Question 1

Evaluate.
-
A) -62
B) -2
C) 170
D) 2

Accepted Answer

The answer of Evaluate.
-
A) -62
B) -2
C) 170
D) 2...

Question 2

Solve using matrices.
-x + 6y = 6 2x - 5y = -5
A) (0, 0)
B) (0, 1)
C) (1, 0)
D) (1, 1)

Accepted Answer

The answer of Solve using matrices.
-x + 6y = 6...

Question 3

Evaluate.
-
A) -73
B) 53
C) -63
D) 63

Accepted Answer

The answer of Evaluate.
-
A) -73
B) 53
C) -63
D) 63...

Question 4

Evaluate.
-
A) -162
B) 162
C) 141
D) -183

Accepted Answer

The answer of Evaluate.
-
A) -162
B) 162
C) 141
D) -183...

Question 5

Evaluate.
-
A) -342
B) 288
C) 315
D) -315

Accepted Answer

The answer of Evaluate.
-
A) -342
B) 288
C) 315
D) -315...

Question 6

Solve using matrices.
-5x - 2y = -1 x + 4y = 35
A) (2, 8)
B) (2, 9)
C) (3, 9)
D) (3, 8)

Accepted Answer

The answer of Solve using matrices.
-5x - 2y = -1...

Question 7

Solve using matrices.
-x + 7y = -2 3x + y = 34
A) (3, 7)
B) (12, -2)
C) (7, 12)
D) (-2, 3)

Accepted Answer

To solve the system of equations using matrices, we can write the equations in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.A = | 1  7 |    | 3  1 |B = | -2 |    | 34 |To find X, we calculate A^-1 * B.First, find the determinant of A: det(A) = (1)(1) - (7)(3) = 1 - 21 = -20.Then, find the inverse of A (A^-1) by swapping the elements on the main diagonal, changing the signs of the off-diagonal elements, and dividing each term by the determinant:A^-1 = 1/-20 * |  1  -7 |               | -3   1 |A^-1 = | -1/20  7/20 |       |  3/20 -1/20 |Multiplying A^-1 by B gives X:X = A^-1 * B = | -1/20  7/20 | * | -2 |               |  3/20 -1/20 |   | 34 |  = | (1/10) + (7*34)/20 |    | (-3*2)/20 + (-34/20) |  = | (1/10) + (238/20) |    | (-6/20) - (34/20) |  = | 0.1 + 11.9 |    | -0.3 - 1.7 |  = | 12 |    | -2 |Therefore, the solution is X = (12, -2), which corresponds to choice B.

Question 8

Evaluate.
-
A) 392
B) -44
C) -4
D) 4

Accepted Answer

The answer of Evaluate.
-
A) 392
B) -44
C) -4
D) 4...

Question 9

Solve using Cramer's rule.
-8a + 3c = 69 5a + 9b - 9c = 30
5a - 9b = -33
A) (7, 7, 7)
B) (6, -7, -7)
C) (6, 7, 7)
D) (7, 5, 7)

Accepted Answer

To solve the system of equations using Cramer's rule, we first need to write the equations in the standard form $Ax = B$, where $A$ is the matrix of coefficients, $x$ is the column matrix of variables, and $B$ is the column matrix of constants.Given equations:1. $8a + 0b + 3c = 69$2. $5a + 9b - 9c = 30$3. $5a - 9b + 0c = -33$Matrix $A$ (coefficients matrix):$$A = \begin{bmatrix}8 & 0 & 3 \5 & 9 & -9 \5 & -9 & 0 \\end{bmatrix}$$Matrix $B$ (constants matrix):$$B = \begin{bmatrix}69 \30 \-33 \\end{bmatrix}$$To find the solution using Cramer's rule, we need to find the determinants of the matrices. However, since the question is multiple choice and we're not provided with the steps to calculate the determinants directly, we can check the given options by substituting them back into the original equations.Option C: (6, 7, 7)Substituting $a = 6$, $b = 7$, and $c = 7$ into the original equations:1. $8(6) + 3(7) = 48 + 21 = 69$2. $5(6) + 9(7) - 9(7) = 30 + 63 - 63 = 30$3. $5(6) - 9(7) = 30 - 63 = -33$All three equations are satisfied by the values in option C, making it the correct answer.

