core connections, course 2 checkpoint materials...checkpoints 5 checkpoint 2 problem 2-120 multiple...

Checkpoints 1

Core Connections, Course 2 Checkpoint Materials

Notes to Students (and their Teachers)

Students master different skills at different speeds. No two students learn exactly the same way at the same time. At some point you will be expected to perform certain skills accurately. Most of the Checkpoint problems incorporate skills that you should have been developing in grades 5 and 6. If you have not mastered these skills yet it does not mean that you will not be successful in this class. However, you may need to do some work outside of class to get caught up on them. Starting in Chapter 1 and finishing in Chapter 9, there are 9 problems designed as Checkpoint problems. Each one is marked with an icon like the one above and numbered according to the chapter that it is in. After you do each of the Checkpoint problems, check your answers by referring to this section. If your answers are incorrect, you may need some extra practice to develop that skill. The practice sets are keyed to each of the Checkpoint problems in the textbook. Each has the topic clearly labeled, followed by the answers to the corresponding Checkpoint problem and then some completed examples. Next, the complete solution to the Checkpoint problem from the text is given, and there are more problems for you to practice with answers included. Remember, looking is not the same as doing! You will never become good at any sport by just watching it, and in the same way, reading through the worked examples and understanding the steps is not the same as being able to do the problems yourself. How many of the extra practice problems do you need to try? That is really up to you. Remember that your goal is to be able to do similar problems on your own confidently and accurately. This is your responsibility. You should not expect your teacher to spend time in class going over the solutions to the Checkpoint problem sets. If you are not confident after reading the examples and trying the problems, you should get help outside of class time or talk to your teacher about working with a tutor.

Checkpoint Topics

1. Area and Perimeter of Polygons 2. Multiple Representations of Portions 3. Multiplying Fractions and Decimals 5 Order of Operations 6. Writing and Evaluating Algebraic Expressions 7A. Simplifying Expressions 7B. Displays of Data: Histograms and Box Plots 8. Solving Multi-Step Equations 9. Unit Rates and Proportions

Fe

Checkpoints 5

Checkpoint 2 Problem 2-120

Multiple Representations of Portions

Answers to problem 2-120: a. 43%, 43100 , b. 910 , 0.9, 90% , c.

39100 , 0.39 , d. 64%, 0.64

Portions of a whole may be represented in various ways as represented by this web. Percent means “per hundred” and the place value of a decimal will determine its name. Change a fraction in an equivalent fraction with 100 parts to name it as a percent. Example 1: Name the given portion as a fraction and as a percent. 0.3 Solution: The digit 3 is in the tenths place so 0.3= three-tenths = 310 .

On a diagram or a hundreds grid, 3 parts out of 10 is equivalent to 30 parts out of 100 so 310 =

30100 = 30% .

Example 2: Name the given portion as a fraction and as a decimal. 35% Solution: 35% = 35100 = thirty-five-hundredths = 0.35 ;

35100 =

720

Now we can go back and solve the original problem. a. 0.43 is forty-three-hundredths or 43100 = 43% b. nine-tenths is 910 =

910 ⋅

1010 =

90100 = 90% ;

910 = 0.9

c. 39% = 39100 = thrity-nine-hundredths = 0.39 d. 1625 =

1625 ⋅

44 =

64100 = 0.64 = 64%

words or

pictures

fraction

decimal percent

Representations of a Portion

e

6 Core Connections, Course 2

Here are some more to try. For each portion of a whole, write it as a percent, fraction, and a decimal.

1. 6% 2. 0.35

3. 14 4. 25

5. 0.16 6. 87%

7. 1325 8. 21%

9. 750 10. 0.050

11. 65% 12. 3.7%

13. 710 14. 0.66

15. 1920 16. 20%

17. 0.23 18. 1.0

19. 135% 20. 77100 Answers:

1. 6100 =350 , 0.06 2. 35%,

35100 =

720

3. 25%, 0.25 4. 40%, 0.4

5. 16%, 16100 =425 6.

