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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Some Basics of Algebra Algebraic Expressions and Their Use Translating to Algebraic Expressions Evaluating Algebraic Expressions Sets of Numbers 1.1
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Slide 1- 2 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Algebraic Expressions and Their Use variable – a letter that is used to represent an unknown value constant – a number having a fixed value
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Slide 1- 3 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Algebraic Expressions An algebraic expression consists of variables, numbers, and operation signs. Examples: When an equals sign is placed between two expressions, an equation is formed.
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Slide 1- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Algebraic Expressions vs. Equations Algebraic Expresions Equations
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Slide 1- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Translating to Algebraic Expressions per of less than more than ratio twicedecreased byincreased by quotient of times minus plus divided byproduct of difference of sum of divide multiply subtract add DivisionMultiplicationSubtractionAddition Key Words
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Slide 1- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Translate each of the following to an algebraic expression: 1.5 minus a number 2.5 less than a number 3.3 less than twice a number 4.2 more than the square of a number 5.5 percent of a number
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Slide 1- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Notation The expression a n, in which n is a counting number means multiply a n times. In a n, a is called the base and n is called the exponent, or power. When no exponent appears, it is assumed to be 1. Thus a 1 = a.
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Slide 1- 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The base of a triangle is 10 feet and the height is 3.1 feet. Find the area of the triangle. Solution 10·3.1 = 15.5 square feet h b Evaluating Algebraic Expressions
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Slide 1- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Slide 1- 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluate the expression Solution 2(x + 3) 2 – 12 x 2 Substituting Simplifying 5 2 and 2 2 Multiplying and Dividing Subtracting = 2(2 + 3) 2 – 12 2 2 Working within parentheses Example
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Slide 1- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Evaluate the expression Solution 4x 2 + 2xy – z = 4·3 2 + 2 · 3 · 2 – 8 = 36 + 12 – 8 = 40 = 4·9 + 2·3·2 – 8 Substituting Simplifying 3 2 Multiplying Adding and Subtracting Example
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Slide 1- 12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Set Notation Roster notation: {2, 4, 6, 8} Set-builder notation: {x | x is an even number between 1 and 9} “The set of all xsuch thatx is an even number between 1 and 9”
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Slide 1- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Using both roster notation and set-builder notation, represent the set consisting of the first 12 odd natural numbers. Solution Using roster notation: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23} Example Using set-builder notation: {n | n is an odd number between 0 and 24}
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Operations and Properties of Real Numbers Absolute Value Inequalities Addition, Subtraction, and Opposites Multiplication, Division, and Reciprocals The Commutative, Associative, and Distributive Laws 1.2
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Slide 1- 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute Value The notation |a|, read “the absolute value of a,” represents the number of units that a is from zero. a. | 4.3| = 4.3 –4.3 is 4.3 units from 0. b. |0| = 0 0 is 0 units from itself. c. |17| = 17 17 is 17 units from 0. Example
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Slide 1- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inequalities < means “is less than” means “is less than or equal to” > means “is greater than” means “is greater than or equal to” For any two numbers on a number line, the one to the left is said to be less than the one to the right. –1 0 1 2 3 –1 < 3 since –1 is to the left of 3
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Slide 1- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3 < 12 “3 is less than 12” is true because 3 is to the left of 12. “0 is less than or equal to – 9 ” is false because 0 is to the right of – 9. 7 > 2 “7 is greater than 2” is true because 7 is to the right of 2. “5 is greater than or equal to 5” is true because 5 = 5. Example Write out the meaning of each inequality and determine whether it is a true statement. Solution
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Slide 1- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Addition of Two Real Numbers 1.Positive numbers: Add the numbers. The result is positive. 2.Negative numbers: Add the absolute values. Make the answer negative. 3.A negative and a positive number: If the numbers have the same absolute value, the answer is 0. Otherwise subtract the smaller absolute value from the larger one. a. If the positive number has the greater absolute value, make the answer positive. b. If the negative number has the greater absolute value, make the answer negative. 4.One number is zero: The sum is the other number.
