• 2022-06-19 问题

    background:url(2、png),url(1、jpg),url(3、png),url(4、jpg);},表示哪张图片处在最上层() A: 2、png B: 1、jpg C: 3、png D: 4、jpg

    background:url(2、png),url(1、jpg),url(3、png),url(4、jpg);},表示哪张图片处在最上层() A: 2、png B: 1、jpg C: 3、png D: 4、jpg

  • 2021-04-14 问题

    background: url(2、png),url(1、jpg),url(3、png),url(4、jpg);},表示哪张图片处在最上层()。

    background: url(2、png),url(1、jpg),url(3、png),url(4、jpg);},表示哪张图片处在最上层()。

  • 2022-06-04 问题

    ∫xe^(x^2)dx=( ) A: 1/2(e^(x^2)) B: 1/2(e^(x^2))+C C: -1/2(e^(x^2)) D: -1/2(e^(x^2))十C

    ∫xe^(x^2)dx=( ) A: 1/2(e^(x^2)) B: 1/2(e^(x^2))+C C: -1/2(e^(x^2)) D: -1/2(e^(x^2))十C

  • 2021-04-14 问题

    【单选题】如图, AB 两点间电压 U AB =______ 。 A . E 1 - E 2 - IR B . E 2 - E 1 - IR C . E 2 - E 1 + IR D . E 1 - E 2 + IR A. E 1 - E 2 - IR B. E 2 - E 1 - IR C. E 2 - E 1 + IR D. E 1 - E 2 + IR

    【单选题】如图, AB 两点间电压 U AB =______ 。 A . E 1 - E 2 - IR B . E 2 - E 1 - IR C . E 2 - E 1 + IR D . E 1 - E 2 + IR A. E 1 - E 2 - IR B. E 2 - E 1 - IR C. E 2 - E 1 + IR D. E 1 - E 2 + IR

  • 2022-07-29 问题

    设矩阵\({A^k} = O \),则\({(E - A)^{ - 1}} = \) A: \(E + A + {A^2} + ... + {A^{k - 1}} \) B: \( A + {A^2} + ... + {A^{k - 1}}\) C: \(E + A + {A^2} + ... + {A^{k }}\) D: \(E + {A^2} + ... + {A^{k - 1}}\)

    设矩阵\({A^k} = O \),则\({(E - A)^{ - 1}} = \) A: \(E + A + {A^2} + ... + {A^{k - 1}} \) B: \( A + {A^2} + ... + {A^{k - 1}}\) C: \(E + A + {A^2} + ... + {A^{k }}\) D: \(E + {A^2} + ... + {A^{k - 1}}\)

  • 2022-06-26 问题

    高分子溶解在良溶剂中,则( )。 A: χ1>1/2, Δμ1 E>0 B: χ1>1/2, Δμ1 E<0 C: χ1<1/2, Δμ1E>0, D: χ1<1/2, Δμ1 E<0

    高分子溶解在良溶剂中,则( )。 A: χ1>1/2, Δμ1 E>0 B: χ1>1/2, Δμ1 E<0 C: χ1<1/2, Δμ1E>0, D: χ1<1/2, Δμ1 E<0

  • 2022-06-11 问题

    <img src="data:image/png;base64,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" />=( ) A: (atan(x^(1/2))+C B: (atan(x)^2+C C: (atan(x^(1/2))^2+C D: (atan(x^2)^(1/2)+C

    <img src="data:image/png;base64,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" />=( ) A: (atan(x^(1/2))+C B: (atan(x)^2+C C: (atan(x^(1/2))^2+C D: (atan(x^2)^(1/2)+C

  • 2022-05-26 问题

    其假设检验的Ho为 A: μ1>μ2 B: μ1<μ2 C: μ1≠μ2 D: μ1=μ2 E: μ1≥μ2

    其假设检验的Ho为 A: μ1>μ2 B: μ1<μ2 C: μ1≠μ2 D: μ1=μ2 E: μ1≥μ2

  • 2022-06-08 问题

    估计积分\(\int_2^0 { { e^ { { x^2} - x}}} dx\)的值为( )。(利用估值定理) A: \([ - 2{e^2}, - 2{e^{ - {1 \over 4}}}]\) B: \([ - 2{e^2}, - 2{e^ { { 1 \over 4}}}]\) C: \([2{e^2},2{e^{ - {1 \over 4}}}]\) D: \([2{e^2},2{e^ { { 1 \over 4}}}]\)

    估计积分\(\int_2^0 { { e^ { { x^2} - x}}} dx\)的值为( )。(利用估值定理) A: \([ - 2{e^2}, - 2{e^{ - {1 \over 4}}}]\) B: \([ - 2{e^2}, - 2{e^ { { 1 \over 4}}}]\) C: \([2{e^2},2{e^{ - {1 \over 4}}}]\) D: \([2{e^2},2{e^ { { 1 \over 4}}}]\)

  • 2022-05-30 问题

    正态分布、t分布、二项分布的参数各有几个? A: 1、1、2 B: 2、2、2 C: 1、1、1 D: 2、1、2 E: 2、2、1

    正态分布、t分布、二项分布的参数各有几个? A: 1、1、2 B: 2、2、2 C: 1、1、1 D: 2、1、2 E: 2、2、1

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