School District of Palm Beach County. Summer Packet Grade 8 Readiness (Outgoing 7th Grade )

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1 School District of Palm Beach County Summer Packet Grade 8 Readiness (Outgoing 7th Grade )

2 Summer 0 Students and Parents, This Summer Packet for Grade 8 Readiness (with a focus on Algebra skills) is designed to provide an opportunity to review and remediate foundational skills from Grade 7 in preparation for success in Grade 8 Pre Algebra. These materials include instruction and problem solving on each worksheet. The focus of each selected worksheet is a foundational skill from Grade 7 designed to prepare students for Mathematics in Grade 8. The source of the worksheets is the Glencoe Middle School Mathematics series. All of the contents of this packet have been copied with permission. We hope you are able to utilize the resources included in this packet to make your summer both educational as well as relaxing. Thank you!

3 - AB Integers and Absolute Value Integers less than zero are negative integers. Integers greater than zero are positive integers. negative integers positive integers zero is neither positive nor negative The absolute value of an integer is the distance the number is from zero on a number line. Two vertical bars are used to represent absolute value. The symbol for absolute value of is. Example Write an integer that represents 60 feet below sea level. Because it represents below sea level, the integer is -60. Evaluate -. On the number line, the point - is units away from 0. So, - = Write an integer for each situation.. C above zero. a loss of $. a gain of 0 pounds. falling 6 feet Evaluate each expression Chapter Course

4 - AC The Coordinate Plane The coordinate plane is used to locate points. The horizontal number line is the x-axis. The vertical number line is the y-axis. Their intersection is the origin. Points are located using ordered pairs. The first number in an ordered pair is the x-coordinate; the second number is the y-coordinate. The coordinate plane is separated into four regions called quadrants. Example Write the ordered pair that corresponds to point P. Then state the quadrant in which P is located. Start at the origin. Move units left along the x-axis. Move units up on the y-axis. The ordered pair for point P is (-, ). P is in the upper left quadrant or Quadrant II. Graph and label point M at (0, -). Start at the origin. Move 0 units along the x-axis. Move units down on the y-axis. Draw a dot and label it M. y --- O x Write the ordered pair corresponding to each point graphed at the right. Then state the quadrant or axis on which each point is located.. P. Q. R. S Graph and label each point on the coordinate plane. 5. A(-, ) 6. B(0, -) 7. C(, ) 8. D(-, -) 9. E(, -) 0. F(, ) y --- O x O x y Chapter 9 Course

5 - AB Add Integers To add integers with the same sign, add their absolute values. The sum is: positive if both integers are positive. negative if both integers are negative. To add integers with different signs, subtract their absolute values. The sum is: positive if the positive integer s absolute value is greater. negative if the negative integer s absolute value is greater. To add integers, it is helpful to use a number line. Example Find + ( 6). Find + ( ). Use a number line. Start at 0. Move units right. Then move 6 units left (-6) = - Use a number line. Start at 0. Move units left. Move another units left (-) = -5 Add (-) (-) (-7) (-0) (-5) (-0) (-) (-6) Write an addition expression to describe each situation. Then find each sum.. HAWK A hawk is in a tree 00 feet above the ground. It flies down to the ground.. RUNNING Leah ran 6 blocks north then back blocks south. Chapter 6 Course

6 - AD Subtract Integers To subtract an integer, add its opposite. Example Find = 6 + (-9) To subtract 9, add -9. = - Simplify. Find -0 - (-) (-) = -0 + To subtract -, add. = Simplify. Example Evaluate a - b if a = - and b = 7. a - b = Replace a with - and b with 7. = - + (-7) To subtract 7, add -7. = -0 Simplify. Subtract (-6) (-5) (-6) (-) (-0) Evaluate each expression if m = -, n = 0, and p = 5.. m n 5. p - (-8) 6. p - m 7. m - n p Chapter Course

7 - C Multiply Integers The product of two integers with different signs is negative. The product of two integers with the same sign is positive. Example Find 5(-). 5(-) = -0 The integers have different signs. The product is negative. Find -(7). -(7) = - The integers have different signs. The product is negative. Example Find -6(-9). -6(-9) = 5 The integers have the same sign. The product is positive. Example Find (-7). (-7) = (-7)(-7) There are factors of -7. = 9 The product is positive. Example 5 Evaluate abc if a =, b =, and c =. abc = ( )() Replace a with, b with, and c with. = 6() Multiply and. = Multiply 6 and. Multiply.. -5(8). -(-7). 0(-8). -8() 5. -(-) 6. (-8) 7. -5(7) 8. (-) 9. -6(-) ALGEBRA Evaluate each expression if a = -, b = -, and c = bc. -b. ac. -a. 9b 5. ab 6. -ac 7. -c 8. abc Chapter Course

