片语

Lyoko minus one减去一区

Minus s one零下一度

One Minus One一减一

doubles minus one两倍再减一

minus one degree Celsius零下一度

TRICK BOX minus one小川洋行

radix-minus-one complement form反码形式

radix-minus-one complement反码

英汉例句

• so the first case is where m k minus one is much bigger than k2.

  所以第一种情况是mk负,比k2大得多。

• k minus one is much bigger than k2, k2 so we can ignore k2 here.

  k负1比k2大得多，所以我们能忽略这的。

• and now, after rearranging your differential equation, putting b in there, k you can rewrite this as k1 plus k minus one times a a0 minus k minus one times b0 plus a0.

  现在，在重新整理你的微分方程后，把b放进这儿，你可以重新把它写作k1加,负1乘以a减k负1乘以b0加。

• we have something that looks like k1 k2 over k minus one times a is the rate of reaction.

  然后我们有看起来是m不再存在,我们有看起来是2,除以k负1乘以,a是反应速率，一级。

• i am doing this for plus one minus one like sodium chloride.

  我正在这样做,因为就像氯化钠那样一正一负。

• a which is k2 times k1 over k minus one times a.

  那是k2乘以k1除以k负1乘以。

• q2 so i「m going to write this as q2 over q1 over minus one plus q2 over q1.

  因此这等于q2除以q1，除以负1加上q2除以,我这么做的目的是。

• this is minus one over r, so i get m mg over r with a minus sign, and that has to be evaluated between infinity and capital r.

  这是r的负值，所以在mmg/r上加-号，它处于,无穷远和r之间。

• verify that the partition number (「minor」) in parted matches /etc/fstab and that grub uses that number minus one in /boot/grub/menu.lst.

  请检验parted中的分区号（「minor」）是否与 /etc/fstab匹配，grub在 /boot/grub/menu.lst 中使用的分区号是该编号减1。

• then f is one minus one plus two, is equal to two.

  那么f等于1减去1再加上2。

• and you get a new differential equation, 1 minus da/dt is equal to k1 times a minus k minus one times b steady state, so we plug this in here. and you get, a so there」s a sitting here.

  你得到一个新的微分方程，负da/dt等于k1乘以a减k负,乘以b稳态，这样我们把这带到这里，你得到，这里有。

• you put k minus one here, k1 k1 here, but it「s the same thing.

  这是k负，这是，但它是一样的。

• rt and then my derivative here,i have minus one over rt.

  然后这里的微分有负1除以。

• or you can write is as a k1, k minus one to b, where k1 could be k forward k f backwards. also written as k sub f, b and this is also written as k sub b.

  或者你可以写作a，k1，b，k负，这里k1可以是k向前,负1可以是k向后，and，k，minus，one，could，be，k,也写为k角标，这也写为k角标。

• and then the other one is m k minus one is much less than k2. m is very small.

  然后另一个是mk负,比k2小得多，m很小。

• the intermediate is what we」re solving a steady state k2 for.so this is k1 k2 a m over m k minus one plus k2.

  我们为这个中间产物解决稳态,所以这是k1，k2，a，m除以mk负1加。