Core Connections: Course Checkpoint Materials Notes to Students (and their Teachers) Students master different skills at different speeds. No two students learn eactl the same wa at the same time. At some point ou will be epected to perform certain skills accuratel. Most of the Checkpoint problems incorporate skills that ou should have been developing in grades and 7. If ou have not mastered these skills et it does not mean that ou will not be successful in this class. However, ou ma need to do some work outside of class to get caught up on them. Starting in Chapter 1 and finishing in Chapter, there are 11 problems designed as Checkpoint problems. Each one is marked with an icon like the one above. After ou do each of the Checkpoint problems, check our answers b referring to this section. If our answers are incorrect, ou ma need some etra practice to develop that skill. The practice sets are keed to each of the Checkpoint problems in the tetbook. Each has the topic clearl labeled, followed b the answers to the corresponding Checkpoint problem and then some completed eamples. Net, the complete solution to the Checkpoint problem from the tet is given, and there are more problems for ou to practice with answers included. Remember, looking is not the same as doing! You will never become good at an sport b just watching it, and in the same wa, reading through the worked eamples and understanding the steps is not the same as being able to do the problems ourself. How man of the etra practice problems do ou need to tr? That is reall up to ou. Remember that our goal is to be able to do similar problems on our own confidentl and accuratel. This is our responsibilit. You should not epect our teacher to spend time in class going over the solutions to the Checkpoint problem sets. If ou are not confident after reading the eamples and tring the problems, ou should get help outside of class time or talk to our teacher about working with a tutor. Checkpoint Topics 1. Computations with Positive Fractions. Evaluating Epressions and Order of Operations. Computations with Positive and Negative Fractions. Unit Rates and Proportions 5. Solving Equations. Multiple Representations 7. Transformations 8. Solving Equations with Fractions (Fraction Busters) 9. Scatter Plots and Association. Laws of Eponents and Scientific Notation 11. Problems Involving Square Roots Checkpoints 89
Checkpoint Number 1 Problem 1-90 Computations with Positive Fractions Answers to problem 1-90: a. 15 = 1 7 15, b. 5 8, c. 7 1 1, d. 1 5 7, e., f. 8 1 To add or subtract fractions a common denominator is needed. Once a common denominator is found, ou either add or subtract the numerators. To add or subtract mied numbers, ou can either add or subtract their parts, or ou can change the mied numbers to fractions greater than one. To multipl fractions, first change an mied numbers to fractions greater than one, then multipl the numerators and multipl the denominators. To divide fractions, multipl the first fraction b the reciprocal of the second. Eample 1: Compute the sum: 8 + 5 Solution: This addition eample shows adding the whole number parts and the fraction parts separatel. The answer is adjusted because the fraction part is greater than one.!!!!!!!!!!8 = 8 +!!!! 5!!!=!!!8 15 5 0!!!!!!!!+ 5!=! + 5!!!!!!= + 8 0!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1 0 = 1 0 Eample : Compute the difference: 1 1 5 Solution: This subtraction eample shows changing the mied numbers to fractions greater than one and then computing in the usual wa. 1! 1 5 " 1! 9 5!!!!!!!!!!!!!!!!!"! 1!#! 5 5!! 9 5!#!!!!!!!!!!!!!!!!!!" 5 0! 5 0 = 11 0 Eample : Multipl 1! 1 Solution: 1! 1 =! 5 =!5! = 5!!5! = 5 = 8 1 Eample : Divide 1 1 1 1 Solution: 1 1 1 1 = =! = 8 9 90 Core Connections: Course
