Course 2 Answer Key. 1.1 Rational & Irrational Numbers. Defining Real Numbers Student Logbook. The Square Root Function Student Logbook


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1 Course Answer Ke. Rational & Irrational Numbers Defining Real Numbers. integers; 0. terminates; repeats 3. two; number 4. ratio; integers 5. terminating; repeating 6. rational; irrational 7. real 8. root 9. radical. The following answers are samples a. ; 8 ; 30 b ; ; 4 0 c. 6 6 ; 4 ; 3 9 d. 9 ; ; 8 8. a b. 0. c. 3.5 d Sample answer: 7, a..646 b c d Working with Radicals d 3. radicand 4., 4, 9, 6, 5 5. a = a a d c b nonnegative square root 7. a b rationalize; denominator; rational. a. 9; 9 b. 65; 5 c. 44;., 44, 69, 96, 5 3. a. 4 0 b c b a. 6 c a. 6 3 b. 5 5 c seconds The Square Root Function.. Sample answer: The slope formula can be used to show that slopes between plotted consecutive points are not equal. 3. Because for each first coordinate there is one and onl one second coordinate 4. interpolate; domain 5. etrapolate; observed 6. nonnegative real numbers 7. nonnegative real numbers 8. parameter 9. steepness; quadrant a b
2 a. 3 b. 4 c. d. e. 5 Unit Assessment. a. rational, terminating decimal b. rational, can be written as 5 4 c. irrational, nonterminating and nonrepeating decimal d. rational, repeating decimal e. rational, can be written as 3 f. irrational, nonterminating and nonrepeating decimal g. rational, repeating decimal h. irrational, nonterminating and nonrepeating decimal. sometimes 3. alwas 4. Irrational; b the Pthagorean theorem, the length of the hpotenuse is given b the square root of 4 5, or the square root of 4. No rational number can be squared to equal 4, so the length of the hpotenuse is given b an irrational number. 5. b and d 6. O (5.9,.43) (4.8,.9) (.7,.64) (5.,.8) (,.4) (3.4,.84) (.3,.4) (0, 0) O c and d 8. a. 5 3 b c. 5 7 d. 4 7 e. 4 3 f. 3 g. 3 5 h. 4 3 i. 3 5 j c 0. a. 4 b. 3 c. d. 5 e. Unit Investigation.. Answers will var. 3. The distance to the horizon when the plane is on the ground 4. about miles 5. about 6,79 feet 6. about 6,874 feet 7. The distance to the horizon decreases. Polnomial Arithmetic Working with Powers. eponent; base. 3. a n 4. reciprocal; opposite 5. base 6. a r + s ; integers 7. a r s ; integers 8. a r s ; integers 9. a n b n ; integer 0. an b n ; integer. a. 3 b. 3 a. 3 4 a O 0,000 0,000 30,000 5,000 5,000 5,000 35, a. b 5 b. 3c c. 5 6 d. 3,or 3 e f
3 4. Mercur: Earth: Mars: Saturn: Pluto: Transforming Equations Using Multiple Operations.. a n ; real number; variable; nonnegative 3. polnomial 4. trinomial; three monomials 5. left; right; descending order 6. left; right; ascending order 7. nonzero value; simplif; identif 8. Their eponents are different. 9. Students answers will var.. No. In the definition of a monomial, a term of the form a n, n is a nonnegative integer. Since n in the epression 3 is negative, this epression is not a monomial.. a. ; binomial b. 4s 3 7s 7 s; trinomial 3. a. 3 0 b. b 4 b 4 c. 7c 3 6c 4. a. 3a a 3 b c. 7c 3 6c 5. a. n 4n b. 3n 7n 3 c. 4n 9n 3 Multipling Polnomials. n. distributive; (n 0)n (n 0) 3. sum; products 4. FOIL stands for First, Outer, Inner, Last: multipl the first terms of each binomial; multipl the outer terms; multiple the inner terms; multiple the last terms. 5. substitute; identit 6. a ab b 7. a ab ab b ; or a ab b 8. a b. a. (n )(n 8); n 0n 6 b. n(n 8); n 8n 0 c. (n )(n 8); n 7n 8 d. (3n )(n 8); 3n 5n 8. (n 3)(4n ); 4n n n 6; 4n 0n 6 3. a. 9b b 4 b n 6n; Sample answer: multipl using the distributive propert, then multipl b, which will double the area. Unit Assessment. b and c. a. 3a 3 b. 5r 6 c , or 4 4, or d e. 8s 3n 6 f a. 7n 0n 5 b. n 3n 5 c. n 5n 4 d. 9n 4n 7 64r 3 343s 3 5. (n 3)(3n 4); 6n 8n 9n ; 6n n 79
