{"id":59657,"date":"2022-06-04T14:23:58","date_gmt":"2022-06-04T08:53:58","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=59657"},"modified":"2022-05-10T14:18:55","modified_gmt":"2022-05-10T08:48:55","slug":"eureka-math-grade-4-module-3-lesson-9","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-4-module-3-lesson-9\/","title":{"rendered":"Eureka Math Grade 4 Module 3 Lesson 9 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 9 Answer Key<\/h2>\n

Eureka Math Grade 4 Module 3 Lesson 9 Problem Set Answer Key<\/h3>\n

Question 1.
\nSolve using each method.
\n\"Eureka
\nAnswer:
\n\"Eureka
\nExplanation:
\nGiven expression as 4 X 34 =
\nPartial Products
\n34
\nX 4
\n<\/span>\u00a0 16
\n+120
\n136 ,
\n<\/span>Here we first write multiplication of 4 X 4 ones, then
\n4 X 3 tens then add as shown above 16 + 120 = 136,
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n34
\nX 4
\n136 ,<\/span>here 4 X 4 ones= 16 ones we write 6 at ones place and take 1 at tens place then 4 x 3 at tens place = 12 tens
\nadd 1 ten to 12 tens we get 13 tens so 4 X 34 = 136.<\/p>\n

\"Eureka
\nAnswer:
\n\"Eureka
\nExplanation:
\nGiven expression as 3 X 224 =
\nPartial Products
\n224
\nX 3
\n<\/span>\u00a0 12
\n60
\n+600
\n672 , <\/span>Here we first write multiplication of 3 X 4 ones, then
\n3 X 2 tens and 3 X 2 hundreds then add as shown above
\n12 + 60 + 600 = 672,
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n224
\nX 3
\n672 ,<\/span>here 3 X 4 ones = 12 ones we write 2 at ones place and
\ntake 1 at tens place then 3 x 2 tens = 6 tens add 1 ten to 6 tens
\nwe get 7 tens and 3 X 2 hundreds = 6 hundreds,\u00a0 so 4 X 244 = 672.<\/p>\n

Question 2.
\nSolve. Use the standard algorithm.
\na.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n\u00a0 \u00a07\u00a0 \u00a0 \u00a05\u00a0 \u00a0 \u00a03\u00a0 \u00a0 \u00a0\u00a0<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n251
\nX 3
\n753 ,<\/span>here 3 X 1 ones = 3 ones then 3 x 5 tens = 15 tens
\nwe write 5 at tens place and take 1 to hundred place and
\n3 X 2 hundreds = 6 hundreds, now adding 1 hundred to
\n6 hundreds we get 7 hundreds , so 3 X 251 = 753.<\/p>\n

b.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n\u00a0 8\u00a0 \u00a0 \u00a0 1\u00a0 \u00a0 \u00a0 0<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1, 3<\/span>
\n135
\nX 6
\n810 ,<\/span>here 6 X 5 ones = 30 ones, we write 0 at ones place and
\ntake 3 to tens places then 6 x 3 tens = 18 tens + 3 tens = 21 tens,
\nwe write 1 at tens place and take 2 to hundred place and
\n6 X 1 hundred = 6 hundreds, now adding 2 hundreds to
\n6 hundreds we get 8 hundreds , so 6 X 135 = 810.<\/p>\n

c.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n2,\u00a0 \u00a07\u00a0 \u00a03\u00a0 \u00a06<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n\u00a03<\/span>
\n304
\nX 9<\/span>
\n2,736<\/span> ,here 9 X 4 ones = 36 ones, we write 6 at ones place and
\ntake 3 to tens places then 9 x 0 ten = 0 ten + 3 tens = 3 tens,
\n9 X 3 hundreds = 27 hundreds, now we write 7 at hundreds place
\nand 2 at thousands place as shown above, So 9 X 304 = 2,736.\u00a0 \u00a0\u00a0
\n<\/span><\/p>\n

