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:
(0, 7/12)
Step-by-step explanation:
to find the y intercept we equate x to 0.
hence,
f(0) = 7/12
the coordinate is (0, 7/12)
Your answer should be B. 58.08 PI m3
Answer: maximum height of the football = 176 feet
Step-by-step explanation:
We want to determine the maximum height of the football from the ground. From the function given,
h(t) = -16t^2+96t +32, it is a quadratic function. Plotting graph if h will result to a parabolic shape. The maximum height of the football = the vertex of the parabola. This vertex is located at time, t
t = -b/2a
b = 96 and a= -16
t = -b/2a = -96/2×-16= 3
Substituting t = 3 into the function if h
h(t) = -16×3^2+96×3 +32
=-16×9 + 96×3 +32
= -144+ 288+32
=176 feet
