Skill Builders. (Extra Practice) Volume I

Transcription

1 Skill Builders (Etra Practice) Volume I 1. Factoring Out Monomial Terms. Laws of Eponents 3. Function Notation 4. Properties of Lines 5. Multiplying Binomials 6. Special Triangles 7. Simplifying and Combining Radicals 8. Similar Triangles 9. Factoring Standard Forms 10. Inverses 11. Algebraic Fractions 1. Solving Quadratics 13. Inequalities

2 1. Factoring Out Monomial Terms Eample: The first thing you should look for when factoring epressions is common factors that appear in every term. These common factors could be numbers or variables, so look for common numerical factors and common variables that appear in every term. 3 y 5! 5y 3 = 3y 3 (y! ) Factor out the largest single term possible from the following epressions y + 7y [ y( + 7y) 3. 4!r 4 h " 0!r h 4 [ 4! r h(6r " 5h 3 ) y 5 z 6! 7 3 y 3 z y z 7 [ 9 3 y z 6 (4 y 3! 3yz z) 7. 7 p 6 q + 45 p 5 q! 54 p 8 q [ 9p 5 q(3p + 5q! 6p 3 ) y y 8 [ 7 y 3 (3 + 5y 5 ) 11. 4!r h + 18!r h + 54!r [ 6! r (4h + 3h + 9) 13. a 5 b 5 c! 6a 3 b 3 c 6 + 7a b 6 c 3 [ a b 3 c(a 3 b! 6ac 5 + 7b 3 c ) 15. 8u 3 v! 3uv uv [ 8uv(u! 4v ). 4a 3 b 3 + 4a b 5 [ 6a b 3 (7a + 4b ) 4. 30b 11 c 6 d b 11 c 7 d 3! 35b 10 c 7 d 4 [ 5b 10 c 6 d 3 (6bd + 3bc! 7cd) 6. 6a 4 b! 7ab 4 + 7a b [ ab (6a 3! 7b + 7a) 8. 8 j 10 k + 0 j 3 k 4 [ 4 j 3 k ( j 7 + 5k ) 10. 4d 3 h d h 5 [ 1d h 3 (d + 3h ) 1. 15mn 4! 33m 3 n [ 3mn(5n 3! 11m ) s t 5! 5st st 5 [ 5st 5 (3s! 5t + 11) y y 5! 3 y + 6 y [ y (3 y + 4y 3! + 3) Pre-Calculus with Trigonometry

3 . Laws of Eponents Eample: Solution: Simplify using the laws of eponents. " # $!3 y y 3!4 % & ' =!4 (!3!1 )!4 (y!3 )!4 =!4 16 y 4 = 16 y 4 16 Simplify the following epressions. 1. ( 3 )( 4 ) [ 7. (y )(y 3 ) [ y 5 3. a 9 a 6 [ a 3 4. p 7 p [ p 5 5. ( 3 ) 3 [ 9 7.! "(z ) 3 [ z 1 # $ 9.! "(z )( 4 )# $ [ z (y ) 5 [ y 10 8.! "(b 3 ) 3 # $ [ b " #(y! )(y 3 ) $ % [ " #(a!3 )(a 5 ) $ % [ a ( 4 ) [ (!3 )( 3 ) + (y 4 )(y!4 ) [ 14. () 3 (3) [ (5) () 5 [ ! (3) 3 () 4 "# [ 79 8 $ %& Skill Builders

4 3. Function Notation Eample: Let f () =! + 3 and g() = 3! 5. Find the indicated values. Solutions: a. f (!) = (!)! (!) + 3 = 15 b. f ( + h) = ( + h)! ( + h) + 3 = ( + h + h! ( + h) + 3 = + 4h + h!! h + 3 c. g( f ()) = g(! + 3) = 3(! + 3)! 5 = 6! Find the indicated values. 1. Given f () = 3 + 5, find f (4). [ 53. Given f () = , find f (!3). [ Given f () = 5 + 3, find f ( + h). [ h + 5h h 4. Given f () = 1! 0, find f ( + h). [ 1 + 4h + 1h! 0! 0h 5. Given f () = and g() = 7 + 3, find f ()g(). [ Given f () = and g() = e, find f ()g(). [ 6 3 e + 7e 7. Given f () = and g() = + + 1, find f ()g() if = 3. [ Given f () = + 5 and g() = 3, find f (g()). [ Given f () = + 5 and g() = 3, find g( f ()). [ Given g() = and h() = 4 + 4, find g(h()). [ Given g() = and h() = 4 + 4, find h(g()). [ Given h() = 16! 3 5, find h(). [ Given f () = 5! 3, find f (3). [ Given f () = e!and! g() = , find f (g()). [ e ( ) 15. Given f () = e and g() = , find g( f ()). [ 3e + e + 6 Pre-Calculus with Trigonometry

