{"id":89464,"date":"2022-05-26T16:40:10","date_gmt":"2022-05-26T11:10:10","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=89464"},"modified":"2022-05-10T14:20:23","modified_gmt":"2022-05-10T08:50:23","slug":"eureka-math-algebra-1-module-1-end-of-module-assessment","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/eureka-math-algebra-1-module-1-end-of-module-assessment\/","title":{"rendered":"Eureka Math Algebra 1 Module 1 End of Module Assessment Answer Key"},"content":{"rendered":"

Engage NY Eureka Math Algebra 1 Module 1 End of Module Assessment Answer Key<\/h2>\n

Eureka Math Albegra 1 Module 1 End of Module Assessment Task Answer Key<\/h3>\n

Question 1.
\nSolve the following equations for x. Write your answer in set notation.<\/p>\n

a. 3x-5=16
\n3x = 21 Solution set: {7}
\nx = 7<\/p>\n

b. 3(x+3)-5=16
\nAnswer:
\n3x + 9 – 5 = 16 Solution set: {4}
\n3x = 12
\nx = 4<\/p>\n

c. 3(2x-3)-5=16
\nAnswer:
\n6x – 9 – 5 = 16 Solution set: {5}
\n6x – 14 = 16
\n6x = 30
\nx = 5<\/p>\n

d. 6(x+3)-10=32
\nAnswer:
\n6x + 18 – 10 = 32 Solution set: {4}
\n6x = 24
\nx = 4<\/p>\n

e. Which two equations above have the same solution set? Write a sentence explaining how the properties of equality can be used to determine the pair without having to find the solution set for each.
\nAnswer:
\nProblems (b) and (d) have the same solution set. The expressions on each side of the equal sign for (d) are twice those for (b). So, if (left side) = (right side) is true for only some x-values, then 2(left side) = 2(right side) will be true for exactly the same x-values. Or simply, applying the multiplicative property of equality does not change the solution set.<\/p>\n

Question 2.
\nLet c and d be real numbers.
\na. If c=42+d is true, then which is greater: c or d or are you not able to tell? Explain how you know your choice is correct.
\nAnswer:
\nc must be greater because c is always 42 more than d.<\/p>\n

b. If c=42-d is true, then which is greater: c or d or are you not able to tell? Explain how you know your choice is correct.
\nAnswer:
\nThere is no way to tell. We only know that the sum of c and d is 42. If d were 10, c would be 32 and, therefore, greater than d. But if d were 40, c would be 2 and, therefore, less than d.<\/p>\n

Question 3.
\nIf a<0 and c>b, circle the expression that is greater:
\na(b – c) or a(c – b)
\nAnswer:
\n\"Eureka<\/p>\n

Use the properties of inequalities to explain your choice.
\nAnswer:
\nSince c > b, it follows that 0 > b – c, and since a < 0, a is negative, and the product of two negatives will be a positive. Since c > b, it follows that c – b > 0. so (c – b) is positive. And since a is negative, the product of a \u2219 (c – b) < a \u2219 (b – c).<\/p>\n

Question 4.
\nSolve for x in each of the equations or inequalities below and name the property and\/or properties used:<\/p>\n

a. \\(\\frac{3}{4}\\) x=9
\nx = 9\u2219 (\\(\\frac{4}{3}\\) ) Multiplication property of equality
\nx = 12<\/p>\n

b. 10+3x=5x
\nAnswer:
\n10 = 2x Addition property of equality
\n5 = x Multiplication property of equality<\/p>\n

c. a+x=b
\nAnswer:
\nx = b-a Addition property of equality<\/p>\n

d. cx=d
\nAnswer:
\nx = \\(\\frac{d}{c}\\) ,c \u2260 0 Multiplication property of equality<\/p>\n

