{"id":83160,"date":"2022-06-02T16:30:19","date_gmt":"2022-06-02T11:00:19","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=83160"},"modified":"2022-05-10T14:19:21","modified_gmt":"2022-05-10T08:49:21","slug":"eureka-math-grade-8-module-7-lesson-11","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-grade-8-module-7-lesson-11\/","title":{"rendered":"Eureka Math Grade 8 Module 7 Lesson 11 Answer Key"},"content":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 11 Answer Key<\/h2>\n

Eureka Math Grade 8 Module 7 Lesson 11 Example Answer Key<\/h3>\n

Example 1.
\nConsider the decimal expansion of \\(\\sqrt{3}\\).
\nFind the first two values of the decimal expansion using the following fact: If c2 <\/sup>< 3 < d2<\/sup> for positive numbers c and d, then c < \\(\\sqrt{3}\\) < d.
\nFirst approximation: Because 1 <3 < 4, we have 1 < \\(\\sqrt{3}\\) < 2.
\nAnswer:
\nWe learned in Lesson 3 that if c and d are positive numbers, then c2 <\/sup>< d2<\/sup> implies c2<\/sup> < d2<\/sup>. It follows from this that if c2 <\/sup>< N < d2<\/sup>, then c<\\(\\sqrt{N}\\) c < \\(\\sqrt{N}\\) < d, then c2 <\/sup>< N < d2<\/sup>.)
\nSince 1 < 3 < 4, we get our first approximation: 1 < \\(\\sqrt{3}\\) < 2.
\nTo get more precise with our estimate of \\(\\sqrt{3}\\), we can look at the tenths between the numbers 1 and 2.<\/p>\n

Second approximation:
\n\"Engage
\nAnswer:
\n\u2192 Is \\(\\sqrt{3}\\) between 1.2 and 1.3?
\n\u2192 If 1.2 < \\(\\sqrt{3}\\) < 1.3, then we should have 1.22 <\/sup>< 3 < 1.32<\/sup>. But we don\u2019t: 1.22<\/sup> = 1.44 and 1.32<\/sup> = 1.69. These squares are too small.
\n\u2192 Is \\(\\sqrt{3}\\) between 1.8 and 1.9?
\nIf 1.8 < \\(\\sqrt{3}\\) < 1.9, then we should have 1.82<\/sup><3<1.92<\/sup>. But we don\u2019t: 1.82<\/sup> = 3.24 and 1.92<\/sup> = 3.81. These squares are too large.<\/p>\n

\u2192 Can you find the right tenth interval in which \\(\\sqrt{3}\\) belongs?
\nAfter some trial and error, students see that \\(\\sqrt{3}\\) lies between 1.7 and 1.8 . We have 1.72<\/sup> = 2.89 and 1.82<\/sup> = 3.24, and so 2.89 < 3 < 3.24.
\nSo the decimal expansion of \\(\\sqrt{3}\\) begins 1.7\u2026. How could we pin down its next decimal place?
\nLook for where \\(\\sqrt{3}\\) lies in the interval between 1.7 and 1.8. Divide that interval into ten parts, too.
\nLet\u2019s do that!<\/p>\n

Third approximation:
\n\"Engage
\nAnswer:
\n\"Engage
\nHave students use trial and error to eventually establish that \\(\\sqrt{3}\\) lies between 1.73 and 1.74: we have 1.732<\/sup> = 2.9929 and 1.742<\/sup> = 3.0276 and 2.9926 < 3 < 3.0276.
\n\u2192 So what are the first two places of the decimal expansion of \\(\\sqrt{3}\\)?
\nWe have \\(\\sqrt{3}\\) = 1.73\u2026.
\n\u2192 What do you think will need to be done to get an even more precise estimate of the number \\(\\sqrt{3}\\)?
\nWe will need to look at the interval between 1.73 and 1.74 more closely and repeat the process we did before.<\/p>\n

