) 4. Rational and Radical Functions. Solutions Key. (3 x 2 y) Holt McDougal Algebra 2. variation functions. check it out!

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1 CHAPTER 5 Rational and Radical Functions Solutions Ke Are ou read? 1. D. A. B. F 5. C ( ) - (-) ( - ) ( )( - ) ( ) z ( ) z 1 z 9. ( ) ( 5 ) ( )( 5 ) (1) a = a a 1a = a The GCF o a and 1a is a. 1. c d = c c d cd = c d d The GCF o c d and cd is cd = 0 = 5 The GCF o 1 and 0 is. 1. ( - 5)( + 1) 19. ( - )( + ) 0. ( + )( + ) 1. ( + )( + ). ( - ). ( - 10)( + ). 5 = 5 = 9 = ± Ç 9 = ±. ( - ) = ( - ) = 1 - = ± ÇÇ 1 = ± = or = 9 = 100 = 5 = ± ÇÇ 5 = ± variation unctions check it out! 1. = k.5 = = k = P 1 s 1 = P s = 75 s 1s = 75(1.5) s =.5 The side length s is.5 in.. L = krl π = k(.5)(1) π = k The radius r is 1. m.. = k = k 10 k = 0 = L = πrl π = πr(5) 1. = r t = k = 150 v 15 = 50 It would take them 1 hours to build a house. a. inverse b. direct 7. V = kt = 0.05(00) = 0 P 1 The volume is 0 L. think and discuss 1. Possible answer: A direct variation equation is in the orm = m + b, with m = k and b = 0.. Possible answer: The length varies inversel as the width, with a constant o variation o Holt McDougal Algebra

2 . Tpe o Variation Equation Graph Direct = k Eample d = rt 9. = = Joint Inverse = kz = k 0 I = Prt I = V R 11. = - 5 eercises guided practice 1. indirect variation. = - -. = λ 5. 1 v 1 = λ v 0 15 = λ (0) = 15λ 1 = λ The wavelength λ is 1 t. d. 1 t 1 = d t = 17.5 t 11.5t = 17.5(15) t = The time t is h. 7. V = klw = k()() 7 = k The length l is t.. = V = klw 10 = 7l(5) = l. C = mtk 1 = m(50)() 0.0 = m C = mtk = 0.0(0)() = The total cost C is about 10 cents t = d r.75 = d 0 5 = d t = d = 5 r 50 = 5.7 It would take the driver 5.7 h. 1. neither 1. inverse 15. direct 1. P = kd t 17 = k(500) = k P = kd = 1.7(700) = t 0 The power P is kilowatts. - practice and problem solving 17. = 1 1. = = Holt McDougal Algebra

3 d 0. 1 m 1 = d m = d (70) = 55d 17 d The dosage d is 17 mg. C 1. 1 w 1 = C w 5. = C 1.5 5(1.5) =.C C There are Cal in the melon.. N = kap 90 = k(700)(70) 0.0 = k N = kap = 0.0(1000)(75) = 1500 The number o bags N is Q = kmt 090 = k(1)(5) 1 = k The mass m is 0. kg.. = = d = k w = k 0 0 = k d = k = 0 w 1 = 5 It would take 5 das. Q = kmt 7 = 1m(10) 0. = m 5. = inverse 9. neither 0 0. direct 1. V = kt P V = kt 0 = k(0) P 15 = 0.05(0) 1 P 0.05 = k P = 1.75 The pressure P is 1.75 atm.. alwas. sometimes. never 5. alwas. never 7a. s = k t 00 = k = k s = 00 t b. s = 00 t 10 = 00 t 0 = t It takes 0 s to complete one lap.. Answers will var. Possible answer: I(d) = 1150, where I is intensit d in milliwatts per square centimeter and d is distance rom the light source in centimeters 9a. I = kpt 1.5 = k(500) ( 1 ) 0.0 = k I = 0.0Pt b. True Federal Bank c. I = kpt = 0.0(000) ( 1 ) = 0 The interest earned in si months is $ ; -1; = 5; z = ; =.. approimatel 9 min. Possible answer: points are needed to determine a line. Since direct variations alwas include (0, 0), onl 1 other point is needed to write the equation.. Possible answer: I the ratios o the coordinates in each ordered pair are the same, the variation is direct. I the products o the coordinates in each ordered pair are the same, the variation is inverse. test prep 5. D. H 7. D. c = k n 50 = k c = k n 00 = 000 n 000 = k n = 0 The number o students n is 0. challenge and etend 9. = kz 19 = k(7)( 9 ) 1 = k = kz = 1 () ( ) = 50a. r = kp 19 = k(1,1,05) k r p 19 Holt McDougal Algebra

4 b. FL: r (15,9,7) 5 IL: r (1,19,9) 19 MI: r (9,9,) 15 c. r = kp p p 1,000,000 The population p is about 1,000, = 7πz ( = ) 7π multipling and dividing rational epressions check it out! 1a b ( + )( - 1) 9 ; 0 1 ; and 1 c a ( + 1)( + ) ( - )( + ) -( - 5) ; - and - -; b. a ( - ) ( - 1)( - ) ; and 1-1 b a. 10( - ) ( - )( - ) 1 + 5( + ) b ( - )( + 1) - )( - 1) ( ( + )( - ) ( + 1)( - 1) ( - )( + 1) - ) ( ( + )( - ) + 1 ( - ) a. b = -7 ( + ) = 5 ( + )( - ) = 5 ( + )( - ) + = -7 - = = -7 = = - = since -, no solution think and discuss 1. Possible answer: Set the denominator o the unsimpliied epression equal to 0, and solve.. Possible answer: Solving a rational equation ma produce etraneous solutions.. Fractions Rational Epressions Simpliing Multipling Dividing eercises guided practice 1 1 = 1 = = = = 9 = = 1 = 0 = = = 15 5 = 5 1. Possible answer: A rational epression is the quotient o polnomials ( ) ( - ) ( + 5)( - 1) ( - 7)( - 1) - ; + 5 ; 1 and ( + )( - 1) -( + ) ( -5)( + ) 1-1 ; - and ( + )( - 1) -( - 1) + - ; 0 and ( - ) ( - )( + ) -1-5 ; 5 and ( + 1) + 1 ; deined or all real values o ) ( ( - ) ( + )( - ) - ( - ) ( + )( + ) 5 - ( + )( - ) Holt McDougal Algebra

5 ( - 1) ( + 5)( - 5) + )( + 5) ( ( + )( - ) ( - 5)( + ) ( + 5) ( - )( + ) ( + 1) - )( - 1) ( ( - )( + ) ( + 1)( - 1) = - = 10 ( - )( + ) ( + 5)( - ) = - = = = 10 = - = 5 since -, no solution = 1 ( + )( - ) = = 1 = -1 practice and problem solving ( - ) ( - ) ; 0 and ( + )( - ) ( - ) + - ; ( - 1) ( + 5)( - 1) ; -5 and ( + ) -( + )( - ) - ; - and ( + 9) ( + 9)( - 9) ( + 5) -( + 5) - ; -9 and 9-9 -; ( - )( - 1) ( - )( + ) ( - )( + ) + )( + ) ( ( + )( - ) ( + )( - ) ( + ) ( - )( + ) ( - 5) ( - ) ( - ) - 5 ( - 5) ( + )( + ) ( + )( + 1) + 1 ( + ) + ( + ) ( - 5)( + 1) + )( - ) ( ( - 1)( - ) ( - 5)( + ) ( - ) - ( + ) ( + 1)( - ) + )( - ) ( ( +1)( - ) ( + 5)( - ) = - = 5 ( + )( + ) = - ( + )( - ) = 5 -( + ) - -( + ) = - + = 5 + = = - =. + - ( + 7)( - ) = πr ; π(5r) ( + 7)( - ) ( + 7)( - ) = -11 πr = πr 1 = -11 π(5r) no solution. - ( - ) ( - ) 5 π( 5r ) = ( - ) ( + 1)( - 1) Holt McDougal Algebra

