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 find 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 find 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
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