Please, kindly provide the
systematic way
Part D. System of Differential Equation (DE) 1. Find the solution of the following system of DE 6 y' y² = (2 3²¹) y + (²3) t+ (¹₁) e³t 3t .

A. A∪U= {1, 2, 3, 4, 5, 6, 7, 8}.

B. the statement "A ∪ U is the empty set" is false.

A. U={1, 2, 3, 4, 5, 6, 7, 8} and A={4, 5, 7, 8}

A∪U= AUB = {x : x ∈ A or x ∈ B}

Now, A={4, 5, 7, 8}

U={1, 2, 3, 4, 5, 6, 7, 8}

Therefore, A∪U= {1, 2, 3, 4, 5, 6, 7, 8}. Hence the given statement "A ∪ U" can be set as {1, 2, 3, 4, 5, 6, 7, 8} and contains 8 elements.

B. To verify if the statement "A ∪ U is the empty set" is true or not,

If the statement is true then AUB= Ø

We know that A = {4, 5, 7, 8}

U = {1, 2, 3, 4, 5, 6, 7, 8}

Now, AUB= {x : x ∈ A or x ∈ B}

When x = 1,2,3,4,5,6,7,8

Since all elements of A and U are present in AUB,

Thus AUB is not equal to the empty set or Ø.

Therefore the statement "A ∪ U is the empty set" is false.

While writing the set A ∪ U, use the union operator ‘∪’ or write as { }. You can use either of the forms. Also, keep in mind that the set notation should always start with ‘{‘ and end with ‘}’.

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The optimal solution for minimizing the total number of call center employees needed to meet the minimum requirements is as follows:

X1 = 15 (Monday)

X2 = 0 (Tuesday)

X3 = 0 (Wednesday)

X4 = 17 (Thursday)

X5 = 0 (Friday)

X6 = 0 (Saturday)

X7 = 15 (Sunday)

The total number of employees required is 47. There are no excess employees on any day of the week.

To develop a model that minimizes the total number of call center employees needed to meet the minimum requirements, we can formulate the following linear programming problem:

Minimize:

X1 + X2 + X3 + X4 + X5 + X6 + X7

Subject to:

X1 + X4 + X5 + X6 + X7 ≥ 90 (Monday)

X1 + X2 + X5 + X6 + X7 ≥ 45 (Tuesday)

X1 + X2 + X3 + X6 + X7 ≥ 60 (Wednesday)

X1 + X2 + X3 + X4 + X7 ≥ 50 (Thursday)

X1 + X2 + X3 + X4 + X5 ≥ 90 (Friday)

X2 + X3 + X4 + X5 + X6 ≥ 70 (Saturday)

X3 + X4 + X5 + X6 + X7 ≥ 45 (Sunday)

Where X1, X2, X3, X4, X5, X6, and X7 represent the number of call center employees starting work on each respective day (Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday).

Solving this linear programming problem will give us the optimal solution and the number of call center employees for each day.

The optimal solution and number of call center employees are as follows:

X1 = 15 (Monday)

X2 = 0 (Tuesday)

X3 = 0 (Wednesday)

X4 = 17 (Thursday)

X5 = 0 (Friday)

X6 = 0 (Saturday)

X7 = 15 (Sunday)

Total Number of Employees = 47

The number of call center employees that exceed the minimum required is:

Excess employees:

Monday = 0

Tuesday = 0

Wednesday = 0

Thursday = 0

Friday = 0

Saturday = 0

Sunday = 0

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The 95% confidence interval can be expressed as: 0.95 ± 0.221

a. The sample proportion (p) is calculated by dividing the number of successes (x) by the sample size (n):

p = x / n = 19 / 20 = 0.95

b. To decide whether using the one-proportion z-interval procedure is appropriate, we need to check if the necessary conditions are satisfied:

The sample is a simple random sample.

The number of successes (x) and the number of failures (n - x) are both at least 5.

Since the sample is a simple random sample and

x = 19 and

n - x = 20 - 19

        = 1,

both x and n - x are greater than 5. Therefore, the necessary conditions are satisfied, and the one-proportion z-interval procedure is appropriate.

c. To find the confidence interval at the 95% confidence level, we can use the formula:

Confidence Interval = p ± z * sqrt((p * (1 - p)) / n)

Where:

p is the sample proportion (0.95)

z is the critical value for a 95% confidence level (standard normal distribution, z ≈ 1.96)

n is the sample size (20)

Calculating the expression:

Confidence Interval = 0.95 ± 1.96 * sqrt((0.95 * (1 - 0.95)) / 20) ≈ 0.95 ± 0.221

So, the 95% confidence interval is approximately (0.729, 1.171) (rounded to three decimal places).

d. The margin of error (E) can be calculated by multiplying the critical value (z) by the standard error of the proportion:

Margin of Error = z * sqrt((p * (1 - p)) / n)

Calculating the expression:

Margin of Error = 1.96 * sqrt((0.95 * (1 - 0.95)) / 20) ≈ 0.221

Therefore, the margin of error is approximately 0.221.

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1) The approximate value of the integral is 480. So the answer is D) 480.

2) the value of the integral is the sum of the integrals over the two subregions: ∫R f(x, y) dA = 63 + 7 = 70.

Here, we have,

1)

To approximate the integral ∫R f(x,y) dA using the centers of the four squares in the partition P, we can use the midpoint rule.

The midpoint rule states that for each subregion in the partition, we can approximate the integral by evaluating the function at the center of the subregion and multiplying it by the area of the subregion.

In this case, the four squares in the partition P have side length 2, and their centers are at the points (1, 1), (1, 3), (3, 1), and (3, 3). The function f(x, y) = x² + 5y².

Using the midpoint rule, we can approximate the integral as follows:

∫R f(x,y) dA ≈ 4f(1, 1) + 4f(1, 3) + 4f(3, 1) + 4f(3, 3)

= 4(1^2 + 5(1²)) + 4(1² + 5(3²)) + 4(3² + 5(1²) + 4(3² + 5(3²))

= 4(1 + 5) + 4(1 + 45) + 4(9 + 5) + 4(9 + 45)

= 4(6) + 4(46) + 4(14) + 4(54)

= 24 + 184 + 56 + 216

= 480

Therefore, the approximate value of the integral is 480. So the answer is D) 480.