6 (5 + 7)(5-7) (5-7) (5 + 7) ( + ) ( + ) ( + 1) + ( + 1) 5 ( + ) ( + )( + 1) ( + 1) - 5 ( - 5) - ( + 1) - ( - 5) - 5 ( + 1)( - ) - - a. square prism: s h = h s 1 ; clinder: πr h = h πr 1 b. square prism: s + sh s(s + h) s + h = = ; s h s(sh) sh clinder: πr + πrh = πr(r + h) = r + h πr h πr(rh) rh (r) + (h) r + h r + h + h c. = r (r)(h) rh rh rh r + h = rh rh r + h = 1 The ratio would be reduced b a actor o 1 1 a. v = d v 0 t + t = at t b. v = v = v at at = + 1 (10)() = 79 The driver s average speed was 79 t/s. 5. Student A; the student didn t leave a 1 in the numerator.. No; it is true or all real numbers ecept =. At this value, the rational epression is undeined. test prep 7. D; + - = 0 ( - 1)( + ) = 0-1 = 0 or + = 0 = 1 or -.. F; 9. A; ( + )( + 5) ( - )( - ) ( + )( + 9) ( - ) ( + )( - ) challenge and etend ( - 1) ( + + 1) ( + )( - ) + ( - 1)( - ) ( + 5)( - 5) ( - 5) ( ) ( + 5)( ) ( - 5) ( ) ( ) ( + )( - ) ( + )( - ) ( + )( - ) - + ( - )( + ) ( - ) - ( - ) ( - ) + ( - ) ( - 1 )( - ) ( - 1) ( + 1)( - 1) ( + 1)( - ) ( + + 1) ( - 1) ( - 1) ( + + 1)( + )( - ) ( + 1)( - 1) ( + 1)( - ) ( + + 1) ( - 1) ( + 1)( + ) + 1 Adding and subtracting rational epressions check it out! 1a ; ± Ç - b ( - - ) ; Holt McDougal Algebra

7 a. 7 = 7 5 = 5 The LCM is b. - = ( + )( - ) = ( + )( + ) The LCM is ( + )( - )( + ). a ( - 1) + - ( - 1) ( - 1) ( ) ( - ) ( - 1) ( - 1) 15 - ( - 1) ; 1 a ( - 1) ( b ) ( + ) ( + ) ) ( ( ) ( - )(5 - ) - ( + 5) ( + 5)(5 - ) ( + 5)(5 - ) ; -5 and ( + 5)(5 - ) 5 b ( - - ) + - ( - )( - ) ( + )( + ) ( + )( - ) + ; a ( + 1)( - 1) ; b ( - 1) c ( 1 )( - ) + ( 1 ) ( - ) ( + - ) ( - ) ( - ) + ( - ) ( + ) ( - ) ( + ). d d 0 + d 5 d(0) d (0) + d 0 5 (0) 70d 9d + d 70d. 17d The average speed is. mi/h. think and discuss 1. Possible answer: Factor each denominator. Find the product o the dierent actors. I the same actor is in both denominators, use the highest power o that actor.. Possible answer: Adding (like denominators): = Rational Epressions Subtracting (unlike denominators): Simpliing a comple raction: - 5 = = = - ( ) - ( - ( ) - ( - ) = ( - ) = ( - ) = ( - + ) ( - ) = ( + 1) ( - 5) - ) 15 Holt McDougal Algebra

8 eercises guided practice 1. Possible answer: A comple raction has 1 or more ractions in its numerator, its denominator, or both.. Possible answer: (5 + ) ; -5 ; ( + ) ; = ( + 5)( - 5) = ( + 5) The LCM is ( + 5) ( - 5) ( - 1-1) ( + + ) ( - )( - 1) + ( - )( + ) ( + )( - 1) ( + )( - 1) ( + )( - 1) 5. = 1 = The LCM is 1. ( + - ) ; - and 1 ( + )( - 1) ( + 1) ( ) ( - 1) ( - - ) ( + 1) - + ( - 1)( - ) ( + )( - ) ( + 1) ; -1 ( + 1) (+ )( - ) ; ± ( + )( - ) ( ) ( - 5-5) ( - 5)( + 1) - ( - 5)( - 5) ( - 5)( + 1) ( - 5)( + 1) ; -1 and 5 ( - 5)( + 1) ( + + ) + - ( + ) ( - )( + ) ( - )( + ) -( + ) ( - )( + ) -1 ; ± ( + 1)( + ) ( ) + - ( + 1)( + 1) ( + )( + 1) ( + )( + 1) ; - and -1 ( + )( + 1) )( - ) - 7 ( ( - )( + ) ( - 7)(5 - ) ( + 5)( - 1) d + d ( ) ( + ) + ( d 7.75 ) ( + ) d(1) ( d (1) + d + ) ( + ) (1) d (+ ) + (+ ) 5d + d d.9 ( + ) 9d The average speed is about.9 t/s ( - ) ; ( - 5) ; ( - 5) + - ; = 1 = 7 The LCM is. 15 Holt McDougal Algebra

9 = ( + 5)( - 5) = ( - 5)( + 1) The LCM is ( - 5)( + 5)( + 1) ( - 1-1) + - 1( + + ) ( - )( - 1) + ( + ) ( + )( - 1) ( + )( - 1) ; - and 1 ( + )( - 1) ( ) + - ( - 7) + ( - ) ( - ) 1-1 ( - ) 7( - ) ( - ) ; ( ) ( 5( + 1) + ( + 1) + 5 ( + 1) ; -1 and 0 ) ( + )( - ) ( + + ) ( - ) - ( - )(+ ) (- )(+ ) ( - )( + ) ( - )( + ) -( + )( - ) ; ± ( - )( + ) ( - - ) ( + + ) ( - ) - ( + 1)( + ) ( - )( + ) ( - )( + ) ; ± ( - )( + ) ( - - ) - ( - )( - ) - - ( - )( - ) - ( - )( - ) 1 ; and ( + )( - ) ( - )( - 1) ( + )( - ) ( + )( - ) ( - )( + ) (5 + 1)( + ) ( + 1) () 50 () () The average rate is 0. C/min. a. d d 00 b. d d d 00 d(700) d (700) + 5d (700) 5900d 10d + 15d 5900d 19 05d The average speed is 19 mi/h. 155 Holt McDougal Algebra

10 ( - - ) + - ( + + ) ( - ) + ( + ) ( + )( - ) ( + )( - ) + - ; - and ( + )( - ) ( + )( - ) ( - - ) + ( + )( - ) (+ )( - ) ( + )( - ) - ; ± ( + )( - ) ( - 5)( + ) ( + + ) + ( + )( + ) ( - 5-5) ( + ) + ( - 5) ( - 5)( + )( + ) ( - 5)( + )( + ) 5-9 ; -, - and 5 ( - 5)( + )( + ) ( - 5) ( - - ) ( - 1-1) ( - ) - 9( - 1) ( - 1)( - ) ; ( - )( - 1) 0 and ( - )( - 1) ; 1 and ( ) - ( + ) ( + ) - ( + ) ( + ) ; - ( + ) ( - - ) - - ( + + ) ( - ) - ( + ) ( + )( - ) ( + )( - ) ( ) ; - and ( + )( - ) ( - )( + ) ( + + ) - ( - 5)( + ) ( - )( + ) ( - )( + ) ; - and ( - )( + ) ( + 7)( + ) ( ) - 10 ( + 1)( + 7) ( + + ) ( + 7)( + 1) - 10( + ) ( + 1)( + 7)( + ) ( + 1)( + 7)( + ) ; -7, - and -1 ( + 1)( + 7)( + ) a s s s ( s + s + 1) (s + 1) - 100s s(s + 1) 19,00 s(s + 1) s + 1 ( s s ) b. 19,00 s(s + 1) = 19,00 ( + 1).7 Each senior has to clean about.7 m more. c. The number o meters that a student can clean in one minute is 15 m = 1.5 m/min 10 min = = The junior class will inish about. min sooner ( + ) (5 + ) ( - )(5 + ) 15 Holt McDougal Algebra

