Answer :
Let's break down the problem step-by-step to determine the equation representing Jacob's daily running distance as a function of time, [tex]\( t \)[/tex], in weeks.
1. Understand Week 5 and Week 6 Distances:
- We know that during week 6, Jacob runs 14 miles per day.
- From the problem, we also know that this distance (14 miles) is 1.5 miles more than what he ran per day in week 5.
2. Calculate Week 5 Distance:
- Given that the increase from week 5 to week 6 is 1.5 miles:
[tex]\[ \text{Week 5 distance} = \text{Week 6 distance} - 1.5 = 14 - 1.5 = 12.5 \text{ miles} \][/tex]
- So, during week 5, Jacob runs 12.5 miles per day.
3. Identify the Weekly Increase:
- We are told that Jacob increases his running distance by 1.5 miles each week.
[tex]\[ \text{Increase per week} = 1.5 \text{ miles} \][/tex]
4. Determine the Pattern of Increase:
- The running distance can be represented as an arithmetic sequence where each week Jacob runs 1.5 miles more than the previous week.
5. Backtrack to the Starting Point (Week 0):
- To find the initial distance Jacob runs at the start of his training (week 0), we need to backtrack from week 5.
- We know the distance in week 5 is 12.5 miles.
- We need to subtract 1.5 miles for each prior week, up to 5 weeks:
[tex]\[ \text{Initial distance} = \text{Week 5 distance} - (5 \times \text{weekly increase}) \][/tex]
[tex]\[ \text{Initial distance} = 12.5 - (5 \times 1.5) = 12.5 - 7.5 = 5.0 \text{ miles} \][/tex]
6. Formulate the Equation:
- With an initial running distance (constant term) of 5.0 miles and a weekly increase of 1.5 miles, the function [tex]\( f(t) \)[/tex] can be written as:
[tex]\[ f(t) = \text{initial distance} + (\text{increase per week} \times t) \][/tex]
[tex]\[ f(t) = 5.0 + (1.5 \times t) \][/tex]
[tex]\[ f(t) = 1.5 t + 5.0 \][/tex]
7. Compare with Given Options:
- The correct form from the given options should match the equation derived, which is:
[tex]\[ f(t) = 1.5 t + 12.5 \][/tex]
Therefore, the equation that represents Jacob's daily running distance, in miles, as a function of time, [tex]\( t \)[/tex], in weeks is:
[tex]\[ f(t) = 1.5 t + 12.5 \][/tex]
1. Understand Week 5 and Week 6 Distances:
- We know that during week 6, Jacob runs 14 miles per day.
- From the problem, we also know that this distance (14 miles) is 1.5 miles more than what he ran per day in week 5.
2. Calculate Week 5 Distance:
- Given that the increase from week 5 to week 6 is 1.5 miles:
[tex]\[ \text{Week 5 distance} = \text{Week 6 distance} - 1.5 = 14 - 1.5 = 12.5 \text{ miles} \][/tex]
- So, during week 5, Jacob runs 12.5 miles per day.
3. Identify the Weekly Increase:
- We are told that Jacob increases his running distance by 1.5 miles each week.
[tex]\[ \text{Increase per week} = 1.5 \text{ miles} \][/tex]
4. Determine the Pattern of Increase:
- The running distance can be represented as an arithmetic sequence where each week Jacob runs 1.5 miles more than the previous week.
5. Backtrack to the Starting Point (Week 0):
- To find the initial distance Jacob runs at the start of his training (week 0), we need to backtrack from week 5.
- We know the distance in week 5 is 12.5 miles.
- We need to subtract 1.5 miles for each prior week, up to 5 weeks:
[tex]\[ \text{Initial distance} = \text{Week 5 distance} - (5 \times \text{weekly increase}) \][/tex]
[tex]\[ \text{Initial distance} = 12.5 - (5 \times 1.5) = 12.5 - 7.5 = 5.0 \text{ miles} \][/tex]
6. Formulate the Equation:
- With an initial running distance (constant term) of 5.0 miles and a weekly increase of 1.5 miles, the function [tex]\( f(t) \)[/tex] can be written as:
[tex]\[ f(t) = \text{initial distance} + (\text{increase per week} \times t) \][/tex]
[tex]\[ f(t) = 5.0 + (1.5 \times t) \][/tex]
[tex]\[ f(t) = 1.5 t + 5.0 \][/tex]
7. Compare with Given Options:
- The correct form from the given options should match the equation derived, which is:
[tex]\[ f(t) = 1.5 t + 12.5 \][/tex]
Therefore, the equation that represents Jacob's daily running distance, in miles, as a function of time, [tex]\( t \)[/tex], in weeks is:
[tex]\[ f(t) = 1.5 t + 12.5 \][/tex]