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Solution Manual for Precalculus: Graphs and Models, A Right Triangle Approach, 6th Edition

Solution Manual for Precalculus: Graphs and Models, A Right Triangle Approach, 6th Edition is your study companion, providing answers to textbook exercises and reinforcing key concepts.

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    S OLUTIONS M ANUAL J UDITH A. P ENNA A LGEBRA & T RIGONOMETRY G RAPHS AND M ODELS S IXTH E DITION P RECALCULUS G RAPHS AND M ODELS , A R IGHT T RIANGLE A PPROACH S IXTH E DITION Marvin L. Bittinger Indiana University Purdue University Indianapolis Judith A. Beecher David J. Ellenbogen Community College of Vermont Judith A. Penna

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    Contents Just-in-Time Review . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . 333 Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . 387 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . 429 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . 483 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 583 Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 667 Chapter R . . . . . . . . . . . . . . . . . . . . . . . . . . 709

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    Just-in-Time Review 1. Real Numbers 1. Rational numbers: 2 3 , 6, − 2 . 45, 18 . 4, − 11, 3 √ 27, 5 1 6 , − 8 7 , 0, √ 16 2. Rational numbers but not integers: 2 3 , − 2 . 45, 18 . 4, 5 1 6 , − 8 7 3. Irrational numbers: √ 3, 6 √ 26, 7 . 151551555 . . . , − √ 35, 5 √ 3 (Although there is a pattern in 7 . 151551555 . . . , there is no repeating block of digits.) 4. Integers: 6, − 11, 3 √ 27, 0, √ 16 5. Whole numbers: 6, 3 √ 27, 0, √ 16 6. Real numbers: All of them 2. Properties of Real Numbers 1. − 24 + 24 = 0 illustrates the additive inverse property. 2. 7( xy ) = (7 x ) y illustrates the associative property of mul- tiplication. 3. 9( r − s ) = 9 r − 9 s illustrates a distributive property. 4. 11 + z = z + 11 illustrates the commutative property of addition. 5. − 20 · 1 = − 20 illustrates the multiplicative identity prop- erty. 6. 5( x + y ) = ( x + y )5 illustrates the commutative property of multiplication. 7. q + 0 = q illustrates the additive identity property. 8. 75 · 1 75 = 1 illustrates the multiplicative inverse property. 9. ( x + y )+ w = x +( y + w ) illustrates the associative property of addition. 10. 8( a + b ) = 8 a + 8 b illustrates a distributive property. 3. Order on the Number Line 1. 9 is to the right of − 9 on the number line, so it is false that 9 < − 9. 2. − 10 is to the left of − 1 on the number line, so it is true that − 10 ≤ − 1. 3. − 5 = − √ 25, and − √ 26 is to the left of − √ 25, or − 5, on the number line. Thus it is true that − √ 26 < − 5. 4. √ 6 = √ 6, so it is true that √ 6 ≤ √ 6. 5. − 30 is to the left of − 25 on the number line, so it is false that − 30 > − 25. 6. − 4 5 = − 16 20 and − 5 4 = − 25 20 ; − 16 20 is to the right of − 25 20 , so it is true that − 4 5 > − 5 4 . 4. Absolute Value 1. | − 98 | = 98 ( | a | = − a , if a < 0 . ) 2. | 0 | = 0 ( | a | = a , if a ≥ 0 . ) 3. | 4 . 7 | = 4 . 7 ( | a | = a, if a ≥ 0 . ) 4. ∣ ∣ ∣ ∣ − 2 3 ∣ ∣ ∣ ∣ = 2 3 ( | a | = − a, if a < 0 . ) 5. | − 7 − 13 | = | − 20 | = 20, or | 13 − ( − 7) | = | 13 + 7 | = | 20 | = 20 6. | 2 − 14 . 6 | = | − 12 . 6 | = 12 . 6, or | 14 . 6 − 2 | = | 12 . 6 | = 12 . 6 7. | − 39 − ( − 28) | = | − 39 + 28 | = | − 11 | = 11, or | − 28 − ( − 39) | = | − 28 + 39 | = | 11 | = 11 8. ∣ ∣ ∣ ∣ − 3 4 − 15 8 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ − 6 8 − 15 8 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ − 21 8 ∣ ∣ ∣ ∣ = 21 8 , or ∣ ∣ ∣ ∣ 15 8 − ( − 3 4 )∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 15 8 + 6 8 ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ 21 8 ∣ ∣ ∣ ∣ = 21 8 5. Operations with Real Numbers 1. 8 − ( − 11) = 8 + 11 = 19 2. − 3 10 · ( − 1 3 ) = 3 · 1 10 · 3 = 3 3 · 1 10 = 1 · 1 10 = 1 10 3. 15 ÷ ( − 3) = − 5 4. − 4 − ( − 1) = − 4 + 1 = − 3 5. 7 · ( − 50) = − 350 6. − 0 . 5 − 5 = − 0 . 5 + ( − 5) = − 5 . 5 7. − 3 + 27 = 24 8. − 400 ÷ − 40 = 10 9. 4 . 2 · ( − 3) = − 12 . 6

