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Sagot :
Sure, let's solve the problem step-by-step:
1. Given Information:
- Jorden drives to the store at 30 miles per hour.
- On her way home, she averages only 20 miles per hour.
- The total driving time is half an hour (0.5 hours).
2. Unknown:
- Distance to the store (and back) which we'll denote as [tex]\( d \)[/tex].
3. Relationships and Formulas:
- Time to store [tex]\( t_1 \)[/tex] and time back home [tex]\( t_2 \)[/tex].
- Using the relationship: [tex]\[ \text{Time} = \frac{\text{Distance}}{\text{Rate}}. \][/tex]
4. Step-by-Step Solution:
- Let the distance to the store be [tex]\( d \)[/tex] miles.
- Time to drive to the store: [tex]\( t_1 = \frac{d}{30} \)[/tex] hours.
- Time to drive back home: [tex]\( t_2 = \frac{d}{20} \)[/tex] hours.
- Total driving time is given as 0.5 hours.
5. Set up the equation:
- The total time to and from the store: [tex]\[ t_1 + t_2 = 0.5. \][/tex]
- Substituting the expressions for [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] we get: [tex]\[ \frac{d}{30} + \frac{d}{20} = 0.5. \][/tex]
6. Solve the equation:
- To solve the equation, find a common denominator for the fractions. The common denominator of 30 and 20 is 60.
- Rewrite the equation: [tex]\[ \frac{2d}{60} + \frac{3d}{60} = 0.5. \][/tex]
- Combine the fractions: [tex]\[ \frac{2d + 3d}{60} = 0.5. \][/tex]
- Simplify: [tex]\[ \frac{5d}{60} = 0.5. \][/tex]
- Multiply both sides by 60 to clear the fraction: [tex]\[ 5d = 0.5 \times 60. \][/tex]
- Simplify: [tex]\[ 5d = 30. \][/tex]
- Solve for [tex]\( d \)[/tex]: [tex]\[ d = \frac{30}{5}. \][/tex]
- Hence, [tex]\( d = 6 \)[/tex] miles.
7. Conclusion:
- Jorden lives 6 miles from the store.
8. Filling the Table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \multicolumn{2}{|c}{Distance} & Rate & Time \\ \hline To store & 6 \text{ miles} & 30 \text{ miles per hour} & \frac{6}{30} = 0.2 \text{ hours} \\ \hline \begin{tabular}{c} Return home \end{tabular} & 6 \text{ miles} & 20 \text{ miles per hour} & \frac{6}{20} = 0.3 \text{ hours} \\ \hline \end{array} \][/tex]
Thus, the table is filled with the calculated distances, rates, and times.
1. Given Information:
- Jorden drives to the store at 30 miles per hour.
- On her way home, she averages only 20 miles per hour.
- The total driving time is half an hour (0.5 hours).
2. Unknown:
- Distance to the store (and back) which we'll denote as [tex]\( d \)[/tex].
3. Relationships and Formulas:
- Time to store [tex]\( t_1 \)[/tex] and time back home [tex]\( t_2 \)[/tex].
- Using the relationship: [tex]\[ \text{Time} = \frac{\text{Distance}}{\text{Rate}}. \][/tex]
4. Step-by-Step Solution:
- Let the distance to the store be [tex]\( d \)[/tex] miles.
- Time to drive to the store: [tex]\( t_1 = \frac{d}{30} \)[/tex] hours.
- Time to drive back home: [tex]\( t_2 = \frac{d}{20} \)[/tex] hours.
- Total driving time is given as 0.5 hours.
5. Set up the equation:
- The total time to and from the store: [tex]\[ t_1 + t_2 = 0.5. \][/tex]
- Substituting the expressions for [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] we get: [tex]\[ \frac{d}{30} + \frac{d}{20} = 0.5. \][/tex]
6. Solve the equation:
- To solve the equation, find a common denominator for the fractions. The common denominator of 30 and 20 is 60.
- Rewrite the equation: [tex]\[ \frac{2d}{60} + \frac{3d}{60} = 0.5. \][/tex]
- Combine the fractions: [tex]\[ \frac{2d + 3d}{60} = 0.5. \][/tex]
- Simplify: [tex]\[ \frac{5d}{60} = 0.5. \][/tex]
- Multiply both sides by 60 to clear the fraction: [tex]\[ 5d = 0.5 \times 60. \][/tex]
- Simplify: [tex]\[ 5d = 30. \][/tex]
- Solve for [tex]\( d \)[/tex]: [tex]\[ d = \frac{30}{5}. \][/tex]
- Hence, [tex]\( d = 6 \)[/tex] miles.
7. Conclusion:
- Jorden lives 6 miles from the store.
8. Filling the Table:
[tex]\[ \begin{array}{|c|c|c|c|} \hline \multicolumn{2}{|c}{Distance} & Rate & Time \\ \hline To store & 6 \text{ miles} & 30 \text{ miles per hour} & \frac{6}{30} = 0.2 \text{ hours} \\ \hline \begin{tabular}{c} Return home \end{tabular} & 6 \text{ miles} & 20 \text{ miles per hour} & \frac{6}{20} = 0.3 \text{ hours} \\ \hline \end{array} \][/tex]
Thus, the table is filled with the calculated distances, rates, and times.
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