# Question 1 of 40

Question 1 of 40

2.5 Points

Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.

x + y – z = -2

2x – y + z = 5

-x + 2y + 2z = 1

A. {(0, -1, -2)}

B. {(2, 0, 2)}

C. {(1, -1, 2)}

D. {(4, -1, 3)}

Question 2 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to each system.

x – 3y + z = 1

-2x + y + 3z = -7

x – 4y + 2z = 0

A. {(2t + 4, t + 1, t)}

B. {(2t + 5, t + 2, t)}

C. {(1t + 3, t + 2, t)}

D. {(3t + 3, t + 1, t)}

Question 3 of 40

2.5 Points

Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.

2x – y – z = 4

x + y – 5z = -4

x – 2y = 4

A. {(2, -1, 1)}

B. {(-2, -3, 0)}

C. {(3, -1, 2)}

D. {(3, -1, 0)}

Question 4 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

x + 2y + 2z = 5

2x + 4y + 7z = 19

-2x – 5y – 2z = 8

A. {(33, -11, 4)}

B. {(13, 12, -3)}

C. {(23, -12, 3)}

D. {(13, -14, 3)}

Question 5 of 40

2.5 Points

Give the order of the following matrix; if A = [aij], identify a32 and a23.

1

0

-2

-5

7

1/2

?

-6

11

e

-?

-1/5

A. 3 * 4; a32 = 1/45; a23 = 6

B. 3 * 4; a32 = 1/2; a23 = -6

C. 3 * 2; a32 = 1/3; a23 = -5

D. 2 * 3; a32 = 1/4; a23 = 4

Question 6 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

2x = 3y + 2

5x = 51 – 4y

A. {(8, 2)}

B. {(3, -4)}

C. {(2, 5)}

D. {(7, 4)}

Question 7 of 40

2.5 Points

Find values for x, y, and z so that the following matrices are equal.

2x

z

y + 7

4

=

-10

6

13

4

A. x = -7; y = 6; z = 2

B. x = 5; y = -6; z = 2

C. x = -3; y = 4; z = 6

D. x = -5; y = 6; z = 6

Question 8 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

3x – 4y = 4

2x + 2y = 12

A. {(3, 1)}

B. {(4, 2)}

C. {(5, 1)}

D. {(2, 1)}

Question 9 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.

5x + 8y – 6z = 14

3x + 4y – 2z = 8

x + 2y – 2z = 3

A. {(-4t + 2, 2t + 1/2, t)}

B. {(-3t + 1, 5t + 1/3, t)}

C. {(2t + -2, t + 1/2, t)}

D. {(-2t + 2, 2t + 1/2, t)}

Question 10 of 40

2.5 Points

Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.

x + 3y = 0

x + y + z = 1

3x – y – z = 11

A. {(3, -1, -1)}

B. {(2, -3, -1)}

C. {(2, -2, -4)}

D. {(2, 0, -1)}

Question 11 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.

8x + 5y + 11z = 30

-x – 4y + 2z = 3

2x – y + 5z = 12

A. {(3 – 3t, 2 + t, t)}

B. {(6 – 3t, 2 + t, t)}

C. {(5 – 2t, -2 + t, t)}

D. {(2 – 1t, -4 + t, t)}

Question 12 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to each system.

x1 + 4×2 + 3×3 – 6×4 = 5

x1 + 3×2 + x3 – 4×4 = 3

2×1 + 8×2 + 7×3 – 5×4 = 11

2×1 + 5×2 – 6×4 = 4

A. {(-47t + 4, 12t, 7t + 1, t)}

B. {(-37t + 2, 16t, -7t + 1, t)}

C. {(-35t + 3, 16t, -6t + 1, t)}

D. {(-27t + 2, 17t, -7t + 1, t)}

Question 13 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to each system.

2x + 3y – 5z = 15

x + 2y – z = 4

A. {(6t + 28, -7t – 6, t)}

B. {(7t + 18, -3t – 7, t)}

C. {(7t + 19, -1t – 9, t)}

D. {(4t + 29, -3t – 2, t)}

Question 14 of 40

2.5 Points

Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.

x + y + z = 4

x – y – z = 0

x – y + z = 2

A. {(3, 1, 0)}

B. {(2, 1, 1)}

C. {(4, 2, 1)}

D. {(2, 1, 0)}

Question 15 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

x + y + z = 0

2x – y + z = -1

-x + 3y – z = -8

A. {(-1, -3, 7)}

B. {(-6, -2, 4)}

C. {(-5, -2, 7)}

D. {(-4, -1, 7)}

Question 16 of 40

2.5 Points

If AB = -BA, then A and B are said to be anticommutative.

Are A =

0

1

-1

0

and B =

1

0

0

-1

anticommutative?

