It is approximately -.4048
H=10^(-pH)
log H = -pH
pH=-logH => so it is affirmed that it is -.40
H=10^(-pH)
log H = -pH
pH=-logH => so it is affirmed that it is -.40
What is the length of the hypotenuse? If necessary, round to the nearest tenth.
1 answer:
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Answer:
The solutions for your four question problem are:
a) t = D^-1(x) = (1/57)*x -(13/57)
b) t = D^-1(x) = 1 h
c) D^-1(x) represents time.
d) The number of hours of driving needed for Sarah to be x km from Tempe. (option (c))
Step-by-step explanation:
a) Determine a formula in terms of x for: t = D^-1(x)
The distance in km from Tempe after t hours of driving is given by
x = D(t) = 13 + 57t.
We just need to find the value t function of x
x = 13 + 57*t
x -13 = 57*t
57*t = x -13
t = (1/57)*x -(13/57)
We can see the plots of both equation in the picture below.
b) Compute D^-1(70)
Once we find the expression for D^-1(x)
We substitute for x = 70 km
t = D^-1(x) = (1/57)*x -(13/57)
t = D^-1(x) = (1/57)*(70) -(13/57)
t = D^-1(x) = (70/57) -(13/57)
t = D^-1(x) = (1.228) -(0.228)
t = D^-1(x) = 1 h
c) In the expression D^-1(x) : what quantity (distance or time) does the x represent? what quantity (distance or time) does the entire D^-1(x) represent?
x represents Distance in both equations (D(t), and D^-1(x))
t represents Time in both equations (D(t), and D^-1(x))
Since t = D^-1(x),
D^-1(x) represents time.
d) Which of the following statements best describes D^-1(x)?
The number of hours of driving needed for Sarah to be x km from Tempe.
Since, t = D^-1(x), and t represents the amount of time elapsed since Sarah, parted from Tempe, the correct answer is option (c)
The expression for D^-1(x) can be found in the previous answers
t = D^-1(x) = (1/57)*x -(13/57)
The input is x (distance) and the output is t (time)
