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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
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