5 4. Properties of Lines Given the points A = (!, 5) and B = (3,!4). Eample 1: Find the equation of the line that goes through A and B. Write your answer in slope-intercept form. Solution: slope = (5!(!4)) (!!3) = 9!5 =! 9 5 Now use the form y = m + b with m =! 9 5 to get the equation y =! b. Substitute the and y from point A or point B and solve for b. 5 =! 9 5 (!) + b 5 = b b = 7 5 y =! Eample : Solution: Find the equation of the line that is perpendicular to line AB and passes through the point (, 3). Perpendicular lines have negative reciprocal slopes. Since the slope of AB is! 9 5, the slope of the line perpendicular to it is 5 9. Point- slope equations have the form y! y 1 = m(! 1 ). Substitute (, 3) for ( 1, y 1 ) and 5 9 for m to get: y! 3 = 5 (! ). 9 Eample 3: Find the distance between points A and B. Solution: ( ) + 5! (!4 ) d =!! 3 ( ) =!5 ( ) + ( 9) = 106 Skill Builders

6 1. Find the equation of the line that is parallel to the line y = and passes through the point ( 4, 3). Write your answer in point-slope form. [ y! 3 = 3( + 4). Find the equation of the line that is parallel to the line y = + 8 and passes through the point ( 4, 3). Write your answer in slope-intercept form. [ y = Find the equation of the line that is parallel to the line y =!6! 8 and passes through the point (8, 3). Write your answer in point-slope form. [ y! 3 =!6(! 8) 4. Find the equation of the line that is parallel to the line y =!5! 9 and passes through the point (7, 1). Write your answer in slope-intercept form. [ y =! Find the equation of the line that contains the points (1, 7) and ( 5, 3). Write your answer in slope-intercept form. [ y = Find the equation of the line that contains the points ( 5, 10) and (9, 6). Write your answer in slope-intercept form. [ y = 8 7! Find the equation of the line that contains the points ( 1, 4) and ( 5, 4). Write your answer in point-slope form. [ y! 4 = ( + 1) or y + 4 = ( + 5) 8. Find the equation of the line that contains the points (5, ) and (8, 11). Write your answer in point-slope form. [ y! = 3(! 5) or y! 11 = 3(! 8) 9. Find the equation of the line that is perpendicular to the line y = and passes through the point (3, 1). Write your answer in point-slope form. [ y + 1 =! 1 7 (! 3) 10. Find the equation of the line that is perpendicular to the line y = 6 + and passes through the point (3, 7). Write your answer in slope-intercept form. [ y =! Find the equation of the line that is perpendicular to the line y =!9! 8 and passes through the point (1,6). Write your answer in slope-intercept form. [ y = Find the equation of the line that is perpendicular to the line y = + 5 and passes through the point ( 6, 7). Write your answer in slope-intercept form. [ y =! 1! Find the distance between the points ( 3, 5) and (5, 4). [ Find the distance between the points ( 1, 6) and (3, 4). [ Find the distance between the points (4, 3) and (18, 14). [ Find the distance between the points (0, 7) and (4, 16). [ 97 Pre-Calculus with Trigonometry