87100 , 0.87

7. 52%, 0.52 8. 21100 , 0.21

9. 14%, 0.14 10. 5%, 5100 =120

11. 1320 , 0.65 12. 371000 , 0.037

13. 70%, 0.7 14. 66%, 66100 =3350

15. 95%, 0.95 16. 20100 =15 , 0.2

17. 23%, 23100 18. 100%,100100 =

11

19. 135100 = 135100 = 1

720 ,1.35 20. 77%, 0.77

Checkpoints 7


Multiplying Fractions and Decimals

Answers to problem 3-110: a. 920 , b. 1

5, c. 4 29 , d. 7

15 , e. 12.308, f. 0.000208

To multiply fractions, multiply the numerators and then multiply the denominators. To multiply mixed numbers, change each mixed number to a fraction greater than one before multiplying. In both cases, simplify by looking for factors than make “one.” To multiply decimals, multiply as with whole numbers. In the product, the number of decimal places is equal to the total number of decimal places in the multiplied numbers. Sometimes zeros need to be added to place the decimal point. Example 1: Multiply

3

8

4

5 Example 2: Multiply

3

1

32

1

2

Solution: Solution: 38 ⋅45 ⇒

3⋅48⋅5 ⇒

3⋅ 42⋅ 4 ⋅5 ⇒

310 3

13 ⋅2

12 ⇒

103 ⋅

52 ⇒

10⋅53⋅2 ⇒

5⋅2 ⋅53⋅2 ⇒

253 or 8

13

Note that we are simplifying using Giant Ones but no longer drawing the Giant One. Example 3: Multiply 12.5 0.36 Solution: Now we can go back and solve the original problem. a. 23 ⋅

25 ⇒

2⋅23⋅5 ⇒

415 b.

710 ⋅

27 ⇒

7 ⋅25⋅2 ⋅ 7 ⇒

15

c. 2 13 ⋅212 ⇒

73 ⋅

52 ⇒

7⋅53⋅2 ⇒

356 or 5

56 d. 1

13 ⋅2

16 ⇒

43 ⋅

136 ⇒

2⋅2 ⋅133⋅2 ⋅3 ⇒

269 or 2

89

e. f.

(one decimal place)

(two decimal places)

(three decimal places)

Toe


Here are some more to try. Multiply the fractions and decimals below.

1. 0.08 4.7 2. 0.21 3.42

3. 47

1

2 4. 5

6

3

8

5. 89

3

4 6. 7

10

3

4

7. 3.07 5.4 8. 6.57 2.8

9. 56

3

20 10. 2.9 0.056

11. 67

4

9 12. 3 1

71 25

13. 23

5

9 14. 3

5

9

13

15. 2.34 2.7 16. 2 134 45

17. 4 35

1

2 18. 3

8

5

9

19. 0.235 0.43 20. 421 0.00005 Answers:

1. 0.376 2. 0.7182

3. 27

4. 516

5. 23

6. 2140

7. 16.578 8. 18.396

9. 18

10. 0.1624

11. 821

12. 4 25

13. 1027

14. 2765

15. 6.318 16. 11 15

17. 2 310

18. 524

19. 0.10105 20. 0.02105

Checkpoints 9


Order of Operations Answers to problem 5-148: a: 20, b: –4

In general, simplify an expression by using the order of operations: • Evaluate each exponential (for example, 52 = 5 ⋅5 = 25 ). • Multiply and divide each term from left to right.

• Combine like terms by adding and subtracting from left to right. But simplify the expressions in parentheses or any other expressions of grouped

numbers first. Numbers above or below a “fraction bar” are considered grouped. A good way to remember is to circle the terms like in the following example. Remember that terms are separated by + and – signs.