Question 10

Solve using Cramer's rule.
-2x + 2y = 22 2x - 2y = 2
A) (5, 6)
B) (-5, 6)
C) (6, 5)
D) (-6, -5)

Accepted Answer

Cramer's rule involves using determinants to solve a system of linear equations. For the given system:1. The determinant of the coefficient matrix (D) is calculated using the coefficients of x and y:$$D = \begin{vmatrix} 2 & 2 \ 2 & -2 \end{vmatrix} = (2)(-2) - (2)(2) = -4 - 4 = -8.$$2. The determinant for x (Dx) replaces the x-column with the constants from the equations:$$D_x = \begin{vmatrix} 22 & 2 \ 2 & -2 \end{vmatrix} = (22)(-2) - (2)(2) = -44 - 4 = -48.$$3. The determinant for y (Dy) replaces the y-column with the constants:$$D_y = \begin{vmatrix} 2 & 22 \ 2 & 2 \end{vmatrix} = (2)(2) - (22)(2) = 4 - 44 = -40.$$4. Solve for x and y:$$x = \frac{D_x}{D} = \frac{-48}{-8} = 6,$$$$y = \frac{D_y}{D} = \frac{-40}{-8} = 5.$$Therefore, the solution is (6, 5).

Question 11

Solve using matrices.
-x + y + z = -5 x - y + 5z = -7
5x + y + z = 7
A) (-3, -5, 3)
B) (3, -5, -3)
C) (-5, -3, 3)
D) (-3, 3, -5)

Accepted Answer

The answer of Solve using matrices.
-x + y + z...

Question 12

Solve using matrices.
-2x + 4y = - 1 7x - y = 4
A) 
B) 
C) 
D)

Accepted Answer

The answer of Solve using matrices.
-2x + 4y = -...

Question 13

Solve using Cramer's rule.
-3x - 7y - z = -40 x - 4y - 5z = -36
-9x + y + z = -44
A) (6, -8, -2)
B) (7, 6, 2)
C) (6, 8, 2)
D) (8, 2, 8)

Accepted Answer

The answer of Solve using Cramer's rule.
-3x - 7y -...

Question 14

Solve using Cramer's rule.
-3x - 7y - 5z = -58 2x + 6y - 7z = 25
-4x - 6y - 2z = -56
A) (3, 5, 3)
B) (2, -7, -3)
C) (7, 3, 7)
D) (2, 7, 3)

Accepted Answer

Using Cramer's rule, the solution to the system of equations is (2, 7, 3).

Question 15

Solve using Cramer's rule.
-2x - 2y = 2 2x + 3y = 37
A) (-7, 8)
B) (-8, -7)
C) (7, 8)
D) (8, 7)

Accepted Answer

Cramer's rule involves using determinants to solve a system of linear equations. For the given system:1. The determinant of the coefficient matrix (D) is calculated using the coefficients of x and y:$$D = \begin{vmatrix} 2 & -2 \ 2 & 3 \end{vmatrix} = (2*3) - (2*(-2)) = 6 + 4 = 10.$$2. The determinant for x (Dx) replaces the x coefficients with the constants from the equations:$$D_x = \begin{vmatrix} 2 & -2 \ 37 & 3 \end{vmatrix} = (2*3) - (37*(-2)) = 6 + 74 = 80.$$3. The determinant for y (Dy) replaces the y coefficients with the constants:$$D_y = \begin{vmatrix} 2 & 2 \ 2 & 37 \end{vmatrix} = (2*37) - (2*2) = 74 - 4 = 70.$$4. Solve for x and y:$$x = \frac{D_x}{D} = \frac{80}{10} = 8,$$$$y = \frac{D_y}{D} = \frac{70}{10} = 7.$$Therefore, the solution is (8, 7).

Question 16

Solve using matrices.
-x - y + 5z = -6 2x + z = -1
X + 5y + z = 4
A) (1, -1, 0)
B) (-1, 0, 1)
C) (-1, 1, 0)
D) (0, 1, -1)

Accepted Answer

The answer of Solve using matrices.
-x - y + 5z...

Question 17

Solve using Cramer's rule.
-5x + 5y = 10 5x + y = -10
A) (5, -3)
B) (-3, 5)
C) (3, -5)
D) (-5, -3)

Accepted Answer

The answer of Solve using Cramer's rule.
-5x + 5y =...

