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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 \(\displaystyle (6,17)\)
Using the point to the left gives \(\displaystyle m_1 = \frac{17-14}{6-4} = \frac{3}{2}\). Using the point to the right gives \(\displaystyle m_2 = \frac{17-22}{6-8} = \frac{5}{2}\). Taking the average of the slopes gives \(\displaystyle \frac{\frac{3}{2}+\frac{5}{2}}{2} = 2\). The estimate of the velocity is \(\displaystyle 2\) 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 $(6,17)$\newline
\begin{tabular}{|c|c|c|c|c|c|}\hline
$t$ seconds & 0 & 2 & 4 & 6 & 8 \\ \hline
$x$ meters & 6 & 9 & 14 & 17 & 22 \\ \hline
\end{tabular}\newline
\soln{9cm}{Using the point to the left gives $m_1 = \frac{17-14}{6-4} = \frac{3}{2}$. Using the point to the right gives $m_2 = \frac{17-22}{6-8} = \frac{5}{2}$. Taking the average of the slopes gives $\frac{\frac{3}{2}+\frac{5}{2}}{2} = 2$. The estimate of the velocity is $2$ 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 (6,17) " src="/equation_images/%20%5Cdisplaystyle%20%286%2C17%29%20" alt="LaTeX: \displaystyle (6,17) " data-equation-content=" \displaystyle (6,17) " /> <br>
</p> </p>
<p> <p>Using the point to the left gives <img class="equation_image" title=" \displaystyle m_1 = \frac{17-14}{6-4} = \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20m_1%20%3D%20%5Cfrac%7B17-14%7D%7B6-4%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle m_1 = \frac{17-14}{6-4} = \frac{3}{2} " data-equation-content=" \displaystyle m_1 = \frac{17-14}{6-4} = \frac{3}{2} " /> . Using the point to the right gives <img class="equation_image" title=" \displaystyle m_2 = \frac{17-22}{6-8} = \frac{5}{2} " src="/equation_images/%20%5Cdisplaystyle%20m_2%20%3D%20%5Cfrac%7B17-22%7D%7B6-8%7D%20%3D%20%5Cfrac%7B5%7D%7B2%7D%20" alt="LaTeX: \displaystyle m_2 = \frac{17-22}{6-8} = \frac{5}{2} " data-equation-content=" \displaystyle m_2 = \frac{17-22}{6-8} = \frac{5}{2} " /> . Taking the average of the slopes gives <img class="equation_image" title=" \displaystyle \frac{\frac{3}{2}+\frac{5}{2}}{2} = 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%5Cfrac%7B3%7D%7B2%7D%2B%5Cfrac%7B5%7D%7B2%7D%7D%7B2%7D%20%3D%202%20" alt="LaTeX: \displaystyle \frac{\frac{3}{2}+\frac{5}{2}}{2} = 2 " data-equation-content=" \displaystyle \frac{\frac{3}{2}+\frac{5}{2}}{2} = 2 " /> . The estimate of the velocity is <img class="equation_image" title=" \displaystyle 2 " src="/equation_images/%20%5Cdisplaystyle%202%20" alt="LaTeX: \displaystyle 2 " data-equation-content=" \displaystyle 2 " /> meters per second.</p> </p>