8 - D Divide Integers The quotient of two integers with different signs is negative. The quotient of two integers with the same sign is positive. Example Find 0 (-5). 0 (-5) The integers have different signs. 0 (-5) = -6 The quotient is negative. Find -00 (-5). -00 (-5) The integers have the same sign. -00 (-5) = 0 The quotient is positive. Divide (-7) (-) (-5) (-) ALGEBRA Evaluate each expression if d = -, e = -, and f = 8.. e. 0 f. d 6. d e 5. f e 6. e f 7. -d e 9. f ef 0. d - e 5 Chapter 8 Course

9 - A Add and Subtract Like Fractions Like fractions are fractions that have the same denominator. To add or subtract like fractions, add or subtract the numerators and write the result over the denominator. Simplify if necessary. Example Find +. Write in simplest form. + = + Add the numerators. = Write the sum over the denominator. 0 = Simplify. Find -. Write in simplest form. - = - = Subtract the numerators. Write the difference over the denominator. 0 Add or subtract. Write in simplest form Chapter 5 Course

10 - C Add and Subtract Unlike Fractions To add or subtract fractions with different denominators, Rename the fractions using the least common denominator (LCD). Add or subtract as with like fractions. If necessary, simplify the sum or difference. Example Find +. Method Use a Model. + Method Use the LCD. + = + = 8 + or Rename using the LCD,. Add the fractions. Add or subtract. Write in simplest form (- 5. 6) (- 6. ) (- 8. ) Chapter Course

11 - D Add and Subtract Mixed Numbers To add or subtract mixed numbers: Add or subtract the fractions. Rename using the LCD if necessary. Then, add or subtract the whole numbers. Simplify if necessary. Example or 8 5 Find Write in simplest form. 0 Add the whole numbers and the fractions separately. Simplify. 8-6 Find Rename the fractions using the LCD. Subtract. Example Find Rename 5 0 as 5 0. Add or subtract. Write in simplest form Subtract the whole numbers and then the fractions Chapter 6 Course

12 - B Multiply Fractions To multiply fractions, multiply the numerators and multiply the denominators = = 5 0 = To multiply mixed numbers, rename each mixed number as an improper fraction. Then multiply the fractions. = 8 5 = 0 = Example Find. Write in simplest form. 5 5 = Multiply the numerators. 5 Multiply the denominators. = 8 5 Simplify. 5 = 5 = 5 = 5 6 Find. Write in simplest form. Rename as an improper fraction, 5. Multiply. Simplify. Multiply. Write in simplest form (- 6. 5) (- 9. ) Chapter Course

13 - D Divide Fractions To divide by a fraction, multiply by its multiplicative inverse or reciprocal. To divide by a mixed number, rename the mixed number as an improper fraction. Example Find. Write in simplest form. 9 9 = 0 9 Rename as an improper fraction. = = 0 9 / / / / = 5 Multiply. Multiply by the reciprocal of 9, which is 9. Divide out common factors. Divide. Write in simplest form (- 5. ) (-0). 8. (- 5. ) 6 6 Chapter 5 Course

14 - E Powers and Exponents exponent = = 8 base common factors The exponent tells you how many times the base is used as a factor. Example Write 6 as a product of the same factor. The base is 6. The exponent means that 6 is used as a factor times. 6 = = = 65 Example Evaluate 5. Write in exponential form. The base is. It is used as a factor 5 times, so the exponent is 5. = ( ) 5 Write each power as a product of the same factor ( ). 5 Evaluate each expression ( Write each product in exponential form ) Chapter 57 Course

15 - D Solve One-Step Addition and Subtraction Equations Remember, equations must always remain balanced. If you subtract the same number from each side of an equation, the two sides remain equal. Also, if you add the same number to each side of an equation, the two sides remain equal. Example Solve x + 5 =. Check your solution. x + 5 = Write the equation. - 5 = -5 Subtract 5 from each side. x = 6 Simplify. Check x + 5 = Write the original equation Replace x with 6. = This sentence is true. The solution is 6. Solve 5 = t -. Check your solution. 5 = t - Write the equation. + = + Add to each side. 7 = t Simplify. Check 5 = t - Write the original equation Replace t with 7. 5 = 5 This sentence is true. The solution is 7. Solve each equation. Check your solution.. h + =. m + 8 =. p + 5 = 5. 7 = y w + = - 6. k + 5 = = + r z = b - = = c - 5. j - = 8. v - = = w -. y - 8 = - 5. = f - 6. = n - Chapter 8 Course

