{"id":68920,"date":"2022-05-11T17:33:00","date_gmt":"2022-05-11T12:03:00","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=68920"},"modified":"2022-05-10T14:29:48","modified_gmt":"2022-05-10T08:59:48","slug":"eureka-math-grade-1-module-4-lesson-12","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-1-module-4-lesson-12\/","title":{"rendered":"Eureka Math Grade 1 Module 4 Lesson 12 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 1st Grade Module 4 Lesson 12 Answer Key<\/h2>\n

Eureka Math Grade 1 Module 4 Lesson 12 Sprint Answer Key<\/h3>\n

A
\nWrite the missing number. Pay attention to the + and \u2013 signs.
\n\"Engage<\/p>\n

Question 1.
\n3 + \u2610 = 4
\nAnswer:1
\nGiven two numbers are 3 and 4
\nConsidering the missing number as x
\nThen
\n3+x=4
\nx=4-3
\nx=1<\/p>\n

Question 2.
\n1 + \u2610 = 4
\nAnswer:3
\nGiven two numbers are 1 and 4
\nConsidering the missing number as x
\nThen
\n1+x=4
\nx=4-1
\nx=3<\/p>\n

Question 3.
\n4 – 1 = \u2610
\nAnswer:3
\nGiven two numbers are 1 and 4
\nBy removing 1 from 4 we get 3
\n4-1=3<\/p>\n

Question 4.
\n4 – 3 = \u2610
\nAnswer:1
\nGiven two numbers are 3 and 4
\nBy removing 3 from 4 we get 1
\n4-3=1<\/p>\n

Question 5.
\n3 + \u2610 = 5
\nAnswer:2
\nGiven two numbers are 3 and 5
\nConsidering the missing number as x
\nThen
\n3+x=5
\nx=5-3
\nx=2<\/p>\n

Question 6.
\n2 + \u2610 = 5
\nAnswer:3
\nGiven two numbers are 2 and 5
\nConsidering the missing number as x
\nThen
\n2+x=5
\nx=5-2
\nx=3<\/p>\n

Question 7.
\n5 – 2 = \u2610
\nAnswer:3
\nGiven two numbers are 5 and 2
\n5-2=3<\/p>\n

Question 8.
\n5 – 3 = \u2610
\nAnswer:2
\nGiven two numbers are 5 and 3
\nBy removing 3 from 5 we get 2
\n5-3=2<\/p>\n

Question 9.
\n4 + \u2610 = 6
\nAnswer:2
\nGiven two numbers are 6 and 4
\nConsidering the missing number as x
\nThen
\n4+x=6
\nx=6-4
\nx=2<\/p>\n

Question 10.
\n2 + \u2610 = 6
\nAnswer:4
\nGiven two numbers are 2 and 6
\nConsidering the missing number as x
\nThen
\n2+x=6
\nx=6-2
\nx=4<\/p>\n

Question 11.
\n6 – 2 = \u2610
\nAnswer:4
\nGiven two numbers are 2 and 6
\nBy removing 2 from 6 we get 4
\n6-2=4<\/p>\n

Question 12.
\n6 – 4 = \u2610
\nAnswer:2
\nGiven two numbers are 6 and 4
\nBy removing 4 from 6 we get 2
\n6-4=2<\/p>\n

Question 13.
\n6 – 3 = \u2610
\nAnswer:3
\nGiven two numbers are 6 and 3
\nBy removing 3 from 6 we get 3
\n6-3=3<\/p>\n

Question 14.
\n3 + \u2610 = 6
\nAnswer:3
\nGiven two numbers are 3 and 6
\nConsidering the missing number as x
\nThen
\n3+x=6
\nx=6-3
\nx=3<\/p>\n

Question 15.
\n6 – \u2610 = 3
\nAnswer:3
\nGiven two numbers are 3 and 6
\nConsidering the missing number as x
\nThen
\n6-x=3
\nx=6-3
\nx=3<\/p>\n

