9514 1404 393
Answer:
3 +3d ≥ 15
d ≥ 4
Explanation:
Since Joe runs 3 miles for each day he runs, the number of miles he runs in d days is 3d. That, added to the distance he has already run, will total the amount for the week. He wants that to be at least (greater than or equal to) 15 miles:
3 + 3d ≥ 15
__
We can divide by 3 to get ...
1 + d ≥ 5
d ≥ 4 . . . . . . . subtract 1
Joe must run at least 4 additional days this week to run at least 15 miles.