求疯切防征谓条普察狂原始人高清的种子,不要枪版.有的上传到知道
你好,下载地址:http://zhidao.baidu.com/question/543867555?&oldq=1#answer-13744384易药者组固买落省收57【103...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
北京时间 **************************,“神舟八号”飞船和“天宫一号”目标飞行器交会对接成功。据此
小题1:C小题2:D小题3:A试题**:由图中可判断如下表节气时间(前后)直射点*置移动方向对应点春分3月21日赤道向北a夏至6月22日北回归线向南b秋分9月23日赤道向南c冬至...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
将下列算式按得数从小到大排列. 18+28 28+58 &n...
18+28=38,28+58=78,1-38=58,58+18=68,58+38=1;因为38<58<68<78<1,所以18+28<1-38<58+18<28+58<58+38...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
在长方体AC1中,AB=4,BC=CC1=22.M是BC1的中点,N是MC1的中点,则异面直线AN与CM所成的角为______
解:以DA为x轴,以DC为y轴,以DD1为z轴,建立空间直角坐标系,∵长方体AC1中,AB=4,BC=CC1=22.M是BC1的中点,N是MC1的中点,∴A(22,0,0),N(...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
数列{a n }是以a为首项,q为公比的等比数列.*********************************************************************************************************
(1)当q=1时,bn=1-(a1*a2*…*an)=1-na,cn=2-(b1*b2*…*bn)=2-[(1-a)*(1-na)]n2=a2n2*(a2-1)n*****≠1时...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
“支撑”话题作文导写文题:阅读下面材料,根据要求作文。有了自尊的脖子才能支撑起自信的头颅;有了不屈
撑起自己的天空夜幕低垂,习习凉风吹拂着我零乱的面颊。我独自一人静静地坐在溪边的岩石上,望着潺潺的流水,我的心里像无边的海面,一会儿波涛汹涌,一会儿寂然无声。正当我的思绪飘忽不定的...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
在______
5a3=a5a,1a=aa,a3=aa,32a=42a,<...展开5a3=a5a,1a=aa,a3=aa,32a=42a,<收起本回答由提问者推荐评论赞0踩0index闦74采...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
(20大2?湛江模拟)已知,如图,在直角梯形COAB中,CB∥OA,以O为原点建立平面直角坐标系,A、B、C的坐标
解:(小)设所求抛物线的解析式为y=ax2+bx+c(a≠0)依题意,得c=0小00a+小0b=0小小a+二b=四,解得a=?小3b=小03c=0,故所求抛物线的解析式为y=-小...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
(20大2?湛江模拟)已知,如图,在直角梯形COAB中,CB∥OA,以O为原点建立平面直角坐标系,A、B、C的坐标
解:(小)设所求抛物线的解析式为y=ax2+bx+c(a≠0)依题意,得c=0小00a+小0b=0小小a+二b=四,解得a=?小3b=小03c=0,故所求抛物线的解析式为y=-小...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
对于函数 ,若存在区间 ,使得 ,则称函数 为“可等域函数”,区间 为函数 的一个“可等域区间”.
B本回答由提问者推荐评论姆姆EK97采纳率:68%擅长:暂未定制为您推荐:varcpro_psid="u*******";varcpro_pswidth="450";varcpr...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
(2010?青州市模拟)如图,在直角梯形**********************************************2,过A作AE⊥CD,垂足
(Ⅰ)取AB中***H,连接GH,FH,又G*AD的中点,∴GH∥BD.又因*GH?平面BCD,BD?平面BCD,∴GH∥平面BCD同理可证FH∥BC,FH∥平面BCD,所以面F...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
已知:抛物线y=ax2+bx+c经过点O(0,0),A(7,4),且对称轴l与x轴交于点B(5,0).(1)求抛物线的表
(1)由题意得?b2a=5c=049a+7b+c=4(1分),解得a=?421b=4021c=0.,∴y=?421x2+4021x.(3分)(2)∵△BOC与△DOC重合,OB=...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)
已知:如图,在半径为4的⊙O中,AB,CD是两条直径,M为OB的中点,CM的延长线交⊙O于点E,且EM>MC.连接D
解:(1)连接AE,BC,∵∠AEC与∠MBC都为AC所对的圆周角,∴∠AEC=∠MBC,又∠AME=∠BMC(对顶角相等),∴△AME∽△CMB,∴AM:CM=EM:MB,即A...
展开阅读全文 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAG1BMVEUAAABuk71wj79vkr1wj79wn79vlL1vk75vk73EmhGbAAAACHRSTlMA0BDfIBDfv/U9wHQAAABISURBVBjTY6AlYBOC0OEJQIK9UQHEZpUQAJJMEmCpQIiYIohiBQlBpKASUCmwBETKQiiw2QFmunOjhAncKhYLkARcCihBWwAA5n0JqdkCrS4AAAAASUVORK5CYII=)
收起 ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABgAAAAYBAMAAAASWSDLAAAAHlBMVEUAAABuk71wj79vlL1wl79wn79vkr1vlL5vk75vk71klGr6AAAACXRSTlMA0BDfIBDfv7/eQcl1AAAARklEQVQY02OgNXA2QbBZLCc7ICQmSpogJIQUJytAOYETFZgkhSBsVhBDcaICRAJEA6XgEjApRhAJkmoAkmxFEK2KCQw0BAA5Lgp0ywp4owAAAABJRU5ErkJggg==)