Seaveyalyssa25idn
Answered

IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Our experts provide prompt and accurate answers to help you make informed decisions on any topic.

Select the correct answer.

The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible for this situation?

\begin{tabular}{|c|c|c|c|c|c|}
\hline Time (s) & A. Distance (m) & B. Distance (m) & C. Distance (m) & D. Distance (m) & E. Distance (m) \\
\hline 0 & 0 & 2.00 & 9.00 & ㅁ⼠ & वᄆ士 \\
\hline 1 & 1.00 & 4.00 & 18.00 & 1.00 & 1.00 \\
\hline 2 & 4.00 & 6.00 & 27.00 & 0.50 & 0.25 \\
\hline 3 & 9.00 & 8.00 & 36.00 & 0.33 & 0.11 \\
\hline 4 & 16.00 & 10.00 & 45.00 & 0.25 & 0.06 \\
\hline 5 & 25.00 & 12.00 & 54.00 & 0.20 & 0.04 \\
\hline 6 & 36.00 & 14.00 & 63.00 & 0.16 & 0.02 \\
\hline
\end{tabular}

A. Column A
B. Column B
C. Column C

Sagot :

GinnyAnsweridn
To determine which column of distance data represents a quadratic relationship, let's analyze the given data:

1. Understanding a Quadratic Relationship:
- A quadratic relationship means that the distance [tex]\( d \)[/tex] is proportional to the square of time [tex]\( t \)[/tex].
- Mathematically, this can be expressed as [tex]\( d = k \cdot t^2 \)[/tex], where [tex]\( k \)[/tex] is a constant.

2. Analyzing Column A:
- Given data for time [tex]\( t \)[/tex] and distance [tex]\( d \)[/tex]:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Time (s)} & \text{A. Distance (m)} \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 2 & 4 \\ \hline 3 & 9 \\ \hline 4 & 16 \\ \hline 5 & 25 \\ \hline 6 & 36 \\ \hline \end{tabular} \][/tex]
- Let's check if this follows [tex]\( d = k \cdot t^2 \)[/tex]:
[tex]\[ \begin{align*} t = 0 & , \quad d = 0 = 0^2 \\ t = 1 & , \quad d = 1 = 1^2 \\ t = 2 & , \quad d = 4 = 2^2 \\ t = 3 & , \quad d = 9 = 3^2 \\ t = 4 & , \quad d = 16 = 4^2 \\ t = 5 & , \quad d = 25 = 5^2 \\ t = 6 & , \quad d = 36 = 6^2 \\ \end{align*} \][/tex]
- Clearly, distance [tex]\( d \)[/tex] is following a quadratic relationship [tex]\( d = t^2 \)[/tex].

3. Analyzing Column B:
- Given data for time [tex]\( t \)[/tex] and distance [tex]\( d \)[/tex]:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Time (s)} & \text{B. Distance (m)} \\ \hline 0 & 2 \\ \hline 1 & 4 \\ \hline 2 & 6 \\ \hline 3 & 8 \\ \hline 4 & 10 \\ \hline 5 & 12 \\ \hline 6 & 14 \\ \hline \end{tabular} \][/tex]
- This data does not follow the quadratic form [tex]\( d = k \cdot t^2 \)[/tex]. It appears to follow a linear relationship instead.

4. Analyzing Column C:
- Given data for time [tex]\( t \)[/tex] and distance [tex]\( d \)[/tex]:
[tex]\[ \begin{tabular}{|c|c|} \hline \text{Time (s)} & \text{C. Distance (m)} \\ \hline 0 & 9 \\ \hline 1 & 18 \\ \hline 2 & 27 \\ \hline 3 & 36 \\ \hline 4 & 45 \\ \hline 5 & 54 \\ \hline 6 & 63 \\ \hline \end{tabular} \][/tex]
- Checking for quadratic relationship:
[tex]\[ \begin{align*} t = 0 & , \quad d = 9 \neq 0^2 \\ t = 1 & , \quad d = 18 \neq 1^2 \\ t = 2 & , \quad d = 27 \neq 2^2 \\ t = 3 & , \quad d = 36 \neq 3^2 \\ t = 4 & , \quad d = 45 \neq 4^2 \\ t = 5 & , \quad d = 54 \neq 5^2 \\ t = 6 & , \quad d = 63 \neq 6^2 \\ \end{align*} \][/tex]
- Clearly, this distance data does not follow a quadratic relationship.

Conclusion:
- Column A represents a quadratic relationship as distance [tex]\( d \)[/tex] is given by [tex]\( d = t^2 \)[/tex].

Thus, the correct answer is:
A. Column A
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.