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# Algebra II Practice Test (page 2)

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Updated on Oct 27, 2011

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

y = ax2 + bx + c

y = 2x2 – 8x + 1

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

y = 2x2 + (–8)x + 1

Thus, a = 2, b = –8, 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 = 2, y will be:

y = (2)(2)2 – (8)(2) + 1

y = (2)(4) – (2)(8) + 1

= 8 – 16 + 1

= –8 + 1

= –7

Thus, the coordinates of the vertex equal (2,–7).

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

y = ax2 + bx + c

y = 2x2 + 16x + 0

3. Thus, a = 2, b = 16, and c = 0.

The x-coordinate of the vertex is given by:

Substitute in the values from the equation:

Thus, the line of symmetry is x = –4.

When x = –4, y will be:

y = (2)(–4)2 + (16)(–4)

y = (2)(16) + (16)(–4)

= 32 – 64

= –32

The vertex is at (–4,–32).

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

y = ax2 + bx + c

y = 5x2 + 0(x) + 0

Thus, a = 5, b = 0, and c = 0.

The x-coordinate of the turning point, or vertex, of the parabola is given by:

Substitute in the preceding values:

When x = 0, y will be:

y = (5)(02)

y = 0

The vertex is (0,0).

5. Because the rule of the sequence is that each term is nine more than the previous term, to find the value of x, multiply the last term, 36, by , then add 9: (36) = 12, 12 + 9 = 21. In the same way, the value of y is 21+ 9 = 7 + 9 = 16. Therefore, the value of yx = 16 – 21 = –5.

6. Because the rule of the sequence is each term is 20 less than five times the previous term, to find the value of x, add 20 to 0 and divide by 5: . In the same way, the value of y is –20. Therefore, the value of x + y = 4 + –20 = –16.

7. Continue the sequence: 28.5 is the fourth term of the sequence. The fifth term is – 2 = 14.25 – 2 = 12.25. The sixth term is – 2 = 6.125 – 2 = 4.125, the seventh term is – 2 = 2.0625 – 2 = 0.0625. Half of this number minus two will yield a negative value, so the eighth term of the sequence is the first term of the sequence that is a negative number.

8. Because the rule of the sequence is each term is 16 more than –4 times the previous term, to find the value of y, subtract 16 from –80 and divide by –4: . In the same way, the value of x is = –2. Therefore, the value of x + y = –2 + 24 = 22.

9. To isolate the variable, subtract 15 from both sides:

5√c + 15 – 15 = 35 – 15

5√c = 20

Next, divide both sides by 5:

c = 4

Last, square both sides:

(√c)2 = 42

c = 16

10. Factor each term in the numerator: √(a2b =(√(a2))(√b)= ab; √(ab2) =(√a)(√(b2)) = ba. Next, multiply the two radicals. Multiply the coefficients of each radical and multiply the radicands of each (ab)(ba) = abab. The expression is now ab Cancel the √ab terms from the numerator and denominator, leaving ab.

11. First, cube the ab term: (ab)3 = a3b3. Next, raise the fraction to the fourth power. Multiply each exponent of the a and b terms by 4: . To divide b12 by b4, subtract the exponents: . Therefore, .

12. First, cube the 4g2 term. Cube the constant, 4, and multiply the exponent of g, 2, by 3: (4g2)3 = 64g6. Next, multiply 64g6 by g4. Add the exponents of the g terms: (64g6)(g4) = 64g10. Finally, take the square root of 64g10: (64g10) = 8g5, because (8g5)( 8g5) = 64g10.

13. Because (a)2 = a, the value of a is equal to the value of a squared. Therefore, a = 62 = 36.

14. (√p)4 = (p)4. Multiply the exponents: (p)4 = p2. Substitute for q: . A fraction with a negative exponent can be rewritten as the reciprocal of the fraction with a positive exponent:()–2=(–3)2 = 9; p2 = 9, and p = –3 or 3.

15. Substitute for a and 9 for b: ; 1 is raised to the power of 3, but the value of the exponent does not matter; 1 raised to any power is 1.

