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Report an Error
A: ( 5 , 1 ) ( 5 , 1 ) B: ( −2 , 4 ) ( −2 , 4 ) C: ( −5 , −1 ) ( −5 , −1 ) D: ( 3 , −2 ) ( 3 , −2 ) E: ( 0 , −5 ) ( 0 , −5 ) F: ( 4 , 0 ) ( 4 , 0 )
A: ( 4 , 2 ) ( 4 , 2 ) B: ( −2 , 3 ) ( −2 , 3 ) C: ( −4 , −4 ) ( −4 , −4 ) D: ( 3 , −5 ) ( 3 , −5 ) E: ( −3 , 0 ) ( −3 , 0 ) F: ( 0 , 2 ) ( 0 , 2 )
0 | ||
2 | 5 |
0 | 1 | |
1 | 7 | |
0 | ||
10 | 0 | |
0 | ||
4 | 0 | |
Answers will vary.
ⓐ yes, yes ⓑ yes, yes
ⓐ no, no ⓑ yes, yes
x - intercept: ( 2 , 0 ) ( 2 , 0 ) ; y - intercept: ( 0 , −2 ) ( 0 , −2 )
x - intercept: ( 3 , 0 ) ( 3 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )
x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , 12 ) ( 0 , 12 )
x - intercept: ( 8 , 0 ) ( 8 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )
x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −3 ) ( 0 , −3 )
x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −2 ) ( 0 , −2 )
− 2 3 − 2 3
− 4 3 − 4 3
− 3 5 − 3 5
− 1 36 − 1 36
− 1 48 − 1 48
slope m = 2 3 m = 2 3 and y -intercept ( 0 , −1 ) ( 0 , −1 )
slope m = 1 2 m = 1 2 and y -intercept ( 0 , 3 ) ( 0 , 3 )
2 5 ; ( 0 , −1 ) 2 5 ; ( 0 , −1 )
− 4 3 ; ( 0 , 1 ) − 4 3 ; ( 0 , 1 )
− 1 4 ; ( 0 , 2 ) − 1 4 ; ( 0 , 2 )
− 3 2 ; ( 0 , 6 ) − 3 2 ; ( 0 , 6 )
ⓐ intercepts ⓑ horizontal line ⓒ slope–intercept ⓓ vertical line
ⓐ vertical line ⓑ slope–intercept ⓒ horizontal line ⓓ intercepts
- ⓐ 50 inches
- ⓑ 66 inches
- ⓒ The slope, 2, means that the height, h , increases by 2 inches when the shoe size, s , increases by 1. The h -intercept means that when the shoe size is 0, the height is 50 inches.
- ⓐ 40 degrees
- ⓑ 65 degrees
- ⓒ The slope, 1 4 1 4 , means that the temperature Fahrenheit ( F ) increases 1 degree when the number of chirps, n , increases by 4. The T -intercept means that when the number of chirps is 0, the temperature is 40 ° 40 ° .
- ⓒ The slope, 0.5, means that the weekly cost, C , increases by $0.50 when the number of miles driven, n, increases by 1. The C -intercept means that when the number of miles driven is 0, the weekly cost is $60
- ⓒ The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n , increases by 1.80. The C -intercept means that when the number of invitations is 0, the weekly cost is $35.;