7 5. Multiplying Binomials Eample: Multiply ( + )(! 5). Each term in the first binomial is distributed over the second binomial, as shown below. Solution: ( + )(! 5) = (! 5) + (! 5) =! 5 +! 10 =! 3! 10 Multiply the following binomials. 1. ( + 3)(! 6) [! 3! (b 3)(b + 5) [ b 3 + 5b 3b (p 11 p 10 )(p 10 p 9 ) [ p 1 p 0 + p (5h 4 + 1)(6h ) [ 30h 5 10h 4 + 6h 9. ( + 3)( + 6)( 7) [ (m 99 + m 98 )(m 97 m 96 ) [ m 196 m (100 j 100 j)(9 j + 1) { } [ 900 j j j 3 j 15. (c + d 3 y)( + c d) [ c + d 3 y + c 4 d + c d 4 y. (a 3 + 1)(a 4) [ a 5 4a 3 + a 4 4. (z 3 + 1)(z 1) [ z 5 z 3 + z 1 6. (q 4 + 4)(q 3 + 3) [ q 7 + 3q 4 + 4q (d 5 d 4 )(d 6 + 1) [ d 11 d 10 + d 5 d ( + 3)( )( + 1) [ (6n 4 + 3n 3 )(6n 3 + 3n ) [ 36n n 6 + 9n (!k 5 + 5k 4 )(!k + 3) [! k ! k k ( f 3 y 3 + g )( fy + g) [ f 4 y 4 + f 3 gy 3 + fg y + g 3 3 Skill Builders

8 6. Special Triangles Eample: The following relationships eist between the sides of the special triangles. Use these relationships to solve each of the following problems. 1. Solve for angles α and β in the triangle at right. [! = 30!, " = 60!. Given a triangle with hypotenuse length 8, find the eact length of the long leg. [ A triangle has a hypotenuse 0.5 inches long. How long is the short leg? [ 0.5 inches 4. Given a triangle whose long leg is 13 cm, how long is the hypotenuse? (Solve eactly.) [ cm 5. Solve for leg lengths a and b. [ Both a and b = Given a triangle with hypotenuse length 13, how long are the legs? [ 13 each 7. If a triangle s legs are 17 mm long, eactly how long is its hypotenuse? [17 mm 8. What is the eact area of the equilateral triangle whose legs are each 6 cm long? [ 9 3 cm Pre-Calculus with Trigonometry

9 9. Find the eact area of the following isosceles triangle. [ units A triangle has angles 30, 60, and 90, and its legs have lengths 7 3, 7, and 14. How long is the side opposite the 60 angle? [ Solve for α and β in the following triangle. [! = 30!, " = 60! 1. If a triangle has a hypotenuse length 100, how long is the short leg? [ A triangle with long leg length 15 3 has a hypotenuse of what length? [ Find the area of the triangle at right. [ Solve for length in terms of c. [ = c 16. Solve for lengths and y in terms of b. [ = b 3, y = b Skill Builders

10 7. Simplifying and Combining Radicals Eample: Simplify 88. When you are asked to write a radical in simpler form, the real task is to reduce the number under the radical sign as much as possible by taking perfect squares outside the radical. Solution: Simplify 88. Find the factorization of 88, choosing perfect squares as factors whenever possible. 88 = 4! 4! 9! 88 = =!! 3! 88 = 1 Rewrite the following as radicals in simpler form [ [ a 3 b 5 [ ab ab 4. 7 b 9 [ 3 b 4 b Eample: Simplify 3 15! 80. Solution: First simplify each radical separately and then combine like terms = = = 16 5 = ! 80 = 15 5! 8 5 = 7 5 Simplify the following epressions [ [ [ ( 99 ) [ ( 1)( 75) [ ( 0)( 80) [ ( 50)( 98) [ 70 Pre-Calculus with Trigonometry

11 Eample: Rationalize 45. Solution: To rationalize the denominator you need to multiply both numerator and denominator by the radical in the denominator. This is equivalent to multiplying by 1 so it does not change the value of the epression. 4 5 = = Rationalize the following epressions [ [ a b 4 a 5 b 9 [ ab ab Eample: Combine and simplify Solution: To combine rational epressions you first need to find a common denominator and write the epression as one fraction. Then, you must rationalize the denominator =! + 1 = 5 = Rewrite the following as single, rationalized epression [ [ 9 4 Skill Builders

12 8. Similar Triangles Eample: Solution: Find the length of. To find you need to put corresponding sides in a ratio. In this triangle 17 corresponds to 7 and 1 corresponds to. So our ratio is: 1 = = 84 = Solve each triangle for the indicated variable. 1. Find the length of. [ = Find the length of y. [ = Find the length of. [ = Find the length of. [ = 18 1 Pre-Calculus with Trigonometry

13 5. Find the length of. [ = Find the length of. [ = 3 7. Find the length of k in terms of. [ k = 8. Find the value of h in terms of y. [ h = 4 y 10! y 9. Find the length of b in terms of. [ b = 10. Find the value of h in terms of. [ h = Skill Builders

