{"id":83203,"date":"2022-06-02T16:30:20","date_gmt":"2022-06-02T11:00:20","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=83203"},"modified":"2022-05-10T14:19:21","modified_gmt":"2022-05-10T08:49:21","slug":"eureka-math-grade-8-module-7-lesson-12","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-7-lesson-12\/","title":{"rendered":"Eureka Math Grade 8 Module 7 Lesson 12 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 12 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 7 Lesson 12 Example Answer Key<\/h3>\n

Example 1.
\nFind the decimal expansion of \\(\\frac{35}{11}\\).
\nAnswer:
\nFor fun, let\u2019s see if we can find the decimal expansion of \\(\\frac{35}{11}\\) without using long division.
\nTo start, can we say between which two integers this number lies?
\n\"Engage
\nThe number \\(\\frac{35}{11}\\) would lie between 3 and 4 on the number line because \\(\\frac{35}{11}\\) = \\(\\frac{33}{11}\\) + \\(\\frac{2}{11}\\) = 3 + \\(\\frac{2}{11}\\).
\nCould we say in which tenth between 3 and 4 the number 3 + \\(\\frac{2}{11}\\) lies? Is this tricky?
\nWe know that \\(\\frac{35}{11}\\) has a decimal expansion beginning with 3 in the ones place because \\(\\frac{35}{11}\\) = 3 + \\(\\frac{2}{11}\\). Now we want to determine the tenths digit, the hundredths digit, and then the thousandths digit.
\n\"Engage
\nOnes Tenths Hundredths Thousandths
\nTo figure out the tenths digit, we will use an inequality based on tenths. We are looking for the consecutive integers, m and m + 1, so that
\n3 + \\(\\frac{m}{10}\\) < \\(\\frac{35}{11}\\) < 3 + \\(\\frac{m + 1}{10}\\) .
\n\"Engage
\nWe can rewrite the middle term:
\n3 + \\(\\frac{m}{10}\\) < 3 + \\(\\frac{2}{11}\\) < 3 + \\(\\frac{m + 1}{10}\\) .<\/p>\n

This means we\u2019re looking at
\n\\(\\frac{m}{10}\\) < \\(\\frac{2}{11}\\) < \\(\\frac{m + 1}{10}\\).<\/p>\n

Give students time to make sense of the inequalities 3 + \\(\\frac{m}{10}\\) < \\(\\frac{35}{11}\\) < 3 + \\(\\frac{m + 1}{10}\\) and \\(\\frac{m}{10}\\) < \\(\\frac{2}{11}\\) < \\(\\frac{m + 1}{10}\\).
\nSince the intervals of tenths are represented by \\(\\frac{m}{10}\\) and \\(\\frac{m + 1}{10}\\), consider using concrete numbers. The chart below may help students make sense of the intervals and the inequalities.
\n\"Engage<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 12 Exercise Answer Key<\/h3>\n

