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
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Answer:
sim eu também preciso desta respota
For lines to be perpendicular their slopes must be negative reciprocals of one another, mathematically:
m1*m2=-1, and in this case:
-m/4=-1
-m=-4
m=4, so the slope of the perpendicular line is 4 so we have thus far:
y=4x+b, using point (-4,3) we can solve for the y-intercept, "b"
3=4(-4)+b
3=-16+b
19=b, so our line is:
y=4x+19
Answer:
So, the length (L) = 7 3/4 ft = 31/4 (as an improper fraction)
the width (W) = (1/3)xL
So the width = 1/3 x (31/4) =
31/12 = 2 7/12 feet
Step-by-step explanation:
its up there
You would recieve $37.95 a year then multiply that by 7 years to get $265.65
