Respuesta :
So we know the total distance is 65 miles.
Let d1 = distance to 1st delivery from distribution center
d2 = distance to 2nd delivery from 1st delivery site
d3 = distance to 3rd delivery from 2nd delivery site
d4 = distance from 3rd delivery site back to distribution center.
So total distance D = d1 + d2 + d3 + d4 = 65
We know that d1 to d2, d2 to d3 and d3 to d4 each differ by 1/2 a mile.
This means that
d2 = d1 + 1/2
d3 = d2 + 1/2 = (d1 + 1/2) + 1/2 = d1 + 1
d4= d3 + 1/2 =d1 + 1 + 1/2 = d1 + 3/2
Adding all these distances:
d1 + d2 + d3 + d4
= d1 + (d1 + 1/2) + (d1 + 1) + (d1 + 3/2)
= 4d1 + 3 = 65
d1 = (65 - 3)/4
So the distance to the first delivery, d1 is (65 - 3)/4.
Does that make sense?
Let d1 = distance to 1st delivery from distribution center
d2 = distance to 2nd delivery from 1st delivery site
d3 = distance to 3rd delivery from 2nd delivery site
d4 = distance from 3rd delivery site back to distribution center.
So total distance D = d1 + d2 + d3 + d4 = 65
We know that d1 to d2, d2 to d3 and d3 to d4 each differ by 1/2 a mile.
This means that
d2 = d1 + 1/2
d3 = d2 + 1/2 = (d1 + 1/2) + 1/2 = d1 + 1
d4= d3 + 1/2 =d1 + 1 + 1/2 = d1 + 3/2
Adding all these distances:
d1 + d2 + d3 + d4
= d1 + (d1 + 1/2) + (d1 + 1) + (d1 + 3/2)
= 4d1 + 3 = 65
d1 = (65 - 3)/4
So the distance to the first delivery, d1 is (65 - 3)/4.
Does that make sense?
Let x be the distance from the Fruit Distribution center to his first delivery, then:
x+0.5 --> second distance
x+0.5+0.5 = x+1 --> third distance
x+1+0.5 = x+1.5 --> fourth distance
x + x+0.5 + x+1 + x+1.5 = 65
4x + 3 = 65
4x = 65 - 3
4x = 62
x = 62/4
x = 15.5
The distance from the Fruit Distribution center to his first delivery is 15.5 miles.
x+0.5 --> second distance
x+0.5+0.5 = x+1 --> third distance
x+1+0.5 = x+1.5 --> fourth distance
x + x+0.5 + x+1 + x+1.5 = 65
4x + 3 = 65
4x = 65 - 3
4x = 62
x = 62/4
x = 15.5
The distance from the Fruit Distribution center to his first delivery is 15.5 miles.