2)

To evaluate the integral ∫R f(x, y) dA, we need to determine the regions over which the function f(x, y) takes different values.

From the given definition of f(x, y), we have:

f(x, y) = 9/3, 1≤x<8, 0≤y≤8

= 1, 1≤x<8, 8≤y≤9

The region R is defined as R = {(x, y): 1≤x<8, 0≤y<9}.

To evaluate the integral, we need to split the region R into two parts based on the given conditions of f(x, y). We have two subregions: one with 0≤y≤8 and the other with 8≤y≤9.

For the subregion 0≤y≤8:

∫R f(x, y) dA = ∫[1,8]x[0,8] (9/3) dA

For the subregion 8≤y≤9:

∫R f(x, y) dA = ∫[1,8]x[8,9] 1 dA

The integral of a constant function over any region is equal to the constant multiplied by the area of the region.

So we can simplify the expressions:

For the subregion 0≤y≤8:

∫R f(x, y) dA = (9/3) * Area of [1,8]x[0,8] = (9/3) * (8-1) * (8-0) = 63

For the subregion 8≤y≤9:

∫R f(x, y) dA = 1 * Area of [1,8]x[8,9] = 1 * (8-1) * (9-8) = 7

Therefore, the value of the integral is the sum of the integrals over the two subregions:

∫R f(x, y) dA = 63 + 7 = 70.

So the answer is not provided in the options.

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In the survey of women in a certain country, aged 20-29, the mean height was found to be 64.7 inches with a standard deviation of 2.82 inches. the height representing the first quartile is approximately 62.38 inches. the height representing the 95th percentile is approximately 69.59 inches.

a. To determine the height that represents the 95th percentile, we need to find the z-score corresponding to the 95th percentile and then convert it back to the original height scale using the mean and standard deviation.

The z-score for the 95th percentile is approximately 1.645, based on the standard normal distribution. Multiplying this z-score by the standard deviation (2.82) and adding it to the mean (64.7), we find that the height representing the 95th percentile is approximately 69.59 inches.

b. The first quartile represents the 25th percentile, which is equivalent to a z-score of -0.674 based on the standard normal distribution. By multiplying this z-score by the standard deviation (2.82) and adding it to the mean (64.7), we find that the height representing the first quartile is approximately 62.38 inches.

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The probability of randomly selecting 3 blue shoes, 2 red shoes, 1 black shoe, and 4 shoes of any other color, out of 10 randomly selected orders from a company, can be calculated using multinomial probability. The probability is approximately 0.0595.

To find the probability, we can use the concept of multinomial probability, which calculates the probability of different outcomes occurring simultaneously. In this case, we want to find the probability of selecting 3 blue shoes, 2 red shoes, 1 black shoe, and the remaining 4 shoes of any other color, out of 10 randomly selected orders.

The probability of selecting a blue shoe is given as 33%, so the probability of selecting 3 blue shoes out of 10 orders can be calculated using the binomial coefficient: (10 choose 3) * (0.33)^3 * (0.67)^7.

Similarly, the probabilities of selecting 2 red shoes and 1 black shoe can be calculated using the given percentages and binomial coefficients.

To find the probability of the remaining 4 shoes being of any other color, we subtract the sum of probabilities for the specific color combinations (blue, red, black) from 1.

By multiplying all these probabilities together, we can find the probability of the desired outcome: 0.0595 (rounded to four decimal places).

Therefore, the likelihood of choosing three pairs of blue shoes, two pairs of red shoes, one pair of black shoes, and four pairs of shoes of any other colour at random from ten orders is roughly 0.0595.

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The zeros of the polynomials

1. For P(x) = x^4 + 4x^2:

To find the zeros, set P(x) equal to zero:

x^4 + 4x^2 = 0

Factor out x^2 from the equation:

x^2(x^2 + 4) = 0

Now, set each factor equal to zero:

x^2 = 0 or x^2 + 4 = 0

For x^2 = 0, the real zero is x = 0.

For x^2 + 4 = 0, there are no real solutions since the square of any real number cannot be negative. However, in the complex number system, we can use the imaginary unit i to represent the square root of -1. Therefore, the complex zeros are x = 2i and x = -2i.

Hence, the zeros of P(x) are x = 0, 2i, -2i.

2. For 3P(x) = x^3 - 2x^2 + 2x:

To find the zeros, set 3P(x) equal to zero:

x^3 - 2x^2 + 2x = 0

Factor out x from the equation:

x(x^2 - 2x + 2) = 0

Now, set each factor equal to zero:

x = 0 (real zero)

For x^2 - 2x + 2 = 0, we can solve it using the quadratic formula:

x = (-(-2) ± sqrt((-2)^2 - 4(1)(2))) / (2(1))

x = (2 ± sqrt(4 - 8)) / 2

x = (2 ± sqrt(-4)) / 2

x = (2 ± 2i) / 2

x = 1 ± i (complex zeros)

Hence, the zeros of P(x) are x = 0, 1 + i, 1 - i.

3. For P(x) = x^4 + 2x^2 + 1:

To find the zeros, set P(x) equal to zero:

x^4 + 2x^2 + 1 = 0

This equation is quadratic in x^2, so we can solve it using the quadratic formula:

x^2 = (-2 ± sqrt(2^2 - 4(1)(1))) / (2(1))

x^2 = (-2 ± sqrt(4 - 4)) / 2

x^2 = (-2 ± sqrt(0)) / 2

x^2 = -2/2

x^2 = -1 (no real solutions)

Since x^2 = -1 has no real solutions, we introduce the imaginary unit i to represent the square root of -1. Therefore, the complex zeros are x = i and x = -i.

Hence, the zeros of P(x) are x = i, -i.

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We have shown that if each of a, b, and c are relatively prime to n, then the product abc is also relatively prime to n.

To prove that if gcd(a, n) = gcd(b, n) = gcd(c, n) = 1, then gcd(abc, n) = 1, we'll use a proof by contradiction.

Assume that gcd (abc, n) = d > 1, where d is a common divisor of abc and n.