16 - B Solve One-Step Multiplication and Division Equations Use the Division Property of Equality to solve multiplication equations and the Multiplication Property of Equality to solve division equations. The Division Property of Equality states that if you divide each side of an equation by the same nonzero number, the two sides remain equal. The Multiplication Property of Equality states that if you multiply each side of an equation by the same number, the two sides remain equal. Example Solve 0 = 6x. 0 = 6x Write the equation. 0 6 = 6x 6 Divide each side of the equation by 6. 5 = x 0 6 = 5. The solution is 5. Solve x 5 =. x = Write the equation. 5 x ( 5) = ( 5) Multiply each side of the equation by 5. 5 x = 0 ( 5) = 0. The solution is 0. Solve each equation. Check your solution.. x =. 9k = 60. 5a = 5. = b 5. x 5 = 6. 6 = a 7. c = y = 9. m 6 = 0. = b 9 Chapter 5 Course

17 - D Solve Equations with Rational Coefficents Multiplicative inverses, or reciprocals, are two numbers whose product is. To solve an equation in which the coefficient is a fraction, multiply each side of the equation by the reciprocal of the coefficient. Example = Find the multiplicative inverse of. Rename the mixed number as an improper fraction. = Multiply by to get the product. The multiplicative inverse of is. ( 5 5 ) Solve x = 8. Check your solution. 5 x = 8 Write the equation. 5 x = ( 5 x = 0 The solution is 0. ) 8 Multiply each side by the reciprocal of Simplify. 5, 5. Find the multiplicative inverse of each number Solve each equation. Check your solution x = 6. 6 = 0 a 7. 9 = 0.n y = 9. = 0.75a 0. = b Chapter Course

18 - B Solve Two-Step Equations To solve a two-step equation, undo the addition or subtraction first. Then undo the multiplication or division. Example Solve 7v - = 5. Check your solution. 7v - = 5 Write the equation. + = + Undo the subtraction by adding to each side. 7v = 8 Simplify. 7v 7 = 8 7 Undo the multiplication by dividing each side by 7. v = Simplify. Check 7v - = 5 Write the original equation. 7() - 5 Replace v with. 8-5 Multiply. 5 = 5 The solution checks. The solution is. Solve -0 = 8 + x. Check your solution. -0 = 8 + x Write the equation. -8 = -8 Undo the addition by subtracting 8 from each side. -8 = x Simplify. -8 = x Undo the multiplication by dividing each side by. -6 = x Simplify. Check -0 = 8 + x Write the original equation (-6) Replace x with (-8) Multiply. -0 = -0 The solution checks. The solution is -6. Solve each equation. Check your solution.. y + =. 6x + = 6. - = 5k + 7. n + = = -c p + = = -5t r + = n = 0. 5 = 7 + b p = 9. 9 = 6 + y 5. = t -. -9x - 0 = = z g = x = = 6q c - = 0. 9y + = Chapter 7 Course

19 - D Solve Equations with Variables on Each Side To solve an equation with variables on each side, use the Properties of Equality to write an equivalent equation with the variables on one side. Then solve the equation. Example 6x = x x = x Express 6x + 0 = x + 8 as another equivalent equation. Subtract 0 from each side. Simplify. x = x + 7 x x = x x + 7 x = 7 x = 7 x = 9 Solve x = x + 7. Write the equation. Subtraction Property of Equality Simplify. Division Property of Equality Simplify. Check your solution. Example x 6 = 5x x x 6 = 5x x Solve x 6 = 5x. 6 = x Simplify. Write the equation. Subtraction Property of Equality 6 + = x + Addition Property of Equality = x Simplify. = x Division Property of Equality 6 = x Simplify. Check your solution. Express each equation as another equivalent equation. Justify your answer.. + x = x. 0 + x = 7x + Solve each equation. Check your solution.. x + 6 = x + 5. x = 8x 5. x = 8 x Chapter Course