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Slide 1- 19 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley a) – 8.3 + 7.5Subtract absolute values. The answer is negative, – 0.8. b) – 14 + (– 9) Add the absolute values. The answer is negative, – 23. c) – 7 + 24Subtract absolute values. The answer is positive, 17. Example Add: Solution
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Slide 1- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Law of Opposites For any two numbers a and –a, a + (– a) = 0. (When opposites are added, their sum is 0.) The opposite of 3.5 is – 3.5. The opposite of 0 is 0. The opposite of is Example
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Slide 1- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute Value (When x is nonnegative, the absolute value of x is x. When x is negative, the absolute value of x is the opposite of x. Thus, |x| is never negative.)
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Slide 1- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley b) –5.1 –(– 2.3) = –5.1 + 2.3 = –2.8 Subtract: Example a)4 – 18 = 4 + ( 18) = 14 c)
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Slide 1- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplication or Division of Two Real Numbers 1. To multiply or divide two numbers with unlike signs, multiply or divide their absolute values. The answer is negative. 2. To multiply or divide two numbers having the same sign, multiply or divide their absolute values. The answer is positive.
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Slide 1- 24 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley c) Multiply or divide as indicated: Example a) d) b)
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Slide 1- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Sign of a Fraction For any number a and any nonzero number b,
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Slide 1- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Law of Reciprocals For any two numbers a and 1/a (When reciprocals are multiplied, their product is 1.)
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Slide 1- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the reciprocal: Example a) b) The reciprocal of is. The reciprocal of –1 is –1.
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Slide 1- 28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Division by Zero We never divide by 0. If asked to divide a nonzero number by 0, we say that the answer is undefined. If asked to divide 0 by 0, we say that the answer is indeterminate.
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Slide 1- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify: Solution Simplifying (–3) 2 Multiplying and Dividing Adding and Subtracting Example
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Slide 1- 30 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Commutative Laws For any real numbers a and b, a + b = b + a; (for Addition) (for Multiplication) The commutative laws provide one way of writing equivalent expressions.
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Slide 1- 31 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equivalent Expressions Two expressions that have the same value for all possible replacements are called equivalent expressions. are equivalent expressions.
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Slide 1- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Associative Laws For any real numbers a, b and c, a + (b + c) = (a + b) + c; (for Addition) (for Multiplication)
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Slide 1- 33 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write an equivalent expression to (8x + y) + 1 using the associative law. Solution 8x + (y + 1) Example (8x + y) + 1 =
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Slide 1- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Distributive Law For any real numbers a, b, and c, a(b + c) = ab + ac.
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Slide 1- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Obtain an expression equivalent to 5y(x + 11) by multiplying. Solution Example
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Slide 1- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Obtain an expression equivalent to 8x – 24 by factoring. Solution 8(x – 3) Example 8x – 24 =
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Equations Equivalent Equations The Addition and Multiplication Principles Combining Like Terms Types of Equations 1.3
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Slide 1- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equivalent Equations Two equations are equivalent if they have the same solution(s). are equivalent equations since they have the same solution. They are only true when x is 1.
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Slide 1- 39 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Addition and Multiplication Principles for Equations For any real numbers a, b, and c: a) a = b is equivalent to a + c = b + c; b) a = b is equivalent to ac = bc, provided
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Slide 1- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Check: Solution Addition principle The law of opposites Simplifying Example
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Slide 1- 41 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Check: Solution Using the multiplication principle, we multiply by the reciprocal of -2/3. Using the law of reciprocals Simplifying Example
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Slide 1- 42 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A term is a number, variable, a product of numbers and/or variables, or a quotient of numbers and/or variables. When terms have exactly the same variable factors, we refer to those terms as like or similar terms. has terms Combining Like Terms
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Slide 1- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Example
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Slide 1- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Check: Solution The solution is – 5. Example
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Slide 1- 45 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Types of Equations An identity is an equation that is true for all replacements that can be used on both sides of the equation. A contradiction is an equation that is never true. A conditional equation is sometimes true and sometimes false, depending on what the replacement is.
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Slide 1- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity. Solution Example
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Slide 1- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction. Solution Example
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Slide 1- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation. Solution Example
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Problem Solving The Five-Step Strategy Problem Solving 1.4
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Slide 1- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Five Steps for Problem Solving with Algebra 1. Familiarize yourself with the problem. 2. Translate to mathematical language. 3. Carry out some mathematical manipulation. 4. Check your possible answer in the original problem. 5. State the answer clearly.
