2)The boy's father went to the teacher's office and talked with him about his son.(3)The teacher let all his students pass the exam.(4)After that, the teacher let the boy pass the exam.(5)The boy's father was not good at history.14. (10分)钢琴家朗朗是世界古典音乐新一代领军人物,被誉为“当今世界最年轻的钢琴大师”.Lang Lang, called “Chinese Mozart”, one of the best-known Chinese pianists in the world, was born in Shenyang in 1982. He began to take piano lessons at the age of three. He played the piano at least 20 hours a week at that time. At the age of five, he won the Shenyang Piano Competition. Four years later, his father took him to Beijing to study at the Beijing Central Music Conservatory(中央音乐学院). When he was 11, he won the first prize at the Fourth International Young Pianists Competition in Germany. In 1996, he went to the United States to study music. Soon he became famous all over the US, and then around the world.The road to success has never been easy. Lang Lang's father even stopped his job in his police office. For over ten years, he followed Lang Lang everywhere he went, not only as a father, but also as a manager, tutor and friend, while his mother stayed in Shenyang to make money. In spite of(尽管) giving on average over 150 performances around the world every year now, he still finds time to practice and learn new works. Lang Lang considers himself lucky and believes he should give something back. He has helped the children in poor areas a lot.(1)How old was Lang Lang when he went to Beijing to study?A . He was threeB . He was fiveC . He was nineD . He was eleven.(2)Why did Lang Lang’s father take him to Beijing?A . They went on vacation to BeijingB . Their family needed money at that time.C . Lang Lang wanted to take part in the Piano CompetitionD . Lang Lang went there to study at a famous music conservatory.(3)Where was the fourth International Youth Piano Competition held?A . In Germany.B . In the United StatesC . In JapanD . In China.(4)Which is NOT TRUE according to the passage?A . Lang Lang went to the United States to study music at the age of 14.B . Lang Lang's father had to stay in Shenyang to make money.C . Lang Lang's parents have done a lot for him to be a famous pianistD . Lang Lang is called “Chinese Mozart” because he plays the piano very well.(5)What can we learn from this passage?A . It's not very difficult to be a pianist.B . A pianist is always popular in the world.C . You have to practice and learn much to be successful.D . Pianist s never hate playing the piano because it’s interesting.15. (25分)阅读下面短文,简要回答所给问题。

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