The graph of y = f(x) is the solid black graph below. Which function represents the dotted graph?

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

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 :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

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

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

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;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

1 year ago

A guy wire is used to support flagpoles in windy areas. These wires are anchored in the ground to provide constant stability. Yo

r-ruslan [8.4K]

Answer: a) Length of wire to replace old one=9ft

b) Length of wire=3.09ft

Leftover wire=0.91ft

Step-by-step explanation: from the right angle triangle, Tan&=20/15

&=53.13°

Length of wire,x=cos 31.13=x/15

15cos31.13

X=9

b) Tan&=10/6

Tan^-110/6

&=59.05

Length= 6cos59.04

X=3.09

4-3.09=.91= leftover wire

8 0

1 year ago

For the diagram below figure b is a reflection of y=2x of the line y =x

Scilla [17]

The equation of the line would be y = (1/2)x.

To find the equation of a line that is reflected through the line y = x, we just switch the x and y. Then, solve for y.

y = 2x
x = 2y
0.5x = y

5 0

2 years ago

Ben Carson was loaned $125.00 to pay $6.00 interest for $100.00 every week ,after 3 weeks how much do he pay back to clear the d

Ghella [55]

Answer:

Total amount pay after 3 week = $147.5

Step-by-step explanation:

Given:

Amount of loan taken = $125

Amount interest per $100 = $6 per week

Find:

Total amount pay after 3 week

Computation:

Amount of interest per week = 125[6/100]

Amount of interest per week = $7.5 per week

Amount of interest in three week = 3 x Amount of interest per week

Amount of interest in three week = 3 x 7.5

Amount of interest in three week = $22.5

Total amount pay after 3 week = Amount of loan taken + Amount of interest in three week

Total amount pay after 3 week = 125 + 22.5

Total amount pay after 3 week = $147.5

8 0

2 years ago

The coordinates of the midpoint of line GH are M(−132,−6) and the coordinates of one endpoint are G(−4, 1). The coordinates of t

Darina [25.2K]

Answer:

Since, the coordinates of the midpoint of line GH are M(\frac{-13}{2}, -6)(

2

−13

,−6) .

The coordinates of endpoint G are (-4,1)

We have to determine the coordinates of endpoint H.

The midpoint of the line segment joining the points (x_1, y_1)(x

1

,y

1

) and (x_2, y_2)(x

2

,y

2

) is given by the formula (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})(

2

x

1

+x

2

,

2

y

1

+y

2

) .

Here, The endpoint G is (-4,1) So, x_1 = -4 , y_1=1x

1

=−4,y

1

=1

Let the endpoint H be (x_2,y_2)(x

2

,y

2

)

The midpoint coordinate M is (\frac{-13}{2}, -6)(

2

−13

,−6) .

So, \frac{-13}{2} = \frac{-4+x_2}{2}

2

−13

=

2

−4+x

2

{-13} = {-4+x_2}−13=−4+x

2

{-13}+4 = {x_2}−13+4=x

2

{x_2}=9x

2

=9

Now, -6 = \frac{1+y_2}{2}−6=

2

1+y

2

-12 = {1+y_2}−12=1+y

2

y_2= -13y

2

=−13

So, the other endpoint H is (-9,-13).

7 0

1 year ago