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Slide 1- 51 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The First Step in Problem Solving with Algebra To familiarize yourself with the problem: 1. If the problem is written, read it carefully. Then read it again, perhaps aloud. Verbalize the problem to yourself. 2. List the information given and restate the question being asked. Select a variable or variables to represent any unknown(s) and clearly state what each variable represents. Be descriptive! 3. Find additional information. Look up formulas or definitions with which you are not familiar.
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Slide 1- 52 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The First Step in Problem Solving(continued) 4. Create a table, using variables, in which both known and unknown information is listed. Look for possible patterns. 5. Make and label a drawing. 6. Estimate an answer and check to see whether it is correct.
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Slide 1- 53 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Second and Third Steps in Problem Solving with Algebra Translate the problem to mathematical language. This is sometimes done by writing an algebraic expression, but most often in this text it is done by translating to an equation. Carry out some mathematical manipulation. If you have translated to an equation, this means to solve the equation.
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Slide 1- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Fourth and Fifth Steps in Problem Solving with Algebra Check your possible answer in the original problem. Make sure that the answer is reasonable and that all the conditions of the original problem have been satisfied. State the answer clearly. Write a complete English sentence stating the solution.
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Slide 1- 55 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The sum of two numbers is 57. If one of the numbers is 6 more than twice the other, what are the numbers? Solution 1. Familiarize. We need to find two numbers. Let x be the first number. First Numberx Second Number6 more than twice the first First + Second = 57 Example
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Slide 1- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Translate. Express the second number using the variable x. First xx Second6 more than twice the first 2x + 6 First + Second = 57 3.Carry out. Solve the equation. First number
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Slide 1- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3.Carry out (continued). First number Second number 4.Check. There are two numbers, 17 and 40. Is the second 6 more than twice the first? The answer is yes. 5. State. The two numbers are 17 and 40. Do the numbers add to 57? 17 + 40 = 57
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Slide 1- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The faculty discount at a bookstore is 15%. If a sweatshirt after the discount was $32.30. What was the original price of the sweatshirt? Solution 1. Familiarize. Note that the discount is calculated from and then subtracted from the original price. Guess $40.00. The discount is (0.15)(40) = $6. Subtracting: $40 – (0.15)($40) = $34. Let S = the sweatshirt’s original price. Example
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Slide 1- 59 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Translate. the sweatshirt price 3.Carry out. Solve the equation. the discount is the sale price minus
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Slide 1- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.Check. Note that 15% of $38 would be (0.15)($38) = $5.70. When this is subtracted from $38 we have Thus $38 checks with the original problem. 5. State. The original price of the sweatshirt was $38.
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Slide 1- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley One angle of a triangle has the same measure as a second angle. The third angle measures 10 o more than three times the measure of the first angle. Find the measures of the angles. Solution 1. Familiarize. We need to find three angles. Let x be the first angle. First Angle x Second Angle The same as the first Third Angle10 o more than three times the first Example
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Slide 1- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 1.Familiarize (continued). First Angle + Second Angle + Third Angle = 180 o First xx Second The same as the firstx Third10 o more than three times the first 3x + 10 2.Translate. Express the second and third angle using the variable x. First Angle + Second Angle + Third Angle = 180 o
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Slide 1- 63 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3.Carry out. first angle second angle third angle
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Slide 1- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.Check. The third angle is 10 o more than three times the first. The angles add to 180 o. 5. State. The three angles are 34 o, 34 o, and 112 o.
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Formulas, Models, and Geometry Solving Formulas Mathematical Models 1.5
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Slide 1- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A formula is an equation that uses letters to represent a relationship between two or more quantities. r The area of a circle of radius r is given by the formula:
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Slide 1- 67 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Formulas Solution Divide by l and h Simplify Example
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Slide 1- 68 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Solve a Formula for a Specified Letter 1.Get all the terms with the specified letter on one side of the equation and all the other terms on the other side, using the addition principle. To do this may require removing parentheses. To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the distributive law. 2.When all the terms with the specified letter are on the same side, factor (if necessary) so that the variable is written only once. 3.Solve for the letter in question by dividing both sides by the multiplier of that letter.
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Slide 1- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Subtract x/a Multiply by b Example
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Slide 1- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Subtracting Factoring Dividing Simplifying Example
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Slide 1- 71 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Mathematical Models A mathematical model can be a formula, or set of formulas, developed to represent a real- world situation.
