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: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)
\(二次型f(x)=x^{T}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}x的秩为\)
\(二次型f(x)=x^{T}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}x的秩为\)
(1)7X=5分之3(2)12分之5x=8分之3(3)X÷9分之4=12分之7(4)3分之2X÷4分之1=9分之8
(1)7X=5分之3(2)12分之5x=8分之3(3)X÷9分之4=12分之7(4)3分之2X÷4分之1=9分之8
set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为? A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为? A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
掷一枚均匀骰子,直到出现的点数小于3为止,记抛掷的次数为X,则以下结果正确的是 A: P(X=2)=2/9 B: P(X≥3)=4/9 C: P(X≤3)=19/27 D: P(X=1)=2/3 E: P(X≤2)=3/4 F: P(X=1)=1/2 G: P(X=2)=1/4 H: P(X<3)=7/8
掷一枚均匀骰子,直到出现的点数小于3为止,记抛掷的次数为X,则以下结果正确的是 A: P(X=2)=2/9 B: P(X≥3)=4/9 C: P(X≤3)=19/27 D: P(X=1)=2/3 E: P(X≤2)=3/4 F: P(X=1)=1/2 G: P(X=2)=1/4 H: P(X<3)=7/8
假设“☆”是一种新的运算,若3☆2=3×4,6☆3=6×7×8,x☆4=840(x>0),那么x等于: A: 2 B: 3 C: 4 D: 5 E: 6 F: 7 G: 8 H: 9
假设“☆”是一种新的运算,若3☆2=3×4,6☆3=6×7×8,x☆4=840(x>0),那么x等于: A: 2 B: 3 C: 4 D: 5 E: 6 F: 7 G: 8 H: 9
若要将一个长度为N=16的序列x(n)重新位倒序,作为某一FFT算法的输入,则位倒序后序列的样本序号为( )。 A: x(15), x(14), x(13), x(12), x(11), x(10), x(9), x(8), x(7), x(6),<br/>x(5), x(4), x(3), x(2), x(1), x(0) B: x(0), x(4), x(2), x(6), x(1), x(5), x(3), x(7), x(8), x(12), x(10),<br/>x(14), x(9), x(13), x(11), x(15) C: x(0), x(2), x(4), x(6), x(8), x(10), x(12), x(14), x(1), x(3), x(5),<br/>x(7), x(9), x(11), x(13), x(15) D: x(0), x(8), x(4), x(12), x(2), x(10), x(6), x(14), x(1), x(9), x(5),<br/>x(13), x(3), x(11), x(7), x(15)
若要将一个长度为N=16的序列x(n)重新位倒序,作为某一FFT算法的输入,则位倒序后序列的样本序号为( )。 A: x(15), x(14), x(13), x(12), x(11), x(10), x(9), x(8), x(7), x(6),<br/>x(5), x(4), x(3), x(2), x(1), x(0) B: x(0), x(4), x(2), x(6), x(1), x(5), x(3), x(7), x(8), x(12), x(10),<br/>x(14), x(9), x(13), x(11), x(15) C: x(0), x(2), x(4), x(6), x(8), x(10), x(12), x(14), x(1), x(3), x(5),<br/>x(7), x(9), x(11), x(13), x(15) D: x(0), x(8), x(4), x(12), x(2), x(10), x(6), x(14), x(1), x(9), x(5),<br/>x(13), x(3), x(11), x(7), x(15)
If<em>g</em>(<em>f</em>(<em>x</em>))=2<em>x</em>+1and<em>f</em>(<em>x</em>)=-1/4<em>x</em>-1,then<em>g</em>(<em>x</em>)= A: -8/7 B: 8<em>x</em>+9 C: 1/4(2<em>x</em>+1) D: 8<em>x</em>-8 E: 8<em>x</em>
If<em>g</em>(<em>f</em>(<em>x</em>))=2<em>x</em>+1and<em>f</em>(<em>x</em>)=-1/4<em>x</em>-1,then<em>g</em>(<em>x</em>)= A: -8/7 B: 8<em>x</em>+9 C: 1/4(2<em>x</em>+1) D: 8<em>x</em>-8 E: 8<em>x</em>
设函数$y = f({x^3})$可导,求函数的二阶导数$y'' = $( ) A: $6xf'({x^3}) + 9{x^4}f''({x^3})$ B: $6f'({x^3}) + 9{x^3}f''({x^3})$ C: $6xf'({x^3}) + 9{x^3}f''({x^3})$ D: $6{x^2}f'({x^3}) + 9{x^3}f''({x^3})$
设函数$y = f({x^3})$可导,求函数的二阶导数$y'' = $( ) A: $6xf'({x^3}) + 9{x^4}f''({x^3})$ B: $6f'({x^3}) + 9{x^3}f''({x^3})$ C: $6xf'({x^3}) + 9{x^3}f''({x^3})$ D: $6{x^2}f'({x^3}) + 9{x^3}f''({x^3})$
【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=
【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=