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    2 Just-in-Time Review 10. − 13 − ( − 33) = − 13 + 33 = 20 11. − 60 + 45 = − 15 12. 1 2 − 2 3 = 1 2 + ( − 2 3 ) = 3 6 + ( − 4 6 ) = − 1 6 13. − 24 ÷ 3 = − 8 14. − 6 + ( − 16) = − 22 15. − 1 2 ÷ ( − 5 8 ) = − 1 2 · ( − 8 5 ) = 1 · 8 2 · 5 = 1 · 2 / · 4 2 / · 5 = 4 5 6. Interval Notation 1. This is a closed interval, so we use brackets. Interval no- tation is [ − 5 , 5]. 2. This is a half-open interval. We use a parenthesis on the left and a bracket on the right. Interval notation is ( − 3 , − 1]. 3. This interval is of unlimited extent in the negative direc- tion, and the endpoint − 2 is included. Interval notation is ( −∞ , − 2]. 4. This interval is of unlimited extent in the positive direc- tion, and the endpoint 3 . 8 is not included. Interval nota- tion is (3 . 8 , ∞ ). 5. { x | 7 < x } , or { x | x > 7 } . This interval is of unlimited extent in the positive direction and the endpoint 7 is not included. Interval notation is (7 , ∞ ). 6. The endpoints − 2 and 2 are not included in the interval, so we use parentheses. Interval notation is ( − 2 , 2). 7. The endpoints − 4 and 5 are not included in the interval, so we use parentheses. Interval notation is ( − 4 , 5). 8. The interval is of unlimited extent in the positive direc- tion, and the endpoint 1.7 is included. Internal notation is [1 . 7 , ∞ ). 9. The endpoint − 5 is not included in the interval, so we use a parenthesis before − 5. The endpoint − 2 is included in the interval, so we use a bracket after − 2. Interval notation is ( − 5 , − 2]. 10. This interval is of unlimited extent in the negative direc- tion, and the endpoint √ 5 is not included. Interval nota- tion is ( −∞ , √ 5). 7. Integers as Exponents 1. 3 − 6 = 1 3 6 Using a − m = 1 a m 2. 1 (0 . 2) − 5 = (0 . 2) 5 Using a − m = 1 a m 3. w − 4 z − 9 = z 9 w 4 Using a − m b − n = b n a m 4. ( z y ) 2 = z 2 y 2 Raising a quotient to a power 5. 100 0 = 1 Using a 0 = 1, a = 0 6. a 5 a − 3 = a 5 − ( − 3) = a 5+3 = a 8 Using the quotient rule 7. (2 xy 3 )( − 3 x − 5 y ) = 2( − 3) x · x − 5 · y 3 · y = − 6 x 1+( − 5) y 3+1 = − 6 x − 4 y 4 , or − 6 y 4 x 4 8. x − 4 · x − 7 = x − 4+( − 7) = x − 11 , or 1 x 11 9. ( mn ) − 6 = m − 6 n − 6 , or 1 m 6 n 6 10. ( t − 5 ) 4 = t − 5 · 4 = t − 20 , or 1 t 20 8. Scientific Notation 1. Convert 18,500,000 to scientific notation. We want the decimal point to be positioned between the 1 and the 8, so we move it 7 places to the left. Since 18,500,000 is greater than 10, the exponent must be posi- tive. 18 , 500 , 000 = 1 . 85 × 10 7 2. Convert 0.000786 to scientific notation. We want the decimal point to be positioned between the 7 and the 8, so we move it 4 places to the right. Since 0.000786 is between 0 and 1, the exponent must be nega- tive. 0 . 000786 = 7 . 86 × 10 − 4 3. Convert 0.0000000023 to scientific notation. We want the decimal point to be positioned between the 2 and the 3, so we move it 9 places to the right. Since 0.0000000023 is between 0 and 1, the exponent must be negative. 0 . 0000000023 = 2 . 3 × 10 − 9 4. Convert 8,927,000,000 to scientific notation. We want the decimal point to be positioned between the 8 and the 9, so we move it 9 places to the left. Since 8,927,000,000 is greater than 10, the exponent must be positive. 8 , 927 , 000 , 000 = 8 . 927 × 10 9 5. Convert 4 . 3 × 10 − 8 to decimal notation. The exponent is negative, so the number is between 0 and 1. We move the decimal point 8 places to the left. 4 . 3 × 10 − 8 = 0 . 000000043

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    Just-in-Time Review 3 6. Convert 5 . 17 × 10 6 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 6 places to the right. 5 . 17 × 10 6 = 5 , 170 , 000 7. Convert 6 . 203 × 10 11 to decimal notation. The exponent is positive, so the number is greater than 10. We move the decimal point 11 places to the right. 6 . 203 × 10 11 = 620 , 300 , 000 , 000 8. Convert 2 . 94 × 10 − 5 to scientific notation. The exponent is negative, so the number is between 0 and 1. We move the decimal point 5 places to the left. 2 . 94 × 10 − 5 = 0 . 0000294 9. Order of Operations 1. 3 + 18 ÷ 6 − 3 = 3 + 3 − 3 Dividing = 6 − 3 = 3 Adding and subtracting 2. = 5 · 3 + 8 · 3 2 + 4(6 − 2) = 5 · 3 + 8 · 3 2 + 4 · 4 Working inside parentheses = 5 · 3 + 8 · 9 + 4 · 4 Evaluating 3 2 = 15 + 72 + 16 Multiplying = 87 + 16 Adding in order = 103 from left to right 3. 5[3 − 8 · 3 2 + 4 · 6 − 2] = 5[3 − 8 · 9 + 4 · 6 − 2] = 5[3 − 72 + 24 − 2] = 5[ − 69 + 24 − 2] = 5[ − 45 − 2] = 5[ − 47] = − 235 4. 16 ÷ 4 · 4 ÷ 2 · 256 = 4 · 4 ÷ 2 · 256 Multiplying and dividing in order from left to right = 16 ÷ 2 · 256 = 8 · 256 = 2048 5. 2 6 · 2 − 3 ÷ 2 10 ÷ 2 − 8 = 2 3 ÷ 2 10 ÷ 2 − 8 = 2 − 7 ÷ 2 − 8 = 2 6. 4(8 − 6) 2 − 4 · 3 + 2 · 8 3 1 + 19 0 = 4 · 2 2 − 4 · 3 + 2 · 8 3 + 1 Calculating in the numerator and in the denominator = 4 · 4 − 4 · 3 + 2 · 8 4 = 16 − 12 + 16 4 = 4 + 16 4 = 20 4 = 5 7. 64 ÷ [( − 4) ÷ ( − 2)] = 64 ÷ 2 = 32 8. 6[9 − (3 − 2)] + 4(2 − 3) = 6[9 − 1] + 4(2 − 3) = 6 · 8 + 4( − 1) = 48 − 4 = 44 10. Introduction to Polynomials 1. 5 − x 6 The term of highest degree is − x 6 , so the degree of the polynomial is 6. 2. x 2 y 5 − x 7 y + 4 The degree of x 2 y 5 is 2 + 5, or 7; the degree of − x 7 y is 7 + 1, or 8; the degree of 4 is 0 (4 = 4 x 0 ). Thus the degree of the polynomial is 8. 3. 2 a 4 − 3 + a 2 The term of highest degree is 2 a 4 , so the degree of the polynomial is 4. 4. − 41 = − 41 x 0 , so the degree of the polynomial is 0. 5. 4 x − x 3 + 0 . 1 x 8 − 2 x 5 The term of highest degree is 0 . 1 x 8 , so the degree of the polynomial is 8. 6. x − 3 has two terms. It is a binomial. 7. 14 y 5 has one term. It is a monomial. 8. 2 y − 1 4 y 2 + 8 has three terms. It is a trinomial. 11. Add and Subtract Polynomials 1. (8 y − 1) − (3 − y ) = (8 y − 1) + ( − 3 + y ) = (8 + 1) y + ( − 1 − 3) = 9 y − 4