Exercises 1\u20133<\/p>\n

Exercise 1.
\nFind the decimal expansion of \\(\\frac{5}{3}\\) without using long division.
\nAnswer:
\n\\(\\frac{5}{3}\\) = \\(\\frac{3}{3}\\) + \\(\\frac{2}{3}\\)
\n= 1 + \\(\\frac{2}{3}\\)
\nThe decimal expansion begins with the integer 1.
\nAmong the intervals of tenths, we are looking for integers m and m + 1 so that
\n1 + \\(\\frac{m}{10}\\) < 1 + \\(\\frac{2}{3}\\) < 1 + \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\n\\(\\frac{m}{10}\\) < \\(\\frac{2}{3}\\) < \\(\\frac{m + 1}{10}\\)
\nm < \\(\\frac{20}{3}\\) < m + 1
\nand
\n\\(\\frac{20}{3}\\) = \\(\\frac{18}{3}\\) + \\(\\frac{2}{3}\\)
\n= 6 + \\(\\frac{2}{3}\\).
\nThe tenths digit is 6.
\nAmong the intervals of hundredths we are looking for integers m and m + 1 so that
\n1 + \\(\\frac{6}{10}\\) + \\(\\frac{m}{100}\\) < \\(\\frac{5}{3}\\) < 1 + \\(\\frac{6}{10}\\) + \\(\\frac{m + 1}{10}\\) ,
\nwhich is equivalent to
\n\\(\\frac{m}{100}\\) < \\(\\frac{2}{3}\\) – \\(\\frac{6}{10}\\) < \\(\\frac{m + 1}{10}\\).
\nNow
\n\\(\\frac{2}{3}\\) – \\(\\frac{6}{10}\\) = \\(\\frac{2}{30}\\) .
\nSo we are looking for integers m and m + 1 where
\n\\(\\frac{m}{100}\\) < \\(\\frac{2}{30}\\) < \\(\\frac{m + 1}{100}\\),
\nwhich is the same as
\nm < \\(\\frac{20}{3}\\) < m + 1 .
\nBut we already know that \\(\\frac{20}{3}\\) = 6 + \\(\\frac{2}{3}\\); therefore, the hundredths digit is 6. We feel like we are repeating our work, so we suspect \\(\\frac{5}{3}\\) = 1.666″\u2026” . To check: 0.6666″\u2026” = 6\/9 = \\(\\frac{2}{3}\\) and 1.6666″\u2026” = 1 + \\(\\frac{2}{3}\\) = \\(\\frac{5}{3}\\). We are correct.<\/p>\n

Exercise 2.
\nFind the decimal expansion of \\(\\frac{5}{11}\\) without using long division.
\nAnswer:
\nIts decimal expansion begins with the integer 0.
\nIn the intervals of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{5}{11}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{50}{11}\\) < m + 1
\n\\(\\frac{50}{11}\\) = \\(\\frac{44}{11}\\) + \\(\\frac{6}{11}\\)
\n= 4 + \\(\\frac{6}{11}\\)
\nThe tenths digit is 4.
\nIn the intervals of hundredths, we are looking for integers m and m + 1 so that
\n\"Engage
\nAs
\n\\(\\frac{60}{11}\\) = \\(\\frac{55}{11}\\) + \\(\\frac{5}{11}\\)
\n= 5 + \\(\\frac{5}{11}\\)
\nwe see that the hundredths digit is 5.
\nThe fraction \\(\\frac{5}{11}\\) has reappeared, which makes us suspect we are in a repeating pattern and we have
\n\\(\\frac{5}{11}\\) = 0.454545″\u2026” . To check: 0.454545″\u2026” = \\(\\frac{45}{99}\\) = \\(\\frac{5}{11}\\). We are correct.<\/p>\n

Exercise 3.
\nFind the decimal expansion of the number \\(\\frac{23}{99}\\) first without using long division and then again using long division.
\nAnswer:
\nThe decimal expansion begins with the integer 0.
\nIn the interval of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{23}{99}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{230}{99}\\) < m + 1.
\nNow
\n\\(\\frac{230}{99}\\) = \\(\\frac{198}{99}\\) + \\(\\frac{32}{99}\\)
\n= 2 + \\(\\frac{32}{99}\\)
\nshowing that the tenths digit is 2.
\nIn the interval of hundredths, we are looking for integers m and m + 1 so that
\n\"Engage
\nNow
\n\\(\\frac{320}{99}\\) = \\(\\frac{297}{99}\\) + \\(\\frac{23}{99}\\)
\n= 3 + \\(\\frac{23}{99}\\)
\nThe hundredths digit is 3. The reappearance of 23\/99 makes us suspect that we\u2019re in a repeating pattern and \\(\\frac{23}{99}\\) = 0.232323″\u2026” . We check that 0.232323″\u2026” does indeed equal \\(\\frac{23}{99}\\) , and we are correct.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 12 Problem Set Answer Key<\/h3>\n