Now we can go back and solve the original problem. a. + 5 =! 5 5 + 5! = 15 + 1 15 = 15 = 1 7 15 b. 1! 8 = 8 8! 8 = 5 8 c. 1 + = + + 1 + = + 1! +! = + 1 + 9 1 = + 1 1 = 7 1 1 d. 5 1! 7 8 = 11! 1 8 = 11 "! 1 8 = 8! 1 8 = 1 8 = 1 5 8 e. 5 1! = 1! =!!! = 1 = f. 7 8 1 1 = 7 8 = 7 8! = 1 = 7 1 Here are some more to tr. Compute each of the following problems 1. + 7. 5! 1. 1 + 7 8. 7! 1 1 1 5. 1 1 + 5. 9! 7 8 7.! 1 8. 5 1 9. 1! 5. 7 1 7 11. 5! 1 1. 1 5!1 1. 5 1!1 1 1. 1 + 1 5 15. 1 5 8 5 1. 9 + 1 17. 1 18. 5 1! 5 19. 1 7 8! 0. 1 5 1 1 1. 8 + 15 7 1. 7 1! 1. 11 1!1 1. 7 8 Checkpoints 91
Answers: 1. 1 0 = 1 11 0. 7 15. 1 8. 1 5. 5 1. 7 7. 11 8. 0 9. 5. 11. = 1. 77 0 = 17 0 1. 5 1 1. 1 0 15. 5 8 1. 5 9 17. 1 1 18. 9 = 1 19. 5 = 1 1 0. 0 = 1 1 0 1. 19. 5 1. 17. 9 9 Core Connections: Course
Checkpoint Number Problem -8 Evaluating Epressions and Order of Operations Answers to problem -8: a. 8, b. 1, c., d. 17, e. 5, f. 15 To evaluate an epression, calculate the value of the epression after each variable has been replaced with its numerical value. Be sure to follow the order of operations when doing the calculations as shown below: Rules for Order of Operations (after substitution) 1. Circle the parenthesis or an other grouped numbers, and then circle the terms inside the grouping. (A fraction bar acts as parenthesis.). Simplif each term in the parenthesis.. Circle the terms in the epression.. Simplif each term until it one number b evaluating each eponential and then multipling and dividing from left to right. 5. Combine terms b adding and subtracting from left to right. (! "1) +! 5! (! "1)! 5!(! "1) +! 5 ( )! 5 Eample: + + 9! 5 =!!! ( ) +! 5! 1! 1! 1! 9 + ( 5)! 5 9 + 0! 5 Eample 1: Evaluate + for =!5 Solution: (!5)! (!5) + " (5)! (!5) + " 50! (!5) + " 50 + 5 + = 57 Eample : Evaluate 5 ( +! ) for =,! = Solution: 5!+"!! ( ) # 5 (!+!! ) # 5 ( 1!5 ) #!1 Checkpoints 9
Now we can go back and solve the original problem. a. + + z! (") + (") + 5! " + "9 + 5 = "8 b.! " (!)! (!) "! + = 1 ( )! "+" 5 c. + z ( )! ( "5 )! ("1) = " 5 d.! + 1 " (!)! (!) + 1 " ()! (!) + 1 " 1! (!) + 1 = 17 e. ( +! ) " (!)(! + (!)! (!)) " (!) (! +! (!)) " (!)(5) =!5 f.!z (1!)! "!5 (1!(!))!!(!) "!5(1!(!))!!(!) "!5(5)!1 = 15 Here are some more to tr. Evaluate each epression using =,! =!,!z =!. 1.!. z + 5. z!.! z 5.! +. z! 8! 7. +! 0 8. ( + +1 ) 9.! + 7. z! 11.! 1. ( + z)! 1 1. z( +! ) 1. + (+1) 15. ( + ) 1. ( + )( + z) 17.! 5 + z 18. + 1z! 19. + 0. ( + z)! 1. z + 8z!.!. z! + + z.! (z!5)! 9 Core Connections: Course
Answers: 1. 5. 1. 5. 5. 0. 9 7. 8. 1 9. 17. 11. 1. 5 1. 0 1.! 15 15. 1. 0 17. 0 18. 1 19. 5 1 0. 1. 5. 7.. 5 Checkpoints 95
Checkpoint Number Problem -1 Computations with Positive and Negative Fractions Answers to problem -1: a.! 17, b. 5, c. 1 5, d.!1 1, e. 7 1, f. 7 Use the same processes with positive and negative fractions as are done with integers (positive and negative whole numbers.) Eample 1: Compute 1 + ( 9 0 ) Solution: When adding a positive number with a negative number, subtract the values and the number further from zero determines the sign. 1 +! 9 0 = 1 " 0 0 +! 9 0 " = 0 0 +! 7 0 =! 7 0 Eample : Compute!1 1 ( 9 ) Solution: Change an subtraction problem to addition of the opposite and then follow the addition process.!1 1! (! 9 ) "!1 1 + 9 =!1 5 0 + 18 0 = 1 0 Eample : Compute 1 1 7 1 Solution: With multiplication or division, if the signs are the same, then the answer is positive. If the signs are different, then the answer is negative.!1 1 7 1 =! 5 15 =! 5 " 15 =! 5 " """ 5 =! 1 Now we can go back and solve the original problem. a. Both numbers are negative so add the values and the sign is negative.! 1 +! 8 =! 1 " 8 8 +! 8 " =! 8 +! 9 =! 17 b. Change the subtraction to addition of the opposite. ( ) = 1 + 1 1 = + 1 " + 1 + 1 " = + + 1+ = 5 1!!1 1 9 Core Connections: Course
c. The signs are the same so the product is positive. Multipl as usual.! 1 5 "! 1 =! 1 5 "! 1 7" "1 = = 7 5" 5 = 1 5 d. The signs are different so the quotient is negative. Divide as usual.! 1 = "! 1 =! " "1 =! =!1 1 e. When adding a positive number with a negative number, subtract the values and the number further from zero determines the sign.!1 + 5 1 =! 7 + 1 =! 7 " + 1 " =! 1 1 + 1 = 1 = 7 1 f. The signs are the same so the quotient is positive. Divide as usual.!!1 1 =! 8! 7 =! 8 "! 8" " = 7 = 1 "7 7 = 7 Here are some more to tr. Compute each of the following problems with fractions. ( ) 1.! + 1.!! 5 1.! 5 7 +.!1 7 + (! ) 5. ( 1 )!(" 5 ).! 1 " 7.! 7 1! 1 8. 1 + (! 8 ) 9.!1 1! (! 1 ). ( 5 9 )!(" 7 ) 11.!! (! 5 7 ) 1.!1 9 1.! 7 9 " 1.! 5!1 1 15.!5 1! 1. 5 8 + (! 1 ) 17. 5 1 5 + (! 15 ) 18. 1! (!1 5 8 ) 19.! 7 9 " 1 7 0.! 1 5! 1. 5 1 1! (! 7 ).! 1 7 "! 5.! 8 1.!! 1 1 5 Checkpoints 97
