Honors Solutions. Honors Lesson1Lesson 1 1. Honors Lesson2Lesson 2. Honors Lesson3Lesson 3. Trigonometry Arithmetic Algebra 3. 4.

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1 πhonors SOLUTIONS Honors Solutions Honors LessonLesson Trigonometry Arithmetic Algebra Calculus Geometry Honors LessonLesson. Make negative: ) Make negative: ) In both cases, changing the variable produces the original equation, so this equation is symmetrical in both axes.. 5 not symmetrical about the -axis) ) 5 5 not symmetrical about the -axis). ) symmetrical about the -axis) not symmetrical about the -axis) ) ) symmetrical around the - and -axis and the origin) 5. ) ) making negative changes the value, 6. while making negative does not; symmetrical around the -axis, but not the -axis or the origin) ) ) changing or would change the value, but changing both does not; symmetrical about the origin, but not the -or -axes) 7. ) ) changing changes the value, but changing does not; symmetrical about the -axis only) Honors LessonLesson round to 6 and add: 59.6º 9. 9 round to and add: 59.7º round to 6 and add:.86 ALGEBRA Honors Lesson - HONORS LESSON

2 Honors Lesson - HONORS LESSON round to and add:. 858 Honors LessonLesson ) ) 8 ) ) ) ) ) ) ) ) 6 a b a b) a a b b ) 8 W 8W W 6) W ) 6) ) 6r 8w r w) 6r 8rw w ) 8 r w) r rw w ) B B ) B 6) B B ) B ) B ) B ) 5 6 W 5W 6 W ) W ) ) ). 5 5 ) ) 5 x ) 5) ) ) ) ) 5) x 5) ) ) ) ). ) ). 8 ) ) a b a b ) a b ) 5. a b a b a b ) n 6. x x n n n n x x n x x n n x ) ) x x x Honors Lesson5Lesson 5 ou may have factored some of these in a different order, but your results should be the same.. x 5 5 ) ) x 5 ) 5) x ) ) ) Honors Lesson 6. tan 6º m.6 θ solutions

3 Honors Lesson 6 - HONORS LESSON m tanθ θ 58º tanθ 6 tan θ. 8 θ 8. 7º done tan º t an º ). 89) meters º 6. º 5'. 5º sin. 5º m º.5º º 5 m tan. 5º 86. Z Z. 7Z 86. Z 7. m Z 8 m m tanº m tan º m tan 5º m m There is more than one way to draw and solve many of these problems. Depending on the level of accuracy needed, you may also chose to round later in your calculations. ou can review significant digits in Algebra or Honors for Algebra. 5 m. º '. 5º 6º 5' 6. 75º 5º º Adding an extra line m above the ground is helpful here. It creates a rectangle with height length. tan tan ) m m above the ground Honors Lesson7Lesson 7. The angle of elevation from the ponds is the same as the angle of depression from the top of the mountain. alternate interior angles) m.5º 6.75º m solutions

4 Honors Lesson 7 - HONORS LESSON 8. tan m. sin 65º sin 6. 5º sin. 75º m m 7 m 6.5º 96. A 9. 6 x sq ft 8 m.75º 5 ft ft 65º 5. tan 5º tan 7º ) ) ft height ft Honors Lesson8Lesson 8. W N º 5' E S 5 ft 7º 85 ft 5º 5 ft solutions

5 Honors Lesson 8 - HONORS LESSON 9.. N N W 8º 5' E W 5º E. S S N W º ' E Honors Lesson9Lesson 9. sin 5. 7º tan5. 7º km km trip back km round trip km N S 65 km W 5.7º E S solutions 5

6 Honors Lesson 9 - HONORS LESSON 9. The turn between a heading of º and º is a 9º angle, so we can use the Pythagorean theorem: 8 H, 59, 9 H 9, 9 H H km Total trip 8 75 km N 8.8º W 95 m 6.7º E N W 8 km S H km E. sin 8. 8º cos 8. 8º m m The Pythagorean theorem may be used instead to find the third side of the triangle.) A 5. 8) 8. ) 9, 6 sq m S. tan. 8º H 6., H H car went mi car time hours 55 bus time hours hrs, or 8. minutes There may be more than one way to draw these problems. N.8º H 75.7º W 6. mi E S 6 solutions

