Let's solve the problem step-by-step.
1. Identify Rates:
- Jorden drives to the store at a rate of [tex]\( 30 \)[/tex] miles per hour.
- Jorden drives back home at a rate of [tex]\( 20 \)[/tex] miles per hour.
2. Total Time:
- The total driving time (to the store and back home) is [tex]\( 0.5 \)[/tex] hours.
3. Time Calculation:
- We need to calculate the time needed for each leg of the trip using the distance and rates.
4. Distance Calculation:
- Let [tex]\( d \)[/tex] be the distance (in miles) from Jorden's home to the store.
- Time for the trip to the store is [tex]\( \frac{d}{30} \)[/tex] hours.
- Time for the return trip home is [tex]\( \frac{d}{20} \)[/tex] hours.
- According to the problem, the total time is the sum of the times for both legs of the trip:
[tex]\[
\frac{d}{30} + \frac{d}{20} = 0.5
\][/tex]
5. Solve for Distance:
- Combine the fractions:
[tex]\[
\frac{d}{30} + \frac{d}{20} = 0.5
\][/tex]
- Find a common denominator (which is [tex]\( 60 \)[/tex]):
[tex]\[
\frac{2d}{60} + \frac{3d}{60} = 0.5
\][/tex]
[tex]\[
\frac{5d}{60} = 0.5
\][/tex]
- Simplify the equation:
[tex]\[
\frac{d}{12} = 0.5
\][/tex]
- Multiply both sides by [tex]\( 12 \)[/tex] to solve for [tex]\( d \)[/tex]:
[tex]\[
d = 0.5 \times 12 = 6 \text{ miles}
\][/tex]
Given these calculations, the table can be filled as follows:
[tex]\[
\begin{array}{c}
a=30 \\
b=20
\end{array}
\][/tex]
\begin{tabular}{|c|c|c|c|}
\hline & Distance & Rate & Time \\
\hline To store & 3 & 30 & \\
\hline \begin{tabular}{l}
Return \\
home
\end{tabular} & 2 & 20 & \\
\hline
\end{tabular}
The value next to the letter a is 30, and the value next to the letter b is 20. The distance to the store is [tex]\( 6 \)[/tex] miles.
