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coldgirl [10]
2 years ago
10

The last rest area you spent $3.25 on water bottles, $11.00 on snacks, and $4.00 on a magazine, and gave the cashier $20.00 How

much money did you spend in all? What was the change that you were given?
Mathematics
2 answers:
photoshop1234 [79]2 years ago
6 0

Answer:

You spent 18.25, and got 1.75 in change

Step-by-step explanation:

3.25 + 11 + 4 = 18.25

20.00 - 18.25 = 1.75

Marizza181 [45]2 years ago
4 0

Answer:

Spending total: $18.25

Change given back: $1.74

Step-by-step explanation:

$3.25+$11.00+$4.00=$18.25

$20.00-$18.25=$1.75

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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]
Answers: 

(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&#10;\\&#10;\\ \indent xdy = \left ( y^2 - y \right )dx&#10;\\&#10;\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}&#10;\\&#10;\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} &#10;\\&#10;\\ \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}&#10;\\&#10;\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} &#10;\\&#10;\\ \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&#10;\\&#10;\\ \indent \Rightarrow (A+B)y - B = 0y + 1&#10;\\&#10;\\ \indent \Rightarrow \begin{cases}&#10; A + B = 0&#10;& \text{(3)}\\-B = 1&#10; & \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}} &#10;\\&#10;\\ \indent \indent \indent \indent = \ln (y-1) - \ln y&#10;\\&#10;\\ \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&#10;\\&#10;\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2&#10;\\&#10;\\ \indent  \frac{y-1}{y} = e^{C_1 - C_2}x&#10;\\&#10;\\ \indent  \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}&#10;\\&#10;\\ \indent  1 - \frac{1}{y} = Cx&#10;\\&#10;\\ \indent \frac{1}{y} = 1 - Cx&#10;\\&#10;\\ \indent \boxed{y = \frac{1}{1 - Cx}}&#10;       (5)

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

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1&#10;&#10;

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}&#10;\\&#10;\\ \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}&#10;\\&#10;\\ \indent 16 = \frac{1}{1 - C(16)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 16C}&#10;\\&#10;\\ \indent 16(1 - 16C) = 1&#10;\\ \indent 16 - 256C = 1&#10;\\ \indent - 256C = -15&#10;\\ \indent \boxed{C = \frac{15}{256}}&#10;&#10;&#10;

By replacing this value of C, the general solution becomes

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent y = \frac{1}{1 - \frac{15}{256}x} &#10;\\ &#10;\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}&#10;\\&#10;\\&#10;\\ \indent \boxed{y = \frac{256}{256 - 15x}}&#10;&#10;&#10;&#10;

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} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - C(4)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 4C} &#10;\\ &#10;\\ \indent 16(1 - 4C) = 1 &#10;\\ \indent 16 - 64C = 1 &#10;\\ \indent - 64C = -15 &#10;\\ \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
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