The key is Esther travelled the same distance - x - in both her morning and evening commute.
45(time she took in the morning, or p) = x
30(time she took in the evening, or q) = x
Therefore 45(p) = 30(q), or divide both sides by 5 and get 9(p) = 6(q). I know you can divide it further, but these numbers are small enough and it's not worth the time.
Since the whole trip took an hour, (p + q) = 60min, and so, p = 60-q.
Therefore 9(60-q) = 6q or 540-9q = 6q. So 540 = 15q, which makes q = 36. If q = 36, then by (p+q)=60, p (the time she took in the morning) must equal 24.
45 miles per hour, her speed in the morning, times (24/60) hours, her time, makes 18 miles travelled in the morning. If you check, 30 miles per hour times (36/60) hours also makes 18 miles in the evening.
<span>Hope that makes a little sense. And I also hope it's right</span>
An = a1 + (n - 1)(d)
Where a1 is the first term and d is the common difference.
First find d, the common difference.
24, ____, 32
a3 a4 a5
Subtract 32-24 = 8
Subtract a5 - a3 = 2
Divide 8/2 = 4
d = 4
Use d and one of the values they give us to find a1.
a3 = 24
24 = a1 + (3 - 1)(4)
24 = a1 + 2(4)
24 = a1 + 8
Subtract 8 from both sides
16 = a1
an = 16 + (n - 1)(4)
Can also be written
an = 16 + 4n - 4
an = 4n + 12
2nd point distance-1st point distant
Answer:
B is the right answer and if you add
Their sum can be greater than two. There is nothing that states that they couldn't be. All it states it that the two factions sums are less than 1. I hoped this helps