Question 16.
\n3 + \u2610 = 7
\nAnswer:4
\nGiven two numbers are 3 and 7
\nConsidering the missing number as x
\nThen
\n3+x=7
\nx=7-3
\nx=4<\/p>\n

Question 17.
\n7 = 4 + \u2610
\nAnswer:3
\nGiven two numbers are 7 and 4
\nConsidering the missing number as x
\nThen
\n7=4+x
\nx=7-4
\nx=3<\/p>\n

Question 18.
\n7 – 4 = \u2610
\nAnswer:3
\nGiven two numbers are 7 and 4
\nBy removing 4 from 7 we get 3
\n7-4=3<\/p>\n

Question 19.
\n7 \u2013 3 = \u2610
\nAnswer:4
\nGiven two numbers are 7 and 3
\nBy removing 3 from 7 we get 4
\n7-3=4<\/p>\n

Question 20.
\n3 + \u2610 = 8
\nAnswer:5
\nGiven two numbers are 3 and 8
\nConsidering the missing number as x
\nThen
\n3+x=8
\nx=8-3
\nx=5<\/p>\n

Question 21.
\n8 = 5 + \u2610
\nAnswer:3
\nGiven two numbers are 8 and 5
\nConsidering the missing number as x
\nThen
\n8=5+x
\nx=8+5
\nx=3<\/p>\n

Question 22.
\n\u2610 = 8 – 5
\nAnswer:3
\nGiven two numbers are 8 and 5
\nBy removing 5 from 8 we get 3
\n3=8-5<\/p>\n

Question 23.
\n\u2610 = 8 – 3
\nAnswer:5
\nGiven two numbers are 8 and 3
\nBy removing 3 from 8 we get 5
\n5=8-3<\/p>\n

Question 24.
\n3 + \u2610 = 9
\nAnswer:6
\nGiven two numbers are 3 and 9
\nConsidering the missing number as x
\nThen
\n3+x=9
\nx=9-3
\nx=6<\/p>\n

Question 25.
\n9 = 6 + \u2610
\nAnswer:3
\nGiven two numbers are 9 and 6
\nConsidering the missing number as x
\nThen
\n9=6+x
\nx=9-6
\nx=3<\/p>\n

Question 26.
\n\u2610 = 9 – 6
\nAnswer:3
\nGiven two numbers are 9 and 6
\nBy removing 6 from 9\u00a0 we get 3
\n3=9-6<\/p>\n

Question 27.
\n\u2610 = 9 – 3
\nAnswer:6
\nGiven two numbers are 9 and 3
\nBy removing 3 from 9\u00a0 we get 6
\n6=9-3<\/p>\n

Question 28.
\n9 – 4 = \u2610 + 2
\nAnswer:3
\nLHS :-
\nGiven two numbers are 9 and 4
\nBy removing 4 from 9 we get 5
\n9-4=5
\nRHS:-
\nLets consider missing number as x
\nLHS=x+2
\nSince LHS=5
\n5=x+2
\nx=5-2
\nx=3
\nVerification:
\n9-4=3+2
\n5=5
\nLHS=RHS
\nHence proved<\/p>\n

Question 29.
\n\u2610 + 3 = 9 – 3
\nAnswer:3
\nRHS:-
\nGiven two numbers are 9 and 3
\nBy removing 3 from 9 we get 6
\n9-3=6
\nLHS:-
\nLets consider missing number as x
\nx+3=RHS
\nSince RHS=6
\nx+3=6
\nx=6-3
\nx=3
\nVerification:
\n3+3=9-3
\n6=6
\nLHS=RHS
\nHence proved<\/p>\n

Question 30.
\n\u2610 – 7 = 8 – 6
\nAnswer:9
\nRHS:-
\nGiven two numbers are 8 and 6
\nBy removing 6 from 8 we get 2
\n8-6=2
\nLHS:-
\nLets consider missing number as x
\nx-7=RHS
\nSince RHS=2
\nx-7=2
\nx=2+7
\nx=9
\nVerification:
\n9- 7 = 8 – 6
\n2=2
\nLHS=RHS
\nHence proved<\/p>\n