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Slide 1- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Oprah is 5 ft 7 in. tall and if she has a body mass index of approximately 23.5. What is her weight? Solution 1. Familiarize. The body mass index I depends on a person’s height and weight. The formula from the text is: where W is the weight in pounds and H is the height in inches. Example
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Slide 1- 73 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Translate. Solve the formula for W: 3.Carry out. 5 ft 7 in. = 67 in.
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Slide 1- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4.Check. 5. State. Oprah weighs about 150 pounds.
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Properties of Exponents The Product and Quotient Rules The Zero Exponent Negative Integers as Exponents Raising Powers to Powers Raising a Product or Quotient to a Power 1.6
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Slide 1- 76 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiplying with Like Bases: The Product Rule For any number a and any positive integers m and n, (When multiplying, if the bases are the same, keep the base and add the exponents.)
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Slide 1- 77 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Multiply and simplify: Example
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Slide 1- 78 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Dividing with Like Bases: The Quotient Rule For any nonzero number a and any positive integers m and n, m > n, (When dividing, if the bases are the same, keep the base and subtract the exponent of the denominator from the exponent of the numerator.)
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Slide 1- 79 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Divide and simplify: Example
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Slide 1- 80 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Zero Exponent For any nonzero real number a, (Any nonzero number raised to the zero power is 1. The expression 0 0 is undefined.)
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Slide 1- 81 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Evaluate each of the following for y = 5: Example
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Slide 1- 82 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Negative Exponents For any real number a that is nonzero and any integer n, (The numbers a -n and a n are reciprocals of each other.)
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Slide 1- 83 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Express each of the following using positive exponents and, if possible, simplify: Solution Example
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Slide 1- 84 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factors and Negative Exponents For any nonzero real numbers a and b and any integers m and n, (A factor can be moved to the other side of the fraction bar if the sign of the exponent is changed.)
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Slide 1- 85 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Write an equivalent expression without negative exponents: Solution Example
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Slide 1- 86 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The product and quotient rules apply for all integer exponents. Solution Simplify: Example
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Slide 1- 87 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Power Rule For any real number a and any integers m and n, for which a m and (a m ) n exist, (To raise a power to a power, multiply the exponents.)
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Slide 1- 88 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify: Solution Example
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Slide 1- 89 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Raising a Product to a Power For any integer n, and any real numbers a and b for which (ab) n exists, (To raise a product to a power, raise each factor to that power.)
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Slide 1- 90 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify: Solution Example
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Slide 1- 91 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Raising a Quotient to a Power For any integer n, and any real numbers a and b for which a/b, a n, and b n exist, (To raise a quotient to a power, raise both the numerator and denominator to that power.)
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Slide 1- 92 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Simplify: Solution Example
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Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Scientific Notation Conversions Significant Digits and Rounding Scientific Notation in Problem Solving 1.7
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Slide 1- 94 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Scientific Notation Scientific notation for a number is an expression of the form, where N is in decimal notation, and m is an integer.
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Slide 1- 95 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Decimal Notation Scientific Notation Examples Convert:
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Slide 1- 96 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Significant Digits and Rounding When two or more measurements written in scientific notation are multiplied or divided, the result should be rounded so that it has the same number of significant digits as the measurement with the fewest significant digits. Rounding should be performed at the end of the calculation.
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Slide 1- 97 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Multiply and write scientific notation for the answer: Solution rounded to 2 significant digits: Example
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Slide 1- 98 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Divide and write scientific notation for the answer: Solution rounded to 2 significant digits: Example
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Slide 1- 99 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Scientific Notation in Problem Solving According to the United States Mint website, there were 6,836,000,000 pennies minted in 2004. Each of the pennies is 0.00155 m thick and has a diameter of 0.01905 m. Assuming that pennies are cylinders, find the volume of the pennies produced in 2004. Example
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Slide 1- 100 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 1. Familiarize. We need to find the volume of each penny and then multiply that amount by the number of pennies produced. where r is the radius and h is the height (thickness). The volume of a cylinder is given by:
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Slide 1- 101 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Translate. Solve the formula for V: Height (thickness) = 0.00155 m Radius = ½ (diameter) = 0.009525 m Number of pennies:
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Slide 1- 102 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3.Carry out. Volume of each penny
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Slide 1- 103 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3.Carry Out (continued) Volume for all 2004 pennies: 5. State. The approximate total volume of all the pennies produced in the 2004 by the United States Mint was 3020 m 3. 4. Check. Recheck the translation and calculations.
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