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
brainly.com/question/8783264
Answer:
False
Step-by-step explanation:
2*2*2*2 is the correct answer
14.2 because if u divide your numbers you get this
Y = 6 because - ( 1 ) + 7 = 6
Answer:
https://www.caverna.k12.ky.us/userfiles/110/Classes/7142//userfiles/110/my%20files/study%20guide%20answer%20key%20(1).pdf?id=4059
Step-by-step explanation:
copy and past this pdf
