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求sin(-1350°)=sin(-1350)=sin(-4π+2/π)把负号提出 得sin=[-(4π-2/π)] 又有诱导公式 sin(-a)=-sina 则sin=[-(4π-2/π)]=-sin(4π-2/π) 又有诱导公式sin(π-a)=sina 那么就等于-sin2/π 想不通为什么最后结果等于1 我觉得应该等于-1 我想的过程是哪出错了吗?
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－sin(4π－π／2)=－sin(－π／2)=sinπ／2=1
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派是180度，sin(-1350)=sin(-8派 派/2)=sin(派/2)=1，sin(-8派 派/2)=-(8派-派/2)=-sin(-派/2)=sin(派/2)=1
扫描下载二维码数学ln（-x）为什么能进行求导？倒过来，e的平方不能得出负数的吧，想不通_百度知道
数学ln（-x）为什么能进行求导？倒过来，e的平方不能得出负数的吧，想不通
数学中lnx当x是负数的时候也就是loge（-x），为什么还能对lnx进行求导？倒过来想的话，e的平方不可能有负数-x的出现呀？求解，我概念不是很清楚...
数学中lnx当x是负数的时候也就是log e （-x），为什么还能对lnx进行求导？倒过来想的话，e的平方不可能有负数-x的出现呀？求解，我概念不是很清楚
nieyunzhao
nieyunzhao
获赞数：42
这个题目是考察你的求导能力，不是考察你对对数函数的定义域的理解，你不要错解了出题老师的原意，另外log e （-x）怎么不能求导，log e （-x）的导数就是X分之1啊
feidao2010
feidao2010
采纳数：21031
获赞数：128218
ln（-x）定义域x&0是复合函数的求导[ln(-x)]'=1/(-x)
*(-x)'=1/(-x) *(-1)=1/x
获赞数：114
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问一个关于不等式的符号问题.如：(m-x)(n+x)＞0变为：（x-m)（x+n）＜0我就是想不通,以前解不等式时,左边除以负数要变号,但这条式子我可以在左右两边都一起除以-1啊,这样就不用变号了么?但是似乎不对啊.
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哈哈,你搞错了!你必须记住：“不等式两边同乘以或除以正数,不等号不变向.而不等式两边同乘以或除以负数,不等号要变向.”这样,你才永远不会做错这类题了!
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如果想不通可以代数字看看啊！比如-8<0两边除以-1得8>0是不是要变号！
扫描下载二维码In recent months, a computer virus spread across networks in China. The virus came with an icon of a lovely panda, hence the name Panda Virus. What makes this virus difficult to handle is that it has many variations.
Unfortunately, our lab's network was also infected with the Panda Virus. As you can see from the above diagram, the computers in our lab are placed in a matrix of M rows and N columns. A computer is only connected with the computers next to it. At the beginning, Tcomputers were infected with the Panda Virus, each with a different variation (Type 1, Type 2... Type T). Each computer in the network has a specific defense level L (0 & L & 1000). The Panda Virus will rapidly spread across the network according to the following rules:
The virus can only spread along the network from the already infected computers to the clean ones.
If a computer has already been infected by one virus variation, it will never be infected by another variation.
The transmission capacity of the Panda Virus will increase each day. In day 1, the virus only infects computers with a defense level 1 provided the virus can spread to that computer, however, a computer with a defense level &1 will stop the transmission along that path. In day D, it can spread to all the computers connected with a defense level &=D, provided that the transmission is not stopped by a computer with a defense level & D along the path.
Within one day, the virus variation of type 1 would spread first and infects all the computers it can reach. And then the virus variation of type 2, then type 3, etc.
The following samples show the infection process described above:
At the beginning, only 2 computers were infected:
So at last, all the computers in the networks were infected by virus.
Your task is to calculate after all the computers are infected, how many computers are infected with some specific virus variations.
The input contains multiple test cases!
On the first line of each test case are two integers M and N (1 &= M, N &= 500), followed by a M * N matrix. A positive integer T in the matrix indicates that the corresponding computer had already been infected by the virus variations of type T at the beginning while a negative integer -L indicates that the computer has a defense level L. Then there is an integer Q indicating the number of queries. Each of the following Q lines has an integer which is the virus variation type we care.
For each query of the input, output an integer in a single line which indicates the number of computers attacked by this type of virus variation.
Sample Input
1 -3 -2 -3
-2 -1 -2 2
-3 -2 -1 -1
Sample Output
我们了解病毒的传播方式后，发现和我们bfs的过程简直就是一样。所以，这道题理所当然的就用bfs了。
当然，我们不能简单的bfs，题目中说了按照天数来，每天的感染是从1号病毒开始的，所以要用到优先队列。
#include &iostream&
#include &queue&
#include &algorithm&
#include &cstdio&
#include &cmath&
#include &cstring&
struct node
bool operator & (const node &i)const
if(day==i.day) return num&i.//在天数相同的情况下，从小号病毒开始传播。
return day&i.//天数不同，从天数小的开始传播。
int mp[505][505];
int vis[250005];
int dx[4]= {1,-1,0,0},dy[4]= {0,0,-1,1};
priority_queue&node& Q;
void bfs()
while(!Q.empty()) Q.pop();
for(i=0; i&n; i++)//这个循环就是找到初始感染源
for(j=0; j&m; j++)
if(mp[i][j]&0)
q.num=mp[i][j];
vis[q.num]++;
Q.push(q);
while(!Q.empty())
int cot=0;
q=Q.top();
for(i=0; i&4; i++)
p.x=q.x+dx[i];
p.y=q.y+dy[i];
if(p.x&=n||p.x&0||p.y&=m||p.y&0||mp[p.x][p.y]&0)
if((-1)*(mp[p.x][p.y])&=q.day)
Q.push(p);
mp[p.x][p.y]=q.
vis[p.num]++;
else//这里是这道题的关键地方。
{//在当前感染点当前天数不找到可感染的电脑后
//找到感染点附件的，且防御等级最低的计算机
if (!cot) cot=mp[p.x][p.y];
cot=max(cot,mp[p.x][p.y]);
{//当前感染点的四个方向搜索完毕，将当前感染点附近防御级别最低的计算机入队
q.day=-//因为防御等级为负数，记得加负号
Q.push(q);
int main()
while(~scanf("%d%d",&n,&m))
for(i=0; i&n; i++)
for(j=0; j&m; j++)
scanf("%d",&mp[i][j]);
memset(vis,0,sizeof(vis));
scanf("%d",&k);
while(k--)
scanf("%d",&ans);
printf("%d\n",vis[ans]);
在一个电脑网络中有n*m台电脑
题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2849#include
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2849 主要考优先级队列和一些剪枝技巧 #include typedef s...
Attack of Panda VirusTime Limit: 3 Seconds
Memory Limit: 32768 KB In recent months, a computer...
Attack of Panda VirusTime Limit: 3 Seconds
Memory Limit: 32768 KB
In recent months, a computer ...
这题目很是坑爹，各种坑爹，一开始SF了不止多少次，后来发现是Type的不仅仅是1，2.而是到1000，还有给出的图 是500 * 500的，可是直接去搜索一直超时，想不通，题目给的是3000ms，早就...
Attack of Panda Virus
Time Limit: 3 Seconds
Memory Limit: 32768 KB
In recent months, a ...
In recent months, a computer virus spread across networks in China. The virus came with an icon of a...
In recent months, a computer virus spread across networks in China. The virus came with an icon of a...
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