開方( (r)^2 + (r)^2 ) =4
求r
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畢氏定理 一題]]
2008-04-06 9:29 am
回答 (2)
2008-04-06 9:55 am
✔ 最佳答案
等式兩邊同時平方得(r)^2 + (r)^2 =16
2(r)^2 =16
(r)^2 =8
r=2乘開方2
2008-04-06 10:04 am
開方( (r)^2 (r)^2 ) =4
開方(2r^2) = 4
2r^2 = 16
r^2 = 8
r = 開方8
= 2開方2
開方(2r^2) = 4
2r^2 = 16
r^2 = 8
r = 開方8
= 2開方2
參考: 代返落去係o岩既
收錄日期: 2021-04-14 19:04:36
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