(-16K^-7) / (8K^5)
更新1:
is it -2K^12 OR -2K^-12
Help algebra?
回答 (8)
2008-03-11 7:31 pm
✔ 最佳答案
- 2----------
(k^7)(k^5)
- 2
-------
k^12
2008-03-11 3:55 pm
1/(-16K^7)(8K^5)
1/(-128K^12)
1/(-128K^12)
2008-03-11 4:01 pm
(-16K^-7)/(8K^5) =
(-16/8)(K^-7.K^-5)=
-2 K^-12
(-16/8)(K^-7.K^-5)=
-2 K^-12
2008-03-11 3:55 pm
hi,
factor numerator and denominator.
and remember that:
a^m/a^n = a^(m-n)
therefore:
(-16K^-7) / (8K^5)
=[(8)(-2)/8] K^-7-5 --> common factor is 8.
= (-2)K^-12 --> note: a^-n = 1/aⁿ
= (-2)(1/K^12)
= -2/K^12. --> answer.
.........
simplified form of (-16K^-7) / (8K^5) is -2/K^12.
hope it helps.
factor numerator and denominator.
and remember that:
a^m/a^n = a^(m-n)
therefore:
(-16K^-7) / (8K^5)
=[(8)(-2)/8] K^-7-5 --> common factor is 8.
= (-2)K^-12 --> note: a^-n = 1/aⁿ
= (-2)(1/K^12)
= -2/K^12. --> answer.
.........
simplified form of (-16K^-7) / (8K^5) is -2/K^12.
hope it helps.
2008-03-11 4:22 pm
-2/K^12
2008-03-11 3:58 pm
here it is -2K^12
2008-03-11 7:11 pm
(-16k^-7)/(8k^5)
= (-16/k^7)/(8k^5)
= -16/k^7 x 1/8k^5
= -16/8k^12
= -2/k^12
= (-16/k^7)/(8k^5)
= -16/k^7 x 1/8k^5
= -16/8k^12
= -2/k^12
2008-03-11 3:58 pm
(-16K^-7)/(8K^5)= - 2K-7/K^5= -2K^-12
收錄日期: 2021-05-01 10:12:49
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