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:
5:7
5/7
5 to 7
10:14 --> 10x2=20 14x2=28
20:28 --> 20x2=40 28x2=56
40:56
(3.5, 16)
Midpoint Formula is
((x1+x2)/2, (y1+y2)/2)
<span>22.6666666667, in other words D. 22 2/3</span>
Answer:
the revenue recognized is $1,275
Step-by-step explanation:
The computation of the revenue recognized is shown below:
= Number of boxes of strawberries sold to customers × price per box
= 85 boxes × $15
= $1,275
Hence, the revenue recognized is $1,275
We simply applied the above formula and the same should be considered
Also
= 130 - 85
= 45 boxes would not be recognized as sale is not made