Question 18

Solve using matrices.
-x - 5y = -4 x + 8y = -4
A) (0, -4)
B) (4, 0)
C) (0, 4)
D) (-4, 0)

Accepted Answer

The answer of Solve using matrices.
-x - 5y = -4...

Question 19

Solve using matrices.
-5x + 6y = 20 2x + 8y = 36
A) (-2, 5)
B) (5, 2)
C) (-2, -5)
D) (5, -2)

Accepted Answer

To solve the system of equations using matrices, we first express it in matrix form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. For the given system:$$A = \begin{pmatrix} 5 & 6 \ 2 & 8 \end{pmatrix}, X = \begin{pmatrix} x \ y \end{pmatrix}, B = \begin{pmatrix} 20 \ 36 \end{pmatrix}$$To find X, we calculate $A^{-1}B$.$$A^{-1} = \frac{1}{(5*8 - 6*2)} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} = \frac{1}{40 - 12} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix}$$$$A^{-1}B = \frac{1}{28} \begin{pmatrix} 8 & -6 \ -2 & 5 \end{pmatrix} \begin{pmatrix} 20 \ 36 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} 160 - 216 \ -40 + 180 \end{pmatrix} = \frac{1}{28} \begin{pmatrix} -56 \ 140 \end{pmatrix} = \begin{pmatrix} -2 \ 5 \end{pmatrix}$$Thus, the solution is $(-2, 5)$, which corresponds to choice A.

Question 20

Solve using matrices.
-2x + y = 6 9x + 2y = 37
A) (-4, -5)
B) (5, -5)
C) (5, -4)
D) (-4, 5)

Accepted Answer

To solve the system of equations using matrices, we can write the equations in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. For the given system:$$ \begin{align*}2x + y &= 6 \9x + 2y &= 37\end{align*} $$A (the coefficient matrix) is:$$ \begin{bmatrix}2 & 1 \9 & 2\end{bmatrix} $$X (the variable matrix) is:$$ \begin{bmatrix}x \y\end{bmatrix} $$B (the constant matrix) is:$$ \begin{bmatrix}6 \37\end{bmatrix} $$To find X, we calculate $A^{-1}B$.First, find the determinant of A:$$ \text{det}(A) = (2)(2) - (9)(1) = 4 - 9 = -5 $$Then, find the inverse of A ($A^{-1}$):$$ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix}2 & -1 \-9 & 2\end{bmatrix}^T = \frac{1}{-5} \begin{bmatrix}2 & -9 \-1 & 2\end{bmatrix} = \begin{bmatrix}-0.4 & 1.8 \0.2 & -0.4\end{bmatrix} $$Multiply $A^{-1}$ by B to find X:$$ X = A^{-1}B = \begin{bmatrix}-0.4 & 1.8 \0.2 & -0.4\end{bmatrix} \begin{bmatrix}6 \37\end{bmatrix} = \begin{bmatrix}5 \-4\end{bmatrix} $$Therefore, the solution is $x = 5$, $y = -4$, which corresponds to choice C.

Quiz 14: Calculating Means and Medians of Data Sets

Evaluate. -

Solve using matrices. -x + 6y = 6 2x - 5y = -5

Evaluate. -

Evaluate. -

Evaluate. -

Solve using matrices. -5x - 2y = -1 x + 4y = 35

Solve using matrices. -x + 7y = -2 3x + y = 34

Evaluate. -

Solve using Cramer's rule. -8a + 3c = 69 5a + 9b - 9c = 30 5a - 9b = -33

Solve using Cramer's rule. -2x + 2y = 22 2x - 2y = 2

Solve using matrices. -x + y + z = -5 x - y + 5z = -7 5x + y + z = 7

Solve using matrices. -2x + 4y = - 1 7x - y = 4

Solve using Cramer's rule. -3x - 7y - z = -40 x - 4y - 5z = -36 -9x + y + z = -44

Solve using Cramer's rule. -3x - 7y - 5z = -58 2x + 6y - 7z = 25 -4x - 6y - 2z = -56

Solve using Cramer's rule. -2x - 2y = 2 2x + 3y = 37

Solve using matrices. -x - y + 5z = -6 2x + z = -1 X + 5y + z = 4

Solve using Cramer's rule. -5x + 5y = 10 5x + y = -10

Solve using matrices. -x - 5y = -4 x + 8y = -4

Solve using matrices. -5x + 6y = 20 2x + 8y = 36

Solve using matrices. -2x + y = 6 9x + 2y = 37