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gigi1226
Joined: 20 Nov 2004Posts: 7
Location: HK
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Posted: Thu Dec 02, 2004 9:25 pm Post subject: Index問題…urgent!
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Simplify the following and express with positive indices. (2-2^0)^(n+1) over 2^-1 = 2(2-1)^n+1 = 2(1)^n+1 = 2
我唔明 2(2-1)^n+1 呢個step點黎…..
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Kenny TM~
Frequent VisitorJoined: 20 Jan 2004Posts: 127
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Posted: Thu Dec 02, 2004 10:43 pm Post subject: Re: Index問題…urgent!
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gigi1226 wrote:
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| Simplify the following and express with positive indices. (2-2^0)^(n+1) over 2^-1 = 2(2-1)^n+1 = 2(1)^n+1 = 2
我唔明 2(2-1)^n+1 呢個step點黎…..
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任何(非零)數的零次方都是 1.
[應是 Junior Math 來的…]
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gigi1226
Joined: 20 Nov 2004Posts: 7
Location: HK
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Posted: Thu Dec 02, 2004 10:58 pm Post subject:
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或者你誤解左我意思. 我知任何數(=/=0) 既0次方係1…. 我只係諗唔到first step括號出面果個2點黎o者….
anyway, thanks…
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tfchan
Joined: 18 Jan 2004Posts: 10
Location: LSKC
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Posted: Thu Dec 02, 2004 11:45 pm Post subject:
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gigi1226 wrote:
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| 或者你誤解左我意思. 我知任何數(=/=0) 既0次方係1…. 我只係諗唔到first step括號出面果個2點黎o者….
anyway, thanks…
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括號出面個 2 係o係 “over 2^-1” 度o黎 ge ….
1/(2^-1) = 2^(-(-1)) = 2
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