2017-06-29 10:29:00hzb53jl55v
exponential function question
標題:
exponential function question
發問:
^3√27=? ^3√-27=? ^3√64=? ^3√125=? 更新: And why teach 我thx 更新 2: √16 the square roots of 16 are____ 更新 3: ^5√32= ( 32 )^1/5 = ( 3x3x3 )^1/5 = ^6√64= 更新 4: thx thx thx 你真係very very good ar 更新 5: ^4√16= the 4th roots of 16are____ ^3√8= the cube root of 8 is____ 更新 6: ^4√25= √16^3= ^3√64/125= ^√-1/32= (^3√-20^3 and explain
最佳解答:
3√27 = 3 3√-27 = -3 3√64 = 4 3√125 = 5 2009-02-02 12:10:44 補充： OK... detailed steps: 3^√27 = ( 27 )^1/3 = ( 3x3x3 )^1/3 = ( 3^3 )^1/3 = 3^(3x1/3) = 3^1 = 3 3^√-27 = ( -27 )^1/3 = ( -3x-3x-3 )^1/3 = ( -3^3 )^1/3 = -3^(3x1/3) = -3^1 = -3 2009-02-02 12:11:03 補充： 3^√64 = ( 64 )^1/3 = ( 4x4x4 )^1/3 = ( 4^3 )^1/3 = 4^(3x1/3) = 4^1 = 4 3^√125 = ( 125 )^1/3 = ( 5x5x5 )^1/3 = ( 5^3 )^1/3 = 5^(3x1/3) = 5^1 = 5 If you don't understand, please let me know.. OK? 2009-02-02 12:18:27 補充： √16 square roots of 16 are 2 and -2. becareful: when the index is even number, it always has 2 roots. but for odd index, it only has 1 root 2009-02-02 12:19:07 補充： please ignore the above answer.. √16 square roots of 16 are 4 and -4. becareful: when the index is even number, it always has 2 roots. but for odd index, it only has 1 root 2009-02-02 12:22:10 補充： 2009-02-02 12:17:27 補充 5^√32= ( 32 )^1/5 = ( 2x2x2x2x2 )^1/5 = (2^5)^1/5 = 2^1 = 2 6^√64= (2x2x2x2x2x2)^1/6 = 2 AND = (-2x-2x-2x-2x-2x-2)^1/6 = -2 2009-02-02 12:23:21 補充： so... now you understand how to do? 2009-02-02 12:24:34 補充： 2009-02-02 12:23:10 補充 4^√16= the 4th roots of 16 are 2 and -2 3^√8= the cube root of 8 is 2 only. 2009-02-02 15:12:16 補充： 2009-02-02 14:00:36 補充 ^4√25 = ±√5... because √5x√5x√5x√5 = 25 AND -√5x-√5x-√5x-√5 = 25 √16^3 = 4^3 = 4x4x4 = 64 AND = -4^3 = -4x-4x-4 = -64 ^3√64/125 = ^3√(4x4x4) / (5x5x5) = 4/5 5^√-1/32 = 5^√(-1/5 x -1/5 x -1/5 x -1/5 x -1/5) = -1/5 ^3√-20^3 = -20^(1/3 x 3) = -20^1 = -20
其他解答:
exponential function question
發問:
^3√27=? ^3√-27=? ^3√64=? ^3√125=? 更新: And why teach 我thx 更新 2: √16 the square roots of 16 are____ 更新 3: ^5√32= ( 32 )^1/5 = ( 3x3x3 )^1/5 = ^6√64= 更新 4: thx thx thx 你真係very very good ar 更新 5: ^4√16= the 4th roots of 16are____ ^3√8= the cube root of 8 is____ 更新 6: ^4√25= √16^3= ^3√64/125= ^√-1/32= (^3√-20^3 and explain
最佳解答:
3√27 = 3 3√-27 = -3 3√64 = 4 3√125 = 5 2009-02-02 12:10:44 補充： OK... detailed steps: 3^√27 = ( 27 )^1/3 = ( 3x3x3 )^1/3 = ( 3^3 )^1/3 = 3^(3x1/3) = 3^1 = 3 3^√-27 = ( -27 )^1/3 = ( -3x-3x-3 )^1/3 = ( -3^3 )^1/3 = -3^(3x1/3) = -3^1 = -3 2009-02-02 12:11:03 補充： 3^√64 = ( 64 )^1/3 = ( 4x4x4 )^1/3 = ( 4^3 )^1/3 = 4^(3x1/3) = 4^1 = 4 3^√125 = ( 125 )^1/3 = ( 5x5x5 )^1/3 = ( 5^3 )^1/3 = 5^(3x1/3) = 5^1 = 5 If you don't understand, please let me know.. OK? 2009-02-02 12:18:27 補充： √16 square roots of 16 are 2 and -2. becareful: when the index is even number, it always has 2 roots. but for odd index, it only has 1 root 2009-02-02 12:19:07 補充： please ignore the above answer.. √16 square roots of 16 are 4 and -4. becareful: when the index is even number, it always has 2 roots. but for odd index, it only has 1 root 2009-02-02 12:22:10 補充： 2009-02-02 12:17:27 補充 5^√32= ( 32 )^1/5 = ( 2x2x2x2x2 )^1/5 = (2^5)^1/5 = 2^1 = 2 6^√64= (2x2x2x2x2x2)^1/6 = 2 AND = (-2x-2x-2x-2x-2x-2)^1/6 = -2 2009-02-02 12:23:21 補充： so... now you understand how to do? 2009-02-02 12:24:34 補充： 2009-02-02 12:23:10 補充 4^√16= the 4th roots of 16 are 2 and -2 3^√8= the cube root of 8 is 2 only. 2009-02-02 15:12:16 補充： 2009-02-02 14:00:36 補充 ^4√25 = ±√5... because √5x√5x√5x√5 = 25 AND -√5x-√5x-√5x-√5 = 25 √16^3 = 4^3 = 4x4x4 = 64 AND = -4^3 = -4x-4x-4 = -64 ^3√64/125 = ^3√(4x4x4) / (5x5x5) = 4/5 5^√-1/32 = 5^√(-1/5 x -1/5 x -1/5 x -1/5 x -1/5) = -1/5 ^3√-20^3 = -20^(1/3 x 3) = -20^1 = -20
其他解答:
此文章來自奇摩知識+如有不便請留言告知