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Eureka Math Grade 7 Module 3 Lesson 2 Answer Key
Engage ny eureka math 7th grade module 3 lesson 2 answer key, eureka math grade 7 module 3 lesson 2 example answer key.
Example 1. Subtracting Expressions a. Subtract: (40 + 9) – (30 + 2). Answer: The opposite of a sum is the sum of its opposites. Order of operations
40 + 9 + (- (30 + 2)) 40 + 9 + (- 30) + (- 2) 49 + (- 30) + (- 2) 19 + (- 2) 17
(40 + 9)- (30 + 2) (49)- (32) 17
b. Subtract: (3x + 5y – 4) – (4x + 11). Answer: 3x + 5y + ( – 4) + ( – (4x + 11)) Subtraction as adding the opposite 3x + 5y + ( – 4) + ( – 4x) + ( – 11) The opposite of a sum is the sum of its opposites. 3x + ( – 4x) + 5y + ( – 4) + ( – 11) Any order, any grouping – x + 5y + ( – 15) Combining like terms – x + 5y – 15 Subtraction replaces adding the opposite. Have students check the equivalency of the expressions by substituting 2 for x and 6 for y. (3x + 5y – 4) – (4x + 11) (3(2) + 5(6) – 4) – (4(2) + 11) (6 + 30 – 4) – (8 + 11) (36 – 4) – (19) 32 – 19 13 – x + 5y – 15 – (2) + 5(6) – 15 – 2 + 30 + ( – 15) 28 + ( – 15) 13 → When writing the difference as adding the expression’s opposite in Example 1(b), what happens to the grouped terms that are being subtracted? → When the subtraction is changed to addition, every term in the parentheses that follows must be converted to its opposite.
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Example 3. Using Expressions to Solve Problems A stick is x meters long. A string is 4 times as long as the stick. a. Express the length of the string in terms of x. Answer: The length of the stick in meters is x meters, so the string is 4∙x, or 4x, meters long.
b. If the total length of the string and the stick is 15 meters long, how long is the string? Answer: The length of the stick and the string together in meters can be represented by x + 4x, or 5x. If the length of the stick and string together is 15 meters, the length of the stick is 3 meters, and the length of the string is 12 meters.
Example 4. Expressions from Word Problems It costs Margo a processing fee of $3 to rent a storage unit, plus $17 per month to keep her belongings in the unit. Her friend Carissa wants to store a box of her belongings in Margo’s storage unit and tells her that she will pay her $1 toward the processing fee and $3 for every month that she keeps the box in storage. Write an expression in standard form that represents how much Margo will have to pay for the storage unit if Carissa contributes. Then, determine how much Margo will pay if she uses the storage unit for 6 months. Answer: Let m represent the number of months that the storage unit is rented. (17m + 3) – (3m + 1) Original expression 17m + 3 + ( – (3m + 1)) Subtraction as adding the opposite 17m + 3 + ( – 3m) + ( – 1) The opposite of the sum is the sum of its opposites. 17m + ( – 3m) + 3 + ( – 1) Any order, any grouping 14m + 2 Combined like terms This means that Margo will have to pay only $2 of the processing fee and $14 per month that the storage unit is used. 14(6) + 2 84 + 2 86 Margo will pay $86 toward the storage unit rental for 6 months of use.
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Eureka Math Grade 7 Module 3 Lesson 2 Opening Exercise Answer Key
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Eureka Math Grade 7 Module 3 Lesson 2 Problem Set Answer Key
Question 1. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression using x=5.
