jacob is training for a marathon. his plan is to run the same distance for 3 days a week, then increase that…

jacob is training for a marathon. his plan is to run the same distance for 3 days a week, then increase that distance by the same amount each week of training. during week 6, jacob runs 14 miles per day, which is 1.5 miles more per day than he ran during week 5. which equation represents the daily running distance, in miles, as a function of time, t, in weeks?\n○ (f(t)=0.5t + 7)\n○ (f(t)=0.5t + 11)\n○ (f(t)=1.5t + 5)\n○ (f(t)=1.5t + 12.5)

jacob is training for a marathon. his plan is to run the same distance for 3 days a week, then increase that distance by the same amount each week of training. during week 6, jacob runs 14 miles per day, which is 1.5 miles more per day than he ran during week 5. which equation represents the daily running distance, in miles, as a function of time, t, in weeks?\n○ (f(t)=0.5t + 7)\n○ (f(t)=0.5t + 11)\n○ (f(t)=1.5t + 5)\n○ (f(t)=1.5t + 12.5)

Answer

Answer:

B. $f(t)=0.5t + 11$

Explanation:

Step1: Find the increase per week

We know that from week 5 to week 6 (a 1 - week increase), the distance increases by 1.5 miles. So the rate of increase per week (the slope $m$) is 1.5 miles per week. But this is for a 1 - week change. If we consider the general linear - function $y = mx + b$ (here $f(t)=mt + b$), and note that the increase is over a 1 - week period, the slope $m = 1.5\div3=0.5$ (since he runs 3 days a week and we want the daily rate of increase).

Step2: Find the initial value

In week 6 ($t = 6$), $f(6)=14$. Using the linear - function formula $f(t)=mt + b$ with $m = 0.5$ and $t = 6$, we substitute into $f(t)=0.5t + b$: $14=0.5\times6 + b$. $14 = 3 + b$. Subtract 3 from both sides: $b=14 - 3=11$. So the function is $f(t)=0.5t + 11$.