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    4 Just-in-Time Review 2. (3 x 2 − 2 x − x 3 + 2) − (5 x 2 − 8 x − x 3 + 4) = (3 x 2 − 2 x − x 3 + 2) + ( − 5 x 2 + 8 x + x 3 − 4) = (3 − 5) x 2 + ( − 2 + 8) x + ( − 1 + 1) x 3 + (2 − 4) = − 2 x 2 + 6 x − 2 3. (2 x + 3 y + z − 7) + (4 x − 2 y − z + 8)+ ( − 3 x + y − 2 z − 4) = (2 + 4 − 3) x + (3 − 2 + 1) y + (1 − 1 − 2) z + ( − 7 + 8 − 4) = 3 x + 2 y − 2 z − 3 4. (3 ab 2 − 4 a 2 b − 2 ab + 6)+ ( − ab 2 − 5 a 2 b + 8 ab + 4) = (3 − 1) ab 2 + ( − 4 − 5) a 2 b + ( − 2 + 8) ab + (6 + 4) = 2 ab 2 − 9 a 2 b + 6 ab + 10 5. (5 x 2 + 4 xy − 3 y 2 + 2) − (9 x 2 − 4 xy + 2 y 2 − 1) = (5 x 2 + 4 xy − 3 y 2 + 2) + ( − 9 x 2 + 4 xy − 2 y 2 + 1) = (5 − 9) x 2 + (4 + 4) xy + ( − 3 − 2) y 2 + (2 + 1) = − 4 x 2 + 8 xy − 5 y 2 + 3 12. Multiply Polynomials 1. (3 a 2 )( − 7 a 4 ) = [3( − 7)]( a 2 · a 4 ) = − 21 a 6 2. ( y − 3)( y + 5) = y 2 + 5 y − 3 y − 15 Using FOIL = y 2 + 2 y − 15 Collecting like terms 3. ( x + 6)( x + 3) = x 2 + 3 x + 6 x + 18 Using FOIL = x 2 + 9 x + 18 Collecting like terms 4. (2 a + 3)( a + 5) = 2 a 2 + 10 a + 3 a + 15 Using FOIL = 2 a 2 + 13 a + 15 Collecting like terms 5. (2 x + 3 y )(2 x + y ) = 4 x 2 + 2 xy + 6 xy + 3 y 2 Using FOIL = 4 x 2 + 8 xy + 3 y 2 6. (11 t − 1)(3 t + 4) = 33 t 2 + 44 t − 3 t − 4 Using FOIL = 33 t 2 + 41 t − 4 13. Special Products of Binomials 1. ( x + 3) 2 = x 2 + 2 · x · 3 + 3 2 [( A + B ) 2 = A 2 + 2 AB + B 2 ] = x 2 + 6 x + 9 2. (5 x − 3) 2 = (5 x ) 2 − 2 · 5 x · 3 + 3 2 [( A − B ) 2 = A 2 − 2 AB + B 2 ] = 25 x 2 − 30 x + 9 3. (2 x + 3 y ) 2 = (2 x ) 2 + 2(2 x )(3 y ) + (3 y ) 2 [( A + B ) 2 = A 2 +2 AB + B 2 ] = 4 x 2 + 12 xy + 9 y 2 4. ( a − 5 b ) 2 = a 2 − 2 · a · 5 b + (5 b ) 2 [( A − B ) 2 = A 2 − 2 AB + B 2 ] = a 2 − 10 ab + 25 b 2 5. ( n + 6)( n − 6) = n 2 − 6 2 [( A + B )( A − B ) = A 2 − B 2 ] = n 2 − 36 6. (3 y + 4)(3 y − 4) = (3 y ) 2 − 4 2 [( A + B )( A − B ) = A 2 − B 2 ] = 9 y 2 − 16 14. Factor Polynomials; The FOIL Method 1. 3 x + 18 = 3 · x + 3 · 6 = 3( x + 6) 2. 2 z 3 − 8 z 2 = 2 z 2 · z − 2 z 2 · 4 = 2 z 2 ( z − 4) 3. 3 x 3 − x 2 + 18 x − 6 = x 2 (3 x − 1) + 6(3 x − 1) = (3 x − 1)( x 2 + 6) 4. t 3 + 6 t 2 − 2 t − 12 = t 2 ( t + 6) − 2( t + 6) = ( t + 6)( t 2 − 2) 5. w 2 − 7 w + 10 We look for two numbers with a product of 10 and a sum of − 7. By trial, we determine that they are − 5 and − 2. w 2 − 7 w + 10 = ( w − 5)( w − 2) 6. t 2 + 8 t + 15 We look for two numbers with a product of 15 and a sum of 8. By trial, we determine that they are 3 and 5. t 2 + 8 t + 15 = ( t + 3)( t + 5) 7. 2 n 2 − 20 n − 48 = 2( n 2 − 10 n − 24) Now factor n 2 − 10 n − 24. We look for two numbers with a product of − 24 and a sum of − 10. By trial, we determine that they are 2 and − 12. Then n 2 − 10 n − 24 = ( n + 2)( n − 12). We must include the common factor, 2, to have a factorization of the original trinomial. 2 n 2 − 20 n − 48 = 2( n + 2)( n − 12)

