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unweighed是什么意思，unweighed中文翻譯，unweighed怎么讀、發(fā)音、用法及例句

2025-09-11 投稿

?unweighed

unweighed 發(fā)音

[[ʌn'weid]]

英：　　美：

unweighed 中文意思翻譯

常見釋義：

[計量]未稱量過的.

unweighed 相似詞語短語

1、reweighed ─── v.重新稱量

2、neighed ─── v.（馬）嘶鳴；（人）發(fā)馬嘶般的聲音；n.馬嘶聲

3、unwished ─── adj.非所希望的；不想要的；v.不再指望…；放棄希望（unwish的過去分詞）

4、unweight ─── vt.移去…的重量；vi.減重

5、unweights ─── vt.移去…的重量；vi.減重

6、unweighted ─── adj.無重負的；[數(shù)]未加權的；無關緊要的；v.減重（unweight的過去式和過去分詞）

7、inveighed ─── vi.痛罵；漫罵；猛烈抨擊

8、unwigged ─── 無鉤

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

unweighed 常見例句（雙語使用場景）

1、And Solomon left all the vessels unweighed because of the very great number; the weight of bronze could not be ascertained. ─── 47因為這一切器具太多，所以所羅門都沒有過秤；銅的重量無法可查。

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次稱量，天平又平衡了，那么我們就知道那個最終沒有稱量的球是異常的球。

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這一切、所羅門都沒有過秤因為甚多．銅的輕重也無法可查。

4、Solomon left all the vessels unweighed since there were so many of them. And so the weight of the bronze was not known. ─── 撒羅滿沒有秤量這一切器具，因為銅太多，重量無法計算。

5、47 And Solomon left all the vessels unweighed, because they were exceeding many: neither was the weight of the brass found out. ─── 這一切、所羅門都沒有過秤因為甚多．銅的輕重也無法可查。

6、And Solomon left all the vessels unweighed, because they were exceeding many: neither was the weight of the brass found out. ─── 這一切所羅門都沒有過秤。因為甚多，銅的輕重也無法可查。

7、And Solomon left all the vessels unweighed， because they were exceeding many： neither was the weight of the brass found out. ─── 47這一切所羅門都沒有過秤。因為甚多，銅的輕重也無法可查。

8、Solomon left all these things unweighed, because there were so many; the weight of the bronze was not determined. ─── 這一切所羅門都沒有過秤。因為甚多，銅的輕重也無法可查。

9、Now you know that the odd ball is one of the unweighed 4. ─── 現(xiàn)在你知道那個異常的球就在剩下的沒有稱量的4個球之中。

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. ─── 步驟2：把三個沒有稱量的球放在A中，另外三個已知正常的球放在B中。