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1.2 Algebra 2 1.2 Algebra 2 Algebraic Expression- an expression that contains one of more variables. Ex’s: 2x - 3 3 – 2p z+y Term- is a number, variable, or a product of a number and one or more variables Ex’s: 4x + 3 has two terms 4x2 + 3x + 4 has three terms Coefficient- the numerical factor in a term. Ex’s: -3k -3 is the coefficient 2x 2 is the coefficient 1.2 Algebra 2 Evaluate- substitute numbers for variables in the expression and then simplify the expression following the order of operations. Example: 2x + 3y when x = 2 and y = -1 2(2) + 3(-1) 4 + -3 =1 1.2 Algebra 2 Algebraic Expressions Lesson 1-2 Algebra 2 Additional Examples Evaluate 7x – 3xy for x = –2 and y = 5. 7x – 3xy = 7(–2) – 3(–2) (5) Substitute –2 for x and 5 for y. = –14 – (–30) Multiply first. = –14 + 30 To subtract, add the opposite. = 16 Add. 1.2 Algebra 2 Evaluate each expression for x = 4 and y = - 2 1.) x + y 2.) 3x – 4y + x – y Algebraic Expressions Lesson 1-2 Algebra 2 Additional Examples Evaluate (k – 18)2 – 4k for k = 6. (k – 18)2 – 4k = (6 – 18)2 – 4(6) Substitute 6 for k. = (–12)2 – 4(6) Subtract within parentheses. = 144 – 4(6) Simplify the power. = 144 – 24 Multiply. = 120 Subtract. Algebraic Expressions Lesson 1-2 Algebra 2 Additional Examples The expression –0.08y 2 + 3y models the percent increase of Hispanic voters in a town from 1990 to 2000. In the expression, y represents the number of years since 1990. Find the approximate percent of increase of Hispanic voters by 1998. Since 1998 – 1990 = 8, y = 8 represents the year 1998. –0.08y2 + 3y = –0.08(8)2 + 3(8) Substitute 8 for y 19 The number of Hispanic voters had increased by about 19%. Algebraic Expressions Lesson 1-2 Algebra 2 Additional Examples Simplify by combining like terms. 2h – 3k + 7(2h – 3k) 2h – 3k + 7(2h – 3k) = 2h – 3k + 14h – 21k Distributive Property = 2h + 14h – 3k – 21k Commutative Property = (2 + 14)h – (3 + 21)k Distributive Property = 16h – 24k Algebraic Expressions Lesson 1-2 Algebra 2 Additional Examples Find the perimeter of this figure. Simplify the answer. c c P = c + 2 + d + (d – c) + d + 2 + c + d c c = c + 2+ d + d – c + d + 2 + c + d c c = 2 + 2 + c + 4d = 2c + c + 4d 2 = c + c + 4d = 2c + 4d 1.2 Algebra 2