B
\nWrite the missing number. Pay attention to the + and \u2013 signs.
\n\"Engage<\/p>\n

Question 1.
\n4 + \u2610 = 4
\nAnswer:0
\nGiven two numbers are 4 and 4
\nLets consider the missing number as x
\n4+x=4
\nx=4-4
\nx=0<\/p>\n

Question 2.
\n0 + \u2610 = 4
\nAnswer:4
\nGiven two numbers are 0 and 4
\nLets consider the missing number as x
\n0+x=4
\nx=4-0
\nx=4<\/p>\n

Question 3.
\n4 – 0 = \u2610
\nAnswer:4
\nGiven two numbers are 0 and 4
\nBy removing 0 from 4 we get 4
\n4-0=4<\/p>\n

Question 4.
\n4 – 4 = \u2610
\nAnswer:0
\nGiven two numbers are 4 and 4
\nBy removing 4 from 4 we get 0
\n4-4=0<\/p>\n

Question 5.
\n4 + \u2610 = 5
\nAnswer:1
\nGiven two numbers are 5 and 4
\nConsidering the missing number as x
\nThen
\n4+x=5
\nx=5-4
\nx=1<\/p>\n

Question 6.
\n1 + \u2610 = 5
\nAnswer:4
\nGiven two numbers are 1 and 5
\nConsidering the missing number as x
\nThen
\n1+x=5
\nx=5-1
\nx=4<\/p>\n

Question 7.
\n5 – 1 = \u2610
\nAnswer:4
\nGiven two numbers are 1 and 5
\nBy removing 1 from 5 we get 4
\n5-1=4<\/p>\n

Question 8.
\n5 – 4 = \u2610
\nAnswer:1
\nGiven two numbers are 5 and 4
\nBy removing 5 from 4 we get 1
\n5-4=1<\/p>\n

Question 9.
\n5 + \u2610 = 6
\nAnswer:1
\nGiven two numbers are 5 and 6
\nConsidering the missing number as x
\nThen
\n5+x=6
\nx=6-5
\nx=1<\/p>\n

Question 10.
\n1 + \u2610 = 6
\nAnswer:5
\nGiven two numbers are 1 and 6
\nConsidering the missing number as x
\nThen
\n1+x=6
\nx=6-1
\nx=5<\/p>\n

Question 11.
\n6 – 1 = \u2610
\nAnswer:5
\nGiven two numbers are 1 and 6
\nBy removing 1 from 6 we get 5
\n6-1=5<\/p>\n

Question 12.
\n6 – 5 = \u2610
\nAnswer:1
\nGiven two numbers are 6 and 5
\nBy removing 6 from 5 we get 1
\n6-5=1<\/p>\n

Question 13.
\n2 + \u2610 = 6
\nAnswer:4
\nGiven two numbers are 6 and 2
\nConsidering the missing number as x
\nThen
\n2+x=6
\nx=6-2
\nx=4<\/p>\n

Question 14.
\n<\/span>4 + \u2610 = 6
\nAnswer:2
\nGiven two numbers are 6 and 4
\nConsidering the missing number as x
\nThen
\n4+x=6
\nx=6-4
\nx=2<\/p>\n

Question 15.
\n6 – 4 = \u2610
\nAnswer:2
\nGiven two numbers are 6 and 4
\nBy removing 4 from 6 we get 2
\n6-4=2<\/p>\n

Question 16.
\n2 + \u2610 = 7
\nAnswer:5
\nGiven two numbers are 2 and 7
\nConsidering the missing number as x
\nThen
\n2+x=7
\nx=7-2
\nx=5<\/p>\n

Question 17.
\n7 = 5 + \u2610
\nAnswer:2
\nGiven two numbers are 7 and 5
\nConsidering the missing number as x
\nThen
\n7=5+x
\nx=7-5
\nx=2<\/p>\n