11 b. ( 1 ) ( - ) + ( ) ( - ) ( ) ( - ) ( - ) + ( - ) (- 1) 7(- ) ( - 1) Length-to-Width Ratio a. Rooms with a Width o 0 t Length (t) a b (ab) 1 (ab) + 1 a b (ab) ab b + a Height (t) : : 5 : 0. 5 : Ç : c. Both the width and the height will double. 7. Possible answer: or Possible answer: Using the LCD reduces the need to simpli the sum o the rational epressions. test prep 9. D; ( ) A; ( + 1) H; ( + + ) - + ( + + ) 5( + ) - ( + ) ( + )( + ) - + ( + )( + ) 7 5. H; d d d d 5. d(.) d 5.5 (.) + d 1.1 (.) + d 5. (.) 791.0d 79.1d + 90.d d 791.0d d challenge and etend ( - - ) + ( + )( - ) - - ( + + ) ( - 1)( - ) + - ( + ) ( - )( + ) ( - )( + ) ( + )( - ) ( ) + -1 ( ) -1 ( ) - -1 ( ) ( + ) - - ( - ) -1 1 ( - ( + ) - ) - 1 ( + - ( - ) - ( + ) ( + ) ( - ) - ( + ) ( - ) 5. ( - ) -1 - ( + ) ( + + ) ( - - ) ( + ) - ( - ) ( - )( + ) ( - )( + ) 5 + ) ( ( - 1) ( - 1) ) + + ( - 1) ( ) 5( - 1) + ( + )( + 5) ( - 1)( + 5) ( + - 5) so, = Holt McDougal Algebra

12 - 5- RAtional unctions check it out! 1a. g is translated units let. g - 0. asmptotes: =, = -5; D: { }; R: { -5}. zeros: -, -1; asmptote: = - - b. zero: ; asmptotes: = -1, = 0, = hole at = Hole at = think and discuss b. g is translated 1 unit up. - - a. zeros: -5, ; asmptote: = c. zeros: - 1, 0; asmptotes: = -, =, = - - g 0 1. Possible answer: The -value o the vertical asmptotes are ecluded rom the domain.. Both tpes o unctions ma have more than one zero. Rational unctions ma have one or more asmptotes, but polnomial unctions do not. The domain o polnomial unctions is all real numbers, but the domain o man rational unctions do not include all real numbers.. Zeros: at each real value o or which Vertical asmptotes: at each real value p() = 0 o or which q() = 0 ( ) = p ( ) q ( ) Horizontal asmptotes: none i degree o p > degree o q; the line = 0 i degree o p < degree o q; the line leading coeicient o p = leading coeicient o q i degree o p = degree o q eercises guided practice 1. discontinuous. g is translated units down Holes: at an point where = b i - b is a actor o both p and q and the line = b is not a vertical asmptote. g is translated 5 units let g is translated 1 unit right and units up asmptotes: = 0, = -1; D: { 0}; R: { -1}. asmptotes: = -, = ; D: { -}; R: { } 7. asmptotes: =, = -; D: { }; R: { -}. zeros: -, vertical asmptote: = zeros: 0, 5 vertical asmptote: = Holt McDougal Algebra

13 10. zero: 0; 11. zeros: -, -1; vertical asmptote: = 1 asmptote: = zero: ; asmptotes: = -, = 0, = hole at = hole at = -5 Hole at = -5 - Hole at = practice and problem solving 17. g is translated 5 units down zero: - 5 ; asmptotes: = -1, = hole at = Hole at = - 1. g is translated units let g is verticall stretched b a actor o 0-0. asmptotes: = -, = 0; D: { -}; R: { 0} 1. asmptotes: = 0, = 5; D: { 0}; R: { 5}. asmptotes: =, = -1; D: { }; R: { -1}. zeros: -, 5; vertical asmptote: = zeros: -, ; vertical asmptote: = zero: ; asmptotes: = -, =, = 0-9. hole at = 0 Hole at = hole at = 7 Hole at = zeros: -, ; vertical asmptote: = zeros: -1, ; vertical asmptote: = 1 -. zeros: - 1, 0; asmptotes: = -1, = 1, = hole at = 1 Hole at = Holt McDougal Algebra

14 a. () = b. Average cost ($) Band members c. (0) = = 00 0 The average cost per person is $00.. zero: -1; asmptotes: = 0, = 1; hole at = zero: 5 ; Hole at = asmptotes: =, = zero: 0; asmptotes: =, = -, = hole at = Hole at = 1. zeros: -, -; asmptotes: = 0, = zeros: ±; asmptotes: =, = -, = ( + 1)( - ) 9. Possible answer: () = ( - ) 0. Possible answer: () = ( + ) ( - )( + ) 1. Possible answer: () = ( + )( + 1) a. a. a b. D: 핉 ; R: { 0 < } c. = 0 Concentration (%) Salt added (g) b. The chemist must add about 17 g salt. Average cost ($) DVDs purchased b. () = 0 + k 55 = k 55 = 0 + k k = 15 c. () = 15 ( ) = = 5 The total cost is $ Possible answer: The graph has onl 1 vertical asmptote, at = -1. There is a hole at = 1 because - 1 is a actor o both the numerator and denominator.. Possible answer: Yes; a rational unction with a denominator that is never equal to zero will have no vertical asmptotes (or eample: () = - ) Holt McDougal Algebra

15 7a. Lap time (s) Average speed (mi/h) b. t = 1; the number o seconds the driver spent at the pit stop 1(00) c. t(00) = = It takes 57 s in total.. The equation is alse or = - because the denominator o the let side o the equation is equal to 0 when = -. Division b 0 is undeined. 9. Possible answer: The domain o a rational unction is all real numbers ecept the -values o vertical asmptotes and holes. test prep 50. A 51. F 5. D challenge and etend 5. holes at = 1, =, = ; Holes at = 1, = and = 5. zeros: -, - ; asmptotes: = -, =, = ; hole at = - - Hole at = 55a. Possible answer: b. Possible answer: 1 c. Possible answer: 0 5. Possible answer: () = ( - 1 ) 57. Possible answer: () = 5-5 ( - 1 ) ( - 9 ) solving rational equations and inequalities check it out! 1a. 10 = + 10 ( ) = () + () 10 = + = = b. + 5 = - 7 = - () = -() = - = - c. = - 1 () = () - 1() = = 0 ( - )( + ) = 0 - = 0 or + = 0 = or = - a. 1 = ( - 1 ) = ( - 1 ) 1 = ( + ) = + = The solution = is etraneous. Thereore there is no solution. 1 b. - 1 = ( 1-1) ( - 1) = ( = + ( - 1) = = 0 ( -1)( + ) = 0-1 = 0 or + = 0 = 1 or = - The solution = 1 is etraneous. The onl solution is = = - c + + c - 1 ) ( - 1) + ( ) ( - 1) 5( - c ) = ( - c - c ) + ( + c - c ) 5( - c ) = ( + c) + ( - c) 0-5c = -5c = -1 c ±1.5 The average speed o the current is 1.5 mi/h.. 1 (11) h (11) = 1 1 (11)(0h) + 1 (11)(0h) = 1(0h) 0 h 11h + 0 = 0h 0 = 9h h It will take Rem about min to mulch the garden when working alone. 5a. < b. = Holt McDougal Algebra

16 a. - - LCD is positive. LCD is negative. ( - ) -( - ) ( - ) -( - ) ; - > 0 > Solution in this case is >. 9 < < > - ; + > 0 > - The solution in this case is > ; - < 0 < Solution in this case is 1. The solution to the inequalit is 1 or >. b. 9 + < LCD is positive. LCD is negative. 9 ( + ) < ( + ) 9 ( + ) > ( + ) > > < - ; + < 0 < - The solution in this case is < -.. The solution to the inequalit is < - or > -. think and discuss 1. Possible answer: The LCD is a multiple o each denominator. Thereore, each denominator is a actor o the LCD.. Possible answer: When ou multipl both sides o an equation b a variable epression, ou ma produce an equation with solutions that make denominators o the original equation equal to 0.. Possible answer: (1) Graph each side o the inequalit. () Multipl both sides b and consider two cases, is positive or is negative.. Possible answer: Deinition: equations that contain rational epressions Characteristics: can be solved b multipling both sides b the LCD o all the terms in the equation; ma generate etraneous Rational solutions when solved Eamples: Equations Noneamples: 1 = 5, + - = - eercises guided practice + =, = 5 1. Possible answer: An equation is a statement that epressions are equal. A rational epression is a quotient o polnomials. A rational equation contains at least 1 rational epression t = 17 t 1 ( t) + (t) = 17 t t (t) t + 1 = 17 t = 1 1. r - 5 = 7 r ( 1 r - 5 ) r(r - 5) = ( = r) r(r - 5) r = 7(r - 5) r = 7r- 5 r = 7. 7 = 1 w - 7(w) = 1 (w) - (w) w 7w = 1 - w 11w = 1 w = 1 11 () = () - 5 () = = 0 ( - )( + 1) = 0 - = 0 or + 1 = 0 = or = - 1. m + 1 m = 7 7. k + 1 m(m) + 1 k = (m) = 7(m) m k(k) + 1 (k) = (k) m k + 1 = 7m m k + 1 = k - 7m + 1 = 0 k - k + 1 = 0 (m - )(m - ) = 0 (k - 1) = 0 m - = 0 or m - = 0 k - 1 = 0 m = or m = k = = + ( - + ) ( + ) + ( ) ( + ) = ( + ) ( + ) - + ( + ) = = 0 ( - )( + ) = 0 = or = - The solution = - is etraneous. The onl solution is = = - ( - ) + - ( - ) = ( - ) - + ( - ) = - 7 = 0 ( - 7) = 0 = 0 or = ( + 1) - 1 = + ( ( + 1) ) ( ( + 1) - 1( + 1) = ( + 1) + ) - ( + 1) = ( + 1) = 0 = 0 or = -1 Both solutions are etraneous. Thereore there is no solution. 1 Holt McDougal Algebra