not parallel; same line
perpendicular
not perpendicular
y = 2 5 x + 4 y = 2 5 x + 4
y = − x − 3 y = − x − 3
y = 3 5 x + 1 y = 3 5 x + 1
y = 4 3 x − 5 y = 4 3 x − 5
y = 5 6 x − 2 y = 5 6 x − 2
y = 2 3 x − 4 y = 2 3 x − 4
y = − 2 5 x − 1 y = − 2 5 x − 1
y = − 3 4 x − 4 y = − 3 4 x − 4
y = 8 y = 8
y = 4 y = 4
y = 5 2 x − 13 2 y = 5 2 x − 13 2
y = − 2 5 x + 22 5 y = − 2 5 x + 22 5
y = 1 3 x − 10 3 y = 1 3 x − 10 3
y = − 2 5 x − 23 5 y = − 2 5 x − 23 5
x = 5 x = 5
x = −4 x = −4
y = 3 x − 10 y = 3 x − 10
y = 1 2 x + 1 y = 1 2 x + 1
y = − 1 3 x + 10 3 y = − 1 3 x + 10 3
y = −2 x + 16 y = −2 x + 16
y = −5 y = −5
y = −1 y = −1
x = −5 x = −5
ⓐ yes ⓑ yes ⓒ yes ⓓ yes ⓔ no
ⓐ yes ⓑ yes ⓒ no ⓓ no ⓔ yes
y ≥ −2 x + 3 y ≥ −2 x + 3
y < 1 2 x − 4 y < 1 2 x − 4
x − 4 y ≤ 8 x − 4 y ≤ 8
3 x − y ≤ 6 3 x − y ≤ 6
Section 4.1 Exercises
A: ( −4 , 1 ) ( −4 , 1 ) B: ( −3 , −4 ) ( −3 , −4 ) C: ( 1 , −3 ) ( 1 , −3 ) D: ( 4 , 3 ) ( 4 , 3 )
A: ( 0 , −2 ) ( 0 , −2 ) B: ( −2 , 0 ) ( −2 , 0 ) C: ( 0 , 5 ) ( 0 , 5 ) D: ( 5 , 0 ) ( 5 , 0 )
0 | ||
2 | 0 | |
0 | 5 | |
3 | 2 | |
7 |
0 | 1 | |
3 | 2 | |
6 | 3 |
0 | ||
2 | ||
1 |
0 | 2 | |
3 | 4 | |
6 | 0 |
0 | ||
10 | 2 | |
5 | 0 |
ⓑ Age and weight are only positive.
Section 4.2 Exercises
ⓐ yes; no ⓑ no; no ⓒ yes; yes ⓓ yes; yes
ⓐ yes; yes ⓑ yes; yes ⓒ yes; yes ⓓ no; no
$722, $850, $978
Section 4.3 Exercises
( 3 , 0 ) , ( 0 , 3 ) ( 3 , 0 ) , ( 0 , 3 )
( 5 , 0 ) , ( 0 , −5 ) ( 5 , 0 ) , ( 0 , −5 )
( −2 , 0 ) , ( 0 , −2 ) ( −2 , 0 ) , ( 0 , −2 )
( −1 , 0 ) , ( 0 , 1 ) ( −1 , 0 ) , ( 0 , 1 )
( 6 , 0 ) , ( 0 , 3 ) ( 6 , 0 ) , ( 0 , 3 )
( 0 , 0 ) ( 0 , 0 )
( 4 , 0 ) , ( 0 , 4 ) ( 4 , 0 ) , ( 0 , 4 )
( −3 , 0 ) , ( 0 , 3 ) ( −3 , 0 ) , ( 0 , 3 )
( 8 , 0 ) , ( 0 , 4 ) ( 8 , 0 ) , ( 0 , 4 )
( 2 , 0 ) , ( 0 , 6 ) ( 2 , 0 ) , ( 0 , 6 )
( 12 , 0 ) , ( 0 , −4 ) ( 12 , 0 ) , ( 0 , −4 )
( 2 , 0 ) , ( 0 , −8 ) ( 2 , 0 ) , ( 0 , −8 )
( 5 , 0 ) , ( 0 , 2 ) ( 5 , 0 ) , ( 0 , 2 )
( 4 , 0 ) , ( 0 , −6 ) ( 4 , 0 ) , ( 0 , −6 )
( 3 , 0 ) , ( 0 , 1 ) ( 3 , 0 ) , ( 0 , 1 )
( −10 , 0 ) , ( 0 , 2 ) ( −10 , 0 ) , ( 0 , 2 )
ⓐ ( 0 , 1000 ) , ( 15 , 0 ) ( 0 , 1000 ) , ( 15 , 0 ) ⓑ At ( 0 , 1000 ) ( 0 , 1000 ) , he has been gone 0 hours and has 1000 miles left. At ( 15 , 0 ) ( 15 , 0 ) , he has been gone 15 hours and has 0 miles left to go.