Example 1: Simplify 12 ÷ 22 − 4 + 3(1+ 2)3

Simplify within the circled terms: Be sure to perform the exponent operations before dividing. 12 ÷ 22 = 12 ÷ 2 ⋅2 = 3

Then perform the exponent operation: 33 = 3⋅3⋅3= 27

Next, multiply and divide left to right: 3(27) = 81

Finally, add and subtract left to right: 3 – 4 = –1 Example 2: Simplify −32 − 2+73 + 8 ÷

12( )

Simplify within the circled terms: −32 = −3⋅3= −9 2+73 =

93 = 3 8 ÷

12 = 8 ⋅

21 = 16

Then add and subtract, left to right. Now we can go back and solve the original problem.

a. 16 − 23 ÷ 8 + 516 − 2 ⋅2 ⋅2 ÷ 8 + 516 − 4 ⋅2 ÷ 8 + 516 − 8 ÷ 8 + 516 −1+ 515 + 520

b. (−2 + 6)2 − 32( ) ⋅14 +1(4)2 − 422 +116 − 21+1−5 +1−4

12 ÷ 22 − 4 + 3(1+ 2)3

3− 4 + 3(3)3

3− 4 + 3(27)

3− 4 + 81

−1+ 8180

−32 − 2+73 + 8 ÷12( )

−9 − 3+16−12 +164


Here are some more to try.

1. 10 ⋅ 12 + (−6)(−3) 2. −5+(−6) 23( )

−3

3. (6 − 8)(9 −10)− (4 + 2)(6 + 3) 4. −8436−12(−5)

5. 12 6 − 2( )2 − 4 ⋅3 6. 3 2 1+ 5( ) + 8 − 32( )

7. 8 +12( ) ÷ 4 − 6 8. −62 + 4 ⋅8

9. 18 ⋅3÷ 33 10. 10 + 52 − 25

11. 20 − (33 ÷ 9) ⋅2 12. 100 − (23 − 6)÷ 2

13. 85 − (4 ⋅2)2 − 3 14. 22 + (3⋅2)2 ÷ 2

15. 16 +11(−2)2 − 25 25( ) 16. 54 ÷ 32 + 43( ) 272( )−12 17. 2 3−1( )3 ÷ 8 18. (72 −1)÷ 4 + 2

19. −3⋅2 ÷ (−2 − 4) 20. −3⋅2 ÷ −2 − 4

21. 12 + 3 8−212−9( )− 2 9−119−15( ) 22. 15 + 4 11−29−6( )− 2 12−418−10( )

23. 32 ÷16 − 8 ⋅25 12( )2 24. 36 +16 ⋅ 14( )− (50 ÷ 25)2 Answers:

1. 23 2. 3

3. –52 4. − 78

5. –4 6. 33

7. –1 8. –4

9. 2 10. 10

11. 14 12. 99

13. 18 14. 40

15. 50 16. 23

17. 2 18. 14

19. 1 20. –1

21. 14 22. 25

23. –48 24. 36

Checkpoints 11


Writing and Evaluating Algebraic Expressions Answers to problem 5-133: a. x + 6 , b. y − 5 , c. 2x + 3 , d. 5y , e. 11; 3; 13; 40 There are some vocabulary words that are frequently used to represent arithmetic operations. Addition is often suggested by: sum, increased, more than, greater than, total

Subtraction is often suggested by: difference, decreased by, less than, smaller than

Multiplication is often suggested by: product, times, twice, double

Division is often suggested by: quotient, divided by, shared evenly Examples Five more than m: Five more than a number, increases a number by 5 so it would be m + 5 . Three less than x: Three less than a number makes the number smaller by 3 so it would be x – 3. Triple m: Tripling a number is the same as multiplying the number by 3 so it would be 3m . Five divided by x: Division is usually written as a fraction so it would be 5x . To evaluate an algebraic expression means to calculate the value of the expression when the variable is replaced by a numerical value. Examples Evaluate 2x − 5 if x = 7 Solution: 2x − 5⇒ 2 ⋅7 − 5 = 14 − 5 = 9 Evaluate 6x + 9 if x = 2 Solution: 6x + 9⇒