\u2192 Would you like to find the next decimal place for \\(\\sqrt{3}\\) just for fun or leave it be for now?
\nGive students the option to find the next decimal place if they wish.
\nHow accurate is our approximation \\(\\sqrt{3}\\) = 1.73\u2026? (If students computed
\n\\(\\sqrt{3}\\) = 1.732\u2026, adjust this question and the answer below appropriately.)
\nWe know \\(\\sqrt{3}\\) = 1.73abc\u2026 for some more digits a, b, c, and so on. Now 1.73 and 1.73abc\u2026 differ by 0.00abc\u2026, which is less than 0.01. A decimal expansion computed to two decimal places gives an approximation that is accurate with an error that is at most 0.01.<\/p>\n

Example 2.
\nFind the first few places of the decimal expansion of \\(\\sqrt{28}\\).<\/p>\n

First approximation:
\nAnswer:
\nBetween which two integers does \\(\\sqrt{28}\\) lie?
\nSince 25 < 28 < 36, we see 5 < \\(\\sqrt{28}\\) < 6 .
\nIn which tenth between 5 and 6 does \\(\\sqrt{28}\\) lie?<\/p>\n

Second approximation:
\n\"Engage
\nAnswer:
\nHow do we determine which interval is correct?
\nWhat if we just square the numbers 5.0, 5.1, and 5.2 and see between which two squares 28 lies? After all, we are hoping to see that 5.3 < \\(\\sqrt{28}\\) < 5.4, in which case we should have 5.32 <\/sup>< 28 <5 .42<\/sup>. (This interval is probably not correct, but we can check!)
\nProvide students time to determine in which interval the number \\(\\sqrt{28}\\) lies. Ask students to share their findings, and then continue the discussion.
\nNow that we know that the number \\(\\sqrt{28}\\) lies between 5.2 and 5.3, let\u2019s look at hundredths.<\/p>\n

Third approximation:
\n\"Engage
\nAnswer:
\n\"Engage
\nCan we be efficient? Since 5.202<\/sup> = 27.04 and 5.302<\/sup> = 28.09, would an interval to the left, to the middle, or to the right likely contain \\(\\sqrt{28}\\)?
\nWe suspect that the interval between 5.29 and 5.30 might contain \\(\\sqrt{28}\\) because 28 is close to 28.09.
\nProvide students time to determine which interval the number \\(\\sqrt{28}\\) will lie between. Ask students to share their findings, and then continue the discussion.
\nNow we know that the number \\(\\sqrt{28}\\) is between 5.29 and 5.30. Let\u2019s go one step further and examine intervals of thousandths.<\/p>\n

Fourth approximation:
\n\"Engage
\nAnswer:
\n\"Engage
\nSince 5.2902<\/sup> = 27.9841 and 5.3002<\/sup> = 28.09, where should we begin our search?
\nWe should begin with the intervals closer to 5.290 and 5.291 because 28 is closer to 27.9841 than to 28.09.
\nProvide students time to determine which interval the number \\(\\sqrt{28}\\) will lie between. Ask students to share their findings, and then finish the discussion.
\nThe number \\(\\sqrt{28}\\) lies between 5.291 and 5.292 because 5.2912<\/sup> = 27.994681 and 5.2922<\/sup> = 28.005264. At this point, we have a fair approximation of the value of \\(\\sqrt{28}\\). It is between 5.291 and 5.292 on the number line:
\n\"Engage
\nWe could continue this process of rational approximation to see that \\(\\sqrt{28}\\) = 5.291502622\u2026.
\nAs before, use an online calculator to show the decimal expansion of \\(\\sqrt{28}\\). Once displayed, ask students to examine the decimal expansion for any patterns, or lack thereof
\nConsider going back to the Opening Exercise to determine whose approximation was the closest.
\nCan we conduct this work to also pin down the location of \\(\\sqrt{121}\\) on the number line?
\nNo need! \\(\\sqrt{121}\\) = 11, so we know where it sits!<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 11 Exercise Answer Key<\/h3>\n