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    Just-in-Time Review 5 8. y 4 − 9 y 3 + 14 y 2 = y 2 ( y 2 − 9 y + 14) Now factor y 2 − 9 y + 14. Look for two numbers with a product of 14 and a sum of − 9. The numbers are − 2 and − 7. Then y 2 − 9 y + 14 = ( y − 2)( y − 7). We must include the common factor, y 2 , in order to have a factorization of the original trinomial. y 4 − 9 y 3 + 14 y 2 = y 2 ( y − 2)( y − 7) 9. 2 n 2 + 9 n − 56 1. There is no common factor other than 1 or − 1. 2. The factorization must be of the form (2 n + )( n + ). 3. Factor the constant term, − 56. The possibilities are − 1 · 56, 1( − 56), − 2 · 28, 2( − 28), − 4 · 16, 4( − 16), − 7 · 8, and 7( − 8). The factors can be written in the opposite order as well: 56( − 1), − 56 · 1, 28( − 2), − 28 · 2, 16( − 4), − 16 · 4, 8( − 7), and − 8 · 7. 4. Find a pair of factors for which the sum of the outer and the inner products is the middle term, 9 n . By trial, we determine that the factorization is (2 n − 7)( n + 8). 10. 2 y 2 + y − 6 1. There is no common factor other than 1 or − 1. 2. The factorization must be of the form (2 y + )( y + ). 3. Factor the constant term, − 6. The possibilities are − 1 · 6, 1( − 6), − 2 · 3, and 2( − 3). The factors can be written in the opposite order as well: 6( − 1), − 6 · 1, 3( − 2) and − 3 · 2. 4. Find a pair of factors for which the sum of the outer and the inner products is the middle term, y . By trial, we determine that the factorization is (2 y − 3)( y + 2). 11. b 2 − 6 bt + 5 t 2 We look for two numbers with a product of 5 and a sum of − 6. By trial, we determine that they are − 1 and − 5. b 2 − 6 bt + 5 t 2 = ( b − t )( b − 5 t ) 12. x 4 − 7 x 2 − 30 = ( x 2 ) 2 − 7 x 2 − 30 We look for two numbers with a product of − 30 and a sum of − 7. By trial, we determine that they are 3 and − 10. x 4 − 7 x 2 − 30 = ( x 2 + 3)( x 2 − 10) 15. Factoring Polynomials; The ac -Method 1. 8 x 2 − 6 x − 9 1. There is no common factor other than 1 or − 1. 2. Multiply the leading coefficient and the constant: 8( − 9) = − 72. 3. Try to factor − 72 so that the sum of the factors is the coefficient of the middle term, − 6. The factors we want are − 12 and 6. 4. Split the middle term using the numbers found in step (3): − 6 x = − 12 x + 6 x 5. Factor by grouping. 8 x 2 − 6 x − 9 = 8 x 2 − 12 x + 6 x − 9 = 4 x (2 x − 3) + 3(2 x − 3) = (2 x − 3)(4 x + 3) 2. 10 t 2 + 4 t − 6 1. Factor out the largest common factor, 2. 10 t 2 + 4 t − 6 = 2(5 t 2 + 2 t − 3) Now factor 5 t 2 + 2 t − 3. 2. Multiply the leading coefficient and the constant: 5( − 3) = − 15. 3. Try to factor − 15 so that the sum of the factors is the coefficient of the middle term, 2. The factors we want are 5 and − 3. 4. Split the middle term using the numbers found in step (3): 2 t = 5 t − 3 t . 5. Factor by grouping. 5 t 2 + 2 t − 3 = 5 t 2 + 5 t − 3 t − 3 = 5 t ( t + 1) − 3( t + 1) = ( t + 1)(5 t − 3) Include the largest common factor in the final factoriza- tion. 10 t 2 + 4 t − 6 = 2( t + 1)(5 t − 3) 3. 18 a 2 − 51 a + 15 1. Factor out the largest common factor, 3. 18 a 2 − 51 a + 15 = 3(6 a 2 − 17 a + 5) Now factor 6 a 2 − 17 a + 5. 2. Multiply the leading coefficient and the constant: 6(5) = 30. 3. Try to factor 30 so that the sum of the factors is the coefficient of the middle term, − 17. The factors we want are − 2 and − 15. 4. Split the middle term using the numbers found in step (3): − 17 a = − 2 a − 15 a . 5. Factor by grouping. 6 a 2 − 17 a + 5 = 6 a 2 − 2 a − 15 a + 5 = 2 a (3 a − 1) − 5(3 a − 1) = (3 a − 1)(2 a − 5) Include the largest common factor in the final factoriza- tion. 18 a 2 − 51 a + 15 = 3(3 a − 1)(2 a − 5)

