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1、学习好资料欢迎下载-1-2011 年内蒙古呼和浩特市中考数学试卷一、选择题（本大题共10 个小题，每小题3 分，共 30 分.在每小题给出的四个选项中，只有一项是符合题目要求的，请把该选项的序号填入题干后面的括号内）1、（2011?呼和浩特）如果a的相反数是2，那么 a 等于（）A、2 B、2 C、D、考点：相反数。分析：因为绝对值相等且符号不同的两个数互为相反数，根据题意可求得a的绝对值，再根据相反数的概念不难求得a 的值解答：解：a 的相反数是2，|a|=|2|=2，a=2故选 A点评：此题主要考查学生对相反数的概念的理解及掌握情况2、（2006?重庆）计算2x2?（3x3）的结果是（）A
2、、6x5B、6x5C、2x6D、2x6考点：同底数幂的乘法；单项式乘单项式。分析：根据单项式乘单项式的法则和同底数幂相乘，底数不变，指数相加计算后选取答案解答：解：2x2?（3x3），=2（3）?（x2?x3），=6x5故选 A点评：本题 主要考查单项式相乘的法则和同底数幂的乘法的性质3、（2011?呼和浩特）已知圆柱的底面半径为1，母线长为2，则圆柱的侧面积为（）A、2 B、4 C、2D、4考点：圆柱的计算。专题：计算题。分析：圆柱侧面积=底面周长 高学习好资料欢迎下载-2-解答：解：圆柱沿一条母线剪开，所得到的侧面展开图是一个矩形，它的长是底面圆的周长，即 2，宽为母线长为2cm，所以它的
3、面积为4cm2故选 D点评：本题考查了圆柱的计算，掌握特殊立体图形的侧面展开图的特点，是解决此类问题的关键4、（2011?呼和浩特）用四舍五入法按要求对0.05049 分别取近似值，其中错误的是（）A、0.1（精确到 0.1）B、0.05（精确到百分位）C、0.05（精确到千分位）D、0.050（精确到0.001）考点：近似数和有效数字。专题：探究型。分析：根据近似数与有效数字的概念对四个选项进行逐一分析即可解答：解：A、0.05049 精确到 0.1 应保留一个有效数字，故是0.1，故本选项正确；B、0.05049 精确到百分位应保留一个有效数字，故是0.05，故本选项正确；C、0.0504
4、9 精确到千分位应是0.050，故本选项错误；D、0.05049 精确到 0.001 应是 0、050，故本选项正确故选 C点评：本题考查的是近似数与有效数字，即从一个数的左边第一个不是0 的数字起到末位数字止，所有的数字都是这个数的有效数字5、（2011?呼和浩特）将如图所示表面带有图案的正方体沿某些棱展开后，得到的图形是（）A、B、C、D、考点：几何体的展开图。文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1
5、C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4
6、 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1
7、C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4
8、 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1
9、C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4
10、 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1
11、C4L3学习好资料欢迎下载-3-专题：几何图形问题。分析：由平面图形的折叠及立体图形的表面展开图的特点解题注意带图案的三个面相交于一点解答：解：由原正方体知，带图案的三个面相交于一点，而通过折叠后A、B 都不符合，且D 折叠后图案的位置正好相反，所以能得到的图形是C故选 C点评：考查了几何体的展开图，解决此类问题，要充分考虑带有各种符号的面的特点及位置6、（2011?呼和浩特）经过某十字路口的汽车，它可能继续直行，也可能向左或向右转若这三种可能性大小相同，则两辆汽车经过该十字路口全部继续直行的概率为（）A、B、C、D、考点：列表法与树状图法。分析：列举出所有情况，看两辆汽车经过这个十字路口全部
12、继续直行的情况占总情况的多少即可解答：解：列表得：一共有9 种情况，两辆汽车经过这个十字路口全部继续直行的有一种，两辆汽车经过这个十字路口全部继续直行的概率是故选 C点评：本题主要考查用列表法与树状图法求概率，用到的知识点为：概率=所求情况数与总情况数之比7、如果等腰三角形两边长是6cm 和 3cm，那么它的周长是（）A、9cm B、12cm C、15cm 或 12cm D、15cm 考点：等腰三角形的性质；三角形三边关系。专题：分类讨论。分析：求等腰三角形的周长，即是确定等腰三角形的腰与底的长求周长根据三角形三边关文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L
13、3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM
14、1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L
15、3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM
16、1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L
17、3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM
18、1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L
19、3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3学习好资料欢迎下载-4-系定理列出不等式，确定是否符合题意解答：解：当 6 为腰，3 为底时，6366+3，能构成等腰三角形，周长为5+5+3=13；当 3 为腰，6 为底时，3+3=6，不能构成三角形故选 D点评：本题从边的方面考查三角形，涉及分类讨论的思想方法求三角形的周长，不能盲目地将三边长相加起来，而应养成检验三边长能否组成三角形的好习惯，把不符合题意的舍去8、（2011?呼和浩特）已知一元二次方程x2+bx3=0 的一根为
20、3，在二次函数y=x2+bx3的图象上有三点、，y1、y2、y3的大小关系是（）A、y1y2 y3B、y2y1y3C、y3 y1 y2D、y1y3y2考点：二次函数图象上点的坐标特征；一元二次方程的解。分析：将 x=3 代入 x2+bx3=0 中，求 b，得出二次函数y=x2+bx 3 的解析式，再根据抛物线的对称轴，开口方向确定增减性，比较y1、y2、y3的大小关系解答：解：把 x=3 代入 x2+bx3=0 中，得 93b3=0，解得 b=2，二次函数解析式为y=x2+2x3，抛物线开口向上，对称轴为x=1，1，且 1（）=，（1）=，而，y1 y2y3故选 A点评：本题考查了二次函数图象
