计算:[tex=7.071x2.214]MqOfsQLAB/zeVSdv1WggGP/8ImYsCrL4mENpKvyIKvDIVd+T8LGLCoDzl+DK77Tcq7/GOls5QuN/gGHzbZWbvM+7sVXeQOT0KVJM7Gh2zhc=[/tex]
计算:[tex=7.071x2.214]MqOfsQLAB/zeVSdv1WggGP/8ImYsCrL4mENpKvyIKvDIVd+T8LGLCoDzl+DK77Tcq7/GOls5QuN/gGHzbZWbvM+7sVXeQOT0KVJM7Gh2zhc=[/tex]
求极限: [tex=7.071x2.214]MqOfsQLAB/zeVSdv1WggGP/8ImYsCrL4mENpKvyIKvDIVd+T8LGLCoDzl+DK77Tcq7/GOls5QuN/gGHzbZWbvM+7sVXeQOT0KVJM7Gh2zhc=[/tex].
求极限: [tex=7.071x2.214]MqOfsQLAB/zeVSdv1WggGP/8ImYsCrL4mENpKvyIKvDIVd+T8LGLCoDzl+DK77Tcq7/GOls5QuN/gGHzbZWbvM+7sVXeQOT0KVJM7Gh2zhc=[/tex].
下列音节拼写上如有错误,请加以改正,并说明修改理由: qun zhong(群众)
下列音节拼写上如有错误,请加以改正,并说明修改理由: qun zhong(群众)
One of the common clothing styles during the Song Dynasty was A: Ao qun (袄裙) B: Beizi (褙子) C: Changpao D: Mianfu(冕服)
One of the common clothing styles during the Song Dynasty was A: Ao qun (袄裙) B: Beizi (褙子) C: Changpao D: Mianfu(冕服)
One of the common clothing styles during the Song Dynasty was ( ) A: Ao qun (袄裙) B: Beizi (褙子) C: Changpao (长袍) D: Mianfu(冕服)
One of the common clothing styles during the Song Dynasty was ( ) A: Ao qun (袄裙) B: Beizi (褙子) C: Changpao (长袍) D: Mianfu(冕服)
One of the common clothing styles during the Song Dynasty was____( )。 A: Mianfu(冕服) B: Beizi (褙子) C: Changpao D: Ao qun (袄裙)
One of the common clothing styles during the Song Dynasty was____( )。 A: Mianfu(冕服) B: Beizi (褙子) C: Changpao D: Ao qun (袄裙)
【单选题】要指定一个绝对位置 5 , 5 的点,应该输入: () @5 , 5 5 》 5 5 , 5 5<5< span=""> A. @5,5 B. 5》5 C. 5,5 D. 5<5
【单选题】要指定一个绝对位置 5 , 5 的点,应该输入: () @5 , 5 5 》 5 5 , 5 5<5< span=""> A. @5,5 B. 5》5 C. 5,5 D. 5<5
根据乘方的意义可知:53=5×5×5,54=5×5×5×5,则53×54=(5×5×5)×(5×5×5×5)=57即53×54=57。想一想:
根据乘方的意义可知:53=5×5×5,54=5×5×5×5,则53×54=(5×5×5)×(5×5×5×5)=57即53×54=57。想一想:
【单选题】下面程序执行结果() #include main() { int i ,j ; for(i=1; i<=5; i++) { for(j=1; j<=i; j++) printf("%2d",i); printf(" ",); } } A. 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 B. 1 2 3 4 5 C. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D. 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
【单选题】下面程序执行结果() #include main() { int i ,j ; for(i=1; i<=5; i++) { for(j=1; j<=i; j++) printf("%2d",i); printf(" ",); } } A. 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 B. 1 2 3 4 5 C. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 D. 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5
5^2 5^3=5^5
5^2 5^3=5^5