Question 1.
\nDraw tape diagrams to show two ways to represent 2 units of 4\\(\\frac{2}{3}\\).
\nWrite a multiplication expression to match each tape diagram.<\/p>\n
Answer:
\n2 \u00d7 4\\(\\frac{2}{3}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n2 \u00d7 4\\(\\frac{2}{3}\\)
\n2 \u00d7 (4 + \\(\\frac{2}{3}\\))
\n(2 \u00d7 4) + (2 \u00d7 \\(\\frac{2}{3}\\))
\n8 + \\(\\frac{4}{3}\\)
\n8 + 1\\(\\frac{1}{3}\\)
\n9\\(\\frac{1}{3}\\)<\/p>\n
Question 2.
\nSolve the following using the distributive property. The first one has been done for you. (As soon as you are ready, you may omit the step that is in line 2.)
\na. 3 \u00d7 6\\(\\frac{4}{5}\\) = 3 \u00d7 (6 + \\(\\frac{4}{5}\\))
\n= (3 \u00d7 6) + (3 \u00d7 \\(\\frac{4}{5}\\))
\n= 18 + \\(\\frac{12}{5}\\)
\n= 18 + 2\\(\\frac{2}{5}\\)
\n= 20\\(\\frac{2}{5}\\)<\/p>\n
b. 2 \u00d7 4\\(\\frac{2}{3}\\)<\/p>\n
Answer:
\n2 \u00d7 4\\(\\frac{2}{3}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n2 \u00d7 4\\(\\frac{2}{3}\\)
\n2 \u00d7 (4 + \\(\\frac{2}{3}\\))
\n(2 \u00d7 4) + (2 \u00d7 \\(\\frac{2}{3}\\))
\n8 + \\(\\frac{4}{3}\\)
\n8 + 1\\(\\frac{1}{3}\\)
\n9\\(\\frac{1}{3}\\)<\/p>\n
c. 3 \u00d7 2\\(\\frac{5}{8}\\)<\/p>\n
Answer:
\n3 \u00d7 2\\(\\frac{5}{8}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n3 \u00d7 2\\(\\frac{5}{8}\\)
\n3 \u00d7 (2 + \\(\\frac{5}{8}\\))
\n(3 \u00d7 2) + (3 \u00d7 \\(\\frac{5}{8}\\))
\n6 + \\(\\frac{15}{8}\\)
\n6 + 1\\(\\frac{7}{8}\\)
\n7\\(\\frac{7}{8}\\)<\/p>\n
d. 2 \u00d7 4\\(\\frac{7}{10}\\)<\/p>\n
Answer:
\n2 \u00d7 4\\(\\frac{7}{10}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n2 \u00d7 4\\(\\frac{7}{10}\\)
\n2 \u00d7 (4 + \\(\\frac{7}{10}\\))
\n(2 \u00d7 4) + (2 \u00d7 \\(\\frac{7}{10}\\))
\n8 + \\(\\frac{14}{10}\\)
\n8 + 1\\(\\frac{4}{10}\\)
\n9\\(\\frac{4}{10}\\)<\/p>\n
e. 3 7\\(\\frac{3}{4}\\)<\/p>\n
Answer:
\n3 \u00d7 7\\(\\frac{3}{4}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n3 \u00d7 7\\(\\frac{3}{4}\\)
\n3 \u00d7 (7 + \\(\\frac{3}{4}\\))
\n(3 \u00d7 7) + (3 \u00d7 \\(\\frac{3}{4}\\))
\n21 + \\(\\frac{9}{4}\\)
\n21 + 2\\(\\frac{1}{4}\\)
\n23\\(\\frac{1}{4}\\)<\/p>\n
f. 6 \u00d7 3\\(\\frac{1}{2}\\)<\/p>\n
Answer:
\n6 \u00d7 3\\(\\frac{1}{2}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n6 \u00d7 3\\(\\frac{1}{2}\\)
\n6 \u00d7 (3 + \\(\\frac{1}{2}\\))
\n(6 \u00d7 3) + (6 \u00d7 \\(\\frac{1}{2}\\))
\n18 + \\(\\frac{6}{2}\\)
\n18 + 2\\(\\frac{2}{2}\\)
\n20\\(\\frac{2}{2}\\)<\/p>\n
g. 4 \u00d7 9\\(\\frac{1}{5}\\)<\/p>\n
Answer:
\n4 \u00d7 9\\(\\frac{1}{5}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n4 \u00d7 9\\(\\frac{1}{5}\\)
\n4 \u00d7 (9 + \\(\\frac{1}{5}\\))
\n(4 \u00d7 9) + (4 \u00d7 \\(\\frac{1}{5}\\))
\n36 + \\(\\frac{4}{5}\\)
\n36 + \\(\\frac{4}{5}\\)
\n36\\(\\frac{4}{5}\\)<\/p>\n
h. 5\\(\\frac{6}{8}\\) \u00d7 4<\/p>\n
Answer:
\n5 \u00d7 4\\(\\frac{6}{8}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n5 \u00d7 4\\(\\frac{6}{8}\\)
\n5 \u00d7 (4 + \\(\\frac{6}{8}\\))
\n(5 \u00d7 4) + (5 \u00d7 \\(\\frac{6}{8}\\))
\n20 + \\(\\frac{30}{8}\\)
\n20 + 3\\(\\frac{6}{8}\\)
\n23\\(\\frac{6}{8}\\)<\/p>\n
Question 3.