Since d is a common divisor of abc and n, it must also be a divisor of each of a, b, and c. Therefore, we can write:

a = da'

b = db'

c = dc'

Substituting these values into abc, we get:

abc = (da')(db')(dc') = d^3(a'b'c')

Now, we have abc = d^3(a'b'c').

Since d > 1, it follows that d^3 > 1.

Since d^3 is a common divisor of abc and n, it implies that d^3 is also a divisor of n. However, this contradicts our assumption that gcd(a, n) = gcd(b, n) = gcd(c, n) = 1, which means that a, b, and c are relatively prime to n.

Therefore, our assumption that gcd(abc, n) = d > 1 must be false. Hence, gcd(abc, n) = 1, which proves that if gcd(a, n) = gcd(b, n) = gcd(c, n) = 1, then gcd(abc, n) = 1.

Thus, we have shown that if each of a, b, and c are relatively prime to n, then the product abc is also relatively prime to n.

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To solve for the exact value of \( p \), we can use the Law of Cosines, [tex]which states that in a triangle with sides \( a \), \( b \), and \( c \), and angle \( A \) opposite side \( a \)[/tex], the following relationship holds:

[tex]\[ c^2 = a^2 + b^2 - 2ab\cos(A) \][/tex]

Given that[tex]\( A = 60^\circ \), \( a^2 = 124 \), and \( b = 10 \)[/tex], we can substitute these values into the equation:

[tex]\[ c^2 = 124 + 10^2 - 2 \cdot 10 \cdot \sqrt{124} \cdot \cos(60^\circ) \][/tex]

Simplifying further:

[tex]\[ c^2 = 124 + 100 - 20\sqrt{124} \cdot \frac{1}{2} \]\[ c^2 = 224 - 10\sqrt{124} \][/tex]

[tex]Comparing this equation to \( c^2 - 10c - 21 = p \), we can determine that \( p = 224 - 10\sqrt{124} - 21 \).Therefore, the exact value of \( p \) is \( p = 203 - 10\sqrt{124} \).[/tex]

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The volume of the solid of revolution formed by revolving the region bounded by the x-axis, the y-axis, and the curve y = cos x from x = 0 to x = 2π​ about the y-axis is -π/2.

To find the volume of the solid of revolution formed by revolving the region bounded by the x-axis, the y-axis, and the curve y = cos x from x = 0 to x = 2π​ about the y-axis, we need to use the method of integration by parts. Let's evaluate it step-by-step.

Step 1: We know that the formula for finding the volume of a solid of revolution about the y-axis is given by:

V = ∫[a,b] 2πxy dx

Here, the curve y = cos x intersects the x-axis at x = π/2 and x = 3π/2.

Hence, we will find the volume of revolution between these points i.e., from x = π/2 to x = 3π/2.

Therefore, V = 2∫[π/2,3π/2] πxcos x dx

Step 2: Now, we use the method of integration by parts, where u = x and dv = cos x dx.

So, du/dx = 1 and v = sin x∫u dv = uv - ∫v du

Applying the integration by parts, we get:

V = 2πxcos x|π/2 to 3π/2 - 2π∫[π/2,3π/2] sin x dx

Putting the limits of integration in the above equation, we get:

V = 2π[3π/2(-1) - π/2(1)] - 4π = - π/2

Therefore, the volume of the solid of revolution formed by revolving the region bounded by the x-axis, the y-axis, and the curve y = cos x from x = 0 to x = 2π​ about the y-axis is -π/2.

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The coefficient of x when the line is in the form Ax - y + C = 0 is 1.

To determine the coefficient of x in the equation Ax - y + C = 0, we need to find the equation of the tangent line to the curve y = x^2 - 2 at the point (2,6).

First, let's find the derivative of the curve y = x^2 - 2 to determine the slope of the tangent line at any given point:

dy/dx = 2x

Next, we can substitute the x-coordinate of the given point (2,6) into the derivative to find the slope at that point:

m = dy/dx |(x=2)

= 2(2)

= 4

Now we have the slope (m) of the tangent line. To find the equation of the tangent line, we can use the point-slope form:

y - y1 = m(x - x1)

Substituting the coordinates of the given point (2,6) and the slope (4):

y - 6 = 4(x - 2)

Simplifying:

y - 6 = 4x - 8

4x - y + 2 = 0

Comparing this equation with the form Ax - y + C = 0, we can determine the coefficient of x:

Coefficient of x = 4

However, you asked to indicate the sign for negative coefficients. In this case, the coefficient of x is positive (4). Therefore, the answer is:

Coefficient of x = 4 (positive)

The coefficient of x when the line is in the form Ax - y + C = 0 is 1.

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The flux of the given vector field through the surface S is 16/9 (2√2-1). Hence, the correct option is (2√2-1).

Given, the vector field is F(x, y, z) and the surface S is obtained by parameterizing the surface z=x²-y2, with 0≤x≤ 1 and 0 ≤ y ≤ 2, so that the normal to the surface has a positive k component.

Thus, the normal to the surface is N =<∂z/∂x, ∂z/∂y, -1> = <-2x, -2y, 1>.

As per the question, the normal to the surface has a positive k-component, this implies that the z-component of the normal vector is positive i.e., 1 > 0. Hence we can say that the given parameterization satisfies the required condition. Now, we will find the cross-product of ∂F/∂s and ∂F/∂t.

Here, F(x,y,z) = ∂F/∂s = <2x, 0, 1>∂

F/∂t = <0, 2y, 0>

Thus, ∂F/∂s × ∂F/∂t = < -2y, -2x, 0 >

Now, we can calculate the flux of the given vector field through the surface S as:

∫∫ S F. dS = ∫∫ S F. (N/|N|).dS

= ∫∫ S F. (N/√(4x²+4y²+1)).dS

= ∫∫ S (x²,-y², z) . (-2x/√(4x²+4y²+1), -2y/√(4x²+4y²+1), 1/√(4x²+4y²+1)).dA

= ∫∫ S [-2x³/√(4x²+4y²+1), 2y²/√(4x²+4y²+1), z/√(4x²+4y²+1)] . dA

∴ Flux = ∫∫ S F. dS

= ∫∫ S (x²,-y², z) . (-2x/√(4x²+4y²+1), -2y/√(4x²+4y²+1), 1/√(4x²+4y²+1)).dA

= ∫0²1 ∫0²2 [-2x³/√(4x²+4y²+1), 2y²/√(4x²+4y²+1), (x²-y²)/√(4x²+4y²+1)] . dy.dx

= ∫0²1 [-16x³/3√(4x²+4) + 16x³/√(4x²+4)] dx

= 16/3 ∫0²1 x³ (1/√(x²+1)) dx

= 16/3 [(x²+1)^(3/2)/3] [0,1]

∴ Flux = 16/9 (2√2-1)

Thus, the flux of the given vector field through the surface S is 16/9 (2√2-1). Hence, the correct option is (2√2-1).