Question 1.
\nWithout using long division, explain why the tenths digit of \\(\\frac{3}{11}\\) is a 2.
\nAnswer:
\nIn the interval of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{3}{11}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{30}{11}\\) < m + 1
\n\\(\\frac{30}{11}\\) = \\(\\frac{22}{11}\\) + \\(\\frac{8}{11}\\)
\n= 2 + \\(\\frac{8}{11}\\)
\nIn looking at the interval of tenths, we see that the number \\(\\frac{3}{11}\\) must be between \\(\\frac{2}{10}\\) and \\(\\frac{3}{10}\\)because \\(\\frac{2}{10}\\) < \\(\\frac{3}{11}\\) < \\(\\frac{3}{10}\\). For this reason, the tenths digit of the decimal expansion of \\(\\frac{3}{11}\\) must be 2.<\/p>\n

Question 2.
\nFind the decimal expansion of \\(\\frac{25}{9}\\) without using long division.
\nAnswer:
\n\\(\\frac{25}{9}\\) = \\(\\frac{18}{9}\\) + \\(\\frac{7}{9}\\)
\n= 2 + \\(\\frac{7}{9}\\)
\nThe ones digit is 2. In the interval of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{7}{9}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{70}{9}\\) < m + 1
\n\\(\\frac{70}{9}\\) = \\(\\frac{63}{9}\\) + \\(\\frac{7}{9}\\)
\n= 7 + \\(\\frac{7}{9}\\)
\nThe tenths digit is 7. The difference between \\(\\frac{7}{9}\\) and \\(\\frac{7}{10}\\) is
\n\\(\\frac{7}{9}\\) – \\(\\frac{7}{10}\\) = \\(\\frac{7}{90}\\).
\nIn the interval of hundredths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{100}\\) < \\(\\frac{7}{90}\\) < \\(\\frac{m + 1}{100}\\),
\nwhich is the same as
\nm < \\(\\frac{70}{9}\\) < m + 1.
\nHowever, we already know that \\(\\frac{70}{9}\\) = 7 + \\(\\frac{7}{9}\\); therefore, the hundredths digit is 7. Because we keep getting \\(\\frac{7}{9}\\), we can assume the digit of 7 will continue to repeat. Therefore, the decimal expansion of \\(\\frac{25}{9}\\) is 2.777\u2026.<\/p>\n

Question 3.
\nFind the decimal expansion of \\(\\frac{11}{41}\\) to at least 5 digits without using long division.
\nAnswer:
\nIn the interval of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{11}{41}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{110}{41}\\) < m + 1
\n\\(\\frac{110}{41}\\) = \\(\\frac{82}{41}\\) + \\(\\frac{28}{41}\\) = 2 + \\(\\frac{28}{41}\\).
\nThe tenths digit is 2. The difference between \\(\\frac{11}{41}\\) and \\(\\frac{2}{10}\\) is
\n\\(\\frac{11}{41}\\) – \\(\\frac{2}{10}\\) = \\(\\frac{28}{410}\\).
\nIn the interval of hundredths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{100}\\) < \\(\\frac{28}{410}\\) < \\(\\frac{m + 1}{100}\\),
\nwhich is the same as
\n\"Eureka
\nIn the interval of ten – thousandths, we are looking for integers m and m + 1 so that
\n\"Eureka
\nThe hundred – thousandths digit is 9. We see again the fraction \\(\\frac{11}{41}\\), so we can expect the decimal digits to repeat at this point. Therefore, the decimal approximation of \\(\\frac{11}{1}\\) is 0.2682926829\u2026.<\/p>\n

