He lives in clover.

allan52022-10-04 11:39:541条回答

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永远的番茄 共回答了30个问题 | 采纳率96.7%
他过着优裕的生活.
live in clover,be in clover有过着优裕生活的意思
1年前

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一道紫罗兰公式解题CLOVER+CROCUS=VIOLET(解题时请列成竖式,方便解题)每个字母分别代表0~9这十个数字
一道紫罗兰公式解题
CLOVER+CROCUS=VIOLET(解题时请列成竖式,方便解题)每个字母分别代表0~9这十个数字,最好有解题过程,有条理的讲述思路.
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flora1983 共回答了18个问题 | 采纳率94.4%
我算了一下
CLOVER 429871
CROCUS 419405
VIOLET 849276
CLOVER=280516
CROCUS=260297
VIOLET=540813
首先去顶C小于等于4
当O=9时,U必然为0
这个时候V+C必然大于10
因为C小于等于4
那么V必然大于等于6
当O=0时,U必然为9
这个时候V+C必然小于10的
因为C小于等于4
可以假设C的值
来大致确定V的取值范围
但是就算这样,一个个探测数字的话
也需要几百次
如果换一种C--1 2 3 4 5 6
C2 L8 o0 V5 E1 R6
+ C2 R6 o0 C2 U9 S7
___________________
= V5 I4 o0 L8 E1 T3把字母分成6个列,找有相同数字最多的列,先解出来:
C3:o + o = o,只有一种可能,那就是 o=0(无论进位否)
同时推出第4列不会进位到第三列来,
进而推出:
C4 第4列两个数字之和< = 9
C5 E + U = E,有两个数字相同,所以U肯定很特别,两种可能:
U= 9 (有进位)
U= 0 (无进位)
所以,U =9,而且 E + U之后一定要进位到第4列的.
进而推出:
C4 第4列两个数字之和 + 1 < = 8 ( 1为第5列进位过来)
( 现在思维一已无法找到更多结果,进入思维二
思维二:找出各列之间有相同字母的,来看它们之间的关系.也是从在各列中出现最多的开始:)
C1C4:有两个字母相同,都有字母C、V,且:
C1:C + C = 2C = V
或者 2C + 1 = V (C2进位到C1时)
C4:V + C + 1 = 3C + 1 = L (前面已知C4必然有C3的进位)
或者 【(2C + 1) + C】 + 1= L (C2进位到C1时)
从而有:
C4 3C + 1 = L 10,
同时,在余下的5个数字中,挑出3个可以组成一个等式的,只有两种情况:
4 + 7 = 11 (R=4,7,S=7,4,T=1)
6 + 7 = 13 (R=6,7,S=7,6,T=3)
由于S和其它列不关联,所以要先确定R:
当R = 4 时,S=7,T=1,I=2=C,不正确;
当R = 6 时,S=7,T=3,I=5,E=1,(全部成立)
当R = 7 时,S=6,T=3,I=1,E=3=T,不正确.
不知你能不能看懂
请【遗传学】大牛进来解释In clover plants,the pattern on the leaves is de
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In clover plants,the pattern on the leaves is determined by asingle gene with multiple alleles that are related in a dominanceseries.7 different alleles of the gene are known; an allelethat determines the absence of a pattern is recessive to the other6 alleles,eah of which produces a different pattern.Allheterozygous combinations of alleles show complete dominance.
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B.What is the largest number of different genotypesthat could be associated with any one phenotype?Is there anyphenotype that could be represented by only a single genotype?
C.In a particular field,you find that the large majority of clover plants lack a pattern on their leaves,enen though you can identify a few plants representative of all possible pattern types.Explain this finding.
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A.7


B.largest – 13 (C1_),smallest – 1 (C7_)


C.Although according to the table,the amounts of clover plants that lack a pattern on their leaves should be thelowest,but actually it exists with highest amounts.This may be due to many reasons:
i.The recessive allele C7 is the ancestor for the other dominantalleles.The presence of other alleles may be is due to some mutations in gene,so they will exist in lower amounts compare to the original allele.
ii.Or due to natural selection.The type best suited to the environment willeasier to survive and has higher amount.
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CLOVER+CROCUS=VIOLET 每个字母只代表一个数字,不能重复使用 怎样才能是这个等式成立
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C--1 2 3 4 5 6
C2 L8 o0 V5 E1 R6
+ C2 R6 o0 C2 U9 S7
___________________
= V5 I4 o0 L8 E1 T3
解法如下:
1-把字母分成6个列,(思维一)找有相同数字最多的列,先解出来:
C3: o + o = o, 只有一种可能,那就是 o=0(无论进位否)
同时推出第4列不会进位到第三列来,
进而推出:
C4 第4列两个数字之和< = 9
C5 E + U = E, 有两个数字相同,所以U肯定很特别,两种可能:
U= 9 (有进位)
U= 0 (无进位)
所以,U =9,而且 E + U之后一定要进位到第4列的.
进而推出:
C4 第4列两个数字之和 + 1 < = 8 ( 1为第5列进位过来)

( 现在思维一已无法找到更多结果,进入思维二
思维二:找出各列之间有相同字母的,来看它们之间的关系.也是从在各列中出现最多的开始:)
C1C4:有两个字母相同,都有字母C、V,且:
C1: C + C = 2C = V
或者 2C + 1 = V (C2进位到C1时)
C4: V + C + 1 = 3C + 1 = L (前面已知C4必然有C3的进位)
或者 【(2C + 1) + C】 + 1= L (C2进位到C1时)
从而有:
C4 3C + 1 = L 10,
同时,在余下的5个数字中,挑出3个可以组成一个等式的,只有两种情况:
4 + 7 = 11 (R=4,7, S=7,4, T=1)
6 + 7 = 13 (R=6,7, S=7,6, T=3)
由于S和其它列不关联,所以要先确定R:
当R = 4 时,S=7, T=1, I=2=C,不正确;
当R = 6 时,S=7, T=3, I=5, E=1,(全部成立)
当R = 7 时,S=6, T=3, I=1, E=3=T,不正确.


2-从整体上判断,哪些列会>10,主要是看一下加数是在5以上或者5以下的可能性大小,其次是看一下会否进位,让高一位上要加1.(这里还看不出来)
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什么数加任何一个数等于它本身CLOVER+CROCUS=VIOLET
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请详阅:
This is clover the first leaf said hope.
四叶草的第一片叶子代表希望.
This is clover the second leaf said confidence.
四叶草的第二片叶子代表信心.
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This is clover the forth leaf said luck.
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紫罗兰算术题 有这样一道算术题 CLOVER +CROCUS —————— VIOLET
紫罗兰算术题 有这样一道算术题 CLOVER +CROCUS —————— VIOLET
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注意:相同的字母表示同一数字,不同的字母表示不同数字.
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