A: 3000N•m
B: 4000N•m
C: 5000N•m
D: 6000N•m
举一反三
- 法兰底座转盘的最大扭矩是（）。 A: A3000N•m B: B4000N•m C: C5000N•m D: D6000N•m
- 交换m,n数值的方法 A: m,n=n,m B: n,m=m,n C: m,n=m,n D: n,m=n,m
- 【单选题】如果m>0,n<0,m<|n|,那么m,n,-m,-n的大小关系是() A. -n＞m＞-m＞n B. m＞n＞-m＞-n C. -n＞m＞n＞-m D. n＞m＞-n＞-m
- 下面程序的功能是用“辗转相除法”求两个正整数的最大公约数。请分析程序填空。#includemain(){intr,m,n;scanf("%d%d",&m,&n);if(m A: 【1】r=m,m=n,n=r;【2】m%n; B: 【1】m%n;【2】r=m,m=n,n=r; C: 【1】r=m,m=n,n=r;【2】n%m; D: 【1】n%m;【2】r=m,m=n,n=r;
- 设A与B为矩阵,AC=CB，C为m×n矩阵，则A与B分别是( )矩阵 A: n×m,m×n B: m×n,n×m C: n×n,m×m D: m×m,n×n
内容
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求给定自然数m和n的最大公约数。请完善程序。 Private Sub Command1_Click() Dim m As Integer, n As Integer, r As Integer m = Text1.Text n = Text2.Text If m < n Then r = m: m = n: n = r Do r = m Mod n m = n n = r Loop Until__________ Text3.Text = m End Sub Private Sub Command2_Click() Dim m As Integer, n As Integer, r As Integer m = Text1.Text n = Text2.Text If m < n Then r = m: m = n: n = r r = m Mod n Do Until _______ m = n n = r r = m Mod n Loop Text3.Text =____ End Sub Private Sub Command3_Click() Dim m As Integer, n As Integer, r As Integer m = Text1.Text n = Text2.Text If m < n Then r = m: m = n: n = r Do r = m Mod n m = n n = r Loop While________ Text3.Text =______ End Sub Private Sub Command4_Click() Dim m As Integer, n As Integer, r As Integer m = Text1.Text n = Text2.Text If m < n Then r = m: m = n: n = r r = m Mod n Do While______ m = n n = r r = m Mod n Loop Text3.Text =____ End Sube73f1fd63959cc8362d8c82d419c2353.png
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求极限[img=144x45]1803d6afb8a008a.png[/img]的MATLAB程序为：syms x m n ay=(x^m-a^m)/(x^n-a^n)limit(y,x,a)计算结果为： A: (a^(m - n)*m)/n B: a^(m - n)*m/n C: 无法求值 D: a^(m - n)*m*n E: a^(m - n)/m/n F: (a^m - n)*m)/n
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有以下程序#includemain(){intm,n;scanf("%d%d",&m,&n);while(m!=n){while(m>n){m=m-n;}while(n>m){n=n-m;}}printf("%d",m);}该程序的功能是 A: 计算m和n的最小公倍数 B: 计算m和n的最大公约数 C: c D: 找出m和n中的较大值
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应力的单位是（）。 A: N B: N•m C: N／m²； D: N／m
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The Lame strain potential has the dimension of ( ). A: N·m B: N C: N/m D: N/m/m