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Algebra 1 Module 1 Answer Key

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Eureka Math Algebra 1 Module 1 Lesson 2 Example Answer Key

General Mathematics Quarter 1 – Module 1 Answer Key

Exploratory ChallengePlot a graphical representation of change in elevation over time for the following graphing story. It is a video of a man jumping from 36 ft. above ground into 1 ft. of water.http://www.youtube.com/watch?v=ZCFBC8aXz-g orhttp://youtu.be/ZCFBC8aXz-g

Allow students to analyze the full video, initially giving them as much control as possible in choosing when to pause the video, play it half speed, and so on. Perhaps conduct this part of a class in a computer lab or display the video on an interactive white board. Then direct students to investigate one of the two portions of the clip below, depending on the level of technology you have available:Portion 1: From time 32 sec. to time 33 sec. Use this to get data for drawing the graph .Portion 2: From time 48 sec. to time 53 sec. Use this for a slow-motion version that can be analyzed easily via YouTube.

Example 2.The table below gives the area of a square with sides of whole number lengths. Have students plot the points in the table on a graph and draw the curve that goes through the points.On the same graph, reflect the curve across the y-axis. This graph is an example of a graph of a quadratic function.

Bring up or ask: On the graph, what do the points between the plotted points from the table represent? Areas of squares with non-whole number side lengths.

Eureka Math Algebra 1 Module 1 Lesson 11 Exit Ticket Answer Key

Question 1.Here is the graphical representation of a set of real numbers:a. Describe this set of real numbers in words.Answer:The set of all real numbers less than or equal to two.

b. Describe this set of real numbers in set notation.Answer:

c. Write an equation or an inequality that has the set above as its solution set.Answer:w-7-5

Question 2.Indicate whether each of the following equations is sure to have a solution set of all real numbers. Explain your answers for each.a. 3=3x+1Answer:No. The two algebraic expressions are not equivalent.

b. x+2=2+xAnswer:Yes. The two expressions are algebraically equivalent by application of the commutative property.

c. 4x=4x+4x2Answer:Yes. The two expressions are algebraically equivalent by application of the distributive property and the commutative property.

d. 3x=72×3

Eureka Math Algebra 1 Module 1 Lesson 6 Problem Set Answer Key

Question 1.Insert parentheses to make each statement true.a. 2+3×42+1=81Answer:×42+1=81

b. 2+3×42+1=85Answer:

c. 2+3×42+1=51Answer:2++1=51

d. 2+3×42+1=53Answer:2+3×=53

Question 2.Using starting symbols of w, q, 2, and -2, which of the following expressions will NOT appear when following the rules of the game played in Exercise 3?a. 7w+3q+e. -2w+2Answer:The expressions in parts and cannot be obtained by following the rules of the game played in Exercise 3. The expression in part appears as w+w+2+2+2, which is equivalent to 2w+6.

Question 3.Luke wants to play the 4-number game with the numbers 1, 2, 3, and 4 and the operations of addition, multiplication, AND subtraction.Leoni responds, Or we just could play the 4-number game with just the operations of addition and multiplication, but now with the numbers -1, -2, -3, -4, 1, 2, 3, and 4 instead.What observation is Leoni trying to point out to Luke?Answer:Subtraction can be viewed as the addition of a negative ). By introducing negative integers, we need not consider subtraction as a new operation.

Question 4.Consider the expression: .a. Draw a picture to represent the expression.Answer:

b. Write an equivalent expression by applying the distributive property.Answer:= x2 y+5xy+6y+x2+5x+6

b. Jill, when she saw the picture, highlighted eight rectangles and squares to conclude:2=x2+4ax+4a2.Which eight rectangles and squares did she highlight?Answer:

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Eureka Math Algebra 1 Module 1 Lesson 5 Example Answer Key

Example 1.Consider the story:Maya and Earl live at opposite ends of the hallway in their apartment building. Their doors are 50 ft. apart. Each starts at his or her own door and walks at a steady pace toward each other and stops when they meet.What would their graphing stories look like if we put them on the same graph? When the two people meet in the hallway, what would be happening on the graph? Sketch a graph that shows their distance from Mayas door.

Eureka Math Algebra 2 Module 1 Lesson 11 Exercise Answer Key

Mathemitics Circulum Lesson 15 Answer Key / Eureka Math Grade 3 Module ...

