2022年2009年中考数学复习教材回归知识讲解 .pdf
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2022年2009年中考数学复习教材回归知识讲解 .pdf
2009 年中考数学复习教材回归知识讲解+例题解析+强化训练一次函数知识讲解1正比例函数的定义一般地,形如 y=kx(k 是常数,k0)的函数,叫做正比例函数,其中 k 叫做比例系数2正比例函数的图像正比例函数y=kx(k 是常数且k 0)的图像是一条经过原点(0,0)和点(1,k)?的直线,我们称它为直线y=kx;当 k0 时,直线 y=kx 经过第一,三象限,y 随着 x 的增大而增大,当k0 时,y 随 x 的增大而增大,此时若b0,则直线y=kx+b 经过第一,二,三象限;若b0,则直线 y=kx+b 经过第一,三,四象限,当k0 时,直线y=kx+b 经过第一,二,四象限;当b0)?个单位得到一次函数 y=kx+b m;一次函数y=kx+b 沿着 x 轴向左(“”)、?右(“”)平移 n(n0)个单位得到一次函数y=k(xn)+b;一次函数沿着y 轴平移与沿着x 轴平移往往是同步进行的只不过是一种情况,两种表示罢了;直线y=kx+b 与 x 轴交点为(bk,0),与 y 轴交点为(0,b),且这两个交点与坐标原点构成的三角形面积为S=12bk b例题解析例 1(2006,江西省)已知直线L1经过点 A(1,0)与点 B(2,3),另一条直线L2经过点 B,且与 x 轴相交于点P(m,0)(1)求直线L1的解析式;(2)若 APB 的面积为3,求 m 的值【分析】函数图像上的两点坐标也即是x,y 的两组对应值,?可用待定系数法求解,求函数与坐标轴所围成的三角形面积关键是求出函数解析式的k,b 的值【解答】(1)设直线L 的解析式为y=kx+b,由题意得0,23.kbkb解得1,1.kb所以,直线L1的解析式为y=x+1(2)当点 P 在点 A 的右侧时,AP=m(1)=m+1,有 SAPC=12(m+1)3=3解得 m=1,此时点P 的坐标为(1,0);当点 P 在点 A 的左侧时,AP=1m,有 S=(m1)3=3,解得 m=3,此时,点 P 的坐标为(3,0)综上所述,m 的值为 1 或 3【点评】先设一次函数的解析式,再代入点的坐标,利用方程组求解,其步骤是:设、代,求、答例 2(2004,黑龙江省)下图表示甲,乙两名选手在一次自行车越野赛中,路程 y(km)随时间 x(min)的变化的图像(全程),根据图像回答下列问题:(1)求比赛开始多少分钟时,两人第一次相遇?(2)求这次比赛全程是多少千米?(3)求比赛开始多少分钟时,两人第二次相遇文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3文档编码:CH1G2L5H5H2 HY4V2M9G4L10 ZN7C8H4U6M3【分析】观察图像知,甲选手的路程y 随时间 x 变化是一个分段函数,第一次相遇时是在 AB 段,故求出15x 33 时的函数关系式;欲求出比赛全程,则需知乙的速度,这可由第一次相遇时的路程与时间的关系求得,要求第二次相遇时间,?即先求甲在BC 段的函数关系式,再求出BC 和 OD 的交点坐标即可【解答】(1)当 15x33 时,设 yAB=k1x+b1,将(15,5)与(33,7)代入得:1111515733kbkb解得1119103kbyAB=19x+103当 y=6 时,有:6=19x+103,解得 x=24比赛进行到24min 时,两人第一次相遇(2)设 yOD=kx,将(24,6)代入得:6=24k,k=14yOD=14x 当 x=48 时,yOD=1448=12 比赛全程为12km(3)当 33x43 时,设 yBC=k2x+b2,将(33,7)和(43,12)代入得:22227331243kbkb文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1解得2212192kbyBC=12x1921192214xyxy解得19238xy比赛进行到38min 时,两人第二次相遇【点评】解答图像应用题的要领是从图像的形状特点、变化趋势、相关位置、相关数据出发,充分发掘图像所蕴含的信息,利用函数、方程(组)、不等式等知识去分析图像以解决问题例 3(2006,贵州铜仁)铜仁某水果销售公司准备从外地购买西瓜31t,柚子 12t,现计划租甲,乙两种货车共10 辆,将这批水果运到铜仁,已知甲种货车可装西瓜4t 和柚子 1t,乙种货车可装西瓜,柚子各2t(1)该公司安排甲,乙两种货车时有几种方案?(2)若甲种货车每辆要付运输费1800 元,乙种货车每辆要付运输费1200 元,?则该公司选择哪种方案运费最少?最少运费是多少元?