Inverse Variation ALGEBRA 1 LESSON 12-1 (For help, go to Lesson 5-5.) Suppose y varies directly with...
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Transcript of Inverse Variation ALGEBRA 1 LESSON 12-1 (For help, go to Lesson 5-5.) Suppose y varies directly with...
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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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
ALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
Inverse VariationInverse Variation
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
ALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
Inverse VariationInverse Variation
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
ALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
Inverse VariationInverse Variation
(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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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.
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 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
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
8 17
8 15
158
12-1
68. (3a – 1)(a + 4)
69. (5x + 2)(3x + 7)
70. (2y – 3)(y + 8)
1517
Inverse VariationInverse VariationALGEBRA 1 LESSON 12-1ALGEBRA 1 LESSON 12-1
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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
Solutions (continued)
4. ƒ(x) = 2x + 1 5. g(x) = – x2
6. y = 2x
12-2
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
(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
ALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
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)
ALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
Step 3: Graph the function.
12-2
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
pages 648–650 Exercises
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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
14. x = 0;
15. x = 0;
16. x = –1;
17. x = 5;
18. x = –4;
19. x = –4;
12-2
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
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
>–
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 12-2
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
Graphing Rational FunctionsGraphing Rational FunctionsALGEBRA 1 LESSON 12-2ALGEBRA 1 LESSON 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. = – = –
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
Solutions (continued)
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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
12-3
pages 654–656 Exercises
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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
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
Simplifying Rational ExpressionsSimplifying Rational ExpressionsALGEBRA 1 LESSON 12-3ALGEBRA 1 LESSON 12-3
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
(For help, go to Lessons 8-3 and 9-6.)
Simplify each expression.
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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4
pages 659–661 Exercises
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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4
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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4
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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4
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
Multiplying and Dividing Rational ExpressionsMultiplying and Dividing Rational ExpressionsALGEBRA 1 LESSON 12-4ALGEBRA 1 LESSON 12-4
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
(For help, go to Lessons 9-1 and 9-3.)
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
ALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
12-5
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
pages 664–666 Exercises
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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 12-5
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
Dividing PolynomialsDividing PolynomialsALGEBRA 1 LESSON 12-5ALGEBRA 1 LESSON 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.)
Simplify each expression.
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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
Solutions (continued)
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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 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
ALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
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)
ALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
12-6
pages 669–671 Exercises
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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
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)
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
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
Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational ExpressionsALGEBRA 1 LESSON 12-6ALGEBRA 1 LESSON 12-6
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
(For help, go to Lessons 4-1 and 12-6.)
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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
(continued)
Check:38
48
142(8)+
78
1416
78
78=
12-7
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
(continued)
Check:632
53 – 1
622
52 – 1
23 =
23
32 =
32
12-7
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
(continued)
Check:4x2
1x + 8=
482
18 + 8
4(–4)2
1–4 + 8
416
14=
4 64
116 =
12-7
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
pages 675–677 Exercises
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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 12-7
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
Solving Rational EquationsSolving Rational EquationsALGEBRA 1 LESSON 12-7ALGEBRA 1 LESSON 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)
ALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
(For help, go to Lessons 12-8 and 4-6.)
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
ALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
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
Shirt 2Tie1Tie 2Tie 3
Shirt 2, Tie 1Shirt 2, Tie 2Shirt 2, Tie 3
Shirt 3Tie1Tie 2Tie 3
Shirt 3, Tie 1Shirt 3, Tie 2Shirt 3, Tie 3
12-8
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 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.
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
pages 682–685 Exercises
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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
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
Counting Methods and PermutationsCounting Methods and PermutationsALGEBRA 1 LESSON 12-8ALGEBRA 1 LESSON 12-8
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
(For help, go to Lessons 12-8 and 4-6.)
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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
(For help, go to Lessons 12-8 and 4-6.)
Solutions
18
59• =
572
13
34• =
1 • 33 • 4 =
14
56• =
910 =
34
3 • 3 • 5 5 • 2 • 3 • 2
32 • 2=
12-9
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
12-9
pages 689–691 Exercises
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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
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
CombinationsCombinationsALGEBRA 1 LESSON 12-9ALGEBRA 1 LESSON 12-9
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
Rational Expressions and FunctionsRational Expressions and FunctionsALGEBRA 1 CHAPTER 12ALGEBRA 1 CHAPTER 12
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
Rational Expressions and FunctionsRational Expressions and FunctionsALGEBRA 1 CHAPTER 12ALGEBRA 1 CHAPTER 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