Suspend=n does not freeze the server waiting for the debug connection, so you may miss the start of the agent.
Suspend=n没有冻结服务器来等待调试连接，所以可以错过代理的启动。
Such laws may be only the start and local night curfews will not stop cats from eating diurnal animals say pato n.
佩顿说，这些法规可能只是开始，当地的宵禁不阻止猫吃白天活动的动物。
So let's start with a case when we have some number of the equations say n equations and n unknown.
所以让我们从一个例子入手，当我们有一些方程式，假定n个方程，n个未知数。
Start this container using the command lxc-start -n vsplain.
使用命令lxc - start - n vsplain启动这个容器。
When you start your containers (for instance by using lxc-start -n vs1), you'll likely get a few audit messages about SELinux access denials.
在启动容器时(例如通过使用命令lxc - start -n vs1)，很可能会收到一些关于SELinux访问拒绝的审计消息。
So instead you'd have to maybe if you start with wavelength, go over there, and then figure out velocity and do something more like kinetic energy equals 1/2 n b squared to get there.
这时你要先从波长开始，到这，然后算出速度,然后像动能等于1/2nb平方得到这。
And unlike n, l can start all the way down at 0, and it increases by integer value, so we go 1, 2, 3, and all the way up.
不像n，l可以从0开始取，然后每次增加一个整数，所以我们可以去1，2，3，一直下去。
On the second day, the start is at this splitting point and the finish is at point N.
第二天，起点是那个中间路口，而终点为路口 N。
So if I start off with a list of length n, how many times can I divide it by 2, until I get to something no more than two left?
我能够除以多少次2呢？，直到我得到的长度不超过2么？，对数次，对吧？就像刚才那位同学说的那样？
So if I start off with a list of length n, how many times can I divide it by 2, until I get to something no more than two left?
我能够除以多少次2呢？，直到我得到的长度不超过2么？，对数次，对吧？就像刚才那位同学说的那样？
应用推荐