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jonny [76]
2 years ago
11

The first 5 terms of a number pattern are shown below.

Mathematics
1 answer:
kotykmax [81]2 years ago
4 0

\bf 4~~,~~\stackrel{4+5}{9}~~,~~\stackrel{9+5}{14}~~,~~\stackrel{14+5}{19}~~,~~\stackrel{19+5}{24}\qquad \impliedby \stackrel{\textit{common difference}}{d=5} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=4\\ d=5 \end{cases} \\\\\\ a_n=4+(n-1)5\implies a_n=4+5n-5\implies a_n=5n-1

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\large\displaystyle\text{$\begin{gathered}\sf 9|x-8| < 36 \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf Divide \ both \ sides \ by \ 9. \end{gathered}$}

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\large\displaystyle\text{$\begin{gathered}\sf Solve \ Absolute \ Value. \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf |x-8| < 4 \end{gathered}$}

\large\displaystyle\text{$\begin{gathered}\sf We \ know \ x-8 < 4 \ and \ x-8 > -4 \end{gathered}$}

<u>                                                                                                                             </u>

         \large\displaystyle\text{$\begin{gathered}\sf x-8 < 4 \ (Condition \ 1) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 < 4+8 \ (Add \ 8 \ to \ both \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x < 12 \end{gathered}$}

<u>                                                                                                                             </u>

           \large\displaystyle\text{$\begin{gathered}\sf x-8 > -4 \ (Condition \ 2) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x-8+8 > -4+8 \ (Add \ 8 \ to \ both \ \ sides) \end{gathered}$}\\\large\displaystyle\text{$\begin{gathered}\sf x > 4 \end{gathered}$}

<u>                                                                                                                             </u>

<u />\underline{\boldsymbol{\sf{Answer}}}

\boxed{\large\displaystyle\text{$\begin{gathered}\sf x < 12 \ and \ x > 4 \end{gathered}$} }

\large\displaystyle\text{$\begin{gathered}\sf Therefore,\bf{\underline{the \  correct \ option}} \  \end{gathered}$}\large\displaystyle\text{$\begin{gathered}\sf is \ \bf{\underline{"A"}}. \end{gathered}$}

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