3.下列盛放物质的方法错误的是 ( )
A.将金属钠保存在煤油中 B.少量的白磷可保存在水中
C.纯碱溶液用带磨口玻璃瓶塞的试剂瓶保存
D.硝酸银溶液盛放在棕色试剂瓶中
1.下列关于化学学习和研究的说法错误的是( )
A.化学模型有助于解释一些化学现象
B质量守恒定律是大量实验事实的总结
C.化学家提出的假设都能被实验证实
D化学基本原理的应用是有一定条件的
2.清蒸大闸蟹由青色变成红色,一同学认为这种红色物质可能象酸碱指示剂一样,遇到酸或碱颜色会发生改变。就这位同学的看法而言,这应该属于科学探究中的( )
A.实验 B、假设 C.观察 D.分类
25.(本小题满分8分)
解:(1)如图①,过作交于点,则四边形是平行四边形.
∵ ,∴ .
∴ .
∴ .
由题意知,当、运动到秒时,
∵ ,∴ .
∴ .即 .
解得,.············································································································· 5分
(3)分三种情况讨论:
① 当时,如图②,即.
∴ .··················································································································· 6分
② 当时,如图③,过作于,于H.
则 ,.
∴ .
∵ ,∴ .
∴ .即 .
∴ .··················································································································· 7分
③ 当时,如图④,过作于点.
则 .
∵,
∴ .
∴ .即 .
∴ . --------------------------------------------------------------------------8分
综上所述,当、或时,为等腰三角形.
24.(本小题满分7分)
解:(1)A(0,2), B(,1).····················································································· 2分
(2)解析式为;·················································································· 3分
顶点为().····································································································· 4分
(3)如图,过点作轴于点M,过点B作轴于点N,过点作
轴于点P.
在Rt△AB′M与Rt△BAN中,
∵ AB=AB′, ∠AB′M=∠BAN=90°-∠B′AM,
∴ Rt△AB′M≌Rt△BAN.
∴ B′M=AN=1,AM=BN=3, ∴ B′(1,).
同理△AC′P≌△CAO,C′P=OA=2,AP=OC=1,
可得点C′(2,1);
将点B′、C′的坐标代入,
可知点B′、C′在抛物线上.····························································································· 7分
(事实上,点P与点N重合)
23.(本小题满分7分)
解:(1)将分别代入中,
得,
∴ .
∴ 反比例函数的表达式为:;
正比例函数的表达式为.·········································································· 2分
(2)观察图象得,在第一象限内,当时,
反比例函数的值大于正比例函数的值.--------------------------------------------4分
(3).
理由:∵ ,
∴ .
即 .
∵ ,
∴ .
即 .
∴ .
∴ .
∴.············································································································· 7分
22.(本小题满分4分)
解:(1)同意.如图,设AD与EF交于点M,
由折叠知,∠BAD=∠CAD,
∠AME=∠AMF=90O. ------------------------------1分
∴ 根据三角形内角和定理得
∠AEF=∠AFE. ------------------------------------2分
∴ △AEF是等腰三角形.···················································································· 3分
(2)图⑤中的大小是22.5o.··················································································· 4分
21.(本小题满分6分)
解:(1)甲种电子钟走时误差的平均数是:
;
乙种电子钟走时误差的平均数是:
.
∴ 两种电子钟走时误差的平均数都是0秒. --------------------------------- 2分
(2);
.
∴ 甲乙两种电子钟走时误差的方差分别是6s2和4.8s2.---------------------------4分
(3)我会用乙种电子钟,因为平均水平相同,且甲的方差比乙的大,说明乙的稳定性更好,故乙种电子钟的质量更优. -----------------------------------------6分
20.(本小题满分5分)
解:设商场第一次购进套运动服,
由题意得: .··············································································· 3分
解这个方程,得.
经检验,是所列方程的根.
.
答:商场两次共购进这种运动服600套.········································································· 5分
19. (本小题满分5分)
(1)证明:如图,连结,则 .
∴ .
∵ AC=BC, ∴ .
∴ .
∵ ∥,∴ .
∵ 于F,∴ .
∴.∴ .
∴ EF是⊙O的切线. ------------------------------------------------------------3分
( 2 ) 连结BG,∵BC是直径, ∴∠BGC=90=∠CFE.
∴ BG∥EF.∴ .
设 ,则 .
在Rt△BGA中,.
在Rt△BGC中, .
∴ .解得 .即 .
在Rt△BGC中, .
∴ sin∠E. --------------------------------------------- --------------------------------5分
18.(本小题满分5分)
解:如图,∵ AC平分∠BAD,
∴ 把△ADC沿AC翻折得△AEC,
∴ AE=AD=9,CE=CD=10=BC.------------------------------------------------------2分
作CF⊥AB于点F.∴ EF=FB=BE=(AB-AE)=6.------------------------3分
在Rt△BFC(或Rt△EFC)中,由勾股定理得 CF=8.----------------------------4分
在Rt△AFC中,由勾股定理得 AC=17.
∴ AC的长为17. -------------------------------------------------------------------------5分
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