29580 + 3440 = 33020. The is about 33,000mi. The answer is d.
The explanation about the graph is illustrated below.
<h3>How to illustrate the graph?</h3>
The graph shows 0 to 30 on the x-axis at increments of 5 and 0 to 12 on the y-axis at increments of 1. The label on the x-axis is Temperature in degrees C, and the label on the y axis is the Number of Sweaters Sold.
If you will draw lines through, (12.5,6) and (15,4), it will best represent a line of best fit, because most of the points on the same and opposite sides of line are equidistant from each other.
Equation of line passing through (12.5,6) and (15,4) A line of best is the line that passes through all or some points or only through a single point, which describes the relationship between x and y values of data set
4x + 5y = 80
5y = -4x + 80.
y = -4x/5 + 16
Slope = -4/5 and y intercept is 16.
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Answer:
60 is the answer
Step-by-step explanation:
Find x you mean?
substract both side by 2x
5x=-35
divide both side by 5
x=-7
->D
Answer:
a. dQ/dt = -kQ
b. 
c. k = 0.178
d. Q = 1.063 mg
Step-by-step explanation:
a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.
Let Q be the quantity of drug left in the body.
Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then
-dQ/dt ∝ Q
-dQ/dt = kQ
dQ/dt = -kQ
This is the required differential equation.
b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.
with t = 0, Q(0) = 9 mg
dQ/dt = -kQ
separating the variables, we have
dQ/Q = -kdt
Integrating we have
∫dQ/Q = ∫-kdt
㏑Q = -kt + c

when t = 0, Q = 9

So, 
c) Use the half-life to find the constant of proportionality k.
At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours
So,

taking natural logarithm of both sides, we have
d) How much of the 9 mg dose is still in the body after 12 hours?
Since k = 0.178,

when t = 12 hours,
