inverse variation

Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1

(For help, go to Lesson 5-5.)

Suppose y varies directly with x. Find each constant of variation.

1. y = 5x 2. y = –7x 3. 3y = x 4. 0.25y = x

Write an equation of the direct variation that includes the given point.

5. (2, 4) 6. (3, 1.5) 7. (–4, 1) 8. (–5, –2)

12-1


1. y = 5x; constant of variation = 5

2. y = –7x; constant of variation = –7

3. 3y = x

• 3y = • x

y = x; constant of variation =

4. 0.25y = x

y = x

4 • y = 4 • x

y = 4x; constant of variation = 4

13

1313

13

1414

Solutions

12-1


Solutions (continued)

5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x.

6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x.

7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – and y = – x.

8. Point (–5, –2) in y = kx: –2 = k(–5), so k = and y = x.

14

14

25

25

12-1

ALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1

Inverse VariationInverse Variation

Suppose y varies inversely with x, and y = 9 when x = 85.

Write an equation for the inverse variation.

xy = k Use the general form for an inverse variation.

(8)(9) = k Substitute 8 for x and 9 for y.

72 = k Multiply to solve for k.

xy = 72 Write an equation. Substitute 72 for k in xy = k.

The equation of the inverse variation is xy = 72 or y = .72x

12-1



The points (5, 6) and (3, y) are two points on the graph of an

inverse variation. Find the missing value.

x1 • y1 = x2 • y2 Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation.

5(6) = 3y2 Substitute 5 for x1, 6 for y1, and 3 for x2.

30 = 3y2 Simplify.

10 = y2 Solve for y2.

The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6).

12-1



Jeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If

Tracy weighs 93 pounds, how far from the fulcrum should she sit in

order to balance the lever?

Relate: A weight of 130 lb is 5 ft from the fulcrum.A weight of 93 lb is x ft from the fulcrum.Weight and distance vary inversely.

Define: Let weight1 = 130 lbLet weight2 = 93 lbLet distance1 = 5 ftLet distance2 = x ft

12-1



(continued)

650 = 93x Simplify.

Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever.

Write: weight1 • distance1 = weight2 • distance2

130 • 5 = 93 • x Substitute.

6.99 = x Simplify.

= x Solve for x.65093

12-1


Decide if each data set represents a direct variation or an

inverse variation. Then write an equation to model the data.

x y

3 10

5 6

10 3

a. The values of y seem to vary inversely with the values of x.

Check each product xy.

xy: 3(10) = 30 5(6) = 30 10(3) = 30

The product of xy is the same for all pairs of data.

So, this is an inverse variation, and k = 30.

The equation is xy = 30.

12-1


(continued)

x y

2 3

4 6

8 12

b. The values of y seem to vary directly with the values of x.

So, this is a direct variation, and k = 1.5.

The equation is y = 1.5x.

64 = 1.5 = 1.5

128

yx = 1.5

32

The ratio is the same for all pairs of data.yx

Check each ratio .yx

12-1


Explain whether each situation represents a direct variation

or an inverse variation.

b. The cost of a $25 birthday present is split among several friends.

Since the total cost is a constant product of $25, this is an inverse variation.

The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs.

Since the ratio is constant at $10 each, this is a direct variation. cost souvenirs

12-1

The cost per person times the number of people equals the total cost of the gift.

a. You buy several souvenirs for $10 each.


pages 640–642 Exercises

1. xy = 18

2. xy = 2

3. xy = 56

4. xy = 1.5

5. xy = 24

6. xy = 7.7

7. xy = 2

8. xy = 0.5

9. xy = 0.06

10. 8

11. 15

12. 6

13. 7

14. 3

15. 130

16. 12

17. 96

18. 3125

19. 2

20.

21. 20

22. 3 h

23. 13.3 mi/h

24. direct variation; y = 0.5x

25. inverse variation; xy = 60

26. inverse variation; xy = 72

27. Direct variation; the ratio is constant at $1.79.

28. Inverse variation; the total number of slices is constant at 8.

29. Inverse variation; the product of the length and width remains constant with an area of 24 square units.

30. 32; xy = 32

31. 1.1; rt = 1.1

32. 2.5; xy = 2.5

cost pound

12-1

16


33. 1; ab = 1

34. 15.6; pq = 15.6

35. 375; xy = 375

36. Direct variation; the ratio of the perimeter to the side length is constant at 3.

37. Inverse variation; the product of the rate and the time is always 150.

38. Direct variation; the ratio of the circumference to the radius is constant at 2 .

39. 121 ft

40. 2.4 days

41. direct variation; y = 0.4x; 8

42. direct variation; y = 70x; 0.9

43. inverse variation; xy = 48; 0.5

44. a. greaterb. greaterc. less

45. a. 16 h; 10 h; 8 h; 4 hb. hr worked, rate of payc. rt = 80

46. Check students’ work.

47. 10.2 L

48. p: y = 0.5x; q: xy = 8

49. a. y is doubled.b. y is halved.

12-1


50. 4; s d = sd2 = k, so s = 4 .

51. a. x4y = k

b. = k

52. C

53. F

54. [2] Direct variation: y = kx,

10 = 5k, k = 2. So when x = 8,

y = 2 • 8 = 16. Inverse variation:

xy = k, 5 • 10 = 50,

So when x = 8, y = , or 6.25.

[1] no work shown OR

one computational error

x4yz

12

2

508

55. [4] a.

b. The variables speed and time are inversely related.

c. 2 h

[3] one computational error[2] one part missing[1] two parts missing

56.

57.

12

8 17

1517

12-1

14

kd2

58.

59.

60.

61.

62. 5.1

63. 12.0

64. 12.0

65. 10.6

66. 2.2

67. 2.5


8 17

8 15

158

12-1

68. (3a – 1)(a + 4)

69. (5x + 2)(3x + 7)

70. (2y – 3)(y + 8)

1517


1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y.

2. Find the constant of variation k for the inverse variation where a = 2.5 when b = 7.

4. Tell whether each situation represents a direct variation or an inverse variation.a. You buy several notebooks for $3 each.

b. The $45 cost of a dinner at a restaurant is split among several people.

3. Write an equation to model the data and complete the table.

0.5

17.5

direct variation

Inverse variation

x y

1

2

6

131619

xy =13 3

1 18

12-1

Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2

(For help, go to Lessons 5-3, 8-7, and 10-1.)

Evaluate each function for x = –2, 0, 3.

1. ƒ(x) = x – 8 2. g(x) = x2 + 4 3. y = 3x

Graph each function.

4. ƒ(x) = 2x + 1 5. g(x) = –x2 6. y = 2x

12-2


1. ƒ(x) = x – 8 for x = –2, 0, 3:ƒ(–2) = –2 – 8 = –10ƒ(0) = 0 – 8 = –8ƒ(3) = 3 – 8 = –5

2. g(x) = x2 + 4 for x = –2, 0, 3:g(–2) = (–2)2 + 4 = 4 + 4 = 8g(0) = 02 + 4 = 0 + 4 = 4g(3) = 32 + 4 = 9 + 4 = 13

3. y = 3x for x = –2, 0, 3:

y = 3–2 = = =

y = 30 = 1

y = 33 = 3 • 3 • 3 = 9 • 3 = 27

1 32

1 3 • 3

19

Solutions

12-2



4. ƒ(x) = 2x + 1 5. g(x) = – x2

6. y = 2x

12-2


The function t = models the time it will take you to travel

70 miles at different rates of speed. Graph this function.