Answers: 1.! 1. 1 1.! 1 1.! 17 8 5.!1 1.!1 1 7. 15 1 8.! 7 8 9. 1 11 1.!1 1 11.! 1 8 1.! 1 1.! 5 1. 11 15. 7 1 1. 8 1 8 17. 1 15 18. 1 8 19.!8 1. 7 79 8 0. 1. 5.! 19.!5 1 98 Core Connections: Course
Checkpoint Number Problem 5-7 Unit Rates and Proportions Answers to problem 5-7: a. $0.8, b. 0., c.! $.50, d. $11.88, e.! 19 shots, f. 1 teaspoons A rate is a ratio comparing two quantities and a unit rate has a denominator of one after simplifing. Unit rates or proportions (ratio equations) ma be used to solve ratio problems. Solutions ma also be approimated b looking at graphs of lines passing through the origin and the given information. Eample 1: Sam paid $.95 for price (dollars per pound)? pound of his favorite trail-mi. What is the unit $.95 Solution: The unit price is. To create a unit rate we need a denominator of one.!pound $.95!!pound! = $.95! 1!pound = $.95!!= $.0 per pound Eample : Based on the table, what is the unit growth rate (meters per ear)? + height (m) 15 17 ears 75 85 Solution: 1!meters!ears =!meters!ears = 5!meter = 1 meter 1!ear 5 ear + Note: This same kind of information might be determined b looking at a graph. Eample : In line at the movies are 1 people in front of ou. If ou count 9 tickets sold in 70 seconds, how long will it take before ou bu our ticket? 9!tickets Solution: The information ma be organized in a proportion 70 seconds = 1!tickets. Solving the proportion 9 70 = 1 ields 9 = 0 so! 115.5 seconds or! 19 minutes. Checkpoints 99
Now we can go back and solve the original problem. a. $1.89! $1.89 1 = 9 pound 9!pound! 9 = $1.89! 9 1!pound = $1.89! = $0.8 per pound 9 b. For ever increase of three grams, the length increases b five centimeters.!grams 5 cm = 5 gram cm c. The graph passes through ( bottles, $5). $5 = $.50 per bottle.!bottles d. 00!vitamins 500 vitamins =! 00 = 7.50! " $11.88 $.75 e. 7!made 85 attempts =! 85 = 0! " 19 made 00 attempts f. 1 teaspoon = cup 5 cups! = 5! = = 1 cups Here are some more to tr. In problems 1 through find the unit rate and in problems 11 through 0 use ratios to solve each problem. 1. Am usuall swims 0 laps in 0 minutes. What is her rate in laps per minute?. For of a pound of peanuts, Jimm paid $.5. What is the unit rate (dollars per pound)?. In 7 minutes, Jack tpes 511 words. How man words does he tpe per minute?. Using the graph below, determine how long it takes to mow an acre of grass. Acres 1 0.5 1.0 1.5 Hours 500 Core Connections: Course
5. The table below illustrates the growth of bab Martha through her first months. Based on the information given, how much did Martha grow each month? Length (in.) Age (months) 19 1. 5.5 1. If Christ and Lucas eat cups of popcorn in 0 minutes, how man minutes does it take them to eat a cup of popcorn? 7. In the past ear, Ana has spent $7, 0 on her rent. What is the unit cost (dollars per month)? 8. While constructing his house, the little pigg found that bundles of sticks contained a total of 08 sticks. How man sticks are in each bundle? 9. Meg knows that it takes 1 minutes to read 7 8 her to read the entire chapter? of the chapter. How long will it take. At the frozen ogurt shop, Colin pas $.8 for his treat, which weighs 8 1 5 ounces. What is the unit cost (cents per ounce)? 11. When Sarah does her math homework, she finishes problems ever 1 minutes. How long will it take for her to complete 5 problems? 1. Ben and his friends are having a TV marathon, and after hours the have watched 5 episodes of the show. About how long will it take to complete the season, which has episodes? 1. The ta on a $00 painting is $. What should be the ta on a $700 painting? 1. Use the table below to write and solve a ratio for how long it will take Isadora to earn $. $ Earned 5 7.50 75 Das Worked 1 15. While baking, Hannah discovered a recipe that required cups of sugar for ever cups of flour. How man cups of sugar will she need for cups of flour? 1 Checkpoints 501