16. Substitute 20 for n: . Cancel the √5 terms and multiply the fraction by 10: .

17. If a2 = b = 4, then a = 2. Substitute 2 for a and 4 for b:.

18. For xy > 5, rearrange so that the y is by itself on the left: xy > 5 becomes x > 5 + y, which equals x – 5 > y, or y x – 5. For y < x – 5, the slope is 1 and the y-intercept is –5. Start at the y-intercept (0,–5) and go up 1 and over 1 (right) to plot points. This line will be dashed because the symbol is <. The area below this line satisfies the condition.

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

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

21. Rearrange the first equation so that the y is by itself on the left: 6x + 3y < 9 becomes 3y < –6x + 9, 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 down 2 and over 1 (right) to plot points. This line will be dashed because the symbol is <. The area below this line satisfies the condition. Rearrange the second equation so that the y is by itself on the left: –x + y ≤ –4 becomes yx + –4. For yx + –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. This line will be solid because the symbol is ≤. The area below this line satisfies the condition. Shade the area that satisfies both conditions:

22. For y ≥ 6, draw a solid line at y = 6. The area above this line satisfies the given condition. For x + y < 3, rearrange so that the y is by itself on the left: x + y < 3 becomes y < –x + 3. For y < –x + 3, the slope is –1 and the y-intercept is 3. Start at the y-intercept (0,3) and go down 1 and over 1 (right) to plot points. This line will be dashed because the symbol is <. The area below this line satisfies the condition. Shade the area that satisfies both conditions:

23. For x ≤ –2, draw a solid line at x = 2. The area to the left satisfies the condition, so shade it:

24. For y ≥ 0, draw a solid line at y = 0. The area above the line satisfies the condition, so shade it:

25. Plugging in (1,1) gives 9 > 9, which is not a true statement.

Plugging in (2,1) gives 9 > 15, which is not a true statement.

Plugging in (1,2) gives 6 > 9, which is not a true statement.

Plugging in (–1,–2) gives 18 > –3, which IS a true statement.

Plugging in (2,–1) gives 15 > 15, which is not a true statement.

Therefore, (–1,–2) is the only correct solution.

26. 18x6; –3x is squared properly to 9x2 and the exponents 4 and 2 are properly added to 6.
27. –56x3y7; –56 is the proper product of –7 and 8 and the exponents are properly added.
28. 8ab3(3a – 4b); this answer has the correct greatest common factor (GCF) and is properly factored.
29. (x + 3)(x – 4); after FOIL, all three terms would be correct.
30. To solve the equation, subtract 5 from both sides of the equation, then divide by 9: 9a + 5 = –22, 9a + 5 – 5 = –22 – 5, 9a = –27, a = –3.
31. –72p7q7; the product of (–6)2 and –2 is –72, and there are seven factors of both p and q.
32. 2x2 – 9x – 18; this is the result of the FOIL method.
33. 12a2 – 7ab – 12b2; this is the result of the FOIL method.
34. 18a2b2c2 (1 + 3ac); this is the proper factorization using the GCF. By redistributing, the original binomial can be obtained.
35. Choice c is the definition of a function. Choice a is incorrect; if a range value is repeated, the function would not fail the vertical line test. Choice b has the words domain and range reversed. Choice d is incorrect; this guarantees that the relation would not be a function.
36. Choice b is correct; it indicates the correct translation of the inequality. Choice a is the equation 2x – 4 < 7 + (x – 2). Choice c represents 2x – 4 > 7(x – 2), and choice d is the equation 2 + x – 4 < 7(x – 2).
37. (x – 4)(x – 6); This is one possible factoring for the given trinomial.
38. All real numbers greater than or equal to 3; the smallest value possible for |x| is 0, which means that the smallest possible range is equal to 3.
39. Because each term in the sequence is –4 times the previous term, y is equal to = 16, and x = = –4. Therefore, xy = (16)(–4) = –64.
40. Every term in the sequence is 3 raised to a power. The first term, 1, is 30. The second term, 3, is 31. The value of the exponent is one less than the position of the term in the sequence. The 100th term of the sequence is equal to 3100 – 1 = 399 and the 101st term in the sequence is equal to 399 + 1 = 3100. To multiply two terms with common bases, add the exponents of the terms: (399)(3100) = 3199.
41. Because the rule of the sequence is each term is two less than three times the previous term, multiply the last term, –53, by 3, then subtract 2: (–53)(3) = –159 – 2 = –161.
42. d.; because each term in the sequence is equal to the sum of the two previous terms, d= b + c. You know that e = c + d, because c and d are the two terms previous to e. If e = c + d, then, by subtracting c from both sides of the equation, d = ec. In the same way, f = d + e, the terms that precede it, and that equation can be rewritten as d = fe; d = b + c, and c = a +b. Therefore, d = b + (a + b), d = a + 2b. However, d is not equal to e – 2b. Also, d = ec, and c = a + b, not 2b, because a is not equal to b.
43. Solve the second equation for n in terms of m: mn = 0, n = m. Substitute this expression for n in the first equation and solve for m:

