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【单选题】
关于柴油机连杆受力情况的论述不正确的是( )。
A.
增压二冲程柴油机连杆受压力作用
B.
.四冲程柴油机连杆受拉压交变作用
C.
气体力使连杆受压,往上的惯性力使连杆受拉
D.
连杆不受弯矩作用
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【单选题】以下程序运行时,若从键盘输入 9,则输出结果 是 ( ) 。 #inlude int main() { int a=1 ; scanf(" %d ",&a); if(a++<9) printf(" %d\n ",a); else printf(" %d\n ",a); return 0; }
A.
8
B.
9
C.
10
D.
11
【单选题】以下程序运行时,若从键盘输入9,则输出结果是_______。 main() { int a=1; scanf("%d",&a); if(a++<9) printf("%d\n",a); else printf("%d\n",a); }
A.
10
B.
11
C.
9
D.
8
【单选题】In doing "aggregate planning" for a firm producing paint, the aggregate planners would most likely deal with:
A.
just gallons of paint, without concern for the different colors and sizes
B.
gallons of paint, but be concerned with the different colors to be produced
C.
gallons, quarts, pints, and all the different sizes to be produced
D.
all the different sizes and all the different colors by size
E.
none of the above
【单选题】Aggregate planners seek to match supply and demand:
A.
at minimum overall cost
B.
by staying within company policy
C.
(a) and (b)
D.
keeping inventories at a minimum
E.
all of the above
【单选题】Aggregate planners attempt to balance:
A.
demand and inventories
B.
demand and costs
C.
capacity and inventories
D.
capacity and costs
E.
capacity and demand
【单选题】The main disadvantage(s) of informal techniques used for aggregate planning is(are):
A.
they are expensive to do
B.
they may not result in the best plan
C.
they take a long time to do
D.
they require use of a computer
E.
lack of formal education of the planners
【单选题】Accommodating peak demands and effectively using labor resources during periods of low demand would be the goal of aggregate planners in 
A.
Manufacturing
B.
Military
C.
Archeology
D.
Libraries
E.
Financial Services
【单选题】以下程序运行时,若从键盘输入9,则输出结果是_______。 #inlude main() { int a=1; scanf(" %d ",&a); if(a++<9) printf(" %d\n ",a); else printf(" %d\n ",a);
A.
10
B.
11
C.
9
D.
8
【判断题】有以下程序(说明:字符0的ASCII码值为48) #include main() {char c1,c2; scanf("%d",&c1); c2=c1+9; printf("%c%c\n",c1,c2); } 若程序运行时从键盘输入48,则输出结果为 09
A.
正确
B.
错误
【单选题】以下程序运行时,若从键盘输入9,则输出结果是( )。#includeint main(){int a=1;scanf("%d",&a);if(a++<9) printf("%d\n",a);else printf("%d\n",a);}
A.
10
B.
11
C.
9
D.
8
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