1. Based on the graph, how man cups of water does Shelb drink in 7 hours? Cups of Water 1 1.0 1.5.0 Hours Passed 17. M brother grew 1 inches in 1 months. How much should he grow in one ear? 18. On his afternoon jog, Chris took minutes to run miles. How man miles can he run in 90 minutes? 19. If Caitlin needs 1 7 cans of paint for each room in her house, how man cans of paint 8 will she need to paint the 9-room house? 0. Stephen receives gumballs for ever two hours of homework he does. If he wants gumballs, how man hours will he need to work? 50 Core Connections: Course
Answers: 1. laps minute. 7 words minute. $.1. hours 5. 1 inches month. 7 1 minutes cup 7. $0 month sticks 8. 5 bundle 9. 18 7 minutes. $0.0 ounce 11. minutes 1. 19 1 5 hours 1. $59.50 1. 19 1 5 das 15. 1 1 cups 1. 11 cups 17. 8 5 inches 18. 8 1 8 miles 19. 1 7 8 cans 0. 7 1 hours Checkpoints 50
Checkpoint Number 5 Problem -111 Solving Equations Answers to problem -111: a.!, b. 1 1, c., d. no solution Equations ma be solved in a variet of was. Commonl, the first steps are to remove parenthesis using the distributive propert and then simplif b combining like terms. Net isolate the variable on one side and the constants on the other. Finall, divide to find the value of the variable. Note: When the process of solving an equation ends with different numbers on each side of the equal sign (for eample, = ), there is no solution to the problem. When the result is the same epression or number on each side of the equation (for eample, + = + ) it means that all numbers are solutions. Eample 1: Solve + = + 9 Solution: +! = + 9!!!!!!!!problem 8! = + 9!!!!!!!!simplif = 1!!!!!!!!!!!!!!!add, subtract on each side =!!!!!!!!!!!!!!!!divide Eample : Solve + ( + 1) = + ( + 5) Solution:! +! (! + 1) =! + (! + 5)!!!!!problem! + +! 1 =!! + 5!!!!!!!!!!remove parenthesis (flip)! + 1 =! +!!!!!!!!!!!!!!!!simplif! = 1!!!!!!!!!!!!!!!!!!!!!!!!!add, subtract 1 to each side =! 1!!!!!!!!!!!!!!!!!!!!!divide Now we can go back and solve the original problem. a. + 7 =!! 1 b. =!8 =! 1! + 5 =!! + =! 9 = 1 1 = c.!! =!! (! 1) d.! + 1 =!! 5 + 5!! =!! + 1! =! 5!! =!5 +! =!5 = 9! "!5 # no solution = 50 Core Connections: Course
Here are some more to tr. Solve each equation. 1.! =! +. + + = + 5.!! = (! ).!! =! 5 5.!( + ) =!.! + =! 5! 7. 1 +! =! + 8. 5! + = + 7 + 9.! 8! =.! + = 11.! + =!! 1. (! ) + = 5 1.!! =! 1. = + 5 + 15.! 1! 1 =!! (!5 + ) 1. +! + = + 17.! +! 1 = + 1 + 18.! 7 =!! 1 19. 7 =!! ( + ) 0. 5 + (!! ) = + Answers: 1.. 1. 1 5. 5. 1. 7 7. 5 8. no solution 9.. 11. 0 1. 1. 1 1. 1 15. 1. 1 17. 18. 19. 1 0. Checkpoints 505
Answers to problem 7-11: Checkpoint Number Problem 7-11 Multiple Representations a. figure # 0 1 b. # of tiles 8 0 1 7 1 # of tiles 8 1 Figure # = + 7 5 1 1 c. = + 1 d. 1 8 1 figure # 0 1 # of tiles =! + 0 If ou know one representation of a linear pattern, it can also be represented in three other different was. Linear representations ma be written as equations in the form = m + b where m is the growth factor and b is the starting value. Graph Table Pattern Rule 50 Core Connections: Course
Eample 1: Using the pattern at right, create an! table, a graph, and write the rule as an equation. Figure 0 Figure 1 Figure Figure Solution: The number of tiles matched with the figure number gives the! table. Plotting those points gives the graph. Using the starting value for b and the growth factor for m gives the information to write the equation in = m + b form. figure # () 0 1 # of tiles () 5 7 9 The starting number is tiles and the pattern grows b tiles each figure so the equation is = +. Total Number of Tiles 8 7 5 1 1 Figure Number Eample : Write an! table and the rule (equation) based on the graph below. 1 8 1 Solution: Place the given points in a table: 0 1 1 9 The starting value is 1 and the growth factor is. The equation is: =! + 1 Checkpoints 507