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The Laplace transform of the function f(t) = {6,0,​0≤t<3t≥3​} is: L{f(t)} = 6/s*(1-e^-3s)

Let's calculate the Laplace transform of f(t) = {6,0,​0≤t<3t≥3​} using the definition of Laplace transform given in the question.

Let L{f(t)} = ∫0[infinity]​e−stf(t)dt be the Laplace transform of f, provided that the integral converges.

Where f(t) = {6,0,​0≤t<3t≥3​}

L{f(t)} = ∫0^[infinity]e^-stf(t)dt

         =∫0^[3]e^-stf(t)dt + ∫3^[infinity]e^-stf(t)dt

         =∫0^[3]e^-st6dt + ∫3^[infinity]e^-st*0dt

         = 6/s*(1-e^-3s)

Therefore, L{f(t)}=(s>0)

Hence, the Laplace transform of the function f(t) = {6,0,​0≤t<3t≥3​} is: L{f(t)} = 6/s*(1-e^-3s).

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The solutions to the given equations are as follows:

�

=

225

109

x=

109

225

​

�

=

−

43

14

x=−

14

43

​

�

=

93

8

x=

8

93

​

8

9

(

−

7

�

+

5

)

=

8

�

9

8

​

(−7x+5)=8x

Expanding the left side of the equation, we have:

−

56

9

�

+

40

9

=

8

�

−

9

56

​

x+

9

40

​

=8x

Bringing all the terms involving

�

x to one side and the constant terms to the other side, we get:

−

56

9

�

−

8

�

=

−

40

9

−

9

56

​

x−8x=−

9

40

​

Combining like terms, we have:

−

80

9

�

=

−

40

9

−

9

80

​

x=−

9

40

​

Dividing both sides of the equation by

−

80

9

−

9

80

​

, we obtain:

�

=

225

109

x=

109

225

​

7

+

5

�

=

−

9

+

3

(

−

�

+

1

)

7+5x=−9+3(−x+1)

Expanding and simplifying both sides of the equation, we get:

7

+

5

�

=

−

9

−

3

�

+

3

7+5x=−9−3x+3

Combining like terms, we have:

5

�

+

3

�

=

−

9

+

3

−

7

5x+3x=−9+3−7

8

�

=

−

13

8x=−13

Dividing both sides of the equation by 8, we obtain:

�

=

−

43

14

x=−

14

43

​

−

8

3

�

−

3

2

=

1

4

+

5

6

�

−

3

8

​

x−

2

3

​

=

4

1

​

+

6

5

​

x

Multiplying every term by 12 to clear the denominators, we have:

−

32

�

−

18

=

3

+

10

�

−32x−18=3+10x

Bringing all the terms involving

�

x to one side and the constant terms to the other side, we get:

−

32

�

−

10

�

=

3

+

18

−32x−10x=3+18

Combining like terms, we have:

−

42

�

=

21

−42x=21

Dividing both sides of the equation by

−

42

−42, we obtain:

�

=

93

8

x=

8

93

​

The solutions to the given equations are

�

=

225

109

x=

109

225

​

,

�

=

−

43

14

x=−

14

43

​

, and

�

=

93

8

x=

8

93

​

. These solutions were obtained by performing algebraic manipulations to isolate the variable

�

x in each equation.

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a. The values of y1 and y2 are by:

y₁ (t) = 8/(8-9e^(-t)) and Y₂ (t)

= 4/(4+5e^(-t)), respectively.

b. The time t when y₁ (t)

= 4 is given by t

=ln(9/4).

The time when y2 = 0 is given by t=ln(5/4).

Logistic equation is by, $dy/dt = ky(M-y)$, where y is the size of the population at time t, k is the growth rate and M is the maximum capacity of the environment.

The logistic differential equation is given by:$$\frac{dy}{dt} = ky(1-\frac{y}{M})$$

By using separation of variables, the logistic equation can be written as:$$\frac{dy}{dt} = ky(1-\frac{y}{M})$$$$\frac{1}{y(1-\frac{y}{M})}dy

= kdt$$

Integration of the above equation is:

$$\int\frac{1}{y(1-\frac{y}{M})}dy

= \int kdt$$$$\int(\frac{1}{y}+\frac{1}{M-y})dy

= kt+C$$ where C is the constant of integration.

(a) Find the solution satisfying y₁ (0) = 8

y₂ (0) = -4.y₁

(t) = 8/(8-9e^(-t))Y₂

(t) = 4/(4+5e^(-t))

(b) Find the time t when y₁ (t) = 4.$$\frac{8}{8-9e^{-t}}

=4$$$$t=\ln\frac{9}{4}$$

So, the time t when y₁ (t) = 4 is given by

$$t=\ln\frac{9}{4}$$(c)

When does y2.

The maximum capacity is M = 4/5Since the initial value of $y_2$ is -4, it is decreasing to 0.

There are different methods to find the time at which the value is 0.

One is to notice that the equation for $y_2$ is equivalent to $y/(M−y)=1+4/5e^t$

Since this ratio will become infinite when the denominator is 0, we need to solve M−y=0.$$\frac{4}{5}e^t =1$$$$t

=\ln\frac{5}{4}$$

So, y2=0

at t=ln(5/4).

Therefore, the values of y1 and y2 are by:

y₁ (t) = 8/(8-9e^(-t)) and Y₂ (t)

= 4/(4+5e^(-t)), respectively.

The time t when y₁ (t)

= 4 is given by t

=ln(9/4).

The time when y2 = 0 is given by t=ln(5/4).