70r

r t 15 4.67 20 3.5 30 2.33 40 1.75 60 1.17

Step 1: Make a table of values. Step 2: Plot the points.

12-2


Identify the vertical asymptote of y = . Then graph the

function.

Step 1: Find the vertical asymptote.

x – 3 = 0 The numerator and denominator have no common factors. Find any value(s) where the denominator equals zero.

x = 3 This is the equation of the vertical asymptote.

12-2

4 x – 3


(continued)

Step 2: Make a table of values. Use values of x near 3, the asymptote.

43–

43

x y 2 –4 1 –2 0 –1 –1 4 4 5 2 6

Step 3: Graph the function.

12-2

Graphing Rational FunctionsGraphing Rational Functions

Identify the asymptotes of y = + 3. Then graph the

function.

4 x + 4


Step 1: From the form of the function, you can see that there is a vertical asymptote at x = –4 and a horizontal asymptote at y = 3. Sketch the asymptotes.

Step 2: Make a table of values using values of x near –4.

13

x y–10 2 –8 2 –6 1 –5 –1 –3 7 –2 5 –1 4 2 3

1323

12-2

Graphing Rational FunctionsGraphing Rational Functions

(continued)


Step 3: Graph the function.

12-2


Describe the graph of each function.

b. y = 6x

The graph is of exponential growth.

c. y = 6x2

The graph is a parabola with axis of symmetry at x = 0.

a. y = x6

The graph is a line with slope and y-intercept 0.16

d. y = | x – 6|

The graph is an absolute value function with a vertex at (6, 0).

12-2


(continued)

f. y = x – 6

The graph is the radical function y = x shifted right 6 units.

g. y = 6x

The graph is a line with slope 6 and y-intercept 0.

The graph is a rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 0.

e. y = 6x

12-2



1.

2.

3.

4.

5.

6. 0

7. 2

8. –2

9. 2

10. x = 2, y = 0

11. x = –1, y = 0

12. x = 1, y = –1

13. x = 0, y = 2

12-2


14. x = 0;

15. x = 0;

16. x = –1;

17. x = 5;

18. x = –4;

19. x = –4;

12-2


20. x = 0, y = –5;

21. x = 0, y = 5;

22. x = 0, y = –6;

23. x = –1, y = 4;

24. x = 3, y = –5;

25. x = 1, y = –2;

12-2


26. line with slope 4, y-int. 1

27. absolute value function with vertex (4, 0)

28. exponential decay

29. line with slope , y-int. 0

30. rational function, with asymptotes x = 0, y = 1

31. radical function; y = x shifted right 4, up 1

32. parabola with axis of symmetry x = 0

33. rational function with asymptotes x = –4, y = –1

34. parabola with axis of symmetry x = –

35. moves graph 1 unit to the left

14

14

36. moves graph 3 units to the right

37. lowers graph 15 units

38. moves graph 12 units left

39. moves the graph up 12 units

40. moves the graph left 3 units

41. moves the graph down 2 units

42. moves the graph 3 units left and 2 units down

12-2


43. x = 0, y = 0;

44. x = 0, y = 0;

45. x = –4, y = 0;

46. x = 0, y = 1;

47. x = –1, y = 4;

48. x = –1, y = –3;

12-2


12-2

49. x = 1, y = 3;

50. x = –5, y = 1;

51. x = 3, y = –2;

52. Answers may vary. Sample: ƒ(x) = + 3, g(x) =

53. 17.8 lumens; 1.97 lumens54. a.

b. x = 0, y = 0; x = 0, y = 0

c. y is any real number except 0; y > 0.

1x

1x


55. a.

d 40

b. 16; 1600; 160,000

c. The signal is extremely strong when you are in the immediate vicinity of a transmitter and it will interfere with the other station.

56. The graph of y = and y = – are both composed of

two curves with asymptotes x = 0 and y = 0. The graph of

y = – is a reflection of the graph of y = over the y-axis.

3x

3x

3x

3x

57.

58.

12-2

>–


59.

60.

61. a. x = –3, y = –2

b. y = – 2

62.

No; ƒ(x) = is equivalent to

g(x) = x + 1 for all values except x = –2.

63. C

64. I

65. A

66. C

(x + 2)(x + 1)x + 2

12-2

1x + 3


67. [2] a. The graph of g(x) = + 5

is a translation of ƒ(x) =

5 units up and 1 unit right.

b. x = 1 and y = 5[1] one part answered correctly

68. xy = 21

69. xy = 16

70. xy = 22

71. xy = 21.08

72. 0

73. 1

74. 2

75. 3(d – 6)(d + 6)

76. 2(m – 12)(m + 5)

77. (t 2 + 3)(t – 1)

4 x – 1

4x

12-2


12-2

3. Describe the graph of each function.

1. Identify the vertical asymptote of y = . Then graph the function. 3 x + 2

2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function.

1 x – 2

a. y = x + 3

b. y =

c. y = + 2

d. y = 3x2

x3

3x


1. Identify the vertical asymptote of y = . Then graph the function. 3 x + 2

2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function.

1 x – 2

x = –2

x = 2; y = –3

12-2


3. Describe the graph of each function.

a. y = x + 3

b. y =

c. y = + 2

d. y = 3x2

x3

3x

radical function y = x shifted left 3 units

line with slope and y-intercept 013

rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 2

parabola with axis of symmetry at x = 0.

12-2

Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3

(For help, go to Lessons 9-5 and 9-6.)

Write each fraction in simplest form.

1. 2. 3.82

1524–

2535

Factor each quadratic expression.

4. x2 + x – 12 5. x2 + 6x + 8 6. x2 – 2x – 15

7. x2 + 8x + 16 8. x2 – x – 12 9. x2 – 7x + 12

12-3

25353. = =

5 • 55 • 7

57

2. = – = –


Solutions

821. = 8 ÷ 2 = 4 15

24–3 • 53 • 8

58

4. Factors of –12 with a sum of 1: 4 and –3.x2 + x – 12 = (x + 4)(x – 3)

5. Factors of 8 with a sum of 6: 2 and 4.x2 + 6x + 8 = (x + 2)(x + 4)

6. Factors of –15 with a sum of –2: 3 and –5.x2 – 2x – 15 = (x + 3)(x – 5)

12-3



7. Factors of 16 with a sum of 8: 4 and 4.x2 + 8x + 16 = (x + 4)(x + 4) or (x + 4)2

8. Factors of –12 with a sum of –1: 3 and –4.x2 – x – 12 = (x + 3)(x – 4)

9. Factors of 12 with a sum of –7: –3 and –4.x2 – 7x + 12 = (x – 3)(x – 4)

12-3


Simplify .3x + 9x + 3

= 3 Simplify.

Factor the numerator. The denominator cannot be factored.

3x + 9x + 3

3(x + 3)x + 3=

Divide out the common factor x + 3.3(x + 3)x + 3

1

1=

12-3


Simplify .4x – 20

x2 – 9x + 20

4x – 20x2 – 9x + 20

4(x – 5)(x – 4) (x – 5)= Factor the numerator and the denominator.

Divide out the common factor x – 5.4(x – 5)(x – 4) (x – 5)

11=

4x – 4= Simplify.

12-3


Simplify .3x – 2781 – x2

3x – 2781 – x2

3(x – 9)(9 – x) (9 + x)= Factor the numerator and the denominator.

39 + x

= – Simplify.

