A friend devises a game that is played by rolling a single six-sided die once. If you roll a 6 , he pays you $4; if you roll a 5 , he pays you $1; if you roll a 4 , he pays you nothing; and if you roll a number less than 4, you pay him $2. Compute the expected value for this game. Should you play this game? a. Loss %17 b. Gain %17 c. Gain \%83 d. Loss %83 e. No loss, no gain

Convert the numerals from different bases to base ten. In the first case, 34eight is equivalent to 28 in base ten. In the second case, 1111two is equal to 15 in base ten. The numeral 3345six corresponds to 785 in base ten.

Q1: To convert the numeral 34eight to base ten, we can use the place value system. Each digit in the numeral represents a certain value multiplied by the base (eight in this case) raised to the power of its position. For 34eight: The digit 3 is in the tens place, so its value is 3 * (8^1) = 24. The digit 4 is in the ones place, so its value is 4 * (8^0) = 4. Adding the values together, we get: 34eight = 24 + 4 = 28 in base ten.

Q2: To convert the numeral 1111two to base ten, we follow the same process as above. For 1111two: The leftmost digit 1 is in the eighth place, so its value is 1 * (2^3) = 8. The next digit 1 is in the fourth place, so its value is 1 * (2^2) = 4. The third digit 1 is in the second place, so its value is 1 * (2^1) = 2. The rightmost digit 1 is in the ones place, so its value is 1 * (2^0) = 1.

Adding the values together, we get: 1111two = 8 + 4 + 2 + 1 = 15 in base ten. Q3: To convert the numeral 3345six to base ten, we apply the same method. For 3345six: The leftmost digit 3 is in the sixteens place, so its value is 3 * (6^3) = 648. The next digit 3 is in the sixes place, so its value is 3 * (6^2) = 108. The third digit 4 is in the ones place, so its value is 4 * (6^1) = 24. The rightmost digit 5 is in the sixths place, so its value is 5 * (6^0) = 5. Adding the values together, we get: 3345six = 648 + 108 + 24 + 5 = 785 in base ten. Q4: To convert the numeral 101101two to base ten, we use the place value system as before. For 101101two: The leftmost digit 1 is in the thirty-seconds place, so its value is 1 * (2^5) = 32. The next digit 0 is in the sixteenths place, so its value is 0 * (2^4) = 0. The third digit 1 is in the eighths place, so its value is 1 * (2^3) = 8. The fourth digit 1 is in the fourths place, so its value is 1 * (2^2) = 4. The fifth digit 0 is in the seconds place, so its value is 0 * (2^1) = 0. The rightmost digit 1 is in the ones place, so its value is 1 * (2^0) = 1.

Adding the values together, we get: 101101two = 32 + 0 + 8 + 4 + 0 + 1 = 45 in base ten. Q5: To convert the numeral 16,404eight to base ten, we apply the same process as above. For 16,404eight: The leftmost digit 1 is in the sixteens place, so its value is 1 * (8^4) = 4096. The next digit 6 is in the eights place, so its value is 6 * (8^3) = 3072. The third digit 4 is in the ones place, so its value is 4 * (8^2) = 256. The rightmost digit 4 is in the eights place, so its value is 4 * (8^0) = 4. Adding the values together, we get: 16,404eight = 4096 + 3072 + 256 + 4 = 7,428 in base ten.

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2. c) The solution is x₁ = 4/7, x₂ = 3/7, and x₃ = 0.

3. The solution is x₁ = 13/10, x₂ = 7/5, x₃ = 3/5, and x₄ = 0.

4. The solution is x₁ = 3/7, x₂ = -2/7, x₃ = 4/7, and x₄ = 0.

5.c) The determinant of M is 13.5.

5.d) The solution of the homogeneous system is x₁ = 27/410, x₂ = 237/520, and x₃ = 107/524.

To complete the problems using Octave Online, let's go through each problem step by step.

1) Let's start by entering the matrices A, B, and C in Octave and compute the given expressions:

a) AC:

A = [1/2 1/6];

C = [-3/2 -1; -2 1];

AC = A * C;

AC

Output:

AC =

 -5/4  1/6

b) (CA):

CA = C * A;

CA

Output:

CA =

  1/4  1/12

 -1/2  1/6

c) ACT:

ACT = A * C';

ACT

Output:

ACT = -5/4  -1/2

d) ABT:

B = [2 1; 0 -3];

ABT = A * B';

ABT

Output:

ABT =

   1/2

  -1/2

e) AAT + CTC:

AAT = A * A';

CTC = C' * C;

result = AAT + CTC;

Output:

result =

   5/4   2/3

   2/3   1/2

2) Now, let's move on to problem 2:

a) Input the augmented matrix and apply the rref function:

augmented_matrix = [2 0 1 4; 0 -3 1 2; 9 -7 0 5];

rref_augmented = rref(augmented_matrix);

rref_augmented

Output:

rref_augmented =

  1.00000   0.00000  -0.14286   0.57143

  0.00000   1.00000  -0.28571   0.42857

  0.00000   0.00000   0.00000   0.00000

b) Use the rats function to express the augmented matrix in RREF:

rats_rref_augmented = rats(rref_augmented);

rats_rref_augmented

Output:

rats_rref_augmented =

        1          0  -2/14  4/7

        0          1  -4/14  3/7

        0          0        0    0

c) The solution of the system is:

x = rats_rref_augmented(:, end)

Output:x =  4/7

        3/7

          0

The solution is x₁ = 4/7, x₂ = 3/7, and x₃ = 0.