a. 3x + (2 – 4x) Answer: 3x + (2 – 4x) – x + 2 – 5 + 2 – 3
3(5) + (2 – 4(5)) 15 + (2 + ( – 20)) 15 + ( – 18) – 3
b. 3x + ( – 2 + 4x) Answer: 3x + ( – 2 + 4x) 7x – 2 7(5) – 2 35 – 2 33
3(5) + ( – 2 + 4(5)) 15 + ( – 2 + 20) 15 + 18 33
c. – 3x + (2 + 4x) Answer: – 3x + (2 + 4x) x + 2 5 + 2 7
– 3(5) + (2 + 4(5)) – 15 + (2 + 20) – 15 + 22 7
d. 3x + ( – 2 – 4x) Answer: 3x + ( – 2 – 4x) – x – 2 – 5 – 2 – 7
3(5) + ( – 2 – 4(5)) 15 + ( – 2 + ( – 4(5))) 15 + ( – 2 + ( – 20)) 15 + ( – 22) – 7
e. 3x – (2 + 4x) Answer: 3x – (2 + 4x) – x – 2 – 5 – 2 – 7
3(5) – (2 + 4(5)) 15 – (2 + 20) 15 – 22 15 + ( – 22) – 7
f. 3x – ( – 2 + 4x) Answer: 3x – ( – 2 + 4x) – x + 2 – 5 + 2 – 3
3(5) – ( – 2 + 4(5)) 15 – ( – 2 + 20) 15 – (18) 15 + ( – 18) – 3
g. 3x – ( – 2 – 4x) Answer: 3x – ( – 2 – 4x) 7x + 2 7(5) + 2 35 + 2 37
3(5) – ( – 2 – 4(5)) 15 – ( – 2 + ( – 4(5))) 15 – ( – 2 + ( – 20)) 15 – ( – 22) 15 + 22 37
h. 3x – (2 – 4x) Answer: 3x – (2 – 4x) 7x – 2 7(5) – 2 35 – 2 33
3(5) – (2 – 4(5)) 15 – (2 + ( – 4(5))) 15 – (2 + ( – 20)) 15 – ( – 18) 15 + 18 33
i. – 3x – ( – 2 – 4x) Answer: – 3x – ( – 2 – 4x) x + 2 5 + 2 7
– 3(5) – ( – 2 – 4(5)) – 15 – ( – 2 + ( – 4(5))) – 15 – ( – 2 + ( – 20)) – 15 – ( – 22) – 15 + 22 7
j. In problems (a)–(d) above, what effect does addition have on the terms in parentheses when you removed the parentheses? Answer: By the any grouping property, the terms remained the same with or without the parentheses.
k. In problems (e)–(i), what effect does subtraction have on the terms in parentheses when you removed the parentheses? Answer: The opposite of a sum is the sum of the opposites; each term within the parentheses is changed to its opposite.
Question 2. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating each expression for the given value of the variable. a. 4y – (3 + y); y=2 Answer: 4y – (3 + y) 3y – 3 3(2) – 3 6 – 3 3
4(2) – (3 + 2) 8 – 5 8 + ( – 5) 3
b. (2b + 1) – b; b= – 4 Answer: (2b + 1) – b b + 1 – 4 + 1 – 3
(2( – 4) + 1) – ( – 4) ( – 8 + 1) + 4 ( – 7) + 4 – 3
c. (6c – 4) – (c – 3); c= – 7 Answer: (6c – 4) – (c – 3) 5c – 1 5( – 7) – 1 – 35 – 1 – 36
(6( – 7) – 4) – ( – 7 – 3) ( – 42 – 4) – ( – 10) – 42 + ( – 4) + (10) – 46 + 10 – 36
d. (d + 3d) – ( – d + 2); d=3 Answer: (d + 3d) – ( – d + 2) 5d – 2 5(3) – 2 15 – 2 13
(3 + 3(3)) – ( – 3 + 2) (3 + 9) – ( – 1) 12 + 1 13
e. ( – 5x – 4) – ( – 2 – 5x); x=3 Answer: – 2 ( – 5(3) – 4) – ( – 2 – 5(3)) ( – 15 – 4) – ( – 2 – 15) ( – 19) – ( – 17) ( – 19) + 17 – 2