3(x – 9)– 1 (x – 9) (9 + x) Factor –1 from 9 – x.=

3(x – 9)– 1 (x – 9) (9 + x)

1

1Divide out the common factor x – 9.=

12-3


The baking time for bread depends, in part, on its size and

shape. A good approximation for the baking time, in minutes, of a

cylindrical loaf is , or , where the radius r and the length

h of the baked loaf are in inches. Find the baking time for a loaf that is

8 inches long and has a radius of 3 inches. Round your answer to the

nearest minute.

60 • volumesurface area

30rhr + h

30rhr + h

30 (3) (8)3 + 8= Substitute 8 for r and 3 for h.

Simplify.72011=

Round to the nearest whole number.65

The baking time is approximately 65 minutes.

12-3


12-3


1.

2.

3.

4.

5. 3x

6.

7.

8.

9.

2a + 34

1 7x1312

x + 2x2

23 2

b + 4 1 m – 7

10.

11.

12.

13.

14. b + 3

15.

16. –1

17.

18. –2

19. –

w w – 7a + 1

5m + 3m + 2c – 4c + 3

1 m – 2

–4 t + 1

12

20. –

21. –

22. 36 min

23. 13 min

24. 13 min

25.

26.

27.

28.

29.

1 w – 4

2r – 1r + 5

1 v + 5

7z + 2z – 1

4a2

2a – 1

5t – 43t – 1

3(z + 4)z3


30.

31. –

32.

33.

34. Answers may vary. Sample:

35. a. i. ii.

b. ;

36. The student canceled terms instead of factors.

37. –3 is not in the domain of .

2s + 1s2

2a + 1a + 3

4 + 3mm – 7

–c(3c + 5)5c + 4

3 (x – 2)(x + 3)

2b + 4hbh

2h + 2rrh

49

49

x2 – 9x + 3

38.

39.

40.

41.

42.

43.

44.

45. sometimes

46. sometimes

47. never

48. C

5w 5w + 614

3y4(y + 4)

t + 3 3(t + 2)

a – 3ba + 4b

6v – 7w3v – 2w

49. I

50. B

51. D

52. C

53. [2] The student put the 4 in the numerator rather than in the denominator.

=

=

[1] no explanation OR incorrectly simplified expression

x – 5 4(x – 5)

x – 5 4x – 20

14

12-3

m – n m + 10n


12-3

54. vertical asymptote: x = 0; horizontal asymptote: y = 2;

55. vertical asymptote: x = 4; horizontal asymptote: y = 0;

56. vertical asymptote: x = 0; horizontal asymptote: y = –4;

57. 10 2

58. a2b3c4 b

59. 2 2

60.

61. y = x2, y = –2x2, y = 3x2

62. y = x2, y = x2, y = x2

63. y = 0.5x2, y = 2x2, y = –4x2

64. y = –x2, y = 2.3x2, y = –3.8x2

2 5m2

14

13

25


Simplify each expression.

6 – 2xx – 3

1. 2. 3.x2 + 8xx2 – 64

x – 636 – x2

4. 5.6x2 – x – 128x2 – 10x – 3

4x2 + xx3

–2x

x – 8–1

x + 6

3x + 44x + 1

4x + 1x2

12-3

Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4



1. r 2 • r 8 2. b3 • b4 3. c7 ÷ c2

4. 3x4 • 2x5 5. 5n2 • n2 6. 15a3 (–3a2)

Factor each polynomial.

7. 2c2 + 15c + 7 8. 15t2 – 26t + 11 9. 2q2 + 11q + 5

12-4


Solutions

1. r 2 • r 8 – r (2 + 8) = r 10 2. b3 • b4 = b(3+4) = b7

3. c7 ÷ c2 = c(7 – 2) = c5 4. 3x4 • 2x5 = (3 • 2)(x4 • x5) = 6x(4 + 5) = 6x9

5. 5n2 • n2 = 5(n2 • n2) = 5n(2 + 2) = 5n4

6. 15a3(–3a2) = 15(–3)(a3 • a2) = –45a(3 + 2) = –45a5

7. 2c2 + 15c + 7 = (2c + 1)(c + 7)Check: (2c + 1)(c + 7) = 2c2 + 14c + 1c + 7

= 2c2 + 15c + 7

8. 15t2 – 26t + 11 = (15t – 11)(t – 1)Check: (15t – 11)(t – 1) = 15t2 – 15t – 11t + 11

= 15t2 – 26t + 11

9. 2q2 + 11q + 5 = (2q + 1)(q + 5)Check: (2q + 1)(q + 5) = 2q2 + 10q + 1q + 5

= 2q2 + 11q + 5

12-4


Multiply.

a. •7y

8y2

7y

8y2• =

56y3

Multiply the numerators and multiply the denominators.

b. •x

x + 5x – 2x – 6

Multiply the numerators and multiply the denominators. Leave the answer in factored form.

xx + 5

x – 2x – 6

x(x – 2)(x + 5) (x – 6)=•

12-4


Multiply and .3x +1

48x

9x2 – 1

3x + 14

8x9x2 – 1• =

3x + 14 •

8x(3x – 1) (3x + 1) Factor denominator.

Divide out the common factors (3x +1) and 4.=

8x(3x – 1) (3x + 1)

1

23x +1

4

1

1•

= Simplify.2x

3x – 1

12-4


Multiply and x2 + 7x + 12.5x + 1

3x + 12

5x + 13x + 12 • (x2 + 7x + 12) =

5x + 13 (x + 4) •

(x + 3) (x + 4)1 Factor.

(5x +1) (x + 3)3= Leave in factored form.

Divide out the common factor x + 4.

5x + 13 (x + 4)

= • (x + 3) (x + 4)1

1

1

12-4


Divide by .x + 8

x2 – 49

=(x + 5) (x + 8)

x – 7(x + 7) (x – 7)

x + 8• Factor.

Leave in factored form.= (x + 5) (x + 7)

Divide out the common factors x + 8 and x – 7.=

(x + 8) (x + 5)x – 7

(x + 7) (x – 7)x + 8•

1

11

1

x2 + 13x +40x – 7

x2 – 49x + 8x + 8

x2 – 49÷ =x2 – 49x + 8•

Multiply by , the

reciprocal of . x + 8 x2 – 49

x2 + 13x +40x – 7

x2 + 13x +40x – 7

12-4


Divide by (8x2 + 16x).x2 + 9x + 14

11x

Factor.1

8x (x + 2)(x + 7) (x + 2)

11x •=

=x + 788x2 Simplify.

Divide out the common factor x + 2.

=1

8x (x + 2)(x + 7) (x + 2)

11x •

1

1

x2 + 9x + 1411x

8x2 + 16x1÷ =

x2 + 9x + 1411x •

18x2 + 16x

Multiply by the reciprocal of 8x2 + 16x

12-4



1.

2.

3.

4.

5.

6.

7.

8.

9.

35x3612t 2

40 3a5

m(m – 2) (m + 2)(m – 1)2x(x – 1)3(x + 1)

12x2

5(x + 1)

2c c – 1

5x4

29t

10.

11.

12.

13. 4(t + 1)(t + 2)

14. 3(2m + 1)(m + 2)

15.

16.

17. –

18.

19.

13123(4x + 1)

x – 1

(x – 1)(x – 2)3

x + 122d – 5

6d 2

1 c2 – 1 1 s + 4

20.

21. 6

22. –

23. –

24.

25.

26.

27.

28.

29. t + 3

x – 1x + 3

1213

2(x + 2)x – 1

n – 3 4n + 53x

1 x + 1

12-4

11 7k – 15


30.