14 11. Kathleen is 6 feet tall. Her shadow is 3 feet long. At the same time of day, a tree casts a shadow that is 1 feet long. How tall is the tree? [ 4 feet 1. Kyle is 1.5 meters tall and casts a shadow 4 meters long. A flagpole casts a shadow that is 30 meters long. How tall is the flagpole? [ 11.5 meters 13. Lisa s dog is.5 feet tall and casts a shadow that is 4 feet long. A fire hydrant casts a shadow that is 6 feet long. How tall is the fire hydrant? [ 3.75 feet 14. Fernando is constructing a scale model of his father s yacht. The actual yacht has a sail that is 5 feet tall and the base of the sail is 15 feet long. If the model s sail is 14 inches tall, how long is its base? [ 8.4 inches 15. Cameron is building a scale model of his sailboat. The right-triangular sail of the actual boat has a base that is 1 feet long. The height of the actual sail is 0 feet long. The base of the model s sail is 8 inches long. How long is the hypotenuse of the model s sail? (The hypotenuse is different from the height!) [ List all of the similar triangles that are in the figure at right. [ ABC, APB, CPB Pre-Calculus with Trigonometry

15 9. Factoring Standard Forms Rules for Factoring Difference of Two Squares: a! b = (a + b)(a! b) Sum of Two Cubes: a 3 + b 3 = (a + b)(a! ab + b ) Difference of Two Cubes: a 3! b 3 = (a! b)(a + ab + b ) Special Products (a + b) = a + ab + b (a! b) = a! ab + b (a + b) 3 = a 3 + 3a b + 3ab + b 3 (a! b) 3 = a 3! 3a b + 3ab! b 3 Factor each of the following epressions. 1.! [ (! 6) 3. 9a + 4a + 16 [ (3a + 4) 5. ( + y)! 4( + y) + 4 [ ( + y! ) 7. b! 16 [ (b! 4)(b + 4) 9. 4! 1 [ (! 1)( + 1) 11. s 6! t 6 [ (s 3! t 3 )(s 3 + t 3 ) 13. a 3! b 3 [ (a! b)(a + ab + b ). z! 6z + 9 [ (z! 3) y + 5y [ (4 + 5y) 6. 4(a! b)! 8(a! b) + 4 [ 4(a! b! 1) ! 7 [ ( ) (! 1)( + 1) 10. (a! b)! (a + b) [!4ab [ ( + 1)(! + 1) p 3 + 7q 3 [ (p + 3q)(4p! 6pq + 9q ) Skill Builders

16 10. Inverses Eample: Find the inverse of f () =! Solution: y =! Start by replacing f() with y. = y! Interchange and y in the equation. y = y! Solve for y. y! y = y(! 1) = y =!1 f 1 () =!1 Replace y with f!1 (). Find the equation of the inverse for each function below. 1. f () = [ f!1 () =! f () = + 5 [ f!1 () = ±! 5 5. f () =!4 5 [ f!1 () = f () = [ f!1 () = 4 3 (! 8) 9. f () = 3 [ f -1 () = log 3. f () = 5! 6 [ f!1 () = f () = 4 [ f!1 () = ± 6. f () = [ f!1 () = 11! f () = 7! [ f!1 () = f() = log [ f -1 () = f () = f () =!4 [ f!1 () =!1 3 [ f!1 () = 4 1! 13. f () = 11!1 [ f!1 () = f () = 10!15 5 [ f!1 () = f () = [ f!1 () = 4 5! f () = 7!4 3 [ f!1 () =!4 3 7 Pre-Calculus with Trigonometry

17 11. Algebraic Fractions Facts about Fractions PK QK = P Q where Q! 0 and K! 0 Multiplication: P Q! R S = PR QS Division: P Q R S = P Q! S R = PS QR Addition: Subtraction: P Q + R Q = P+R Q P Q! R Q = P!R Q Simplify the following fractions a b [ 1 ab. 1 a 1 b [ b a [ [!( + 5) 6. [! 6!1 a 3 +b 3 a b [ a! ab+ b a! b Eample: Solution: Use Fraction Busters to solve =. Multiply every term on both sides of the equation by all the terms in the denominator. Now you have an equation without fractions. Use the Quadratic Formula to solve. 3!! 3 + 3!! 5 =! 3! /3!! /3 + 3! /! 5 / = = 6 " = 0 =3±i 6 Skill Builders