Exercise 1.Find the solutions of = 0.Answer:The solutions to are the solutions of x2 9 = 0 combined with the solutions of x2 16 = 0.These solutions are 3, 3, 4, and 4.

Exercise 2.Find the zeros of the following polynomial functions, with their multiplicities.a. f = Answer:

p = 3 6 10The degree of p is 20.

Exercise 4.Is there more than one polynomial function that has the same zeros and multiplicities as the one you found in Exercise 3?Answer:Yes. Consider = 3 6 10. Since there are no real solutions to x2 + 5 = 0, adding this factor does not produce a new zero. Thus p and q have the same zeros and multiplicities but are different functions.

Exercise 5.Can you find a rule that relates the multiplicities of the zeros to the degree of the polynomial function?Answer:Yes. If p is a polynomial function of degree n, then the sum of the multiplicities of all of the zeros is less than or equal to n. If p can be factored into linear terms, then the sum of the multiplicities of all of the zeros is exactly equal to n.

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Eureka Math Algebra 1 Module 1 Lesson 1 Exit Ticket Answer Key

Students may describe the intervals in words. Do not worry about the endpoints of the intervals in this lesson.

The graph in the Exploratory Challenge is made by combining pieces of nine linear functions . Each linear function is defined over an interval of time, represented on the horizontal axis. List those nine time intervals.Answer:Between 0 and 3 sec. between 3 and 5.5 sec. between 5.5 and 7 sec. between 7 and 8.5 sec. between 8.5 and 9 sec. between 9 and 11 sec. between 11 and 12.7 sec. between 12.7 and 13 sec. and 13 sec. onward.

Eureka Math Algebra 2 Module 1 Lesson 11 Problem Set Answer Key

For Problems 1 4, find all solutions to the given equations.

Question 1. = 0Answer:

Find the zeros with multiplicity for the function p = .Answer:We can factor p to give p = x3 = x3 2 .Then, 0 is a zero of multiplicity 3, 2 is a zero of multiplicity 1, and 2 is a zero of multiplicity 2.

Question 7.Find two different polynomial functions that have zeros at 1, 3, and 5 of multiplicity 1.Answer:p = and q =

Question 8.Find two different polynomial functions that have a zero at 2 of multIplicity 5 and a zero at 4 of multiplIcity 3.Answer:p = 5 3 and q = 53

Question 9.Find three solutions to the equation = 0.Answer:From Lesson 6, we know that is a factor of , so three solutions are 3, 3, and 2.

Question 10.Find two solutions to the equation = 0.Answer:From Lesson 6, we know that is a factor of , and is a factor of , so two solutions are 1 and 4.

Question 11.If p, q, r, s are nonzero numbers, find the solutions to the equation = 0 in terms of p, q, r, s.Answer:Setting each factor equal to zero gives solutions \ and \.

Use the identity a2 b2 = to solve the equations given in Problems 12 13.

Question 12.2 = 2Answer:Using algebra, we have 2 2 = 0. Applying the difference of squares formula, we have ) + ) = 0. Combining like terms gives = 0, so the solutions are \ and \.

Question 13.2 = 2Answer:Using algebra, we have 2 2 = 0. Then ) + ) = 0,so we have = 0. Thus the solutions are \ and 3.

b. Evaluate P.

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Eureka Math Algebra 2 Module 1 Lesson 11 Example Answer Key

Example 1.Suppose we know that the polynomial equation 4×3 12×2 + 3x + 5 = 0 has three real solutions and that one of the factors of 4×3 12×2 + 3x + 5 is . How can we find all three solutions to the given equation?Answer:Steer the discussion to help students conjecture that the polynomial 4×3 12×2 + 3x + 5 must be the product of and some quadratic polynomial.

Since is a factor, and we know how to divide polynomials, we can find the quadratic polynomial by dividing:\ = 4×2 8x 5.

Now we know that 4×3 12×2 + 3x + 5 = , and we also know that 4×2 8x 5 is a quadratic polynomial that has linear factors and .

Therefore, 4×3 12×2 + 3x + 5 = 0 has the same solutions as = 0, which has the same solutions as = 0.In this factored form, the solutions of f = 0 are readily apparent: \, 1, and \.

