• 2021-04-14 问题

    设f(cos x) = 3 − cos 2x,则f(sin x) =()

    设f(cos x) = 3 − cos 2x,则f(sin x) =()

  • 2022-06-04 问题

    设函数f(x)=cos(x+π3

    设函数f(x)=cos(x+π3

  • 2022-06-09 问题

    设\(z = f(x,y)\),\(x = \sin t\),\(y = {t^3}\),则全导数\( { { dz} \over {dt}} = \) A: \({f'_x} \sin t+ 3{t^2}{f'_y}\) B: \({f'_x} \cos t+ {t^2}{f'_y}\) C: \({f'_x} \cos t+ 3{t^2}{f'_y}\) D: \({f'_y} \cos t+ 3{t^2}{f'_x}\)

    设\(z = f(x,y)\),\(x = \sin t\),\(y = {t^3}\),则全导数\( { { dz} \over {dt}} = \) A: \({f'_x} \sin t+ 3{t^2}{f'_y}\) B: \({f'_x} \cos t+ {t^2}{f'_y}\) C: \({f'_x} \cos t+ 3{t^2}{f'_y}\) D: \({f'_y} \cos t+ 3{t^2}{f'_x}\)

  • 2022-05-30 问题

    将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)

    将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)

  • 2022-06-05 问题

    f(x)=cos(2x-丌/3)化为sin

    f(x)=cos(2x-丌/3)化为sin

  • 2022-06-06 问题

    Which two pairings match the application with the correct Cos value?() A: CoS 0 = Best-Effort Data B: CoS 1 = Best-Effort Data C: CoS 3 = High Priority D: CoS 4 = High Priority Data E: CoS 5 = Voice Bearer F: CoS 7 = Voice Bearer

    Which two pairings match the application with the correct Cos value?() A: CoS 0 = Best-Effort Data B: CoS 1 = Best-Effort Data C: CoS 3 = High Priority D: CoS 4 = High Priority Data E: CoS 5 = Voice Bearer F: CoS 7 = Voice Bearer

  • 2022-11-02 问题

    该二元阵的阵因子为( )(以观察方向与阵轴的夹角δ为自变量)。 A: F(δ)=|cos[π(cosδ-1)/4]| B: F(δ)=|cos[π(sinδ-1)/2]| C: F(δ)=|cos[π(cosδ-1)]| D: F(δ)=|cos[π(sinδ-1)/4]|

    该二元阵的阵因子为( )(以观察方向与阵轴的夹角δ为自变量)。 A: F(δ)=|cos[π(cosδ-1)/4]| B: F(δ)=|cos[π(sinδ-1)/2]| C: F(δ)=|cos[π(cosδ-1)]| D: F(δ)=|cos[π(sinδ-1)/4]|

  • 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

  • 1 2 3 4 5 6 7 8 9 10