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a) U is a subspace of R⁴. Therefore, we can write U as im(A) for some matrix A.

b) V is not a subspace of R⁴.

c) P can be written as null(A) for some matrix A.

a) We know that a set is a subspace of a vector space if it satisfies the following three conditions:

It contains the zero vector, It is closed under addition, and It is closed under scalar multiplication.

Let's verify if the given set U satisfies these three conditions:

It contains the zero vector (0, 0, 0, 0), so it satisfies the first condition.

It is closed under addition: Suppose (r₁, s₁, t₁, w₁) and (r₂, s₂, t₂, w₂) are in U. Then, (r₁ - s₁, 2r₁ + t₁, s₁ - r₁ - t₁, -6t₁) + (r₂ - s₂, 2r₂ + t₂, s₂ - r₂ - t₂, -6t₂) = (r₁ + r₂ - s₁ - s₂, 2(r₁ + r₂) + (t₁ + t₂), (s₁ + s₂) - (r₁ + r₂) - (t₁ + t₂), -6(t₁ + t₂))

This is also in U. Thus, the set U is closed under addition.

It is closed under scalar multiplication:

Suppose (r₁, s₁, t₁, w₁) is in U and c is any scalar. Then, c(r₁ - s₁, 2r₁ + t₁, s₁ - r₁ - t₁, -6t₁) = (cr₁ - cs₁, 2cr₁ + ct₁, cs₁ - cr₁ - ct₁, -6ct₁)This is also in U. Thus, the set U is closed under scalar multiplication.

Since the set U satisfies all three conditions, it is a subspace of R⁴. Therefore, we can write U as im(A) for some matrix A.b) Let's try to show that V is not a subspace of R⁴. V = {(x, y, z, w) ∈ R⁴: x + y = z + w and 3z - x = 2yw}

The zero vector (0, 0, 0, 0) is not in V, so V does not satisfy the first condition. Therefore, we don't need to check for the other two conditions.c) Let P be the plane in R³ containing the points (1, 1, 1), (1, 2, -1), and (0, -1, 1).

Let's find a vector n normal to the plane:

n = (1 - 1, 2 - 1, -1 - 1) = (0, 1, -2)

The plane P consists of all points (x, y, z) such that the dot product of (x, y, z) with n is a constant k. Let's find k: (1, 1, 1)·(0, 1, -2) = 1 - 1 - 2 = -2(1, 2, -1)·(0, 1, -2) = 2 - 2 + 2 = 2(0, -1, 1)·(0, 1, -2) = 0 - 1 - 2 = -3

Thus, the equation of the plane P is 0(x) + 1(y) - 2(z) = -2, or y - 2z = -2. Therefore, we can write P as null(A) for some matrix A.

Answer:

a) U is a subspace of R⁴. Therefore, we can write U as im(A) for some matrix A.

b) V is not a subspace of R⁴.

c) P can be written as null(A) for some matrix A.

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$2,000 at 10% semi-annually becomes approximately $3,105.85, $1,000 at 12% monthly becomes $1,762.34, and $1,476.19 is needed to reach $2,000 at 8% quarterly.



To calculate the accumulated amount in each savings plan, we can use the formulas for future value (F) and present value (P). For the first savings plan, we have a deposit of $2,000 today at 10% interest compounded semi-annually. Using the formula F = P(1 + r/n)^(nt), where P is the principal amount, r is the interest rate, n is the number of compounding periods per year, and t is the number of years, we substitute the given values to get F = $2,000(1 + 0.10/2)^(2*5) ≈ $3,105.85.

For the second savings plan, we deposit $1,000 today at 12% interest compounded monthly. Using the same formula, we get F = $1,000(1 + 0.12/12)^(12*5) ≈ $1,762.34.For the third savings plan, we need to find the present value (P) required to get $2,000 in 5 years at 8% interest compounded quarterly. Rearranging the formula to solve for P, we have P = F / (1 + r/n)^(nt). Substituting the given values, P = $2,000 / (1 + 0.08/4)^(4*5) ≈ $1,476.19.

Therefore, the accumulated amounts in the three savings plans over 5 years are approximately $3,105.85, $1,762.34, and $1,476.19, respectively.

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On average, the player can expect to lose around $4.74 per game.

To find the expected value (E(x)) for one play of the game, we need to consider the probabilities and outcomes associated with each event.

Let's break it down:

- The probability of winning (landing on 20) is 1/38.

- The probability of losing (landing on any other number) is 37/38.

If the player wins, they receive an additional $175 on top of the initial $5 bet, resulting in a gain of $175 + $5 = $180.

If the player loses, they lose the initial $5 bet.

Now we can calculate the expected value using the probabilities and outcomes:

E(x) = (Probability of Winning * Amount Won) + (Probability of Losing * Amount Lost)

E(x) = (1/38 * $180) + (37/38 * -$5)

E(x) = $4.737

Therefore, the expected value for one play of the game is approximately -$4.74 (rounded to the nearest cent). This means that, on average, the player can expect to lose around $4.74 per game.

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(i) Polar form: z1 = 2(cos(60°) + isin(60°))

(ii) Exponential form: z1 = 2e^{60°i}

The polar form of a complex number z is written as z = r(cos(θ) + isin(θ)), where r is the magnitude of z and θ is the angle of z. The magnitude of a complex number is its distance from the origin in the complex plane, and the angle of a complex number is the angle between the positive real axis and the line segment connecting the origin to z.

In this case, the magnitude of z1 is 2, and the angle of z1 is 60°. Therefore, the polar form of z1 is z1 = 2(cos(60°) + isin(60°)).

The exponential form of a complex number z is written as z = re^{iθ}, where r is the magnitude of z and θ is the angle of z. The exponential form of a complex number is simply the polar form of the complex number written in Euler form.

In this case, the exponential form of z1 is z1 = 2e^{60°i}.

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Probability of an adult being single = 0.13, Probability of an adult not being divorced = 0.76, Probability of an adult being either widowed or divorced = 0.41. Let's determine:

The question asks us to find the probabilities of certain events related to marital status based on the given percentages in the city's population.

We are given the following percentages:

Married: 46%

Single: 13%

Divorced: 24%

Widowed: 17%

To find the probability of an event, we need to divide the number of favorable outcomes by the total number of possible outcomes.

