unweighed是什么意思、unweighed中文翻译、怎么读、发音、用法及例句

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  unweighed

  【【ʌnweid】】

  英:

  美:

  常见释义:

  【计量】未称量过的.

  1、reweighed ─── v.重新称量

  2、neighed ─── v.(马)嘶鸣;(人)发马嘶般的声音;n.马嘶声

  3、unwished ─── adj.非所希望的;不想要的;v.不再指望…;放弃希望(unwish的过去分词)

  4、unweight ─── vt.移去…的重量;vi.减重

  5、unweights ─── vt.移去…的重量;vi.减重

  6、unweighted ─── adj.无重负的;【数】未加权的;无关紧要的;v.减重(unweight的过去式和过去分词)

  7、inveighed ─── vi.痛骂;漫骂;猛烈抨击

  8、unwigged ─── 无钩

  9、outweighed ─── vt.比……重(在重量上);比……重要;比……有价值

  1、And Solomon left all the vessels

  unweighed

  because of the very great number; the weight of bronze could not be ascertained. ─── 47因为这一切器具太多,所以所罗门都没有过秤;铜的重量无法可查。

  2、I. If on this second weighing, the scale balances again, we know that the final

  unweighed

  ball is the odd one. ─── 如果第2次称量,天平又平衡了,那么我们就知道那个最终没有称量的球是异常的球。

  3、Solomon left all the utensils

  unweighed

  , because they were too many; the weight of the bronze could not be ascertained. ─── 王上7:47这一切、所罗门都没有过秤因为甚多.铜的轻重也无法可查。

  4、Solomon left all the vessels

  unweighed

  since there were so many of them. And so the weight of the bronze was not known. ─── 撒罗满没有秤量这一切器具,因为铜太多,重量无法计算。

  5、47 And Solomon left all the vessels

  unweighed

  , because they were exceeding many: neither was the weight of the brass found out. ─── 这一切、所罗门都没有过秤因为甚多.铜的轻重也无法可查。

  6、And Solomon left all the vessels

  unweighed

  , because they were exceeding many: neither was the weight of the brass found out. ─── 这一切所罗门都没有过秤。因为甚多,铜的轻重也无法可查。

  7、And Solomon left all the vessels

  unweighed

  , because they were exceeding many: neither was the weight of the brass found out. ─── 47这一切所罗门都没有过秤。因为甚多,铜的轻重也无法可查。

  8、Solomon left all these things

  unweighed

  , because there were so many; the weight of the bronze was not determined. ─── 这一切所罗门都没有过秤。因为甚多,铜的轻重也无法可查。

  9、Now you know that the odd ball is one of the

  unweighed

  4. ─── 现在你知道那个异常的球就在剩下的没有称量的4个球之中。

  10、Step 2: Put three of the

  unweighed

  balls on the Side A; put three balls that are known to be normal on Side B. ─── 步骤2:把三个没有称量的球放在A中,另外三个已知正常的球放在B中。

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