已知T(1)=9,T(2)=8,T(0)=5,Total=T(1)+T(2)+T(0),则Total=()。 A: 9 B: 22 C: 8 D: 5
已知T(1)=9,T(2)=8,T(0)=5,Total=T(1)+T(2)+T(0),则Total=()。 A: 9 B: 22 C: 8 D: 5
已知T(1)=9,T(2)=8,T(0)=5,Total=T(1)+T(2)+T(0),则Total=()。
已知T(1)=9,T(2)=8,T(0)=5,Total=T(1)+T(2)+T(0),则Total=()。
计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。 A: \(2\pi {a^7}\) B: \(2\pi {a^6}\) C: \(2\pi {a^5}\) D: \(2\pi {a^8}\)
计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。 A: \(2\pi {a^7}\) B: \(2\pi {a^6}\) C: \(2\pi {a^5}\) D: \(2\pi {a^8}\)
8、信号f(t)=3cos(4t+4π/3)的周期是( ) A: 2π B: π C: π/2 D: π/4
8、信号f(t)=3cos(4t+4π/3)的周期是( ) A: 2π B: π C: π/2 D: π/4
8、信号f(t)=3cos(4t+π/3)的周期是( ) A: 2π B: π C: π/2 D: π/4
8、信号f(t)=3cos(4t+π/3)的周期是( ) A: 2π B: π C: π/2 D: π/4
求向量组a1=(1,2,-1,4)T,a2=(9,100,10,4)T,a3=(-2,-4,2,-8
求向量组a1=(1,2,-1,4)T,a2=(9,100,10,4)T,a3=(-2,-4,2,-8
Fill in the blankFor the expressionf(t)=tε(t)+2ε(t−2)−tε(t−2),f(3)=______ .
Fill in the blankFor the expressionf(t)=tε(t)+2ε(t−2)−tε(t−2),f(3)=______ .
(3B6D.14)()16()=(?)()2()(1100100101.0101)()2()=(?)()8()(239.375)()10()=(?)()2()(365.5)()8()=(?)()10()(115.4375)()10()=(?)()2()=(?)()8()=(?)()16()(2A.3C)()16()=(?)()10()=(?)()2()(126.75)()8()=(?)()16()=(?)()2()(1001101.10101)()2()=(?)()16()=(?)()8
(3B6D.14)()16()=(?)()2()(1100100101.0101)()2()=(?)()8()(239.375)()10()=(?)()2()(365.5)()8()=(?)()10()(115.4375)()10()=(?)()2()=(?)()8()=(?)()16()(2A.3C)()16()=(?)()10()=(?)()2()(126.75)()8()=(?)()16()=(?)()2()(1001101.10101)()2()=(?)()16()=(?)()8
一阶常微分方程[img=152x26]1802e4d6075ee4f.png[/img]的通解为 A: sin(2*t)/5-cos(2*t)/10+C*exp(-4*t) B: sin(2*t)/7+cos(2*t)/5-C*exp(-3*t) C: sin(2*t)/7-C*cos(2*t)/10+C*exp(-2*t) D: sin(2*t)/7-cos(2*t)/7+C*exp(-5*t)
一阶常微分方程[img=152x26]1802e4d6075ee4f.png[/img]的通解为 A: sin(2*t)/5-cos(2*t)/10+C*exp(-4*t) B: sin(2*t)/7+cos(2*t)/5-C*exp(-3*t) C: sin(2*t)/7-C*cos(2*t)/10+C*exp(-2*t) D: sin(2*t)/7-cos(2*t)/7+C*exp(-5*t)
假设检验的拒绝域是()。 A: (-∞,-z<sub>α/2</sub>]∪[z<sub>α/2</sub>,+∞) B: (-∞,-t<sub>α/2</sub>]∪[t<sub>α/2</sub>,+∞),t<sub>α/2</sub>=t<sub>α/2</sub>(n) C: (-∞,-t<sub>α/2</sub>]∪[t<sub>α/2</sub>,+∞),t<sub>α/2</sub>=t<sub>α/2</sub>(n-1) D: (t<sub>α</sub>,+∞)
假设检验的拒绝域是()。 A: (-∞,-z<sub>α/2</sub>]∪[z<sub>α/2</sub>,+∞) B: (-∞,-t<sub>α/2</sub>]∪[t<sub>α/2</sub>,+∞),t<sub>α/2</sub>=t<sub>α/2</sub>(n) C: (-∞,-t<sub>α/2</sub>]∪[t<sub>α/2</sub>,+∞),t<sub>α/2</sub>=t<sub>α/2</sub>(n-1) D: (t<sub>α</sub>,+∞)