Section 4.4 Exercises
−3 2 = − 3 2 −3 2 = − 3 2
− 1 3 − 1 3
− 3 4 − 3 4
− 5 2 − 5 2
− 8 7 − 8 7
ⓐ 1 3 1 3 ⓑ 4 12 pitch or 4-in-12 pitch
3 50 3 50 ; rise = 3, run = 50
ⓐ 288 inches (24 feet) ⓑ Models will vary.
When the slope is a positive number the line goes up from left to right. When the slope is a negative number the line goes down from left to right.
A vertical line has 0 run and since division by 0 is undefined the slope is undefined.
Section 4.5 Exercises
slope m = 4 m = 4 and y -intercept ( 0 , −2 ) ( 0 , −2 )
slope m = −3 m = −3 and y -intercept ( 0 , 1 ) ( 0 , 1 )
slope m = − 2 5 m = − 2 5 and y -intercept ( 0 , 3 ) ( 0 , 3 )
−9 ; ( 0 , 7 ) −9 ; ( 0 , 7 )
4 ; ( 0 , −10 ) 4 ; ( 0 , −10 )
−4 ; ( 0 , 8 ) −4 ; ( 0 , 8 )
− 8 3 ; ( 0 , 4 ) − 8 3 ; ( 0 , 4 )
7 3 ; ( 0 , −3 ) 7 3 ; ( 0 , −3 )
horizontal line
vertical line
slope–intercept
- ⓒ The slope, 2.54, means that Randy’s payment, P , increases by $2.54 when the number of units of water he used, w, increases by 1. The P –intercept means that if the number units of water Randy used was 0, the payment would be $28.
- ⓒ The slope, 0.32, means that the cost, C , increases by $0.32 when the number of miles driven, m, increases by 1. The C -intercept means that if Janelle drives 0 miles one day, the cost would be $15.
- ⓒ The slope, 0.09, means that Patel’s salary, S , increases by $0.09 for every $1 increase in his sales. The S -intercept means that when his sales are $0, his salary is $750.
- ⓒ The slope, 42, means that the cost, C , increases by $42 for when the number of guests increases by 1. The C -intercept means that when the number of guests is 0, the cost would be $750.
not parallel
- ⓐ For every increase of one degree Fahrenheit, the number of chirps increases by four.
- ⓑ There would be −160 −160 chirps when the Fahrenheit temperature is 0 ° 0 ° . (Notice that this does not make sense; this model cannot be used for all possible temperatures.)
Section 4.6 Exercises
y = 4 x + 1 y = 4 x + 1
y = 8 x − 6 y = 8 x − 6
y = − x + 7 y = − x + 7
y = −3 x − 1 y = −3 x − 1
y = 1 5 x − 5 y = 1 5 x − 5
y = − 2 3 x − 3 y = − 2 3 x − 3
y = 2 y = 2
y = −4 x y = −4 x
y = −2 x + 4 y = −2 x + 4
y = 3 4 x + 2 y = 3 4 x + 2
y = − 3 2 x − 1 y = − 3 2 x − 1
y = 6 y = 6
y = 3 8 x − 1 y = 3 8 x − 1
y = 5 6 x + 2 y = 5 6 x + 2
y = − 3 5 x + 1 y = − 3 5 x + 1
y = − 1 3 x − 11 y = − 1 3 x − 11
y = −7 y = −7
y = − 5 2 x − 22 y = − 5 2 x − 22
y = −4 x − 11 y = −4 x − 11
y = −8 y = −8
y = −4 x + 13 y = −4 x + 13
y = x + 5 y = x + 5
y = − 1 3 x − 14 3 y = − 1 3 x − 14 3
y = 7 x + 22 y = 7 x + 22
y = − 6 7 x + 4 7 y = − 6 7 x + 4 7
y = 1 5 x − 2 y = 1 5 x − 2
x = 4 x = 4
x = −2 x = −2
y = −3 y = −3
y = 4 x y = 4 x
y = 1 2 x + 3 2 y = 1 2 x + 3 2
y = 5 y = 5
y = 3 x − 1 y = 3 x − 1
y = −3 x + 3 y = −3 x + 3
y = 2 x − 6 y = 2 x − 6
y = − 2 3 x + 5 y = − 2 3 x + 5
x = −3 x = −3
y = −4 y = −4
y = x y = x
y = − 3 4 x − 1 4 y = − 3 4 x − 1 4
y = 5 4 x y = 5 4 x
y = 1 y = 1
y = x + 2 y = x + 2
y = 3 4 x y = 3 4 x
y = 1.2 x + 5.2 y = 1.2 x + 5.2
Section 4.7 Exercises
ⓐ yes ⓑ no ⓒ no ⓓ yes ⓔ no
ⓐ yes ⓑ no ⓒ no ⓓ yes ⓔ yes
ⓐ no ⓑ no ⓒ no ⓓ yes ⓔ yes
y < 2 x − 4 y < 2 x − 4
y ≤ − 1 3 x − 2 y ≤ − 1 3 x − 2
x + y ≥ 3 x + y ≥ 3
x + 2 y ≥ −2 x + 2 y ≥ −2
2 x − y < 4 2 x − y < 4
4 x − 3 y > 12 4 x − 3 y > 12
- ⓑ Answers will vary.