Question 4.
\nWhich number is larger, \\(\\sqrt{10}\\) or \\(\\frac{28}{9}\\)? Answer this question without using long division.
\nAnswer:
\nThe number \\(\\sqrt{10}\\) is between 3 and 4. In the sequence of tenths, \\(\\sqrt{10}\\) is between 3.1 and 3.2 because
\n3.12<\/sup> < (\\(\\sqrt{10}\\))2<\/sup> < 3.22<\/sup>. In the sequence of hundredths, \\(\\sqrt{10}\\) is between 3.16 and 3.17 because
\n3.162<\/sup> < (\\(\\sqrt{10}\\))2<\/sup> < 3.172<\/sup>. In the sequence of thousandths, \\(\\sqrt{10}\\) is between 3.162 and 3.163 because
\n3.1622<\/sup> < (\\(\\sqrt{10}\\))2<\/sup> < 3.1632<\/sup>. The decimal expansion of \\(\\sqrt{10}\\) is approximately 3.162\u2026.
\n\\(\\frac{28}{9}\\) = \\(\\frac{27}{9}\\) + \\(\\frac{1}{9}\\)
\n= 3 + \\(\\frac{1}{9}\\)
\nIn the interval of tenths, we are looking for the integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{1}{9}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{10}{9}\\) < m + 1
\n\\(\\frac{10}{9}\\) = \\(\\frac{9}{9}\\) + \\(\\frac{1}{9}\\)
\n= 1 + \\(\\frac{1}{9}\\)
\nThe tenths digit is 1. Since the fraction \\(\\frac{1}{9}\\) has reappeared, we can assume that the next digit is also 1, and the work will continue to repeat. Therefore, the decimal expansion of \\(\\frac{28}{9}\\)9 is 3.1111\u2026.
\nTherefore, \\(\\frac{28}{9}\\) < \\(\\sqrt{10}\\).<\/p>\n

Alternatively: (\\(\\sqrt{10}\\))2<\/sup> = 10 and (\\(\\frac{28}{9}\\))2<\/sup> = 784\/81, which is less than \\(\\frac{810}{81}\\) or 10. Thus, \\(\\frac{28}{9}\\) is the smaller number.<\/p>\n

Question 5.
\nSam says that \\(\\frac{7}{11}\\) = 0.63, and Jaylen says that \\(\\frac{7}{11}\\) = 0.636. Who is correct? Why?
\nAnswer:
\nIn the interval of tenths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{10}\\) < \\(\\frac{7}{11}\\) < \\(\\frac{m + 1}{10}\\),
\nwhich is the same as
\nm < \\(\\frac{70}{11}\\) < \\(\\frac{m + 1}{10}\\)
\n\\(\\frac{70}{11}\\) = \\(\\frac{66}{11}\\) + \\(\\frac{4}{11}\\)
\n= 6 + \\(\\frac{4}{11}\\)
\nThe tenths digit is 6. The difference between \\(\\frac{7}{11}\\) and \\(\\frac{6}{10}\\) is
\n\\(\\frac{7}{11}\\) – \\(\\frac{6}{10}\\) = \\(\\frac{4}{110}\\).
\nIn the interval of hundredths, we are looking for integers m and m + 1 so that
\n\\(\\frac{m}{100}\\) < \\(\\frac{4}{110}\\) < \\(\\frac{m + 1}{100}\\),
\nwhich is the same as
\nm < \\(\\frac{40}{11}\\) < m + 1
\n\\(\\frac{40}{11}\\) = \\(\\frac{33}{11}\\) + \\(\\frac{7}{11}\\)
\n= 3 + \\(\\frac{7}{11}\\)
\nThe hundredths digit is 3. Again, we see the fraction \\(\\frac{7}{11}\\), which means the next decimal digit will be 6, as it was in the tenths place. This means we will again see the fraction \\(\\frac{4}{11}\\), meaning we will have another digit of 3. Therefore, the decimal expansion of \\(\\frac{7}{11}\\) is 0.6363\u2026.<\/p>\n