31.

32.

33.

34.

35. The student forgot to rewrite the expression using the reciprocal before canceling.

36. Answers may vary. Sample: • ;

37. 0, 4, –4

38. $88.71

c + 1c – 13t – 5

7t 2

5(2x – 5)x – 5

x – 2x – 3x – 5

x

9(m + 1)3m2

m + 2 3(m + 1)

m + 2m2

39. $132.96

40. a. $100,000b. 360 paymentsc. $599.55d. $215,838

41.

42.

43.

44.

45. She wrote w5 as a fraction so she could easily see what she could cancel.

2m2(m + 2)(m – 1)(m + 4)

x – 2 4(x + 7)

12-4

2 a + 5 r + 3 (r – 1)(r + 1)2


12-4

46. a.

b.

47.

48. 1

49.

50.

51.

52.

53.

54. B

55. G

x(3x + 2) 4(2x + 1)2

x(x – 2)2(x – 1)

m – 2 2m(m – 1)

x 2 4(2x + 1)2

1 (w + 2)(w + 3)

–(2a + 3b)(a + 2b)(5a + b)(2a – 3b)

9m 2(m + 1) 2

56. D

57. G

58. [2] • =

= 3(x + 1)

[1] one computational error OR answer with no work shown

59.

60.

61.

62.

63.

x 2 – 1 x

3x x – 1

(x + 1)(x – 1)(3x)x(x – 1)

b – 52

3 4k73

q4

4 5t 2 – 9

8

x y + 5


12-4

64.

65.

66.

67.

68. 8.2

69. 5.3

70. 5

71. 11

72. 0.2

73. 7.1

1 2a2 – 3

2z + 3z + 1

2c – 92c + 8

76.74.

75.

m2 2 – 3m


Multiply or divide.

7x 2

51514x•

1. 2.

3. 4.

5.

6.

6x + 3x + 6 •

x2 + 9x + 182x + 1

x + 3x + 1 ÷ (x2 + 5x + 6) 4x + 8

3x •9x 2

x + 2

2x + 4x2 + 11x + 18 ÷

x + 1x2 + 14x + 45

(x2 + 12x + 11) •x + 9

x2 + 20x + 99

3x2

3(x + 3)

1(x + 1)(x + 2) 12x

2(x + 5)x + 1

x + 1

12-4

Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5


Write each polynomial in standard form.

1. 9a – 4a2 + 1

2. 3x2 – 6 + 5x – x3

3. –2 + 8t

Find each product.

4. (2x + 4)(x + 3)

5. (–3n – 4)(n – 5)

6. (3a2 + 1)(2a – 7)

12-5

Dividing PolynomialsDividing Polynomials

1. 9a – 4a2 + 1 = –4a2 + 9a + 1

2. 3x2 – 6 + 5x – x3 = –x3 + 3x2 + 5x – 6

3. –2 + 8t = 8t – 2

4. (2x + 4)(x + 3) = (2x)(x) + (2x)(3) + (4)(x) + (4)(3)= 2x2 + 6x + 4x + 12 = 2x2 + 10x + 12

5. (–3n – 4)(n – 5) = (–3n)(n) + (–3n)(–5) + (–4)(n) + (–4)(–5)= –3n2 + 15n – 4n + 20 = –3n2 + 11n + 20

6. (3a2 + 1)(2a – 7) = (3a2)(2a) + (3a2)(–7) + (1)(2a) + (1)(–7)= 6a3 – 21a2 + 2a – 7

Solutions


12-5


Divide (18x3 + 9x2 – 15x) by 3x2.

(18x3 + 9x2 – 15x) 3x2÷ = (18x3 + 9x2 – 15x) • .1

3x2

Multiply by the reciprocal of 3x2.

= + –18x3

3x2

9x2

3x2

15x3x2 Use the Distributive Property.

= 6x1 + 3x0 –5x Use the division rules for exponents.

= 6x + 3 –5x Simplify.

12-5


Divide (5x2 + 2x – 3) by (x + 2)

x + 2 5x2 + 2x – 3 Divide: Think 5x2 ÷ x = 5x.

Step 1: Begin the long division process.

5x

Align terms by their degree. So put 5x above 2x of the dividend.

– 8x – 3 Bring down – 3.

5x2 + 10x Multiply: 5x(x + 2) = 5x2 + 10x. Then subtract.

12-5


(continued)

Step 2: Repeat the process: Divide, multiply, subtract, bring down.

13 The remainder is 13.

The answer is 5x – 8 + .13

x + 2

5x – 8

x + 2 5x2 + 2x – 3

5x2 + 10x

Divide: –8x ÷ x = – 8– 8x – 3

Multiply: – 8(x + 2) = – 8x – 16. Then subtract.– 8x – 16

12-5


The width and area of a rectangle are shown in the figure

below. What is the length?

Since A = w, divide the area by the width to find the length.

The length of the rectangle is (3x2 + 2x + 3) in.

2x – 3 6x3 – 5x2 + 0x – 9 Rewrite the dividend with 0x.

6x3 – 9x2

4x2 + 0x

3x2

4x2 + 6x–6x – 9

+ 2x + 3

–6x – 90

12-5


Divide (–8x – 2 + 6x2) by (–1 + x).

Rewrite –8x – 2 + 6x2 as 6x2 – 8x – 2 and –1 + x as x – 1.Then divide.

The answer is 6x – 2 – .4

x – 1

6x

x – 1 6x2 – 8x – 2

6x2 – 6x

– 2

–2x –2

–2x + 2

–4

12-5



1. x4 – x3 + x2

2. 3x4 –

3. 3c2 + 2c –

4. n2 – 18n + 3

5. 4 –

6. –t 3 + 2t 2 – 4t + 5

7. x – 3

8. 2t + 9 +

9. n – 1

10. y – 3 +

13

2x

16q

11. 3x – 1

12. –2q – 10 +

13. 5t – 50

14. 2w2 + 2w + 5 –

15. b2 – 3b – 1 +

16. c2 –

17. t 2 – 2t – 2

18. n2 – 2n – 21 –

19. (r 2 + 5r + 1) cm

20. (4c2 – 8c + 16) ft

21. b + 12 +

22 2q + 1

3 3b – 1

1 c – 1

8 n + 2

12-5

10 w – 1

1 b + 4

8 y + 2

16 t – 3


12-5

22. a – 1 –

23. 10w – 681 +

24. t –

25. 2x2 + 5x + 2

26. 3q 2 + 2q + 3 +

27. 3x + 2 –

28. c2 + 11c – 15 +

29. 2b2 + 2b + 10 +

30. y2 + 5y + 29 +

31. 28a – 12

32. 5t 3 – 25t 2 + 115t – 575 +

1 2x

8c

12 q – 2

2 a + 4

49,046w + 72

10 b – 1

138 y – 5

9 t + 4

2881t + 5

33. k 2 – 0.3k – 0.4

34. 3s – 8 +

35. –2z2 + 3z – 4 +

36. 6m2 – 24m + 99 –

37. –16c2 – 20c – 25

38. 2r 4 + r 2 – 7

39. t – 1 +

40. z3 – 3z2 + 10z – 30 +

41. a. Answers may vary. Sample: (c3 + 3c2 – 2c – 4); (c + 1)

b. (c3 + 3c2 – 2c – 4) ÷ (c + 1) = c2 + 2c – 4

29 2s + 3

5 z + 1

326 m + 4

2t 2t 3 + 1

88 z + 3


12-5

42. a. y = 2 –

b. Answers may vary. Sample:x –2 –1 0 1 2

y 1

c. vertical asymptote: x = –3horizontal asymptote: y = 2

43. The binomial is a factor of the polynomial if there is no remainder from the division.

1 x + 3 44. m2 + 5m + 4

45. a. d – 2 +

b. d 2 – 2d + 3 –

c. d 3 – 2d 2 + 3d – 4 +

d. Answers may vary. Sample:

d 4 – 2d 3 + 3d 2 – 4d + 5 –

e. d 4 – 2d 3 + 3d 2 – 4d + 5 –

46. 12

47. a. t =

b. t 2 – 7t + 12

48. 2a2b2 – 3ab3 + 5ab2

49. 3x + 2y

32

53

74

95

3 d + 1

4 d + 1

5 d + 1

6 d + 1 6 6 d + 1

dr


12-5

50. 10r 5 + 2r 4 + 5r 2

51. 2b3 – 2b2 + 3

52. a. x – 3 +

b. ƒ(x) = x – 3 +

c. y = x – 3

d.

