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Questions: Algebra BusinessCalculus
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Use the table below to estimate the velocity at \(\displaystyle t=9\) by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to \(\displaystyle (9,28)\)
Using the point to the left gives \(\displaystyle m_1 = \frac{28-20}{9-6} = \frac{8}{3}\). Using the point to the right gives \(\displaystyle m_2 = \frac{28-36}{9-12} = \frac{8}{3}\). Taking the average of the slopes gives \(\displaystyle \frac{\frac{8}{3}+\frac{8}{3}}{2} = \frac{8}{3}\). The estimate of the velocity is \(\displaystyle \frac{8}{3}\) meters per second.
\begin{question}Use the table below to estimate the velocity at $t=9$ by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to $(9,28)$\newline
\begin{tabular}{|c|c|c|c|c|c|}\hline
$t$ seconds & 0 & 3 & 6 & 9 & 12 \\ \hline
$x$ meters & 8 & 11 & 20 & 28 & 36 \\ \hline
\end{tabular}\newline
\soln{9cm}{Using the point to the left gives $m_1 = \frac{28-20}{9-6} = \frac{8}{3}$. Using the point to the right gives $m_2 = \frac{28-36}{9-12} = \frac{8}{3}$. Taking the average of the slopes gives $\frac{\frac{8}{3}+\frac{8}{3}}{2} = \frac{8}{3}$. The estimate of the velocity is $\frac{8}{3}$ 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=9 " src="/equation_images/%20%5Cdisplaystyle%20t%3D9%20" alt="LaTeX: \displaystyle t=9 " data-equation-content=" \displaystyle t=9 " /> by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to <img class="equation_image" title=" \displaystyle (9,28) " src="/equation_images/%20%5Cdisplaystyle%20%289%2C28%29%20" alt="LaTeX: \displaystyle (9,28) " data-equation-content=" \displaystyle (9,28) " /> <br>
</p> </p>
<p> <p>Using the point to the left gives <img class="equation_image" title=" \displaystyle m_1 = \frac{28-20}{9-6} = \frac{8}{3} " src="/equation_images/%20%5Cdisplaystyle%20m_1%20%3D%20%5Cfrac%7B28-20%7D%7B9-6%7D%20%3D%20%5Cfrac%7B8%7D%7B3%7D%20" alt="LaTeX: \displaystyle m_1 = \frac{28-20}{9-6} = \frac{8}{3} " data-equation-content=" \displaystyle m_1 = \frac{28-20}{9-6} = \frac{8}{3} " /> . Using the point to the right gives <img class="equation_image" title=" \displaystyle m_2 = \frac{28-36}{9-12} = \frac{8}{3} " src="/equation_images/%20%5Cdisplaystyle%20m_2%20%3D%20%5Cfrac%7B28-36%7D%7B9-12%7D%20%3D%20%5Cfrac%7B8%7D%7B3%7D%20" alt="LaTeX: \displaystyle m_2 = \frac{28-36}{9-12} = \frac{8}{3} " data-equation-content=" \displaystyle m_2 = \frac{28-36}{9-12} = \frac{8}{3} " /> . Taking the average of the slopes gives <img class="equation_image" title=" \displaystyle \frac{\frac{8}{3}+\frac{8}{3}}{2} = \frac{8}{3} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%5Cfrac%7B8%7D%7B3%7D%2B%5Cfrac%7B8%7D%7B3%7D%7D%7B2%7D%20%3D%20%5Cfrac%7B8%7D%7B3%7D%20" alt="LaTeX: \displaystyle \frac{\frac{8}{3}+\frac{8}{3}}{2} = \frac{8}{3} " data-equation-content=" \displaystyle \frac{\frac{8}{3}+\frac{8}{3}}{2} = \frac{8}{3} " /> . The estimate of the velocity is <img class="equation_image" title=" \displaystyle \frac{8}{3} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B8%7D%7B3%7D%20" alt="LaTeX: \displaystyle \frac{8}{3} " data-equation-content=" \displaystyle \frac{8}{3} " /> meters per second.</p> </p>