Technically, Sam and Jaylen are incorrect because the fraction \\(\\frac{7}{11}\\) is an infinite decimal. However, Sam is correct to the first two decimal digits of the number, and Jaylen is correct to the first three decimal digits of the number.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 12 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nFind the decimal expansion of \\(\\frac{41}{6}\\) without using long division.
\nAnswer:
\n\\(\\frac{41}{6}\\) = \\(\\frac{36}{6}\\) + \\(\\frac{5}{6}\\)
\n= 6 + \\(\\frac{5}{6}\\)
\nThe ones digit is 6.
\nTo determine in which interval of tenths the fraction lies, we look for integers m and m + 1 so that
\n\"Eureka
\nThe tenths digit is 8.
\nTo determine in which interval of hundredths the fraction lies, we look for integers m and m + 1 so that
\n\"Eureka
\nWe are a third over a whole number of tenths and a third over a whole number of hundredths. We suspect we are in a repeating pattern and that \\(\\frac{41}{6}\\) = 6.83333\u2026.
\nTo check:
\nx = 6.8333\u2026
\n10x = 68.3333\u2026
\n10x = 68 + 0.3333\u2026
\n10x = 68 + \\(\\frac{1}{3}\\)
\n10x = \\(\\frac{204}{3}\\) + \\(\\frac{1}{3}\\)
\n10x = \\(\\frac{205}{3}\\)
\nx = \\(\\frac{205}{30}\\)
\nx = \\(\\frac{41}{6}\\)
\nWe are correct.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 12 Area and Volume I Answer Key<\/h3>\n

Question 1.
\nFind the area of the square shown below.
\n\"Eureka
\nAnswer:
\nA = (4 m)2<\/sup>
\n= 16 m2<\/sup><\/p>\n

Question 2.
\nFind the volume of the cube shown below.
\n\"Eureka
\nAnswer:
\nV = (4 m)3<\/sup>
\n= 64 m3<\/sup><\/p>\n

Question 3.
\nFind the area of the rectangle shown below.
\n\"Eureka
\nAnswer:
\nA = (8 cm)(4 cm)
\n= 32 cm2<\/sup><\/p>\n

Question 4.
\nFind the volume of the rectangular prism shown below.
\n\"Eureka
\nAnswer:
\nV = (32 cm2<\/sup>)(6 cm)
\n= 192 cm3<\/sup><\/p>\n

Question 5.
\nFind the area of the circle shown below.
\n\"Eureka
\nAnswer:
\nA = (7 m)2<\/sup> \u03c0
\n= 49\u03c0 m2<\/sup><\/p>\n

Question 6.
\nFind the volume of the cylinder shown below.
\n\"Eureka
\nAnswer:
\nV = (49\u03c0 m2<\/sup>)(12 m)
\n= 588\u03c0 m3<\/sup><\/p>\n

Question 7.
\nFind the area of the circle shown below.
\n\"Eureka
\nAnswer:
\nA = (6 in.)2<\/sup> \u03c0
\n= 36\u03c0 in2<\/sup><\/p>\n

Question 8.
\nFind the volume of the cone shown below.
\n\"Eureka
\nAnswer:
\nV = (\\(\\frac{1}{3}\\))(36\u03c0 in2<\/sup> )(10 in.)
\n= 120\u03c0 in3<\/sup><\/p>\n

Question 9.
\nFind the area of the circle shown below.
\n\"Eureka
\nAnswer:
\nA = (8 mm)2<\/sup> \u03c0
\n= 64\u03c0 mm2<\/sup><\/p>\n

Question 10.
\nFind the volume of the sphere shown below.
\n\"Eureka
\nAnswer:
\nV = (\\(\\frac{4}{3}\\))\u03c0(64 mm2<\/sup> )(8 mm)
\n= \\(\\frac{2048}{2}\\) \u03c0 mm3<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 12 Answer Key Eureka Math Grade 8 Module 7 Lesson 12 Example Answer Key Example 1. Find the decimal expansion of . Answer: For fun, let\u2019s see if we can find the decimal expansion of without using long division. To start, can we say between which … 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\/83203"}],"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=83203"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83203\/revisions"}],"predecessor-version":[{"id":137283,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83203\/revisions\/137283"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=83203"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=83203"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=83203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}