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
The whole wall area including the window is 150 ft squared (15 ft x 10 ft). The window is 50 square feet (5 x 10). Take the whole area and subtract the window area and you have 100 sq ft thats paintable. One can is 50 sq ft so 100/50 is 2 cans of paint needed. Your answer is H.
Answer: 12/5
Step-by-step explanation:
<h2>Answer:</h2><h2>A. 4 13/24</h2><h2 /><h2 /><h2>Hope this helps!!</h2>
