Drag each function to the correct location on the chart. Match the functions with their periods.

[tex]\[
\begin{array}{llll}
y = 5 \cot \pi - 8 & y = \frac{2}{5} \sin \left(\frac{2}{z} - \pi\right) & y = -\frac{1}{3} \csc 2x & y = 5 \sec (2x + 2\pi) \\
y = -6 \cot x - 10 & y = -3 \tan \frac{x}{2}
\end{array}
\][/tex]

\begin{tabular}{|l|l|}
\hline
[tex]$\pi$[/tex] & [tex]$2\pi$[/tex] \\
\hline
& \\
\hline
& \\
\hline
\end{tabular}



Answer :

To match each function with its respective period, follow these steps:

Step 1: Identify the periods of each function based on the given results:

1. \( y = 5 \cot \pi - 8 \) has a period of \(\pi\).
2. \( y = \frac{2}{5} \sin \left( \frac{2}{z} - \pi \right) \) has a period of \(2\pi\).
3. \( y = -\frac{1}{3} \csc 2x \) has a period of \(\pi\).
4. \( y = 5 \sec (2x + 2\pi) \) has a period of \(\pi\).
5. \( y = -6 \cot x - 10 \) has a period of \(\pi\).
6. \( y = -3 \tan \frac{x}{2} \) has a period of \(4\pi\).

Step 2: Place each function into the appropriate category on the chart.

### Chart:
[tex]\[ \begin{array}{|l|l|} \hline \pi & 2\pi \\ \hline y = 5 \cot \pi - 8 & y = \frac{2}{5} \sin \left( \frac{2}{z} - \pi \right) \\ y = -\frac{1}{3} \csc 2x & \\ y = 5 \sec (2x + 2\pi) & \\ y = -6 \cot x - 10 & \\ & \\ \hline \end{array} \][/tex]

### Final Result:

[tex]\[ \begin{array}{|l|l|} \hline \pi & 2\pi \\ \hline y = 5 \cot \pi - 8 & y = \frac{2}{5} \sin \left( \frac{2}{z} - \pi \right) \\ y = -\frac{1}{3} \csc 2x & \\ y = 5 \sec (2x + 2\pi) & \\ y = -6 \cot x - 10 & \\ & \\ \hline \end{array} \][/tex]

In summary, the functions have been matched with their respective periods as follows:

- \(\pi\):
- \( y = 5 \cot \pi - 8 \)
- \( y = -\frac{1}{3} \csc 2 x \)
- \( y = 5 \sec (2 x + 2 \pi) \)
- \( y = -6 \cot x - 10 \)

- \(2\pi\):
- \( y = \frac{2}{5} \sin \left( \frac{2}{z} - \pi \right) \)

The function [tex]\( y = -3 \tan \frac{x}{2} \)[/tex] was assigned a period of [tex]\(4\pi\)[/tex] earlier and is not included in the [tex]\(\pi\)[/tex] or [tex]\(2\pi\)[/tex] columns.