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Introductory Algebra 2 Practice Test (page 2)

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Updated on Oct 27, 2011

Answers

  1. First, compare the given equation to the y = ax2 + bx + c formula:

    y = ax2 + bx + c

    y = 3x2 – 3x + 1

    The a and the c are clear, but to clearly see what b equals, convert the subtraction to add the opposite:

    y = 3x2 + (–3)x + 1

    Thus, a = 3, b = –3, and c = 1. The x-coordinate of the turning point, or vertex, of the parabola is given by:

    Substitute in the values from the equation:

    When x = 0.5, y will be:

    y = (3)(0.25) – (3)(0.5) + 1

    = 0.75 – 1.5 + 1

    = –0.75 + 1

    = 0.25

    Thus, the coordinates of the vertex are (0.5,0.25).

  2. First, compare the given equation to the y = ax2 + bx + c formula:

    y = ax2 + bx + c

    y = 2x2 + 16x + 1

    Thus, a = 2, b = 16, and c = 1. The line of symmetry is given by:

    Substitute in the values from the equation:

    Thus, the line of symmetry is x = –4.

  3. First, compare the given equation to the y = ax2 + bx + c formula:

    y = ax2 + bx + c

    y = 3x2 + 0(x) + 0

    Thus, a = 3, b = 0, and c = 0. The line of symmetry of the parabola is given by:

    Thus, the line of symmetry is x = 0.

  4. Factor √(a3) into two radicals; a2 is a perfect square, so factor √a3 into √aa2 = aa. Multiply the coefficient of the given expression by aa: (a3)(aa) = a4a .

  5. Factor √4g into two radicals; 4 is a perfect square, so factor √4g into √4g = 2√gg. Simplify the fraction by dividing the numerator by the denominator. Cancel the √g terms from the numerator and denominator. That leaves .

  6. First, find the square root of 9pr. √9pr = √9pr = 3√prpr. The denominator has a negative exponent, so it can be rewritten in the numerator with a positive exponent. √pr can be written as because a value raised to the exponent is another way of representing the square root of the value. The expression is now To multiply the pr terms, add the exponents. =3(pr)2 = 3p2r2.

  7. First, square Next, look at the term. A fraction with a negative exponent can be rewritten as the reciprocal of the fraction with a positive exponent. Multiply the fractions in the numerator by adding the exponents of the fractions: Finally, divide this fraction by xy:

  8. If y = –x, then y = –2. Substitute 2 for x and –2 for y: (((2)(–2))–2)2 = .

  9. First, cross multiply: g(g108) = √3, g2108 = √3. Divide both sides of the equation by Take the square root of both sides of the equation to find the value of g: Simplify the fraction by multiplying it by

  10. The n term in the denominator has a negative exponent. It can be placed in the numerator with a positive exponent, because The numerator of the fraction is now and the denominator of the fraction is 1. To multiply terms with like bases, keep the base and add the exponents: = n. Therefore, nm = 5, and m = .

  11. For x > 5, draw a dashed line at x = 5. The area to the right satisfies the condition, so shade it:

    Pretest

  12. For y < –4, draw a dashed line at y = –4. The area under the line satisfies the condition, so shade it:

    Pretest

  13. For –4x + 2y ≥ 6, rearrange so that the y is by itself on the left: –4x + 2y ≥ 6 becomes 2y ≥ 4x + 6, which equals y ≥ 2x + 3. For y ≥ 2x + 3, the slope is 2 and the y-intercept is 3. Start at the y-intercept (0,3) and go up 2 and over 1 (right) to plot points. This line will be solid because the symbol is ≥. The area above this line satisfies the condition.

    Pretest

  14. For y = x + 4, the slope is 1 and the y-intercept is 4. Start at the y-intercept (0,4) and go up 1 and over 1 (right) to plot points.

    Pretest

  15. For y = 2x + 3, the slope is 2 and the y-intercept is 3. Start at the y-intercept (0,3) and go up 2 and over 1 (right) to plot points.