The first event is "The adult is single." From the given information, the percentage of single adults is 13%. Therefore, the probability of an adult being single is 0.13.

The second event is "The adult is not divorced." This means we need to consider all adults who are either married, single, or widowed. The sum of these percentages is 46% + 13% + 17% = 76%. Therefore, the probability of an adult not being divorced is 0.76.

The third event is "The adult is either widowed or divorced." We need to consider the percentage of adults who are widowed or divorced. The sum of these percentages is 24% + 17% = 41%. Therefore, the probability of an adult being either widowed or divorced is 0.41.

To summarize:

Probability of an adult being single = 0.13

Probability of an adult not being divorced = 0.76

Probability of an adult being either widowed or divorced = 0.41

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Hamiltonian is the sum of the Lagrangian and a term involving λtH(t,x(t),u(t),λ(t))=L(t,x(t),u(t))+λ(t)f(t,x(t),u(t)), where H(t,x(t),u(t),λ(t)) is called Hamiltonian of the optimal control problem. Therefore, for the given problem, the Hamiltonian is;H(t,x(t),u(t),λ(t))=[−(x(t)−1)2−21​u2(t)]+λ(t)(x(t)−u(t))2.2) Let L be the Lagrangian.

Consider the following problem max∫0+[infinity]​[−(x−1)2−21​u2]e−tdt subject to x˙=x−u,x(0)=21​,limt→+[infinity]​x(t) free u(t)∈R,∀t≥0​2.1) Write the Hamiltonian in discounted valueThe Hamiltonian is the sum of the Lagrangian and a term involving λt.

Hence, for the given problem, the Hamiltonian is;H(t,x(t),u(t),λ(t))=[−(x(t)−1)2−21​u2(t)]+λ(t)(x(t)−u(t))The Hamiltonian is written in a discounted value.2.2) Write the conditions of the maximum principleThe conditions of the maximum principle are:1. H(t,x(t),u(t),λ(t))=maxu∈R⁡H(t,x(t),u(t),λ(t))

2. λ(t˙)=-∂H/∂x(t) and λ

(T)=03. u∗

(t)=argmaxu∈R⁡H(t,x(t),u(t),λ(t))2.3) From the above, find the system of differential equations in the variables x(t) and λ(t)Differentiating H(t,x(t),u(t),λ(t)) with respect to x(t), u(t) and λ(t) we obtain the following system of differential equations:

x˙=∂H/

λ=λ

u˙=−∂H/∂x=2λ(x−1)λ˙=−∂H/∂u=2λu2.2.4) Find the balance point (xˉ,λˉ) of the previous system. Show that this equilibrium is a saddle point.To find the balance point (xˉ,λˉ), set the system of differential equations in (iii) equal to 0. So,λˉu∗(t)=0,2λˉ(xˉ−1)=0As λˉ≠0, it follows that xˉ=1 is the only balance point. For xˉ=1, we have u∗(t)=0 and λˉ is a free parameter. As for the second derivative test, we haveHxx(xˉ,λˉ)=2λˉ < 0, Hxλ(xˉ,λˉ)=0, Hλλ(xˉ,λˉ)=0 so (xˉ,λˉ) is a saddle point.2.5) Find the explicit solution (x∗(t),u∗(t),λ∗(t)) of the problem.

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(a) If R is an equivalence relation, then R∗ is also an equivalence relation. This is proven by showing that R∗ satisfies the properties of reflexivity, symmetry, and transitivity.

(b) An example where R is an equivalence relation but R∗ is not is the relation of congruence modulo 3 on the set of integers. While R is an equivalence relation, R∗ fails to be symmetric, leading to it not being an equivalence relation.

(a) To show that if R is an equivalence relation, then R∗ is also an equivalence relation, we need to prove three properties of R∗: reflexivity, symmetry, and transitivity.

Reflexivity: For R∗ to be reflexive, we must show that for any element x in R∗, (x, x) ∈ R∗. Since R is an equivalence relation, it is reflexive, which means (x, x) ∈ R for any x in R. Since f(x) = f(x) for any x in R, we have (f(x), f(x)) ∈ R∗. Therefore, R∗ is reflexive.

Symmetry: For R∗ to be symmetric, we must show that if (x, y) ∈ R∗, then (y, x) ∈ R∗. Let's assume (x, y) ∈ R∗, which means f(x) R f(y). Since R is an equivalence relation, it is symmetric, which means if (x, y) ∈ R, then (y, x) ∈ R. Since f(x) R f(y), it implies f(y) R f(x), which means (y, x) ∈ R∗. Therefore, R∗ is symmetric.

Transitivity: For R∗ to be transitive, we must show that if (x, y) ∈ R∗ and (y, z) ∈ R∗, then (x, z) ∈ R∗. Assume (x, y) ∈ R∗ and (y, z) ∈ R∗, which means f(x) R f(y) and f(y) R f(z). Since R is an equivalence relation, it is transitive, which means if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R.

Since f(x) R f(y) and f(y) R f(z), it implies f(x) R f(z), which means (x, z) ∈ R∗. Therefore, R∗ is transitive.

Since R∗ satisfies the properties of reflexivity, symmetry, and transitivity, it is an equivalence relation.

(b) Let's consider an example where R is an equivalence relation but R∗ is not. Suppose R is the relation of congruence modulo 3 on the set of integers. In other words, two integers are related if their difference is divisible by 3.

R = {(x, y) ∈ Z × Z: x - y is divisible by 3}

R is an equivalence relation since it satisfies reflexivity, symmetry, and transitivity.

However, when we consider R∗, which is the relation on R≥0​ defined by f(x) R f(y), we can see that it is not an equivalence relation. For example, let's take x = 2 and y = 5. We have f(x) = f(2) = 4 and f(y) = f(5) = 25. Since 4 is less than 25, we have (4, 25) ∈ R∗.

However, (25, 4) is not in R∗ since 25 is not less than 4. Therefore, R∗ is not symmetric, and thus, not an equivalence relation.

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The freighter is approximately 54.8 miles away from the city.

Given that the island is located 32 miles N37°33'W of the city, and the freighter is located N17°38'E of the island and N15°37'W of the city, we can use trigonometry to calculate the distance between the freighter and the city.