4 6. d 7. See students work. After substituting and simplifing a. n n b. 3n 3n 4 c. n 4n 4 Unit Investigation. Students diagrams will var. Dimensions should be reasonable.. Students diagrams will var. Dimensions should be reasonable. 3. Students diagrams will var. Area and dimensions should be reasonable. 4. Students answers will var.. Factoring Polnomials Finding Common Factors. positive integers; ; itself. It has eactl one factor (itself), not two. 3. composite number 4. prime factors; product 5. Factor the variables. The greatest common factor of the variable terms is equal to the variable term with the lower eponent. 6. degree 7. n 8. the product of two or more polnomials 9. highest; monomial. a. ()()(3)(5) b. (5)(3)()() c. ()()()()(3)(3)(n)(n). a. 8 3 b. 4a 3 3. a. b. 3 ( ) c. If = 4, 3(4)[()(4) ] = (6)(6) (); ()(9) = (96) ; 08 = a. 4n(3n 5) b. 8 3 (9 5) c. ( )( 5) d. 3(m 7)(m ) Factoring Quadratic Trinomials. 4; 0. ( 4)( 6) 3. quadratic term 4. linear term 5. constant term 6. Yes. It is of form a b c; a, b, c are, 0, 4 which are real numbers and a opposite 8. ( 3)( 4) 9. (r 3)(r ) 0. (n 5)(3n ). a. b. (g ) (g 5) c. Sample answer: if g, ( )[()() 5] (4)(9); a. s 5s b. s c. 5s d. 3. a. ( )( 3) b. (d 8)(d 4) c. (p )(p 3) d. (3 4)( ) e. 3(f )(f 3) Special Cases. (a b) g. difference; two squares 3. ( 3)( 3) g g S 80
5 4. (a b)(a b) 5. (5k ) (5k ) 6. es; 4 and ( 8)( 8) 8. prime 9. a. common factors; distributive b. perfect square trinomial; difference of two squares 0. prime. Factored form Trinomial epression Special Case ( 9) 8 8 perfect square trinomial ( 0) perfect square trinomial ( 3) 6 9 perfect square trinomial ( 5)( 5) 5 difference of squares ( 7)( 7) 49 difference of squares prime, not factorable 8 sum of squares ( 0) or 4( 0) perfect square trinomial 7. a.(g 0)(g ) b. (k 6) c. (p 8)(p ) d. ( 3)( 5) e. prime f. (4a 5)(4a 5) 8. a. 4(p 4) b. (h 0) c. (3d 8) d. 4( ) Unit Investigation. S S.. a. (a b) is the square of the differences and is in factored form and equal to a ab b. (a b ) is the difference of squares and is in standard form and factors to (a b)(a b). b. Solutions will var. Eample: let a and b ; is not equal to 3 when the are substituted. Unit Assessment. The net step would be to eliminate all multiples of 3 from the table, then multiples of 4, etc. The numbers remaining would not be multiples of an numbers other than themselves and and are therefore prime a. Solutions will var. Eample: 8 and, GCF is 4 4. t 4 5. b. Solutions will var. Eample: 5 and 8, GCF is 3 Terms Greatest common factor 6, m, 3m, 96m 3m 4, a. ( 3)( ) 4 3 b. Solutions will var. Eample: Let = 5, 3. epanded form: c 4c 4; factored form: (c ) 4. a. n n n b. epanded form: c nc n ; factored form: (c n) 5. a perfect square trinomial 6. Floor plans will var. 7. Answers will var. Eample: If there is 6 feet of tile on one wall and foot of tile on the other, then ( 6)( ) The polnomial in epanded form clearl shows the regions and the areas of the regions that will have carpet and those that will have tile. The quadratic term is the carpet and the linear and constant terms are the tiled areas. The factored form of the area gives the lengths of the sides of the room. 8
