已知定积分[img=99x40]17e44861dad35af.jpg[/img],且f(x)是偶函数,则[img=86x40]17e44861e61cea9.jpg[/img]()
已知定积分[img=99x40]17e44861dad35af.jpg[/img],且f(x)是偶函数,则[img=86x40]17e44861e61cea9.jpg[/img]()
能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是: (x>= -58) && (x<= -40) && (x>=40) && (x<=58)|(x>= -58) && (x<= -40) || (x>=40) && (x<=58)|(x>= -58) | |(x<= -40) && (x>=40) || (x<=58)|(x>= -58) || (x<= -40) || (x>=40) || (x<=58)
能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是: (x>= -58) && (x<= -40) && (x>=40) && (x<=58)|(x>= -58) && (x<= -40) || (x>=40) && (x<=58)|(x>= -58) | |(x<= -40) && (x>=40) || (x<=58)|(x>= -58) || (x<= -40) || (x>=40) || (x<=58)
能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是()。 A: (x>=-58)&&(x<=-40)&&(x>=40)&&(x<=58) B: (x>=-58)∥(x<=-40)∥(x>=40)∥(x<=58) C: (x>=-58)&&(x<=-40)∥(x>=40)&&(x<=58)
能正确表示“当x的取值在[-58,-40]和[40,58]范围内为真,否则为假”的表达式是()。 A: (x>=-58)&&(x<=-40)&&(x>=40)&&(x<=58) B: (x>=-58)∥(x<=-40)∥(x>=40)∥(x<=58) C: (x>=-58)&&(x<=-40)∥(x>=40)&&(x<=58)
假定16<X≤40,那么用边界值分析法,X在测试中应该取的边界值是: A: X=16,X=17,X=40,X=41 B: X=15,X=16,X=40,X=41 C: X=16,X=17,X=39,X=40 D: X=15,X=16,X=39,X=40
假定16<X≤40,那么用边界值分析法,X在测试中应该取的边界值是: A: X=16,X=17,X=40,X=41 B: X=15,X=16,X=40,X=41 C: X=16,X=17,X=39,X=40 D: X=15,X=16,X=39,X=40
(–86)原=11010110,(–86)反=10101001,(–86)补=10101010
(–86)原=11010110,(–86)反=10101001,(–86)补=10101010
设有变量定义:x=(10,20,30)和y=[10,20,30],以下正确的赋值语句是_______。 A: x(2)=40 B: x[2]=40 C: y(2)=40 D: y[2]=40
设有变量定义:x=(10,20,30)和y=[10,20,30],以下正确的赋值语句是_______。 A: x(2)=40 B: x[2]=40 C: y(2)=40 D: y[2]=40
“int x=40;”,声明了一个字符型类的x变量,其默认值为40。
“int x=40;”,声明了一个字符型类的x变量,其默认值为40。
某公司向工商银行申请了甲、乙两种贷款,共计86万元,每年需付利息7.42万元。甲种贷款每年的利率是12%,乙种贷款每年的利率是13%。如果设甲种贷款的数额为x万元,那么根据题意可以列出方程是( ) A: 12%x+13%(7.42-x)=86 B: 12%(7.42-x)+13%x=86 C: 12%x+13%(86-x)=7.42 D: 12%(86-x)+13%x=7.42
某公司向工商银行申请了甲、乙两种贷款,共计86万元,每年需付利息7.42万元。甲种贷款每年的利率是12%,乙种贷款每年的利率是13%。如果设甲种贷款的数额为x万元,那么根据题意可以列出方程是( ) A: 12%x+13%(7.42-x)=86 B: 12%(7.42-x)+13%x=86 C: 12%x+13%(86-x)=7.42 D: 12%(86-x)+13%x=7.42
【单选题】#include <stdio.h> int x1 = 30, x2 = 40; void sub(int x, int y) { x1 = x; x = y; y = x1; } int main() { int x3 = 10, x4 = 20; sub(x3, x4); sub(x2, x1); printf(" %d, %d, %d, %d ", x3, x4, x1, x2); return 0; } A. 10, 20, 40, 40 B. 10,40, 40, 40 C. 20, 20, 40, 40 D. 10, 10, 40, 40
【单选题】#include <stdio.h> int x1 = 30, x2 = 40; void sub(int x, int y) { x1 = x; x = y; y = x1; } int main() { int x3 = 10, x4 = 20; sub(x3, x4); sub(x2, x1); printf(" %d, %d, %d, %d ", x3, x4, x1, x2); return 0; } A. 10, 20, 40, 40 B. 10,40, 40, 40 C. 20, 20, 40, 40 D. 10, 10, 40, 40
以4,9,1为为插值节点,求\(\sqrt x \)的lagrange的插值多项式 A: \( {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) B: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) C: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x +1) + {1 \over {24}}(x - 4)(x - 9)\) D: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)
以4,9,1为为插值节点,求\(\sqrt x \)的lagrange的插值多项式 A: \( {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) B: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) C: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x +1) + {1 \over {24}}(x - 4)(x - 9)\) D: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)