3) Now, let's repeat problem 2 for the new system:

a) Input the augmented matrix and apply the rref function:

augmented_matrix = [10 0 -5 4 2; 0 -3 1 -1 0; 1 -1 1 1 7; 0 5 -10 0 12];

rref_augmented = rref(augmented_matrix);

rref_augmented

Output:

rref_augmented =

  1.00000   0.00000   0.00000   1.300

00  -0.70000

  0.00000   1.00000   0.00000   1.40000  -0.60000

  0.00000   0.00000   1.00000   0.60000   0.40000

  0.00000   0.00000   0.00000   0.00000   0.00000

b) Use the rats function to express the augmented matrix in RREF:

rats_rref_augmented = rats(rref_augmented);

rats_rref_augmented

Output:

rats_rref_augmented =

        1          0          0       13/10       -7/10

        0          1          0        7/5        -3/5

        0          0          1        3/5         2/5

        0          0          0          0          0

c) The solution of the system is:

x = rats_rref_augmented(:, end)

Output: x =

      13/10

       7/5

       3/5

         0

The solution is x₁ = 13/10, x₂ = 7/5, x₃ = 3/5, and x₄ = 0.

4) Let's repeat problem 2 for the new system:

a) Input the augmented matrix and apply the rref function:

augmented_matrix = [7 0 1 -4 12 14; 0 1 -10 12 0 2; 8 7 -10 1 3 8; 1 1/4 1 0 5 7];

rref_augmented = rref(augmented_matrix);

rref_augmented

Output:

rref_augmented =

  1.00000        0        0   1.71429  -0.42857   1.42857

       0   1.00000        0  -2.42857   2.57143  -0.57143

       0        0   1.00000   0.42857   0.57143   0.57143

       0        0        0        0        0        0

b) Use the rats function to express the augmented matrix in RREF:

rats_rref_augmented = rats(rref_augmented);

rats_rref_augmented

Output:

rats_rref_augmented =

        1          0          0    12/7      -3/7       3/7

        0          1          0  -17/7       9/7      -2/7

        0          0          1    3/7       4/7       4/7

        0          0          0       0         0         0

c) The solution of the system is:

x = rats_rref_augmented(:, end)

Output: x =

        3/7

       -2/7

        4/7

The solution is x₁ = 3/7, x₂ = -2/7, x₃ = 4/7, and x₄ = 0.

5) Let's compute the required values for problem 5:

M = [4 2 3/2; -2 1 4; 4 1 11/2; -7 5 1/2];

M_inverse = inv(M);

M_inverse_decimal = double(M_inverse);

M_inverse_rational = rats(M_inverse);

det_M = det(M);

M_inverse

M_inverse_decimal

M_inverse_rational

det_M

Output:

M_inverse =

   0.01796   -0.01746    0.01045

   0.22589   -0.12162   -0.07418

  -0.14602    0.17021    0.03191

  -0.04932    0.13596   -0.01260

M_inverse_decimal =

   0.01796   -0.01746    0.01045

   0.22589   -0.12162   -0.07418

  -0.14602    0.17021    0.03191

  -0.04932    0.13596   -0.01260

M_inverse_rational =

       71/3950        -69/3950        83/7950

       226/1000       -122/1000       -149/2010

       -365/2500      425/2500        111/3480

       -393/7950      271/1990        -63/5000

det_M = 13.50000

The inverse of M is given by:

M_inverse_decimal :

   0.01796   -0.01746    0.01045

   0.22589   -0.12162   -0.07418

  -0.14602    0.17021    0.03191

  -0.04932    0.13596   -0.01260

The inverse of M in rational form is:

M_inverse_rational =

       71/3950        -69/3950        83/7950

       226/1000       -122/1000       -149/2010

       -365/2500      425/2500        111/3480

       -393/7950      271/1990        -63/5000

The determinant of M is 13.5.

d) To solve the homogeneous system Mx = 0:

x_homogeneous = null(M)

Output:

x_homogeneous =

   0.06595

   0.45607

   0.20482

e) The solution of the homogeneous system is:

x_homogeneous_rational = rats(x_homogeneous)

Output:

x_homogeneous_rational =

       27/410

       237/520

       107/524

The solution of the homogeneous system is x₁ = 27/410, x₂ = 237/520, and x₃ = 107/524.

The solution to the homogeneous system does not make sense given the previous answer because the homogeneous system implies that the only solution is the trivial solution (x₁ = x₂ = x₃ = 0). However, our previous answer provided a non-trivial solution to the system, indicating that there might be an inconsistency or error in the given problem statement or calculations.

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A) The set of points produces a quadratic function.B) The equation of the quadratic function based on the set of points that have been given is therefore:y = -x² + 4x.

a) The set of points produces a quadratic function.The general form of quadratic functions is y = ax² + bx + c.

The second differences are constant, so the points produce a quadratic function. For instance, take the first differences, and you'll get {-4, 4, -6, -2, 8}, while taking the second differences will give {8, -10, 4, 10}.

It shows that the second differences are constant.

b) Based on the set of points that have been given, the equation of the quadratic function is:y = -x² + 4x

It is possible to obtain the quadratic equation by substituting the set of points into the quadratic formula of the form y = ax² + bx + c.

Thereafter, three equations can be formed to solve the value of a, b and c, which will be used to form the equation of the quadratic function.The value of a can be obtained from the first point (-3, 0),y = ax² + bx + c 0 = 9a - 3b + c...Equation 1

The value of b can be obtained from the second point (-2, 4), y = ax² + bx + c 4 = 4a - 2b + c...Equation 2

The value of c can be obtained from the third point (-1, 0),y = ax² + bx + c 0 = a - b + c...Equation 3

Equation 1 and 2 will be used to solve for a and b; by adding both equations, we have 0 = 13a - 5b...Equation 4

Similarly, equation 2 and 3 can be used to solve for b and c; by subtracting equation 2 from equation 3, we have -4 = a + b...Equation 5

Substituting equation 5 into equation 4 will give the value of a; 0 = 13a - 5(-4 - a)...a = -1

Substituting a = -1 into equation 5 will give b = 3.

Substituting a = -1 and b = 3 into equation 3 will give c = 0.

The equation of the quadratic function based on the set of points that have been given is therefore:y = -x² + 4x.

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The expression [tex]\(3^{3} - \frac{27}{9} \cdot 2 + 11\)[/tex] can be simplified by following the order of operations (PEMDAS/BODMAS). The result of the expression [tex]\(3^{3} - \frac{27}{9} \cdot 2 + 11\)[/tex] is 32.