62 + 9 = 3+ 9 = 12

Now we can go back and solve the original problem. Part (e) is included with the writing of the expression. a. Six more suggests adding 6 so it would be x + 6 ; if x = 5 then x + 6⇒ 5 + 6 = 11 . b. Five less suggests subtracting 5 so it would be y − 5 ; if y = 8 then y − 5⇒ 8 − 5 = 3 . c. Twice a number suggests multiplication by 2 and then increasing by 3 means to add 3 to the

previous product so it would be 2x + 3 ; if x = 5 then 2x + 3⇒ 2 ⋅5 + 3= 13 . d. Product suggests multiply so it would be 5y ; if y = 8 then 5y⇒ 5 ⋅8 = 40 .


Here are some more to try. For problems 1 through 16 write an algebraic expression. For problems 17 through 24 evaluate the expression for the given values.

1. 5 greater than m 2. Double s

3. 7 less than t 4. 6 more than y

5. 14 divided by b 6. m subtracted from a

7. x divided among 5 8. The product of 7 and e

9. The sum of d and l 10. 6 times c

11. The product of x, y, and w 12. h times 8, added to j

13. 4 divided by p, increased by 7 14. Half of r

15. Three times q increased by 5 16. Two less than triple n

17. 4x − 3 if x = 5 18. 7d if d = 10

19. 2xy +1 if x = 3, y = 4 20. 4 + 8g if g = 6

21. 4b +12 if b = 11 22. 9g27 if g = 3

23. mw + h if m = 5,w = 8, h = 6 24. yz + 7 if y = 4, z = 2 Answers:

1. m + 5 2. 2s

3. t − 7 4. y + 6

5. 14b 6. a −m

7. x5 8. 7e

9. d + l 10. 6c

11. xyw 12. 8h + j

13. 4p + 7 14. r2

15. 3q + 5 16. 3n − 2

17. 17 18. 70

19. 25 20. 52

21. 56 22. 1

23. 46 24. 9

Checkpoints 13

Checkpoint 7A Problem 7-50

Simplifying Expressions

Answers to problem 7-50: a: 2x2 + x −10 , b: x2 − 9x +11 Like terms are two or more terms that are exactly the same except for their coefficients. That is, they have the same variable(s), with corresponding variable(s) raised to the same power. Like terms can be combined into one quantity by adding and/or subtracting the coefficients of the terms. Terms are usually listed in the order of decreasing powers of the variable. Combining like terms, one way of simplifying expressions, using algebra tiles is shown in the first two examples. Example 1: Simplify (2x2 + 4x + 5)+ (x2 + x + 3) means combine 2x2 + 4x + 5 with x2 + x + 3 .

(2x2 + 4x + 5)+ (x2 + x + 3) = 3x2 + 5x + 8 Example 2:

Simplify x2 + 3x − 4 + 2(2x2 − x)+ 3 . Now we can go back and solve the original problem. a. 4x2 + 3x − 7 + −2x2 − 2x + (−3)

(4x2 − 2x2 )+ (3x − 2x)+ (−7 + (−3))2x2 + x −10

b. −3x2 − 2x + 5 + 4x2 − 7x + 6(−3x2 + 4x2 )+ (−2x − 7x)+ (5 + 6)

x2 − 9x +11

= +1 = –1

Remember:

0

= x 2 x 2

x 2

x x x

x x 2

x 2 x 2

x

= 5x2 + x −1

x 2 x 2

x

x 2

x 2

x 2 x

x 2

x

x x

x 2

x



1. (x2 + 3x + 4)+ (x2 + 3x + 2) 2. x2 + 4x + 3+ x2 + 2x + 5

3. 2x2 + 2x −1+ x2 − 4x + 5 4. 3x2 − x + 7 − 3x2 + 2x + (−4)

5. 2x2 + 4x + (−3)+ x2 − 3x + 5 6. 4x2 + 2x − 8 + (−2x2 )+ 5x +1

7. −4x2 + 2x + 8 − 3x2 + 5x − 3 8. 3x2 + 2(4x +1)− 2x2 + 4x + 5

9. 3(5x2 + 4x) – 7 + 2x + 3 10. 3x2 – 4x + 2 + 4(2x2 + 2x)

