一、选择题: 题1 号 答A 案 二、填空题: 11.-4<x<10 12.三、解答题:
3?π32 3 4 5 6 7 8 9 10 D B C A C A B B B 13.22 14.
21.(8分)22、m=-3,舍去m=1; 22.(8分)
解:(1)过点A作AE⊥CD于点E, 根据题意,得?DBC????60°,?DAE????30°,
D A 甲
?AE?BC,EC?AB?36米, ····················· (2分)
设DE?x,则DC?DE?EC?x?36,
?tan30°?DEAEE
乙 在Rt△AED中,tan?DAE?AE?3x,?BC?AE?,
? C
B
3x,
DCBC,?3?x?363x在Rt△DCB中,tan?DBC?tan60°?,
?3x?x?36,x?18,?DC?54(米).········································(6分)
(2)?BC?BC??AE?3x,x?18,
分)
3?18?18?1.732≈31.18(米). ····································(8
22.(10分) 解:(1)根据题意得??65k?b?55,?75k?b?45.解得k??1,b?120.
所求一次函数的表达式为y??x?120. ····································(2分) (2)W
?(x?60)?(?x?120) ??x?180x?7200
2??(x?90)?900, ·························································(4
2分)
?抛物线的开口向下,?当x?90时,W随x的增大而增大,
而60≤x≤87,
?当x?87时,W??(87?90)?900?891.
2?当销售单价定为87元时,商场可获得最大利润,最大利润是891
元. ··························································································(6分) (3)由W?500,得500??x2?180x?7200,
··············(7?110.
分)
整理得,x2?180x?7700?0,解得,x1?70,x2由图象可知,要使该商场获得利润不低于500元,销售单价应在70元到110元之间,而60≤x≤87,所以,销售单价x的范围是70≤x≤87.
································································································ (10分) 24.1,4,10,??
点的个数 3 4 5 ?? n 可连成三角形个数 1=S4=S3??3?2?164?3?26?6 410=S?? Sn?5?4?35n(n-1)(n-2)6 推理:平面上有n个点,过不在同一条直线上的三个点可以确定一个三角形,取第一个点A有n种方法,取第二个点有B有(n-1)种取法,取第三个点C有(n-2)种取法,所以一共可以作n(n-1)(n-2)个三角形,但?ABC、?ACB、?BAC、?BCA、?CAB、?CBA是同一个三角形,故应除以6,即S结论:S
nn?n(n-1)(n-2)6。
?n(n-1)(n-2)6
23.(12分) 解:(1)①∵t?1秒, ∴BP?CQ?3?1?3厘米,
D A ∵AB?10厘米,点D为AB的中点, ∴BD?5厘米. 又∵PC∴PC∴PC?BC?BP,BC?8厘米,
Q P
C B ?8?3?5?BD厘米,
. ,
又∵AB?AC∴?B??C,
∴△BPD≌△CQP. ···································································(4分) ②∵vP?vQ, ∴BP?CQ,
又∵△BPD≌△CQP,?B??C,则BP?PC?4,CQ?BD?5, ∴点P,点Q运动的时间t?∴vQ?CQt?543?154BP3?43秒,
厘米/秒. ·····················································(7分)
(2)设经过x秒后点P与点Q第一次相遇, 由题意,得解得x?803154x?3x?2?10,
秒.
803?3?80厘米.
∴点P共运动了
∵80?2?28?24,
∴点P、点Q在AB边上相遇, ∴经过
803秒点P与点Q第一次在边AB上相遇. ··················· (12分)
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