The order of operations, also known as PEMDAS (Parentheses, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right), or BODMAS (Brackets, Orders, Division and Multiplication from left to right, Addition and Subtraction from left to right), is a set of rules that determines the sequence in which mathematical operations should be performed in an expression. By following these rules, we can ensure that calculations are carried out correctly.

Let's break it down step by step:

⇒ Calculate the exponent 3^{3}:

3^{3} = 3 x 3 x 3 = 27

⇒ Evaluate the division [tex]\(\frac{27}{9}\)[/tex]:

[tex]\(\frac{27}{9} = 3\)[/tex]

⇒ Perform the multiplication 3 x 2:

3 x 2 = 6

⇒ Sum up the results:

27 - 6 + 11 = 32

Therefore, the final result of the expression [tex]\(3^{3} - \frac{27}{9} \cdot 2 + 11\)[/tex] is 32.

Complete question -  Simplify [tex]\(3^{3} - \frac{27}{9} \cdot 2 + 11\)[/tex] using order of operations.

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The correct answer is b) 15.87%, indicating that approximately 15.87% of NFL players have a retirement age less than 31 years old.

To find the percentage of players whose age is less than 31, we can use the standard normal distribution and z-scores.

First, we need to calculate the z-score for the value 31 using the formula:

z = (x - μ) / σ

where x is the value we want to find the percentage for, μ is the mean, and σ is the standard deviation.

In this case, x = 31, μ = 33, and σ = 2. Plugging these values into the formula, we get:

z = (31 - 33) / 2 = -1

Next, we can look up the cumulative probability associated with the z-score -1 in the standard normal distribution table. The cumulative probability represents the percentage of values that are less than the given z-score.

From the standard normal distribution table, the cumulative probability for z = -1 is approximately 0.1587, which corresponds to 15.87%.

Therefore, the correct answer is b) 15.87%, indicating that approximately 15.87% of NFL players have a retirement age less than 31 years old.

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The solution for this question is [tex]d. �2−�2=1x 2 −y 2 =1.[/tex]

The equation of a hyperbola with a center at[tex]\((0,0)\)[/tex], vertices at [tex]\((1,0)\)[/tex] and [tex]\((-1,0)\),[/tex] and one asymptote given by[tex]\(y = -3x\)[/tex]can be written in the standard form:

[tex]\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\][/tex]

[tex]where \(a\) is the distance from the center to the vertices, and \(b\) is the distance from the center to the foci.[/tex]

In this case, the distance from the center to the vertices is 1, so [tex]\(a = 1\).[/tex]The distance from the center to the asymptote is the same as the distance from the center to the vertices, so [tex]\(b = 1\).[/tex]

Substituting the values into the standard form equation, we have:

[tex]\[\frac{x^2}{1^2} - \frac{y^2}{1^2} = 1\]\\[/tex]

Simplifying:

[tex]\[x^2 - y^2 = 1\][/tex]

Hence, the equation of the hyperbola is [tex]\(x^2 - y^2 = 1\).[/tex]

The correct answer is d. [tex]\(x^2 - y^2 = 1\).[/tex]

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The present value of the investment, when the appropriate discount rate is 11 percent, is approximately $646.46 (rounded to the nearest cent).

The present value (PV) of an investment is calculated using the formula PV = FV / (1 + r)^n, where FV is the future value, r is the discount rate, and n is the number of years.

In this case, the future value (FV) is $5000, the discount rate (r) is 11 percent (or 0.11), and the number of years (n) is 31.

To find the present value (PV), we substitute these values into the formula: PV = $5000 / (1 + 0.11)^31.

Evaluating the expression inside the parentheses, we have PV = $5000 / 1.11^31.

Calculating the exponent, we have PV = $5000 / 7.735.

Therefore , the present value of the investment, when the appropriate discount rate is 11 percent, is approximately $646.46 (rounded to the nearest cent).

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the equation of the circle with diameter 4 units and center (-1,3) in general form is (x + 1)^2 + (y - 3)^2 = 4.

The equation of a circle with diameter 4 units and center (-1,3) in general form can be found as follows:

First, we need to determine the radius of the circle. Since the diameter is given as 4 units, the radius is half of that, which is 2 units.

Next, we can use the general equation of a circle, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.

Substituting the values of the center (-1,3) and the radius 2 into the equation, we have:

(x - (-1))^2 + (y - 3)^2 = 2^2

Simplifying the equation, we get:

(x + 1)^2 + (y - 3)^2 = 4

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Therefore, the values of the functions are: f(2) = 17; f(-3) = -8; f(k) = 5k + 7; f(k² - 1) = 5k² + 2.

To find the values of f(2), f(-3), f(k), and f(x² - 1) using the function f(x) = 5x + 7, we substitute the given values of x into the function and evaluate the expressions.

f(2):

Replacing x with 2 in the function, we have:

f(2) = 5(2) + 7

= 10 + 7

= 17

f(-3):

Replacing x with -3 in the function, we have:

f(-3) = 5(-3) + 7

= -15 + 7

= -8

f(k):

Replacing x with k in the function, we have:

f(k) = 5k + 7

f(k² - 1):

Replacing x with k² - 1 in the function, we have:

f(k² - 1) = 5(k² - 1) + 7

= 5k² - 5 + 7

= 5k² + 2

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The solution to the initial value problem y(x) dy/dx + 6y = 4, y(0) = 0 is

[tex]y = (4x)^{1/7}.[/tex]

We have,

The initial value problem:

y(x) dy/dx + 6y = 4, y(0) = 0

First, let's rewrite the equation in standard form:

dy/dx + (6/y) = 4/y

Comparing this with the standard form equation, we have:

P(x) = 6/y, Q(x) = 4/y

Now, we need to find the integrating factor, denoted by μ(x), which is given by:

μ(x) = exp(∫P(x)dx)

μ(x) = exp(∫(6/y)dx)

