No. For example, f(0) = 3, but f 1 (3) 0. Kent did not follow the order of operations when undoing. The correct inverse is f 1 (x) = x 3

Transcription

1 Lesson a: Each laer has 7 cubes, so the volume is 42 cubic units. b: = 98 square units c: (1) V = 20 units 3, SA = 58 units 2 (2) V = 24 units 3, SA = 60 units 2 (3) V = 60 units 3, SA = 94 units a: = 9 b: = 2( + 3) 2 25; verte: ( 3, 25); -ints: appro. (0.54, 0) and ( 6.54, 0); -int: (0, 7) a: Step function and/or piecewise-defined function. b: See graph at right. c: It will be much cheaper ($13.38 vs $22.42) to mail them in one package No. For eample, f(0) = 3, but f 1 (3) 0. Kent did not follow the order of operations when undoing. The correct inverse is f 1 () = 3 2, or f 1 () = D Cost ($) Weight (lb) The angles, from smallest to largest, measure 64, 90, 116, 130, and 140, so the probabilit is a: 64 units 2 b: 27.0 units 2 c: units 2

2 Lesson ( 4) 2 + ( + 3) 2 = 16, the center is (4, 3) and the radius is 4 units See graph at right. The graph should include a circle with radius 5, center (0, 0), and a line with slope 1 and -intercept (0, 1). Intersection points are (3, 4) and ( 4, 3) a: 1 3 b: = 2 ± or 4.65 c: < 9 or > 4 d: = 2.5 or See graph at right. a: ( 2.5, 0) and (3, 0) b: The graph of = (2 2 15) is a reflection of = across the -ais because each -value has the opposite sign. The -intercepts are the same a: 2 5i b: i c: i V = (16)(16)(16) = 4096 units 3 ; SA = (6)(16)(16) = 1536 units a: See diagram at right. b: P(both blue) = 5 6 c: = % (49π ) square cm d: = 12 1, 12 1 (360 ) = a: 4( 3) b: 3( + 1) 2 c: m(2m + 1)(m + 3) d: (3 2)( + 2) blue red Core Connections Integrated II

3 Lesson Da a: = 3 b: m = 10 c: p = 4 or 2 3 d: = Possible function below, where f(t) = value of the account after t ears < t 0.25 f (t) = 200(1.02) t 0.25 < t A slice of circular pizza has area ft 2 (or a slice of square pizza has area 18 in 2 ) so ou should order a slice of the square pizza º a: 1: eponential; 2: quadratic; 3: linear b: The linear model does not fit the data well, as shown b a pattern in the residual plots. The linear model is predicting values too low at the ends of the data and too high in the middle. Both the eponential and quadratic look reasonable for this range of data however the eponential makes more sense if ou wish to etrapolate the data. Notice the verte of the quadratic model has been reached so the parabola will begin rising, meaning that training beond that distance would predict slower (longer) 5K times using the quadratic model. c: 1: min; 2: min; 3: min d: There is a strong association between the distances run in training and 5K race times. However, this stud cannot show cause and effect. Perhaps people who are natural runners enjo running more so the run more in training. It is not possible to tell which is causing which. There are also man other variables that could be a source of the faster 5K race times tied to training distance like age, eperience, equipment, and gender a: b: c: ( + 5i)( 5i) a: 30º b: 5 cm c: 0.5 d: C

4 Lesson Da a: b: ( 6) or + 6 c: = ( + 6) 2 or = a: (2, 3); r = 5 b: ( + 1) 2 + ( + 3) 2 = 16; ( 1, 3); r = a: f 1 () = 3( + 2) b: g 1 () = 2( 5) a: 1 is the starting area of bacteria. Baile s epression shows that the area is multiplied b 2 (ever 20 minutes), and because there are si 20-minute periods in two hours, the multiplier needs to be applied 6 times. Carmen s epression shows that the area of bacteria will multipl b 2 1/20 per minute, and there are 120 minutes in two hours. Demetri s epression shows that the area of bacteria will multipl b 2 3 ever hour during the two-hour period. There will be 64 sq. cm of bacteria after two hours (assuming that the dish is not completel covered before that time). b: Possible epressions: 1(2 1/2 ) 1 or 1(2 1/20 ) 10 or 1(2) 1/ sq. cm a: = f(); f(0) = 3 and f(1) = 2, both function values are smaller than the minimum value of g(0) = 9. b: ( + 3i)( 3i) a: one solution; Eplanations will var. b: no real solutions; 2 must be positive or zero. c: two solutions; Eplanations will var. d: no solution; Absolute value must be positive or zero a: a = 120º, b = 108º, so a is greater C b: Not enough information is given since it is not known if the lines are parallel. c: Third side is approimatel 8.9 units, so b is opposite the greater side and must be greater than a. d: a is three more than b, so a must be greater. e: sin 23º = cos 67º, so a 7 = b 7 and a = b. Core Connections Integrated II

