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tekilochka [14]
2 years ago
7

RIP OpenStudy ;(

Mathematics
2 answers:
dalvyx [7]2 years ago
6 0
Hi steve ;) 
 
you just have to apply simple exponent rule:
 \frac{x^n}{x^y} =x^{n-m}

& RIP OS ;-; :( 
#os<3
klemol [59]2 years ago
3 0
First note that \frac{2^n+1}{2^{n+1}} =  \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2^{n+1}}

If you take limit, then you have \lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}



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The answer is:  [B]:  " 49 " .
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2.03 * 24  =  48.72 ;  round to "2 significant digits" ==> 49 .

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