farmer joe is enclosing a rectangular area on his arm for his chicken with 200 feet of fencing that he recently acquired in a tr
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The number of calories per ounce of soda is 10
<h3>Part A: Represent the relationship between the number of calories and the number of ounces</h3>The given parameters are:
Calories = 50
Ounces = 5
Let the number of calories be y and the ounces be x.
So, we have:
y = kx
Substitute y = 50 and x = 5
50 = 5k
Divide by 5
k = 10
Substitute k = 10 in y = kx
y = 10x
See attachment for the graph of the relationship between the number of calories and the number of ounces
<h3>Part B: What is the number of calories per ounce of soda?</h3>In (a), we have:
k = 10
This means that the number of calories per ounce of soda is 10
<h3>Part C: How does the unit rate relate to the slope of the line in the graph above? </h3>The unit rate and the slope represent the same and they have the same value
Read more about linear graphs at:
#SPJ1
Answers:
(a)
, for any constant C
(b) Solution does not exist
(c)
(d)
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
(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:
(2)
Since equation (2) has the same denominator, the numerator has to be equal. So,
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,
So,
Now, equation (1) becomes
(5)
At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have
Hence, for any constant C, the following solution will pass thru (0, 1):
(b) Using equation (5) in problem (a),
(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
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
By replacing this value of C, the general solution becomes
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
Now, we replace C using the derived value in the general solution. Then,
(a)
(b) Solution does not exist
(c)
(d)
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
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:
Since equation (2) has the same denominator, the numerator has to be equal. So,
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,
So,
Now, equation (1) becomes
At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have
Hence, for any constant C, the following solution will pass thru (0, 1):
(b) Using equation (5) in problem (a),
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
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
By replacing this value of C, the general solution becomes
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
Now, we replace C using the derived value in the general solution. Then,