m(n + 1) = 2

m (m + 1) = 2

m2 + m = 2

m2 + m – 2 = 0

(m + 2)(m – 1) = 0

m + 2 = 0, m = –2

m – 1 = 0, m = 1

44. Solve the second equation for c in terms of d: c – 6d = 0, c = 6d. Substitute this expression for c in the first equation and solve for d:

= 0

= 0

– 2 = 0

d – 2 = 0

d = 2

Substitute the value of d into the second equation and solve for c:

c – 6(2) = 0

c – 12 = 0

c = 12

Because c = 12 and d = 2, the value of = 6.

45. Divide the second equation by 2 and add it to the first equation. The b terms will drop out, and you can solve for a:

= 3a – 6b = –3

a = 3

Substitute the value of a into the first equation and solve for b:

4(3) + 6b = 24

12 + 6b = 24

6b = 12

b = 2

Because a = 3 and b = 2, the value of a + b = 3 + 2 = 5.

46. Multiply the first equation by –6 and add it to the second equation. The x terms will drop out, and you can solve for y:

–6( – 2y = 14) = –2x + 12y = –84

47. y = –5

Substitute the value of y into the second equation and solve for x:

2x + 6(–5) = –6

2x – 30 = –6

2x = 24

x = 12

Because x = 12 and y = –5, the value of xy = 12 – (–5) = 12 + 5 = 17.

48. First, multiply (x + 2) by 4: 4(x + 2) = 4x + 8. Then, subtract 3x from both sides of the inequality and subtract 8 from both sides of the inequality:

3x – 6 ≤ 4x + 8

3x – 6 – 3x ≤ 4x + 8 – 3x

<

–6 ≤ x + 8

–6 – 8 ≤ x + 8 – 8

x ≥ –14

49. First, combine like terms on each side of the equation: 6x – 4x = 2x and 4 – 9 = –5. Now, subtract 2x from both sides of the equation and add 5 to both sides of the equation:

2x + 9 = 6x – 5

2x – 2x + 9 = 6x – 2x – 5

9 = 4x – 5

9 + 5 = 4x – 5 + 5

14 = 4x

Finally, divide both sides of the equation by 4: .

50. First, multiply (x + 3) by –8 and multiply (–2x + 10) by 2:

–8(x + 3) = –8x – 24,

2(–2x + 10) = –4x + 20.

Then, add 8x to both sides of the inequality and subtract 20 from both sides of the inequality:

–8x – 24 ≤ –4x + 20

–8x – 24 + 8x ≤ –4x + 20 + 8x

–24 ≤ 4x + 20

–24 – 20 ≤ 4x + 20 – 20

–44 ≤ 4x

Finally, divide both sides of the inequality by 4: , x ≥ –11.

51. First, reduce the fraction by dividing the numerator and denominator by 3c: = . Now, subtract 9 from both sides of the equation and then multiply both sides of the equation by 2:

+ 9 = 15

+ 9 – 9 = 15 – 9

= 6

(2)() = (6)(2)

c = 12

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