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3 years ago

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Triphasil-28 birth control tablets are taken sequentially, 1 tablet per day for 28 days, with the tablets containing the followi

ng: Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol Then 7 inert tablets (no drug). How many total milligrams each of levonorgestrel and ethinyl estradiol are taken during the 28-day period?

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

Step-by-step explanation:

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3}$

In which $L_{1}, L_{2}$ and $L_{3}$ are the number of miligrams of levonorgestrel taken in each phase.

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3}$

In which $E_{1},E_{2}$ and $E_{3}$ are the number of miligrams of ethinyl estradiol taken in each phase.

We can solve this by phase.

Phase 1: 6 tablets, each containing 0.050 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$6*0.050mg = 0.30mg$ of levonorgestrel and $6*0.030mg = 0.18mg$ of ethinyl estradiol are taken.

So $L_{1} = 0.30$ and $E_{1} = 0.18$

Phase 2: 5 tablets, each containing 0.075 mg of levonorgestrel and 0.040 mg ethinyl estradiol

In this phase,

$5*0.075mg = 0.375mg$ of levonorgestrel and $5*0.040mg = 0.20mg$ of ethinyl estradiol are taken.

So $L_{2} = 0.375$ and $E_{2} = 0.20$

Phase 3: 10 tablets, each containing 0.125 mg of levonorgestrel and 0.030 mg ethinyl estradiol

In this phase,

$10*0.125mg = 1.25mg$ of levonorgestrel and $10*0.030mg = 0.30mg$ of ethinyl estradiol are taken.

So $L_{3} = 1.25$ and $E_{3} = 0.30$

The total milligrams of levonorgestrel is:

$L = L_{1} + L_{2} + L_{3} = 0.30 + 0.375 + 1.25 = 1.925$

The total milligrams of ethinyl estradiol is:

$E = E_{1} + E_{2} + E_{3} = 0.18 + 0.20 + 0.30 = 0.68$

1.925mg of levonorgestrel and 0.68mg of ethinyl estradiol are taken during the 28-day period.

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Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d

Leni [432]

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(a) $y = \frac{1}{1 - Cx}$ , for any constant C

(b) Solution does not exist

(c) $y = \frac{256}{256 - 15x}$

(d) $y = \frac{64}{64 - 15x}$

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

$x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}$ (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

$\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }$ (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

$1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)} \end{cases}$

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1

Hence,

$\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}$

So,

$\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}$

Now, equation (1) becomes

$\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent 1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
$ (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

$

Hence, for any constant C, the following solution will pass thru (0, 1):

$\boxed{y = \frac{1}{1 - Cx}}$

(b) Using equation (5) in problem (a),

$y = \frac{1}{1 - Cx}$ (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that

$y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1$

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C.

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

$y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}


$

By replacing this value of C, the general solution becomes

$y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}



$

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
- Substitute the values of x and y to the general solution.
- Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that

$y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}$

Now, we replace C using the derived value in the general solution. Then,

$y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}$

5 0

3 years ago

How do i do this question ........................,,,,

swat32

You equal each to zero, so -6,0,3,7

3 0

3 years ago