5 Lesson a: 70 b: 50 c: See solution graph at right. a: C: (0, 0); r = 4.5 b: C: (0, 0); r = c: C: (3, 0); r = 1 d: C: (2, 1); r = = 24, so the tower is 24 blocks tall (4, 3) and ( 4, 3). Solving algebraicall involves working with fractions (or decimals). If ou graph the line precisel using the slope starting from the -intercept at the origin, ou ma notice that moving up 3 units and right 4 units ields an eact intersection point, as does moving down 3 units and left 4 units a: Investment A: linear, growing $10/month; Investment B: eponential, growth rate of 1%; Investment C: quadratic, growing b $1, $3, $5, $7, $9, b: A: = $10/mo; B: = $10.45/mo; C: = $10/mo c: Answers var. Investment C will surpass Investment B before the end of the ear. Investment B will eventuall grow larger than Investment C, but onl after 588 months, or about 49 ears a: = 45 4 = b: = 10 or = 10 c: = 1.3 d: = 2 ± i a: See diagram at right. b: 2(3w 2) + 2w = 100 and w(3w 2) = 481 c: width is 13 feet, length is 37 feet w P = 100 ft l = 3w M is a midpoint of Given P L Given Definition of midpoint KML QMP Vertical angles are congruent ΔKLM ΔQPM AAS

6 Lesson a: 64 b: 128 c: 64 d: 180 e: 128 f: a: See diagram at right. B b: 108 ; interior angle of a regular polgon c: 72 ; central angle A O C d: 216 ; This measure can be calculated as 2(m EDC) or as 3(m BOC). E D a: C: ( 5, 0), r = 10 b: C: (3,1), r = a: = 16 5 b: no soln. c: = 11 or 3 d: = a: f 1 () = +2 7 b: Yes, it should be See graph at right. a: b: 4 c: = (4 ) 2 or = Original: A = 135 sq. units, P = 48 units New: A = 15 sq. units, P = 16 units a: If is the number of guests, f() = b: p() = 50 ( ), or p() = c: > 100; at least 10 guests Core Connections Integrated II

7 Lesson a: 124 b: 25π units 2 c: 12.3 units a: 3 b: 6 c: 2 d: 1 e: 4 f: Point X is the circumcenter of TRA. It is equidistant from the vertices of TRA and is the center of the circumscribed circle of the triangle. It is the point of concurrenc of the perpendicular bisectors of the sides of TRA cm a: = b: = 2 or = 1 2 c: 3 = 5 d: + = 180º a: 2 3 b: = 4 or 2 c: 5 < < 4 d: = 3 2 ± 1 2 i a: no intersection points b: (1, 4) and ( 2, 7) a: See graph at right. b: 4 c: f( 4) = 1

8 Lesson MA = = 39 feet, MB = 6 feet, so AB = ER and ER = feet See graph at right. -intercepts: (4, 0) and ( 4, 0); -intercepts: (0, 2) and (0, 8) a: Using properties of isosceles triangles and trigonometr, AC b: a: 360º 9 = 40º b: mad = 2(97º ) = 194º ; m C = 0.5(194º) = 97º c: mab = 125º and the length of AB = 125º 360º (16π ) 17.5 units ; area = 125º 360º (64π ) 69.8 sq units ( + 3) 2 = 26 2 ; = a: 2 b: 3 c: ( π (3)2 (2)) 7.07 in = 360º; 25.7º Core Connections Integrated II

9 Lesson a: = 270 b: = 132º, 15.7 c: 3( + 2) = 6, = ( 4) 2 + ( 2) 2 = a: 50 b: 50 c: 67 d: 126 e: 54 f: a: 13 b: 6.5 c: 67.4º d: 134.8º a: = 2 or 5 b: = 2 3 ± i 26 3 c: = 8±2 7 2 = 4 ± or 6.65 d: = 3 or If t = hours, t = 12 ; 40 hours Since the perimeter is 100, each side is 10. The central angle is = 36. The right triangle has acute angles 18 and 72. Area = units ; One method: The sum of the angles in PSR is 180, so m SPR + SRP = 40. Then add the 40 and 35 of QPS and QRS to get m QPR + QRP = 115. Thus, m Q = = 65.