20 5- B Equations and Functions The solution of an equation with two variables consists of two numbers, one for each variable that makes the equation true. When a relationship assigns exactly one output value for each input value, it is called a function. Function tables help to organize input numbers, output numbers, and function rules. Example Complete a function table for y = 5x. Then identify the domain and range. Choose four values for x. Substitute the values for x into the expression. Then evaluate to find the y value. x 5x y 0 5(0) 0 5() 5 5() 0 5() 5 The domain is {0,,, }. The range is {0, 5, 0, 5}. Complete the following function tables. Then identify the domain and range.. y = x. y = 0x x x y x 0x y 0. y = -0.5x. y = x x -0.5x y x x y 5 0 Chapter 5 Course

21 5- C Functions and Graphs The solution of an equation with two variables consists of two numbers, one for each variable, that make the equation true. The solution is usually written as an ordered pair (x, y), which can be graphed. If the graph for an equation is a straight line, then the equation is a linear equation. Example Graph y = x -. Select any four values for the input x. We chose,, 0, and -. Substitute these values for x to find the output y. x x - y (x, y) () - (, ) () - (, ) 0 (0) - - (0, -) - (-) - -5 (-, -5) Four solutions are (, ), (, ), (0, -), and (-, -5). The graph is shown at the right. y O x Graph each equation.. y = x -. y = x +. y = -x y y O x O x. y = x 5. y = x + 6. y = x y y O x O x y O y O x x Chapter 5 9 Course

22 5- B Constant Rate of Change A rate of change is a rate that describes how one quantity changes in relation to another. A constant rate of change is the rate of change of a linear relationship. Example Find the constant rate of change for the table. Students Number of Textbooks The change in the number of textbooks is 5. The change in the number of students is 5. change in number of textbooks change in number of students = 5 textbooks 5 students The number of textbooks increased by 5 for every 5 students. = textbooks Write as a unit rate. student So, the number of textbooks increases by textbooks per student. The graph represents the number of T-shirts sold at a band concert. Use the graph to find the constant rate of change in number per hour. To find the rate of change, pick any two points on the line, such as (8, 5) and (0, 5). change in number change in time = (5 5) (0 8) = 0 Find the each constant rate of change.. Side Length Perimeter 8 6 or 5 T-shirts per hour. Miles T-shirts P.M. 9 P.M. 0 P.M. Time 0 0 A.M. P.M. P.M. Time Chapter 5 6 Course

23 5- C Slope Slope is the rate of change between any two points on a line. slope = change in y change in x = vertical change horizontal change or rise run Example The table shows the length of a patio as blocks are added. Number of Patio Blocks 0 Length (in.) Graph the data. Then find the slope of the line. Explain what the slope represents. change in y slope = Definition of slope change in x = 8 Use (, 8) and (, ). = 6 length number = 8 Simplify. So, for every 8 inches, there is patio block. Length (in.) Number Draw a graph and find the slope of the line. Explain what the slope represents.. The table shows the number of juice bottles per case. Cases Juice Bottles 6 8. At 6 A.M., the retention pond had 8 inches of water in it. The water receded so that at 0 A.M. there were 6 inches of water left. Bottles Inches Cases A.M 8 A.M 0 A.M P.M Time Chapter 5 Course

24 5- A Example Problem-Solving Investigation: Use a Graph The graph shows the results of a survey of teachers years of experience and number of honors level classes taught at a high school. How many honors classes would you predict a teacher with years experience would teach? Number of Honors Classes 6 5 Understand Plan Solve Check Years Experience You know the number of honors level classes taught and the number of years of experience from the graph. Look at the trends in the data on the graph. Using the line, you can predict that a teacher with years experience would teach 5 honors level classes. Draw a line that is as close to as many of the points as possible. The estimate is close to the line, so the prediction is reasonable. Use the graph below. Each point on the graph represents one person in a group training for a long-distance bicycle ride. The point shows the number of miles that person cycles each day and the number of weeks that person has been in training.. Draw a line that is close to as many of the points as possible. Distance (mi) Weeks Training. Does the number of miles bicycled each day increase as the number of weeks in training increases?. Predict the number of miles bicycled each day for someone who has been training for 9 weeks. Chapter 5 6 Course