11. −2(3x2 – x + 2)+ 3x –1 12. 2(x2 – 2x)+ 7 + 5(x2 + 4x) – 3

13. 2x2 – 3(x – 3)+ 5x2 – 4x 14. –3x + 3(x2 + 2)+ (–x) – 4

15. (x2 + 3x + 2)− (x2 − 3x − 2) 16. (2x2 − 5x)− (x2 + 7x)

17. (5x + 6)− (x2 − 5x + 6) 18. 4a + 2b + 3c − 6a + 3b − 6c

19. 3c + 4a − 7c + 5b − (−4a)+ 7 20. −5a + 6b − 7c − (3c − 4b − a) Answers:

1. 2x2 + 6x + 6 2. 2x2 + 6x + 8

3. 3x2 − 2x + 4 4. x + 3

5. 3x2 + x + 2 6. 2x2 + 7x − 7

7. −7x2 + 7x + 5 8. x2 +12x + 7

9. 15x2 +14x − 4 10. 11x2 + 4x + 2

11. −6x2 + 5x − 5 12. 7x2 +16x + 4

13. 7x2 − 7x + 9 14. 3x2 − 4x + 2

15. 6x + 4 16. x2 −12x

17. −x2 +10x 18. −2a + 5b − 3c

19. 8a + 5b − 7c + 7 20. −4a +10b −10c

Checkpoints 19


Solving Multi-Step Equations Answers to problem 8-109: a: x = 7, b: x = –2, c: x = 0, d: x = 23 A general strategy for solving equations is to first simplify each side of the equation. Next isolate the variable on one side and the constants on the other by adding equal values on both sides of the equation or removing balanced sets or zeros. Finally determine the value of the variable–usually by division. Note: When the process of solving an equation ends with different numbers on each side of the equal sign (for example, 2 = 4), there is no solution to the problem. When the result is the same expression or number on each side of the equation (for example, x + 2 = x + 2 ) it means that all numbers are solutions. Example 1: Solve 3x + 3x – 1 = 4x + 9

3x + 3x −1= 4x + 96x −1= 4x + 92x = 10x = 5

Example 2: Solve –2x + 1 + 3(x – 1) = ––4 + –– x – 2

−2x +1+ 3(x −1) = −4 + −x − 2−2x +1+ 3x − 3= −x − 6

x − 2 = −x − 62x = −4x = −2

Now we can go back and solve the original problem. a. 24 = 3x + 3

21= 3x7 = x

b. 6x +12 = −x − 26x = −x −147x = −14x = −2

c. 3x + 3− x + 2 = x + 52x + 5 = x + 52x = xx = 0

d. 5(x −1) = 5(4x − 3)5x − 5 = 20x −155x = 20x −10

−15x = −10x = −10−15 =

23

Solution

Solution

F



1. 2x + 3= −7 2. −3x − 2 = −1

3. 3x + 2 + x = x + 5 4. 4x − 2 − 2x = x − 5

5. 2x − 3= −x + 3 6. 1+ 3x − x = x − 4 + 2x

7. 4 − 3x = 2x − 6 8. 3+ 3x − x + 2 = 3x + 4

9. −x − 3= 2x − 6 10. −4 + 3x −1= 2x +1+ 2x

11. −x + 3= 6 12. 5x − 3+ 2x = x + 2 + x

13. 2x − 7 = −x −1 14. −2 + 3x = x − 2 − 4x

15. −3x + 7 = x −1 16. 1+ 2x − 4 = −3+ x

17. 3(x + 2) = x + 2 18. 2(x − 2)+ x = 5

19. 10 = x + 5 + x 20. −x + 2 = x − 5 − 3x

21. 3x + 2 − x = x − 2 + x 22. 4x +12 = 2x − 8

23. 3(4 + x) = x + 6 24. 6 − x − 3= 4(x − 2) Answers:

