\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Use the table below to estimate the velocity at \(\displaystyle t=2\) by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to \(\displaystyle (2,13)\)
Using the point to the left gives \(\displaystyle m_1 = \frac{13-8}{2-0} = \frac{5}{2}\). Using the point to the right gives \(\displaystyle m_2 = \frac{13-20}{2-4} = \frac{7}{2}\). Taking the average of the slopes gives \(\displaystyle \frac{\frac{5}{2}+\frac{7}{2}}{2} = 3\). The estimate of the velocity is \(\displaystyle 3\) meters per second.
\begin{question}Use the table below to estimate the velocity at $t=2$ by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to $(2,13)$\newline
\begin{tabular}{|c|c|c|c|c|c|}\hline
$t$ seconds & 0 & 2 & 4 & 6 & 8 \\ \hline
$x$ meters & 8 & 13 & 20 & 23 & 26 \\ \hline
\end{tabular}\newline
\soln{9cm}{Using the point to the left gives $m_1 = \frac{13-8}{2-0} = \frac{5}{2}$. Using the point to the right gives $m_2 = \frac{13-20}{2-4} = \frac{7}{2}$. Taking the average of the slopes gives $\frac{\frac{5}{2}+\frac{7}{2}}{2} = 3$. The estimate of the velocity is $3$ meters per second.}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\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=2 " src="/equation_images/%20%5Cdisplaystyle%20t%3D2%20" alt="LaTeX: \displaystyle t=2 " data-equation-content=" \displaystyle t=2 " /> by averaging the slopes of the two adjacent secant lines corresponding to the two points closest to <img class="equation_image" title=" \displaystyle (2,13) " src="/equation_images/%20%5Cdisplaystyle%20%282%2C13%29%20" alt="LaTeX: \displaystyle (2,13) " data-equation-content=" \displaystyle (2,13) " /> <br>
</p> </p>
<p> <p>Using the point to the left gives <img class="equation_image" title=" \displaystyle m_1 = \frac{13-8}{2-0} = \frac{5}{2} " src="/equation_images/%20%5Cdisplaystyle%20m_1%20%3D%20%5Cfrac%7B13-8%7D%7B2-0%7D%20%3D%20%5Cfrac%7B5%7D%7B2%7D%20" alt="LaTeX: \displaystyle m_1 = \frac{13-8}{2-0} = \frac{5}{2} " data-equation-content=" \displaystyle m_1 = \frac{13-8}{2-0} = \frac{5}{2} " /> . Using the point to the right gives <img class="equation_image" title=" \displaystyle m_2 = \frac{13-20}{2-4} = \frac{7}{2} " src="/equation_images/%20%5Cdisplaystyle%20m_2%20%3D%20%5Cfrac%7B13-20%7D%7B2-4%7D%20%3D%20%5Cfrac%7B7%7D%7B2%7D%20" alt="LaTeX: \displaystyle m_2 = \frac{13-20}{2-4} = \frac{7}{2} " data-equation-content=" \displaystyle m_2 = \frac{13-20}{2-4} = \frac{7}{2} " /> . Taking the average of the slopes gives <img class="equation_image" title=" \displaystyle \frac{\frac{5}{2}+\frac{7}{2}}{2} = 3 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B%5Cfrac%7B5%7D%7B2%7D%2B%5Cfrac%7B7%7D%7B2%7D%7D%7B2%7D%20%3D%203%20" alt="LaTeX: \displaystyle \frac{\frac{5}{2}+\frac{7}{2}}{2} = 3 " data-equation-content=" \displaystyle \frac{\frac{5}{2}+\frac{7}{2}}{2} = 3 " /> . The estimate of the velocity is <img class="equation_image" title=" \displaystyle 3 " src="/equation_images/%20%5Cdisplaystyle%203%20" alt="LaTeX: \displaystyle 3 " data-equation-content=" \displaystyle 3 " /> meters per second.</p> </p>