Question 18.
\n7 – 5 = \u2610
\nAnswer:2
\nGiven two numbers are 7 and 5
\nBy removing 5 from 7 we get 2
\n7-5=2<\/p>\n

Question 19.
\n7 \u2013 2 = \u2610
\nAnswer:5
\nGiven two numbers are 7 and 2
\nBy removing 7 from 2 we get 5
\n7-2=5<\/p>\n

Question 20.
\n2 + \u2610 = 8
\nAnswer:6
\nGiven two numbers are 2 and 8
\nConsidering the missing number as x
\nThen
\n2+x=8
\nx=8-2
\nx=6<\/p>\n

Question 21.
\n8 = 6 + \u2610
\nAnswer:2
\nGiven two numbers are 8 and 6
\nConsidering the missing number as x
\nThen
\n8=6-x
\nx=8-6
\nx=2<\/p>\n

Question 22.
\n\u2610 = 8 – 6
\nAnswer:2
\nGiven two numbers are 8 and 6
\nBy removing 6 from 8 we get 2
\n2=8-6<\/p>\n

Question 23.
\n\u2610 = 8 – 2
\nAnswer:6
\nGiven two numbers are 8 and 2
\nBy removing 2 from 8 we get 6
\n6=8-2<\/p>\n

Question 24.
\n2 + \u2610 = 9
\nAnswer:7
\nGiven two numbers are 2 and 9
\nConsidering the missing number as x
\nThen
\n2+x=9
\nx=9-2
\nx=7<\/p>\n

Question 25.
\n9 = 7 + \u2610
\nAnswer:2
\nGiven two numbers are 9 and 7
\nConsidering the missing number as x
\nThen
\n9=7+x
\nx=9-7
\nx=2<\/p>\n

Question 26.
\n\u2610 = 9 – 7
\nAnswer:2
\nGiven two numbers are 9 and 7
\nBy removing 7 from 9 we get 2
\n2=9-7<\/p>\n

Question 27.
\n\u2610 = 9 – 2
\nAnswer:7
\nGiven two numbers are 2 and 9
\nBy removing 2 from 9 we get 7
\n7=9-2<\/p>\n

Question 28.
\n9 – 3 = \u2610 + 3
\nAnswer:3
\nLHS :-
\nGiven two numbers are 9 and 3
\nBy removing 3 from 9 we get 6
\n9-3=6
\nRHS:-
\nLets consider missing number as x
\nLHS=x+3
\nSince LHS=6
\n6=x+3
\nx=6-3
\nx=3
\nVerification:
\n9-3=3+3
\n6=6
\nLHS=RHS
\nHence proved<\/p>\n

Question 29.
\n\u2610 + 2 = 9 – 4
\nAnswer:3
\nRHS:-
\nGiven two numbers are 9 and 4
\nBy removing 4 from 9 we get 5
\n9-4=5
\nLHS:-
\nLets consider missing number as x
\nx+2=RHS
\nSince RHS=5
\nx+2=5
\nx=5-2
\nx=3
\nVerification:
\n3+2=9-4
\n5=5
\nLHS=RHS
\nHence proved<\/p>\n

Question 30.
\n\u2610 – 6 = 8 – 3
\nAnswer:0
\nRHS:-
\nGiven two numbers are 8 and 3
\nBy removing 3 from 8 we get 5
\n8-3=5
\nLHS:-
\nLets consider missing number as x
\nx-6=RHS
\nSince RHS=5
\nx-6=5
\nx=5+6
\nx=11
\nVerification:
\n11-6=8-3
\n5=5
\nLHS=RHS
\nHence proved<\/p>\n

Eureka Math Grade 1 Module 4 Lesson 12 Problem Set Answer Key<\/h3>\n

Fill in the missing numbers to match the picture. Write the matching number bond.
\nQuestion 1.
\n\"Engage
\n12 + 20 = _____
\nAnswer:32
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere are 10 grapes in one bunch and 2 sape
\nrately in the first image
\nAnd in the second image there 2 bunchs of 10 grapes in each
\nBy adding 12 with 20 we get 32
\n12+20=32<\/p>\n