e. \\(\\frac{1}{2}\\) x-g<m
\nAnswer:
\n\\(\\frac{1}{2}\\) x < m + g Addition property of equality
\nx < 2 \u2219 (m + g ) Multiplication property of equality f. q+5x=7x-r q + r = 2x Addition property of equality \\(\\frac{(q+r)}{2}\\) = x Multiplication property of equality g. \\(\\frac{3}{4}\\) (x+2)=6(x+12) Answer: 3 \u2219 (x + 2) = 24 \u2219 (x + 12) Multiplication property of equality 3x + 6 = 24x + 288 Distributive property –\\(\\frac{282}{21}\\) = x Addition property of equality and multiplication –\\(\\frac{94}{7}\\) = x Property of equality –\\(\\frac{94}{7}\\) = x h. 3(5 \u2013 5x)> 5x
\nAnswer:
\n15-15x > 5x Distributive property
\n15 > 20x Addition property of inequality
\n\\(\\frac{3}{4}\\) > x Multiplication property of equality<\/p>\n

Question 5.
\nThe equation, 3x+4=5x-4, has the solution set {4}.
\na. Explain why the equation, (3x+4)+4=(5x-4)+4, also has the solution set {4}.
\nAnswer:
\nSince the new equation can be created by applying the addition property of equality, the solution set does not change.
\nOR
\nEach side of this equation is 4 more than the sides of the original equation. Whatever value(s) make 3x + 4 = 5x – 4 true would also make 4 more than 3x + 4 equal to 4 more than 5x – 4.<\/p>\n

b. In part (a), the expression (3x+4)+4 is equivalent to the expression 3x+8. What is the definition of equivalent algebraic expressions? Describe why changing an expression on one side of an equation to an equivalent expression leaves the solution set unchanged?
\nAnswer:
\nAlgebraic expressions are equivalent if (possibly repeated) use of the distributive, associative, and commutative properties and\/or the properties of rational exponents can be applied to one expression to convert it to the other expression.
\nWhen two expressions are equivalent, assigning the same value to x in both expressions will give an equivalent numerical expression, which then evaluates to the same number. Therefore, changing the expression to something equivalent will not change the truth value of the equation once values are assigned to x.<\/p>\n

c. When we square both sides of the original equation, we get the following new equation:
\n(3x+4)2<\/sup>=(5x-4)2<\/sup>.
\nShow that 4 is still a solution to the new equation. Show that 0 is also a solution to the new equation but is not a solution to the original equation. Write a sentence that describes how the solution set to an equation may change when both sides of the equation are squared.
\nAnswer:
\n(3 \u2219 4 + 4 )2<\/sup> = (5 \u2219 4 – 4 )2<\/sup> gives \u301616 \u30172<\/sup> = \u301616 \u3017^6 , which is true.
\n(3 \u2219 0 + 4 )2<\/sup> = (5 \u2219 0 – 4 )2<\/sup> gives 4 2<\/sup> = (-4 )2<\/sup> , which is true.
\nBut, (3 \u2219 0 + 4 ) = (5 \u2219 0 -4 ) gives 4 = -4 , which is false.
\nWhen both sides are squared, you might introduce new numbers to the solution set because statements like 4 =-4 are false, but statements like 4 2<\/sup> = (-4 )2<\/sup> are true.<\/p>\n

d. When we replace x by x2<\/sup> in the original equation, we get the following new equation:
\n3x2<\/sup>+4=5x2<\/sup>-4.
\nUse the fact that the solution set to the original equation is {4} to find the solution set to this new equation.
\nAnswer:
\nSince the original equation 3x + 4 = 5x – 4 was true when x = 4, the new equation 3 x 2<\/sup> + 4 = 5 x 2<\/sup> -4 should be true when x 2<\/sup> = 4. And, x 2<\/sup> = 4 when x = 2 , so the solution set to the new equation is {-2,2 }.<\/p>\n

Question 6.
\nThe Zonda Information and Telephone Company calculates a customer\u2019s total monthly cell phone charge using the formula,
\nC=(b+rm)(1+t),
\nwhere C is the total cell phone charge, b is a basic monthly fee, r is the rate per minute, m is the number of minutes used that month, and t is the tax rate.
\nSolve for m, the number of minutes the customer used that month.
\nAnswer:
\nC = b + bt + rm + rmt
\nC-b-bt = m \u2219 (r + rt )
\n\\(\\frac{c-b-b t}{r+r t}\\) = m t \u2260 -1
\nr \u2260 0<\/p>\n

