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Questions: Algebra BusinessCalculus
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Use the table below to estimate the velocity at \(\displaystyle t=6\) by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to \(\displaystyle (6,20)\)
Using the point to the left gives \(\displaystyle m_1 = \frac{20-10}{6-4} = 5\). Using the point to the right gives \(\displaystyle m_2 = \frac{20-28}{6-8} = 4\). Taking the average of the slopes gives \(\displaystyle \frac{5+4}{2} = \frac{9}{2}\). The estimate of the velocity is \(\displaystyle \frac{9}{2}\) meters per second.
\begin{question}Use the table below to estimate the velocity at $t=6$ by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to $(6,20)$\newline
\begin{tabular}{|c|c|c|c|c|c|}\hline
$t$ seconds & 0 & 2 & 4 & 6 & 8 \\ \hline
$x$ meters & 4 & 7 & 10 & 20 & 28 \\ \hline
\end{tabular}\newline
\soln{9cm}{Using the point to the left gives $m_1 = \frac{20-10}{6-4} = 5$. Using the point to the right gives $m_2 = \frac{20-28}{6-8} = 4$. Taking the average of the slopes gives $\frac{5+4}{2} = \frac{9}{2}$. The estimate of the velocity is $\frac{9}{2}$ meters per second.}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use the table below to estimate the velocity at <img class="equation_image" title=" \displaystyle t=6 " src="/equation_images/%20%5Cdisplaystyle%20t%3D6%20" alt="LaTeX: \displaystyle t=6 " data-equation-content=" \displaystyle t=6 " /> by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to <img class="equation_image" title=" \displaystyle (6,20) " src="/equation_images/%20%5Cdisplaystyle%20%286%2C20%29%20" alt="LaTeX: \displaystyle (6,20) " data-equation-content=" \displaystyle (6,20) " /> <br>
</p> </p>
<p> <p>Using the point to the left gives <img class="equation_image" title=" \displaystyle m_1 = \frac{20-10}{6-4} = 5 " src="/equation_images/%20%5Cdisplaystyle%20m_1%20%3D%20%5Cfrac%7B20-10%7D%7B6-4%7D%20%3D%205%20" alt="LaTeX: \displaystyle m_1 = \frac{20-10}{6-4} = 5 " data-equation-content=" \displaystyle m_1 = \frac{20-10}{6-4} = 5 " /> . Using the point to the right gives <img class="equation_image" title=" \displaystyle m_2 = \frac{20-28}{6-8} = 4 " src="/equation_images/%20%5Cdisplaystyle%20m_2%20%3D%20%5Cfrac%7B20-28%7D%7B6-8%7D%20%3D%204%20" alt="LaTeX: \displaystyle m_2 = \frac{20-28}{6-8} = 4 " data-equation-content=" \displaystyle m_2 = \frac{20-28}{6-8} = 4 " /> . Taking the average of the slopes gives <img class="equation_image" title=" \displaystyle \frac{5+4}{2} = \frac{9}{2} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B5%2B4%7D%7B2%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20" alt="LaTeX: \displaystyle \frac{5+4}{2} = \frac{9}{2} " data-equation-content=" \displaystyle \frac{5+4}{2} = \frac{9}{2} " /> . The estimate of the velocity is <img class="equation_image" title=" \displaystyle \frac{9}{2} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B9%7D%7B2%7D%20" alt="LaTeX: \displaystyle \frac{9}{2} " data-equation-content=" \displaystyle \frac{9}{2} " /> meters per second.</p> </p>