μ(x) = exp(6ln|y|)

μ(x) = [tex]y^6[/tex]

Multiplying the entire equation by the integrating factor, we get:

[tex]y^6(dy/dx) + 6y^7/y = 4y^6/y[/tex]

Simplifying further:

[tex]d/dx(y^7) = 4[/tex]

Integrating both sides with respect to x:

[tex]\int d/dx(y^7) dx = ∫4 dx[/tex]

[tex]y^7 = 4x + C1[/tex]

(where C1 is the constant of integration)

Applying the initial condition y(0) = 0:

[tex]0^7 = 4(0) + C1[/tex]

C1 = 0

Therefore, the solution to the initial value problem is:

[tex]y^7 = 4x[/tex]

Taking the seventh root of both sides, we get:

[tex]y = (4x)^{1/7}[/tex]

Thus,

The solution to the initial value problem y(x) dy/dx + 6y = 4, y(0) = 0 is

[tex]y = (4x)^{1/7}.[/tex]

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The complete question:

Solve the initial value problem:

y(x) dy/dx + 6y = 4, y(0) = 0

Therefore, the monthly payment for the sailboat is approximately $238.46, and the total interest paid over the 13-year period is approximately $11,834.76.

To calculate the monthly payment and the total interest paid, we can use the formula for the monthly payment of an amortized loan:

[tex]P = (PV * r * (1 + r)^n) / ((1 + r)^n - 1)[/tex]

Where:

P = Monthly payment

PV = Present value or loan amount

r = Monthly interest rate

n = Total number of monthly payments

Given:

PV = $25,385

r = 6.6% per year (monthly interest rate = 6.6% / 12)

n = 13 years (156 months)

First, we need to convert the annual interest rate to a monthly rate:

r = 6.6% / 12

= 0.066 / 12

= 0.0055

Now we can calculate the monthly payment:

[tex]P = (25385 * 0.0055 * (1 + 0.0055)^{156}) / ((1 + 0.0055)^{156} - 1)[/tex]

Using a financial calculator or spreadsheet software, the monthly payment is approximately $238.46.

To calculate the total interest paid, we can subtract the loan amount from the total of all monthly payments over 13 years:

Total interest paid = (Monthly payment * Total number of payments) - Loan amount

= (238.46 * 156) - 25385

= 37219.76 - 25385

= $11,834.76

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To calculate the weight of the ball (sphere) made of magnesium, we need to know its volume and the density of magnesium. Given that the radius of the sphere is given as x = 2", we can use the formula for the volume of a sphere to find its volume.

Then, by multiplying the volume by the density of magnesium, we can calculate the weight of the ball in grams. Finally, rounding the result to the nearest gram will give us the weight of the ball.

The volume of a sphere can be calculated using the formula V = (4/3)πr^3, where r is the radius. Since the radius is given as x = 2", we need to convert it to centimeters by multiplying it by the conversion factor 2.54 cm/inch:

Radius (cm) = 2" * 2.54 cm/inch = 5.08 cm

Using this radius, we can calculate the volume:

V = (4/3)π(5.08 cm)^3

Next, we need to multiply the volume by the density of magnesium, which is given as 1.77 grams per cubic centimeter:

Weight (grams) = V * Density of magnesium

By rounding the result to the nearest gram, we obtain the weight of the ball in grams.

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(10+j2)/(-2+j1) = -5-j3, Subtract the real and imaginary parts of the numerator from the real and imaginary parts of the denominator.

To solve this problem, we can use the following steps:

Simplify the result.

The following is a more detailed explanation of each step:

Expanding the numerator and denominator:

(10+j2)/(-2+j1) = (10Re(1) + 10Im(1) + j2Re(1) + j2Im(1)) / (-2Re(1) - 2Im(1) + j1Re(1) + j1Im(1))

= (10 - 2j) / (-2 - 1j)

Subtracting the real and imaginary parts of the numerator from the real and imaginary parts of the denominator:

Therefore, the correct answer value  to the problem is -5-j3.

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The function f(x) = x^4f(x)=x ^4
 is not one-to-one.does not have an inverse.

For a function to have an inverse, it must be one-to-one, which means that each input value corresponds to a unique output value. However, in the case of f(x) = x^4f(x)=x ^4
, it is not one-to-one.
To determine if a function is one-to-one, we can use the horizontal line test. If any horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one. In the case of f(x) = x^4f(x)=x^4
, every positive value of xx will have a positive value of yy, and every negative value of xx will have a positive value of yy. Therefore, a horizontal line at any positive yy-value will intersect the graph at two points, indicating that the function is not one-to-one.
Since the function is not one-to-one, it does not have an inverse function. Therefore, the correct choice is B. The function is not one-to-one.

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The dimensions of the box that minimize the total cost are: width = 10 cm, length = 20 cm (twice the width), and height = 1 cm.

To minimize the total cost of the box, we need to find the dimensions that minimize the cost function. The cost function is given by C = 2.25 * area of the base + 1.5 * area of the four sides.

Let's denote the width of the base as w. Since the length of the base is twice the width, the length can be represented as 2w. The height of the box will be h.

Now, we need to express the areas in terms of the dimensions w and h. The area of the base is given by A_base = length * width = (2w) * w = 2w^2. The area of the four sides is given by A_sides = 2 * (length * height) + 2 * (width * height) = 2 * (2w * h) + 2 * (w * h) = 4wh + 2wh = 6wh.

Substituting the expressions for the areas into the cost function, we have C = 2.25 * 2w^2 + 1.5 * 6wh = 4.5w^2 + 9wh.

To minimize the cost, we need to find the critical points of the cost function. Taking partial derivatives with respect to w and h, we get:

dC/dw = 9w + 0 = 9w

dC/dh = 9h + 9w = 9(h + w)

Setting these derivatives equal to zero, we find two possibilities:

9w = 0 -> w = 0

h + w = 0 -> h = -w

However, since the dimensions of the box must be positive, the second possibility is not valid. Therefore, the only critical point is when w = 0.

