比项多一笔的字偏旁加了一笔这究竟是什么字,望大家帮我找找看,

lirongme2022-10-04 11:39:541条回答

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axiangyaofa 共回答了21个问题 | 采纳率100%
顼(顼)


姓(原为古帝颛顼的省称)
赵顼: 大宋皇帝
---------------------------------
顼(顼)

ㄒㄩˋ
〔颛~〕见“
(顼)颛”.
郑码:CGO,U:987C,GBK:E7EF
笔画数:10,部首:页,笔顺编号:1121132534
1年前

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根据对话内容,从对话后的选项中选出能填入空白处的最佳选项, 并将答案填涂在二卷对应位置 (否则不予给分),选项中有两项多
根据对话内容,从对话后的选项中选出能填入空白处的最佳选项, 并将答案填涂在二卷对应位置 (否则不予给分),选项中有两项多余选项。
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(14+9)-(30-12)
=23-18、
=5人
第二节 根据对话内容,从对话后的选项中选出能填入空白处的最佳选项,并将答案写在本题下面的横线上。选项中有两项多

第二节 根据对话内容,从对话后的选项中选出能填入空白处的最佳选项,并将答案写在本题下面的横线上。选项中有两项多余选项。
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—It finished in a draw. 64
—It was quite good, but I missed the beginning of it because I had to eat first.
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已知等差数列an的前7项的和为42,且前9项之和比第4项多57求公差d
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等差数列an的前7项的和为42;
Sn=na1+[n(n-1)d]/2==>S7=7*a1+7*(7-1)*d/2=7a1+21d=42---(1);
S9=9*a1+(9-1)*9d/2=9a1+36d;
a4=a1+(4-1)d=a1+3d;
S9-a4=57;==>9a1+36d-a1-3d=57==>8a1+33d=57;----(2)
联立(1)(2)解得:d=1;a1=3;
已知等差数列an的前7项和为42,且前9项之和比第4项多57,求公差d
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等差数列an的前7项的和为42;
Sn=na1+[n(n-1)d]/2==>S7=7*a1+7*(7-1)*d/2=7a1+21d=42---(1);
S9=9*a1+(9-1)*9d/2=9a1+36d;
a4=a1+(4-1)d=a1+3d;
S9-a4=57;==>9a1+36d-a1-3d=57==>8a1+33d=57;----(2)
联立(1)(2)解得:d=1;a1=3;
所以公差d=1
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等差数列前7项为42,前9项的和比第4项多57,求公差?
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第四项为 42/7=6
前九项和为57+6=63
所以第五项为63/9=7
公差7-6=1