Exercise 1.
\nIn which interval of hundredths does \\(\\sqrt{14}\\) lie? Show your work.
\nAnswer:
\nThe number \\(\\sqrt{14}\\) is between integers 3 and 4 because 9<14<16. Then, \\(\\sqrt{14}\\) must be checked for the interval of tenths between 3 and 4. Since \\(\\sqrt{14}\\) is closer to 4, we will begin with the interval from 3.9 to 4.0. The number \\(\\sqrt{14}\\) is between 3.7 and 3.8 because 3.72<\/sup> = 13.69 and 3.82<\/sup> = 14.44. Now, we must look at the interval of hundredths between 3.7 and 3.8. Since \\(\\sqrt{14}\\) is closer to 3.7, we will begin with the interval 3.70 to 3.71. The number \\(\\sqrt{14}\\) is between 3.74 and 3.75 because 3.742<\/sup> = 13.9876 and 3.752<\/sup> = 14.0625.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 11 Problem Set Answer Key<\/h3>\n

Question 1.
\nIn which hundredth interval of the number line does \\(\\sqrt{84}\\) lie?
\nAnswer:
\nThe number \\(\\sqrt{84}\\) is between 9 and 10 but closer to 9. Looking at the interval of tenths, beginning with 9.0 to 9.1, the number \\(\\sqrt{84}\\) lies between 9.1 and 9.2 because 9.12<\/sup> = 82.81 and 9.22<\/sup> = 84.64 but is closer to 9.2. In the interval of hundredths, the number \\(\\sqrt{84}\\) lies between 9.16 and 9.17 because 9.162<\/sup> = 83.9056 and 9.172<\/sup> = 84.0889.<\/p>\n

Question 2.
\nDetermine the three-decimal digit approximation of the number \\(\\sqrt{34}\\).
\nAnswer:
\nThe number \\(\\sqrt{34}\\) is between 5 and 6 but closer to 6. Looking at the interval of tenths, beginning with 5.9 to 6.0, the number \\(\\sqrt{34}\\) lies between 5.8 and 5.9 because 5.82<\/sup> = 33.64 and 5.92<\/sup> = 34.81 and is closer to 5.8. In the interval of hundredths, the number \\(\\sqrt{34}\\) lies between 5.83 and 5.84 because 5.832<\/sup> = 33.9889 and 5.842<\/sup> = 34.1056 and is closer to 5.83. In the interval of thousandths, the number \\(\\sqrt{34}\\) lies between 5.830 and 5.831 because 5.8302<\/sup> = 33.9889 and 5.8312<\/sup> = 34.000 \u2006 561 but is closer to 5.831. Since 34 is closer to 5.8312<\/sup> than 5.8302<\/sup>, then the three-decimal digit approximation of the number is 5.831.<\/p>\n

Question 3.
\nWrite the decimal expansion of \\(\\sqrt{47}\\) to at least two-decimal digits.
\nAnswer:
\nThe number \\(\\sqrt{47}\\) is between 6 and 7 but closer to 7 because 62<\/sup><47<72<\/sup>. In the interval of tenths, the number \\(\\sqrt{47}\\) is between 6.8 and 6.9 because 6.82<\/sup> = 46.24 and 6.92<\/sup> = 47.61. In the interval of hundredths, the number \\(\\sqrt{47}\\) is between 6.85 and 6.86 because 6.852<\/sup> = 46.9225 and 6.862<\/sup> = 47.0596. Therefore, to two-decimal digits, the number \\(\\sqrt{47}\\) is approximately 6.85<\/p>\n

