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一、选择题(本大题共12小题,每小题3分,共36分)1-5ADBCD6-10BDCAB11-12CC二、填空题(本大题共4小题,每小题3分,共12分)【解析】连接AC,交BD于点O,过点E作EG⊥OD于点G∵四边形ABCD是菱形∴AO=AD2-OD2=4,EG∥AO∵E是AD的中点 1DEEG 1==2三、解答题(本大题共8小题,共72分.解答应写出文字说明、证明过程或演算步骤) (2)设□为x,则:6-2x-5<-13∴指针应指向8所在区域 7分 2分正确解答:①×2,得:2x-4y=8③③-②,得:-y=7把y=-7代入①,得:x-2×(-7)=4解得:x=-10·····························································································4分∴原方程组的解为···········································································5分(2)联立得:······························································································6分解得:〈ly=8·······································································································(2)∵△ABC≌△ADE∵F为BD的中点∴AF⊥BD···········································································································8分20.(1)5036····································································································2分解23(元)························································4分众数为10元···································································································5分学生人数为50人,从小到大排列后位于第25第26名捐款为15元所以中位数是15元··························································································6分(3)设李老师的捐款金额为x,目前捐款总人数为51人,中位数位于第26位,捐款数仍是15元则15··································································································7分李老师的捐款金额为115元····················································································8分21.解1)30°·········································································································2分(2)方法一:如下图,点M即为所求5分5分提示:∵∠CMD=2∠CED∴作出DE的垂直平分线交CE于点M即可方法二:如下图,点M即为所求5分提示:∵∠CMD=2∠CED=60°∴过点D作CD的垂线交CE于点M即可(答案不唯一,作法正确即可)(3)如图1,若⊙O与CD,DE两边相切,则点O在∠CDE的平分线上如图2,若⊙O与BC,DE两边相切∴C是⊙O与BC边的切点∴D是⊙O与DE边的切点综上,OE的长为33或43··················································································9分34J······················································3分3J/(kg·℃)·······················································································5分34,解得m∴冷冻的猪肉的质量为1.68kg·················································································9分23.解1)90°·········································································································3分(2)由题知点A'在以B为圆心,以BA长为半径的圆上,当B,A',D共线时,DA'的值最小····4分∴DA'的值最小为10-6=4·····················································································5分∴DF=5·············································································································8分 1 1若A'在直线BC上方,α=90°-30°=60°,则边BA扫过区域的面积为若A'在直线BC下方,α=90°+30°=120°,则边BA扫过区域的面积为24.解1)将A(3,1B(0,-2代入y=x2+bx+c 4分(2)①把x=5代入y=x2-2x-2中,y=13≠6∴点P(5,6)不在图象C1上 6分②根据平移规律可得新抛物线解析式为:yx-1-n)2-3当C2经过点P(5,6)时,则有65-1-n)2-3解得:n=1或n=7····························································································8分(3)①设直线AB的解析式为y=kx+a将A(3,1B(0,-2)代入得1解得∴直线AB的解析式为y=x-2设直线GG'的解析式为y=x+d,过G(1,-3)∴直线GG'的解析式为y=x-4,G'在x轴上,G'的坐标为(4,0)∴G移动的距离GG'为··································································10分②G
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