Marvin and Alicia are putting new carpet in their living room. The carpet costs $4.15 per square foot and $97 to have it install
2 answers:
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Answer:
a)
b)
c) the steady state mass of the drug is 2000 mg
d) t ≅ 153.51 minutes
Step-by-step explanation:
From the given information;
At time t= 0
an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500
The inflow rate is 0.06 L/min.
Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.
The objective of the question is to calculate the following :
a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.
From above information given :
Therefore;
with respect to x(0) = 0
b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.
By Using Integration Method:
However; if x(0) = 0 ;
Then
C = -2000
Therefore
c) What is the steady-state mass of the drug in the blood?
the steady-state mass of the drug in the blood when t = infinity
x = 2000 - 0
x = 2000
Thus; the steady state mass of the drug is 2000 mg
d) After how many minutes does the drug mass reach 90% of its stead-state level?
After 90% of its steady state level; the mas of the drug is 90% × 2000
= 0.9 × 2000
= 1800
Hence;
t = 153.5056729
t ≅ 153.51 minutes
To write the system we need the slope of each line and at least one point on the line. The two lines to consider will be the lines connecting the location of each plane to the airport they are flying to. It is also worth noting that the coordinates of the airport represent the point of intersection of the two lines and thus the solution to the system.
1. slope of the line connecting airplane one and the airport: m = 2 you can see this clearly if you graph the two points. From airplane 1 location we rise 8 units and move to the right 4 units to get to the airport. Slope is defined as rise over run: so 8 divided by 4 = 2(the slope) Now substitute the slope and the point (2,4) into point-slope form of a line:
y - 4 = 2(x -4) the standard form of this equation is 2x - y = 0
2. slope of the line connecting airplane 2 and the airport: m = -
To find this slope, simply observe the vertical change of down 3 and a horizontal shift of right 9 from the airport to airplane 2. Now substitute this slope and and the point (15,9) into point-slope form of a line:
y - 9 =
(x - 15) the standard form of this equation is:
x + 3y = 42
Let's write the system:
2x - y = 0
x + 3y = 42
Multiply the first equation by 3 to get the new system
6x - 3y = 0
x + 3y = 42 add these two equations to get an equation in terms of x
7x = 42 thus x = 6 and substituting this value into 2x - y = 0 we see y = 12
In other words, we have proven that the location of the airport is in fact the solution to our system.
PS: You just have to do a little algebra to get from point-slope form of the two equations to standard form. I did not show this process, but if you need it just let me know... thanks
1. slope of the line connecting airplane one and the airport: m = 2 you can see this clearly if you graph the two points. From airplane 1 location we rise 8 units and move to the right 4 units to get to the airport. Slope is defined as rise over run: so 8 divided by 4 = 2(the slope) Now substitute the slope and the point (2,4) into point-slope form of a line:
y - 4 = 2(x -4) the standard form of this equation is 2x - y = 0
2. slope of the line connecting airplane 2 and the airport: m = -
y - 9 =
x + 3y = 42
Let's write the system:
2x - y = 0
x + 3y = 42
Multiply the first equation by 3 to get the new system
6x - 3y = 0
x + 3y = 42 add these two equations to get an equation in terms of x
7x = 42 thus x = 6 and substituting this value into 2x - y = 0 we see y = 12
In other words, we have proven that the location of the airport is in fact the solution to our system.
PS: You just have to do a little algebra to get from point-slope form of the two equations to standard form. I did not show this process, but if you need it just let me know... thanks