    Pretest

  16. For y < x + 2, graph the line. The slope is 1 and the y-intercept is 2. Start at the y-intercept (0,2) and go up 1 and over 1 (right) to plot points. This line will be dashed because the symbol is <. The area under this line satisfies the condition. For y < –x + 4, graph the line. The slope is –1 and the y-intercept is 4. Start at the y-intercept (0,4) and go down 1 and over 1 (right) to plot points. This line will also be dashed because the symbol is <. The area under this line satisfies the condition. Shade the area common to both equations:

    Pretest

  17. Rearrange the first equation so that the y is by itself on the left:

    x + y > 5 becomes y > –1x + 5. For y > –1x + 5, the slope is –1 and the y-intercept is 5. Start at the y-intercept (0,5) and go down 1 and over 1 (right) to plot points. This line will be dashed because the symbol is >. The area above this line satisfies the condition. Rearrange the second equation so that the y is by itself on the left: –2x + y > 3 becomes y > 2x + 3. For y > 2x + 3, the slope is 2 and the y-intercept is 3. Start at the y-intercept (0,3) and go up 2 and over 1 (right) to plot points. This line will also be dashed because the symbol is >. The area above this line satisfies the condition. Shade the region common to both graphs:

    Pretest

  18. Using the zero product rule, you know that either x + 4 = 0 or that x – 2 = 0.

    So solve both of these possible equations:

    x + 4 = 0 x – 2 = 0
    x + 4 – 4 = 0 – 4 x – 2 + 2 = 0 + 2
    x = –4 x = 2

    So, you know that both x = –4 and x = 2 will make (x + 4)(x – 2) = 0. The zero product rule is useful when solving quadratic equations because you can rewrite a quadratic equation as equal to zero and take advantage of the fact that one of the factors of the quadratic equation is thus equal to 0.

  19. Add 1 to both sides of the equation. 2x2 – 33 + 1 = –1 + 1

    Simplify both sides of the equation. 2x2 – 32 = 0

    Take out the common factor. 2(x2 – 16) = 0

    Factor the difference of two squares. 2(x – 4)(x + 4) = 0

    Disregard the 2 and set the other factors equal to zero: x – 4 = 0 and x + 4 = 0

    Solve the first equation. x – 4 = 0

    Add 4 to both sides of the equation. x – 4 + 4 = 0 + 4

    Simplify both sides of the equation. x = 4

    Solve the second equation. x + 4 = 0

    Subtract 4 from both sides of the equation. x + 4 – 4 = 0 – 4

    Simplify both sides of the equation. x = –4

    The solutions are 4 and –4.

  20. To solve for d, isolate the variable:

    6√d – 10 = 32

    6√d – 10 + 10 = 32 + 10

    6√d = 42

    d = 7

    d2 = 72

    d = 49

  21. The fourth term in the sequence is You are looking for the eighth term, which is four terms after the fourth term. Because each term is more than the previous term, the eighth term will be more than Because the number of terms is reasonable, you can check your answer by repeatedly adding

  22. The term that precedes x is 5. Therefore, the value of x is 5 – 7 = –2, and the value of y is –2 – 7 = –9. Therefore, xy = –2 – (–9) = –2 + 9 = 7.

  23. The fourth term in the sequence is You are looking for the seventh term, which is three terms after the fourth term. You must multiply by three times, so the seventh term will be Alternatively, every term in the sequence is 18 times raised to a power. The first term, 18, is 18 × ()0. The second term, 12, is 18 × ()1. The value of the exponent is one less than the position of the term in the sequence. The seventh term of the sequence is equal to

  24. Every term in the sequence is 5 raised to a power. The first term, The second term, The value of the exponent is four less than the position of the term in the sequence. The 20th term of the sequence is equal to 520 – 4 = 516.

  25. The fourth term in the sequence is 74. You are looking for the ninth term, which is five terms after the fourth term. Because each term is nine less than the previous term, the ninth term will be 5(9) = 45 less than 74: 74 – 45 = 29. Because the number of terms is reasonable, you can check your answer by repeatedly subtracting 9: 74 – 9 = 65, 65 – 9 = 56, 56 – 9 = 47, 47 – 9 = 38, 38 – 9 = 29.

  26. The term that follows z is 7. Because each term is 6 more than the previous term, z must be 6 less than 7. Therefore, z = 7 – 6 = 1. In the same way, y is 6 less than z and x is 6 less than y: y = 1 – 6 = –5 and x = –5 – 6 = –11. The sum of x + z is equal to –11 + 1 = –10.

  27. The first term in the sequence is 2. The next term in the sequence, a, is more than b is more than a, c is more than d is more than c, . Add the values of a, b, c, and

  28. The fourth term in the sequence is –24. You are looking for the seventh term, which is three terms after the fourth term. You must multiply by –2 three times, so the seventh term will be (–2)3 = –8 times –24: (–24)(–8) = 192. Because the number of terms is reasonable, you can check your answer by repeatedly multiplying by –2: (–24)(–2) = 48, (48)(–2) = –96, (–96)(–2) = 192.

