記得寫下做法 我需要^^
或者可以加我即時,在用塗鴉板說也可以^^
先說生謝謝囉XD
圖片參考：http://imgcld.yimg.com/8/n/AD01667384/o/151201250501113871873160.jpg
數學log+三角函數問題~~~幫忙解答^^
2012-01-26 4:57 am
回答 (2)
2012-01-26 9:14 am
2012-01-26 5:55 am
1 考慮log 5^100 = 100log5 = 69.89 所以是69位數 (A)
2 a(2b + c)
= (2i + 3j)(2i - 6j + 3i + j)
= (2i + 3j)(5i - 5j)
= 10 - 15
= -5 (A)
3 (1 + sinθ)/(1 + cosθ) * (1 + secθ)/(1 + cscθ)
= (1 + sinθ)/(1 + cosθ) * (cosθ + 1)sinθ/(sinθ + 1)cosθ
= sinθ/cosθ
= tanθ (D)
2012-01-25 23:13:58 補充：
i * i = j * j = 1
i * j = 0
2 a(2b + c)
= (2i + 3j)(2i - 6j + 3i + j)
= (2i + 3j)(5i - 5j)
= 10 - 15
= -5 (A)
3 (1 + sinθ)/(1 + cosθ) * (1 + secθ)/(1 + cscθ)
= (1 + sinθ)/(1 + cosθ) * (cosθ + 1)sinθ/(sinθ + 1)cosθ
= sinθ/cosθ
= tanθ (D)
2012-01-25 23:13:58 補充：
i * i = j * j = 1
i * j = 0
收錄日期: 2021-04-26 14:56:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120125000015KK05011