\nFor one dance costume, Saisha needs 4\\(\\frac{2}{3}\\) feet of ribbon. How much ribbon does she need for 5 identical costumes?<\/p>\n
Answer:
\nThe ribbon does she need for 5 identical costumes = 13 feet.<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\nFor one dance costume, Saisha needs 4(2\/3) feet of ribbon.
\n4(2\/3) = 4 x 2\/3.
\n4 x 2 = 8\/3.
\n8\/3 = 2.6.
\n2.6 x 5 = 13.<\/p>\n
Multiply. Write each product as a mixed number.
\nQuestion 1.
\n4 \u00d7 5\\(\\frac{3}{8}\\)<\/p>\n
Answer:
\n4 \u00d7 5\\(\\frac{3}{8}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n4 \u00d7 5\\(\\frac{3}{8}\\)
\n4 \u00d7 (5 + \\(\\frac{3}{8}\\))
\n(4 \u00d7 5) + (4 \u00d7 \\(\\frac{3}{8}\\))
\n20 + \\(\\frac{12}{8}\\)
\n20 + 1\\(\\frac{4}{8}\\)
\n21\\(\\frac{4}{8}\\)<\/p>\n
Question 2.
\n4\\(\\frac{3}{10}\\) \u00d7 3<\/p>\n
Answer:
\n4 \u00d7 3\\(\\frac{3}{10}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n4 \u00d7 3\\(\\frac{3}{10}\\)
\n4 \u00d7 (3 + \\(\\frac{3}{10}\\))
\n(4 \u00d7 3) + (4 \u00d7 \\(\\frac{3}{10}\\))
\n12 + \\(\\frac{12}{10}\\)
\n12 + 1\\(\\frac{2}{10}\\)
\n13\\(\\frac{2}{10}\\)<\/p>\n
Question 1.
\nDraw tape diagrams to show two ways to represent 3 units of 5\\(\\frac{1}{12}\\)” .
\nWrite a multiplication expression to match each tape diagram.<\/p>\n
Answer:
\n3 \u00d7 5\\(\\frac{1}{12}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n3 \u00d7 5\\(\\frac{1}{12}\\)
\n3 \u00d7 (5 + \\(\\frac{1}{12}\\))
\n(3 \u00d7 5) + (3 \u00d7 \\(\\frac{1}{12}\\))
\n15 + \\(\\frac{3}{12}\\)
\n15 + \\(\\frac{1}{4}\\)
\n15\\(\\frac{1}{4}\\)<\/p>\n
Question 2.
\nSolve the following using the distributive property. The first one has been done for you. (As soon as you are ready, you may omit the step that is in line 2.)