Eureka Math Algebra 1 Module 1 Lesson 6 Exercise Answer Key

Mathematics 7 Quarter 1 – Module 1 ANSWER KEY Part1 of 4

Exercise 1.The following is known as the 4-number game. It challenges students to write each positive integer as a combination of the digits 1, 2, 3, and 4 each used at most once, combined via the operations of addition and multiplication only, as well as grouping symbols. For example, 24 can be expressed as . Students may use parentheses or not, at their own discretion . Digits may not be juxtaposed to represent larger whole numbers, so the numerals 1 and 2 cannot be used to create the number 12 for instance.

Play the 4-number game as a competition within pairs. Give students three minutes to express the longest list of numbers they can, each written in terms of the digits 1, 2, 3, and 4. Students may want to use small dry erase boards to play this game or pencil and paper. Optionally, consider splitting up the tasks and assign them to different groups.

Below are some sample expressions students may build and a suggested structure for displaying possible expressions on the board as students call out what they have created.

When reviewing the game, it is likely that students will have different expressions for the same number . Share alternative expressions on the board, and discuss as a class the validity of the expressions.

Exercises

Exercise 3.Define the rules of a game as follows:a. Begin by choosing an initial set of symbols, variable or numeric, as a starting set of expressions.Answer:3, x, y, and a

Draw a picture to represent the expression ×.Answer:

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Eureka Math Algebra 2 Module 1 Lesson 11 Exit Ticket Answer Key

Suppose that a polynomial function p can be factored into seven factors: , , and 5 factors of . What are its zeros with multiplicity, and what Is the degree of the polynomial? Explain how you know.Answer:Zeros: 3 with multiplicity 1 1 with multiplicity 1 2 with multiplicity 5The polynomial has degree seven. There are seven linear factors as given above, so p = 5.If the factors were multiplied out, the leading term would be x7, so the degree of p is 7.

Eureka Math Algebra 1 Module 1 Lesson 11 Problem Set Answer Key

For each solution set graphed below, describe the solution set in words, describe the solution set in set notation, and write an equation or an inequality that has the given solution set.

Question 1.a. The set of all real numbers equal to 5b.

\+\=\Answer:No. The solution set is . If we changed it to \+\=8x/15, it would have a solution set of all real numbers.

Question 23.×2 +2×3+3×4=6x9Answer:No. Neither the coefficients nor the exponents are added correctly. One way it could have a solution set of all real numbers would be x2 +2×3+3×4=x2 .

Question 24.×3+4×2 +4x=x2Answer:Yes. Distributive.

Question 25.Solve for w: \2. Describe the solution set in set notation.Answer:\)}

Question 26.Edwina has two sticks: one 2 yards long and the other 2 meters long. She is going to use them, with a third stick of some positive length, to make a triangle. She has decided to measure the length of the third stick in units of feet.a. What is the solution set to the statement: Sticks of lengths 2 yards, 2 meters, and L feet make a triangle? Describe the solution set in words and through a graphical representation.Answer:One meter is equivalent, to two decimal places, to 3.28 feet. We have that L must be a positive length greater than 0.56 feet and less than 12.56 feet. Within these values, the sum of any two sides will be greater than the third side.

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Eureka Math Algebra 1 Module 1 Lesson 11 Exercise Answer Key

Exercise 1.Solve for a: a2 =-25. Present the solution set in words, in set notation, and graphically.Answer:IN WORDS: The solution set to this equation is the empty set. There are no real values to assign to a to make the equation true.IN SET NOTATION: The solution set is .IN A GRAPHICAL REPRESENTATION: The solution set is

Exercise 2.Depict the solution set of 7+p=12 in words, in set notation, and graphically.Answer:IN WORDS: 7+p=12 has the solution p=5.IN SET NOTATION: The solution set is .IN A GRAPHICAL REPRESENTATION:

Exercise 3.Solve \=1 for x over the set of all nonzero real numbers. Describe the solution set in words, in set notation, and graphically.Answer:IN WORDS: The solution set is the set of all nonzero real numbers.IN SET NOTATION: IN A GRAPHICAL REPRESENTATION:

Exercise 4Solve for : +2 =. Describe carefully the reasoning that justifies your solution. Describe the solution set in words, in set notation, and graphically.Answer:IN WORDS: By the distributive property, we have +2 =. This is a true numerical statement no matter what value we assign to . By the commutative property of addition, we have +2 =, which is a true numerical statement no matter what real value we assign to .The solution set is the set of all real numbers.IN SET NOTATION: The solution set is IN GRAPHICAL REPRESENTATION:

