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
Polygons, Number of sides, Measure is angles, and angle sums
Answer:
d
Step-by-step explanation:
Draw a right triangle to represent the problem.
The vertical height of the triangle is 9 ft, and it represents the tree.
The horizontal length, at the bottom of the tree is ground level and has a length of 13 ft.
Let x = angle of elevation.
By definition,
tan x = 9/13 = 0.6923
x = arctan(0.6923) = 34.7 deg. = 35 deg (approx)
Answer: 35°