7 Honors Lesson - HONORS LESSON Honors Lesson Lesson. 8 x 5 mi º tan mi H, 9, 6. 9 H 5, 5. 9 H H mph 5. mph tan A 5. 5 A º 5º ' C 5 mph H.5º A B 5 mi A.5º mi. 5 H tan A 5 9, 65 H A 76. º º 6' 9, 65 H H. mph 7º tan º or º ' 5 5 H H 5 mph The plane is blown º' toward the the west, so its new direction of travel will be: 7º ' º ' 658 º ' 5 mph Heading is 65º' H A mph 7.7º x H 5 mph 5 mph.º solutions 7

8 Honors Lesson - HONORS LESSON 5. tan º 7 6 H H º. 5º 655. º Direction is s 65.5º E at 7.7 knots º 8. cos A cos B sina sin B cos A B) cosacos B sin Asin B cos A B) cosa cosb sina sinb cos A B) Challenge question: The same drawings can be used, but start with two smaller angles and add up to the larger one. ou should end up with this: cosa cos Bsin A sinb cos A B) 78º 7 knots 6 knots Honors LessonLesson. done. done. cos B, sinb). cos B, sinb) 5. CD ) ) cos Bcos A) sinb sin A) cos B cosbcos A cos A sin B sinbsina sin A Combine sin B cos B to make and combine sin A cos A to make.) cosbcos A sinbsin A 6. done 7. FE ) ) sin A B) ) cos A B) ) sin A B) cos A B) cos A B) Combine sin A B) cos A B) to make.) cos A B) 8 solutions

9 Honors Lesson - HONORS LESSON Honors Lesson, ), ), ) 5º 5º º, ) 9º, ) 6º, ) ) 5º º, ), ) 8º º 6º, ), ) Lesson...., ) all sin, csc tan, cot cos, sec º 5º º, ) 7º, ) º 5º º, ), ), ) Honors LessonLesson. ) 5) 6. )sin 5º 969. cm. ). 5) 8. )sin 69º 8. m. ) 8) )sin º 96. cm. 8º 6º 6º) 6º ) 7 ) 7 )sin 6 º. mm solutions 9

10 Honors Lesson - HONORS LESSON Honors LessonLesson. P { ±, ± } poss. roots { ±, ± } q { ± } try : Lessoni ) ) ) yes is a root, so ) must be a factor. ou may use different roots for the initial tries, but final answers should be the same.. ) ) roots are, and p { ±, ± } poss. roots { ±, ± } q { ± } try : Lessonii ) ) ) 7 7 yes is a root, so ) must be a factor. ) ) roots are, and. 5 p { ±, ±, ± } poss. roots { ±, ±, ± } q { ± } try : ) 5 ) 5 yes is a root, so ) must be a factor. 5 5 R epeat with the results of the division: p { ±, ± ± } poss. roots { ±, ±, ± } q { ± } try : ) ) ) 8 8 yes is a root, so ) must be a factor. 6 ) ) all roots are,, and solutions

11 Lessoncontinued) p 8 { ±, ±, ±, ± 6, ± 9, ± 8} q { ±, ± } poss. roots: { ±, ±, ±, ± ± ± ± ± ± },,, 6, 9, 8 try : ) 6 ) 5 ) 8 try : no ) 6 ) 5 ) 8 8 yes is a root, so ) must be a factor ) ) ) roots are,, Honors Lesson5Lesson r. Honors Lesson - HONORS LESSON 6.. r r r r r Honors Lesson6Lesson 6.. has multiplicity of has multiplicity of has multiplicity of ) ) ) ) ) ) roots : has a multiplicity of has a multiplicity of has a multiplicity of solutions

12 Honors Lesson 6 - HONORS LESSON 8. roots : i has a multiplicity of i has a multiplicity of has a multiplicity of 6 has a multiplicity of. multiply out the four factors indicated by the given roots: i) i) ) ) Honors Lesson7Lesson ) 5 ) ) ) ) ) ) ) ) no Honors Lesson8Lesson 8. Airplane : rcosθ cos 9º º). 68) 57. r sinθ sin 5º. 766) 6. kph kph Wind : rcosθ cos ). 988) 9. 5 r sinθ sin ). 76) , 6.) yes yes θ tanθ. r tan θ. 59 r 7, 86 θ 6. º r 5. 8 kph 5.8 kph, n.6º E) solutions

