2n+4−2(2n)2(2n+3)when simplified is = ?2n+4−2(2n)2(2n+3)when simplified is = ?\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}} \cr & {\text{when simplified is = ?}} \cr} " /> 2n+4−2(2n)2(2n+3)when simplified is = ?2n+4−2(2n)2(2n+3)when simplified is = ?\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}} \cr & {\text{when simplified is = ?}} \cr} " /> 2n+4−2(2n)2(2n+3)when simplified is = ?2n+4−2(2n)2(2n+3)when simplified is = ?\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}} \cr & {\text{when simplified is = ?}} \cr} " />

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$\begin{aligned} \frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} \\ when simplified is = ? \end{aligned}$ $\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}} \cr & {\text{when simplified is = ?}} \cr}$

12⁽ⁿ⁺¹⁾ - ¹/₈
2-2⁽ⁿ⁺¹⁾
31 - 2ⁿ
4⁷/₈

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} \\ = \frac{2^{n + 4} - 2^{n + 1}}{2^{n + 4}} \\ = \frac{2^{n + 4}}{2^{n + 4}} - \frac{2^{n + 1}}{2^{n + 4}} \\ = 1 - 2^{(n + 1) - (n + 4)} \\ = 1 - 2^{- 3} \\ = 1 - \frac{1}{8} \\ = \frac{7}{8} \end{aligned}

$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr}$

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", "text": "

\begin{aligned} \frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} \\ when simplified is  =  ? \end{aligned}

$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}} \cr & {\text{when simplified is = ?}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} \\ = \frac{2^{n + 4} - 2^{n + 1}}{2^{n + 4}} \\ = \frac{2^{n + 4}}{2^{n + 4}} - \frac{2^{n + 1}}{2^{n + 4}} \\ = 1 - 2^{(n + 1) - (n + 4)} \\ = 1 - 2^{- 3} \\ = 1 - \frac{1}{8} \\ = \frac{7}{8} \end{aligned}

$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{2^{n + 4} - 2 (2^{n})}{2 (2^{n + 3})} \\ = \frac{2^{n + 4} - 2^{n + 1}}{2^{n + 4}} \\ = \frac{2^{n + 4}}{2^{n + 4}} - \frac{2^{n + 1}}{2^{n + 4}} \\ = 1 - 2^{(n + 1) - (n + 4)} \\ = 1 - 2^{- 3} \\ = 1 - \frac{1}{8} \\ = \frac{7}{8} \end{aligned}

$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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