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    1．南宁市狮山公园计划在健身区铺设广场砖．现有甲.乙两个工程队参加竞标.甲工程队铺设广场砖的造价(元)与铺设面积的函数关系如图12所示,乙工程队铺设广场砖的造价(元)与铺设面积满足函数关系式:． (1)根据图12写出甲工程队铺设广场砖的造价(元)与铺设面积的函数关系式, (2)如果狮山公园铺设广场砖的面积为.那么公园应选择哪个工程队施工更合算? 解:(1)当时.设.把代入上式得: ·················································································································· 2分 当时.设.把.代入上式得: ··································································································· 3分 解得:·········································································································· 4分 ···················································································· 5分 (2)当时.················································· 6分 ·············································································· 7分 当时.即: 得:··················································································································· 8分 当时.即: 得:············································································································· 9分 当时.即. 答:当时.选择甲工程队更合算.当时.选择乙工程队更合算.当时.选择两个工程队的花费一样．···························································································································· 10分 2小王购买了一套经济适用房.他准备将地面铺上地砖.地面结构如图所示．根据图中的数据(单位:).解答下列问题: (1)写出用含x.y的代数式表示的地面总面积, (2)已知客厅面积比卫生间面积多212.且地面总面积是卫生间面积的15倍.铺12地砖的平均费用为80元.求铺地砖的总费用为多少元? 解:(1)地面总面积为:(6x+2y+18)2,···························································· 4分 (2)由题意.得······················································· 6分 解之.得····················································································· 8分 ∴地面总面积为:6x+2y+18＝6×4+2×+18＝45(2)．·············· 9分 ∵铺12地砖的平均费用为80元. ∴铺地砖的总费用为:45×80＝3600(元)．·········································· 10分 3为迎接“建国60周年 国庆.我市准备用灯饰美化红旗路.需采用A.B两种不同类型的灯笼200个.且B灯笼的个数是A灯笼的. (1)求A.B两种灯笼各需多少个? (2)已知A.B两种灯笼的单价分别为40元.60元.则这次美化工程购置灯笼需多少费用? (1)设需种灯笼个.种灯笼个.根据题意得: ··············································································································· 4分 解得·············································································································· 6分 (2)120×40+80×60=9600(元)．············································································· 8分 4是一扇半开着的办公室门的照片.门框镶嵌在墙体中间.门是向室内开的．图(2)画的是它的一个横断面．虚线表示门完全关好和开到最大限度(由于受到墙角的阻碍.再也开不动了)时的两种情形.这时二者的夹角为120°.从室内看门框露在外面部分的宽为4cm.求室内露出的墙的厚度a的值．(假设该门无论开到什么角度.门和门框之间基本都是无缝的．精确到0.1cm.) 解从图中可以看出.在室内厚为acm的墙面.宽 为4cm的门框及开成120°的门之间构成了一 个直角三角形.且其中有一个角为60°．·········· 3分 从而 a=4×tan60° ·············································· 6分 =4×≈6.9(cm)．·································· 8分 即室内露出的墙的厚度约为6.9cm． 5铭润超市用5000元购进一批新品种的苹果进行试销.由于销售状况良好.超市又调拨11000元资金购进该品种苹果.但这次的进货价比试销时每千克多了0.5元.购进苹果数量是试销时的2倍． (1)试销时该品种苹果的进货价是每千克多少元? (2)如果超市将该品种苹果按每千克7元的定价出售.当大部分苹果售出后.余下的400千克按定价的七折售完.那么超市在这两次苹果销售中共盈利多少元? 解:(1)设试销时这种苹果的进货价是每千克元.依题意.得······························· ··············································································· 解之.得 5···················································································· 经检验.5是原方程的解．······························································· (2)试销时进苹果的数量为: 第二次进苹果的数量为:2×10002000········································· 盈利为: 2600×7+400×7×0.7-5000-110004160(元) ······················ 答:试销时苹果的进货价是每千克5元.商场在两次苹果销售中共盈利4160元． ·············································· 6如图.要设计一个等腰梯形的花坛.花坛上底长米.下底长米.上下底相距米.在两腰中点连线处有一条横向甬道.上下底之间有两条纵向甬道.各甬道的宽度相等．设甬道的宽为米． (1)用含的式子表示横向甬道的面积, (2)当三条甬道的面积是梯形面积的八分之一时.求甬道的宽, (3)根据设计的要求.甬道的宽不能超过6米.如果修建甬道的总费用与甬道的宽度成正比例关系.比例系数是5.7.花坛其余部分的绿化费用为每平方米0.02万元.那么当甬道的宽度为多少米时.所建花坛的总费用最少?最少费用是多少万元? ．