53. C

54. G

55. B

56. [2] x2 + 3x + 2

2x – 1 2x3 + 5x2 + x – 22x3 – x2

6x2 + x6x2 – 3x

4x – 24x – 2

0x + 2 x2 + 3x + 2

x2 + 2xx + 2

x + 2 0

The width is x + 1.

[1] one computational error OR correct answer with no work shown

10x + 5

10x + 5

x + 1


57. [4] a. 3x + 4 3x + 10

3x + 12 –2

y = 3 –

b. Tables may vary. Sample:

vertical asymptote: x = –4 horizontal asymptote: y = 3;

[3] appropriate methods, but with one computational error[2] no asymptotes OR no graph[1] one part only

2 x + 4

58. n + 2

59.

60.

61.

62. 563. 9.464. 1565. 17.966. 12.067. 16.268. 5.3869. 1770. –12.771. 63.2572. 6.32

(t – 5)(3t + 1)(2t + 11)(2t – 55)(t + 1)(3t)

3c + 82c + 7(x + 5)(x + 4)2

(x + 7)(x + 8)2

12-5


Divide.

1. (x8 – x6 + x4) ÷ x2 2. (4x2 – 2x – 6) ÷ (x + 1)

3. (6x3 + 5x2 + 11) ÷ (2x + 3) 4. (29 + 64x3) ÷ (4x + 3)

x6 – x4 + x2 4x – 6

3x2 – 2x + 3 +2

2x + 3 16x2 – 12x + 9 +2

4x + 3

12-5

Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6

(For help, go to Lessons 1-5 and 906.)


1. 49

29+ 2. 3

757– 3. 1

2 +52–

4. 56

29+ 5. 1

413– 6. 5

1234–

4x9

2x9+7. 7x

12x

12–8. 9. 712y

112y–

Factor each quadratic expression.

10. x2 + 3x + 2 11. y2 + 7y + 12 12. t 2 – 4t + 4

12-6


Solutions

1. 49

29+ =

4 + 29 =

69

3 • 23 • 3= =

23

2. 37

57– =

3 – 57 =

–27 = –

27

3. 12 +

52– =

1 + (– 5)7 =

–42 = –2

4. 56

29+

5 • 36 • 3

2 • 29 • 2+

1518= + =

1918 or

418 1

118=

5. 14

13–

1 • 34 • 3

1 • 43 • 4– = – =

412=

312

3 – 412 =

–112 = –

112

6. 512

34–

3 • 34 • 3– = – =

912=

512

5 – 912 =

–412 = –

13

512

12-6



4x9

2x9+7. =

4x + 2x9 =

6x9 =

3 • 2 • x3 • 3 =

2x3 or

23

x

7x12

x12–8. =

7x – x12 =

6x12 = =

x2 or

12

x6 • x6 • 2

9. 712y

112y– =

7 – 112y = = =

6 • 16 • 2y

612y

12y

10. Factors of 2 with a sum of 3: 1 and 2.x2 + 3x + 2 = (x + 1)(x + 2)

11. Factors of 12 with a sum of 7: 3 and 4.y2 + 7y + 12 = (y + 3)(y + 4)

12. Factors of 4 with a sum of –4: –2 and –2.t2 – 4t + 4 = (t – 2)(t – 2) or (t – 2)2

12-6


Add and .4

x + 32

x + 3

4x + 3

2x + 3+ =

4 + 2x + 3 Add the numerators.

=6

x + 3 Simplify the numerator.

12-6


Subtract from .3x + 5

3x2 + 2x – 84x + 7

3x2 + 2x – 8

3x + 53x2 + 2x – 8

4x + 73x2 + 2x – 8 – = 4x + 7 – (3x + 5)

3x2 + 2x – 8Subtract the numerators.

Use the Distributive Property.4x + 7 – 3x + 5

3x2 + 2x – 8=

Simplify.1

3x – 4=

Simplify the numerator.x + 2

3x2 + 2x – 8=

Factor the denominator. Divide out the common factor x + 2.

x + 2(3x – 4) (x + 2)=

1

1

12-6


Add + .34x

18

4x = 2 • 2 • x Factor each denominator.

8 = 2 • 2 • 2

LCD = 2 • 2 • 2 • x = 8x

Step 1: Find the LCD of and .34x

18

Step 2: Rewrite using the LCD and add.

Simplify numerators and denominators.68x

x8x+=

Add the numerators.=6 + x

8x

Rewrite each fraction using the LCD.34x

18+

2 • 32 • 4x

1 • x8 • x+=

12-6


Add and .7

x + 43

x – 5

Step 1: Find the LCD of x + 4 and x – 5. Since there are no common factors, the LCD is (x + 4)(x – 5).

Step 2: Rewrite using the LCD and add.

7x + 4

3x – 5+ =

7 (x – 5)(x + 4) (x – 5) +

3 (x + 4)(x + 4) (x – 5)

Rewrite the fractions using the LCD.

Simplify each numerator.7x – 35

(x + 4) (x – 5)3x + 12

(x + 4) (x – 5)= +

Add the numerators.7x – 35 + 3x + 12

(x + 4) (x – 5)=

Simplify the numerator.=10x – 23

(x + 4) (x – 5)

12-6

Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational Expressions

The distance between Seattle, Washington, and Miami,

Florida, is about 5415 miles. The ground speed for jet traffic from

Seattle to Miami can be about 14% faster than the ground speed from

Miami to Seattle. Use r for the jet’s ground speed. Write and simplify

an expression for the round-trip air time.

Miami to Seattle time:5415

r time =distance

rate


14% more than a number is 114% of the number.

Seattle to Miami time:54151.14r time =

distancerate

12-6

Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational Expressions

(continued)


An expression for the total time is +5415

r54151.14r

Add the numerators.=115881.14r

5415r

54151.14r+

61731.14r

54151.14r+ Rewrite using the LCD, 1.14r.

Simplify.10165

r=

12-6


12-6


1.

2.

3.

4.

5.

6.

7.

8. –

9.

9 2m 7 6t – 1n + 2n + 3 14 c – 52s2 + 14s2 + 26c – 282c + 7 –3 2 – b

1 t 2 + 1

–2t 2t – 3

10. 1

11. 2

12. –1

13. 2x2

14. 18

15. 7z

16. 35b3c

17.

18.

19.

20.

35 + 6a15a

12 – 2x3x

18 + 20x2 15x8

9 + 2m24m3

21.

22.

23.

24.

25.

26.

27.

28.