A. AB = -AB so they are not anticommutative.

B. AB = BA so they are anticommutative.

C. BA = -BA so they are not anticommutative.

D. AB = -BA so they are anticommutative.

Question 17 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

x + y = 7

x – y = 3

A. {(7, 2)}

B. {(8, -2)}

C. {(5, 2)}

D. {(9, 3)}

Question 18 of 40

2.5 Points

Solve the following system of equations using matrices. Use Gaussian elimination with back substitution or Gauss-Jordan elimination.

x + 2y = z – 1

x = 4 + y – z

x + y – 3z = -2

A. {(3, -1, 0)}

B. {(2, -1, 0)}

C. {(3, -2, 1)}

D. {(2, -1, 1)}

Question 19 of 40

2.5 Points

Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.

3x + 4y + 2z = 3

4x – 2y – 8z = -4

x + y – z = 3

A. {(-2, 1, 2)}

B. {(-3, 4, -2)}

C. {(5, -4, -2)}

D. {(-2, 0, -1)}

Question 20 of 40

2.5 Points

Use Cramer’s Rule to solve the following system.

4x – 5y – 6z = -1

x – 2y – 5z = -12

2x – y = 7

A. {(2, -3, 4)}

B. {(5, -7, 4)}

C. {(3, -3, 3)}

D. {(1, -3, 5)}

Question 21 of 40

2.5 Points

Convert each equation to standard form by completing the square on x or y. Then ï¬nd the vertex, focus, and directrix of the parabola.

x2 – 2x – 4y + 9 = 0

A. (x – 4)2 = 4(y – 2); vertex: (1, 4); focus: (1, 3) ; directrix: y = 1

B. (x – 2)2 = 4(y – 3); vertex: (1, 2); focus: (1, 3) ; directrix: y = 3

C. (x – 1)2 = 4(y – 2); vertex: (1, 2); focus: (1, 3) ; directrix: y = 1

D. (x – 1)2 = 2(y – 2); vertex: (1, 3); focus: (1, 2) ; directrix: y = 5

Question 22 of 40

2.5 Points

Find the focus and directrix of the parabola with the given equation.

8×2 + 4y = 0

A. Focus: (0, -1/4); directrix: y = 1/4

B. Focus: (0, -1/6); directrix: y = 1/6

C. Focus: (0, -1/8); directrix: y = 1/8

D. Focus: (0, -1/2); directrix: y = 1/2

Question 23 of 40

2.5 Points

Locate the foci and find the equations of the asymptotes.

x2/9 – y2/25 = 1

A. Foci: ({±v36, 0) ;asymptotes: y = ±5/3x

B. Foci: ({±v38, 0) ;asymptotes: y = ±5/3x

C. Foci: ({±v34, 0) ;asymptotes: y = ±5/3x

D. Foci: ({±v54, 0) ;asymptotes: y = ±6/3x

Question 24 of 40

2.5 Points

Find the vertex, focus, and directrix of each parabola with the given equation.

(y + 1)2 = -8x

A. Vertex: (0, -1); focus: (-2, -1); directrix: x = 2

B. Vertex: (0, -1); focus: (-3, -1); directrix: x = 3

C. Vertex: (0, -1); focus: (2, -1); directrix: x = 1

D. Vertex: (0, -3); focus: (-2, -1); directrix: x = 5

Question 25 of 40

2.5 Points

Find the standard form of the equation of each hyperbola satisfying the given conditions.

Foci: (-4, 0), (4, 0)

Vertices: (-3, 0), (3, 0)

A. x2/4 – y2/6 = 1

B. x2/6 – y2/7 = 1

C. x2/6 – y2/7 = 1

D. x2/9 – y2/7 = 1

Question 26 of 40

2.5 Points

Find the vertex, focus, and directrix of each parabola with the given equation.

(x + 1)2 = -8(y + 1)

A. Vertex: (-1, -2); focus: (-1, -2); directrix: y = 1

B. Vertex: (-1, -1); focus: (-1, -3); directrix: y = 1

C. Vertex: (-3, -1); focus: (-2, -3); directrix: y = 1

D. Vertex: (-4, -1); focus: (-2, -3); directrix: y = 1

Question 27 of 40

2.5 Points

Find the vertices and locate the foci of each hyperbola with the given equation.

x2/4 – y2/1 =1

A.

Vertices at (2, 0) and (-2, 0); foci at (v5, 0) and (-v5, 0)

B.

Vertices at (3, 0) and (-3 0); foci at (12, 0) and (-12, 0)

C. Vertices at (4, 0) and (-4, 0); foci at (16, 0) and (-16, 0)

D. Vertices at (5, 0) and (-5, 0); foci at (11, 0) and (-11, 0)

Question 28 of 40

2.5 Points

Convert each equation to standard form by completing the square on x or y. Then ï¬nd the vertex, focus, and directrix of the parabola.

y2 – 2y + 12x – 35 = 0

A. (y – 2)2 = -10(x – 3); vertex: (3, 1); focus: (0, 1); directrix: x = 9

B. (y – 1)2 = -12(x – 3); vertex: (3, 1); focus: (0, 1); directrix: x = 6

C. (y – 5)2 = -14(x – 3); vertex: (2, 1); focus: (0, 1); directrix: x = 6

D. (y – 2)2 = -12(x – 3); vertex: (3, 1); focus: (0, 1); directrix: x = 8

Question 29 of 40

2.5 Points

Locate the foci and find the equations of the asymptotes.

x2/100 – y2/64 = 1

A. Foci: ({= ±2v21, 0); asymptotes: y = ±2/5x

B. Foci: ({= ±2v31, 0); asymptotes: y = ±4/7x

C. Foci: ({= ±2v41, 0); asymptotes: y = ±4/7x

D. Foci: ({= ±2v41, 0); asymptotes: y = ±4/5x

Question 30 of 40

2.5 Points

Locate the foci of the ellipse of the following equation.