d.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n1,\u00a0 6\u00a0 \u00a0 2\u00a0 \u00a00<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n2<\/span>
\n405
\nX 4<\/span>
\n1,620<\/span> ,here 4 X 5 ones = 20 ones, we write 0 at ones place and
\ntake 2 to tens places then 4 x 0 ten = 0 ten + 2 tens = 2 tens,
\n4 X 4 hundreds = 16 hundreds, now we write 6 at hundreds place
\nand 1 at thousand place as shown above, So 4 X 405 = 1,620.<\/p>\n

e.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n1,\u00a0 5\u00a0 \u00a0 8\u00a0 \u00a00<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n3<\/span>
\n316
\nX 5<\/span>
\n1,580<\/span> ,here 5 X 6 ones = 30 ones, we write 0 at ones place and
\ntake 3 to tens places then 5 x 1 ten = 5 tens + 3 tens = 8 tens,
\n5 X 3 hundreds = 15 hundreds, now we write 5 at hundreds place
\nand 1 at thousand place as shown above, So 5 X 316 = 1,580.<\/p>\n

f.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n2,\u00a0 3\u00a0 \u00a05\u00a0 \u00a0 2<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n5,1<\/span>
\n392
\nX 6<\/span>
\n2,352<\/span> ,here 6 X 2 ones = 12 ones, we write 2 at ones place and
\ntake 1 to tens places then 6 x 9 tens = 54 tens + 1 tens = 55 tens,
\nwe write 5 at tens place and take another 5 to hundreds place,
\n6 X 3 hundreds = 18 hundreds,18 hundreds + 5 hundreds =
\n23 hundreds now we write 3 at hundreds place
\nand 2 at thousands place as shown above, So 6 X 392 = 2,352.<\/p>\n

Question 3.
\nThe product of 7 and 86 is ________.
\nAnswer:
\nThe product of 7 and 86 is 602,<\/p>\n

Explanation:
\nGive to find the product of 7 and 86 is
\n4<\/span>
\n86
\nX 7
\n602
\n<\/span>first we multiply 7 with 6 ones = 42 ones we write 2 at ones
\nplace and take 4 to tens place now 7 X 8 = 56 tens adding
\n4 tens we get 60 tens, So 7 X 86 = 602.<\/p>\n

Question 4.
\n9 times as many as 457 is _________.
\nAnswer:
\n9 times as many as 457 is 4,113,<\/p>\n

Explanation:
\nGiven to find 9 times as many as 457 is
\n6
\n457
\nX 9<\/span>
\n4,113<\/span> ,here 9 X 7 ones = 63 ones, we write 3 at ones place and
\ntake 6 to tens places then 9 x 5 tens = 45 tens + 6 tens = 51 tens,
\nwe write 1 at tens place and take 5 to hundreds place,
\n9 X 4 hundreds = 36 hundreds, 36 hundreds + 5 hundreds =
\n41 hundreds now we write 1 at hundred place
\nand 4 at thousands place as shown above, So 9 X 457 = 4,113.<\/p>\n

Question 5.
\nJashawn wants to make 5 airplane propellers. He needs 18 centimeters of wood for each propeller. How many centimeters of wood will he use?
\n\"Eureka
\nAnswer:
\nJashawn needs 90 centimeters of wood to make
\n5 airplane propellers,<\/p>\n

Explanation:
\nGiven Jashawn wants to make 5 airplane propellers.
\nHe needs 18 centimeters of wood for each propeller.
\nSo number of centimeters of wood will he use is 5 X 18 cms =
\n4<\/span>
\n18
\nX 5
\n90
\n<\/span>here first we multiply 5 X 8 ones = 40 ones, we write 0
\nat ones place and take 4 to tens place then 5 X 1 ten =
\n5 tens+ 4 tens = 9 tens, So 5 X 18 = 90 centimeters,
\ntherefore, Jashawn needs 90 centimeters of wood to make
\n5 airplane propellers.<\/p>\n

Question 6.
\nOne game system costs $238. How much will 4 game systems cost?
\nAnswer:
\n4 game systems will cost $952,<\/p>\n

