The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
- Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
- Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.
Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'
D = 180t + 10 + 160t
D = 340t + 10
<h3>Number of minutes before they are 1870 feet</h3>
Making t subject of the formula, we have
t = (D - 10)/340
Since they are 1870 feet apart after t minutes, D = 1870 feet.
t = (D - 10)/340
t = (1870 - 10)/340
t = 1860/340
t = 5.47 minutes
So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes
Learn more about minutes of distance apart here
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x-3x= -4
-2x= -4
x=2
y= -3.2
y= -6
Answer:

Step-by-step explanation:

|x-10| when x = -8
| - 8 - 10 | = |-18| = 18 <==
Answer:
MAYA (6 minutes)
Amy (7 minutes)
Step-by-step explanation:
MAYA
Takes Maya 30 minutes to solve 5 Puzzle:
Let's express it as,
30 mins = 5 Puzzle
1 min = x Puzzle
Cross multiply to solve for the number of questions Maya solves in a minute.
It becomes:
30*x = 5*1
30x = 5
x = 30/5
x = 6mins.
It means it takes Maya 6 mins to solve one question.
FOR AMY
Takes Amy 28 mins to solve 4 puzzles.
Let's represent it as:
28 mins = 4 Puzzles
x min = 1 puzzle
Let's solve for x
(28 mins) * (1 puzzle) = 4 * x
28 = 4x
x = 28/4
x = 7 mins
It means, it takes Amy 7 minutes to solve 1 puzzle