17 = c + c 1.5( - c ) = 0 ( - c - c ) + 0 ( + c - c ) 1.5( - c ) = 0( + c) + 0( - c) c = c = -9 c ±. The average speed o the current is. mi/h. This is close to mi/h, so the barge would take about 0 mi, or 10 h, to travel upstream - mi/h and about 0 mi, or h, to travel downstream. + mi/h The trip should take about 1 h, which is close to the given time, so the answer is reasonable (0) m (0) = 1 1 (0)(50m) + 1 (0)(50m) = 1(50m) 50 m 0m = 50m 1500 = 0m 75 = m The job will take 75 min i the large copier is broken < < < 0 or > < LCD is positive. ( + 1) < ( + 1) + 1 < + 0 < > 0; + 1 > 0 > -1 The solution in this case is > = LCD is negative. ( + 1) > ( + 1) + 1 > + 0 > < 0; + 1 < 0 < -1 The solution in this case is < -1. The solution to the inequalit is < -1 or > LCD is positive. LCD is negative. 1 ( - ) ( - ) 1 ( - ) ( - ) ; ; - > 0 - < 0 > < The solution in this case The solution in this case is. is <. The solution to the inequalit is < or > LCD is positive. LCD is negative. 10 ( + ) > ( + ) 10 ( + ) < ( + ) > < > - < < -; > -; + > 0 + < 0 > - < - The solution in this case No solution in this case. is - < < -. The solution to the inequalit is - < < -. practice and problem solving = () + 1 () = = n - n - 10 () + = 10 = = = a - 7 = 1 = (a - 7) 1 = a - 7 = a. 1 z = 9 - z 1 z. + = () + z - 9z + 1 = 0 (z - )(z - 7) = 0 z = or z = 7 () = () + = - + = 0 ( - ) = 0 - = 0 = 5(n - ) = (n - ) 5n - 0 = n - 1 n = 1 (z) = 9(z) - z(z) 1 = 9z - z () - () = () - = + - = 0 ( - 1)( + ) = 0 = 1 or = - 1 Holt McDougal Algebra

18 = 1 - ( - ) ( - ) + ( ) ( - ) = ( 1 + ( - ) = = 0 ( - )( + ) = 0 = or = - The solution = is etraneous. The onl solution is = -. - ) ( - ). + 1 = ( + 1) = - 1 ( + 1) + 1 = - 1 = -1 The solution = -1 is etraneous. Thereore there is no solution. 7. ( - 1) = ( ( - 1) ) ( - 1) = 1(( - 1)) + ( - 1) (- 1) = ( - 1) = 0 ( - 1)( + ) = 0 = 1 or = - The solution = 1 is etraneous. The onl solution is = -.. (550 - v) = 17(550 + v) v = v 750 = 9v 71 v The average speed o the wind is about 71 mi/h. This is close to 70 mi/h. The plane travels about (550-70), or 10,50, mi on the light to Bomba and about ( ), or 10,50, mi on the light to Los Angeles. Because the distances are approimatel equal, the answer is reasonable (1.5) + 1 h (1.5) = 1 1 (1.5)(h) + 1 (1.5)(h) = 1(h) h 1.5h + = h = 0.5h = h It would take the apprentice h < < = < < LCD is positive. 1 () < () 1 < > 1 ; > 0 > 0 The solution in this case is > 1. LCD is negative. 1 () > () 1 > < 1 ; < 0 < 0 The solution in this case is < 0. The solution to the inequalit is < 0 or > LCD is positive. LCD is negative. 9 ( - ) -( - ) 9 ( - ) -( - ) ;.5; - > 0 - < 0 > < The solution in this case The solution in this case is >. is.5. The solution to the inequalit is.5 or > > LCD is positive. LCD is negative. 9 ( + 10) > ( + 10) 9 ( + 10) < ( + 10) > < > -1 < < -7; > -7; + 10 > < 0 > -10 < -10 The solution in this case is No solution in this case. -10 < < -7. The solution to the inequalit is -10 < < -7. a. P w + ( 17,000 w ) or equivalent inequalit b. No; substituting 00 or P in the inequalit results in nonreal values o w. 1 Holt McDougal Algebra

19 7a. 00 b. Possible answer: 1 + h + h = 191 ; 1 hits 1 c h = h h = h 0.5h = 5 h = hits; He had more at bats than hits, and this will sta constant with each hit. He would need to reach a total o hits rom 15 hits, so the answer is reasonable.. 15n n - = 5 n n n - ( n - ) = 5 ( n - ) - (n - ) n - 15n = 5 - (n - ) 15n = 5 - n + n = 9 n = 9 9. z z + 1 = z z - z(z - ) = z(z + 1) z - z = z + z -z = z z = = 1 ( ) + ( ) = 1 ( ) + = = 0 = - ± ÇÇÇÇÇÇ - ()(-1) = - ± ÇÇ 10 () = - 1 ( ) ( - 1) - ( ) ( - 1) = ( - 1) ( - 1) ( - 1) - ( - 1) = 5-5 = = ( + ). - = - ( + ) = + = =. 1 a a + 1 = 7 a ( a - 1 a - 1 ) + ( a + 1 a - 1 ) = 7 ( a - 1 ) (a + 1) + (a - 1) = 7 5a - = 7 5a = 10 a = a r LCD is positive. LCD is negative. (r) 5 ( r) (r) 5 ( r) r r.5r.5r r.; r.; The solution in this case No solution in this case. is 0 < r.. The solution to the inequalit is 0 < r > LCD is positive. ( + 1) > ( + 1) + 1 > + > < 1; + 1 > 0 > -1 The solution in this case LCD is negative. ( + 1) < ( + 1) + 1 < + < > 1; + 1 < 0 < -1 No solution in this case. is -1 < < 1. The solution to the inequalit is -1 < < 1.. LCD is positive. () () - or ; The solution in this case is. LCD is negative. () () - ; The solution in this case is - < 0. The solution to the inequalit is - < 0 or. 7. ±0.5. ± , = 1 = - = - 51a. 001 winner: 500 s ; 00 winner: 500 s + 5 b. Possible answer: 500 s = 500 s ; 500s(s + 5) 15( + s(s + 5) s ) ( = s(s + 5) ) s 7500s + s(s + 5) = 7500(s + 5) s + 00s = 0 s = -00 ± 00 ÇÇÇÇÇÇÇÇÇ - ()(-17500) () s 11 or -1 The average speed o the 001 winner is 11 mi/h. 5. depending on the values o a, b, and c, either, 1, or 0 15 Holt McDougal Algebra