Review Exercises
ⓐ ( 2 , 0 ) ( 2 , 0 ) ⓑ ( 0 , −5 ) ( 0 , −5 ) ⓒ ( −4.0 ) ( −4.0 ) ⓓ ( 0 , 3 ) ( 0 , 3 )
0 | 3 | |
4 | 1 | (4, 1) |
4 |
0 | ||
2 | 0 | |
ⓐ yes; yes ⓑ yes; no
( 6 , 0 ) , ( 0 , 4 ) ( 6 , 0 ) , ( 0 , 4 )
− 1 2 − 1 2
slope m = − 2 3 m = − 2 3 and y -intercept ( 0 , 4 ) ( 0 , 4 )
5 3 ; ( 0 , −6 ) 5 3 ; ( 0 , −6 )
4 5 ; ( 0 , − 8 5 ) 4 5 ; ( 0 , − 8 5 )
plotting points
ⓐ −$250 ⓑ $450 ⓒ The slope, 35, means that Marjorie’s weekly profit, P , increases by $35 for each additional student lesson she teaches. The P –intercept means that when the number of lessons is 0, Marjorie loses $250. ⓓ
y = −5 x − 3 y = −5 x − 3
y = −2 x y = −2 x
y = −3 x + 5 y = −3 x + 5
y = 3 5 x y = 3 5 x
y = −2 x − 5 y = −2 x − 5
y = 1 2 x − 5 2 y = 1 2 x − 5 2
y = − 2 5 x + 8 y = − 2 5 x + 8
y = 3 y = 3
y = − 3 2 x − 6 y = − 3 2 x − 6
ⓐ yes ⓑ no ⓒ yes ⓓ yes ⓔ no
y > 2 3 x − 3 y > 2 3 x − 3
x − 2 y ≥ 6 x − 2 y ≥ 6
Practice Test
ⓐ yes ⓑ yes ⓒ no
( 3 , 0 ) , ( 0 , −4 ) ( 3 , 0 ) , ( 0 , −4 )
y = − 3 4 x − 2 y = − 3 4 x − 2
y = 1 2 x − 4 y = 1 2 x − 4
y = − 4 5 x − 5 y = − 4 5 x − 5
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- Book title: Elementary Algebra
- Publication date: Feb 22, 2017
- Location: Houston, Texas
- Book URL: https://openstax.org/books/elementary-algebra/pages/1-introduction
- Section URL: https://openstax.org/books/elementary-algebra/pages/chapter-4
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Eureka Math Grade 5 Module 4 Lesson 7 Answer Key
Engage ny eureka math 5th grade module 4 lesson 7 answer key, eureka math grade 5 module 4 lesson 7 problem set answer key.
Solve using a tape diagram. a. \(\frac{1}{3}\) of 18 Answer: 6
b. \(\frac{1}{3}\) of 36 Answer: 11
c. \(\frac{3}{4}\) × 24 Answer: 18
d. \(\frac{3}{8}\) × 24 Answer: 9
e. \(\frac{4}{5}\) × 25 Answer: 20
f. \(\frac{1}{7}\) × 140 Answer: 20
g. \(\frac{1}{4}\) × 9 Answer: 2 1/4
h. \(\frac{2}{5}\) × 12
i. \(\frac{2}{3}\) of a number is 10. What’s the number?
j. \(\frac{3}{4}\) of a number is 24. What’s the number?
