紫外法做定性鉴别时,常用的特征数据有()A、λmax,α,肩峰B、λmax,λmin,A1/A2,E1%cmC、λmax/λmin,末端吸收,E1%cmD、λmax/λmin,肩峰,末端吸收E、λmax,λmin,ΣA
紫外法做定性鉴别时,常用的特征数据有()
- A、λmax,α,肩峰
- B、λmax,λmin,A1/A2,E1%cm
- C、λmax/λmin,末端吸收,E1%cm
- D、λmax/λmin,肩峰,末端吸收
- E、λmax,λmin,ΣA
相关考题:
以下函数 findmax 拟实现在数组中查找最大值并作为函数值返回 , 但程序中有错导致不能实现预定功能#define MIN -2147483647int findmax (int x[],int n){ int i,max;for(i=0;in;i++){ max=MIN;if(maxx[i]) max=x[i];}return max;}造成错误的原因是A) 定义语句 int i,max; 中 max 未赋初值B) 赋值语句 max=MIN; 中,不应给 max 赋 MIN 值C) 语句 if(maxx[i]) max=x[i]; 中判断条件设置错误D) 赋值语句 max=MIN; 放错了位置
X线摄影线量最强波长的计算公式为A.λmin=1.24kVp(nm)B.λmax=1.5λmin(nm)C.λmax=2.5λmin(nm)D.λmax=2λmin(nm)E.λmin=12.4/kVp(nm)
x线最强波长(λmax)与最短波长(λmin)的关系是A.λmax =0.5λminB.λmax =λminC.λmax =1.5λminD.λmax =2λminE.λmax =2.5λmin
fun函数的功能是首先对a所指的N行N列的矩阵找出各行中最大的数,再求这N个最大值中最小的那个数作为函数值返回,请填空。#include stdio.h#define N 100int fun(int(*a)[N]){ int row,col,max,min; for(row=0;rowN;row++) { for(max=a[row][0],col=1;colN;col++) if() max=a[row][col]; if(row==0) min=max; else if() min=max; } return min;}
以下函数findmax拟实现在数组中查找最大值并作为函数值返回,但程序中有错导致不能实现预定功能。#d越lie MIN-2147483647int findmax(int X[],int n){ int i,max;for(i=0;in;i++){max=MIN;if(maxx[i])max=x[i];}return max;}造成错误的原因是A.定义语句int i,max;中max未赋初值B.赋值语句max=MIN;中,不应给max赋MIN值C.语句if(maxx[i])max=X[i];中判断条件设置错误D.赋值语句max=MIN;放错了位置
对于一循环应力,以σmin表示最小应力,σmax表示最大应力,则此循环应力的静力成分σm为()。A:σm=σminB:σm=1/2(σmin+σmax)C:σm=1/2(σmax-σmin)D:σm=σmax
A.零件名称,AVG(单价),MAX(单价)?MIN(单价)B.供应商,AVG(单价),MAX(单价)?MIN(单价)C.零件名称,AVG单价,MAX单价?MIN单价D.供应商,AVG单价,MAX单价?MIN单价
紫外法做定性鉴别时,常用的特征数据有()Aλmax,α,肩峰Bλmax,λmin,A1/A2,E1%cmCλmax/λmin,末端吸收,E1%cmDλmax/λmin,肩峰,末端吸收Eλmax,λmin,ΣA
认沽期权涨跌幅为()A、max{0.001,min[(2×正股价-行权价),正股价]×10%}B、max{0.001,min[(2×行权价-正股价),正股价]×10%}C、max{0.001,min[(2×行权价-正股价),正股价]×5%}D、max{0.001,min[(2×正股价-行权价),正股价]×5%}
X线摄影用最短波长的计算公式为()A、λmin=1.24/kVp(nm)B、λmax=1.2~1.5λmin(nm)C、λmax=2.5λmin(nm)D、λmax=2λmin(nm)E、λmin=12.4/kVp(nm)
Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) JOB_CAT VARCHAR2(30) SALARY NUMBER(8,2) Which statement shows the department ID, minimum salary, and maximum salary paid in that department, only of the minimum salary is less then 5000 and the maximum salary is more than 15000?()A、SELECT dept_id, MIN(salary(, MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX (salary) 15000;B、SELECT dept_id, MIN(salary), MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX(salary) 15000 GROUP BY dept_id;C、SELECT dept_id, MIN(salary), MAX(salary) FROM employees HAVING MIN(salary) 5000 AND MAX (salary) 15000;D、SELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id HAVING MIN (salary) 5000 AND MAX(salary)E、SELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id, salary HAVING MIN (salary) 5000 AND MAX (salary) 15000;
单选题Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) JOB_CAT VARCHAR2(30) SALARY NUMBER(8,2) Which statement shows the department ID, minimum salary, and maximum salary paid in that department, only of the minimum salary is less then 5000 and the maximum salary is more than 15000?()ASELECT dept_id, MIN(salary(, MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX (salary) 15000;BSELECT dept_id, MIN(salary), MAX(salary) FROM employees WHERE MIN(salary) 5000 AND MAX(salary) 15000 GROUP BY dept_id;CSELECT dept_id, MIN(salary), MAX(salary) FROM employees HAVING MIN(salary) 5000 AND MAX (salary) 15000;DSELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id HAVING MIN (salary) 5000 AND MAX(salary)ESELECT dept_id, MIN(salary), MAX(salary) FROM employees GROUP BY dept_id, salary HAVING MIN (salary) 5000 AND MAX (salary) 15000;
单选题以下函数findmax拟实现在数组中查找最大值并作为函数值返回,但程序中有错导致不能实现预定功能。#define MIN -2147483647int fingmax (int x[],int n){ int i,max; for(i=0;iA定义语句int i,max;中,max未赋初值B赋值语句max=MIN;中,不应给max赋MIN值C语句if(maxx[i])max=x[i];中,判断条件设置错误D赋值语句max=MIN;放错了位置
单选题X线摄影线量最强波长的计算公式为( )。Aλmin=1.24kVp(nm)Bλmax=1.5λmin(nm)Cλmax=2.5λmin(nm)Dλmax=2λmin(nm)Eλmin=12.4/kVp(nm)
单选题有如下程序:#include struct pair{ int first,second;};struct pair get_min_max(int*array, int len){ int i; struct pair res; res.first=array[0]; res.second=array[0]; for(i=1;ires.second) res.second=array[i]; } return res;}main(){ int array[5]={9,1,3,4}; struct pair min_max = get_min_max(array,5); printf(min=%d,max=%d, min_max.first, min_max.second);}程序运行后的输出结果是( )。Amin=1,max=9Bmin=0,max=9Cmin=1,max=4Dmin=0,max=4