29. a. +

b.

c. about 0.8 h

189 – 9n7n3

45 + 36x2 20x2

17m – 47 (m + 2)(m – 7)a2 + 9a + 12 (a + 3)(a + 5)a2 + 12a + 15

4(a + 3)c2 + 7c + 20 (c + 5)(c + 3)4t 2 + 5t + 5

t 2(t + 1) 18a + 3 (2a + 1)(2a – 1)

1r

1 0.7r

177r


12-6

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

–y 2 + 2y + 2 3y + 1

h 2 + h + 1 2t 2 – 7

r – 2k – 69 + p3

–3 – x – zxy 2z

k + 3km2m2

12c – 15aabc

c 3 – a3

abc10x + 15

x + 2–21t + 33

2t – 3 6x (x – 3)(x + 3)2

40.

41. The student added the terms in the denominators.

42. a. + ;

b. + ;

c. Yes; they both represent the time it takes to makea round trip.

43. Answers may vary. Sample:Not always; the numerator may contain a factor of the LCD.

44. Answers may vary.

Sample: , ;

k – 1k – 6

2r

2 1.25r

185r

2d

2 0.8d

9 2d

2w w + 3

3w 2 w – 3

3w3 + 11w2 – 6w (w + 3)(w – 3)

45.

46. 8

47.

48.

49.

50.

51. –

52.

53.

54.

55. D

8x 2 – 1x

– 3x – 5x(x – 5) 32x x – 5x – 5

4x 1 2x(x – 5)

d + 3d + 4

–x3 + 6x2 + 35x – 50(a + 12)(a – 5)

x x + 4

5a – 8 (a + 2)(a – 5)


12-6

56. H

57. D

58. [2] a. + = total time;

= =

=

b. = =

= =

The ride took 1 hours.

[1] one computational error OR no work shown

10r

10 r + 3

10(r + 3) + 10rr(r + 3)

10r + 30 + 10rr(r + 3)

20r + 30r(r + 3)

10(2r + 3)r(r + 3)

10(2(12) + 3)12(12 + 3)

10(24 + 3)12(15)

10(27)180

270180

32

59. + 2x – 1

60. 5b2 – 10b + 30 –

61. 6

62. 3

63. no solution

64. ± 6.9

65. ± 3.9

66. no solution

x2

2 60 b + 2

12


1. Add + .25x

19 2. Add + .

23x + 4

6x + 2

3. Subtract – .8

x2 – 43

x2 – 44. Subtract – .

6y – 7y + 2

2y – 3y + 2

5. Add 8 + . x – 4x + 3

18 + 5x45x

4(5x + 7)(3x + 4)(x + 2)

5x2 – 4

4(y – 1)y + 2

9x + 20x + 3

12-6

Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7


Solve each proportion.

1x

35=1. 3

t52=2. m

327m=3.

Find the LCD of each group of expressions.

4. 34n

12

2n; ; 5. 1

3x25

43x; ; 6. 1

8y1y 2

56; ;

12-7


Solutions

1. 3.2.1x

35=

3x = 5

x = 53 or 1

23

3t

52=

5t = 6

t =65 or 1

15 m = ± 81 = ± 9

m3

27m=

m2 = 81

4. 4n = 2 • 2 • n; 2 = 2; n = n; LCD = 2 • 2 • n = 4n

5. 3x = 3 • x; 5 = 5; 3x = 3 • x; LCD = 3 • 5 • x = 15x

6. 8y = 2 • 2 • 2 • y; y2 = y • y; 6 = 2 • 3; LCD = 2 • 2 • 2 • 3 • y • y = 24y2

12-7


Solve + = .38

4x

142x

38

4x

142x+ =

The denominators are 8, x, and 2x. The LCD is 8x.

8x = 8x38

4x+ 14

2xMultiply each side by the LCD.

8x = 8x38

4x+ 14

2x

1

1

8x1

1 1

4

Use the Distributive Property.

No rational expressions.Now you can solve.

3x + 32 = 56

3x = 24 Subtract 32 from each side.

x = 8 Divide each side by 3, then simplify.

12-7


(continued)

Check:38

48

142(8)+

78

1416

78

78=

12-7


Solve = – 1. Check the solution.6x2

5x

5x – 16

x2x2 x2= Multiply each side by the LCD, x2.

5x – x2 (1)6

x2x2 x2=1

1 1

xUse the Distributive Property.

6 = 5x – x2 Simplify.

x2 – 5x + 6 = 0 Collect like terms on one side.

(x – 3)(x – 2) = 0 Factor the quadratic expression.

x – 3 = 0 or x – 2 = 0 Use the Zero-Product Property.

x = 3 or x = 2 Solve.

12-7


(continued)

Check:632

53 – 1

622

52 – 1

23 =

23

32 =

32

12-7


Renee can mow the lawn in 20 minutes. Joanne can do the

same job in 30 minutes. How long will it take them if they work together?

Define: Let n = the time to complete the job if they work together (in minutes).

Person Work Rate Time Worked Part of(part of job/min.) (min) Job Done

Renee n

Joanne n

1201

30

n20n

30

Relate: Renee’s part done + Joanne’s part done = complete job.

12-7


(continued)

Relate: Renee’s part done + Joanne’s part done = complete job.

n 20

n 30+

Write:n

20n

30 = 1+

60 = 60(1) Multiply each side by the LCD, 60.

3n + 2n = 60 Use the Distributive Property.

5n = 60 Simplify.

n = 12 Simplify.

It will take two of them 12 minutes to mow the lawn working together.

120Check: Renee will do • = of the job, and Joanne will do

of the job. Together, they will do = 1, or the whole job.

121

35

121

25

130• =

35

25+

12-7


Solve = . Check the solution.4x2

1x + 8

4x2

1x + 8=

4(x + 8) = x2(1) Write cross products.

4x + 32 = x2 Use the Distributive Property.

x2 – 4x – 32 = 0 Collect terms on one side.

(x – 8)(x + 4) = 0 Factor the quadratic expression.

x – 8 = 0 or x + 4 = 0 Use the Zero-Product Property.

x = 8 or x = –4 Solve.

12-7


(continued)

Check:4x2

1x + 8=

482

18 + 8

4(–4)2

1–4 + 8

416

14=

4 64

116 =

12-7


Solve = .x + 3x – 1

4x – 1

x + 3x – 1

4x – 1=

(x + 3) (x – 1) = 4(x – 1) Write the cross products.

x2 – x + 3x – 3 = 4x – 4 Use the Distributive Property.

x2 + 2x – 3 = 4x – 4 Combine like terms.

x2 – 2x + 1 = 0 Subtract 4x – 4 from each side.

(x – 1) (x – 1) = 0 Factor.

x – 1 = 0 Use the Zero-Product Property.

x = 1 Simplify.

12-7


(continued)

Check: 1 + 31 – 1

41 – 1

40

40= Undefined! There is no division by 0.

The equation has no solution because 1 makes a denominator equal 0.

12-7



1. –2

2. 3

3. –1

4. 6, –1

5. –

6. –2, 4

7. 1, 4

8. 5

9. 1, 3

10.

11.

13

13163

12. –4

13. –2

14. –1

15. –

16. 1 h

17. 12.7 min

18. 3

19. 10, –10

20. – , 4

21. 4

22. no solution

23. 6

23

32

57

24. –14

25. , 2

26. 3

27. –5, 2

28. –1

29.

30. – , –1

31. 0, 2

32. 12 h

33. a. 32b. Answers may vary. Sample: Cross-multiplying;

I think it's quicker.