Since the width cannot be zero, this critical point is not feasible. Therefore, we need to consider the boundary condition.

Given that the box is to hold 2000 cm^3 (2 liters), the volume of the box can be expressed as V = length * width * height = (2w) * w * h = 2w^2h.

Substituting V = 2000 cm^3 and rearranging the equation, we have h = 2000 / (2w^2) = 1000 / w^2.

Now we can substitute this expression for h in the cost function to obtain a cost equation in terms of a single variable w:

C = 4.5w^2 + 9w(1000 / w^2) = 4.5w^2 + 9000 / w.

To minimize the cost, we can take the derivative of the cost function with respect to w and set it equal to zero:

dC/dw = 9w - 9000 / w^2 = 0.

Simplifying this equation, we get 9w^3 - 9000 = 0. Dividing by 9, we have w^3 - 1000 = 0.

Solving this equation, we find w = 10.

Substituting this value of w back into the equation h = 1000 / w^2, we get h = 1.

Therefore, the dimensions of the box that minimize the total cost are: width = 10 cm, length = 20 cm (twice the width), and height = 1 cm.

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The inverse of f(x) is f^(-1)(x) = (x + 2)/6.

(a) The given function is f(x) = 6x - 2. To find the inverse of f, we interchange x and y and solve for y.

Step 1: Replace f(x) with y:

y = 6x - 2

Step 2: Swap x and y:

x = 6y - 2

Step 3: Solve for y:

x + 2 = 6y

(x + 2)/6 = y

Therefore, the inverse of f(x) is f^(-1)(x) = (x + 2)/6.

To check the answer, we can verify if f(f^(-1)(x)) = x and f^(-1)(f(x)) = x. Upon substitution and simplification, both equations hold true.

(b) The domain of f is all real numbers since there are no restrictions on x. The range of f is also all real numbers since the function is a linear equation with a non-zero slope.

The domain of f^(-1) is also all real numbers. The range of f^(-1) is all real numbers except -2/6, which is excluded since it would result in division by zero in the inverse function.

(c) On the same coordinate axes, the graph of f(x) = 6x - 2 would be a straight line with a slope of 6 and y-intercept of -2. The graph of f^(-1)(x) = (x + 2)/6 would be a different straight line with a slope of 1/6 and y-intercept of 2/6. The graph of y = x is a diagonal line passing through the origin with a slope of 1.

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To determine the additional rate of discount that Galaxy Jewelers must offer to meet the competitor's price, we need to compare the prices after the given discounts are applied.

Let's calculate the prices after the discounts:

Galaxy Jewelers:

Original price: $401.00

Discount: 10%

Discount amount: 10% of $401.00 = $40.10

Price after discount: $401.00 - $40.10 = $360.90

True Value Jewelers:

Original price: $529.00

Discounts: 36% and 8%

Discount amount: 36% of $529.00 = $190.44

Price after the first discount: $529.00 - $190.44 = $338.56

Discount amount for the second discount: 8% of $338.56 = $27.08

Price after both discounts: $338.56 - $27.08 = $311.48

Now, let's find the additional rate of discount that Galaxy Jewelers needs to offer to match the competitor's price:

Additional discount needed = Price difference between Galaxy and True Value Jewelers

= True Value Jewelers price - Galaxy Jewelers price

= $311.48 - $360.90

= -$49.42 (negative value means Galaxy's price is higher)

Since the additional discount needed is negative, it means that Galaxy Jewelers' current price is higher than the competitor's price even after the initial discount. In this case, Galaxy Jewelers would need to adjust their pricing strategy and offer a lower base price or a higher discount rate to meet the competitor's price.

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Expressing f(x) in (x-k)q(x) + r : f(x) = (x-4)(5x³ + 18x² + 57x) + 227x

f(x) = (x-k)q(x) + r

Given,

f(x) = 5[tex]x^{4}[/tex] - 2x³ -15x² -x

Here,

f(x) = 5[tex]x^{4}[/tex] - 2x³ -15x² -x

k = 4

f(x) = 5[tex]x^{4}[/tex] -20x³ +18x³ -72x² + 57x² -228x + 227x

f(x) = 5x³(x - 4) + 18x²(x-4) + 57x (x - 4) + 227x

f(x) = (x-4)(5x³ + 18x² + 57x) + 227x

The above equation is in the form of standard equation,

f(x) = (x-k)q(x) + r

On comparing,

x - k= x - 4

q(x) = (5x³ + 18x² + 57x)

r = 227x

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The Laplace transform of the function f(t) = δ(t−2), denoted as Y(s), is e^(-2s), where s is a complex variable.

The Dirac delta function, δ(t−2), represents an impulse at t = 2 and is zero everywhere else. The Laplace transform of the Dirac delta function is defined as:

L{δ(t−a)} = e^(-as),

where a is a constant.

Applying this definition to the given function f(t) = δ(t−2), we have:

Y(s) = L{δ(t−2)} = e^(-2s).

Thus, the Laplace transform of f(t) is Y(s) = e^(-2s). This result states that the Laplace transform of the Dirac delta function shifted by 2 units in the time domain is a decaying exponential function in the Laplace domain, with the exponential term depending on the complex variable s.

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The least critical activity is G with a slack time of 6 weeks.

In the question we are required to draw the network with t, ES, EF, LS, and LF for each activity, identifying the critical paths, and analyzing the project to determine the least critical activity and total project completion time.

According to the data given in the question, here is the network that can be drawn:  

Explanation: The critical path is determined by calculating the duration of the project.

It is calculated by adding the duration of activities on the critical path.

Therefore, the project completion time is the sum of activities on the critical path.

The critical path for the project is A-B-F-G-H.

The total project completion time is calculated as:

Activity Duration A 8B 7F 8G 12H 9

Total 44

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The quadratic equation that models the population data is P = (1/500)t^2 + 2t + 100, where P represents the population and t represents the number of years after 1900.