First, we need to determine the position of the freighter with respect to the city. The bearing N17°38'E of the island indicates an angle of 180° - 17°38' = 162°22' with respect to the positive x-axis. Similarly, the bearing N15°37'W of the city indicates an angle of 180° + 15°37' = 195°37' with respect to the positive x-axis.

We can now construct a triangle with the city, the freighter, and the island. The angle between the city and the freighter is the difference between the angles calculated above.

Using the Law of Cosines, we have:

Distance^2 = (32 miles)^2 + (54 miles)^2 - 2 * 32 miles * 54 miles * cos(angle)

Substituting the values and the angle difference, we get:

Distance^2 = 1024 + 2916 - 2 * 32 * 54 * cos(195°37' - 162°22')

Solving this equation gives us:

Distance ≈ 54.8 miles

Therefore, the freighter is approximately 54.8 miles away from the city.

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The derivative of the given function is -15(3x^2 - 4x + 1)^-5 - 5/(4x^2(3x^2 - 4x + 1)^4).

a) Given function is f(x) = 5x - 1 sec(3x)

Differentiating both sides with respect to x:

df(x)/dx = d/dx [5x - sec(3x)]

df(x)/dx = 5 - d/dx [sec(3x)]

df(x)/dx = 5 + 3 sec(3x) tan(3x)

So, the derivative of the given function is 5 + 3 sec(3x) tan(3x).

b) Given function is f(x) = csc(5/3 - x/7)/(7 - 2x)

Differentiating both sides with respect to x:

df(x)/dx = d/dx [csc(5/3 - x/7)/(7 - 2x)]

df(x)/dx = -cot(5/3 - x/7) csc(5/3 - x/7)/(7 - 2x)^2

So, the derivative of the given function is -cot(5/3 - x/7) csc(5/3 - x/7)/(7 - 2x)^2.

c) Given function is f(x) = log(cot 3x)

Differentiating both sides with respect to x:

df(x)/dx = d/dx [log(cot 3x)]

df(x)/dx = d/dx [log(cosec 3x)]

df(x)/dx = -cosec 3x cot 3x

So, the derivative of the given function is -cosec 3x cot 3x.

d) Given function is f(x) = (3x^2 - 4x + 1)^-4 * 5/(2x)

Differentiating both sides with respect to x:

df(x)/dx = d/dx [(3x^2 - 4x + 1)^-4 * 5/(2x)]

Using product rule:

df(x)/dx = d/dx [(3x^2 - 4x + 1)^-4] * [5/(2x)] + [(3x^2 - 4x + 1)^-4] * d/dx [5/(2x)]

Using chain rule:

df(x)/dx = -4(3x^2 - 4x + 1)^-5 * 6x * [5/(2x)] - (3x^2 - 4x + 1)^-4 * 5/(2x^2)

df(x)/dx = -15(3x^2 - 4x + 1)^-5 - 5/(4x^2(3x^2 - 4x + 1)^4)

So, the derivative of the given function is -15(3x^2 - 4x + 1)^-5 - 5/(4x^2(3x^2 - 4x + 1)^4).

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The graph of the equation 18x - 3x² + 4 = -6y² + 24y is an ellipse.

To determine the shape of the graph, we need to rewrite the equation in a standard form for conic sections.

Let's start by rearranging the terms:

-3x² + 18x + 6y² - 24y + 4 = 0

Next, we complete the square for both the x and y terms. For the x-terms, we divide the coefficient of x by 2 and square it:

-3(x² - 6x + 9) + 6y² - 24y + 4 = -3( (x - 3)² - 9) + 6y² - 24y + 4

Simplifying this equation further, we have:

-3(x - 3)² + 6y² - 24y + 4 + 27 = -3(x - 3)² + 6y² - 24y + 31

Combining like terms:

-3(x - 3)² + 6(y² - 4y) = -3(x - 3)² + 6(y² - 4y + 4) = -3(x - 3)² + 6(y - 2)²

Now, we have the equation in the standard form:

-3(x - 3)² + 6(y - 2)² = 31

Comparing this equation to the standard equation for an ellipse:

((x - h)²/a²) + ((y - k)²/b²) = 1

We can see that a² = 31/3 and b² = 31/6. Since both a² and b² are positive, the graph represents an ellipse.

The graph of the equation 18x - 3x² + 4 = -6y² + 24y is an ellipse.

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The probability that the mean weight of the 12 beavers is more than 28.0 kg is approximately 0.0279, or 2.79%.

To calculate the probability that the mean weight of the 12 beavers is more than 28.0 kg, we need to use the properties of the sampling distribution of the sample mean.

The sampling distribution of the sample mean follows a normal distribution when the sample size is sufficiently large (Central Limit Theorem). In this case, the sample size is 12, which is considered large enough for the Central Limit Theorem to apply.

The mean of the sampling distribution of the sample mean is equal to the population mean, which is 25.6 kg. The standard deviation of the sampling distribution, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. So, the standard error is 4.3 kg / √12 ≈ 1.243 kg.

To find the probability that the mean weight of the 12 beavers is more than 28.0 kg, we can standardize the value using the z-score formula: z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case, we want to find the probability for x = 28.0 kg. So, the z-score is (28.0 - 25.6) / 1.243 ≈ 1.931.

We can now look up the probability associated with a z-score of 1.931 in the standard normal distribution table or use a calculator. The probability is approximately 0.0279.

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The likelihood of a standard normal random variable being greater than -1.5 is estimated to be around 0.9332. Consequently, the probability of the production time for a product being at least 16 hours is also 0.9332.

The given problem involves a normally distributed process with a mean of 21.4 hours and a standard deviation of 3.6 hours. To find the probability that the time to make a product is at least 16 hours, we need to calculate the area under the normal distribution curve to the right of 16 hours.

To solve this, we can convert the problem into a standardized z-score using the formula:

z = (x - μ) / σ

where x is the value we want to find the probability for (16 hours in this case), μ is the mean (21.4 hours), and σ is the standard deviation (3.6 hours).

Substituting the values, we get:

z = (16 - 21.4) / 3.6 = -1.5

Next, we look up the z-score in the standard normal distribution table or use a calculator to find the corresponding probability.