6 3. Graphing Quadratic Functions & Equations Graphing Parabolas. a second degree polnomial function. value; value 3. up 4. down 5. the least value of on the graph of a parabolic function 6. the greatest value of on the graph of a parabolic function 7. ais of smmetr verte. a. a; c. a. concave down b. concave up c. concave up 3. a. b and c 4. c b. a 5. Students graphs should depict a parabola in quadrants I and II whose verte is the origin. The curve is concave up and should pass through points (, ), (, ), (, 8), (, 8), (3, 8), ( 3, 8), and so on. a. domain: all real numbers; range: 0 b. 0 c. the origin, (0, 0) d. concave up e. 0 (minimum) Analzing Properties of Parabolas. the intercept. down; verte;, ; real numbers 4. incomplete; 0 5. midpoint; h parabola; ais; verte 9. intercepts; 0. a. 5 b. 0 c. 67. b, c 3. Students parabolas should open upward, the lowest value of (the minimum) should be 3, and the parabola should be centered around the ais of smmetr,. The parabolas ma have an width, provided these conditions are met. 4. (8, 5) 5. a. maimum 400 b. the height of the cliff, or the distance between the cliff and the ground Solving Quadratic Equations b Graphing. trajector. quadratic; linear; 0; 6 3. verte 4. root; solution 5. horizontal intercepts 6. real roots real root 9. intercept 8
7 . a. 5. b. feet (at d 0) c. Since d cannot be negative in this problem, onl the segment in Quadrant corresponds to the ball s trajector. Unit Assessment. a. 3 b. 5 c. 7 d. equation is of the form a b c e. the coefficient of the quadratic term, 3 f. If the quadratic term is positive, the parabola is concave up; if the coefficient is negative, the parabola is concave down. g. the constant term, 7. a. maimum b. minimum c. minimum 3. a. b. 4. a. real roots b. real roots Equation Roots h = 0.5d + 0 = d = .9t = Concavit Concave up Concave down Concave down Concave up Ma / min (0,) (,4) (0,0) (0.5, 36) c. 0 real roots d. real root t 0 to 5 inclusive 7. 0 to 70 inclusive. Unit Investigation. h and h are the same for all values of t; this shows that the objects were alwas at the same height and that the both hit the ground at the same time.. Each object was in the air for 3. seconds; this is seen b the fact that h (height) becomes 0 when t (time) reaches One covered 50 meters; the other covered 5 meters. 4. The coordinates of the points are: (0, 50), (, 45.), (, 30.4), (3, 5.9), (3., 0); a curve or a parabola; students graphs should show a parabolic curve that passes through all the points. 5. h 4.9t 6. The horizontal intercept increases. 7. The object would remain in the air for a longer period of time (because the horizontal intercept, which represents the time the object hits the ground, is greater). 8. The coordinates for the first trial are: (0, 50), (5.65, 45.), (3.30, 30.4), 46.96, 5.9), (50, 0). The coordinates for the second trial are: (0, 50), (7.83, 45.), (5.65, 30.4), (3.48, 5.9), (5, 0). 9. curves or parabolas 0. h is identical to Trial, and h is identical to Trial.. 50 and 5; the horizontal distance covered b each object. Trial ; The object in trial must have had a greater initial velocit because it covered a greater horizontal distance in the same time period
8 3. Solving Quadratic Equations Using Algebra Factoring & The Zero Product. Zero product theorem. two 3. horizontal intercepts or intercepts 4. ais of smmetr, verte 5. 0, 0 6. I 7., 8. double root 9. intercepts; parabolic function. a. two b (0.5 ) (0.5 ) or or or 4. (), ( 4) 3. a. 0, b. (, 3) c. 4. a. (6 )(6 ), or 4(3 ) b. 3 c The Square Root Method & Completing the Square. square root propert. a ( a) 0, (a) a 0 3. sum 4. perfect square trinomial 5. 3; or 3 8. square root propert 9. completing; square 0. irrational. two. a. 3 b. c. 3. b 4. a. 36 b. 00 c ( ) , 5 7 The Quadratic Formula b 4ac a. b. ; 8; real 5. b 4ac 6. radicand 7. no real 8. one real 9. two real
