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study help
mathematics
college algebra
College Algebra 12th edition Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels - Solutions
Evaluate the discriminant for the equation. Then use it to determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. (Do not solve the equation.)x2 + 4x + 4 = 0
Evaluate the discriminant for the equation. Then use it to determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers. (Do not solve the equation.)x2 - 8x + 16 = 0
For the equation, (a) solve for x in terms of y, and (b) solve for y in terms of x.5x2 - 6xy + 2y2 = 1
For the equation, (a) solve for x in terms of y, and (b) solve for y in terms of x.2x2 + 4xy - 3y2 = 2
For the equation, (a) solve for x in terms of y, and (b) solve for y in terms of x.3y2 + 4xy - 9x2 = -1
For the equation, (a) solve for x in terms of y, and (b) solve for y in terms of x.4x2 - 2xy + 3y2 = 2
Solve the equation for the specified variable. (Assume no denominators are 0.)S = 2πrh + 2pr2, for r
Solve the equation for the specified variable. (Assume no denominators are 0.)h = -16t2 + v0 t + s0, for t
Solve the equation for the specified variable. (Assume no denominators are 0.)s = s0 + gt2 + k, for t
Solve the equation for the specified variable. (Assume no denominators are 0.)r = r0 +1/2 at2, for t
Solve the equation for the specified variable. (Assume no denominators are 0.)E = e2k/2r, for e
Solve the equation for the specified variable. (Assume no denominators are 0.)F = kMv2/r, for v
Solve the equation for the specified variable. (Assume no denominators are 0.)A = πr2, for r
Solve the equation for the specified variable. (Assume no denominators are 0.)s = 1/2 gt2, for t
Solve the cubic equation using factoring and the quadratic formula.x3 + 64 = 0
Solve the cubic equation using factoring and the quadratic formula.x3 + 27 = 0
Solve the cubic equation using factoring and the quadratic formula.x3 - 27 = 0
Solve the cubic equation using factoring and the quadratic formula.x3 - 8 = 0
Why do the following two equations have the same solution set? (Do not solve.)-2x2 + 3x - 6 = 0 and 2x2 - 3x + 6 = 0
Solve each equation using the quadratic formula.(x - 9)(x - 1) = -16
Solve each equation using the quadratic formula.(3x + 2)(x - 1) = 3x
Solve each equation using the quadratic formula.(4x - 1)(x + 2) = 4x
Solve each equation using the quadratic formula.0.1x2 - 0.1x = 0.3
Solve each equation using the quadratic formula.0.2x2 + 0.4x - 0.3 = 0
Solve each equation using the quadratic formula.2/3 x2 + 1/4 x = 3
Solve each equation using the quadratic formula.1/2 x2 + 1/4 x - 3 = 0
Solve each equation using the quadratic formula.-6x2 = 3x + 2
Solve each equation using the quadratic formula.-4x2 = -12x + 11
Solve each equation using the quadratic formula.x2 = 2x - 10
Solve the equation using the quadratic formula.x2 = 2x - 5
Solve the equation using the quadratic formula.x2 - 4x = -1
Solve each equation using the quadratic formula.x2 – 6x = –7
Solve the equation using the quadratic formula.x2 – 3x – 2 = 0
Solve the equation using the quadratic formula.x2 – x – 1 = 0
Francesca, Francisco’s twin sister, claimed that the equationx2 - 19 = 0cannot be solved by the quadratic formula since there is no value for b. Is she correct?
Answer the question.Francisco claimed that the equationx2 – 8x = 0cannot be solved by the quadratic formula since there is no value for c. Is he correct?
Solve the equation using completing the square.–3x2 + 9x = 7
Solve the equation using completing the square.–4x2 + 8x = 7
Solve the equation using completing the square.–3x2 + 6x + 5 = 0
Solve the equation using completing the square.–2x2 + 4x + 3 = 0
Solve the equation using completing the square.3x2 + 2x = 5
Solve the equation using completing the square.2x2 + x = 10
Solve the equation using completing the square.x2 - 10x + 18 = 0
Solve the equation using completing the square.x2 - 2x - 2 = 0
Solve the equation using completing the square.4x2 – 3x – 10 = 0
Solve the equation using completing the square.2x2 – x – 28 = 0
Solve the equation using completing the square.x2 – 7x + 12 = 0
Solve the equation using completing the square.x2 – 4x + 3 = 0
Solve the equation using the square root property.(–2x + 5)2 = –8
Solve the equation using the square root property.(5x – 3)2 = –3
Solve the equation using the square root property.(x – 4)2 = –5
Solve the equation using the square root property.(x + 5)2 = –3
Solve the equation using the square root property.(4x + 1)2 = 20
Solve the equation using the square root property.(3x – 1)2 = 12
Solve the equation using the square root property.x2 = –400
Solve the equation using the square root property.x2 = –81
Solve the equation using the square root property.48 – x2 = 0
Solve the equation using the square root property.27 – x2 = 0
Solve the equation using the square root property.x2 = 121
Solve the equation using the square root property.x2 = 16
Solve the equation using the zero-factor property.36x2 + 60x + 25 = 0
Solve the equation using the zero-factor property.25x2 + 30x + 9 = 0
Solve the equation using the zero-factor property.9x2 – 12x + 4 = 0
Solve the equation using the zero-factor property.4x2 – 4x + 1 = 0
Solve the equation using the zero-factor property.x2 – 64 = 0
Solve the equation using the zero-factor property.x2 – 100 = 0
Solve the equation using the zero-factor property.–6x2 + 7x = –10
Solve the equation using the zero-factor property.–4x2 + x = –3
Solve the equation using the zero-factor property.2x2 - x - 15 = 0
Solve the equation using the zero-factor property.5x2 - 3x - 2 = 0
Solve the equation using the zero-factor property.x2 + 2x - 8 = 0
Solve each equation using the zero-factor property.x2 – 5x + 6 = 0
Use Choices A–D to answer each question.A. 3x2 - 17x - 6 = 0 B. (2x + 5)2 = 7C. x2 + x = 12 D. (3x - 1)(x - 7) = 0Only one of the equations is set up so that the values of a, b, and c can be determined immediately. Which one is it? Solve it.