Now we can go back and solve the original problem. a. Start b making a table from the tile pattern. Since the starting value is and the growth factor is the equation is = +. b. Make a table b replacing in the rule with 0, 1,, and computing. See the table above. Plot the points from the table to get the graph shown above. Looking at the table, the Plot the points from the table to get the graph shown above. c. Looking at the table, the -value changes from 1 to in three growth factors so each one must be three. Since the starting value is 1 and the growth factor is, the table and equation are: See the graph above. = + 1 0 1 1 7 figure # # of tiles 0 1 8 d. Again the table ma be determined from the points on the graph. Looking at the table, the starting value is and the growth factor is so the equation is =! +. Here are some more to tr. For each situation, complete the multiple representations web b finding the missing! table, graph, and/or rule. Since there are man possible patterns, it is not necessar to create one. 1.. 1 8 1 0 11. =!. Figure 1 Figure Figure 1 5.. =! + 1 1.5 508 Core Connections: Course
1 8 7. 8. Figure 1 Figure Figure 1 9. =! + 7. 1 1 1 Answers: 1.. = + 1 5 7 1 8 1 = +.. 0 1 1 7 0 1 5 8 11 1 8 1 1 8 1 = + Checkpoints 509
5.. 1 = 1 + 1 1 8 0 1 1 8 1 0 7. 8. 0 1 0 0 1 1 5 7 = 1 8 1 = + 1 9.. 8 1 0 1 7 5 1 8 = + 5 1 5 Core Connections: Course
Checkpoint Number 7 Problem 8- Transformations Answers to problem 8-: (Given in the order X, Y, Z) a. (, ), (, 0), (1, ), b. (,!1),!(,!),!(1,!1), c. (!1, ),!(!, ),!(!1,1), d. (!8,!),!(!8, ),!(!, ) Rigid transformations are was to move an object while not changing its shape or size. A translation (slide) moves an object verticall, horizontall or both. A reflection (flip) moves an object across a line of reflection as in a mirror. A rotation (turn) moves an object clockwise or counterclockwise around a point. A dilation is a non-rigid transformation. It produces a similar figure to the original b proportionall shrinking or stretching the figure from a point. Eample 1: Translate (slide)!abc left si units and down three units. Give the coordinates of the new!xyz. Solution: The original vertices are A (0, ), B (, 5), and C (5, 1). The new vertices are X (, 1), Y (, ), and Z ( 1, ). Eample : Reflect (flip)!abc across the -ais to get!pqr. Give the coordinates of the new!pqr. X Y Z A B C Now reflect (flip)!abc across the -ais to get!xyz. Give the coordinates of the new!xyz. Solution: The ke is that the reflection is the same distance from the ais as the original figure. For the first reflection the points are P!(,!), Q!(1,!), and R!(,!). For the second reflection the points are X!(!, ), Y!(!1, ), and Z!(!, ). X Z Y B Q C A P R Eample : Rotate (turn)!abc counterclockwise 90 about the origin to get!mno. Give the coordinates of the new!mno. Then rotate!mno counterclockwise another 90 to get!xyz. Give the coordinates of the new!xyz. Solution: After the first 90 rotation, the coordinates of A (,0), B (,0), and C (5, ) became M (0, ), N!(0, ), and O (, 5). Note X A B that each original point (, ) became (!, ). After the net 90 Z rotation, the coordinates of the vertices are now X!(!, 0), Y (, 0), and Z ( 5, ). After the 180 rotation each of the points (, ) in the original!abc became (!,!). Similarl a 70 counterclockwise rotation or a 90 clockwise rotation about the origin takes each point (, ) to the point (,!). Y O N M C Checkpoints 511