解:(1)横向甬道的面积为:··········································· 2分 (2)依题意:·············································· 4分 整理得: ······································································· 6分 甬道的宽为5米． (3)设建设花坛的总费用为万元． ············································ 7分 当时.的值最小．··························································· 8分 因为根据设计的要求.甬道的宽不能超过6米. 米时.总费用最少．······················································································ 9分 最少费用为:万元·················································· 10分 7为奖励在演讲比赛中获奖的同学.班主任派学习委员小明为获奖同学买奖品.要求每人一件．小明到文具店看了商品后.决定奖品在钢笔和笔记本中选择．如果买4个笔记本和2支钢笔.则需86元,如果买3个笔记本和1支钢笔.则需57元． (1)求购买每个笔记本和钢笔分别为多少元? (2)售货员提示.买钢笔有优惠.具体方法是:如果买钢笔超过10支.那么超出部分可以享受8折优惠.若买支钢笔需要花元.请你求出与的函数关系式, 的条件下.小明决定买同一种奖品.数量超过10个.请帮小明判断买哪种奖品省钱． 解(1)解:设每个笔记本元.每支钢笔元．·························································· 1分 ············································································································· 2分 解得 答:每个笔记本14元.每支钢笔15元．······································································· 5分 ······························································· 6分 ······························································· 7分 (2) (3)当时., 当时., 当时.．···················································································· 8分 综上.当买超过10件但少于15件商品时.买笔记本省钱, 当买15件奖品时.买笔记本和钢笔一样, 当买奖品超过15件时.买钢笔省钱．·········································································· 10分 8如图.等腰梯形花圃ABCD的底边AD靠墙.另三边用长为40米的铁栏杆围成.设该花圃的腰AB的长为x米. (1)请求出底边BC的长, (2)若∠BAD=60°, 该花圃的面积为S米2. ①求S与x之间的函数关系式(要指出自变量x的取值范围).并求当S=时x的值, ②如果墙长为24米.试问S有最大值还是最小值?这个值是多少? 解:(1)∵AB=CD=x米.∴BC=40-AB-CD= -------------------- (2)①如图.过点B.C分别作BE⊥AD于E.CF⊥AD于F.在Rt△ABE中.AB=x,∠BAE=60° ∴AE=x,BE=x.同理DF=x,CF=x 又EF=BC=40-2x ∴AD=AE+EF+DF=x+40-2x+x=40-x----------- ∴S= ·x=x = ------------- 当S=时.= 解得:x1=6,x2=.∴x=6------------ ②由题意.得40-x≤24,解得x≥16. 结合①得16≤x＜20------------------------ 由①.S== ∵a=＜0 ∴函数图象为开口向下的抛物线的一段. 其对称轴为x=.∵16＞,由左图可知. 当16≤x＜20时.S随x的增大而减小----------- ∴当x=16时.S取得最大值.--------------- 此时S最大值=.------- 9水产公司有一种海产品共2 104千克.为寻求合适的销售价格.进行了8天试销.试销情况如下: 第1天 第2天 第3天 第4天 第5天 第6天 第7天 第8天 售价x 400 250 240 200 150 125 120 销售量y 30 40 48 60 80 96 100 观察表中数据.发现可以用反比例函数刻画这种海产品的每天销售量y与销售价格x之间的关系．现假定在这批海产品的销售中.每天的销售量y与销售价格x之间都满足这一关系． (1) 写出这个反比例函数的解析式.并补全表格, (2) 在试销8天后.公司决定将这种海产品的销售价格定为150元/千克.并且每天都按这个价格销售.那么余下的这些海产品预计再用多少天可以全部售出? 解:(1) 函数解析式为． --2分 填表如下: 第1天 第2天 第3天 第4天 第5天 第6天 第7天 第8天 售价x 400 300 250 240 200 150 125 120 销售量y 30 40 48 50 60 80 96 100 --2分 (2) 2 104-(30+40+48+50+60+80+96+100)=1 600. 即8天试销后.余下的海产品还有1 600千克． --1分 当x=150时.=80． --2分 1 600÷80=20.所以余下的这些海产品预计再用20天可以全部售出． --1分 102009年5月17日至21日.甲型H1N1流感在日本迅速蔓延.每天的新增病例和累计确诊病例人数如图所示． (1) 在5月17日至5月21日这5天中.日本新增甲型H1N1流感病例最多的是哪一天?该天增加了多少人? (2) 在5月17日至5月21日这5天中.日本平均每天新增加甲型H1N1流感确诊病例多少人?如果接下来的5天中.继续按这个平均数增加.那么到5月26日.日本甲型H1N1流感累计确诊病例将会达到多少人? (3) 甲型H1N1流感病毒的传染性极强.某地因1人患了甲型H1N1流感没有及时隔离治疗.经过两天传染后共有9人患了甲型H1N1流感.每天传染中平均一个人传染了几个人?如果按照这个传染速度.再经过5天的传染后.这个地区一共将会有多少人患甲型H1N1流感? 解:(1) 18日新增甲型H1N1流感病例最多.增加了75人, --4分 (2) 平均每天新增加人. --2分 继续按这个平均数增加.到5月26日可达52.6×5+267=530人, --2分 (3) 设每天传染中平均一个人传染了x个人.则 .. 解得(x = -4舍去)． --2分 再经过5天的传染后.这个地区患甲型H1N1流感的人数为 (1+2)7=2 187(或1+2+6+18+54+162+486+1 458=2 187). 即一共将会有2 187人患甲型H1N1流感． --2分 【查看更多】

     

    题目列表(包括答案和解析)

    （2009•南宁）南宁市狮山公园计划在健身区铺设广场砖．现有甲、乙两个工程队参加竞标，甲工程队铺设广场砖的造价y甲（元）与铺设面积x（m2）的函数关系如图所示；乙工程队铺设广场砖的造价y乙（元）与铺设面积x（m2）满足函数关系式：y乙=kx．
    （1）根据图写出甲工程队铺设广场砖的造价y甲（元）与铺设面积x（m2）的函数关系式；
    （2）如果狮山公园铺设广场砖的面积为1600m2，那么公园应选择哪个工程队施工更合算？

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    （2009•南宁）南宁市狮山公园计划在健身区铺设广场砖．现有甲、乙两个工程队参加竞标，甲工程队铺设广场砖的造价y甲（元）与铺设面积x（m2）的函数关系如图所示；乙工程队铺设广场砖的造价y乙（元）与铺设面积x（m2）满足函数关系式：y乙=kx．
    （1）根据图写出甲工程队铺设广场砖的造价y甲（元）与铺设面积x（m2）的函数关系式；
    （2）如果狮山公园铺设广场砖的面积为1600m2，那么公园应选择哪个工程队施工更合算？

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