There was a salt-marsh that bounded part of the mill-pond.

沙滩龙2022-10-04 11:39:541条回答

There was a salt-marsh that bounded part of the mill-pond.
There was a salt-marsh that bounded part of the mill-pond,on the edge of which,at high water,we used to stand to fish for minnows.句子哪些部分是插入语?
另外感觉句中有2个完整的句子,好像不太对,分别是：There was a salt-marsh that bounded part of the mill-pond 和we used to stand to fish for minnows.请帮我分析一下!

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high8rr 共回答了17个问题 | 采纳率82.4%: 1.没有插入语,on the edge of which 引导非限制性定语从句,修饰mill-pound..
at high water 做时间状语
不懂追问!; 1年前

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应该是材料应该为含镍的碳化钨.

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1.y=4x,x=7,x-axis
∫4xdx|[0,7]=2x^2|[0,7]=98
2、x+8y=8,y-axis,x-axis
y=1-x/8,0≤x≤8
∫（1-x/8）dx|[0,8]=x-1/16*x^2|[0,8]=4
3、 y=7x,y=28,y-axis
x=y/7
∫y/7dy|[0,28]=y^2/14|[0,28]=56
4、y=x^2-16x,x-axis
x^2-16x=0,x=0或16
∫（x^2-16x)dx|[16,0]=1/3*x^3-8x^2|[16,0]=2048/3
5、16x-x^3,x-axis,第一象限
16x-x^3＞0,x（x-4)(x+4)＜0
x＞0,所以积分范围为0＜x＜4
∫（16x-x^3）dx|[0,4]=8x^2-1/4*x^4|[0,4]=64
6、y=26x-x^2,y=x
代入得
x=26x-x^2,解得x1=25,x2=0,所以积分范围为[0,25]
∫(26x-x^2-x)dx|[0,25]=25/2*x^2-1/3x^3=15625/6

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There was a salt -marsh that bounded part of the mill-pond.句 There was a salt -marsh that bounded part of the mill-pond.句子中的that? 怎么我觉得bounded 前面好像少了个be动词,但读起来又好像不太通顺. 如果句子改成There was a salt-marsh which is bounded part of the mill-pound可以吗 shu_01330661年前4: 刘阳爱你共回答了18个问题 | 采纳率94.4%

这里bound是动词把,翻译成束缚了,妨碍了
不懂追问!

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There was a salt-marsh that bounded part of the mill-pond.

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