• 2022-06-07 问题

    Which one of the following sequences is not covergent? A: un=∑nk=1sink2k,n=1,2,⋯. B: un=cos(1!)1⋅2+cos(2!)2⋅3+cos(3!)3⋅4+⋯+cos(n!)n⋅(n+1),n=1,2,⋯. C: un=∑nk=1(−1)k−11k,n=1,2,⋯. D: un=(1+3n(−1)n)1/n,n=1,2,⋯.

    Which one of the following sequences is not covergent? A: un=∑nk=1sink2k,n=1,2,⋯. B: un=cos(1!)1⋅2+cos(2!)2⋅3+cos(3!)3⋅4+⋯+cos(n!)n⋅(n+1),n=1,2,⋯. C: un=∑nk=1(−1)k−11k,n=1,2,⋯. D: un=(1+3n(−1)n)1/n,n=1,2,⋯.

  • 2021-04-14 问题

    <img src="http://edu-image.nosdn.127.net/2507E32A7888F1F05F34CD6088FE894F.png?imageView&thumbnail=890x0&quality=100" />? AC+AB×cosθ1=BC×cosθ3; AB×sinθ1=BC×sinθ3<br >|AC+AB×cosθ1=BC×cosθ3; AB×cosθ1=BCcos×θ3<br >|AB×sinθ1=BC×cosθ3; AC+AB×cosθ1=BC×sinθ3|;AB×cosθ1=BC×cosθ3; AC+AB×sinθ1=BC×sinθ3<br >

    <img src="http://edu-image.nosdn.127.net/2507E32A7888F1F05F34CD6088FE894F.png?imageView&thumbnail=890x0&quality=100" />? AC+AB×cosθ1=BC×cosθ3; AB×sinθ1=BC×sinθ3<br >|AC+AB×cosθ1=BC×cosθ3; AB×cosθ1=BCcos×θ3<br >|AB×sinθ1=BC×cosθ3; AC+AB×cosθ1=BC×sinθ3|;AB×cosθ1=BC×cosθ3; AC+AB×sinθ1=BC×sinθ3<br >

  • 2022-06-16 问题

    已知向量a=(2,2,1),则a的方向余弦为(). A: cosα=2/3,cosβ=2/3,cosγ=1/3 B: cosα=2/5,cosβ=2/5,cosγ=1/5

    已知向量a=(2,2,1),则a的方向余弦为(). A: cosα=2/3,cosβ=2/3,cosγ=1/3 B: cosα=2/5,cosβ=2/5,cosγ=1/5

  • 2022-06-06 问题

    复数12-32i的三角形式是(  ) A: cos(-π3)+isin(-π3) B: cosπ3+isinπ3 C: cosπ3-isinπ3 D: cosπ3+isin5π6

    复数12-32i的三角形式是(  ) A: cos(-π3)+isin(-π3) B: cosπ3+isinπ3 C: cosπ3-isinπ3 D: cosπ3+isin5π6

  • 2022-06-05 问题

    以\( (2,2,1) \)为起点,以\( (1,3,0) \)为终点的向量的方向余弦为( ). A: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = {1 \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) B: \( \cos \alpha = {1 \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) C: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) D: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = {1 \over {\sqrt 3 }} \)

    以\( (2,2,1) \)为起点,以\( (1,3,0) \)为终点的向量的方向余弦为( ). A: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = {1 \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) B: \( \cos \alpha = {1 \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) C: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) D: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = {1 \over {\sqrt 3 }} \)

  • 2022-05-27 问题

    (4)$A$矢量的方向余弦(与三个坐标轴的夹角余弦)的大小是: A: $cos\alpha=3/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ B: $cos\alpha=4/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ C: $cos\alpha=2/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ D: $cos\alpha=3/\sqrt{14},cos\beta=9/\sqrt{14},cos\gamma=3/\sqrt{14}$

    (4)$A$矢量的方向余弦(与三个坐标轴的夹角余弦)的大小是: A: $cos\alpha=3/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ B: $cos\alpha=4/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ C: $cos\alpha=2/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$ D: $cos\alpha=3/\sqrt{14},cos\beta=9/\sqrt{14},cos\gamma=3/\sqrt{14}$

  • 2022-07-24 问题

    求微分方程[img=634x60]17da653955cf9e7.png[/img]的特解。 ( ) A: sin(2*x)/3 - cos(x) - cos(x)/3 B: sin(2*x)/3 - cos(x) - sin(x)/3 C: cos(2*x)/3 - cos(x) - sin(x)/3 D: sin(2*x)/3 - sin(x) - sin(x)/3

    求微分方程[img=634x60]17da653955cf9e7.png[/img]的特解。 ( ) A: sin(2*x)/3 - cos(x) - cos(x)/3 B: sin(2*x)/3 - cos(x) - sin(x)/3 C: cos(2*x)/3 - cos(x) - sin(x)/3 D: sin(2*x)/3 - sin(x) - sin(x)/3

  • 2022-06-06 问题

    生成{Cos&#91;1&#93;,Cos&#91;2&#93;,Cos&#91;3&#93;}。 A: Table[Cos(k),{k,3}] B: a={1,2,3}; Cos[a] C: Table[Cos[k],{k,3}] D: Table[{Cos[k]},{k,3}]

    生成{Cos&#91;1&#93;,Cos&#91;2&#93;,Cos&#91;3&#93;}。 A: Table[Cos(k),{k,3}] B: a={1,2,3}; Cos[a] C: Table[Cos[k],{k,3}] D: Table[{Cos[k]},{k,3}]

  • 2021-04-14 问题

    【计算题】已知sinα+cosα=1,求:(1)sinαcosα; (2)sin α-cos α; (3)sin α-cos α

    【计算题】已知sinα+cosα=1,求:(1)sinαcosα; (2)sin α-cos α; (3)sin α-cos α

  • 2022-06-03 问题

    已知\( y = \sin (2 + \tan 3) \),则\( y' \)为( ). A: 0 B: \( \cos (2 + \tan 3) \) C: \( \tan 3\cos (2 + \tan 3) \) D: \( {\sec ^2}3\cos (2 + \tan 3) \)

    已知\( y = \sin (2 + \tan 3) \),则\( y' \)为( ). A: 0 B: \( \cos (2 + \tan 3) \) C: \( \tan 3\cos (2 + \tan 3) \) D: \( {\sec ^2}3\cos (2 + \tan 3) \)

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