Respuesta :
Given:
Week 6 = 14 miles
Week 5 = 1.5 miles less than week 6.
1.5 miles is the common difference.
Week 6 = 14 miles
Week 5 = 12.5 miles
Week 4 = 11 miles
Week 3 = 9.5 miles
Week 2 = 8 miles
Week 1 = 6.5 miles
f(t) = first term + common difference(t-1)
f(t) = 6.5 miles + 1.5 miles (t-1)
f(1) = 6.5 + 1.5(1-1)
f(1) = 6.5 miles
f(2) = 6.5 + 1.5(2-1)
f(2) = 6.5 + 1.5
f(2) = 8 miles
Week 6 = 14 miles
Week 5 = 1.5 miles less than week 6.
1.5 miles is the common difference.
Week 6 = 14 miles
Week 5 = 12.5 miles
Week 4 = 11 miles
Week 3 = 9.5 miles
Week 2 = 8 miles
Week 1 = 6.5 miles
f(t) = first term + common difference(t-1)
f(t) = 6.5 miles + 1.5 miles (t-1)
f(1) = 6.5 + 1.5(1-1)
f(1) = 6.5 miles
f(2) = 6.5 + 1.5(2-1)
f(2) = 6.5 + 1.5
f(2) = 8 miles
The daily running distance, in miles, as a function of time, t, in weeks is
15+4.5t .
What is a Function ?
A function is a law that defines relation between a dependent variable and a independent variable.
It comes with a defined range and domain.
It is given that
Let jacob run x distance each day ,
First week he travels 3x .
The distance which he increases every week is given by d ,
Then the distance travelled for each week t is given by
3x + (t-1)d
The distance travelled in 6th week is
3x +5d
It is given that
During week 6, Jacob runs 14 miles per day, 14*3 = 42
3x+5d = 42
which is 1.5 miles more per day than he ran during week 5.
1.5*3 = 4.5 more for the week
The distance travelled in 6th week is 3x+4d
3x+4d +4.5 = 3x+5d
d = 4.5 miles.
3x +5 *4.5 = 42
x = 6.5 miles.
The daily running distance, in miles, as a function of time, t, in weeks is 3x + (t-1)d
=19.5 +(t-1) *4.5
=19.5-4.5 +4.5t
=15+4.5t
To know more about Function
https://brainly.com/question/21145944
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