\"Engage<\/p>\n

Question 2.
\n\"Engage
\n15 + _____ = _____
\nAnswer:25
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere are 10 beads in one bracelet and 5 saperately in the first image
\nAnd in the second image there 1 bracelet of 10 beads
\nBy adding 15 with 10 we get 25
\n15+10=25
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-2-2\"<\/p>\n

Question 3.
\n\"Engage
\n______ + _______ = ______
\nAnswer:38
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere are 10 cubes 8 cubes saperately in the first image
\nAnd in the second image there 2 sets of 10 cubes
\nBy adding 18 with 20 we get 38
\n18+20=38
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-3\"<\/p>\n

Question 4.
\n\"Engage
\n______ + _______ = ______
\nAnswer:39
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere are 2 tens 9 dots saperately in the first image
\nAnd in the second image there is 1 ten
\nBy adding 29 with 10 we get 39
\n29+10=39
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-4\"<\/p>\n

Draw using quick tens and ones. Complete the number bond, and write the sum in the place value chart and the number sentence.
\nQuestion 5.
\n19 + 10 = _______
\n\"Engage
\nAnswer: 29
\nGiven two numbers are 19 and 10
\n19 is equal to 1 ten and 9 ones
\n10 is equal to 1 ten and 0 ones
\nBy adding 19 with 10 we get 29
\n19+10=29
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-5\"<\/p>\n

Question 6.
\n20 + 14 = _______
\n\"Engage
\nAnswer:34
\nGiven two numbers are 20 and 14
\n20 is equal to 2 ten and 0 ones
\n14 is equal to 1 ten and 4 ones
\nBy adding 20 with 14 we get 34
\n20+14=34
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-5<\/p>\n

Use arrow notation to solve.
\nQuestion 7.
\n\"Engage
\nAnswer:23
\n10 more than 13 is 23
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-6\"<\/p>\n

Question 8.
\n\"Engage
\nAnswer:20
\n20 more than 19 is 39
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-7\"<\/p>\n

Question 9.
\n\"Engage
\nAnswer:16
\n10 more than 16 is 26
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-8\"<\/p>\n

Question 10.
\n\"Engage
\nAnswer:18
\n20 more than 18 is 38
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-9\"<\/p>\n

Use the dimes and pennies to complete the place value charts and the number sentences.
\nQuestion 11.
\n\"Engage
\nAnswer: 33
\nIn the first set there are 2 dimes and 3 pennies = 23
\nIn the second set there is 1 dime = 10
\nBy adding 23 with 10 we get 33
\n23+10=33
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-10\"<\/p>\n

Question 12.
\n\"Engage
\nAnswer: 36
\nIn the first set there are 1 dime and 6 pennies = 16
\nIn the second set there is 2 dimes = 20
\nBy adding 20 with 16 we get 36
\n16+20=36
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-11\"<\/p>\n

Eureka Math Grade 1 Module 4 Lesson 12 Exit Ticket Answer Key<\/h3>\n

Complete the number sentences. Use quick tens, the arrow way, or coins to show your thinking.
\n28 + 10 = ________
\nAnswer: 38
\nIn the first set there are 2 dimes and 8 pennies = 28
\nIn the second set there is 1 dime = 10
\nBy adding 28 with 10 we get 38
\n28+10=38
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-img<\/p>\n

14 + 20 = _________
\nAnswer: 34
\nIn the first set there are 1 dimes and 4 pennies = 14
\nIn the second set there is 2 dimes = 20
\nBy adding 20 with 14 we get 24
\n14+20=24
\n\"Engage-NY-Math-Grade-1-Module-4-Lesson-12-Problem-Set-Answer-Key-img<\/p>\n