13 Honors Lesson 8 - HONORS LESSON , 9.5) θ θ tanθ tan θ 6. º r 8, 5 r r 6. º, 67.7) º, 6.º 67.7) 5. Bob: 7cos º sin º. 6 steve: cos 65º. 68 sin 65º 7. 9 r r 8,. r 9. 6 tanθ θ 5. 7º 9. 6, 57. º) Rounds to: 9, 6º) Honors Lesson9Lesson 9 ou may have used other values for the charts.). ) º 7 º solutions

14 Honors Lesson 9 - HONORS LESSON. ). translated right two places. translated left four places ; ). ; ) 8. ; ) < ; ) < ; ) < ; ) done translated down two places 7. translated down three places Honors LessonLesson. done. translated right three places solutions

15 Honors Lesson - HONORS LESSON 8-. inverse: ; 7. ) Honors LessonLesson. yes. no. yes. no 5. no 6. ) inverse: ; function inverse: ; function Honors LessonLesson. ) e. 86 ) ln ) log ). 99 and e e. ) e ) ln ) log ). 86 and e e solutions 5

16 Honors Lesson - HONORS LESSON Honors LessonLesson. ) 7. h sin π 6 ) continuous when h sin π. ) 5 6 h. 5) First, find values of that will make the h 5. denominator equal to zero: h. 5. h sin π ) 6 solve the equation using the h ) quadratic formula: h ) ± ) ) ) h 5 ). h sin π ) ± 9 8) 6 6 h ) ± 7 h Continuous for all except 5. h sin π 6 8 ) 7 and 7. h. 866) h ) 6 h Continuous where > h sin π 6 9 ) If 5, the denominator is, if < 5, h ) the denominator is imaginary. h. ) h 8 ; otherwise the numerator is imaginary. 7. h sin π 6 ) 8, because 8 makes the denominator. h ) The first restriction includes the second. h h Honors LessonLesson. hsin π 6 ) h sin h ) h h 8. h sin π 6 5 ) h ) h h 5 9. This means that the highest point of the graph is three units above a line representing the midpoint of the graph. The lowest point of the graph is three units below the line. so, there are six meters between high and low tide. 6 solutions

17 Honors Lesson - HONORS LESSON 6. There are hours from one high tide to the next. Honors Lesson 5. ) parabola : horizontally widens or narrows the parabola. ) parabola : translates the graph up or down. ) 6 hyperbola : horizontally widens or narrows the hyperbola. ) sin sine curve: translates the graph up or down 5. ) sin sine curve: vertically stretches or shrinks the curve 6. ) sin sine curve: horizontally widens or narrows the curve Honors Lesson6Lesson 6. on graph z x.,, ),, ) d ) ) ) midpoint,, ),, ),, ). on graph 5.,, ),, ) d ) ) ) midpoint,, ),, ) 7. ) ) Z ) 8 6 Z Z Z 8 Z 8. Z Z Group each variable together, and put the constant on the right: Z Z Complete the square in each variable: Z Z ) ) Z ) center :,, ) radius : P C Q y D solutions 7

18 Honors Lesson 7 - HONORS LESSON 9 Honors Lesson7Lesson 7. ) ) ) ) ) ). ) ) ln ) 5 )) 5 5 ) ) ) 5 ) ) ) )) 7 5 ) ln. 69. ) ) ) ) ) ln ln. 69 Honors Lesson8Lesson 8. < < < < 9 < < 8, 8) [, ] [ 8, 6] < ) ) < We are looking for values of that yield negative products, so the factors must come from the area where the signs are different, so < < or, ) 5. 5) ) signs must be the same: 5 and, 5] [, ) 5 Honors Lesson9Lesson 9. try : try : ) ) ) 6 no no. It signifies that the value of is not included in the solution. 8 solutions

19 Honors Lesson 9 - HONORS LESSON 9. It signifies that the value of is included in the solution If <, then would be negative. In that case, multiplying by would have reversed the direction of the inequality. < 6 < 5 < 5 < < 5 < < < 5 5, ) 5 6 > 6 > > > > > > <, ) solutions 9

20 Honors Lesson - HONORS LESSON Honors LessonLesson. A) A A 6 A h) A h) A h) 6 A Ah h ) A h) 6 A Ah h A h) 6 ' A) lim [ A Ah h A h) 6][ A A 6] h h lim Ah h h h h lim A h h A. ) 5 h) h) 5 h) h h ) 5 5h 6h h 5 5h ' A) lim [ 6h h 5 5h ][ 5 ] h h lim 6h h 5h h h lim 6 h 5 h 6 5 When you study calculus, you will find that there are shortcuts for problems such as these. 5 solutions