25 5- AC Direct Variation When two variable quantities have a constant ratio, their relationship is called a direct variation. The constant ratio is called the constant of variation. Example The time it takes Lucia to pick pints of blackberries is shown in the graph. Determine the rate in minutes per pint. Since the graph forms a line, the rate of change is constant. Use the graph to find the constant ratio. minutes number of pints = 5 0 or 5 5 or 5 It takes 5 minutes for Lucia to pick pint of blackberries. Minutes Pints There are trading cards in a package. Make a table and graph to show the number of cards in,,, and packages. Is there a constant rate? a direct variation? Numbers of Packages Number of Cards 6 8 Because there is a constant increase of cards, there is a constant rate of change. The equation relating the variables is y = x, where y is the number of cards and x is the number of packages. This is a direct variation. The constant of variation is.. SOAP Wilhema bought 6 bars of soap for $. The next day, Sophia bought 0 bars of the same kind of soap for $0. What is the cost of bar of soap?. COOKING Franklin is cooking a -pound turkey breast for 6 people. If the number of pounds of turkey varies directly with the number of people, make a table to show the number of pounds of turkey for,, and 8 people. Cards Packages Chapter 5 Course

26 5- E Inverse Variation An inverse variation is a relationship where when x increases in value, y decreases in value, or as x decreases in value, y increases in value. The product of x and y is a constant k. An inverse variation function is of the form xy = k or y = k x. Example TICKETS When tickets to the picnic cost $6 each, 50 people attend. When tickets are reduced to $ each, 75 people attend. Complete a table and graph for prices of $0, $5, $, and $. Find the value of k by using the equation y = k x. y = k x 50 = k 6 50(6) = ( k Write the equation. Replace y with 50 and x with 6. (6) 6) Multiply both sides by = k Simplify. Use the value for k to complete the table. Cost $0 $5 $ $ Number of Tickets The graph shows that this is an inverse variation. Number of Tickets Exercise Make a table and graph to complete Exercise.. INVESTMENT The time it takes to double the balance in an account varies inversely with the interest rate. If you invest $,000 at 6% it will take years to double your money. Find the time it will take to double your money at %. Time (yrs) Cost ($) Interest Rate (%) Chapter 5 7 Course

27 6- B Percent of a Number To find the percent of a number, you can write the percent as a fraction and then multiply or write the percent as a decimal and then multiply. Example Find 5% of 80. 5% = 5 00 or Write 5% as a fraction, and reduce to lowest terms. of 80 = 80 or 0 Multiply. So, 5% of 80 is 0. What number is 5% of 00? 5% of 00 = 5% 00 Write a multiplication expression. So, 5% of 00 is 0. = Write 5% as a decimal. = 0 Multiply. Find each number.. Find 0% of 50.. What is 55% of $00?. 5% of,500 is what number?. Find 90% of What is % of $500? 6. 8% of $00 is how much? 7. What is.5% of 60? 8. Find 0.% of Find % of $ What is 0.5% of 80?. 0.5% of is what number?. What is 0.0% of 80? Chapter 6 Course

28 6- B The Percent Proportion A percent proportion compares part of a quantity to a whole quantity for one ratio and lists the percent as a number over 00 for the other ratio. part whole = percent 00 Example What percent of is 8? part whole = percent 00 Let n represent the percent. 8 = n 00 Percent proportion Write the proportion = n Find the cross products.,800 = n Simplify.,800 = n Divide each side by. 75 = n So, 8 is 75% of. What number is 60% of 50? part whole = percent 00 Let a represent the percent. a 50 = a 00 = a = 9,000 00a 00 = 9, a = 90 So, 90 is 60% of 50. Percent proportion Write the proportion. Find the cross products. Simplify. Divide each side by 00. Find each number. Round to the nearest tenth if necessary.. What number is 5% of 0?. What percent of 50 is 0?. 0 is 75% of what number?. 0% of what number is 6? 5. What number is 0% of 65? 6. is what percent of 0? Chapter 6 5 Course

29 6- C The Percent Equation To solve any type of percent problem, you can use the percent equation, part = percent whole, where the percent is written as a decimal. 600 is what percent of 750? 600 is the part and 750 is the whole. Let n represent the percent. part = percent whole 600 = n 750 Write the percent equation = 750n 750 Divide each side by = n Simplify. 80% = n Write 0.8 as a percent. So, 600 is 80% of 750. Example 5 is 90% of what number? 5 is the part and 90% or 0.9 is the percent. Let n represent the whole. part = percent whole 5 = 0.9 n Write the percent equation = 0.9n 0.9 Divide each side by = n Simplify. So, 5 is 90% of 50. Write an equation for each problem. Then solve. Round to the nearest tenth if necessary.. What percent of 56 is?. 6 is what percent of 0?. 80 is 0% of what number?. 65% of what number is 78? 5. What percent of,000 is 8? 6. What is 0% of 80? is what percent of 70? 8. Find 0% of 70. Chapter 6 Course