7×2 = 35 – 5y2

A. Foci at (0, -v2) and (0, v2)

B. Foci at (0, -v1) and (0, v1)

C. Foci at (0, -v7) and (0, v7)

D. Foci at (0, -v5) and (0, v5)

Question 31 of 40

2.5 Points

Locate the foci of the ellipse of the following equation.

x2/16 + y2/4 = 1

A. Foci at (-2v3, 0) and (2v3, 0)

B. Foci at (5v3, 0) and (2v3, 0)

C. Foci at (-2v3, 0) and (5v3, 0)

D. Foci at (-7v2, 0) and (5v2, 0)

Question 32 of 40

2.5 Points

Find the standard form of the equation of the following ellipse satisfying the given conditions.

Foci: (-2, 0), (2, 0)

Y-intercepts: -3 and 3

A. x2/23 + y2/6 = 1

B. x2/24 + y2/2 = 1

C. x2/13 + y2/9 = 1

D. x2/28 + y2/19 = 1

Question 33 of 40

2.5 Points

Find the standard form of the equation of each hyperbola satisfying the given conditions.

Foci: (0, -3), (0, 3)

Vertices: (0, -1), (0, 1)

A. y2 – x2/4 = 0

B. y2 – x2/8 = 1

C. y2 – x2/3 = 1

D. y2 – x2/2 = 0

Question 34 of 40

2.5 Points

Find the standard form of the equation of the following ellipse satisfying the given conditions.

Foci: (0, -4), (0, 4)

Vertices: (0, -7), (0, 7)

A. x2/43 + y2/28 = 1

B. x2/33 + y2/49 = 1

C. x2/53 + y2/21 = 1

D. x2/13 + y2/39 = 1

Question 35 of 40

2.5 Points

Find the standard form of the equation of the ellipse satisfying the given conditions.

Major axis vertical with length = 10

Length of minor axis = 4

Center: (-2, 3)

A. (x + 2)2/4 + (y – 3)2/25 = 1

B. (x + 4)2/4 + (y – 2)2/25 = 1

C. (x + 3)2/4 + (y – 2)2/25 = 1

D. (x + 5)2/4 + (y – 2)2/25 = 1

Question 36 of 40

2.5 Points

Find the vertex, focus, and directrix of each parabola with the given equation.

(x – 2)2 = 8(y – 1)

A. Vertex: (3, 1); focus: (1, 3); directrix: y = -1

B. Vertex: (2, 1); focus: (2, 3); directrix: y = -1

C. Vertex: (1, 1); focus: (2, 4); directrix: y = -1

D. Vertex: (2, 3); focus: (4, 3); directrix: y = -1

Question 37 of 40

2.5 Points

Find the standard form of the equation of the ellipse satisfying the given conditions.

Endpoints of major axis: (7, 9) and (7, 3)

Endpoints of minor axis: (5, 6) and (9, 6)

A. (x – 7)2/6 + (y – 6)2/7 = 1

B. (x – 7)2/5 + (y – 6)2/6 = 1

C. (x – 7)2/4 + (y – 6)2/9 = 1

D. (x – 5)2/4 + (y – 4)2/9 = 1

uestion 38 of 40

2.5 Points

Find the standard form of the equation of the following ellipse satisfying the given conditions.

Foci: (-5, 0), (5, 0)

Vertices: (-8, 0), (8, 0)

A. x2/49 + y2/ 25 = 1

B. x2/64 + y2/39 = 1

C. x2/56 + y2/29 = 1

D. x2/36 + y2/27 = 1

Question 39 of 40

2.5 Points

Convert each equation to standard form by completing the square on x and y.

9×2 + 25y2 – 36x + 50y – 164 = 0

A. (x – 2)2/25 + (y + 1)2/9 = 1

B. (x – 2)2/24 + (y + 1)2/36 = 1

C. (x – 2)2/35 + (y + 1)2/25 = 1

D. (x – 2)2/22 + (y + 1)2/50 = 1

Question 40 of 40

2.5 Points

Find the standard form of the equation of each hyperbola satisfying the given conditions.

Endpoints of transverse axis: (0, -6), (0, 6)

Asymptote: y = 2x

A. y2/6 – x2/9 = 1

B. y2/36 – x2/9 = 1

C. y2/37 – x2/27 = 1

D. y2/9 – x2/6 = 1

File #1 