Explanation:
\nGiven one game system costs $238. So for 4 game
\nsystems it will cost 4 X $238 =
\n\u00a01, 3<\/span>
\n$238
\nX\u00a0 4
\n952
\n<\/span>here first we multiply 4 X 8 ones = 32 ones,
\nwe write 2 at ones place and take 3 to tens places
\nthen 4 x 3 tens = 12 tens + 3 tens = 15 tens,
\nwe write 5 at tens place and take 1 to hundred place,
\n4 X 2 hundreds = 8 hundreds, now 8 hundreds + 1 hundred =
\n9 hundreds, So 5 X $238 = 952, therefore 4 game systems will cost $952.<\/p>\n

Question 7.
\nA small bag of chips weighs 48 grams. A large bag of chips weighs three times as much as the small bag. How much will 7 large bags of chips weigh?
\n\"Eureka
\nAnswer:
\n7 large bags of chips weigh 1,008 grams,<\/p>\n

Explanation:
\nGiven a\u00a0small bag of chips weighs 48 grams.
\nA large bag of chips weighs three times as much as
\nthe small bag means 3 X 48 grams=
\n2<\/span>
\n48
\nX 3
\n144
\n<\/span>here first we multiply 3 X 8 ones = 24 ones, we write 4
\nat ones place and take 2 to tens place then 3 X 4 ten =
\n12 tens+ 2 tens = 14 tens, we write 14 tens as 4 at tens place
\nand 1 at hundreds place, So 3 X 48 grams = 144 grams,
\nNow 7 large bags of chips weigh 7 X 144 grams =
\n\u00a03,2<\/span>
\n144
\nX 7
\n1,008
\n<\/span>here first we multiply 7 X 4 ones = 28 ones, we write 8
\nat ones place and take 2 to tens place then 7 X 4 ten =
\n28 tens+ 2 tens = 30 tens, we write 30 tens as 0 at tens place
\nand other 3 at hundreds place,now 7 X 1 hundred = 7 hundred,
\n7 hundred + 3 hundreds = 10 hundreds we write 0 at
\nhundreds place at 1 at thousands place, So 7 X 144 = 1,008 grams,
\ntherefore, 7 large bags of chips weigh 1,008 grams.
\n<\/span><\/p>\n

Eureka Math Grade 4 Module 3 Lesson 9 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nSolve using the standard algorithm.
\na.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n5,\u00a0 4\u00a0 \u00a0 \u00a07\u00a0 \u00a02
\nExplanation:
\nIn standard algorithm we add same time of multiplying as
\n7<\/span>
\n608
\nX 9<\/span>
\n5,472<\/span> ,here 9 X 8 ones = 72 ones, we write 2 at ones place and
\ntake 7 to tens places then 9 x 0 tens = 0 tens + 7 tens = 7 tens,
\nwe write 7 at tens place and 9 X 6 hundreds = 54 hundreds,
\nnow we write 4 at hundreds place and 5 at thousands place as shown above, So 9 X 608 = 5,472.<\/p>\n

b.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n4, 0 1 8
\nExplanation:
\nIn standard algorithm we add same time of multiplying as
\n2<\/span>
\n574
\nX 7<\/span>
\n4,018<\/span> ,here 7 X 4 ones = 28 ones, we write 8 at ones place and
\ntake 2 to tens places then 7 x 7 tens = 49 tens + 2 tens = 51 tens,
\nwe write 1 at tens place and 5 at hundreds place and
\n7 X 5 hundreds = 35 hundreds, 35 hundreds + 5 hundreds =
\n40 hundreds,now we write 0 at hundreds place and 4
\nat thousands place as shown above, So 7 X 574 = 4,018.<\/p>\n

Question 2.
\nMorgan is 23 years old. Her grandfather is 4 times as old. How old is her grandfather?
\nAnswer:
\nMorgan’s grandfather is 92 years old,<\/p>\n