18 Use Fraction Busters to solve for = 4 [ = 1,! ! +1 5 = [! = + 6! 3 [ =! = [ 1± 881 = 144 Simplify the following epressions [ ! 1 y 1 y [ y!1 y 13. 1! 1! 3 [!! a + b a + 1 b [ a+ b b+ a 15.!1 +1! +1!1! !1 [! a+b a a!b 4a [ (a+ b) a(a! b) Pre-Calculus with Trigonometry

19 1. Solving Quadratics Solve the quadratic equations. Eample 1: 4! 1 = 0 Solution: This type of problem is easily factored into 4(! 3) = 0. Now you can get the solutions 4 = 0 or! 3 = 0, = 0 or! 3 = 0. Eample :! + 13 = 0 Solution: This equation cannot be factored so you should use the Quadratic Formula. = ±!48 = ±4i 3 = 1 ± i Eample 3: + 1 = + Solution: When there are square roots in the problem you will need to square both sides. ( + ) = ( + 1) + = = Since there is still a square root, move all the other terms away from the root. 1 = Now square both sides = Make sure you plug the answer back into the original equation because sometimes etra solutions appear when you square equations. 1.5 = 1.5 Skill Builders

20 Solve for the indicated variable. 1. 9! 1 = 0 [ = 0, ! 1 = 1 [ =!3, 4 5. (y + 6)(y! ) =!7 [ y =!5, 1 7. (y! 5) = 9 [ y =, ! = 0 [ = ± = + 1 [ = = 0 [ =!5± i ! + 3 = 0 [ = 1± i 11. r + 1 = 15r [ r =! 1 5, ! 4 3 =! 1 3 [ = 1,! =! 4 [ = 7 8. (y + 3) = 18 [ y =!3 ± = 9 [ =!,, ± =! [ = = 0 [ = ±i 16. =!(5 + ) [ =!1± i 19 Pre-Calculus with Trigonometry

21 13. Inequalities Eample 1: Solve: 3! 5 " 13. Solution: Start by writing the inequality.!13 " 3! 5 " 13 Add 5 to all three members.!8 " 3 " 18 Divide all three members by 3.! 8 3 " " 6 The answer written in interval notation is " #! 8 3, 6 $ %. Remember that if you multiply or divide an inequality by a negative number you must reverse the inequality. For eample, solve! " 18,! "9. Dividing both sides by reverses the inequality. Eample : Solve:! > 6 Solution: As with equations, we need to have a polynomial equal to zero so that we can factor. Subtract 6 from both sides.!! 6 > 0 Factor: (! 3)( + ) > 0 Find the zero s. = 3, =! Test the intervals to find where the quadratic is greater than zero < > - 3 Here the interval is positive for values less than or greater than 3. It is negative between and 3. Epress the solution using inequalities <! or > 3. Skill Builders

22 Solve the inequality. Write your answer in either interval or inequality notation. 1.! " [ all real numbers.! 3 < 7 [!4 < < 10 3.! 3 + < 6 [!1 < < ! 7 [! " and # ! 6 > 0 [ <!3, > 6.!! 1 " 0 [!3 " " < 5! 1 [ 1 3 < < 1 8. (! )(! 3)(! 4) > 0 [ < < 3, > 4 Solve the inequality and graph the solution set. 9.! 1 <!3 [ no solution ! 6 [! "5 and # "1 Eample: Graph the solution y + 5 >. y Solution: Begin by treating the epression as an equality and sketch the graph. So begin by graphing y =! 5. Since the inequality is y greater than! 5 the line is dotted because it is not included. Net you should choose a test point on one side of the line and plug it into the inequality. If it makes the inequality true, then shade that side of the graph. If it makes the inequality false, than shade the other side of the graph. A good point to use (as long as it is not on the graph) is (0, 0). If we substitute = 0 and y = 0 into the inequality, then it becomes > 10, which is false so shade the other side of the line Pre-Calculus with Trigonometry

23 Graph the solution. 11.! y < ! "6 y y! 6 14.! y >!3 y y y! y! 7 y y Skill Builders