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    6 Just-in-Time Review 16. Special Factorizations 1. z 2 − 81 = z 2 − 9 2 = ( z + 9)( z − 9) 2. 16 x 2 − 9 = (4 x ) 2 − 3 2 = (4 x + 3)(4 x − 3) 3. 7 pq 4 − 7 py 4 = 7 p ( q 4 − y 4 ) = 7 p [( q 2 ) 2 − ( y 2 ) 2 ] = 7 p ( q 2 + y 2 )( q 2 − y 2 ) = 7 p ( q 2 + y 2 )( q + y )( q − y ) 4. x 2 + 12 x + 36 = x 2 + 2 · x · 6 + 6 2 = ( x + 6) 2 5. 9 z 2 − 12 z + 4 = (3 z ) 2 − 2 · 3 z · 2 + 2 2 = (3 z − 2) 2 6. a 3 + 24 a 2 + 144 a = a ( a 2 + 24 a + 144) = a ( a 2 + 2 · a · 12 + 12 2 ) = a ( a + 12) 2 7. x 3 + 64 = x 3 + 4 3 = ( x + 4)( x 2 − 4 x + 16) 8. m 3 − 216 = m 3 − 6 3 = ( m − 6)( m 2 + 6 m + 36) 9. 3 a 5 − 24 a 2 = 3 a 2 ( a 3 − 8) = 3 a 2 ( a 3 − 2 3 ) = 3 a 2 ( a − 2)( a 2 + 2 a + 4) 10. t 6 + 1 = ( t 2 ) 3 + 1 3 = ( t 2 + 1)( t 4 − t 2 + 1) 17. Equation-Solving Principles 1. 7 t = 70 t = 10 Dividing by 7 The solution is 10. 2. x − 5 = 7 x = 12 Adding 5 The solution is 12. 3. 3 x + 4 = − 8 3 x = − 12 Subtracting 4 x = − 4 Dividing by 3 The solution is − 4. 4. 6 x − 15 = 45 6 x = 60 Adding 15 x = 10 Dividing by 6 The solution is 10. 5. 7 y − 1 = 23 − 5 y 12 y − 1 = 23 Adding 5 y 12 y = 24 Adding 1 y = 2 Dividing by 12 The solution is 2. 6. 3 m − 7 = − 13 + m 2 m − 7 = − 13 Subtracting m 2 m = − 6 Adding 7 m = − 3 Dividing by 2 The solution is − 3. 7. 2( x + 7) = 5 x + 14 2 x + 14 = 5 x + 14 − 3 x + 14 = 14 Subtracting 5 x − 3 x = 0 Subtracting 14 x = 0 The solution is 0. 8. 5 y − (2 y − 10) = 25 5 y − 2 y + 10 = 25 3 y + 10 = 25 Collecting like terms 3 y = 15 Subtracting 10 y = 5 Dividing by 3 The solution is 5. 18. Inequality-Solving Principles 1. p + 25 ≥ − 100 p ≥ − 125 Subtracting 25 The solution set is [ − 125 , ∞ ). 2. − 2 3 x > 6 x < − 3 2 · 6 Multiplying by − 3 2 and reversing the inequality symbol x < − 9 The solution set is ( −∞ , − 9). 3. 9 x − 1 < 17 9 x < 18 Adding 1 x < 2 Dividing by 9 The solution set is ( −∞ , 2). 4. − x − 16 ≥ 40 − x ≥ 56 Adding 6 x ≤ − 56 Multiplying by − 1 and reversing the inequality symbol The solution set is ( −∞ , − 56].

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    Just-in-Time Review 7 5. 1 3 y − 6 < 3 1 3 y < 9 Adding 6 y < 27 Multiplying by 3 The solution set is ( −∞ , 27). 6. 8 − 2 w ≤ − 14 − 2 w ≤ − 22 Subtracting 8 w ≥ 11 Dividing by − 2 and reversing the inequality symbol The solution set is [11 , ∞ ). 19. The Principle of Zero Products 1. 2 y 2 + 42 y = 0 2 y ( y + 21) = 0 2 y = 0 or y + 21 = 0 y = 0 or y = − 21 The solutions are 0 and − 21. 2. ( a + 7)( a − 1) = 0 a + 7 = 0 or a − 1 = 0 a = − 7 or a = 1 The solutions are − 7 and 1. 3. (5 y + 3)( y − 4) = 0 5 y + 3 = 0 or y − 4 = 0 5 y = − 3 or y = 4 y = − 3 5 or y = 4 The solutions are − 3 5 and 4. 4. 6 x 2 + 7 x − 5 = 0 (3 x + 5)(2 x − 1) = 0 3 x + 5 = 0 or 2 x − 1 = 0 3 x = − 5 or 2 x = 1 x = − 5 3 or x = 1 2 The solutions are − 5 3 and 1 2 . 5. t ( t − 8) = 0 t = 0 or t − 8 = 0 t = 0 or t = 8 The solutions are 0 and 8. 6. x 2 − 8 x − 33 = 0 ( x + 3)( x − 11) = 0 x + 3 = 0 or x − 11 = 0 x = − 3 or x = 11 The solutions are − 3 and 11. 7. x 2 + 13 x = 30 x 2 + 13 x − 30 = 0 ( x + 15)( x − 2) = 0 x + 15 = 0 or x − 2 = 0 x = − 15 or x = 2 The solutions are − 15 and 2. 8. 12 x 2 − 7 x − 12 = 0 (4 x + 3)(3 x − 4) = 0 4 x + 3 = 0 or 3 x − 4 = 0 4 x = − 3 or 3 x = 4 x = − 3 4 or x = 4 3 The solutions are − 3 4 and 4 3 . 20. The Principle of Square Roots 1. x 2 − 36 = 0 x 2 = 36 x = √ 36 or x = − √ 36 x = 6 or x = − 6 The solutions are 6 and − 6, or ± 6. 2. 2 y 2 − 20 = 0 2 y 2 = 20 y 2 = 10 y = √ 10 or y = − √ 10 The solutions are √ 10 and − √ 10, or ± √ 10. 3. 6 z 2 = 18 z 2 = 3 z = √ 3 or z = − √ 3 The solutions are √ 3 and − √ 3, or ± √ 3. 4. 3 t 2 − 15 = 0 3 t 2 = 15 t 2 = 5 t = √ 5 or t = − √ 5 The solutions are √ 5 and − √ 5, or ± √ 5. 5. z 2 − 1 = 24 z 2 = 25 z = √ 25 or z = − √ 25 The solutions are 5 and − 5, or ± 5. 6. 5 x 2 − 75 = 0 5 x 2 = 75 x 2 = 15 x = √ 15 or x = − √ 15 The solutions are √ 15 and − √ 15, or ± √ 15.