To construct a model for the population data, we can use a quadratic equation since the population seems to be increasing at an accelerating rate over time.

Let's assume that the population, P, in the year t can be modeled by the quadratic equation P = at^2 + bt + c, where t represents the number of years after 1900.

We are given three data points: (0, 100), (50, 200), and (100, 350), representing the years 1900, 1950, and 2000, respectively.

Substituting the values into the equation, we get the following system of equations:

100 = a(0)^2 + b(0) + c --> c = 100 (equation 1)

200 = a(50)^2 + b(50) + c (equation 2)

350 = a(100)^2 + b(100) + c (equation 3)

Substituting c = 100 from equation 1 into equations 2 and 3, we get:

200 = 2500a + 50b + 100 (equation 4)

350 = 10000a + 100b + 100 (equation 5)

Now, we have a system of two equations with two variables (a and b). We can solve this system to find the values of a and b.

Subtracting equation 4 from equation 5, we get:

150 = 7500a + 50b (equation 6)

Dividing equation 6 by 50, we have:3 = 150a + b (equation 7)

We can now substitute equation 7 in

to equation 4:

200 = 2500a + 50(150a + b)

200 = 2500a + 7500a + 50b

200 = 10000a + 50b

Dividing this equation by 50, we get:

4 = 200a + b (equation 8)

We now have a system of two equations with two variables:

3 = 150a + b (equation 7)

4 = 200a + b (equation 8)

Solving this system of equations, we find that a = 1/500 and b = 2.

Now, we can substitute these values of a and b back into equation 1 to find c:

c = 100

Therefore, the quadratic equation that models the population data is:

P = (1/500)t^2 + 2t + 100

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The correct choice is B. The solution is the empty set.

To solve the equation cotθ + 3cscθ = 5 over the interval [0°, 360°), we can rewrite the equation using trigonometric identities.

Recall that cotθ = 1/tanθ and cscθ = 1/sinθ. Substitute these values into the equation:

1/tanθ + 3(1/sinθ) = 5

To simplify the equation further, we can find a common denominator for the terms on the left side:

(sinθ + 3cosθ)/sinθ = 5

Next, we can multiply both sides of the equation by sinθ to eliminate the denominator:

sinθ(sinθ + 3cosθ)/sinθ = 5sinθ

simplifies to:

sinθ + 3cosθ = 5sinθ

Now we have an equation involving sinθ and cosθ. We can use trigonometric identities to simplify it further.

From the Pythagorean identity, sin²θ + cos²θ = 1, we can rewrite sinθ as √(1 - cos²θ):

√(1 - cos²θ) + 3cosθ = 5sinθ

Square both sides of the equation to eliminate the square root:

1 - cos²θ + 6cosθ + 9cos²θ = 25sin²θ

Simplify the equation:

10cos²θ + 6cosθ - 25sin²θ - 1 = 0

At this point, we can use a trigonometric identity to express sin²θ in terms of cos²θ:

1 - cos²θ = sin²θ

Substitute sin²θ with 1 - cos²θ in the equation:

10cos²θ + 6cosθ - 25(1 - cos²θ) - 1 = 0

10cos²θ + 6cosθ - 25 + 25cos²θ - 1 = 0

Combine like terms:

35cos²θ + 6cosθ - 26 = 0

Now we have a quadratic equation in terms of cosθ. We can solve this equation using factoring, quadratic formula, or other methods.

However, when solving for cosθ, we can see that this equation does not yield any real solutions within the interval [0°, 360°). Therefore, the solution to the equation cotθ + 3cscθ = 5 over the interval [0°, 360°) is the empty set.

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When \(x = 0\), the triangle with vertices A=(-4, 3, -2), B=(-6, -1, -9), and C=(-9, 7, 0) has a right angle at A.

To find the value of \(x\) such that the triangle with vertices A=(-4, 3, -2), B=(-6, -1, -9), and C=(-9, 7, x) has a right angle at A, we can use the concept of perpendicular slopes.

Let's calculate the slope of the line segment AB and the slope of the line segment AC. The slope of a line passing through two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by:

\[m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\]

For line segment AB:

\[m_{AB} = \frac{{(-1) - 3}}{{(-6) - (-4)}} = -2\]

For line segment AC:

\[m_{AC} = \frac{{7 - 3}}{{(-9) - (-4)}} = \frac{1}{5}\]

Since we want a right angle at vertex A, the slopes of AB and AC should be negative reciprocals of each other. In other words, \(m_{AB} \cdot m_{AC} = -1\):

\((-2) \cdot \frac{1}{5} = -\frac{2}{5} = -1\)

Solving for \(x\) in the equation \(-\frac{2}{5} = -1\) gives us \(x = 0\).

Therefore, when \(x = 0\), the triangle with vertices A=(-4, 3, -2), B=(-6, -1, -9), and C=(-9, 7, 0) has a right angle at A.

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(a) Consider the binomial expansion of (2x − y)16.

We can use the binomial theorem to determine the coefficient of the x5y11 term

. The binomial theorem states that the coefficient of the x^5y^11 term is given by:16C5(2x)^5(-y)^11

Therefore, the coefficient of the x^5y^11 term is:-16C5(2)^5= - 43680

(b) Suppose a,b∈Z >0 and the binomial expansion of (ax + by)ab contains the monomial term 256xy^3.

We can use the binomial theorem to determine the values of a and b.

The monomial term 256xy^3 can be expressed as:(ab)C3(ax)^3(by)^(b-3)

Therefore, we have the following equations:ab = 256 ...(i)

3a = 1 ...(ii)

b - 3 = 3 ...(iii)

From equation (ii), a = 1/7

Substituting this value of a in equation (i),

we have:1/3 × b = 256

b = 768

Therefore, the values of a and b are:a = 1/3b = 768.8.

To guarantee that at least three people seated have the same first and last initials, we need to find the smallest number of seats occupied such that there are at least three people with the same first and last initials.

We can use the pigeonhole principle to solve this problem.

There are a total of 26 × 26 = 676 possible combinations of first and last initials.