12

12

65

12-7


35. (continued)c. Yes; the x-values are solutions

to the original equation since both sides are equal.

36. 66.6Ω

37. 20Ω

38. 3.75Ω

39. 20Ω

40. Answers may vary.

Sample: =

41. 40 mi/h

42. 9

43. 0,

44. –1

2b b + 2

6b 4b + 3

12-7

33 (continued)c. No; it only works for rational

equations that are proportions.

34. a = 4, b = , c = 11, d = –

35. a. y1 = + 1, y2 =

b. (–9.53, 1.07), (–4.16, 1.35), (–1.12, 5.76), (0.81, 10.16)

13

6 x2

(x + 7)2

6

727

12


12-7

45. 1

46. 23.5 min

47. a. 0.80s

b. 50 – s

c. 0.30(50 – s)

d. 0.8s + 0.3(50 – s) = (0.62)(50)

e. 32

f. 32 L of 80% solution and 18 L of 30% solution

48. 11 h

49. B

50. G

51. C

52. [2] = ;

2(3 + x) = 16 + x

6 + 2x = 16 + x

x = 10

=

[1] appropriate method, but with one computational error

53. –

54.

55.

3 + x 16 + x

12

3 + 10 16 + 10

1212

1326

3 x 2y 2z

3h 2 + 2ht + 4h2(t – 2)(t + 2)

–4k – 61 (k – 4)(k + 10)

13


56.

57.

58.

59.

60.

61.

62. –18, –5

63. 8, 11

64. –23, 1

65. –8, 6

66. –13, –4

67. –91, –1

12-7


1. Solve = .

2. Solve = .

3. Solve = .

4. Solve + = .

5. Juanita can wash the car in 30 minutes. Gabe can wash the car in 40 minutes. Working together, how long will it take?

x4x + 3

2x + 1

25

1x

13

–23(2 + x)

12x

12

4x + 3

–2

1

–4

1, 3

17 min17

12-7

Counting Methods and PermutationsCounting Methods and Permutations

You roll a number cube. Find each probability.

1. P(even number) 2. P(prime number)

3. P(a number greater than 5) 4. P(a negative number)

You roll a blue number cube and a yellow number cube.Find each probability.

5. P(blue 1 and yellow 2) 6. P(blue even and yellow odd)



12-8

Counting Methods and PermutationsCounting Methods and Permutations

Solutions

1. P(even number) = P(2, 4, or 6) = =

2. P(prime number) = P(2, 3, or 5) = =

3. P(a number greater than 5) = P(6) =

4. P(a negative number) = = 0

5. P(blue 1 and yellow 2) = P(blue 1) • P(yellow 2) =

6. P(blue even and yellow odd) = P(blue even) • P(yellow odd) =

36

12

36

12

16

06

16

16 =

136•

36

36•

936= =

14


12-8

Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8

Suppose you have three shirts and three ties that coordinate.

Make a tree diagram to find the number of possible outfits you have.

There are nine possible outfits.

Shirts Ties Outfits

Shirt 1Tie1Tie 2Tie 3

Shirt 1, Tie 1Shirt 1, Tie 2Shirt 1, Tie 3





12-8


Suppose there are two routes you can drive to get from

Austin, Texas to Dallas, Texas, and four routes from Dallas, Texas to

Tulsa, Oklahoma. How many possible routes are there from Austin to

Tulsa through Dallas?

There are eight possible routes from Austin to Tulsa.

2 • 4 = 8 Routes from Austin to Tulsa through Dallas

Routes from Austin to Dallas

Routes from Dallas to Tulsa

12-8


In how many ways can 11 students enter a classroom?

There are 11 choices for the first student, 10 for the second, 9 for the third, and so on.

11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 39,916,800 Use a calculator.

There are 39,916,800 possible ways in which the students can enter the classroom.

12-8

Enter the first factor,8.

Use MATH to select nPR in the PRB screen.

Input 5, since there are 5 factors. Press enter.


Simplify 8P5.

Method 1: Use pencil and paper,

8P5 = 8 • 7 • 6 • 5 • 4 The first factor is 8 and there are 5 factors.

= 6720 Simplify.

Method 2: Use a graphing calculator.

8P5 = 6720

12-8


Suppose you use five different letters from the 26 letters of

the alphabet to make a password. Find the number of possible five-

letter passwords.

There are 26 letters in the alphabet. You are finding the number of permutations of 26 letter arranged 5 at a time.

8P5 = 26 • 25 • 24 • 23 • 22 Use a calculator.

= 7,893,600

There are 7,893,600 five-letter passwords in which letters do not repeat.

12-8



1. 10 choicesShirt 1 Tie 1 S1, T1

Tie 2 S1, T2Tie 3 S1, T3Tie 4 S1, T4Tie 5 S1, T5

Shirt 2 Tie 1 S2, T1Tie 2 S2, T2Tie 3 S2, T3Tie 4 S2, T4Tie 5 S2, T5

2. 12 menusMain

Salad Soup Course MenuC SVC

V B SVBS S SVS

C SCCC B SCB

S SCSC CVC

V B CVBC S CVS

C CCCC B CCB

S CCS

12-8


12-8

3. a. 8, 10, 10, 10b. 8,000,000 telephone numbers

4. a. 6b. 12

5. 3,628,800 orders

6. 120 arrangements

7. 1680

8. 3024

9. 360

10. 120

11. 5040

12. 5040

13. 2520

14. 42

15. 5040

16. 12,144

17. 8P6

18. 9P7

19. 8P4

20. a. 2; 2b. 4c. No; number of consonants •

number of vowels = number of vowels • number of consonants.

21. a. 24

b. 1 24


27. a. 260,000 license platesb. 23,920,000 license plates

28. a. Check students’ work.b. Check students’ work.

29. a. 17,576 codesb. 17,526 codes

30. a. 35,152 call lettersb. 913,952 call letters

31. a. 18,278 companiesb. 12,338,352 companiesc. NASDAQ; 12,320,074

more companies

32. a. 2b. 6c. (n – 1)!

12-8

22. a. 24

b.

c. Answers may vary. Sample: No; if someone tries to guess your password, they’ll probably try your name or initials first.

23. 3

24. 2

25. 5

26. a. 10,000b. With repetition; there are more

permutations when repetition is allowed.

1 14,950


33. 7

34. a. 60 numbersb. 6 numbers

c.

b

35. False; if a = 3 and b = 2, then (a – b)! = 1! = 1, but a! – b! = 3! – 2! = 4.

36. 72

37. 16

38. 4

39. 24

40. 3 10

41. 3

42. –8, 6

43. –10, 1

44. –8, 5

45. , –5

46. AB = 7, AC = 5

47. BC 34, AC 54

48. AB 21, BC 5

49. AB 48, AC 7

50. –6 ± 35

51. no solution

52. –4 ± 3 3

12

53. – ±

54. –2 ± 5

55. 5, 13

72

972

12-8

110 910


1. Jeff has five shirts, three pairs of pants, and two ties. How many possible outfits can he make?

2. Three of 18 students are to be selected for the student government positions of president, vice-president, and treasurer. In how many different ways can the three positions be filled?

3. Calculate 6P3.

4. An Italian restaurant offers four different choices of pasta, three choices of sauce, and two choices of meat. How many different dishes can you order that each consist of one pasta choice, one sauce, and one meat?

30 outfits

4896 ways

120

24

12-8

CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9


Evaluate each expression.