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    8 Just-in-Time Review 21. Simplify Rational Expressions 1. 3 x − 3 x ( x − 1) The denominator is 0 when the factor x = 0 and also when x − 1 = 0, or x = 1. The domain is the set of all real numbers except 0 and 1. 2. y + 6 y 2 + 4 y − 21 = y + 6 ( y + 7)( y − 3) The denominator is 0 when y = − 7 or y = 3. The domain is the set of all real numbers except − 7 and 3. 3. x 2 − 4 x 2 − 4 x + 4 = ( x + 2)( x − 2) ✏ ( x − 2)( x − 2) ✏ = x + 2 x − 2 4. x 2 + 2 x − 3 x 2 − 9 = ( x − 1)( x +3) ✏ ( x +3) ✏ ( x − 3) = x − 1 x − 3 5. x 3 − 6 x 2 + 9 x x 3 − 3 x 2 = x ( x 2 − 6 x + 9) x 2 ( x − 3) = x /( x − 3) ✏ ( x − 3) x / · x ( x − 3) ✏ = x − 3 x 6. 6 y 2 + 12 y − 48 3 y 2 − 9 y + 6 = 6( y 2 + 2 y − 8) 3( y 2 − 3 y + 2) = 2 · 3 / · ( y + 4)( y − 2) ✏ 3 /( y − 1)( y − 2) ✏ = 2( y + 4) y − 1 22. Multiply and Divide Rational Expressions 1. r − s r + s · r 2 − s 2 ( r − s ) 2 = ( r − s )( r 2 − s 2 ) ( r + s )( r − s ) 2 = ( r − s ) ✏ ( r − s ) ✏ ( r + s ) ✏ · 1 ( r + s ) ✏ ( r − s ) ✏ ( r − s ) ✏ = 1 2. m 2 − n 2 r + s ÷ m − n r + s = m 2 − n 2 r + s · r + s m − n = ( m + n )( m − n ) ✏ ( r + s ) ✏ ( r + s ) ✏ ( m − n ) ✏ = m + n 3. 4 x 2 + 9 x + 2 x 2 + x − 2 · x 2 − 1 3 x 2 + x − 2 = (4 x + 1)( x +2) ✏ ( x +1) ✏ ( x − 1) ✏ ( x +2) ✏ ( x − 1) ✏ (3 x − 2)( x +1) ✏ = 4 x + 1 3 x − 2 4. 3 x + 12 2 x − 8 ÷ ( x + 4) 2 ( x − 4) 2 = 3 x + 12 2 x − 8 · ( x − 4) 2 ( x + 4) 2 = 3( x +4) ✏ ( x − 4) ✏ ( x − 4) 2( x − 4) ✏ ( x +4) ✏ ( x + 4) = 3( x − 4) 2( x + 4) 5. a 2 − a − 2 a 2 − a − 6 ÷ a 2 − 2 a 2 a + a 2 = a 2 − a − 2 a 2 − a − 6 · 2 a + a 2 a 2 − 2 a = ( a − 2) ✏ ( a + 1) ( a ) ✧ (2+ a ) ✏ ( a − 3)( a +2) ✏ ( a ) ✧ ( a − 2) ✏ = a + 1 a − 3 6. x 2 − y 2 x 3 − y 3 · x 2 + xy + y 2 x 2 + 2 xy + y 2 = ( x + y )( x − y )( x 2 + xy + y 2 ) ( x − y )( x 2 + xy + y 2 )( x + y )( x + y ) = 1 x + y · ( x + y )( x − y )( x 2 + xy + y 2 ) ( x + y )( x − y )( x 2 + xy + y 2 ) = 1 x + y · 1 Removing a factor of 1 = 1 x + y 23. Add and Subtract Rational Expressions 1. a − 3 b a + b + a + 5 b a + b = 2 a + 2 b a + b = 2( a + b ) ✏ 1 · ( a + b ) ✏ = 2 2. x 2 − 5 3 x 2 − 5 x − 2 + x + 1 3 x − 6 = x 2 − 5 (3 x + 1)( x − 2) + x + 1 3( x − 2) = x 2 − 5 (3 x + 1)( x − 2) · 3 3 + x + 1 3( x − 2) · 3 x + 1 3 x + 1 = 3( x 2 − 5) + ( x + 1)(3 x + 1) 3(3 x + 1)( x − 2) = 3 x 2 − 15 + 3 x 2 + 4 x + 1 3(3 x + 1)( x − 2) = 6 x 2 + 4 x − 14 3(3 x + 1)( x − 2)

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    Just-in-Time Review 9 3. a 2 + 1 a 2 − 1 − a − 1 a + 1 = a 2 + 1 ( a + 1)( a − 1) − a − 1 a + 1 , LCD is ( a + 1)( a − 1) = a 2 + 1 − ( a − 1)( a − 1) ( a + 1)( a − 1) = a 2 + 1 − a 2 + 2 a − 1 ( a + 1)( a − 1) = 2 a ( a + 1)( a − 1) 4. 9 x + 2 3 x 2 − 2 x − 8 + 7 3 x 2 + x − 4 = 9 x + 2 (3 x + 4)( x − 2) + 7 (3 x + 4)( x − 1) , LCD is (3 x + 4)( x − 2)( x − 1) = 9 x + 2 (3 x +4)( x − 2) · x − 1 x − 1 + 7 (3 x +4)( x − 1) · x − 2 x − 2 = 9 x 2 − 7 x − 2 (3 x +4)( x − 2)( x − 1) + 7 x − 14 (3 x +4)( x − 1)( x − 2) = 9 x 2 − 16 (3 x + 4)( x − 2)( x − 1) = (3 x +4) ✘ (3 x − 4) (3 x +4) ✘ ( x − 2)( x − 1) = 3 x − 4 ( x − 2)( x − 1) 5. y y 2 − y − 20 − 2 y + 4 = y ( y + 4)( y − 5) − 2 y + 4 , LCD is ( y + 4)( y − 5) = y ( y + 4)( y − 5) − 2 y + 4 · y − 5 y − 5 = y ( y + 4)( y − 5) − 2 y − 10 ( y + 4)( y − 5) = y − (2 y − 10) ( y + 4)( y − 5) = y − 2 y + 10 ( y + 4)( y − 5) = − y + 10 ( y + 4)( y − 5) 6. 3 y y 2 − 7 y + 10 − 2 y y 2 − 8 y + 15 = 3 y ( y − 2)( y − 5) − 2 y ( y − 5)( y − 3) , LCD is ( y − 2)( y − 5)( y − 3) = 3 y ( y − 3) − 2 y ( y − 2) ( y − 2)( y − 5)( y − 3) = 3 y 2 − 9 y − 2 y 2 + 4 y ( y − 2)( y − 5)( y − 3) = y 2 − 5 y ( y − 2)( y − 5)( y − 3) = y ( y − 5) ✏ ( y − 2)( y − 5) ✏ ( y − 3) = y ( y − 2)( y − 3) 24. Simplify Complex Rational Expressions 1. x y − y x 1 y + 1 x = x y − y x 1 y + 1 x · xy xy , LCM is xy = ( x y − y x ) ( xy ) ( 1 y + 1 x ) ( xy ) = x 2 − y 2 x + y = ( x + y ) ✏ ( x − y ) ( x + y ) ✏ · 1 = x − y 2. a − b b a 2 − b 2 ab = a − b b · ab a 2 − b 2 = a − b b · ab ( a + b )( a − b ) = a b ✧ ( a − b ) ✏ b / ( a + b )( a − b ) ✏ = a a + b