【解答】(1)设安排甲种货车x 辆,则安排乙种货车为(10 x)辆,依题意,得42(10)312(10)12xxxx解这个不等式组,得5.5x 8x 是整数,x 可取 6,7,8即安排甲,乙两种货车有三种方案:甲种货车6 辆,乙种货车4 辆甲种货车7 辆,乙种货车3 辆甲种货车8 辆,乙种货车2 辆(2)设运费为y 元,则 y=1800 x+1200(10 x)=600 x+12000 当 x 取 6 时,运费最少,最少运费是:15600 元【点评】本例需要考生构建一元一次不等式和一次函数来解决实际问题,以考查学生运文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1用综合知识,分析、解决问题的能力强化训练一、填空题1(2006,绍兴)如图所示,一次函数y=x+5 的图像经过点P(a,b),Q(c,d),?则 a(cd)b(cd)的值为 _2(2005,重庆市)直线y=43x+8 与 x 轴,y 轴分别交于点A 和点 B,M 是 OB 上的一点,?若将 ABM 沿 AM 折叠,点 B 恰好落在x 轴上的点B处,则直线AM 的解析式为_3(2006,白云区)关于x 的一次函数y=(a3)x+2a5 的图像与y 轴的交点不在x?轴的下方,且y 随 x 的增大而减小,则a的取值范围是_4已知一次函数y=kx+b(k 0)的图像经过点(0,1),且 y 随 x 的增大而增大,?请你写出一个符合上述条件的函数关系式_5(2005,黑龙江省)一次函数y=kx+3?的图像与坐标轴的两个交点之间的距离为5,则 k 的值为 _6(2005,包头市)若一次函数y=ax+1 a 中,y 随 x 的增大而增大,且它的图像与y 轴交于正半轴,则a1+2a=_7(2005,四川省)如果记y=221xx=f(x),并且 f(1)表示当x=1 时 y 的值,即 f(1)=2211 1=12;f(12)表示当x=12时 y 的值,即f(12)=22()112(1)2=15;如果 f(1)+f(2)+f(12)+f(3)+f(13)+f(n)+f(1n)=_(结果用含n 的代数式表示,n 为正整数)8如图所示,点M 是直线y=2x+3 上的动点,过点M 作 MN垂直 x 轴于点 N,y 轴上是否存在点P,使以 M,N,P为顶点的三角形为等腰直角三角形小明发现:当动点 M 运动到(1,1)时,y 轴上存在点P(0,1),此时有 MN=MP,文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1能使 NMP 为等腰直角三角形在y 轴和直线上还存在符合条件的点P 和点 M请你写出其他符合条件的点P 的坐标 _二、选择题9(2006,南安)如图所示,一个蓄水桶,60min 可匀速将一满桶水放干其中,水位h(cm)随着放水时间t(min)的变化而变化 h 与 t 的函数的大致图像为()10(2005,杭州市)已知一次函数y=kx k,若 y 随 x 的增大而减小,则该函数的图像经过()A第一,二,三象限B第一,二,四象限C第二,三,四象限D第一,三,四象限11(2008,济南)济南市某储运部紧急调拨一批物资,调进物资共用4h,调进物资2h 后开始调出物资(调进物资与调出物资的速度均保持不变)储运部库存物资S(t)?与时间 t(h)之间的函数关系如图5 35 所示,?这批物资从开始调进到全部调出所需要的时间是()A 4h B4.4h C4.8h D5h 12(2006,泉州)小明所在学校离家距离为2km,某天他放学后骑自行车回家,行驶了5min 后,因故停留10min,继续骑了5min 到家,下面哪一个图像能大致描述他回家过程中离家的距离 s(km)与所用时间t(min)之间的关系()13(2006,黄冈)如图所示,在光明中学学生体力测试比赛中,甲,?乙两学生测试的路程s(m)与时间t(s)之间的函数关系图像分别为折线OABC 和线段 OD,?下列说法正文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1确的()A乙比甲先到达终点B乙测试的速度随时间增加而增大C比赛进行到29.7s时,两人出发后第一次相遇D比赛全程甲的测试速度始终比乙的测试速度快14(2005,黄冈市)有一个装有进,出水管的容器,单位时间内进,?出的水量都是一定的已知容器的容积为600L,又知单开进水管10min 可把空容器注满若同时打开进,出水管,20min 可把满容器的水放完现已知水池内有水200L,先打开进水管5min,再打开出水管,两管同时开放,直至把容器中的水放完,则能正确反映这一过程中容器的水量 Q(L)随时间t(min)变化的图像是下图中的()15(2005,重庆市)为了增强抗旱能力,保证今年夏粮丰收,某村新修建了一个蓄水池,这个蓄水池安装了两个进水管和一个出水管(两个进水管的进水速度相同),一个进水管和一个出水管的进出水速度如图a,b 所示,某天0 点到 6 点(?至少打开一个水管),该蓄水池的蓄水量如图c 所示,并给出以下3 个论断:0 点到 1 点不进水,只出水;1点到 4点不进水,不出水;4 点到 6 点只进水,不出水,则一定正确的论断是()(a)(b)(c)ABCD16(2008,重庆)如图所示,在直角梯形ABCD中,DCAB,A=90,AB=28cm,DC=24cm,AD=4cm,点 M 从点 D 出发,以1cm/s 的速度向点C 运动,点N 从点 B 同文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E2G3S1Y3 HJ1S1F9L1J10 ZQ8V7Y5S1R1文档编码:CL10E