1.已知复数,则
等于 ( )
A.2i B.-2i C.2 D.-2
41. 解:(1),所以不能在60天内售完这些椪柑,
(千克)
即60天后还有库存5000千克,总毛利润为
W=;
(2)
要在2月份售完这些椪柑,售价x必须满足不等式
解得
所以要在2月份售完这些椪柑,销售价最高可定为1.4元/千克。
40. 解:(1)在△OAB中,
|
|
|
|
过点A´作A´D垂直于y轴,垂足为D。
在Rt△OD A´中
|
|
|
OD=OA´·
∴A´点的坐标为(,
)
(2)点B的坐标为(,1),点B´的坐标为(0,2),设所求的解析式为
,则
解得,
,∴
当时,
∴A´(,
)在直线BB´上。
39. 解:设y与x之间的关系为y=kx+b,由题意得,解得
.
所以y与x之间的关系式是y=-20x+1000.
(2)当x=0时,y=m=-20×0+1000=1000.
所以m=1000.
38. 解:(1) 根据题意可知:y=4+1.5(x-2) ,
∴ y=1.5x+1(x≥2) ······························································ 4分
(2)依题意得:7.5≤1.5x+1<8.5 ····································································· 6分
∴ ≤x<5············································································ 8分
37.
解:(1)去超市购买所需费用
即··········································································································· 1分
去超市购买所需费用
即········································································································· 2分
当时,即
当时,即
当时,即
·························································································································· 4分
综上所述:当时,去
超市购买更合算;当
时,去
超市或
超市购买一样;当
时,去
超市购买更合算.···················································································································· 5分
(2)当时,即购买10副球拍应配120个乒乓球
若只去超市购买的费用为:
(元)············································································· 6分
若在超市购买10副球拍,去
超市购买余下的乒乓球的费用为:
(元)··············································································· 7分
最佳方案为:只在
超市购买10副球拍,同时获得送30个乒乓球,然后去
超市按九折购买90个乒乓球. 8分
36.
解:(1)当时,设路程
与时间
之间的函数关系式为
,依题意可得:
解得
所以,········································································································· 3分
当时,解得
,
即王师傅开车通过雪峰山隧道的时间为7.4分钟;························································· 4分
(2)当时,王师傅开车的速度为0.8千米/分钟,
当时,王师傅开车的速度为1千米/分钟.··························································· 6分
设王师傅开车从第分钟开始连续2分钟恰好走了1.8千米,
则有,解得
,
即进隧道1分钟后,连续2分钟恰好走了1.8千米. 8分
35.
解:
(1)s=2t
(2)在0< t < 1时,甲的行驶速度小于乙的行驶速度;在t > 1时,甲的行驶速度大于乙的行驶速度.
(3)只要说法合乎情理即可给分。如:乙在第三小时追上甲
34.. (1)1.9 …………………………………………………2分
(2) 设直线EF的解析式为乙=kx+b
∵点E(1.25,0)、点F(7.25,480)均在直线EF上
∴………………………………………………3分
解得∴直线EF的解析式是y乙=80X-100……………4分
∵点C在直线EF上,且点C的横坐标为6,
∴点C的纵坐标为80×6-100=380
∴点C的坐标是(6,380)………………………………………5分
设直线BD的解析式为y甲 = mx+n
∵点C(6,380)、点D(7,480)在直线BD上
∴…………………………………………………6分
解得 ∴BD的解析式是y甲=100X -220 ……………7分
∵B点在直线BD上且点B的横坐标为4.9,代入y甲得B(4.9,270)
∴甲组在排除故障时,距出发点的路程是270千米。……………8分
(3)符合约定
由图像可知:甲、乙两组第一次相遇后在B和D相距最远。
在点B处有y乙-y甲=80×4.9-100-(100×4.9-220)=22千米<25千米
…………………………10分
在点D有y甲-y乙=100×7-220-(80×7-100)=20千米<25千米
…………………………11分
∴按图像所表示的走法符合约定。………………………………12分
33.(1)由题意,知B(0,6),C(8,0)
设直线的解析式为
,则
,解得
则的解析式为
。
(2)解法一:如图,过P作于D,则
由题意,知OA=2,OB=6,OC=8
解法二:如图,过Q作轴于D,则
由题意,知OA=2,OB=6,OC=8
(3)要想使为等腰三角形,需满足CP=CQ,或QC=QP,或PC=PQ。
①当CP=CQ时(如图①),得10-t=t。解,得t=5。
②当QC=QP时(如图②),过Q作轴于D,则
③当PC=PQ时(如图③),过P作于D,则
综上所述,当t=5,或,或
时,
为等腰三角形。
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