Question 7.
\nStudents and adults purchased tickets for a recent basketball playoff game. All tickets were sold at the ticket booth\u2014season passes, discounts, etc., were not allowed.<\/p>\n

Student tickets cost $5 each, and adult tickets cost $10 each. A total of $4,500 was collected.
\n700 tickets were sold.<\/p>\n

a. Write a system of equations that can be used to find the number of student tickets, s, and the number of adult tickets, a, that were sold at the playoff game.
\nAnswer:
\n5s + 10a = 4500
\ns + a = 700<\/p>\n

b. Assuming that the number of students and adults attending would not change, how much more money could have been collected at the playoff game if the ticket booth charged students and adults the same price of $10 per ticket?
\nAnswer:
\n700 \u00d7 $10 = $7000
\n$7000-$4500 = $2500 more<\/p>\n

c. Assuming that the number of students and adults attending would not change, how much more money could have been collected at the playoff game if the student price was kept at $5 per ticket and adults were charged $15 per ticket instead of $10?
\nAnswer:
\nFirst solve for a and s
\n5s + 10a = 4500
\n-5s-5a = -3500
\n5a = 1000
\na = 200
\ns = 500<\/p>\n

$5 \u2219 (500) + $15 \u2219 (200) = $5500
\n$1,000 more
\nOR
\n$5 more per adult ticket (200 \u2219 $5 = $1000 more)<\/p>\n

Question 8.
\nAlexus is modeling the growth of bacteria for an experiment in science. She assumes that there are B bacteria in a Petri dish at 12:00 noon. In reality, each bacterium in the Petri dish subdivides into two new bacteria approximately every 20 minutes. However, for the purposes of the model, Alexus assumes that each bacterium subdivides into two new bacteria exactly every 20 minutes.<\/p>\n

a. Create a table that shows the total number of bacteria in the Petri dish at 1\/3 hour intervals for 2 hours starting with time 0 to represent 12:00 noon.
\nAnswer:
\n\"Eureka<\/p>\n

b. Write an equation that describes the relationship between total number of bacteria T and time h in hours, assuming there are B bacteria in the Petri dish at h=0.
\nAnswer:
\nT = B \u2219 (2 )3h<\/sup> or T = B \u2219 8h<\/sup><\/p>\n

c. If Alexus starts with 100 bacteria in the Petri dish, draw a graph that displays the total number of bacteria with respect to time from 12:00 noon (h=0) to 4:00 p.m. (h=4). Label points on your graph at time h=0,1,2,3,4.
\nAnswer:
\n\"Eureka<\/p>\n

d. For her experiment, Alexus plans to add an anti-bacterial chemical to the Petri dish at 4:00 p.m. that is supposed to kill 99.9% of the bacteria instantaneously. If she started with 100 bacteria at
\n12:00 noon, how many live bacteria might Alexus expect to find in the Petri dish right after she adds the anti-bacterial chemical?
\nAnswer:
\n(1 -0.999 ) \u2219 409 600 = 409.6
\nabout 410 live bacteria<\/p>\n

Question 9.
\nJack is 27 years older than Susan. In 5 years time, he will be 4 times as old as she is.<\/p>\n

a. Find the present ages of Jack and Susan.
\nAnswer:
\nJ = S + 27
\nJ + 5 = 4 \u2219 (S + 5 )<\/p>\n

S + 27 + 5 = 4S + 20
\nS + 32 = 4S + 20
\n12 = 3S
\nS = 4<\/p>\n

J = 4 + 27
\nJ = 31<\/p>\n

Jack is 31 and Susan is 4.<\/p>\n

b. What calculations would you do to check if your answer is correct?
\nAnswer:
\nIs Jack\u2019s age \u2013 Susan\u2019s age = 27?
\nAdd 5 years to Jack\u2019s and Susan\u2019s ages, and see if that makes Jack 4 times as old as Susan.<\/p>\n

