Gaithersburg High School Summer 2018 Math Packet For Rising Algebra 2/Honors Algebra 2 Students

Transcription

1 Gaithersburg High School Math Packet For Rising Algebra 2/Honors Algebra 2 Students 1 This packet is an optional review of the skills that will help you be successful in Algebra 2 in the fall. By completing this packet over the summer, you will keep your brain mathematically active and you will also be able to identify skills that you need to strengthen. Complete the practice problems and then check your answers with the Answer Key. If you struggle with any of the exercises, please seek help from a friend, parent, sibling, book, or online resource (some suggestions have been provided for you). Enjoy your math review and we look forward to meeting you in August! Items marked with * are for rising Honors Algebra 2 students only. I. Operations on Numbers A. Absolute Value value.html Simplify. Evaluate = = 3. x if x c if c 10 B. Rational Numbers numbers.html Simplify (90) = = = = = = 12. 6(6 2) 10 (2) = = = y 12 3 =

2 2 Evaluate cd if c 1 2,d c if c 6 C. Radicals radicals.htm Simplify = = = = = = D. Exponents expressions.htm#powers Simplify = 25. (4) 2 = = = E. Order of Operations expressions.htm#order Simplify each expression using PEMDAS. 28. (12 14) Evaluate. 50 (8 9) b 2 4ac if a 3,b 5,c mx b if m 2 3,b, x II. Linear Equations in One Variable Solve each linear equation. A solution is a value for the variable that makes the equation true. You should check each solution to verify that it makes the left side of the equation equal to the right side w b 1 3 2

3 c h g k 10 k n 8 1 n (2d 4) 5(6 2d) (2m 16) 1 (2m 4) 41. 2(a 8) 7 5(a 2) 3a x Solve for x in terms of b and c. 2x b c 44. Solve for z in terms of a and b. 45. Solve for w in terms of y. b 4z 7 a 2w y 7w Solve for x in terms of y. Eight less than a number x is three more than a number y. III. Linear Equations in Two Variables A. Slope Find the slope of the line that passes through the two points. If the slope does not exist, write no slope. 47. (14, 8) and (7, 6) 48. (4, 3) and (8, 3) 49. ( 2,4) and ( 2,9) B. Slope Intercept Form 50. Write the equation of the line whose slope is 3 and whose y intercept is 5. 2 State the slope and y intercept then graph each line. Label the y intercept and a second point on each line. 51. y 5x y 2 5 x 4

4 4 Graph each horizontal or vertical line. Label two points on each line. 53. x y x y Use slope intercept form to write the equation of the line whose x intercept is 3 and whose y intercept is Use slope intercept form to write the equation of the line that passes through the points ( 1,6) and (3, 2). IV. Systems of Linear Equations equations.htm#addition A solution to a system of equations is an ordered pair ( x, y) that makes both equations true. 59. How many solutions can a system of linear equations have? 60. Is ( 2,1) a solution to the system 4x 2y 6? Why or why not? x y 3 Solve each system using the indicated method. If there is no solution or an infinite number of solutions, state so and explain why. 61. x y 1 y 1 3 x 5 graphing 62. 2x 2y 7 x 2y 1 substitution 63. 3x 2y 1 4x 2y 6 elimination 64. 4x 5y 6 6x 7y 20 elimination V. Power Rules exponents.htm Simplify. 65. (6) 0 = c c c = 67. (4x 3 )(5x 7 ) = 68. (n 2 ) 5 = 69. (7x 6 ) 2 = 70. (5n) 3 =

5 (4a 3 b) 2 (b 3 ) = 72. (18m 2 n) 2 6 mn2 = = y 7 14y 5 = = m 15m 3 = 76. x 5 y 3 xy 7 = 2 = 79. 4a 2 b 3 ab 2 = = 81. 3x 3 = 82. g 7 g 4 = = x6 y 8 5xy 11 = VI. Simplifying Polynomials A. Definitions Define a) monomial, b) binomial, c) trinomial, and d) polynomial. 86. Define a) degree of a monomial and b) degree of a polynomial. 87. Write the polynomial x 3x 4 21x 2 6 x 3 in standard form. Then state the degree of each term and the degree of the polynomial. B. Simplify. 88. ( 2m 2 5m 1) (4m 2 8m 6) 89. (n 2 3n 2) (2n 2 6n 2) 90. 4x(2x 3 2x 3) 91. 2b(b 2 4b 8) 3b(3b 2 9b 18) 5b 2 C. Find each product using the FOIL or box method. trinomial.htm 92. (x 5)(x 7) 93. (x 6)(x 2) 94. (x 8)(x 5) 95. (x 9)(x 3) 96. (x 4)(x 4) 97. (x 4)(x 4) 98. (c 8) 2 *99. (a b) 2 *100. (a b) (3d 1) (2x 1)(x 5) 103. (7g 4)(7g 4)

6 VII. Factoring Polynomials 6 A. Find the GCF of each set of monomials , , , r 2 s 2, 36rs fg 5, 56 f 3 g Factor each polynomial using the Distributive Property. In other words, by factoring out the GCF x y 3 28y 2 y Factor out 1 as the GCF so that the squared term is positive x 2 5x x 2 6x 4 Factor each polynomial by grouping y 2 4y 3y a 2 3a 8a 2 B. Trinomials in the Form ax 2 bx c, where a 1. Factor each trinomial into two binomials. If not factorable, write PRIME m 2 12m 32 = 116. r 2 3r 2 = 117. x 2 4x 21 = 118. x 2 8x 16 = 119. m 2 4m 12 = 120. d d = g g 2 =

7 7 *C. Trinomials in the Form ax 2 bx c, where a 1. Factor each trinomial into two binomials. If not factorable, write PRIME x 2 5x 3 = m 2 8m 3 = c 2 19c 21 = t 2 11t 15 = n 2 8n 5 = x 2 3x 6 = D. Factor each difference of squares. If not factorable, write PRIME x 2 81 = n 2 25 = c 2 = 131. d 2 4e 2 = *E. A perfect square trinomial has the form a 2 2ab b 2, where a and bare real numbers, and is the result of squaring a b. Tell whether each trinomial is a perfect square. If so, factor it x 2 22x 121 ; 133. p 2 8p 64 ; 134. w 2 18w 81 ; c 2 10c 1 ; F. A prime number is a number that has exactly two factors, one and itself. Likewise, a prime polynomial is a polynomial that has exactly two factors, one and itself. Factor each polynomial completely so that each factor in your answer is prime. Always start with factoring out a GCF, if one exists b 2 27b 24 =

8 a 2 14a 18 = *138. 6g 3 14g 2 4g = m 3 300m = 140. a 3 4a = x 2 54x 81 = VIII. Quadratic Formula Solve each equation using the Quadratic Formula. If ax 2 bx c 0, then x b b2 4ac. 2a x 2 3x x 2 4x x 2 2x 9

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