Eample : Dilate (enlarge/reduce)!abc b a scale factor of 1 from the origin to get!xyz. Give the coordinates of the new!xyz. Solution: Multipling the coordinates of the vertices of!abc b the scale factor gives the coordinates of the vertices of the new!xyz. The are X!(!, 0), Y!(0, ) and Z!(1,!1). A B Y X Z C Now we can go back and solve the original problem. a. Sliding the triangle two units to the right and three units up increases the -values b and the -values b. See the answers above. b. Reflecting the triangle across the -ais changes the -values to the opposite and keeps the -values the same. See the answers above. c. Rotating the triangle 90 clockwise moves the triangle into the second quadrant. The original -value becomes the new -value and the opposite of the original -value becomes the -value. See the answers above. d. Dilating the triangle b a scale factor of two multiplies all of the coordinates b two. See the answers above. Here some more to tr. For the following problems, refer to the figures below: Figure A Figure B Figure C B C C A B A B A C State the new coordinates after each transformation. 1. Slide figure A left units and down units.. Rotate figure C 180 counterclockwise about the origin.. Flip figure B across the -ais.. Flip figure B across the -ais. 5. Slide figure A right units and up 1 unit.. Flip figure C across the -ais. 51 Core Connections: Course
7. Rotate figure B 70 counterclockwise about the origin. 8. Slide figure C left 1 unit and up units. 9. Rotate figure A 90 clockwise about the origin.. Flip figure B across the line =. 11. Dilate figure B b a scale factor of. 1. Rotate figure A 180 counterclockwise about the origin. 1. Slide figure B units down and units to the right. 1. Dilate figure A b a scale factor of. 15. Rotate figure C 90 clockwise about the origin. 1. Flip figure A across the -ais. 17. Dilate figure C b a scale factor of 1. 18. Slide figure C units right and units down. 19. Rotate figure C 180 about the origin clockwise. 0. Dilate figure A b a scale factor of. Answers: 1. ( 1, ), (1, ), (,0). (, ), (, ), (, ). (5, ), (1, ), (0, 5). ( 5, ), ( 1, ), (0, 5) 5. (5, 1), (7, 5), (9, ). (, ), (, ), (, ) 7. (, ), (, 1), (5, 0) 8. ( 5, ), (, ), (, 1) 9. (0, 1), (, ), (, 5). (5, ), (1, ), (0, 1) 11. ( 15, ), (, ), (0, 15) 1. ( 1, 0), (, ), ( 5, ) 1. ( 1, ), (, ), (, 1) 1. (, 0), (1, 1), (0, 8) 15. (, ), (, ), (, ) 1. ( 1, 0), (, ), ( 5, ) 17. (, 1), (, 1), ( 1, 1.5) 18. ( 1, 0), (7, 0), (1, 5) 19. (, ), (, ), (, ) 0. (, 0), (, 8), (, ) Checkpoints 51
Checkpoint Number 8 Problem 8- Solving Equations with Fractions (Fraction Busters) Answers to problem 8-: a. 15 =, b.! $1.7, c.!5, d. (1, 1) Equations are usuall easier to solve if there are no fractions. To eliminate fractions from an equation, multipl both sides of the equation b the common denominator. Eample 1: Solve 1 + = Solution: Start b multipling both sides of the equation b 1 (the common denominator.) Simplif and then solve as usual. ( ) = 1 ( ) 1 1 + + 8 = 9 = 1 = 1 Eample : Solve 1.1 + 0.5 = 9 Solution: Two decimal places means that the common denominator is 0. Multipl both sides b 0 and solve as usual. 0(1.1 + 0.5) = 0(9) 1 + 5 = 900 15 = 900 = 0 Now we can go back and solve the original problem. 1 a. 5 + 1 = b. ( ) = 15() 15 1 5 + 1 + 5 = 0 8 = 0 = 0 8 = 15 = + 0.15 = $ 0( + 0.15) = 0($) 0 + 15 = $00 115 = $00! $1.7 + =! 7 c. d. ( ) = 1(! ) 1 + 7 7( + ) = (! ) 7 + 1 =! =!0 =!5 = + 8 = 1 + + 8 = 1 + ( ) = ( 1 + ) + 8 = + 0 + 8 = 1 = (1) + 8 = 1 (1,!1) 51 Core Connections: Course
Here are some more to tr. Solve each equation or sstem of equations. 1. 1 + = 5. = + 1 = 80 +. 15 =! 0. = =! 9 5.! =.! = +! 7. + 5 1 =! 1 8.!! 1 =! 9. 0. + = 0. = + 1 =! 9 11. =! + 1. + = + 1 =! 15 + 1. + =! 1.5 1. +1 = 5 1 15. = 1 + 8 1. = 7 + =! 1! =! 1 17. =! 1 = + 1 18. =! =! 19.!1 = 7 8 0. =! Checkpoints 515
Answers: 1. =. = 1,! =. = 18. = 9,! = 5. =. = 0 7. = 5 8. = 7 9. = 5. =,! = 11. =,! =! 1. = 1 1. = 9 1. = 15. =!1, = 1. = 1,! = 5 1 17. =,! = 1 18. =,! =!1 19. =.5 0. = 51 Core Connections: Course