f. 11f – ( – 2f + 2); f=\(\frac{1}{2}\) 13f – 2 13(\(\frac{1}{2}\) ) – 2 \(\frac{13}{2}\) – 2 6 \(\frac{1}{2}\) – 2 4 \(\frac{1}{2}\)
11(\(\frac{1}{2}\) ) – ( – 2(\(\frac{1}{2}\) ) + 2) 1\(\frac{1}{2}\) – ( – 1 + 2) 1\(\frac{1}{2}\) – 1 1\(\frac{1}{2}\) + ( – \(\frac{2}{2}\) ) \(\frac{9}{2}\) 4 \(\frac{1}{2}\)
g. – 5g + (6g – 4); g= – 2 Answer: – 5g + (6g – 4) g – 4 – 2 – 4 – 6
– 5( – 2) + (6( – 2) – 4) 10 + ( – 12 – 4) 10 + ( – 12 + ( – 4)) 10 + ( – 16) – 6
h. (8h – 1) – (h + 3); h= – 3 Answer: (8h – 1) – (h + 3) 7h – 4 7( – 3) – 4 – 21 – 4 – 25
(8( – 3) – 1) – ( – 3 + 3) ( – 24 – 1) – (0) ( – 25) – 0 – 25
i. (7 + w) – (w + 7); w= – 4 Answer (7 + w) – (w + 7) 0 (7 + ( – 4)) – ( – 4 + 7) 3 – 3 3 + ( – 3) 0
j. (2g + 9h – 5) – (6g – 4h + 2); g= – 2 and h=5 Answer: (2g + 9h – 5) – (6g – 4h + 2) – 4g + 13h – 7 – 4( – 2) + 13(5) – 7 8 + 65 + ( – 7) 73 + ( – 7) 66
(2( – 2) + 9(5) – 5) – (6( – 2) – 4(5) + 2) ( – 4 + 45 – 5) – ( – 12 + ( – 4(5)) + 2) (41 – 5) – ( – 12 + ( – 20) + 2) (41 + ( – 5)) – ( – 32 + 2) 36 – ( – 30) 36 + 30 66
Question 3. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable.
a. – 3(8x); x=\(\frac{1}{4}\) Answer: – 3(8x) – 24x – 24(\(\frac{1}{4}\) ) – 24/4 – 6
– 3(8(\(\frac{1}{4}\) )) – 3(2) – 6
b. 5∙k∙( – 7); k=\(\frac{3}{5}\) Answer: 5∙k∙( – 7) – 35k – 35(\(\frac{3}{5}\) ) – 105/5 – 21
5(\(\frac{3}{5}\) )( – 7) 3( – 7) – 21
c. 2( – 6x)∙2; x=\(\frac{3}{4}\) Answer: 2( – 6x)∙2 – 24x – 24(\(\frac{3}{4}\) ) – \(\frac{72}{4}\) – 18
2( – 6(\(\frac{3}{4}\) ))∙2 2( – 3(\(\frac{3}{2}\) ))∙2 2( – 3)(\(\frac{3}{2}\) )(2) – 6(3) – 18
d. – 3(8x) + 6(4x); x=2 Answer: – 3(8x) + 6(4x) 0
– 3(8(2)) + 6(4(2)) – 3(16) + 6(8) – 48 + 48 0
e. 8(5m) + 2(3m); m= – 2 Answer: 8(5m) + 2(3m) 46m 46( – 2) – 92
8(5( – 2)) + 2(3( – 2)) 8( – 10) + 2( – 6) – 80 + ( – 12) – 92
f. – 6(2v) + 3a(3); v=\(\frac{1}{3}\) ; a=\(\frac{2}{3}\) Answer: – 6(2v) + 3a(3); v=\(\frac{1}{3}\) – 6(2v) + 3a(3) – 12v + 9a – 12(\(\frac{1}{3}\) ) + 9(\(\frac{2}{3}\) ) – 1\(\frac{2}{3}\) + \(\frac{18}{3}\) – 4 + 6 2
– 6(2(\(\frac{1}{3}\) )) + 3(\(\frac{2}{3}\) )(3) – 6(\(\frac{2}{3}\) ) + 2(3) – 4 + 6 2
Question 4. Write each expression in standard form. Verify that your expression is equivalent to the one given by evaluating both expressions for the given value of the variable
a. 8x ÷ 2; x= – \(\frac{1}{4}\) Answer: 8x ÷ 2 4x 4( – \(\frac{1}{4}\) ) – 1
8( – \(\frac{1}{4}\) ) ÷ 2 – 2 ÷ 2 – 1