9 . a. square root propert b. completing the square c. zero product theorem h 4fj f. g h 3. a. 8 0; ; ; 8 b ; 3; ; 5 c ; 8; ; 7 d. 5 0; ; 5; 4. b b 4ac a ( 5) ( 5) 4(5)() Unit Assessment a. The equation has two solutions. This makes sense because the rabbit touches the ground ( 0) at two points, the point that it jumps from and the point where it lands. b. 0 or c. distance 0 ft.m 9 or m 9 3.s 0 or s ( ) b ( 9) (5) 0 or 9 6. b b 4ac a 7 7 4()(5) () or , The discriminant can tell ou how man real number solutions the equation has, as well as how often the parabola described b the equation crosses the ais a, b Unit Investigation..5 seconds. 3.5 seconds 3. a. (3.5, 0) b. (0, 60) 4. When 0,. Since is defined as the height in feet above the plaer s shoulders, the initial height of the ball is feet above the plaer s shoulders. 5. See Question 4 above feet 7..8 feet 8. Plaer C s; For Plaer B, the height is 8.5 feet; for Plaer C, the height is 8.96 feet. 9. Plaer B. Students ma refer to answers for Questions 6 and Radical Equations & Functions Solving Radical Equations. radical equation. a b Quadratic equation a = 0 b = 0 c = 0 d = 0 3. square root 4. radical; square Discriminant If the sstem of equations that represents the left and right sides of the radical equation has a point of intersection, then the  coordinate is the solution to the original radical equation. 48 Nature of roots no real roots no real roots two real roots two real roots 85
10 6. etraneous root true 9. etraneous root. 5. No. m indicates the principal, or positive, root, so it cannot have a negative value. Therefore, it cannot equal r r r 5 8 [(r 5) ] 8 r 5 64 r a. one solution domain. Yes. For ever point on the graph, each second coordinate is paired with one and onl one first coordinate and each first coordinate is paired with one and onl one second coordinate.. a. b. The domain is. c. The range is f() 0. d. f () e. 0 f() f. See part (a). g. See part (a) for graph;. b. 0 ( )( ) c. is an etraneous solution. d. The Inverse of the Square Root Function. onetoone. inverse 3. interchange; function; inverse 4. f () 5. range 6. domain Unit Assessment. a, ( a ), a 44. k 0, k, no solution 3. d 5 5 0, d 5 5 ( d 5) 5, d 5 5, d z z 7 5 z 5 z 7 (z 5) 5 ( z 7) z 0z 5 z 7 z z 8 0 (z )(z 9) 0 z 0, z 9 0, z, z 9 9 is a solution, and is an etraneous root. 5. one solution 86
11 6. f 5 f f f 5 ( f ) f_ 5 f_ f 5 f_ f 5 f_ 5 () 5 f_ f f 5 5 5f 5 f 0f 5 0 f 5f 50 0 (f 5)(f 0) f 5 or f 0, both are solutions 7. No; It is not a onetoone function because there are two values for ever positive value. 8. a. The domain is. b. The range is 0. c. f () d. 0 e. f() The rock hits the ground when s 0: t = 6 6 s, t , t 4 So, the rock hits the ground after 4 seconds. 6. Answers will var. Sample answer: Since the domain of the function is 0 s 56, the range of the function is 0 t 4. When s is 0, t is 0, and when s is 56, t is Yes. Answers will var, but students should indicate that the function is a onetoone function because for ever value of s, there is one and onl one value for t. 4. Rational Epressions, Equations, & Functions Rational Operations. rational epression; one. c a d b Investigating Gravit Student Investigation. The rock is 9 feet above the ground. t = 6 6 s, 6 6 s () 6 6 s s 6 s 9 s. No. After 3 seconds, the rock is feet above the ground. Since the bird is fling at a level of 0 feet, it will be higher than the rock. Graphs should support this answer s about.8 seconds 56 s 8 t 6 6 (8) a 4. ecluded value 5. Yes; the quotient of two rational epressions can be epressed as the product of two rational epressions. a c 6. b ad cb 7. b d 8. denominator; 9. b d. a. b. 0 c. d. 5. a. b, 0, ; b(b ) b. c 0; c. d 0; 4c c 5d 4d d. h 0, 6; h 6 h(h 6) e. k 0, ; 5k 4k k(k ) 87