Checkpoint Number 9 Problem 9- Scatter Plots and Association Answers to problem 9-: a. bo plot, b. scatterplot, c. See graph at right, d. strong negative association, e. =! + 5, f. $17,000, g. A slope of means the car is losing $000 in value each ear, -intercept of 5 means the cost when new was $5,000. Listed Cost (in 00s) 0 An association (relationship) between two numerical variables on a graph can be described b its form, direction, strength, and outliers. When the association has a linear form, a line a best fit can be drawn and its equation can be used to make predictions about other data. 5 Age of Car (ears) Eample 1: Describe the association in the scatterplot at right. Solution: Looking from left to right, ecept for point (e), the points are fairl linear and increasing. This is a moderate, positive linear association. The point (e) is an outlier e Eample : For the scatterplot, draw a line of best fit and determine the equation of the line. Solution: Use a ruler or straightedge to draw the line that approimates the trend of the points. If it is not a perfect line, approimatel the same number of points will be above and below the line of best fit. 15 5 To determine the equation of the line, draw in a slope triangle and determine the ratio of vertical side to horizontal side. In this case it is!0 =!. Estimate the -intercept b looking at 5 where the line intersects the -ais. In this case, it is approimatel 0. The equation of an non-vertical line ma be written in the form = m + b where m is the slope and b is the -intercept. 15 5 0 5 =! + 0 Checkpoints 517
Now we can go back and solve the original problem. a. Since the costs are a single set of data, a bo plot is convenient to show the distribution. b. Age and cost are two sets of related data so a scatterplot is appropriate. Listed Cost (in 00 s) 0 0 c. Reading from left to right, the scatterplot is decreasing, linear and the points are close to the line of best fit. This is a strong, linear, negative association. d. Looking at the line of best fit, the slope triangle has a ratio of! and the -intercept is approimatel 5. 1 Placing that information into the equation of a line = m + b ields =! + 5. e. Substituting = into the equation of part (c) ields =!() + 5 = 17. f. Slope represents the rate of change. A rate of change of means that the value is decreasing b units (in this case each unit is $00) per ear. The -intercept represents the value at ear zero or a new car. Here are some more to tr. In problems 1 through, describe the association. 5 Age of Car (ears) 1.. Height (cm) Shoe Size.. Height (feet) Price (Thousand $) Mileage (mpg) 50 Horsepower Age (ears) 1 1 5 Car Weight (00 s of lbs.) 518 Core Connections: Course
In problems 5 through 8 plot the data, draw a line of best fit and approimate the equation of the line. Distance to airport 5 15 0 5 0 5. Cost of shuttle ($) 1 17 1 1 0. Hours of eercise/month 9 1 15 18 Rate of heart attack/00 1 18 1 0 7. 8. Hours spent studing 0.5.8.5 5 Score on test 5 70 70 85 80 95 0 Hours since purchase 0 8 Number of cookies 0 1 11 5 0 Answers: 1. strong positive association. no association. strong positive association. strong negative association 5.. Cost ($) Miles to airport Rate of heart attack (per 00) Hours of eercise per month! = 1 + 8! = "1.5 + 0 7. 0 8. Score on Test 75 50 Number of Cookies 18 1 1 Hours Studied 5 5 Time since Purchase (Hours)! = +! =. + Checkpoints 519
Checkpoint Number Problem -9 Laws of Eponents and Scientific Notation Answers to problem -9: a. 7, b. 1, c.! = 1 9 f. 8! ", d. 9 =.9, e. 1.8!, The laws of eponents summarize several rules for simplifing epressions that have eponents. The rules are true for an base if! 0. a! b = (a+b) ( a ) b = ab a 0 = 1!a = 1 a b = (a!b) Scientific notation is a wa of writing a number as a product of two factors separated b a multiplication sign. The first factor must be less than and greater than or equal to 1. The second factor has a base of and an integer eponent. Eample 1: Simplif! - Solution: In a multiplication problem, if the bases are the same, add the eponents and keep the base. If the answer ends with a negative eponent, take the reciprocal and change the eponent to positive.! " = " = 1 = 1 1 Eample : Simplif ( )!5 -!5 Solution: Separate the fraction into two fractions with bases and 5. With an eponent on an eponent, multipl the eponents. Net, to divide epressions with eponents and the same base, subtract the eponents. ( )!5 "!5 = ( ) "! 5 5 = "! 5 5 1 = 8! 5 The number in standard form is ver large so it is common to give the answer using eponents. 50 Core Connections: Course