1. x = –5 2. x = − 13

3. x = 1 4. x = –3

5. x = 2 6. x = 5

7. x = 2 8. x = 1

9. x = 1 10. x = –6

11. x = –3 12. x = 1

13. x = 2 14. x = 0

15. x = 2 16. x = 0

17. x = –2 18. x = 3

19. x = 2 12 20. x = –7

21. no solution 22. x = –10

23. x = –3 24. x = 2 15

Checkpoints 21


Unit Rates and Proportions Answers to problem 9-92: a: 24 mpg; b: $0.19; c: x = 8 ; d: m = 32.5 A rate is a ratio comparing two quantities and a unit rate has a denominator of one after simplifying. Unit rates or proportions may be used to solve ratio problems. Solutions may also be approximated by looking at graphs of lines passing through the origin and the given information. Example 1: Judy’s grape vine grew 15 inches in 6 weeks. What is the unit growth rate

(inches per week)? Solution: The growth rate is 15 inches

6 weeks. To create a unit rate we need a denominator of “one.”

15 inches

6 weeks=x inches

1 week. Solve by using a Giant One: 15 inches

6 weeks=6

6

x inches

1 week2.5 inches

week

Example 2: Bob’s favorite oatmeal raisin cookie recipe use 3 cups of raisins for 5 dozen

cookies. How many cups are needed for 40 dozen cookies? Solution: The rate is 3 cups

5 dozen so the problem may be written as this proportion: 3

5=

c

40.

One method of solving the proportion is to use a Giant One:

3

5=

c

40

3

5

8

8=24

40c = 24

Another method is to think about unit rates. Since the unit rate is 3

5cup per dozen, one could

also take the unit rate and multiply by the number of units needed: 3540 = 24 .

Using either method the answer is 24 cups of raisins. Now we can go back and solve the original problem. a. 108 miles

4.5 gallons=4.54.5

x miles1 gallon

24 milesgallon

b. $3.2317 oranges

=1717

x

1 orange0.19 $

orange

c. Using a Giant One: d. Using the Giant One

x12 ⋅2.52.5 =

2.5x30 =

2030 ⇒ 20 = 2.5x⇒ x = 8

1340 ⋅

2.52.5 =

32.5100 ⇒ m = 32.5


Here are some more to try. For problems 1 through 8 find the unit rate. For problems 17 through 24 solve the proportion.

1. Typing 544 words in 17 minutes ( words per minute )

2. Taking 92 minutes to run 10 miles ( minutes per mile )

3. Reading 258 pages in 86 minutes ( pages per minute )

4. Falling 385 feet in 35 seconds ( feet per second )

5. Buying 15 boxes of cereal for $39.75 ( $ per box)

6. Drinking 28 bottles of water in 8 days ( bottles per day )

7. Scoring 98 points in a 40 minute game (points per minute)

8. Planting 76 flowers in 4 hours (flowers per hour)

9. 38=

x

50 10. 2

5=

x

75 11. 7

9=14

x 12. 24

25=96

x

13. 159=12

x 14. 45

60=

x

4 15. 4

7=18

x 16. 8

9=72

x

17. 35=

x

17 18. 17

30=51

x 19. 5

8=16

x 20. 3

22=15

x

21. 15=

x

27 22. x

11=8

15 23. 14

17=

x

34 24. 12

15=36

x

Answers:

1. 32 wordsminute

2. 9.2 minutesmile 3. 3

pages

minute 4. 11 feet

second

5. 2.65 dollarsbox

6. 3.5 bottlesday 7. 2.45

points

minute 8. 19 flowers

hour

9. x = 18.75 10. x = 30 11. x = 18 12. x = 100 13. x = 7.2 14. x = 3 15. x = 31.5 16. x = 81 17. x = 10.2 18. x = 90 19. x = 25.6 20. x = 110

21. x = 5.4 22. x = 5 1315 23. x = 28 24. x = 45

core connections, course 2 checkpoint materials...checkpoints 5 checkpoint 2 problem 2-120 multiple...

Documents