\na. 3 \u00d7 6\\(\\frac{4}{5}\\) = 3 \u00d7 (6 + \\(\\frac{4}{5}\\))
\n= (3 \u00d7 6) + (3 \u00d7 \\(\\frac{4}{5}\\))
\n= 18 + \\(\\frac{12}{5}\\)
\n= 18 + 2\\(\\frac{2}{5}\\)
\n= 20\\(\\frac{2}{5}\\)<\/p>\n
b. 5 \u00d7 4\\(\\frac{1}{6}\\)<\/p>\n
Answer:
\n5 \u00d7 4\\(\\frac{1}{6}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n5 \u00d7 4\\(\\frac{1}{6}\\)
\n5 \u00d7 (4 + \\(\\frac{1}{6}\\))
\n(5 \u00d7 4) + (5 \u00d7 \\(\\frac{1}{6}\\))
\n20 + \\(\\frac{5}{6}\\)
\n20 + \\(\\frac{5}{6}\\)
\n20\\(\\frac{5}{6}\\)<\/p>\n
c. 6 \u00d7 2\\(\\frac{3}{5}\\)<\/p>\n
Answer:
\n6 \u00d7 2\\(\\frac{3}{5}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n6 \u00d7 2\\(\\frac{3}{5}\\)
\n6 \u00d7 (2 + \\(\\frac{3}{5}\\))
\n(6 \u00d7 2) + (6 \u00d7 \\(\\frac{3}{5}\\))
\n12 + \\(\\frac{18}{5}\\)
\n12 + 3\\(\\frac{3}{5}\\)
\n15\\(\\frac{3}{5}\\)<\/p>\n
d. 2 \u00d7 7\\(\\frac{3}{10}\\)<\/p>\n
Answer:
\n2 \u00d7 7\\(\\frac{3}{10}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n2 \u00d7 7\\(\\frac{3}{10}\\)
\n2 \u00d7 (7 + \\(\\frac{3}{10}\\))
\n(2\u00d7 7) + (2 \u00d7 \\(\\frac{3}{10}\\))
\n14 + \\(\\frac{6}{10}\\)
\n14 + \\(\\frac{3}{5}\\)
\n14\\(\\frac{3}{5}\\)<\/p>\n
e. 8 \u00d7 7\\(\\frac{1}{4}\\)<\/p>\n
Answer:
\n8 \u00d7 7\\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n8 \u00d7 7\\(\\frac{1}{4}\\)
\n8 \u00d7 (7 + \\(\\frac{1}{4}\\))
\n(8 \u00d7 7) + (8 \u00d7 \\(\\frac{1}{4}\\))
\n56 + \\(\\frac{8}{4}\\)
\n56 + \\(\\frac{2}{0}\\)
\n56\\(\\frac{2}{0}\\)<\/p>\n
f. 3\\(\\frac{3}{8}\\) \u00d7 12<\/p>\n
Answer:
\n3 \u00d7 12\\(\\frac{3}{8}\\)<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\n3 \u00d7 12\\(\\frac{3}{8}\\)
\n3 \u00d7 (12 + \\(\\frac{3}{8}\\))
\n(3 \u00d7 12) + (3 \u00d7 \\(\\frac{3}{8}\\))
\n36 + \\(\\frac{9}{8}\\)
\n36 + 1\\(\\frac{1}{8}\\)
\n37\\(\\frac{1}{8}\\)<\/p>\n
Question 3.
\nSara\u2019s street is 2\\(\\frac{3}{10}\\) miles long. She ran the length of the street 6 times. How far did she run?<\/p>\n
Answer:
\nThe far did she run = 16.8<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\nSara’s street is 2(3\/10) miles long.
\nshe ran the length of the street 6 times.
\n2(3\/10).
\n6\/10 = 0.6.<\/p>\n
Question 4.
\nKelly\u2019s new puppy weighed 4\\(\\frac{7}{10}\\) pounds when she brought him home. Now, he weighs six times as much. How much does he weigh now?<\/p>\n
Answer:
\nThe weight now = 16.8<\/p>\n
Explanation:
\nIn the above-given question,
\ngiven that,
\nKelly’s new puppy weighed 4(7\/10) pounds when she brought him home.
\nhe weighs six times as much.
\n4(7\/10) = 28\/10.
\n28\/10 = 2.8.
\n2.8 x 6 = 16.8.<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math 4th Grade Module 5 Lesson 37 Answer Key Eureka Math Grade 4 Module 5 Lesson 37 Problem Set Answer Key Question 1. Draw tape diagrams to show two ways to represent 2 units of 4. Write a multiplication expression to match each tape diagram. Answer: 2 \u00d7 4 Explanation: In the … 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\/60689"}],"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=60689"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/60689\/revisions"}],"predecessor-version":[{"id":136674,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/60689\/revisions\/136674"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=60689"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=60689"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=60689"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}