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11-12学年高二数学:必修5综合模块测试 16(人教B版必修5)

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单元测试试题 

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教材版本新人教B版

使用学科数学

使用年级高二

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更新时间2011-11-23

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基本信息

1.在△abc中,已知c= 10b = 60°,a =75°,则b =                         

    a10         b5         c10         d5

2在等差数列{an}中,若a1a5a9 = 2,则a1a3a5a7a9等于               

    a10             b3              c.-10/3         d10/3

3.根据下列5个图形及相应点的个数变化规律,试猜测第n个图形中有几个点                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       

 

 

 

 

 

    a2n -1           b2n–1 +1         cn2 -n+1         d2n-1

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