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Most animals protect(保护) their young until they are old enough to take care of themselves. A fish that swims in the rivers of South America protects its young in a very strange way. This fish, a kind of catfish, uses its mouth to keep its babies from danger(危险). When this catfish lays(产卵) its eggs, it carries them in its mouth. From what we know, it does not eat during this time. After the eggs hatch(孵化), the fish swims with its young to protect them. It opens its mouth when danger appears(出现), and the tiny fish swim inside. They swim out only when it’s safe again. These catfish may have a peculiar way to protect their young, but it seems(好像) to work. 小题1:The best title(标题) is _____________. A.Staying Away From Danger B.How A Fish Protect Its Young C.Laying Eggs D.A Cat That Swims 小题2: The catfish in the story live in _________________. A.lakes B.the sea C.rivers D.North America 小题3:The young fish are ____________________. A.tiny B.large C.yellow and gold D.very brave 小题4:If the young fish were not near their parents, they would be ______________. A.safer B.in greater danger C.happier D.less hungry 小题5: The word “peculiar” in the story means ________________. A.easy B.useless C.strange D.dangerous
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举一反三
【单选题】下列不是面团的面筋质所起的作用的是( )
A.
提高面团的保气能力
B.
阻止二氧化碳气体的溢出
C.
是面包制品形成膨胀、松软的重要条件
D.
提高面团的吸水力
【单选题】从学术视角来看,大数据与数据挖掘之间什么关系?
A.
大数据包含数据挖掘
B.
大数据就是数据挖掘
C.
大数据是数据挖掘之后的新阶段
D.
数据挖掘是大数据之后的新阶段
【单选题】Tom, please pass ______ the glasses. I want to read newspaper.
A.
you
B.
me
C.
him
D.
her
【单选题】下列不是面团的面筋质所起的作用的是( )。
A.
承受面团发酵过程中二氧化硫气体的膨胀
B.
提高面团的保气能力
C.
提高面团的可塑性
D.
阻止二氧化硫气体的溢出
【单选题】律诗的第二联(三、四句)被称为
A.
颔联
B.
颈联
C.
起联
D.
尾联
【单选题】Tom, please pass______ the glasses. I want to read the newspapers.
A.
You
B.
me
C.
him
D.
her
【多选题】关于连锁企业信息化战略重要性的说法正确的有()
A.
信息技术的有效运用可以有力支持企业的竞争战略
B.
连锁企业规模越大,地域分布越广,信息化管理的迫切性就越大
C.
快速反应系统能有效地降低商品的引入时间
D.
RFID技术目前已越来越受到人们的重视
E.
信息技术已经使连锁企业制定更好和更有效的决策成为可能
【单选题】下列不是面团的面筋质所起的作用的是( )
A.
承受面团发酵过程中二氧化碳气体的膨胀
B.
提高面团的保气能力
C.
提高面团的可塑性
D.
阻止二氧化碳气体的溢出
【单选题】信息化战争以()决定效能
A.
数量
B.
信息
C.
技术
D.
质量
【单选题】P61.设栈的顺序存储空间为S(1:50),初始状态为top=0。先经过一些列入栈与退栈运算后,top=20,则当前栈中的元素个数为
A.
30
B.
29
C.
20
D.
19
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