Question 2. Solve using tape diagrams. a. There are 48 students going on a field trip. One-fourth are girls. How many boys are going on the trip? Answer: 36 boys are going on the field trip.
Explanation: As 48 students are going on a field trip in this \(\frac{1}{4}\) are girls. Hence 36 boys are going on the field trip. \(\frac{1}{4}\) × 48 = \(\frac{48}{4}\) = 12 12 × 3 = 36
Explanation: The smallest angle given is \(\frac{3}{8}\) and the largest angle is 160°. 160 × \(\frac{3}{8}\) = \(\frac{480}{8}\) = 60 160 + 60 = 220 360-220= 140 Hence the value of angle a is 140°
c. Abbie spent \(\frac{5}{8}\) of her money and saved the rest. If she spent $45, how much money did she have at first? Answer: Abbie had $72.
Explanation: 45 ÷ \(\frac{5}{8}\) = 45 × \(\frac{8}{5}\) = \(\frac{360}{5}\) =72
d. Mrs. Harrison used 16 ounces of dark chocolate while baking. She used \(\frac{2}{5}\) of the chocolate to make some frosting and used the rest to make brownies. How much more chocolate did Mrs. Harrison use in the brownies than in the frosting? Answer: Mrs. Harrison use 3 × \(\frac{1}{5}\) chocolate for the brownie
Explanation: Harrison used 16 ounces of dark chocolate. She used \(\frac{2}{5}\) of the chocolate to make some frosting and the remaining to make brownies. 16 × \(\frac{1}{5}\) = \(\frac{16}{5}\) = 3 × \(\frac{1}{5}\) Hence, Mrs. Harrison use 3 × \(\frac{1}{5}\) chocolate for the brownie.
Eureka Math Grade 5 Module 4 Lesson 7 Exit Ticket Answer Key
Solve using a tape diagram.
a. \(\frac{3}{5}\) of 30
b. \(\frac{3}{5}\) of a number is 30. What’s the number?
c. Mrs. Johnson baked 2 dozen cookies. Two-thirds of the cookies were oatmeal. How many oatmeal cookies did Mrs. Johnson bake?
Eureka Math Grade 5 Module 4 Lesson 7 Homework Answer Key
Question 1. Solve using a tape diagram. a. \(\frac{1}{4}\) of 24
b. \(\frac{1}{4}\) of 48
c. \(\frac{2}{3}\) × 18
d. \(\frac{2}{6}\) × 18
e. \(\frac{3}{7}\) × 49
f. \(\frac{3}{10}\) × 120
g. \(\frac{1}{3}\) × 31
h. \(\frac{2}{5}\) × 20
i. \(\frac{1}{4}\) × 25
j. \(\frac{3}{4}\) × 25
k. \(\frac{3}{4}\) of a number is 27. What’s the number?
i. \(\frac{2}{5}\) of a number is 14. What’s the number?
Question 2. Solve using tape diagrams. a. A skating rink sold 66 tickets. Of these, \(\frac{2}{3}\) were children’s tickets, and the rest were adult tickets. What total number of adult tickets were sold?
c. Annabel and Eric made 17 ounces of pizza dough. They used \(\frac{5}{8}\) of the dough to make a pizza and used the rest to make calzones. What is the difference between the amount of dough they used to make pizza and the amount of dough they used to make calzones?
d. The New York Rangers hockey team won \(\frac{3}{4}\) of their games last season. If they lost 21 games, how many games did they play in the entire season?
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Eureka Math Grade 4 Module 7 Lesson 14 Problem Set Answer Key Use RDW to solve the following problems. Question 1.