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    10 Just-in-Time Review 3. w + 8 w 2 1 + 2 w = w · w 2 w 2 + 8 w 2 1 · w w + 2 w = w 3 + 8 w 2 w + 2 w = w 3 + 8 w 2 · w w + 2 = ( w +2) ✏ ( w 2 − 2 w + 4) w / w / · w ( w +2) ✏ = w 2 − 2 w + 4 w 4. x 2 − y 2 xy x − y y = x 2 − y 2 xy · y x − y = ( x + y )( x − y ) ✏ y ✧ x y ✧ ( x − y ) ✏ = x + y x 5. a b − b a 1 a − 1 b = a 2 − b 2 b − a Multiplying by ab ab = ( a + b )( a − b ) b − a = ( a + b )( a − b ) ✏ − 1 · ( a − b ) ✏ = − a − b 25. Simplify Radical Expressions 1. √ ( − 21) 2 = | − 21 | = 21 2. √ 9 y 2 = √ (3 y ) 2 = | 3 y | = 3 y 3. √ ( a − 2) 2 = a − 2 4. 3 √ − 27 x 3 = 3 √ ( − 3 x ) 3 = − 3 x 5. 4 √ 81 x 8 = 4 √ (3 x 2 ) 4 = 3 x 2 6. 5 √ 32 = 5 √ 2 5 = 2 7. 4 √ 48 x 6 y 4 = 4 √ 16 x 4 y 4 · 3 x 2 = 2 xy 4 √ 3 x 2 = 2 xy 4 √ 3 x 2 8. √ 15 √ 35 = √ 15 · 35 = √ 3 · 5 · 5 · 7 = √ 5 2 · 3 · 7 = √ 5 2 · √ 3 · 7 = 5 √ 21 9. √ 40 xy √ 8 x = √ 40 xy 8 x = √ 5 y 10. 3 √ 3 x 2 3 √ 24 x 5 = 3 √ 3 x 2 24 x 5 = 3 √ 1 8 x 3 = 1 2 x 11. √ x 2 − 4 x + 4 = √ ( x − 2) 2 = x − 2 12. √ 2 x 3 y √ 12 xy = √ 24 x 4 y 2 = √ 4 x 4 y 2 · 6 = 2 x 2 y √ 6 13. 3 √ 3 x 2 y 3 √ 36 x = 3 √ 108 x 3 y = 3 √ 27 x 3 · 4 y = 3 x 3 √ 4 y 14. 5 √ 2 + 3 √ 32 = 5 √ 2 + 3 √ 16 · 2 = 5 √ 2 + 3 · 4 √ 2 = 5 √ 2 + 12 √ 2 = (5 + 12) √ 2 = 17 √ 2 15. 7 √ 12 − 2 √ 3 = 7 · 2 √ 3 − 2 √ 3 = 14 √ 3 − 2 √ 3 = 12 √ 3 16. 2 √ 32 + 3 √ 8 − 4 √ 18 = 2 · 4 √ 2 + 3 · 2 √ 2 − 4 · 3 √ 2 = 8 √ 2 + 6 √ 2 − 12 √ 2 = 2 √ 2 17. 6 √ 20 − 4 √ 45 + √ 80 = 6 √ 4 · 5 − 4 √ 9 · 5 + √ 16 · 5 = 6 · 2 √ 5 − 4 · 3 √ 5 + 4 √ 5 = 12 √ 5 − 12 √ 5 + 4 √ 5 = (12 − 12 + 4) √ 5 = 4 √ 5 18. (2 + √ 3)(5 + 2 √ 3) = 2 · 5 + 2 · 2 √ 3 + √ 3 · 5 + √ 3 · 2 √ 3 = 10 + 4 √ 3 + 5 √ 3 + 3 · 2 = 10 + 9 √ 3 + 6 = 16 + 9 √ 3 19. ( √ 8 + 2 √ 5 )( √ 8 − 2 √ 5 ) = ( √ 8 ) 2 − ( 2 √ 5 ) 2 = 8 − 4 · 5 = 8 − 20 = − 12 20. (1 + √ 3) 2 = 1 2 + 2 · 1 · √ 3 + ( √ 3) 2 = 1 + 2 √ 3 + 3 = 4 + 2 √ 3 26. Rationalizing Denominators 1. 4 √ 11 = 4 √ 11 · √ 11 √ 11 = 4 √ 11 11 2. √ 3 7 = √ 3 7 · 7 7 = √ 21 49 = √ 21 √ 49 = √ 21 7 3. 3 √ 7 3 √ 2 = 3 √ 7 3 √ 2 · 3 √ 4 3 √ 4 = 3 √ 28 3 √ 8 = 3 √ 28 2 4. 3 √ 16 9 = 3 √ 16 9 · 3 3 = 3 √ 48 27 = 3 √ 48 3 √ 27 = 3 √ 8 · 6 3 = 2 3 √ 6 3