Question 4.
\nWrite the decimal expansion of \\(\\sqrt{46}\\) to at least two-decimal digits.
\nAnswer:
\nThe number \\(\\sqrt{46}\\) is between integers 6 and 7 because 62<\/sup><46<72<\/sup>. Since \\(\\sqrt{46}\\) is closer to 7, I will start checking the tenths intervals between 6.9 and 7. \\(\\sqrt{46}\\) is between 6.7 and 6.8 since 6.72<\/sup> = 44.89 and 6.82<\/sup> = 46.24. Checking the hundredths interval, \\(\\sqrt{46}\\) is between 6.78 and 6.79 since 6.782<\/sup> = 45.9684 and 6.792<\/sup> = 46.1041. The two-decimal approximation \\(\\sqrt{46}\\) is 6.78.<\/p>\n

Question 5.
\nExplain how to improve the accuracy of the decimal expansion of an irrational number.
\nAnswer:
\nIn order to improve the accuracy of the decimal expansion of an irrational number, you must examine increasingly smaller increments on the number line. Specifically, examine increments of decreasing powers of 10. The basic inequality allows us to determine which interval a number is between. We begin by determining which two integers the number lies between and then decreasing the power of 10 to look at the interval of tenths. Again using the basic inequality, we can narrow down the approximation to a specific interval of tenths. Then, we look at the interval of hundredths and use the basic inequality to determine which interval of hundredths the number would lie between. Then, we examine the interval of thousandths. Again, the basic inequality allows us to narrow down the approximation to thousandths. The more intervals we examine, the more accurate the decimal expansion of an irrational number will be.<\/p>\n

Question 6.
\nIs the number \\(\\sqrt{144}\\) rational or irrational? Explain.
\nAnswer:
\nThe number \\(\\sqrt{144}\\) is 12, a rational number.<\/p>\n

Question 7.
\nIs the number \\(0 . \\overline{64}\\) = 0.646464646\u2026 rational or irrational? Explain.
\nAnswer:
\nWe have seen that every number that has a repeating decimal expansion is a fraction; that is, it is a rational number. In this case, 0.646 \u2006 464 \u2006 646\u2026 = \\(\\frac{64}{99}\\), and is therefore a rational number.<\/p>\n

Question 8.
\nHenri computed the first 100 decimal digits of the number \\(\\frac{652}{541}\\) and got 0.650646950092421441774491682070240295748613678373382624768946
\n39556377079482439926062846580406654343807763401109057301294\u2026.
\nHe saw no repeating pattern to the decimal and so concluded that the number is irrational. Do you agree with Henri\u2019s conclusion? If not, what would you say to Henri?
\nAnswer:
\nThe fraction \\(\\frac{352}{541}\\) is certainly a rational number, and so it will have a repeating decimal expansion. One probably has to go beyond 100 decimal places to see the digits repeat.
\n(This decimal actually repeats after the 540th decimal place.)<\/p>\n

Question 9.
\nUse a calculator to determine the decimal expansion of \\(\\sqrt{35}\\). Does the number appear to be rational or irrational? Explain.
\nAnswer:
\nBased on the decimal expansion, the number \\(\\sqrt{35}\\) appears to be irrational. The decimal expansion is infinite and does not appear to have a repeating pattern.<\/p>\n

Question 10.
\nUse a calculator to determine the decimal expansion of \\(\\sqrt{101}\\). Does the number appear to be rational or irrational? Explain.
\nBased on the decimal expansion, the number \\(\\sqrt{101}\\) appears to be irrational. The decimal expansion is infinite and does not appear to have a repeating pattern.<\/p>\n

Question 11.
\nUse a calculator to determine the decimal expansion of \\(\\sqrt{7}\\). Does the number appear to be rational or irrational? Explain.
\nAnswer:
\nBased on the decimal expansion, the number \\(\\sqrt{7}\\) appears to be irrational. The decimal expansion is infinite and does not appear to have a repeating pattern.<\/p>\n