Explanation:
\nGiven Morgan is 23 years old and her grandfather is 4 times
\nas old. So Morgan’s grandfather’s is 4 X 23 years =
\n1<\/span>
\n23
\nX 4<\/span>
\n92, <\/span>here 4 X 3 ones = 12 ones, we write 2 at ones place and
\ntake 1 to tens places then 4 x 2 tens = 8 tens, 8 tens + 1 ten =
\n9 tens, So Morgan’s grandfather is 92 years old.<\/p>\n

Eureka Math Grade 4 Module 3 Lesson 9 Homework Answer Key<\/h3>\n

Question 1.
\nSolve using each method.
\n\"Eureka
\nAnswer:
\n\"Eureka<\/p>\n

Explanation:
\nGiven expression as 2 X 46 =
\nPartial Products
\n46
\nX 2
\n<\/span>\u00a0 12
\n+80
\n92 , <\/span>Here we first write multiplication of 2 X 6 ones, then
\n2 X 4 tens then add as shown above 12 + 80 = 92,
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n46
\nX 2
\n92 ,<\/span>here 2 X 6 ones= 12 ones we write 2 at ones place and take 1
\nat tens place then 2 x 4 at tens place = 8 tens
\nadd 1 ten to 8 tens we get 9 tens so 2 X 46 = 92.<\/p>\n

\"Eureka
\nAnswer:
\n\"Eureka
\nExplanation:
\nGiven expression as 4 X 315 =
\nPartial Products
\n315
\nX 4
\n<\/span>\u00a020
\n40
\n+1200
\n1,260 , <\/span>Here we first write multiplication of 4 X 5 ones, then
\n4 X 1 ten and 4 X 3 hundreds then add as shown above
\n20 + 40 + 1,200 = 1,260,
\nIn standard algorithm we add same time of multiplying as
\n2<\/span>
\n315
\nX 4
\n1,260 ,<\/span>here 4 X 5 ones = 20 ones we write 0 at ones place and
\ntake 2 at tens place then 4 x 1 tens = 4 tens add 2 tens to 4 tens
\nwe get 6 tens and 4 X 3 hundreds = 12 hundreds, we write 2 at
\nhundreds and 1 at thousands places,\u00a0 so 4 X 315 = 1,260.<\/p>\n

Question 2.
\nSolve using the standard algorithm.
\na.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n\u00a0 9\u00a0 \u00a0 \u00a0 \u00a0 \u00a02\u00a0 \u00a0 \u00a0 \u00a08\u00a0\u00a0<\/span><\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n232
\nX 4
\n928 ,<\/span>here 4 X 2 ones = 8 ones then 4 x 3 tens = 12 tens
\nwe write 2 at tens place and take 1 to hundred place and
\n4 X 2 hundreds = 8 hundreds, now adding 1 hundred to
\n8 hundreds we get 9 hundreds , so 4 X 232 = 928.<\/p>\n

b.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n\u00a0 \u00a0 \u00a08\u00a0 \u00a0 \u00a0 5\u00a0 \u00a0 \u00a0 \u00a02<\/span>
\nExplanation:
\nIn standard algorithm we add same time of multiplying as
\n2, 1<\/span>
\n142
\nX 6
\n852 ,<\/span>here 6 X 2 ones = 12 ones, we write 2 at ones place and
\ntake 1 to tens places then 6 x 4 tens = 24 tens + 1 tens = 25 tens,
\nwe write 5 at tens place and take 2 to hundred place and
\n6 X 1 hundred = 6 hundreds, now adding 2 hundreds to
\n6 hundreds we get 8 hundreds , so 6 X 142 = 852.<\/p>\n

c.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n\u00a02,\u00a0 \u00a0 \u00a0 1\u00a0 \u00a0 \u00a0 9\u00a0 \u00a0 8
\n<\/span>Explanation:
\nIn standard algorithm we add same time of multiplying as
\n\u00a02<\/span>
\n314
\nX 7<\/span>
\n2,198<\/span> ,here 7 X 4 ones = 28 ones, we write 8 at ones place and
\ntake 2 to tens places then 7 x 1 ten = 7 tens + 2 tens = 9 tens,
\n7 X 3 hundreds = 21 hundreds, now we write 1 at hundreds place
\nand 2 at thousands place as shown above, So 7 X 314 = 2,198.\u00a0 \u00a0\u00a0
\n<\/span><\/p>\n

d.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n1,\u00a0 \u00a03\u00a0 \u00a0 \u00a02\u00a0 \u00a0 \u00a00<\/p>\n