1. 5P3 2. 6P3 3. 7P3 4. 7P4

A and B are independent events. Find P(A and B) for the given probabilities.

5. P(A) = , P(B) = 6. P(A) = , P(B) =

7. P(A) = , P(B) = 8. P(A) = 0.35, P(B) = 0.2

13

34

18

59

56

910

12-9

1. 5P3 = 5 • 4 • 3 = 60 2. 6P3 = 6 • 5 • 4 = 120

3. 7P3 = 7 • 6 • 5 = 210 4. 7P4 = 7 • 6 • 5 • 4 = 840

5. P(A and B) = P(A) • P(B) =

6. P(A and B) = P(A) • P(B) =

7. P(A and B) = P(A) • P(B) =

8. P(A and B) = P(A) • P(B) = (0.35)(0.2) = 0.07



Solutions

18

59• =

572

13

34• =

1 • 33 • 4 =

14

56• =

910 =

34

3 • 3 • 5 5 • 2 • 3 • 2

32 • 2=

12-9


Simplify 9C6.

9C6 = 9P6

6P6

Write using permutation notation.

Simplify.= 84

Write the product represented by the notation.

9 • 8 • 7 • 6 • 5 • 4 6 • 5 • 4 • 3 • 2 • 1

=

12-9


Eighteen people enter a talent contest. Awards will be given

to the top ten finishers. How many different groups of ten winners can

be chosen?

The order in which the top ten winners are listed once they are chosen does not distinguish one group of winners from another. You need the number of combinations of 18 potential winners chosen 10 at a time. Evaluate 18C10.

Method 1: Use pencil and paper.

18C10 =18P10

10P10

Write using permutation notation.

= 43,758 Use a calculator.

=18 • 17 • 16 • 15 • 14 • 13 • 12 • 11 • 10 • 9

10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1

12-9


(continued)

Method 2: Use a graphing calculator.

There are 43,758 different ten-person groups of winners that can be chosen from a group of 18 people.

Enter the first factor, 18.

Use to select nCr in the PRB screen.

Input 10, since there are ten factors. Press .

18C10 = 43,758.

12-9


Suppose you have eight new CDs (three rock, two jazz, and

three country). If you choose the CDs at random, what is the

probability that the first one is country and the second one is jazz?

Three of the eight CDs are country.

probability the first choice is country =38

Two of the remaining CDs are jazz.

probability the second choice is jazz =27

P(country, then jazz) =38

27• Multiply the probabilities.

656= Multiply.

=3

28 Simplify.

The probability that you choose country, then jazz is .3

28

12-9


Suppose you have eight red pens and four black pens in a

box. You choose five pens without looking. What is the probability that

all the pens you choose are red?

There are 12 pens in all. Eight of the pens are red.

number of favorable outcomes = 8C5

number of ways to choose 5 pens from the 8 red pens

number of possible outcomes = 12C5

number of ways to choose 5 red pens from 12 possible pens

12-9


(continued)

=56

792 Simplify each expression.

=7

99 Simplify.

The probability that you will chose five red pens is , or about 7%.7

99

Use the definition of probability.

P(5 red pens) =total number of outcomes

number of favorable outcomes

8C5

12C5

= Substitute.

12-9


12-9


1. 1

2. 6

3. 15

4. 20

5. 15

6. 6

7. 28

8. 28

9. 21

10. 21

11. 220

12. 56

13.

14. a. 45

b. 3

c.

d.

15. a. 56

b. 1

c.

d.

16. 10

17. 1

18. 35

19. 4

20. combination, since order is not important

21. permutation, since orderis important

22. a.

b. 6

c. 6

d. 45

2 15

1 15

1 56 5 28

15040


12-9

23. a. Answers may vary. Sample: It is a combination problem because order is not important.

b. 45

c. Yes; Each line segment joins two points and each handshake connects two people.

24. a. 59,280 sequences

b. 1482 sequences

c.

d. Answers may vary. Sample: It is unlikely someone will guess the right sequence with more than 59,000 possibilities.

1 40

25. Answers may vary. Sample: Both permutations and combinations are arrangements of some or all of a group of objects. However, permutations take into account order, and combinations do not.

26. 4

27. 8

28. a. 24

b.

29. Check students’ work.

30. a. 792

b. 36

c.

14

1 22


12-9

31. always

32. sometimes

33. always

34.

35. a. 15b. 6; 3c. 20d. 300

36. a.

b. Answers may vary. Sample:

It is the function ƒ(x) = ,

1 28

x(x – 1) 2

36. (continued)which uses the combination

formula xC2 = for x.

This represents the number of combination groups of 2.

c. Groups can only be made from sets of objects, which means they must be integers.

37. B

38. F

39. C

40. [2] There are 10 possible pizzas; combination; order does not matter;

x! 2!(x – 2)!

49.

50. 5

51. –2

52. 0, 16

53. –6.81, 0.81

54. 0.30, 6.70

55. –5.46, 1.46

56. –1.24, 1.35

57. –6, 3

58. –0.05, 13.38


12-9

40. [2] (continued)

5C2 = = = 10

[1] incorrect explanation ORminor error

41. 12

42. 5040

43. 840

44. 9

45. 23

46. 92,610,000 license plates

47. 2

48. no solution

1 10

12

5P2

5P2

5 • 42 • 1


1. Simplify 30C4.

2. You have just received a box of chocolates. There are three turtles, four caramels, and three chocolate-covered cherries. If you choose two chocolates at random, what is the probability that the first one is caramel and the second is a turtle?

3. There are twelve congressmen who want to be on the same committee. Eight are Republicans and four are Democrats. If three people are chosen for the committee, what is the probability that three Republicans are chosen?

4. A friend has loaned you five books and you plan to read all of them. In how many ways could you read the five books?

27,405

215

1455

120 ways

12-9

Rational Expressions and FunctionsRational Expressions and FunctionsALGEBRA 1 CHAPTER 12ALGEBRA 1 CHAPTER 12

1. 30

2. 7.8

3. –24

4. 48.23

5. D

6. 1.2 h

7. x = 0; y = 0

8. x = 0; y = 0

9. x = 0; y = 3

10. x = 0; y = 3

12-A


12-A

11. Answers may vary. Sample: In direct variation and in inverse variation, the variables are related to each other by a constant. But in direct variation, that number is the ratio of any corresponding pair of input and output values. Distance traveled varies directly with average speed. In inverse variation, that number is the product of any corresponding pair of input and output values. The cost per person of splitting a $14 pizza varies inversely with the number of people who are sharing it.

12.

13.

14.

15.

16. Answers may vary. Sample:

17. 4x + 3 –

18. x3 – 2x2 + 4x – 8

19. 2x3 – 2x2 – x +

20. 2x2 – 5x – 2 +

21. 1 h

22. –6, –2

23. 4

x + 24

1 6x2

14 3w – 5 2c(3c – 1) 3(c + 5)(c – 3)

x + 2 (x – 6)(x – 3)

103x

7 2x – 1

17 3x + 2

12


12-A

24. no solution

25.

26.

27.

28.

29. Permutation; order is important; 870 different pairs.

30. Combination; order is not important; 15 different ways.

31. Combination; order is not important; 15 different pairs of toppings.

32. 15

33. 70

t 2 + 5t + 5t(t + 1)

n + 9 n(n + 1) 1 y + 37b 2 + 6b – 4

3b(b + 2)

34. 5

35. 35

36. 4

37. 20,160

38. 604,800

39. 10

40. 20 different combinations

41. 3391

inverse variation

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