Question 12.
\nUse a calculator to determine the decimal expansion of \\(\\sqrt{8720}\\). Does the number appear to be rational or irrational? Explain.
\nAnswer:
\nBased on the decimal expansion, the number \\(\\sqrt{8720}\\) appears to be irrational. The decimal expansion is infinite and does not appear to have a repeating pattern.<\/p>\n

Question 13.
\nUse a calculator to determine the decimal expansion of \\(\\sqrt{17956}\\). Does the number appear to be rational or irrational? Explain.
\nAnswer:
\nBased on the decimal expansion, the number \\(\\sqrt{17956}\\) is rational because it is equivalent to 134.<\/p>\n

Question 14.
\nSince the number \\(\\frac{3}{5}\\) is rational, must the number (3\/5)2<\/sup>be rational as well? Explain.
\nAnswer:
\nYes, since \\(\\frac{3}{5}\\) is rational it makes sense that (\\(\\frac{3}{5}\\))2<\/sup>would also be rational since (\\(\\frac{3}{5}\\))2<\/sup> = \\(\\frac{9}{25}\\) is a ratio of integers.<\/p>\n

Question 15.
\nIf a number x is rational, must the number x2<\/sup> be rational as well? Explain.
\nAnswer:
\nIf x is rational, then we can write x = \\(\\frac{a}{b}\\) for some integers a and b. This means that x2<\/sup> = \\(\\frac{a^{2}}{b^{2}}\\) and so is necessarily rational as well.<\/p>\n

Question 16.
\nChallenge: Determine the two-decimal digit approximation of the number \\(\\sqrt [ 3 ]{ 9 }\\).
\nAnswer:
\nThe number \\(\\sqrt [ 3 ]{ 9 }\\) is between integers 2 and 3 because 23<\/sup><9<33<\/sup>. Since \\(\\sqrt [ 3 ]{ 9 }\\) is closer to 2, I will start checking the tenths intervals between 2 and 3. \\(\\sqrt [ 3 ]{ 9 }\\) is between 2 and 2.1 since 23<\/sup> = 8 and 2.13<\/sup> = 9.261. Checking the hundredths interval, \\(\\sqrt [ 3 ]{ 9 }\\) is between 2.08 and 2.09 since 2.083<\/sup> = 8.998912 and 2.093<\/sup> = 9.129329. The two-decimal digit approximation \\(\\sqrt [ 3 ]{ 9 }\\) is 2.08.<\/p>\n

Eureka Math Grade 8 Module 7 Lesson 11 Exit Ticket Answer Key<\/h3>\n

Question 1.
\nDetermine the three-decimal digit approximation of the number \\(\\sqrt{17}\\).
\nAnswer:
\nThe number \\(\\sqrt{17}\\) is between integers 4 and 5 because 16<17<25. Since \\(\\sqrt{17}\\) is closer to 4, I will start checking the tenths intervals closer to 4. \\(\\sqrt{17}\\) is between 4.1 and 4.2 since 4.12<\/sup> = 16.81 and 4.22<\/sup> = 17.64. Checking the hundredths interval, \\(\\sqrt{17}\\) is between 4.12 and 4.13 since 4.122<\/sup> = 16.9744 and 4.132<\/sup> = 17.0569. Checking the thousandths interval, \\(\\sqrt{17}\\) is between 4.123 and 4.124 since 4.1232<\/sup> = 16.999129 and 4.1242<\/sup> = 17.007376.
\nThe three-decimal digit approximation is 4.123.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math 8th Grade Module 7 Lesson 11 Answer Key Eureka Math Grade 8 Module 7 Lesson 11 Example Answer Key Example 1. Consider the decimal expansion of . Find the first two values of the decimal expansion using the following fact: If c2 < 3 < d2 for positive numbers c and … 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\/83160"}],"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=83160"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83160\/revisions"}],"predecessor-version":[{"id":137284,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/83160\/revisions\/137284"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=83160"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=83160"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=83160"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}