21、上点的坐标特点，一元二次方程解的意义关键是求二次函数解析式，根据二次函数的对称轴，开口方向判断函数值的大小9、（2011?呼和浩特）如图所示，四边形ABCD中，DCAB，BC=1，AB=AC=AD=2 则BD 的长为（）A、B、C、D、文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2
22、C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档
23、编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2
24、C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档
25、编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2
26、C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档
27、编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3学习好资料欢迎下载-5-考点：勾股定理。专题：计算题。分析：以 A 为圆心，AB 长为半径作圆，延长BA 交 A 于 F，连
28、接 DF在 BDF 中，由勾股定理即可求出BD 的长解答：解：以 A 为圆心，AB 长为半径作圆，延长BA 交 A 于 F，连接 DF可证 FDB=90，F=CBF，DF=CB=2，BF=2+2=4，BD=故选 B点评：本题考查了勾股定理，解题的关键是作出以A 为圆心，AB 长为半径的圆，构建直角三角形，从而求解10、（2011?呼和浩特）下列判断正确的有（）顺次连接对角线互相垂直且相等的四边形的各边中点一定构成正方形；中心投影的投影线彼此平行；在周长为定值的扇形中，当半径为时扇形的面积最大；相等的角是对顶角的逆命题是真命题A、4 个B、3 个C、2 个D、1 个考点：二次函数的最值；对顶角、
29、邻补角；正方形的判定；弧长的计算；扇形面积的计算；命题与定理；中心投影。专题：综合题。分析：根据对顶角的性质、扇形面积的计算、中心投影、二次函数的最值等知识点判断各命题的真假，即可得出答案解答：解：顺次连接对角线互相垂直且相等的四边形的各边中点一定构成正方形，此命题正确，故正确；文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3
30、文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1
31、F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3
32、文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1
33、F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3
34、文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1
35、F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3学习好资料欢迎下载-6-中心投影与原物体所对应点的连线都相交于一点，平行投影与原
36、物体所对应点的连线都相互平行，故错误；在周长为定值的扇形中，当半径为时扇形的面积最大；设 a 为扇形圆心角，2r+2 ra/2 =2r+ar=a=（2r）/r，s=r=（2r2+r），根据二次函数极值公式当 r=时扇形的面积最大，故正确；相等的角是对顶角的逆命题是：若两个角是对顶角，则这两个角相等，为真命题故正确故选 C点评：本题主要考查了二次函数的最值、对顶角的性质、扇形面积的计算、中心投影等知识点，考查了学生对综合知识的掌握程度，属于中档题二、填空题（本大题共6 个小题，每小题 3 分，共 18 分.本题要求把正确结果填在每题的横线上，不需要解答过程）11、（2011?呼和浩特）函数中，自
37、变量x 的取值范围x 3考点：函数自变量的取值范围。专题：计算题。分析：根据二次根式的性质和分式的意义，被开方数大于等于0，分母不等于0，就可以求解解答：解：根据二次根式有意义，分式有意义得：x+30且 x+30，解得：x 3故答案为：x 3点评：本题考查的知识点为：分式有意义，分母不为0；二次根式的被开方数是非负数文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F
38、2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文
39、档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F
40、2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文
41、档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F
42、2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文
43、档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3学习好资料欢迎下载-7-12、（2011?
44、呼和浩特）已知关于x 的一次函数y=mx+n 的图象如图所示，则可化简为n考点：二次根式的性质与化简；一次函数图象与系数的关系。专题：数形结合。分析：根据一次函数图象与系数的关系，确定m、n 的符号，然后由绝对值、二次根式的化简运算法则解得即可解答：解：根据图示知，关于x 的一次函数y=mx+n 的图象经过第一、二、四象限，m0；又关于x 的一次函数y=mx+n 的图象与y 轴交与正半轴，n0；=nm（m）=n故答案是：n点评：本题主要考查了二次根式的性质与化简、一次函数图象与系数的关系一次函数y=kx+b（k0）的图象，当k0 时，经过第一、二、三象限；当k 0 时，经过第一、二、四象限13
45、、一个样本为1、3、2、2、a，b，c已知这个样本的众数为3，平均数为2，那么这个样本的方差为考点：方差。分析：因为众数为3，表示 3 的个数最多，因为2 出现的次数为二，所以3 的个数最少为三个，则可设a，b，c 中有两个数值为3另一个未知利用平均数定义求得，从而根据方差公式求方差解答：解：因为众数为3，可设 a=3，b=3，c 未知文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y
46、2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT
47、6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y
48、2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT
49、6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y
50、2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT6I3S1C4L3文档编码:CX7Y9Y2K3Q4 HM1F2C8L1C7 ZT