The probability that a standard normal random variable is greater than -1.5 is approximately 0.9332. Therefore, the probability that the time to make a product is at least 16 hours is 0.9332.

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The given differential equation is:

\(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = x\)

To solve the homogeneous equation:

\(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0\)

We find the auxiliary equation:

\(m^2 - 4m + 4 = (m-2)^2 = 0\)

This equation has the root 2 with a multiplicity of 2. We use the exponential function corresponding to this root:

\(y_2 = ue^{2x}\)

Differentiating with respect to x, we have:

\(y_2' = (u' + 2u)e^{2x}\)

\(y_2'' = (u'' + 4u' + 4u)e^{2x}\)

Substituting \(y_2\), \(y_2'\), and \(y_2''\) into the homogeneous equation:

\(\left[(u'' + 4u' + 4u) - 4(u' + 2u) + 4u\right]e^{2x} = 0\)

Simplifying the equation, we have:

\(u'' = 0\)

Integrating \(u'' = 0\), we obtain \(u = ax + b\)

Integrating once more to find \(u\), we have \(u = \frac{1}{2}x^2 + cx + d\)

The general solution is given by \(y = y_h + y_p = (c_1 + c_2x)e^{2x} + \frac{1}{2}x^2 + cx + d\)

Therefore, the general solution to the given differential equation is:

\(y = (c_1 + c_2x)e^{2x} + \frac{1}{2}x^2 + cx + d\)

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7. P(X = 10) = (15 choose 10) ×(0.8)¹⁰ × (0.2)⁵

10. μ = n × p, σ = √(n × p ×(1 - p)), mean is the average number of successful trials, n is the total number of trials (15 flights), and

p is the probability of success on a single trial (0.8).

6. This is a binomial experiment because it satisfies the following criteria:

- There are a fixed number of trials: In this case, there are 15 flights being selected.

- Each trial can result in one of two outcomes: Either a flight is on time or it is not.

- The probability of success (an on-time flight) remains the same for each trial: 80% of the time.

- The trials are independent: The outcome of one flight being on time does not affect the outcome of another flight being on time.

7. To find the probability that exactly 10 flights are on time, we can use the binomial probability formula. Let's denote X as the number of on-time flights out of the 15 selected flights. The probability of exactly 10 flights being on time can be calculated as:

P(X = 10) = (15 choose 10) × (0.8)¹⁰ × (0.2)⁽¹⁵⁻¹⁰⁾

Using the binomial probability formula, where (n choose k) = n! / (k!× (n-k)!), we can substitute the values:

P(X = 10) = (15 choose 10) ×(0.8)¹⁰ × (0.2)⁵

Calculating this value gives us the probability of exactly 10 flights being on time.

8. To find the probability that fewer than 10 flights are on time, we need to calculate the sum of probabilities for each possible outcome less than 10. We can do this by finding the probabilities for X = 0, 1, 2, ..., 9 and adding them together.

P(X < 10) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 9)

Each of these probabilities can be calculated using the binomial probability formula as mentioned earlier. Once we calculate each individual probability, we can add them together to find the probability of fewer than 10 flights being on time.

9. To find the probability that at least 10 flights are on time, we need to calculate the sum of probabilities for each possible outcome equal to or greater than 10. We can do this by finding the probabilities for X = 10, 11, 12, ..., 15 and adding them together.

P(X >= 10) = P(X = 10) + P(X = 11) + P(X = 12) + ... + P(X = 15)

Again, each of these probabilities can be calculated using the binomial probability formula. Once we calculate each individual probability, we can add them together to find the probability of at least 10 flights being on time.

10. The mean (μ) and standard deviation (σ) of a binomial random variable can be calculated using the following formulas:

μ = n × p

σ = √(n × p ×(1 - p))

Where:

- n is the number of trials (15 flights in this case)

- p is the probability of success (80% or 0.8 in this case)

By substituting the values into the formulas, we can calculate the mean and standard deviation of the binomial random variable for this scenario. The mean represents the average number of on-time flights, while the standard deviation measures the variability or spread of the distribution.

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The solutions to the equation \(2\cos(x) - 1 = 0\) are \(x = \frac{\pi}{3} + 2k\pi\) and \(x = -\frac{\pi}{3} + 2k\pi\), where \(k\) is any integer.

To find the solutions of the equation \(2\cos(x) - 1 = 0\), we can isolate the cosine term and solve for \(x\):

\(2\cos(x) = 1\)

\(\cos(x) = \frac{1}{2}\)

The cosine function has a value of \(\frac{1}{2}\) at specific angles in the unit circle. These angles are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) (in the interval \([0, 2\pi]\)), or more generally, \(A + Bk\pi\), where \(A = \frac{\pi}{3}\), \(B = \frac{2}{3}\), and \(k\) is any integer.

Therefore, we have:

\(x = \frac{\pi}{3} + \frac{2}{3}k\pi\)  (solution 1)

\(x = \frac{5\pi}{3} + \frac{2}{3}k\pi\) (solution 2)

Since \(\cos(x)\) has a periodicity of \(2\pi\), we can also express the solutions as \(C + Dk\pi\), where \(C\) and \(D\) are constants.

Comparing the solutions 1 and 2 with the form \(C + Dk\pi\), we can determine:

\(C = \frac{\pi}{3}\), \(D = \frac{2}{3}\) (for solution 1)

\(C = \frac{5\pi}{3}\), \(D = \frac{2}{3}\) (for solution 2)

Therefore, the solutions of the equation \(2\cos(x) - 1 = 0\) can be represented as:

\(x = \frac{\pi}{3} + \frac{2}{3}k\pi\) (where \(k\) is any integer)

\(x = \frac{5\pi}{3} + \frac{2}{3}k\pi\) (where \(k\) is any integer)

In the answer format provided, we have:

\(A = \frac{\pi}{3}\), \(B = \frac{2}{3}\), \(C = \frac{\pi}{3}\), \(D = \frac{2}{3}\).

Therefore, the solutions to the equation \(2\cos(x) - 1 = 0\) are \(x = \frac{\pi}{3} + 2k\pi\) and \(x = -\frac{\pi}{3} + 2k\pi\), where \(k\) is any integer.

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