Use Choices A–D to answer each question.A. 3x2 - 17x - 6 = 0 B. (2x + 5)2 = 7C. x2 + x = 12 D. (3x - 1)(x - 7) = 0Only one of the equations does not require Step 1 of the method for completing the square described in this section. Which one is it? Solve it.
Use Choices A–D to answer each question.A. 3x2 - 17x - 6 = 0 B. (2x + 5)2 = 7C. x2 + x = 12 D. (3x - 1)(x - 7) = 0Which equation is set up for direct use of the square root property? Solve it.
Use Choices A–D to answer each question.A. 3x2 - 17x - 6 = 0 B. (2x + 5)2 = 7C. x2 + x = 12 D. (3x - 1)(x - 7) = 0Which equation is set up for direct use of the zero-factor property? Solve it.
Solve each inequality. Give the solution set in interval notation. < 5 -4 < 2.
Solve the inequality. Give the solution set in interval notation.2 > -6x + 3 > -3
Solve the inequality. Give the solution set in interval notation.-11 > -3x + 1 > -17
Solve the inequality. Give the solution set in interval notation.-6 ≤ 6x + 3 ≤ 21
Solve the inequality. Give the solution set in interval notation.10 ≤ 2x + 4 ≤ 16
Solve the inequality. Give the solution set in interval notation.-7 < 2 + 3x < 5
Solve each inequality. Give the solution set in interval notation.-5 < 5 + 2x < 11
Find all intervals where each product will at least break even.The cost to produce x units of briefcases is C = 70x + 500, while the revenue is R = 60x.
Find all intervals where each product will at least break even.The cost to produce x units of coffee cups is C = 105x + 900, while the revenue is R = 85x.
Find all intervals where each product will at least break even.The cost to produce x units of picture frames is C = 50x + 5000, while the revenue is R = 60x.
Solve the inequality. Give the solution set in interval notation.2 - 4x + 5(x - 1) < -6(x - 2)
Solve the inequality. Give the solution set in interval notation.8x - 3x + 2 < 2(x + 7)
Solve the inequality. Give the solution set in interval notation.6x - (2x + 3) ≥ 4x - 5
Solve the inequality. Give the solution set in interval notation.3(x + 5) + 1 ≥ 5 + 3x
Solve the inequality. Give the solution set in interval notation.-4x + 3 ≥ -2 + x
Solve the inequality. Give the solution set in interval notation.-2x - 2 ≤ 1 + x
Solve the inequality. Give the solution set in interval notation.-3x - 8 ≤ 7
Solve the inequality. Give the solution set in interval notation.-2x + 8 ≤ 16
The three-part inequality a 6 x 6 b means “a is less than x and x is less than b.” Which inequality is not satisfied by some real number x?A. -3 < x < 10 B. 0 < x < 6C. -3 < x < -1 D. -8 < x < -10
Explain how to determine whether to use a parenthesis or a square bracket when writing the solution set of a linear inequality in interval notation.
Match the inequality in each exercise in Column I with its equivalent interval notation in Column II.A. (-2, 6]B. [-2, 6)C. (-∞, -6)D. [6, ∞)E. (-∞, -3) ⋃ (3, ∞)F. (-∞, -6)G. (0, 8)H. (-∞, ∞)I. (-6, ∞)J. (-∞, 6) -6
In this section we introduced methods of solving equations quadratic in form by substitution and solving equations involving radicals by raising each side of the equation to a power. Suppose we wish to solve x - √x - 12 = 0. We can solve this equation using either of the two methods. To see how
In this section we introduced methods of solving equations quadratic in form by substitution and solving equations involving radicals by raising each side of the equation to a power. Suppose we wish to solve x - √x - 12 = 0. We can solve this equation using either of the two methods. To see how
In this section we introduced methods of solving equations quadratic in form by substitution and solving equations involving radicals by raising each side of the equation to a power. Suppose we wish to solve x - √x - 12 = 0. We can solve this equation using either of the two methods. To see how
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