Eample : Multipl and give the answer in scientific notation. (8! ) "(.5! - ) Solution: Separate the number parts and the eponent parts. Multipl the number parts normall and the eponent part b adding the eponents. If this answer is not in scientific notation, change it appropriatel. (8! ) "(.5! # ) = (8!.5) " (! # ) =! = (.! 1 )! =.! Now we can be back and solve the original problem. a.! 5 = ( +5) = 7 b. (5 0 ) = 5 0! = 5 0 = 1 c.!5 " = (!5+) =! = 1 = 1 9 d.!1 " 7 = 1 " 9 = 9 =.9 e. (8! 5 ) "(1.! # ) = 1.8! = (1.8! 1 )! = 1.8! f.! 5! 5 = 5! ("5) = 0.8! " = (8! "1 )! " = 8! " Here are some more to tr. Simplif each epression. In problems 19 through write the final answer in scientific notation. 1. 5! 5 "1.!!. 7!(7 ) ".! " 1 " 5.! 5! 1 ( ). (! " )! 7 7.! " 0 8. 19 1! 19 19! " 9. 7!9 9!7. 1 1! " 11.! (! ) 0 1. ( )! 1. 5 ( 5 )!1 1. (7! 7 ) 15.!! " "! 1. 1 ( )! " " 0 Checkpoints 51
7 ( ) (! )!1 18. 9!9 "5 17. 7! 5 19. (.5! ) "(! 5 ) 0. (1.! ) " (7.1! # ) 1. (.9! 7 ) "(! ). (5.! " ) # (! "7 ). (! " )..7!.! " 9 0 Answers: 1. 5. 1. 7!.!1 = 1 5.. 1! "17 7.! = 1 7 8. 19!1 "! = 1 19" 9. 7 9. 15 11. 1 1. 1. 5 1. 7 15. 1. 9 17. 9! 7 5 18. 9! 19. 8.5! 8 0. 8.5! 1..07!..5! "1..! "5. 8.75! 5 5 Core Connections: Course
Checkpoint Number 11 Problem -88 Problems Involving Square Roots Answers to problem -88: a. 189! 1.75, b. 15! 1.0, c.!.7 cm, d. See graph at right. If the lengths of an two sides of a right triangle are known, the third side ma be found using the Pthagorean theorem: (leg#1) + (leg#) = hpotenuse. Equations of the form (variable) = number ma be solved using square roots. Eample 1: Determine the length of the missing side in the triangle at right. Solution: The hpotenuse (0) is the side opposite the right angle and the legs are and b. Then substituting this information into the Pthagorean theorem: + b = 0 0 + b = 00 b = 00! b = 00 " 17.!units Eample : The distance (d) in meters an object falls under gravit in (t) seconds is found using the formula d =.9t. How long does it take an object to fall 0 meters? Solution: Substitute into the formula and solve for t b first dividing to undo the multiplication b.9 and then square rooting to undo the squaring. 0 =.9t 0.9 = t! t = 0.9 ".5!seconds Checkpoints 5
Now we can go back and solve the original problem. a. Substituting the given information into the Pthagorean theorem: + = 17 0 + = 89!! = 189!! = 189 " 1.75 b. Graph the two points and draw the segment (d) determined b them. Use that segment as the hpotenuse of a right triangle and determine the length of the legs. Use that information in the Pthagorean theorem to determine the distance. 9 9 + 8 = d 81 + = d!! d = 15! d = 15 " 1.0 d 8 c. Use the formula for area of a circle: A =!r, substitute the value of A, and solve for r. 0 =!r 0! = r!" r = 0! #.7!cm d. Create an! table, fill in values for, and compute the values of. Note that for all <!, is a square root of a negative number, which is not a real number. 1 0 1 E 0 1 1. 1.7 See graph above. 5 Core Connections: Course
Here are some more to tr. Solve each problem. In problems 1 through find the length of the missing side. 1.. a 9.. 0 8 b 7 15 c 5.. 7 7 180 a b In problems 7 through 1 find the distance between the two points. 7. (1, ) and (7,!) 8. (, ) and (, 1) 9. ( 1, ) and (, 1). (, ) and ( 5, 1) In problems 11 through 1 find the required information. 11. A circle has area 0 ft, what is the radius? 1. A ball falls from a second stor window to the ground, a total of 15 feet. How long did it take? 1. + 7 = 11 1.! 8 = 0 15. If a circle has an area of 75 square meters, what is the diameter? 1. A parachutist falls 00 feet before opening her parachute. How long was she in free fall? In problems 17 through 0 make an! table and draw the graph. 17. =! 18. = + 19. = + 0. =! 1 Checkpoints 55
Answers: 1. c =. a = 19!.. c = 9! 1.05. b = 17! 1.7 5. a = 5. b = 1 7. 8. 89! 9. 9.! 5.. 1! 1.7 11. About 5. ft 1. About 1.75 seconds 1. = 1. = 15. About.89 meters 1. About 7.8 seconds 17. 0 1 5 1 0.59 0.7 0 0. 0.5 18. 1 0 1 0 0.5 0.71 0.87 1 1.1 1. 19. 0 1 5.1 5 5.5 5.8.1. 0. 1 5 0.8..7 5 Core Connections: Course