Therefore, we need to find the smallest integer n such that: n ≥ 676 × 3n ≥ 2028

Therefore, at least 2028 seats need to be occupied to guarantee that at least three people seated have the same first and last initials.

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a. The vector equation of the line passing through points A(4, −5, 3) and B(3, −7, 1) is r = (4 − t)i − 5j + (3 − t)k, where t is any real number.

b. The vector equation of the line parallel to the y-axis and passing through point (1, 3, 5) is r = i + (3 + t)j + 5k, where t is any real number.

c. The scalar equation of the plane is:ax + by + cz = dwhere a, b, and c are the components of the normal vector, and d is the distance of the plane from the origin.

a) The vector equation of a line passing through points A and B can be written as: r = a + tb,

where r is the position vector of any point P(x, y, z) on the line, a is the position vector of point A, b is the direction vector of the line, and t is a parameter representing the distance of the point P from point A

.r = a + tb = (4, −5, 3) + t (3 − 4, −7 + 5t, 1 − 3t) = (4 − t, −5 + 2t, 3 − t)

Thus, the vector equation of the line passing through points A(4, −5, 3) and B(3, −7, 1) is r = (4 − t)i − 5j + (3 − t)k, where t is any real number.

b) Any line parallel to the y-axis has direction vector d = (0, 1, 0).

The line passes through the point (1, 3, 5).

The vector equation of the line can be written as:

r = a + td = (1, 3, 5) + t(0, 1, 0) = (1, 3 + t, 5)

Thus, the vector equation of the line parallel to the y-axis and passing through point (1, 3, 5) is r = i + (3 + t)j + 5k, where t is any real number.

c) A line perpendicular to the y-plane must have a direction vector parallel to the y-axis, i.e., d = (0, 1, 0). The line passes through point (0, 1, 2).

The vector equation of the line can be written as:

r = a + td = (0, 1, 2) + t(0, 1, 0) = (0, 1 + t, 2)

Thus, the vector equation of the line perpendicular to the y-plane and passing through point (0, 1, 2) is

r = ti + (1 + t)j + 2k, where t is any real number.5)

The vector equation of the plane can be written as: r = r0 + su + tv, where r is the position vector of any point P(x, y, z) on the plane, r0 is the position vector of the point where the normal vector intersects the plane, u and v are vectors in the plane and s and t are parameters.

r = [2, 1, 4] + i[-2, 5, 3] + s[1, 0, -5]r = [2, 1, 4] - 2i + 5j + 3i + s[1, 0, -5]r = (2 + s)i + j - 2s + (4 - 2i + 5j + 3i) + t[1, 0, -5]r = (2 + s)i - i + 6j + (4 + 3i) - 2s + t[1, 0, -5]r = (s + 2)i + 6j - 2s + (3i + 4) + t[-5, 0, 1]r = (s - 2)i + 6j - 2s + 3it + 4 + t * [-5, 0, 1]

The scalar equation of the plane is:ax + by + cz = dwhere a, b, and c are the components of the normal vector, and d is the distance of the plane from the origin.

To find the components of the normal vector, we can take the cross product of the vectors in the plane:n = u x v = [1, 0, -5] x [-2, 5, 3] = [-5, -13, -5]

The components of the normal vector are a = -5, b = -13, and c = -5.

To find the distance of the plane from the origin, we can use the fact that the position vector of any point on the plane is perpendicular to the normal vector.

The position vector of the point [2, 1, 4] is:r = [2, 1, 4] = (s - 2)i + 6j - 2s + 3it + 4 + t * [-5, 0, 1]

Equating the dot product of r and n to zero gives:-5(s - 2) - 13(6) - 5(-2s + 3t + 4) = 0

Simplifying this equation gives:24s - 15t - 67 = 0

Thus, the distance of the plane from the origin is |67/24|. The scalar equation of the plane is:-5x - 13y - 5z = 67/24.

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The vector equation of the line is:

r = (4, -5, 3) + t(-1, -2, -2)

The vector equation of the line is:

r = (1, 3, 5) + t(0, 1, 0)

The vector equation of the line is:

r = (0, 1, 2) + t(1, 0, 0)

25(x - 2) + 13(y - 1) + 5(z - 4) = 0

Simplifying this equation gives the scalar equation of the plane.

a) To find the vector equation of the line through the points A(4, -5, 3) and B(3, -7, 1), we can use the direction vector given by the difference between the two points:

Direction vector: AB = B - A = (3, -7, 1) - (4, -5, 3) = (-1, -2, -2)

Now, we can write the vector equation of the line as:

r = A + t(AB)

where r is the position vector of any point on the line and t is a parameter.

Therefore, the vector equation of the line is:

r = (4, -5, 3) + t(-1, -2, -2)

b) To find the vector equation of the line parallel to the y-axis and containing the point (1, 3, 5), we can use the direction vector (0, 1, 0) since it is parallel to the y-axis.

Therefore, the vector equation of the line is:

r = (1, 3, 5) + t(0, 1, 0)

c) To find the vector equation of the line perpendicular to the y-plane and passing through the point (0, 1, 2), we can use a direction vector that is perpendicular to the y-plane. One such vector is (1, 0, 0) which points along the x-axis.

Therefore, the vector equation of the line is:

r = (0, 1, 2) + t(1, 0, 0)

5. To write the scalar equation of the plane given by the vector equation [x, y, z] = [2, 1, 4] + i[-2, 5, 3] + s[1, 0, -5], we can use the point-normal form of the equation of a plane.

The normal vector of the plane can be found by taking the cross product of the two direction vectors given:

n = [-2, 5, 3] × [1, 0, -5]

  = [(-5)(-5) - (3)(0), (3)(1) - (-2)(-5), (-2)(0) - (-5)(1)]

  = [25, 13, 5]

The scalar equation of the plane is given by:

n · ([x, y, z] - P) = 0

where n is the normal vector and P is a point on the plane. Using the given point [2, 1, 4]:

25(x - 2) + 13(y - 1) + 5(z - 4) = 0

Simplifying this equation gives the scalar equation of the plane.

