Billy is walking from the front door of his house to his bus stop, which is 960 feet away from his front door. as billy walks out his front door, he walks in a straight path toward his bus stop at a constant rate of 7.5 feet per second.

(a) illustrate the situation with a diagram and define variables to represent the values of the relevant varying quantitities. (label the variables on your picture.)
(b) define a function f to determine billy's distance from his bus stop in terms of the number of seconds he has been walking.
(c) what is the independent quantity and what is the domain of f (the values the independent quantity can take on)?
(d) what is the dependent quantity and what is the range of f (the values the dependent quantity can take on)?
(e) what do each of the following represent: f(0) and f(60.25)?

b) 690 - 7.5*t

c) 0 < t < 92s time (t) is independent quantity

d) 0 < s < 690ft distance from bus stop (s) is dependent quantity

e) f(0) = 690 ft away from bus stop , f(60.25) = 238.125 ft away from bus stop

Step-by-step explanation:

Part a - see diagram

part b

initial distance from bus stop s0 = 690 ft

distance covered = 7.5*t

s = s0 - distance covered

s = 690 - 7.5*t = f(t)

part c

s = 0 or s = 690

0 = 690 -7.5*t

t = 92 s

Hence domain : 0 < t < 92s time (t) is independent quantity

part d

s = 0 or s = 690

Hence range : 0 < s < 690ft distance from bus stop (s) is dependent quantity because it depends on time (t)

part e

f(0) is s @t = 0

f(0) = 690 ft away from bus stop

f(60.25) is s @t = 60.25

f(60.25) = 690 - 7.5*60.25 = 238.125 ft away from bus stop.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

the answer is d. 75%