Explanation:
\nIn standard algorithm we add same time of multiplying as
\n1<\/span>
\n440
\nX 3<\/span>
\n1,320<\/span> ,here 3 X 0 ones = 0 ones, we write 0 at ones place,
\n3 x 4 tens = 12 tens we write 2 at tens place and take 1
\nat hundreds place, 3 X 4 hundreds = 12 hundreds,
\n12 hundreds + 1 hundred = 13 hundreds ,now we write 3
\nat hundreds place and 1 at thousand place as shown above,
\nSo 3 X 440 = 1,320.<\/p>\n

e.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n4,\u00a0 \u00a0 0\u00a0 \u00a0 \u00a05\u00a0 \u00a0 \u00a06
\nExplanation:
\nIn standard algorithm we add same time of multiplying as
\n5<\/span>
\n507
\nX 8<\/span>
\n4,056<\/span> ,here 8 X 7 ones = 56 ones, we write 6 at ones place and
\ntake 5 to tens places then 8 x 0 tens = 0 tens + 5 tens = 5 tens,
\nwe write 5 at tens place, 8 X 5 hundreds = 40 hundreds,now we write
\n0 at hundreds place and 4 at thousands place as shown above,
\nSo 8 X 507 = 4,056.<\/p>\n

f.
\n\"Eureka
\nAnswer:
\n\"Eureka
\n3,\u00a0 \u00a0 4\u00a0 \u00a0 \u00a0 5\u00a0 \u00a0 \u00a0 6
\nExplanation:
\nIn standard algorithm we add same time of multiplying as
\n7, 3<\/span>
\n384
\nX 9<\/span>
\n3,456<\/span> ,here 9 X 4 ones = 36 ones, we write 6 at ones place and
\ntake 3 to tens places then 9 x 8 tens = 72 tens + 3 tens = 75 tens,
\nwe write 5 at tens places 7 at hundreds place and
\n9 X 3 hundreds = 27 hundreds, 27 hundreds + 7 hundreds =
\n34 hundreds now we write 4\u00a0at hundreds place and
\n3 at thousands place as shown above,
\nSo 9 X 384 = 3,456.<\/p>\n

Question 3.
\nWhat is the product of 8 and 54?
\nAnswer:
\nThe product of 8 and 54 is 432,<\/p>\n

Explanation:
\nGive to find the product of 8 and 54 is
\n3<\/span>
\n54
\nX 8
\n432
\n<\/span>first we multiply 8 with 4 ones = 32 ones we write 2 at ones
\nplace and take 3 to tens place now 8 X 5 = 40 tens adding
\n3 tens we get 43 tens, So 8 X 54 = 432.<\/p>\n

Question 4.
\nIsabel earned 350 points while she was playing Blasting Robot. Isabel\u2019s mom earned 3 times as many points as Isabel. How many points did Isabel\u2019s mom earn?
\nAnswer:
\nIsabel\u2019s mom earn 1050 points,<\/p>\n

Explanation:
\nGiven Isabel earned 350 points while she was playing Blasting Robot. Isabel\u2019s mom earned 3 times as many points as Isabel. So \u00a0points did Isabel\u2019s mom earn is 3 X 350 =
\n1<\/span>
\n350
\nX 3
\n1050,
\n<\/span>first we multiply 3 with 0 ones = 0 ones ,
\n3 X 5 = 15 tens we write 15 tens as 5 at tens place and
\n1 at hundreds place and 3 X 3 = 9 hundreds + 1 hundred =
\n10 hundreds we write 10 hundreds as 0 at hundreds place and
\n1 at thousands place, So 3 X 350 = 1050.
\nTherefore, Isabel\u2019s mom earn 1050 points.<\/p>\n