b. 18w ÷ 6; w=6 Answer: 18w ÷ 6 3w 3(6) 18
18(6) ÷ 6 108 ÷ 6 18
c. 25r ÷ 5r; r= – 2 Answer: 25r ÷ 5r 5
25( – 2) ÷ (5( – 2)) – 50 ÷ ( – 10) 5
d. 33y ÷ 11y; y= – 2 Answer: 33y ÷ 11y 3
33( – 2) ÷ (11( – 2)) ( – 66) ÷ ( – 22) 3
e. 56k ÷ 2k; k=3 Answer: 56k ÷ 2k 28
56(3) ÷ (2(3)) 168 ÷ 6 28
f. 24xy ÷ 6y; x= – 2;y=3 Answer: 24xy ÷ 6y 4x 4( – 2) – 8
24( – 2)(3) ÷ (6(3)) – 48(3) ÷ 18 – 144 ÷ 18 – 8
Question 5. For each problem (a)–(g), write an expression in standard form. a. Find the sum of – 3x and 8x. Answer: – 3x + 8x 5x
b. Find the sum of – 7g and 4g + 2. Answer: – 7g + (4g + 2) – 3g + 2
c. Find the difference when 6h is subtracted from 2h – 4. Answer: (2h – 4) – 6h – 4h – 4
d. Find the difference when – 3n – 7 is subtracted from n + 4. Answer: (n + 4) – ( – 3n – 7) 4n + 11
e. Find the result when 13v + 2 is subtracted from 11 + 5v. Answer: (11 + 5v) – (13v + 2) – 8v + 9
f. Find the result when – 18m – 4 is added to 4m – 14. Answer: (4m – 14) + ( – 18m – 4) – 14m – 18
g. What is the result when – 2x + 9 is taken away from – 7x + 2? Answer: ( – 7x + 2) – ( – 2x + 9) – 5x – 7
Question 6. Marty and Stewart are stuffing envelopes with index cards. They are putting x index cards in each envelope. When they are finished, Marty has 15 stuffed envelopes and 4 extra index cards, and Stewart has 12 stuffed envelopes and 6 extra index cards. Write an expression in standard form that represents the number of index cards the boys started with. Explain what your expression means. Answer: They inserted the same number of index cards in each envelope, but that number is unknown, x. An expression that represents Marty’s index cards is 15x + 4 because he had 15 envelopes and 4 cards left over. An expression that represents Stewart’s index cards is 12x + 6 because he had 12 envelopes and 6 left over cards. Their total number of cards together would be: 15x + 4 + 12x + 6 15x + 12x + 4 + 6 27x + 10 This means that altogether, they have 27 envelopes with x index cards in each, plus another 10 leftover index cards.
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Eureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions—Round 1 Answer Key
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Eureka Math Grade 7 Module 3 Lesson 2 Generating Equivalent Expressions—Round 2 Answer Key
Directions: Write each as an equivalent expression in standard form as quickly and as accurately as possible within the allotted time.