12 Rational Functions. rational equation. hperbola 3. vertical ais, horizontal ais asmptote 6. discontinuous 7. ecluded value, vertical asmptote 8. horizontall 9. asmptotes. a. b. c. d.. 3. a. b.. a. k 8k b. w 36w c. p 5p 6. p.8 3. b hours rt 5. a. t 30 b. 8:0 A.M minutes Unit Assessment. a. h 0 b. g 3 c. d. a. w 6 b. z 5z 3 w z a. c. 4 u d f()  undefined 3 4 or 5 6 or 3 7 Rational Equations. common denominator rate, time, work 5. 7 b. 6. t r d 7. etraneous root 8. ecluded values 9. rational equation 88
13 5. 6 t h t, so if t 4, h hours 6. a. 5 miles b. 8 mph c..5 hours Unit Investigation. It decreases.. It decreases. 3. swing per second 4. 4 seconds 5. Students graphs should resemble the graph for f T but should be shifted awa from the asmptotes. 6. Students graphs should resemble the graph for Question 5 but should be shifted even farther awa from the asmptotes. 7. Increasing the wavelength of a wave causes the period of a wave to increase for a given speed. 5. Graphical Displas Stem & Leaf Plots & Bo Plots. collection; displa; analsis. histogram; intervals 3. stem; leaf 4. difference 5. skewed 6. median 7. four 8. second quartile 9. bo plot or boandwhisker plot 0. first; third. outlier. stem leaf a. 3.3 b c d ; 8.6; Scatter Plots & Linear BestFit Graphs. two. bestfit 3. medianmedian 4. No; the line M M 3 does not represent the data in the center group because most of the data points in the center group lie below the line M M parallel; 3 ; 3 6. trend 7. predictions.. Winning time (seconds) M Year. M (93, 58.); M (960, 55.); M 3 (980, 50.40); 3. a b M c M seconds 89
14 Unit Assessment. stem leaf 5 5mph a mph b. 5 mph and 70 mph c. 45 d. Yes; 40.9 represents the central tendenc in the data set because it divides the data set into two equal halves, both consisting of five data points.. a. 3 mph; 4 mph; 45 mph b. 3 mph c. The interquartile range shows that approimatel 50% of the animals listed in the table have speeds between 3 mph and 45 mph. 3. a. onedimensional b. onedimensional Outliers include Saguaro (5 ears) and Yellowstone (7 ears) A stemandleaf plot is the best wa to displa the desired information. The maimum is 7 ears, and the minimum is 5 ears. The mean of the data is 6.8. There are 5 values below the mean and 7 values above the mean. Thus, the data are onl slightl skewed, and the mean does represent a central tendenc. stem leaf /8 8 ears A scatter plot is the best wa to displa a general trend in the data. Accept all reasonable predictions for 000. Sample: The prediction ma not be reasonable because over the last ten ears the number of visitors has not increased at regular intervals. c. twodimensional 4. a. M (937.5, 64.55) M (960, 73.), M 3 (98.5, 77.8) b. m 0.94 c d. 8. ears Unit Investigation Number of visitors (thousands) M 3 M M A bo plot is the best wa to displa the desired information. The range of ears that the middle 50% of the parks have been established is the interquartile range. To find this range, each quartile must be found: Q Q ears ears Time (ears) Q ears The interquartile range is Q 3 Q 3, or ears. 90
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