  29. Solve the first equation for y in terms of x: 2x + y = 6, y = 6 – 2x. Substitute this expression for y in the second equation and solve for x:

    3 – x + 4x = 12

    3x + 3 = 12

    3x = 9

    x = 3

  30. Add the two equations together. The b terms will drop out, and you can solve for a:

    Substitute –4 for a in the first equation and solve for b:

    5(–4) + 3b = –2

    –20 + 3b = –2

    3b = 18

    b = 6

  31. Solve the first equation for x in terms of y: Substitute this expression for x in the second equation and solve for y:

    2xy = 0

    y2 = 64

    y = –8, y = 8

  32. In the first equation, multiply the (x + 4) term by 3: 3(x + 4) = 3x + 12. Then, subtract 12 from both sides of the equation, and the first equation becomes 3x – 2y = –7. Add the two equations together, reordering them so the variables line up. The y terms will drop out, and you can solve for a:

  33. Solve the second equation for a in terms of b: b + a = 13, a = 13 – b. Substitute this expression for a in the first equation and solve for b:

  34. Solve the second equation for x in terms of y: Substitute this expression for x in the first equation and solve for y:

    3x + 7y = 19

    3(4y) + 7y = 19

    12y + 7y = 19

    19y = 19

    y = 1

  35. In the first equation, multiply the (m + n) term by 2 and add m: 2(m + n) + m = 2m + 2n + m = 3m + 2n. Subtract the second equation from the first equation. The m terms will drop out, and you can solve for n:

  36. In the first equation, multiply the (b + 4) term by –2: –2(b + 4) = –2b – 8. Add 8 to both sides of the equation, and the first equation becomes 9a – 2b = 38. Multiply the second equation by 2 and subtract it from the first equation. The a terms will drop out, and you can solve for b:

  37. Solve the second equation for q in terms of p: 4p – 2q = –14, –2q = –4p – 14, q = 2p + 7. Substitute this expression for q in the first equation and solve for p:

    4pq – 6 = 10

    4p(2p + 7) – 6 = 10

    8p2 + 28p – 6 = 10

    8p2 + 28p – 16 = 0

    2p2 + 7p – 4 = 0

    (2p – 1)(p + 4) = 0

    p + 4 = 0, p = –4

  38. Solve the second equation for b in terms of a: b + 2a = –4, b = –2a – 4. Substitute this expression for b in the first equation and solve for a:

    7(2a + 3(–2a – 4)) = 56

    7(2a + –6a – 12) = 56

    7(–4a – 12) = 56

    –28a – 84 = 56

    –28a = 140

    a = –5

  39. Multiply the first equation by 8 and add it to the second equation. The x terms will drop out, and you can solve for y:

  40. Subtract the like terms by subtracting the coefficients of the terms: 9a – 5a = 4a. Because 4a and 12a2 are not like terms, they cannot be combined any further; 9a + 12a2 – 5a = 12a2 + 4a.

  41. Multiply the coefficients of the terms in the numerator, and add the exponents of the bases: (3a)(4a) = 12a2. Do the same with the terms in the denominator: 6(6a2) = 36a2. Finally, divide the numerator by the denominator. Divide the coefficients of the terms and subtract the exponents of the bases:

  42. The terms 5a and 7b have unlike bases; they cannot be combined any further. Add the terms in the denominator: b + 2b = 3b. Divide the b term in the numerator by the 3b in the denominator:

  43. Multiply 2x2 and 4y2 by multiplying the coefficients of the terms: (2x2)(4y2) = 8x2y2. Because 8x2y2 and 6x2y2 have like bases, they can be added. Add the coefficients: 8x2y2 + 6x2y2 = 14x2y2.

  44. To solve the equation, add 12 to both sides of the equation: a – 12 = 12, a – 12 + 12 = 12 + 12, a = 24.

  45. To solve the inequality, divide both sides of the inequality by 6:

  46. To solve the equation, subtract 10 from both sides of the equation: x + 10 = 5, x + 10 – 10 = 5 – 10, x = –5.

  47. To solve the equation, multiply both sides of the equation by 8:

  48. To solve the inequality, divide both sides of the inequality by –3: Remember, when multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality symbol: n > –4.

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