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    Just-in-Time Review 11 5. 3 √ 30 − 4 = 3 √ 30 − 4 · √ 30 + 4 √ 30 + 4 = 3 √ 30 + 12 ( √ 30) 2 − 4 2 = 3 √ 30 + 12 30 − 16 = 3 √ 30 + 12 14 6. 4 √ 7 − √ 3 = 4 √ 7 − √ 3 · √ 7 + √ 3 √ 7 + √ 3 = 4 √ 7 + 4 √ 3 ( √ 7) 2 − ( √ 3) 2 = 4 √ 7 + 4 √ 3 7 − 3 = 4 √ 7 + 4 √ 3 4 = 4( √ 7 + √ 3) 4 = √ 7 + √ 3 7. 6 √ m − √ n = 6 √ m − √ n · √ m + √ n √ m + √ n = 6( √ m + √ n ) ( √ m ) 2 − ( √ n ) 2 = 6 √ m + 6 √ n m − n 8. 1 − √ 2 √ 3 − √ 6 = 1 − √ 2 √ 3 − √ 6 · √ 3 + √ 6 √ 3 + √ 6 = √ 3 + √ 6 − √ 6 − √ 12 3 − 6 = √ 3 + √ 6 − √ 6 − 2 √ 3 3 − 6 = − √ 3 − 3 = √ 3 3 27. Rational Exponents 1. y 5 / 6 = 6 √ y 5 2. x 2 / 3 = 3 √ x 2 3. 16 3 / 4 = (16 1 / 4 ) 3 = ( 4 √ 16 ) 3 = 2 3 = 8 4. 4 7 / 2 = ( √ 4) 7 = 2 7 = 128 5. 125 − 1 / 3 = 1 125 1 / 3 = 1 3 √ 125 = 1 5 6. 32 − 4 / 5 = ( 5 √ 32 ) − 4 = 2 − 4 = 1 16 7. 12 √ y 4 = y 4 / 12 = y 1 / 3 8. √ x 5 = x 5 / 2 9. x 1 / 2 · x 2 / 3 = x 1 / 2+2 / 3 = x 3 / 6+4 / 6 = x 7 / 6 = 6 √ x 7 = x 6 √ x 10. ( a − 2) 9 / 4 ( a − 2) − 1 / 4 = ( a − 2) 9 / 4+( − 1 / 4) = ( a − 2) 8 / 4 = ( a − 2) 2 11. ( m 1 / 2 n 5 / 2 ) 2 / 3 = m 1 2 · 2 3 n 5 2 · 2 3 = m 1 / 3 n 5 / 3 = 3 √ m 3 √ n 5 = 3 √ mn 5 = n 3 √ mn 2 28. The Pythagorean Theorem 1. a 2 + b 2 = c 2 8 2 + 15 2 = c 2 64 + 225 = c 2 289 = c 2 17 = c 2. a 2 + b 2 = c 2 4 2 + 4 2 = c 2 16 + 16 = c 2 32 = c 2 √ 32 = c 5 . 657 ≈ c 3. a 2 + b 2 = c 2 5 2 + b 2 = 13 2 25 + b 2 = 169 b 2 = 144 b = 12 4. a 2 + b 2 = c 2 a 2 + 12 2 = 13 2 a 2 + 144 = 169 a 2 = 25 a = 5 5. a 2 + b 2 = c 2 ( √ 5) 2 + b 2 = 6 2 5 + b 2 = 36 b 2 = 31 b = √ 31 ≈ 5 . 568

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    4 2  2  4 4 2  2  4 x y (  1, 4) (  3,  5) (0, 2) (4, 0) (2,  2) 2 2 4 2 2 4 2 2 2 4 2 4 y x ( 2 5, 0) (2, 2 4) (1, 4) (0, 3) ( 2 4, 2 2) 4 2  2  4 4  2  4 x y (  5, 1) (2, 3) (0, 1) (5, 1) (2,  1) 2 2 4 2 2 4 2 2 2 4 2 4 y x ( 2 5, 2) (4, 2 3) (0, 2 1) (2, 0) ( 2 1, 2 5) Chapter 1 Graphs, Functions, and Models Exercise Set 1.1 1. Point A is located 5 units to the left of the y -axis and 4 units up from the x -axis, so its coordinates are ( − 5 , 4). Point B is located 2 units to the right of the y -axis and 2 units down from the x -axis, so its coordinates are (2 , − 2). Point Cis located 0 units to the right or left of the y -axis and 5 units down from the x -axis, so its coordinates are (0 , − 5). Point D is located 3 units to the right of the y -axis and 5 units up from the x -axis, so its coordinates are (3 , 5). Point E is located 5 units to the left of the y -axis and 4 units down from the x -axis, so its coordinates are ( − 5 , − 4). Point F is located 3 units to the right of the y -axis and 0 units up or down from the x -axis, so its coordinates are (3 , 0). 2. G: (2 , 1); H: (0 , 0); I: (4 , − 3); J: ( − 4 , 0); K: ( − 2 , 3); L: (0 , 5) 3. To graph (4 , 0) we move from the origin 4 units to the right of the y -axis. Since the second coordinate is 0, we do not move up or down from the x -axis. To graph ( − 3 , − 5) we move from the origin 3 units to the left of the y -axis. Then we move 5 units down from the x -axis. To graph ( − 1 , 4) we move from the origin 1 unit to the left of the y -axis. Then we move 4 units up from the x -axis. To graph (0 , 2) we do not move to the right or the left of the y -axis since the first coordinate is 0. From the origin we move 2 units up. To graph (2 , − 2) we move from the origin 2 units to the right of the y -axis. Then we move 2 units down from the x -axis. 4. 5. To graph ( − 5 , 1) we move from the origin 5 units to the left of the y -axis. Then we move 1 unit up from the x -axis. To graph (5 , 1) we move from the origin 5 units to the right of the y -axis. Then we move 1 unit up from the x -axis. To graph (2 , 3) we move from the origin 2 units to the right of the y -axis. Then we move 3 units up from the x -axis. To graph (2 , − 1) we move from the origin 2 units to the right of the y -axis. Then we move 1 unit down from the x -axis. To graph (0 , 1) we do not move to the right or the left of the y -axis since the first coordinate is 0. From the origin we move 1 unit up. 6. 7. The first coordinate represents the year and the corre- sponding second coordinate represents the number of cities served by Southwest Airlines. The ordered pairs are (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72), and (2014, 96). 8. The first coordinate represents the year and the second coordinate represents the percent of Marines who are women. The ordered pairs are (1960, 1%), (1970, 0.9%), (1980, 3.6%), (1990, 4.9%), (2000, 6.1%), (2011, 6.8%), and (2014, 7.6%). 9. To determine whether ( − 1 , − 9) is a solution, substitute − 1 for x and − 9 for y . y = 7 x − 2 − 9 ? 7( − 1) − 2 ∣ ∣ ∣ − 7 − 2 − 9 ∣ ∣ − 9 TRUE The equation − 9 = − 9 is true, so ( − 1 , − 9) is a solution. Copyright c © 2017

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