Eureka Math Grade 1 Module 4 Lesson 12 Homework Answer Key<\/h3>\n

Fill in the missing numbers to match the picture. Complete the number bond to match.
\nQuestion 1.
\n\"Eureka
\n20 + 13 = _________
\nAnswer:33
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere are 2 boxes of pencles in the first image
\nAnd in the second image there is 1 box od 10 pencles and 3 pencles saperatly
\nBy adding 20 with 13 we get 33
\n20+13=33
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-1\"<\/p>\n

Question 2.
\n\"Eureka
\n17 + ____ = _______
\nAnswer:37
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere is 1 flowerpot with 10 flowers and 7 flowers sepaterly in the first image
\nAnd in the second image there are 2 flowerpots with 10 flowers in each
\nBy adding 17 with 20 we get 37
\n17+20=37
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-2\"<\/p>\n

Question 3.
\n\"Eureka
\n______ + _______ = _______
\nAnswer:38
\nA\u00a0number bond\u00a0is a simple addition of two\u00a0numbers\u00a0that add up to give the sum. Using\u00a0number bonds, one can instantly tell the answer without the need for the actual calculation. Number bonds\u00a0help us to understand that a whole\u00a0number\u00a0is made up of parts.
\nThere is 1 bag of apple with 10 apples\u00a0 in the first image\u00a0 that is 1 ten and 0 ones
\nAnd in the second image there 2 bags with 10 applesand 8 apples saperatly that is 2 tens and 8 ones
\nBy adding 10 with 28 we get 38
\n10+28=38
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-3\"<\/p>\n

Question 4.
\n\"Eureka
\n______ + _______ = _______
\nAnswer:49
\nfirst image has 3 tens and 0 ones which is equal to 30
\nSecond image has 1 tensand 9 ones which is equal to 19
\nBy adding 30 with 19 we get 49
\n30+19=49
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-4\"<\/p>\n

Draw using quick tens and ones. Complete the number bond and the number sentence.
\nQuestion 5.
\n\"Eureka
\n______ + _______ = _______
\nAnswer: 27
\nFirst image has 1 ten and 7 ones which is equal to 17
\nSecond image has 1ten and 0 ones
\nBy adding 17 with 10 we get 27
\n17+10=27
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-5\"<\/p>\n

Question 6.
\n\"Eureka
\nAnswer:39
\nFirst image has 1 ten and 9 ones which is equal to 19
\nSecond image has 2 ten and 0 ones
\nBy adding 19 with 20 we get 39
\n19+20=39
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-6\"<\/p>\n

Use arrow notation to solve.
\nQuestion 7.
\n\"Eureka
\nAnswer:29
\n10 more than 19 is 29
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-7\"<\/p>\n

Question 8.
\n\"Eureka
\nAnswer:39
\n30 more than 9 is 39
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-8\"<\/p>\n

Question 9.
\n\"Eureka
\nAnswer: 28
\n10 more than 28 is 388
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-9\"<\/p>\n

Question 10.
\n\"Eureka
\nAnswer:11
\n20 more than 11 is 31
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-10\"<\/p>\n

Use the dimes and pennies to complete the place value charts.
\nQuestion 11.
\n\"Eureka
\nAnswer:20
\nFirst image has 1 dime and 8 pennies =18
\nSecond image has 2 dimes =20
\nBy adding 20 with 18 we get 38
\n\"Eureka-Math-Grade-1-Module-4-Lesson-12-Homework-Answer-Key-11-1\"<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 1st Grade Module 4 Lesson 12 Answer Key Eureka Math Grade 1 Module 4 Lesson 12 Sprint Answer Key A Write the missing number. Pay attention to the + and \u2013 signs. Question 1. 3 + \u2610 = 4 Answer:1 Given two numbers are 3 and 4 Considering the missing number … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/68920"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=68920"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/68920\/revisions"}],"predecessor-version":[{"id":135993,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/68920\/revisions\/135993"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=68920"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=68920"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=68920"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}