20 5. Multipl each term b the LCD, 5. Divide out common actors, to get = Simpli, to get = 0. Use the quadratic ormula to solve or = -5 ± ÇÇ 0. test prep 5. A; = ( 1 ) ( + ) + ( 55. G; = B + ) ( + ) = ( ) ( + ) ( + ) + = (+ ) + = = - 15 = ( - - ) - 1 ( - ) = ( - ) ( + ) - ( - ) = + = 0 ( + 1) = 0 = 0 or + 1 = 0 = 0 or = -1 The solution = 0 is etraneous. The onl solution is = a. Possible answer: 1 (7) h (7) = 1 (7)(15h) + 1 (7)(15h) = 1(15h) h 7h = 15h 105 = h 1 h It would take about 1 h to ill the tank. b challenge and etend 5. = ( - )( + )( - ) = 7( - )( + )( - ) ( - )( + ) ( - )( + ) ( - ) = 7( - ) + 1 = 0 ( + ) = 0 = 0 or + = 0 = 0 or = = ( )( ) ( )( ) ( ) = = - 1 ( ) ( ) = = all real numbers ecept -, 0, and LCD is positive. ( + )( + ) - ( + )( + ) 7( + )( + ) + + ( + ) - ( + ) 7( + )( + ) ( + )( + 5) and ; ( + )( + ) > 0 + > 0 and + > 0 > - or < - The solution in this case is -5 < -. LCD is negative. ( + )( + ) - ( + )( + ) 7( + )( + ) + + ( + ) - ( + ) 7( + )( + ) ( + )( + 5) and or -5 ( + )( + ) < 0 + < 0 and + > 0 - < < - The solution in this case is - < -. The solution to the inequalit is -5 < - or - < > + 5 LCD is positive. ( - ) ( - ) > ( ) ( - ) + (5)( - ) > ( - ) + 0( - ) < 0 ( - )( + 1) < 0 - < 0 and + 1 > 0-1 < < ; - > 0 > The solution in this case is < <. LCD is negative. ( - ) ( - ) < ( ) ( - ) + (5)( - ) < ( - ) + 0( - ) > 0 ( - )( + 1) > 0 - > 0 and + 1 > 0 > or < -1; - < 0 < The solution in this case is < -1. The solution to the inequalit is < -1 or < <. 1 Holt McDougal Algebra

21 . Let h be the number o hours needed or Marcus to paint the barn individuall, then h will be the number o hours needed or Will. 1 () + 1 () = 1 h h 1 ()(h) + 1 ()(h) = 1 h h (h) 1 + = h h = 7 1 ( 1-1 ) 1 7 = 7 = 1 It will take Marcus 1 additional hours to complete. read to go on? Section A Quiz m 1. 1 V 1 = m V = m m =.7(500) m = 9.15 The mass o the statue is 9.15 kg.. t = k n = k k = t = k = n 10 =. It will take 10 workers. h to clean the rides ( + 1) ( - )( + 1) ( + )( + 1) + 1 ; ; and 0 - and ( - ) ( - )( + ) ( - ) ( + )( - ) -1 ; - and ( - ) + ( + )( - ) ( - 9) + )( - ) ( ( - )( + ) ( + )( - ) ( + )( - ) ( - )( + ) - - ( + ) ( + 5) ; ) ( - 5-5) - + ( - 5) ( + 5)( - 5) ( + 5)( - 5) ( + 5)( - ) ( + 5)( - 5) - ; -5 and ( + + ( - - ) ( + ) - ( - ) ( - )( + ) + + ; - and ( - )( + ) 1. d d d 0 d(50) d (50) + d (50) 700d d + 55d 700d 9d The average speed or the entire trip is mi/h. 1. g is translated units right zeros:, -; vertical asmptote: = g is translated 1 unit let and units up zero: 0; vertical asmptotes: = -, =, horizontal asmptote: = Holt McDougal Algebra

22 = () - 10 () = () - 10 = = 0 ( - 5)( + ) = 0-5 = 0 or + = 0 = 5 or = - 1. = ( - ) = - ( - ) - = - = The solution = is etraneous. Thereore there is no solution = 1 - ( + 9) ( + 9) = 1( + 9) + 9 -( + 9) - ( + 15) = = 0 -( + )( + ) = 0 + = 0 or + = 0 = - or = () + 1 h () = 1 1 ()(h) + 1 ()(h) = 1(h) h h + 1 = h h = 1 It would take the small oven 1 hours to bake the bread RAdical epressions and rational eponents check it out! 1a. no real roots b. ±1 c. 5 a. ÇÇ 1 ÇÇÇ Ç Ç b. ÇÇ ÇÇÇ Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç ÇÇ 7 c. Ç 7 Ç ÇÇÇ 7 Ç 9 ÇÇÇÇÇ Ç Ç Ç 5 b. ( Ç ) 5 () 5 1 a. ( ÇÇ ) 1 () 1 c. 5 ( ÇÇ 5 ) (5) 15 a. 1 b. ÇÇ = 1000 c. Ç 5 5a ÇÇ b. (-) c (-) ÇÇ ( - n 1) - = ( 1) = ( -1 ) = ( 1 ) = The ret should be placed cm rom the bridge. think and discuss 1. Possible answer: When a and n are natural numbers, the epression n Ç a n can be written with a rational eponent: a n n. The eponent simpliies to 1, so the epression becomes a 1, or a. 1 Holt McDougal Algebra

23 . Possible answer: ( ) = ; 5 5 = 5 = ( Product o Powers: Power o a Product: 1 ) ( eercises guided practice 1 ); ( ) 1 1 = ( Properties o Rational Eponents 1 )( ) ( 1... ± ÇÇ. ÇÇ ÇÇÇ Ç 5 Ç Ç Ç 7. ÇÇÇ 15 ÇÇÇ 15 Ç ÇÇÇÇÇ 5 Ç Ç 5 Ç Ç Ç 5 Ç 5 Ç Ç Ç Ç Ç 5 Ç Ç 5 ÇÇ Quotient o Powers: 1 1 = ; 5 1 = Power o a Quotient: ) 1 1 = ; 1 ( ) 1 1 = 1 ÇÇÇ Ç Ç Ç Ç Ç. ÇÇ 50 ÇÇÇÇÇ 5 Ç 5 Ç Ç 5 5 Ç Ç 9. Ç Ç ÇÇÇ ÇÇÇ Ç Ç Ç Ç Ç Ç 11. ÇÇ 0 ÇÇ - ÇÇÇÇÇÇÇ 10 ÇÇÇ (-) 10 Ç Ç Ç ÇÇ - 10 ÇÇ - - ÇÇ ( ÇÇ ) () (-7) ÇÇ ÇÇ 5 ÇÇÇ Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç ÇÇ Ç ÇÇ ÇÇÇ 1. 1 ÇÇÇÇÇÇ Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç Ç 7 ÇÇ ( 5 ÇÇ ) () ( Ç ) () ÇÇ = = Holt McDougal Algebra

24 ( ) (-15) 1 1 (-15) Ç 9 ÇÇ 1 or Ç. ( 7 ) 1 Ç 7 ÇÇ Ç ( 1 ) 1 9. ÇÇ 50.. t =. 1 in in. The side length o the cube is about in. practice and problem solving no real roots. Ç 9 ÇÇ 9 ÇÇÇÇ ÇÇÇ Ç Ç. ÇÇÇ ÇÇÇÇÇÇ Ç Ç Ç Ç Ç Ç 5. ÇÇ 50 ÇÇÇ ÇÇ 50 Ç Ç ÇÇ 50 ÇÇ 50 ÇÇ 50 Ç Ç Ç ÇÇ 1000 Ç ÇÇ 5 9 ÇÇÇÇÇÇÇÇ 7 Ç 7 Ç Ç Ç Ç Ç 7 Ç Ç 7 5 Ç 5 ÇÇÇ 7 5 ÇÇÇÇÇ Ç 5 5 Ç 5 5 Ç 5 Ç 5 Ç ÇÇÇÇ 9 ÇÇÇÇÇÇÇÇÇÇÇ (-). ÇÇ 5 5 ÇÇÇÇ ÇÇ 5 Ç Ç Ç ÇÇ 5 Ç ÇÇ 5 Ç ÇÇ 5 5 Ç 5 Ç Ç 5 ÇÇ 5 Ç ÇÇ 10 Ç ÇÇ 10 Ç ÇÇÇ Ç Ç Ç ÇÇÇ (-) Ç Ç Ç Ç Ç (-) Ç - Ç. 1 ( ÇÇ 1 ) (). Ç Ç Ç Ç. (-1000) ( ÇÇÇ ) (-10) 10, ÇÇ (-) 5 7. ( ÇÇ 1 ) = ÇÇ 170 Holt McDougal Algebra