30 6- B Percent of Change A percent of change is a ratio that compares the change in quantity to the original amount. If the original quantity is increased, it is a percent of increase. If the original quantity is decreased, it is a percent of decrease. Example Last year,,76 people attended the rodeo. This year, attendance was,950. What was the percent of change in attendance to the nearest whole percent? Since this year s attendance is greater than last year s attendance, this is a percent of increase. The amount of change is,950,76 or 57. amount of change percent of change = original amount = 57 Substitution,76 0. or % The percent of change is about %. Simplify. Che s grade on the first math exam was 9. His grade on the second math exam was 86. What was the percent of change in Che s grade to the nearest whole percent? Since the second grade is less than the first grade, this is a percent of decrease. The amount of change is 86-9 or -8. amount of change percent of change = original amount = or -9% The percent of change is -9%. Substitution Simplify. Find each percent of change. Round to the nearest whole percent if necessary. State whether the percent of change is an increase or decrease.. original:. original:.0. original: 5. original: $0 new: 5 new:. new: new: $8 5. original: original: original: original: 9.6 new: 6 new: 6 new: 0.5 new: 5.9 Chapter 6 Course

31 6- C Sales Tax and Tips Sales Tax is a percent of the purchase price and is an amount paid in addition to the purchase price. Tip, or gratuity, is a small amount of money in return for service. Example SOCCER Find the total cost of a $7.75 soccer ball if the sales tax is 6%. Method First, find the sales tax. 6% of $7.75 = The sales tax is $.07. Next, add the sales tax to the regular price = 8.8 The total cost of the soccer ball is $8.8. Method 00% + 6% = 06% Add the percent of tax to 00%. The total cost is 06% of the regular price. 06% of $7.75 = MEAL A customer wants to leave a 5% tip on a bill for $8.50 at a restaurant. Method Add tip to regular price. First, find the tip. 5% of $8.50 = =.78 Next, add the tip to the bill total. $ $.78 = $.8 The total cost of the bill is $.8. Find the total cost to the nearest cent. Method Add the percent of tip to 00%. 00% + 5% = 5% Add the percent of tip to 00%. The total cost is 5% of the bill. 5% of $8.50 = =.8. $.95 shirt, 6% tax. $ lunch, 5% tip. $0.85 book, % tax. $97.55 business breakfast, 8% tip 5. $59.99 DVD box set, 6.5% tax 6. $7.65 dinner, 5% tip Chapter 6 6 Course

32 6- D Discount Discount is the amount by which the regular price of an item is reduced. The sale price is the regular price minus the discount. Example TENNIS Find the price of a $69.50 tennis racket that is on sale for 0% off. Method : Subtract the discount from the regular price. First, find the amount of the discount. 0% of $69.50 = 0. $69.50 Write 0% as a decimal. = $.90 The discount is $.90. Next, subtract the discount from the regular price. $ $.90 = $ Method : Subtract the percent of discount from 00%. 00% - 0% = 80% Subtract the discount from 00%. The sale price is 80% of the regular price. 80% of $69.50 = = The sale price of the tennis racket is $ Find the sale price to the nearest cent.. $.5 shirt; 5% discount. $8.79 watch; 0% discount. $0.00 jeans; 0% discount. $7.00 sweatshirt; 5% discount 5. $8.00 basketball; 50% discount 6. $98.00 tent; 0% discount Chapter 6 5 Course

33 6- E Simple Interest Simple interest is the amount of money paid or earned for the use of money. To find simple interest I, use the formula I = prt. Principal p is the amount of money deposited or invested. Rate r is the annual interest rate written as a decimal. Time t is the amount of time the money is invested in years. Example I = prt Find the simple interest earned in a savings account where $6 is deposited for years if the interest rate is 7.5% per year. Formula for simple interest I = Replace p with $6, r with 0.075, and t with. I = 0.0 Simplify. The simple interest earned is $0.0. Find the simple interest for $600 invested at 8.5% for 6 months. 6 months = 6 or 0.5 year Write the time in years. I = prt Formula for simple interest I = p = $600, r = 0.085, t = 0.5 I = 5.50 Simplify. The simple interest is $5.50. Find the simple interest earned to the nearest cent for each principal, interest rate, and time.. $00, 5%, years. $650, 8%, years. $575,.5%, years. $75, 7%, years 5. $,665, 6.75%, years 6. $,05, %, years 7. $90, 8.75%, 8 months 8. $,75, 9%, months Chapter 6 56 Course