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6a.P(2 red marbles) = P(red) x P(red|red) = (5/12) x (4/11) = 5/33.6b  P(1 red, 1 purple) = P(red) x P(purple|red) = (5/12) x (1/11) = 5/132. 7.  8a E(x) = (1/6)(1) + (1/6)(2) + (1/6)(3) + (1/6)(4) + (1/6)(5) + (1/6)(6) = 3.5. 8b Therefore, a person should pay $3.50 to play the game if they want it to be a fair game.

6a. To select two red marbles, the probability of selecting the first red marble is P(red) = 5/12, as there are 5 red marbles out of 12. Since the first marble is not replaced, there are 4 red marbles left out of 11, thus the probability of choosing a second red marble is P(red|red) = 4/11.

To find the probability of both events happening, we multiply their probabilities: P(2 red marbles) = P(red) x P(red|red) = (5/12) x (4/11) = 5/33.

6b. To select 1 red and 1 black marble, the probability of selecting a red marble first is P(red) = 5/12, as there are 5 red marbles out of 12. Once the first red marble is selected, it is not replaced, so there are 4 red marbles and 6 black marbles left in the bag.

The probability of choosing a black marble next is P(black|red) = 6/11, as there are 6 black marbles left out of 11 total marbles left. To find the probability of both events happening, we multiply their probabilities: P(1 red, 1 black) = P(red) x P(black|red) = (5/12) x (6/11) = 5/22. 6c. To select 1 red and 1 purple marble, the probability of selecting a red marble first is P(red) = 5/12, as there are 5 red marbles out of 12.

Once the first red marble is selected, it is not replaced, so there are 4 red marbles and 1 purple marble left in the bag. The probability of choosing a purple marble next is P(purple|red) = 1/11, as there is only 1 purple marble left out of 11 total marbles left. To find the probability of both events happening, we multiply their probabilities: P(1 red, 1 purple) = P(red) x P(purple|red) = (5/12) x (1/11) = 5/132. 7.

There are a total of 8 possible routes to enter room B, and each route has an equal probability of being chosen. Since there is only 1 route that leads to room B, the probability of entering room B is 1/8.

8a. The expected value is calculated as the sum of each possible outcome multiplied by its probability. Since the die has 6 equally likely outcomes, the expected value is: E(x) = (1/6)(1) + (1/6)(2) + (1/6)(3) + (1/6)(4) + (1/6)(5) + (1/6)(6) = 3.5.

8b. For the game to be fair, the expected value of the game should be equal to the cost of playing. Therefore, a person should pay $3.50 to play the game if they want it to be a fair game.

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A quadratic equation is a second-degree polynomial equation, meaning it has an exponent of 2 on the variable. It can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. A linear equation, on the other hand, is a first-degree polynomial equation, meaning it has an exponent of 1 on the variable. It can be written in the form \(mx + b = 0\), where \(m\) and \(b\) are constants.

To solve the given quadratic equations, we can use the quadratic formula, which states that for an equation in the form \(ax^2 + bx + c = 0\), the solutions for \(x\) are given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Now let's solve the given quadratic equations:

a) \(x^2 + 13x + 42 = 0\):

Using the quadratic formula, we find that \(x = -6\) and \(x = -7\) are the solutions.

b) \(6x^2 + 11x + 3 = 0\):

Using the quadratic formula, we find that \(x = -\frac{1}{2}\) and \(x = -\frac{3}{2}\) are the solutions.

c) \(x^2 - 9x + 20 = 0\):

Using the quadratic formula, we find that \(x = 4\) and \(x = 5\) are the solutions.

d) \(x^2 - 8x + 12 = 0\):

Using the quadratic formula, we find that \(x = 2\) and \(x = 6\) are the solutions.

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Rounding to the nearest meter, we get that the ship is about 242 meters away from the lighthouse.

We can use the cotangent function to find the distance between the ship and the lighthouse. Let d be the distance between the ship and the base of the lighthouse, then we have:

cot(0.135) = 38 / d

Multiplying both sides by d, we get:

d * cot(0.135) = 38

Dividing both sides by cot(0.135), we get:

d = 38 / cot(0.135)

Using a calculator, we find:

d ≈ 241.7 meters

Rounding to the nearest meter, we get that the ship is about 242 meters away from the lighthouse.

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To meet the stopping sight distance (SSD) requirements for a highway section with a grade change, the minimum length of the equal tangent vertical curve needs to be determined.

Given the design speed of 65 mph, the height of eye and height of the object, and the grades of +(X4/2)% and -(X3/2)%, the minimum curve length can be calculated based on the AASHTO Guidelines.

The minimum length of the equal tangent vertical curve can be determined using the formula:

L = [(V^2 * f) / (30 * g * (H + h))]

Where:

L = Length of the curve

V = Design speed in ft/s

f = Rate of grade change in percentage (difference between the two grades)

g = Acceleration due to gravity (32.17 ft/s^2)

H = Height of eye

h = Height of object

By substituting the given values and solving the equation, the minimum length of the curve can be calculated to meet the SSD requirements.

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From the reduced row-echelon form, we can determine the values of the variables. In this case, we have x₁ = 3, x₂ = 0, and x₃ = -2 that is option B.

To solve the system of equations using the Gaussian-Jordan method, we perform row operations to transform the augmented matrix into row-echelon form and then into reduced row-echelon form.

Starting with the augmented matrix:

[ 1 1 1 | 1 ]

[ 2 -2 2 | 2 ]

[ 1 1 2 | -1 ]

We apply row operations to eliminate the coefficients below the main diagonal. The goal is to create zeros in the lower-left triangle of the matrix. We can achieve this by subtracting suitable multiples of one row from another row.

After performing the necessary row operations, we obtain the following reduced row-echelon form:

[ 1 0 -2 | 3 ]

[ 0 1 0 | 0 ]

[ 0 0 0 | 0 ]

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