Question 10.
\na. Find the product: (x2<\/sup>-x+1)(2x2<\/sup>+3x+2)
\nAnswer:
\n2 x4<\/sup> + 3 x3<\/sup> + 2 x 2<\/sup> -2 x3<\/sup> -3 x 2<\/sup> -2x + 2 x 2<\/sup> + 3x + 2
\n2 x4<\/sup> + x3<\/sup> + x 2<\/sup> + x + 2<\/p>\n

b. Use the results of part (a) to factor 21,112 as a product of a two-digit number and a three-digit number.
\nAnswer:
\n(100 -10 + 1 ) \u2219 (200 + 30 + 2 )
\n(91 ) \u2219 (232 )<\/p>\n

Question 11.
\nConsider the following system of equations with the solution x=3, y=4.
\nEquation A1: y=x+1
\nEquation A2: y=-2x+10
\n\"Eureka<\/p>\n

a. Write a unique system of two linear equations with the same solution set. This time make both linear equations have positive slope.
\nEquation B1: _______
\nAnswer:
\ny = \\(\\frac{4}{3}\\)x
\nEquation B2: ________
\nAnswer:
\ny = x + 1<\/p>\n

b. The following system of equations was obtained from the original system by adding a multiple of equation A2 to equation A1.<\/p>\n

Equation C1: y=x+1<\/p>\n

Equation C2: 3y=-3x+21<\/p>\n

What multiple of A2 was added to A1?
\nAnswer:
\n2 times A2 was added to A1.<\/p>\n

c. What is the solution to the system given in part (b)?
\nAnswer:
\n(3,4)<\/p>\n

d. For any real number m, the line y=m(x-3)+4 passes through the point (3,4).<\/p>\n

Is it certain then that the system of equations:<\/p>\n

Equation D1: y=x+1<\/p>\n

Equation D2: y=m(x-3)+4<\/p>\n

has only the solution x=3, y=4? Explain.
\nAnswer:
\nNo. If m = 1, then the two lines have the same slope. Both lines pass through the point (3,4), and the lines are parallel; therefore, they coincide. There are infinite solutions. The solution set is all the points on the line. Any other nonzero value of m would create a system with the only solution of (3,4).<\/p>\n

Question 12.
\nThe local theater in Jamie\u2019s home town has a maximum capacity of 160 people. Jamie shared with Venus the following graph and said that the shaded region represented all the possible combinations of adult and child tickets that could be sold for one show.
\n\"Eureka
\na. Venus objected and said there was more than one reason that Jamie\u2019s thinking was flawed. What reasons could Venus be thinking of?
\nAnswer:
\n1. The graph implies that the number of tickets sold could be a fractional amount, but really it only makes sense to sell whole number tickets. x and y must be whole numbers.
\n2. The graph also shows that negative ticket amounts could be sold, which does not make sense.<\/p>\n

b. Use equations, inequalities, graphs, and\/or words to describe for Jamie the set of all possible combinations of adult and child tickets that could be sold for one show.
\nAnswer:
\nThe system would be \"Eureka where a and c are whole numbers.
\n\"Eureka<\/p>\n

c. The theater charges $9 for each adult ticket and $6 for each child ticket. The theater sold 144 tickets for the first showing of the new release. The total money collected from ticket sales for that show was $1,164. Write a system of equations that could be used to find the number of child tickets and the number of adult tickets sold, and solve the system algebraically. Summarize your findings using the context of the problem.
\nAnswer:
\na: the number of adult tickets sold (must be a whole number)
\nc: the number of child tickets sold (must be a whole number)
\n9a + 6c = 1164
\na + c = 144<\/p>\n

9a + 6c = 1164
\n-6a-6c = -864
\n3a = 300
\na = 100,c = 44<\/p>\n

In all, 100 adult tickets and 44 child tickets were sold.<\/p>\n","protected":false},"excerpt":{"rendered":"

Engage NY Eureka Math Algebra 1 Module 1 End of Module Assessment Answer Key Eureka Math Albegra 1 Module 1 End of Module Assessment Task Answer Key Question 1. Solve the following equations for x. Write your answer in set notation. a. 3x-5=16 3x = 21 Solution set: {7} x = 7 b. 3(x+3)-5=16 Answer: … 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\/89464"}],"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=89464"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/89464\/revisions"}],"predecessor-version":[{"id":136980,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/89464\/revisions\/136980"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=89464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=89464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=89464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}