25 . 9. ( ) ÇÇ ( ) 1 or Ç or Ç ( ) 7 ( 7 1 ) 1 9 ( Ç ) ( ÇÇ ) () () 1 1 7) ( ( ) ÇÇ 1) ( 9 1 ( ) 1 ÇÇ 9 1 ÇÇ ( ) or 9 Ç a() = 1000( ) 119 The amount will be about $119 ater ears. 5a. P = 7. ÇÇ A cheetah has a metabolism rate o 10 Cal/da. b. House cat: P = 7. ÇÇ.5 7 Lion: P = 7. ÇÇ = A lion needs to consume about more Calories each da than a house cat. 0 ) 1% About 1% o the sample will remain. 59a. 100 ( 1 t h ) = 100 ( 1 b. Iodine-11: ( 1 ) Iodine-15: ( ) = 1. About 1. g more iodine-15 than iodine-11 will remain ater 0 das. 0. W = (0) (5) (0)(5) 5 The wind chill is about F. 1a. π Ç L g = π L Ç Ç g = π Ç L Ç g Ç g Ç g = π ÇÇ Lg g b. t = π ÇÇÇÇÇ (0.5)(9.) It will take about 1. s to complete one swing.. ÇÇ 7 0. (5) 1 ( 0 ) 1 ( ) ( 5 ÇÇ -9 Ç ) ( 5 ÇÇ (-9) -9 ) ( Ç ) 5. ( - 1 ) (-) ( 1 ) ((-) ) (-). ( ) 5 (-1) 5 ( 15 ) 9 5 ÇÇÇ (-1) (-) ÇÇÇÇÇÇ (-) 9 5 ÇÇÇ (-) 5 5 ÇÇÇÇ (-)(-) ÇÇ 5 5. ( ÇÇ 11 ) ( 11 ) 11 ( ) (5) 5 ÇÇ (5) 5 ÇÇÇ ( a b ) 1 ( a b b ) 1 1 ( a b ) ( b ) ( a b ) 1 b b ÇÇ a b Holt McDougal Algebra

26 70. ( a 1 b ) 1 ( a ) 71. a ( b ) 1 1 b ( a ) 1 ( b ) 1 ( b ) 1 a 1 ( a b ) b 1 Ç b a Ç b Ç b b a ÇÇÇ b Ç b b Ç Ç b Ç b Ç b Ç a b b 7. m() = The mass is about 1 kg ater h. 7. Alwas; possible answer: Ç = = or all real numbers Never; possible answer: i is positive, () is 1 positive and (-) is negative; thereore, 1 () 1 (-) 1. I is negative, () is negative 1 and (-) is 1 positive; thereore, () 1 (-). 1 1 ( a ) ( b b ) 75. Never; possible answer: simpli the let side o the equation: - Ç = - = -. The epression - is negative or all nonzero real numbers.. The epression 1 is positive or all nonzero real Simpli the right side o the equation: - = 1 numbers. Thereore, - Ç Sometimes; possible answer: i is positive, - Ç is less than 0. I is negative, - Ç is greater than 0. Thereore, the inequalit is true onl or positive values o. 77. and ; about. 7. and ; about and -; about -.1 0a. P = 1.7(10) (50) 1. The air pressure in Denver is about 1. psi. b. P = 1.7(10) (90) 5. The air pressure at the top o Mount Everest is about 5. psi. 1. A is incorrect. The Quotient o Powers Propert is 1 incorrectl applied. To simpli 5 5, ou should subtract the eponents rather than dividing them.. values; 1,,,. Possible answer: Raise 10 to the 1 power, or use the Ç unction to ind the sith root o Possible answer: The eponent. is rational because it can be epressed as the ratio o integers. For eample, the eponent can be epressed as the ratio o the integers and 10 because. = 10. test prep 5. A. H; 1 (π) ((V)) 1 (π) (V) 1 (π) (V) = S 7. A; a Ç = Ç = = = Ç challenge and etend ( ( 0 ) ) ( 1 = 0 ) 1 = Ç a > a a is positive. 1 > Ç a 1 > a -1 < a < 1 The solution in this case. F; ÇÇÇ 5a 7 ÇÇ a ÇÇÇÇÇ a a Ç Ç a Ç a a ( ) 1 ( ) = a is negative. 9 1 < Ç a 1 < a a < -1 or a > 1 The solution in this case is a < -1. is 0 < a < 1. The solution to the inequalit is a < -1 or 0 < a < = = 1; -1 + i Ç ( ) = (-1 + i Ç )(-1 + i Ç )(-1 + i Ç ) (- - i Ç = ) (-1 + i Ç ) = = 1; -1 - i Ç ( ) = (-1 - i Ç )(-1 - i Ç )(-1 - i Ç ) (- + i = Ç ) (-1 - i Ç ) = = 1 17 Holt McDougal Algebra

27 RAdical unctions check it out! 1a. D: { 핉 }; R: { 핉 } a. g is translated 1 unit up. g 0 a. g is relected across the -ais and translated units up. g g() = - Ç + 1 b. D: { -1}; R: { 0} 0 b. g is verticall compressed b a actor o 1. 0 b. g is verticall stretched b a actor o, relected across the -ais, and translated 1 unit down h() = ÇÇÇÇ 5 = ÇÇÇ 5 ; 5 h(50) = ÇÇÇÇ 5 (50) 5 The downward velocit is about t/s. a think and discuss b. - g 0 1. Possible answer: No; an asmptote is a line that a curve approaches as or becomes ver large. There is no such line in the graphs o radical unctions.. Possible answer: The radicand must be nonnegative, so + must be greater than or equal to 0. Solve this inequalit or The domain o the unction is -1. g. Possible answer: Transormation Equation Graph Vertical translation Horizontal translation Relection Vertical stretch eercises guided practice g ( ) = + g ( ) = - g ( ) = - g ( ) = Possible answer: The unction rule is a radical epression that contains a variable in the radicand.. D: { -}; R: { 0} D: { }; R: { 0} 0 -. D: { 핉 }; R: { 핉 } D: { 0}; R: { -1} 0-5. D: { 핉 }; R: { 핉 } D: { 핉 }; R: { 핉 } Holt McDougal Algebra

28 . g is translated 7 units down j is translated 5 units right h is stretched horizontall b a actor o and translated units let. h g is relected across the -ais, verticall stretched b a actor o, and then translated units down g j g 9. h is verticall stretched b a actor o g is compressed verticall b a actor o 1 and translated 1 unit down j is relected across the -ais and then translated units right. j h is relected across the -ais, horizontall compressed b a actor o 1, and then translated units let. h j is verticall stretched b a actor o and then translated units up and units let. j h g 19. g() = 5 ÇÇ 5 ; 9 g() = 5 ÇÇ 5 (). 9 The distance to the horizon is about. mi practice and problem solving. D: { }; R: { 0} 0 -. D: { -1}; R: { -} D: { 핉 }; R: { 핉 } g is translated units up. 0 g D: { 0}; R: { 0} D: { 핉 }; R: { 핉 } - 9. D: { 핉 }; R: { 핉 } h is translated units right. 0 h 17. g() = ÇÇÇ ( + 5) - 1. g() = ÇÇÇÇ -( - 7) 17 Holt McDougal Algebra

29 . j is verticall compressed b a actor o h is verticall compressed b a actor o 1 and then relected across the -ais. h g is relected across the -ais, verticall stretched b a actor o, and then translated 1 unit up. 0 - g - j. g is horizontall compressed b a actor o 1 and then translated 5 units let j is translated units let and 1 unit down. j - g 0 7. h is relected across the -ais, verticall stretched b a actor o, and then translated units up. - h - 0. j is relected across the -ais, verticall compressed b a actor o 1, and then translated units let. j g() = 1 ÇÇÇ + 0. g() = ÇÇÇÇÇ - 1 ( - ) 1. g() = - ÇÇÇ ÇÇ ;. g() = ÇÇÇÇ 0 5 = g(1) = ÇÇ 1 = The radius o the can is cm a. h(0.01) = 1(0.01) 7 The resting heart rate is about 7 beats/min. - 1 b. h(00) = 1(00) 5 The resting heart rate is about 5 beats/min.. a vertical stretch b a actor o ollowed b a translation 1 unit let 9. a vertical stretch b a actor o ollowed b a translation 1 unit right and 9 units down 50. a relection across the -ais ollowed b a translation units right and 7 units down 51. D 5. B 5. A 5. C 55a. D(11000) =.5 ÇÇÇ The approimate distance is 7 km. b. A = = 7000 D(7000) =.5 ÇÇ = 7 It will appear to decrease b about 7 km. 5. The speed is about 150 mi/h. Speed (mi/h) Depth (t) ÇÇÇÇÇÇ 57. Yes; M j = 100 (- + 7) = 100 and so 1000 M > M j. 5. v Earth = ÇÇÇÇÇÇÇ 909()(90) 91. v Moon = ÇÇÇÇÇÇÇÇ 909(1/)()(100) The vehicle need to travel about 19, mi/h aster on Earth than on the Moon. 175 Holt McDougal Algebra