Exercise 6.Create an expression for the right side of each equation such that the solution set for the equation will be all real numbers. a. 2x-5= ___

Eureka Math Algebra 1 Module 1 Lesson 2 Problem Set Answer Key

Eureka Math Lesson 18 Homework Answer Key

Question 1.Here is an elevation-versus-time graph of a ball rolling down a ramp. The first section of the graph is slightly curved.a. From the time of about 1.7 sec. onward, the graph is a flat horizontal line. If Ken puts his foot on the ball at time 2 sec. to stop the ball from rolling, how will this graph of elevation versus time change?Answer:Even if the ball is at rest on the floor, its elevation remains 0 in. and does not change. The elevation-versus-time graph does not change.

b. Estimate the number of inches of change in elevation of the ball from 0 sec. to 0.5 sec. Also estimate the change in elevation of the ball between 1.0 sec. and 1.5 sec.Answer:Between 0 and 0.5 sec., the change in elevation was about -2 in. Between 1.0 and 1.5 sec., the change in elevation was about -5.5 in.

c. At what point is the speed of the ball the fastest, near the top of the ramp at the beginning of its journey or near the bottom of the ramp? How does your answer to part support what you say?Answer:The speed of the ball is the fastest near the bottom of the ramp. During the \ sec. from 1.0 sec. to 1.5 sec., the balls change in elevation is greater than the \ sec. at the beginning of its journey. It must have traversed a greater length of the ramp during this \ sec. and so was traveling faster. Its speed is greater near the bottom of the ramp.

Question 3.Use the table below to answer the following questions.a. Plot the points in this table on a graph .Answer:

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Eureka Math Algebra 1 Module 1 Lesson 11 Example Answer Key

Example 1.Consider the equation, x2 =3x+4, where x represents a real number.a. Are the expressions x2 and 3x+4 algebraically equivalent?Answer:No.

Then, we cannot guarantee there will be any real value of x that will make the equation true.

b. The following table shows how we might sift through various values to assign to the variable symbol x in the hunt for values that would make the equation true.Answer: Of course, we should sift through ALL the real numbers if we are seeking all values that make the equation x2 =3x+4 true. So far, we have found that setting x equal to 4 yields a true statement. Look at the image in your student materials. Can you see what is happening here and how it relates to what we have been discussing? The numbers are going down the road and being accepted into the solution set or rejected based on whether or not the equation is true. There happens to be just one more value we can assign to x that makes x2 =3x+4 a true statement. Would you like to continue the search to find it? x=-1

Consider the equation 7+p=12.Answer: Here is a table that could be used to hunt for the value of p that make the equation true:

But is a table necessary for this question? Is it obvious what value we could assign to p to make the equation true?

It seems that each and every positive real number is a solution to this equation.IN WORDS: The solution set is the set of all positive real numbers.IN SET NOTATION: This is .IN A GRAPHICAL REPRESENTATION:

Congratulations You Have Finished Module 1

After working with this module, what are the new ideas about mathematics did youlearn? What is it about mathematics that might have changed your thoughts about it? Whatis most useful about mathematics for humankind?

You are encouraged to provide a lecture notebook where you can write all yourresponses and solutions to the activities and SAAs. Answers to SAAs are provided at thebottom part of the page. If you have difficulty in obtaining the correct answer, you can goover again with the examples. To be successful in mathematics, you have to do mathematics.Do it without the fear of facing more problems and questions to solve. For furtherunderstanding, use the references, suggested readings, and other materials indicated in themodule.

References Aufmann, R., Lockwood, J., et, Mathematics in the Modern World, Rex Bookstore, Inc., 2018. Lerner, K., Lerner, B., Real-life Math, Vol. 2, Thomson Gale, 2006. Nocon, R., Nocon, E., Essential Mathematics for the Modern World, C & E Publishing, Inc. 2018. Post, T., The Role of Manipulative Materials in the Learning Mathematical Concepts. Retrieved from: cehd.umm/ci/rationalnumberproject/81_4.html

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Other Materials vimeo/ htpps:youtu/pb0MSMGSley

Suggested Readings Stewart, Ian, Natures Numbers Adam, John A., Mathematics in Nature: Modeling Patterns in the natural World Adam, John A., A Mathematical Nature Walk Akiyama & Ruis, A Days Adventure in Math Wonderland Enzensberger, The Number Devil

Note To Students

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