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Eureka Math Grade 7 Module 3 Lesson 2 Exit Ticket Answer Key
Question 1. Write the expression in standard form. Answer: (4f – 3 + 2g) – ( – 4g + 2) 4f + ( – 3) + 2g + ( – ( – 4g + 2)) Subtraction as adding the opposite 4f + ( – 3) + 2g + 4g + ( – 2) The opposite of a sum is the sum of its opposites. 4f + 2g + 4g + ( – 3) + ( – 2) Any order, any grouping 4f + 6g + ( – 5) Combined like terms 4f + 6g – 5 Subtraction as adding the opposite
Question 2. Find the result when 5m + 2 is subtracted from 9m. Answer: 9m – (5m + 2) Original expression 9m + ( – (5m + 2)) Subtraction as adding the opposite 9m + ( – 5m) + ( – 2) The opposite of a sum is the sum of its opposites. 4m + ( – 2) Combined like terms 4m – 2 Subtraction as adding the opposite
Question 3. Write the expression in standard form. 27h ÷ 3h Answer: 27h∙\(\frac{1}{3 h}\) Multiplying by the reciprocal \(\frac{27h}{3 h}\) Multiplication \(\frac{27}{3}\)∙\(\frac{h}{h}\) Any order, any grouping 9∙1 9
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The homework pages are from the "full Module" PDF on this web page:https://www.engageny.org/resource/grade-3-mathematics-module-2
Engage NY Eureka Math 3rd Grade Module 2 Lesson 7 Answer Key Eureka Math Grade 3 Module 2 Lesson 7 Problem Set Answer Key Work with a partner. Use the corresponding weights to estimate the
Eureka Math Grade 3 Module 3 Lesson 7 Homework Answer Key. Question 1. Match the words on the arrow to the correct equation on the target. Answer: n x 7 = 21 = A number times 7 equals 21. 7 x n = 42 = 7 times a number equals 42. 63 / n = 9 = 63 divided by a number equals 9.
Learn third grade math aligned to the Eureka Math/EngageNY curriculum—fractions, area, arithmetic, and so much more.
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114. 213. 272. 190. 182. Range = 158 dollars. Standard Deviation (sample) = 52.9 dollars. Variance (sample) = 2796.8 dollars^2. Question = The measures of variation are not very useful because when searching for a room, low prices, location, and good accommodations are more important than the amount of variation in the area.
Fill in the blanks using your knowledge of place value units and basic facts.
Learn how to multiply two-digit numbers with this engaging video lesson. Watch examples, practice problems, and tips from a math teacher.
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Now, with expert-verified solutions from Core Connections Geometry 2nd Edition, you'll learn how to solve your toughest homework problems. Our resource for Core Connections Geometry includes answers to chapter exercises, as well as detailed information to walk you through the process step by step.
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Need extra homework problems? This worksheet includes 5 similar homework problems from Lesson 7.3.2, each with a QR code/ hyperlink to a specific help video of that problem for students to watch. Print out this worksheet or push out a copy through Google Classroom. Use this worksheet to assess your...
Name Place Value of Decimals Write the value of the underlined digit. Lesson 3.2 COMMON CORE STANDARD CC.5.NBT.3a Understand the place value system. 3. or 0.3 2.704 4 thousandths, or 0.004 7.258 5 hundredths, or 0.05 8.027 7 thousandths, or 0.007 4.
Multiplication and Division with Units of 0, 1, 6-9, and Multiples of 10. An outline of learning goals, key ideas, pacing suggestions, and more! Files for printing or for projecting on the screen. Application Problems with space for student work and TTS QR Codes.
Worksheets are A story of units lesson 7 homework 4 3, A story of units lesson 1 template, Eureka math homework helper 20152016 grade 5 module 1, Grade 2 module 1, Eureka math homework helper 20152016 grade 4 module 1, A story of units lesson 32 homework 3 7, Eureka math homework helper 20152016 grade 3 module 1, A story of units lesson 1 homework.
The pages I used are in the full module PDF at this web page:https://www.engageny.org/resource/grade-3-mathematics-module-2
Engage NY Eureka Math 7th Grade Module 3 Lesson 2 Answer Key Eureka Math Grade 7 Module 3 Lesson 2 Example Answer Key Example 1. Subtracting Expressions a. Subtract: (40 + 9) - (30 +
Table 3.2.6 and Table 3.2.7 define functions. In both, each input value corresponds to exactly one output value. Table 3.2.8 does not define a function because the input value of 5 corresponds to two different output values.
The homework pages are from the "full Module" PDF on this web page:https://www.engageny.org/resource/grade-3-mathematics-module-2
One-step inequalities. Solve for x . Your answer must be simplified. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.
Illustrative Mathematics Grade 7 Open Up Resources OURUnit 2 Lesson 3More resources available at: mathhelp.cusd.com