30 59a. Period (s) 0 Length (m) b. Possible answer: The graph o T is a vertical stretch o the graph o b a actor o π and a horizontal stretch o the graph o b a actor o 9.. c. b a actor o 0. sometimes 1. sometimes. alwas. never. no. es 7a Speed o sound in air (m/s) Temperature ( C) 5. es b. The speed o sound at 5 C is about m/s. c. Possible answer: Because k is positive, the value o the unction is 0 onl when T = 0. The value o T that makes this equation true is C. a. s is translated 1 unit up. 0 b. s(5) = ÇÇ samples should be taken. 9. t 1 = ÇÇ 0.9 t = ÇÇ 100 = = 1. it will take about 1. s longer. 70. Possible answer: A vertical compression o the parent unction b a actor o 1 can be represented b g() = 1 Ç. A horizontal stretch o the parent unction b a actor o can be represented b h() = ÇÇ 1. The epression ÇÇ 1 can be simpliied as ollows: ÇÇ 1 = Ç1 Ç = 1 Ç. Thereore, g() = h() = 1 Ç. 71. Possible answer: Onl nonnegative radicands have real square roots; thereore, the domain o squareroot unctions is limited to values o that make the radicand nonnegative. B contrast, all real radicands have cube roots; thereore, the domain o cube-root unctions is not limited. 7. Possible answer: A horizontal translation aects the domain, but not the range. The domain o the translated unction is h, where h is the number o units the unction is translated horizontall. A vertical translation aects the range, but not the domain. The range o the translated unction is () k, where k is the number o units the unction is translated verticall. test prep 7. D 7. J 75. A 7. J 77. Possible answer: The graph was relected across the -ais and then translated units right; g() = ÇÇÇÇ -( - ). challenge and etend 7. g() = - ÇÇÇ () = - ÇÇÇÇ ( + ) + 5 or equivalent unction - 5- solving radical equations and inequalities check it out! 1a. + ÇÇÇ - 1 = 5 ÇÇÇ - 1 = 1 ( ÇÇÇ - 1 ) = 1-1 = 1 = c. ÇÇÇ + 10 = ÇÇÇ + 10 = 7 ( ÇÇÇ + 10 ) = = 9 = 9 1 b. ÇÇÇ - = ( ÇÇÇ - ) = - = = 1 = a. ÇÇÇ + = Ç ( ÇÇÇ + = 9 = + ) = ( Ç ) 17 Holt McDougal Algebra

31 b. ÇÇÇ + = ÇÇÇ - 1 ( ÇÇÇ + ) = ( ÇÇÇ + = ( - 1) 1 = 7 = - 1 ) a. ÇÇÇÇ + 1 = + ( ÇÇÇÇ + 1 ) = ( + ) + 1 = = = ( - 1)( + 5) - 1 = 0 or + 5 = 0 = 1 or = -5 = -5 is etraneous, the onl solution is = 1. b. -9 ÇÇÇÇ + = - + ( ÇÇÇÇ -9 + ) = (- + ) -9 + = = = ( - )( + ) - = 0 or + = 0 = or = - 1 a. ( + 5) = ÇÇÇ + 5 = ( ÇÇÇ + 5 ) = + 5 = 7 = 1 c. ( + ) = 9 ÇÇÇ + = 9 ( ÇÇÇ + ) = 9 9( + ) = 1 + = 9 = b. ÇÇÇ + 1 ( ÇÇÇ + ) b. ( + 15) = ÇÇÇÇ + 15 = ( ÇÇÇÇ + 15 ) = () + 15 = 0 = = ( - 5)( + ) - 5 = 0 or + = 0 = 5 or = - 5a. ÇÇÇ ÇÇÇ - ( ÇÇÇ - ) () - 9 1; - 0 ; 1. Possible answer: s = ÇÇ 0d 0 = ÇÇÇÇ 0(0.7)d 0 = ÇÇ 1d (0) = ÇÇ 1d 900 = 1d d I the car were traveling 0 mi/h, its skid marks would have measured about t. Because the actual skid marks measure less than t, the car was not speeding. think and discuss 1. Possible answer: The equation can be solved algebraicall b squaring both sides, or it can be solved b graphing both sides o the equation.. Possible answer: To solve = a, take the square root o each side. To solve Ç = b, square each side. The operations used to solve each equation are inverses o each other.. Possible answer: 1. Isolate the radical epression. eercises. Raise both sides o the equation to the power equal to the inde o the radical. guided practice. Simpli and solve. Check solutions in original equation. a. I true, the solution is a solution o the radical equation. b. I alse, the solution is etraneous. 1. No; the epression under the radical does not contain a variable.. ÇÇÇ - 9 = 5 ( ÇÇÇ - 9 ) = 5-9 = 5 =. ÇÇÇ - = ( ÇÇÇ - ) = - = = 10. Ç = ÇÇÇ + 9 ( Ç ) = ( ÇÇÇ = + 9 = 9 =. Ç = ÇÇÇ + 7 ( Ç = = 7 = ) ) = ( ÇÇÇ + 7 ) 10. ÇÇÇ + 1 = ÇÇÇ + ( ÇÇÇ + 1 ) = ( ÇÇÇ 1( + 1) = 9( + ) 7 = = ÇÇÇ + 5 = ( ÇÇÇ + 5 ) =. Ç = ( Ç ) = = = 1 5. ÇÇÇ - 1 = ÇÇÇ + ( ÇÇÇ - 1 = + = 5 + ) + 5 = 0 = = ( - )( + 7) - = 0 or + 7 = 0 = - 1 ) = ( ÇÇÇ + ) 7. 5 ÇÇÇ + = 5 ÇÇÇ - ( 5 ÇÇÇ + = - = = + ) 5 = ( 5 ÇÇÇ - ) 5 9. ÇÇÇ + - ÇÇÇ - = 0 ÇÇÇ + = ÇÇÇ - ( ÇÇÇ + ) = ( ÇÇÇ + = - = 10 - ) 177 Holt McDougal Algebra

32 1. ÇÇÇ + 1 = - ( ÇÇÇ + 1 ) = ( - ) + 1 = = = ( - 7)( + ) - 7 = 0 or + = 0 = 7 ( - 0) 1. ÇÇÇÇ - 11 = - ( ÇÇÇÇ - 11 ) = ( - ) - 11 = = = ( - )( - 5) - = 0 or - 5 = 0 = or = 5 1. ÇÇÇ + - = ÇÇÇ + = + ( ÇÇÇ + ) = ( + ) + = = = ( + )( + 5) + = 0 or + 5 = 0 = - ( + 0) ÇÇÇ - 1 = + 1 ( ÇÇÇ ) = ( + 1) = = = ( + 1)( + ) + 1 = 0 or + = 0 = -1 ( + 1 0) 1. ÇÇÇÇ = + ( ÇÇÇÇ ) = ( + ) ( - 5) = = = ( - 1)( + 1) - 1 = 0 or + 1 = 0 = 1 or = - 1 = ÇÇÇ - 5 = ( ÇÇÇ - 5 ) = - 5 = 9 = ( + 5) = ÇÇÇ + 5 = ( ÇÇÇ + 5 ) = + 5 = 0 = = ( - 5)( + 1) - 5 = 0 or + 1 = 0 = 5 ( 0) 1 1. ( + 1) = ÇÇÇ + 1 = ( ÇÇÇ + 1 ) = + 1 = = 7 = ( - 50) = -10 ÇÇÇ - 50 = -10 ( ÇÇÇ - 50 ) = (-10) ( - 50) = = -15 = (5-9) = - 5 ÇÇÇÇ 5-9 = - 5 ( ÇÇÇÇ 5-9 ) = ( - 5) 5-9 = = = ( - 5)( + ) - 5 = 0 or + = 0 = 5 ( - 5 0). Ç + 10 Ç ( Ç ) 1 ; 0 0; ( + 1) = 1 ÇÇÇ + 1 = 1 ( ÇÇÇ + 1 ) = 1 ( + 1) = 1 = - = -. ÇÇÇ ÇÇÇ ÇÇÇ + 5 < 5. s = ÇÇ 1d = ÇÇ 1d = ( ÇÇ 1d ) 09 = 1d 195 d = 5 There will be about 5 t distance. practice and problem solving 7. ÇÇÇ - 1 = 9 ( ÇÇÇ - 1 ) = 9-1 = 1 = ÇÇÇ + 7 = 5 ÇÇÇ + 7 = 5 ( ÇÇÇ + 7 ) = = 5 = 1 1. = 1 ÇÇÇÇ = ÇÇÇÇ = ( ÇÇÇÇ + 0 ) 1 = = = ( ÇÇÇ + 5 ) ; ; -5 0 ( ÇÇÇ + 5 ) < < 5 < 0 < 10; ; - 5 < 10. ÇÇÇ = 0 ÇÇÇ + 1 = ( ÇÇÇ + 1 ) = + 1 = 7 = = 1 0. ÇÇÇ + = ( ÇÇÇ + ) = + = 1 = 10 = 5. - = ÇÇÇ = ÇÇÇ - 7 = ( ÇÇÇ - 7 ) = = 17 Holt McDougal Algebra