Eureka Math Grade 4 Module 7 Answer Key | Engage NY Math 4th Grade Module 7 Answer Key March 27, 2021 / By Sachin
3, 6, 9, 12, 15, 18, 21, 24, 27, 30; multiply the number of yards times 3
Homework Perpendicular lines accurately traced. Answers will vary. Pe ma ; b. No right angles. cu ately identified an DO ⊥ OG gh e. No right angles. Module 4: � � ON ON ; ⊥ NM NM g. No right angles. h. Right angles accurately identified and marked; UT ⊥ TZ TZ ⊥ ZY ZY ⊥ YX YX ⊥ XW WV ⊥ ; VU ; ; Right angles accurately ; identified
Lesson 14 Homework 4 Name Date Use RDW to solve the following problems. 1. Molly baked a pie for 1 hour and 45 minutes. Then, she baked banana bread for 35 minutes less than the
Eureka Math Grade 4 Module 7 Lesson 12 Problem Set Answer Key. Question 1. Draw a tape diagram to show 1 yard divided into 3 equal parts. Question 2. Draw a tape diagram to show 2 23 yards = 8 feet. Question 3. Draw a tape diagram to show 3 4 gallon = 3 quarts. Question 4.
Homework a. - f. Figure drawn accurately g. Answers will vary. a. - g. Figure drawn accurately h. Answers will vary. a. Points labeled; labels will vary.
Lesson 4.7. Practice Level C. 1.1. x 5 9, y 5 11 2. x 5 6, y 5 13 3. x 5 3.5, y 5 9 4. x 5 12, y 5 5 5. x 5 6, y 5 7. 6. x 5 3, y 5 9.5 7. x 5 20.5, y 5 6 8. x 5 8, y 5 3 9. x 5 7, cannot determine y; could find y if it was given that 9y 2 10 is equal to 5y 2 8.
A STORY OF UNITS. n. Homework 4•74. Use the information in the chart about Melissa's school supplies to answer the f. ll. wing questions: a. On Wednesdays, Melissa packs only two notebooks and a bin. r into her backpack. How much does her full backpac. Textbook. o. Laptop5 lb 12 ozb. On Thursdays, Melissa puts her laptop, supply case, two ...
Use the RDW process to solve. Write your answer as a decimal. r bill, 2 dimes, and 7 pennies. John has 2 dollar ls, 3 quarters, and 9 pennies. How much m dollars 13 cents to buy a book. In her wallet, she fin s 3 dollar 14 pennies. How much more money does Suilin need to buy the book? essa has 6 dimes and 2 pennies. Joachim has
Division Drying To find the answer to the riddle, complete each division. Then use the KEY to find the answer to the riddle. 78 6 1. 58 3
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Signing Naturally Unit 4 Lesson 7 Learn with flashcards, games, and more — for free.
Engage NY Eureka Math 4th Grade Module 7 Lesson 7 Answer Key Eureka Math Grade 4 Module 7 Lesson 7 Problem Set Answer Key Question 1. Determine the following sums and differences. Show your work.
Find Math, English language arts (ELA) resources to practice & prepare lesson plans online with pdf, answer key, videos, apps, and worksheets for grades 3-8 on Lumos Learning.
This free textbook is an OpenStax resource written to increase student access to high-quality, peer-reviewed learning materials.
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. OM Lesson 14 Homework 4•7 4. The candy shop puts 10 ounces of gummy bears in ach box. How many box do be 5. Mom can make 10 brownies from a 12 ounce package. make 50 bro
Engage NY Eureka Math 4th Grade Module 7 Lesson 13 Answer Key Eureka Math Grade 4 Module 7 Lesson 13 Problem Set Answer Key Question 1. Solve. a. pound = _______ ounce b. pound =
Homework a. Strip diagram drawn and labeled b. Strip diagram drawn and labeled a. 10 × 2 fifths = 20 fifths b. 3 × 5 sixths = 15 sixths c. 9 × 4 ninths = 36 ninths d. 7 × 3 fourths = 21 fourths Module 4: a. 28 9 b. 18 5 c. 24 4 d. 48 8 e. 84 10 f. 162 100
Engage NY Eureka Math 4th Grade Module 7 Lesson 8 Answer Key Eureka Math Grade 4 Module 7 Lesson 8 Problem Set Answer Key Question 1. Determine the following sums and differences. Show your work.
TEKS EDITION Special thanks go to the Gordon A. Cain Center and to the Department of Mathematics at Louisiana State University for their support in the development of Eureka Math .
Eureka Math Grade 5 Module 4 Lesson 7 Homework Answer Key. Question 1. Solve using a tape diagram. b. 14 of 48. c. 23 × 18. d. 26 × 18. e. 37 × 49.