Question 5.
\nTo get enough money to go on a field trip, every student in a club has to raise $53 by selling chocolate bars. There are 9 students in the club. How much money does the club need to raise to go on the field trip?
\nAnswer:
\nThe club needs to raise $477 to go on the field trip,<\/p>\n

Explanation:
\nGiven to get enough money to go on a field trip, every student in a club has to raise $53 by selling chocolate bars.
\nThere are 9 students in the club. So money does the club need to raise to go on the field trip is 9 X $53 =
\n2<\/span>
\n$53
\nX 9
\n$477
\n<\/span>first we multiply 9 with 3 ones = 27 ones , we write 7 at
\nones place and 2 at tens place, 9 X 5 tens= 45 tens,
\n45 tens + 2 tens = 47 tens we write 47 tens as
\n7 at tens place and 4 at hundreds place so 9 X $53 = $477,
\ntherefore,the club needs to raise $477 to go on the field trip.<\/p>\n

Question 6.
\nMr. Meyers wants to order 4 tablets for his classroom. Each tablet costs $329. How much will all four tablets cost?
\nAnswer:
\nTotal cost for all four tablets are $1,316,<\/p>\n

Explanation:
\nGiven Mr. Meyers wants to order 4 tablets for his classroom.
\nEach tablet costs $329. So total cost for four tablets are
\n4 X $329 =
\n1,3<\/span>
\n$329
\nX 4
\n$ 1,316
\n<\/span>first we multiply 4 with 9 ones = 36 ones , we write 6 at
\nones place and 3 at tens place, 4 X 2 tens= 8 tens,
\n8 tens + 3 tens = 11 tens we write 11 tens as
\n1 at tens place and 1 at hundreds place, Now 4 X 3 hundreds =
\n12 hundreds, 12 hundreds + 1 hundred = 13 hundreds, we write
\n3 at hundreds place and 1 at thousands place, So 4 X $329 = $1,316.
\n<\/span><\/p>\n

Question 7.
\nAmaya read 64 pages last week. Amaya\u2019s older brother, Rogelio, read twice as many pages in the same amount of time. Their big sister, Elianna, is in high school and read 4 times as many pages as Rogelio did. How many pages did Elianna read last week?
\nAnswer:
\nElianna read 512 pages last week,<\/p>\n

Explanation:
\nGiven Amaya read 64 pages last week. Amaya\u2019s older brother, Rogelio, read twice as many pages in the same amount of time.
\nSo Rogelio reads 2 X 64 pages =
\n64
\nX 2
\n128
\n<\/span>2 X 4 ones = 8 \u00a0and 2 X 6 tens= 120, So 2 X 64 = 8 + 120 = 128,
\nSo Rogelio reads 128 pages,
\nTheir big sister, Elianna, is in high school and read 4 times as
\nmany pages as Rogelio did. So number of pages did
\nElianna read last week are 4 X 128 pages=
\n1, 3<\/span>
\n128
\nx 4
\n512
\n<\/span>here we first we multiply 4 with 8 ones = 32 ones ,
\nwe write 2 at ones place and 3 at tens place, 4 X 2 tens= 8 tens,
\n8 tens + 3 tens = 11 tens we write 11 tens as
\n1 at tens place and 1 at hundreds place and 4 X 1 hundred =
\n4 hundreds, 4 hundreds +1 hundred = 5 hundreds,
\nSo
\n4 X 128 pages = 512 pages,
\ntherefore, Elianna read 512 pages last week.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 4th Grade Module 3 Lesson 9 Answer Key Eureka Math Grade 4 Module 3 Lesson 9 Problem Set Answer Key Question 1. Solve using each method. Answer: Explanation: Given expression as 4 X 34 = Partial Products 34 X 4 \u00a0 16 +120 136 , Here we first write multiplication of … 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\/59657"}],"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=59657"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59657\/revisions"}],"predecessor-version":[{"id":137376,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/59657\/revisions\/137376"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=59657"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=59657"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=59657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}