33 . ÇÇÇÇ + 1 = Ç ( ÇÇÇÇ + 1 = 1 = = + 1 ) = ( Ç ) 5. Ç = ÇÇÇ + 7 ( Ç ) = ( ÇÇÇ + 7 ) = + 7 = 7 = 7 7. ÇÇÇÇ = ÇÇÇÇ + 1 = - ( ÇÇÇÇ + 1 ) = ( - ). 5 ÇÇÇ - 1 = ÇÇÇ ) = ( ÇÇÇ + 1 ) (5 ÇÇÇ 5( - 1) = + 1 = = = ÇÇÇ = = = ( - )( + 1) - = 0 or + 1 = 0 = ( 0) ( + ) = ( ÇÇÇ + 5 ) = = 0 ( + 1)( + ) = = 0 or + = 0 = -1( + 0) 1 9. ( - 9) = ÇÇÇ - 9 =. ÇÇÇ + - = - ÇÇÇ + = - ( ÇÇÇ + ) = ( - ) + = = = ( - 1)( - ) - 1 = 0 or - = 0 = ( - 0) 1 0. (5 + 1) 1 = 1. ( + ) = ÇÇÇ = ( ÇÇÇ ) = = 5 5 = 55 = 51. ÇÇÇ + ( ÇÇÇ + ) + 11; ; ( ÇÇÇ - 9 ) = - 9 = 1 = 5 ÇÇÇÇ + = ( ÇÇÇÇ + ) = + = 0 = = ( - 7)( + ) - 7 = 0 or + = 0 = 7 ( 0). ÇÇÇ - ( ÇÇÇ - ) ; - 0 ; 19. ÇÇÇ ( ÇÇÇ + 1 ) ; ; ÇÇ 15w 5. d = π 1.5 = ÇÇ 15w π 1.5π = ÇÇ 15w (1.5π) = ( ÇÇ 15w ). 15w 1.5 About 1.5 tons weight can be lited. a. d = l ÇÇÇÇÇÇ + w + h 1 = ÇÇÇÇÇÇ h 1 = ( ÇÇÇÇ 19 + h ) = 19 + h 10 = h 11. h The height o the prism is about 11. cm. b. The length o the diagonal will double. 7. r = Ç A π r = ( Ç A r = A π A = πr π ) 9. v = ÇÇ E m E m ) v = ( ÇÇ v = E m E = mv E = 1 mv 50a. V = k(f + ) 11 = k( + ) 11 = k() 11 = k 1 k. r = ÇÇ V π r = ( ÇÇ V π ) r = V π V = πr V = πr b. V = 1( + ) 17 The minimum wind velocit is about 17 mi/h. 179 Holt McDougal Algebra

34 c. 00 = 1(F + ). ÇÇÇÇ (F + ). ( ÇÇÇÇ (F + ) ) 17 (F + ) 1 F + 10 F The wind velocit is in the F10 categor. 51a. v = Ç ar 1 = ÇÇÇ 9.r 1 = ( ÇÇÇ 9.r ) 19 = 9.r 5 = r The smallest radius is 5 m. b. v = Ç ar = ÇÇ.5a = ( ÇÇ.5a ) =.5a 5. = a The acceleration is 5. m/s. L 5a. T = π ÇÇ 9. L. = π ÇÇ 9. L 0.5 ÇÇ 9. L 0.5 ( ÇÇ 9. ) 0.15 L L The length o the pendulum is about 1.0 m. L b. T = π ÇÇ = π ÇÇ L 9. L 0.0 ÇÇ 9. L 0.0 ( ÇÇ 9. ) 0.00 L L The length o the pendulum is about 0.0 m. 5a. πr A r A π r Ç A π b. No; i A = 0, then r must be less than or equal to about 5.0. Thereore, 0 would not be a reasonable value o r. 5. Solution B is incorrect; possible answer: in the irst step, the coeicient o the radical should have been squared along with the rest o the equation or a. d = 1.11 Ç h = 1.11 ÇÇ 15.7 The captain can see about.7 mi. b. d = 1.11 Ç h = 1.11 ÇÇ =. The sailor can see about. mi arther. c.. mi 10 mi/h = 0. h 5 min The sailor will see the pirate ship about 5 min sooner than the captain will. m 59. s = Ç ρ s = ( s = m ρ m Ç ρ ) ρs = m m gold - m lead = 19.(5) - 11.(5) = 995 The mass o the cube o gold is 995 g greater. 0. Possible answer: Subtracting 5 rom each side results in the equation ÇÇÇÇ = -7. Because the Ç smbol indicates the principal, or nonnegative, square root, the value o the let side o the equation cannot be negative. Thereore, the equation has no real solution. 1. Possible answer: When solving both tpes o equations, ou must check or etraneous solutions. test prep. D; ÇÇÇ + = ( ÇÇÇ + ) = + = 7 = = H; - 1 = ÇÇÇ 5-9 ( - 1) = ( ÇÇÇ 5-9 ) = = 0 ( - )( - 5) = 0 - = 0 or - 5 = 0 = or = 5. B; 0 = π ÇÇÇÇ + h 1.7 ÇÇÇÇ + h 1.7 ( ÇÇÇÇ + h ) 1 + h 9 h 10 h 5. F; V = ( A ) ( A ( A V = ÇÇ ) ) ) V = ( ÇÇ V = ( A ) ÇÇ V = A V = A 10 Holt McDougal Algebra

35 1. ( - ) = ÇÇÇ - = ( ÇÇÇ - ) = - = 1 = = challenge and etend 7. alwas true. never true 9. alwas true 70. never true 71. Ç = 9 Ç 7. ÇÇÇ + = Ç Ç = 9 ( Ç ) = 9 = 9 7. ÇÇÇÇ - = - ( ÇÇÇÇ - ) = ( - ) - = = 0 = 10 ( ÇÇÇ + ) = ÇÇÇ + = 1 ( ÇÇÇ + ) = 1 + = 5 = 5 7. S 17 = ÇÇÇÇÇÇÇÇÇÇ (1 + 75%)h(1 + 50%)m = ÇÇÇÇ 7.75hm = ÇÇÇ 7.75 ÇÇ hm.1s.1s - S 100% = 11% S The surace area increased b about 11%. read to go on? Section B Quiz 1. ÇÇ ÇÇÇÇÇÇÇ Ç Ç Ç Ç Ç Ç. ÇÇÇ 1 z ÇÇÇÇÇÇÇÇÇÇÇ z z Ç Ç Ç Ç Ç Ç z Ç z z z z. ÇÇ a 9 Ç a Ç a Ç Ç Ç Ç a Ç a Ç ( ÇÇ 1 ) 5 () 5. Ç ÇÇ. (-7) ( ÇÇ -7 ) (-) (-1000) 10. t = 7 = 1 1 n(1) = The population is about 1150 ater 1 week. 11. D: { 0}; R: { } g() = ÇÇÇÇ ( - ) ; 15. g(10) = ÇÇÇÇÇ (10 - ) = 1 The speed is 1 t/s ÇÇÇ 5-5 = -10 ÇÇÇ 5-5 = 5 ( ÇÇÇ 5-5 ) = = 15 5 = 10 = 5 1. D: 핉 ; R: 핉 g() = - ÇÇÇ ÇÇÇ + = - ( ÇÇÇ + ) = ( - ) + = = = ( - 5)( - 1) - 5 = 0 or - 1 = 0 = 1 ( - 0) 11 Holt McDougal Algebra