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Eureka Math Algebra 1 Module 1 Lesson 5 Answer Key

Engage ny eureka math algebra 1 module 1 lesson 5 answer key, eureka math algebra 1 module 1 lesson 5 example answer key.

$Eureka Math Algebra 1 Module 1 Lesson 5 Example Answer Key 1$

Eureka Math Algebra 1 Module 1 Lesson 5 Exercise Answer Key

Exploratory Challenge/Exercises 1–4 Watch the following graphing story.

The video shows a man and a girl walking on the same stairway.

$Eureka Math Algebra 1 Module 1 Lesson 5 Exercise Answer Key 2$

Exercise 2. Add the girl’s elevation to the same graph. How did you account for the fact that the two people did not start at the same time? Answer: Students’ answers should look something like the graph to the right. Student work should also include scales.

Exercise 3. Suppose the two graphs intersect at the point P(24,4). What is the meaning of this point in this situation? Answer: Many students will respond that P is where the two people pass each other on the stairway.

Lead a discussion that highlights these more subtle points before proceeding. → We have two elevation-versus-time graphs, one for each of the two people (and that time is being measured in the same way for both people). → The point P lies on the elevation-versus-time graph for the first person, and it also lies on the elevation-versus-time graph for the second person. → We know the coordinates of the point P. These coordinates mean that since the first person is at an elevation of 4 ft. at 24 sec., the second person is also at an elevation of 4 ft. at 24 sec.

Exercise 4. Is it possible for two people, walking in stairwells, to produce the same graphs you have been using and not pass each other at time 12 sec.? Explain your reasoning. Answer: Yes, they could be walking in separate stairwells. They would still have the same elevation of 4 ft. at time 24 sec. but in different locations.

Give students time to revise their work after discussing this with the entire class.

Example 2/Exercises 5–7 (10 minutes) Exercise 6 has a similar scenario to Example 1. After students work this exercise in small groups, have each group share their results as time permits. Circulate around the classroom providing assistance to groups as needed. Use the results of the exercises in Example 2 to close this session.

Example 2/Exercises 5–7 Consider the story: Duke starts at the base of a ramp and walks up it at a constant rate. His elevation increases by 3 ft. every second. Just as Duke starts walking up the ramp, Shirley starts at the top of the same 25 ft. high ramp and begins walking down the ramp at a constant rate. Her elevation decreases 2 ft. every second.

$Eureka Math Algebra 1 Module 1 Lesson 5 Exercise Answer Key 5$

Exercise 6. What are the coordinates of the point of intersection of the two graphs? At what time do Duke and Shirley pass each other? Answer: (5,15) t=5

Exercise 7. Write down the equation of the line that represents Duke’s motion as he moves up the ramp and the equation of the line that represents Shirley’s motion as she moves down the ramp. Show that the coordinates of the point you found in the question above satisfy both equations. Answer: If y represents elevation in feet and t represents time in seconds, then Duke’s elevation is represented by y=3t and Shirley’s elevation is represented by y=25-2t. The lines intersect at (5,15), and this point does indeed lie on both lines. Duke: 15=3(5) Shirley: 15=25-2(5)

Eureka Math Algebra 1 Module 1 Lesson 5 Problem Set Answer Key

Question 1. Reread the story about Maya and Earl from Example 1. Suppose that Maya walks at a constant rate of 3 ft. every second and Earl walks at a constant rate of 4 ft. every second starting from 50 ft. away. Create equations for each person’s distance from Maya’s door and determine exactly when they meet in the hallway. How far are they from Maya’s door at this time? Answer: Maya’s Equation: y=3t Earl’s Equation: y=50-4t Solving the equation 3t=50-4t gives the solution =7 $\frac{1}{7}$. The two meet at exactly this time at a distance of 3(7 $\frac{1}{7}$)=21$\frac{3}{7}$ ft. from Maya’s door.

$Eureka Math Algebra 1 Module 1 Lesson 5 Problem Set Answer Key 60$

b. Create linear equations that represent each girl’s mileage in terms of time in minutes. You will need two equations for July since her pace changes after 4 laps (1 mi.). Answer: Equations for May, June, and July are shown below. Notice that July has two equations since her speed changes after her first mile, which occurs 13 min. after May starts running. May: d=$\frac{1}{11}$ t June: d=$\frac{1}{9}$(t-5) July: d=$\frac{1}{6}$ (t-7)”,” t≤13 “and” d=$\frac{1}{12}$ (t-13)+1″,” t>13

c. Who was the first person to run 3 mi.? Answer: June at time 32 min.

d. Did June and July pass May on the track? If they did, when and at what mileage? Answer: Assuming that they started at the same place, June passes May at time 27.5 min. at the 2.5 mi. marker. July does not pass May.

e. Did July pass June on the track? If she did, when and at what mileage? Answer: July passes June at time 11 min. at the $\frac{2}{3}$ mi. marker.

$Eureka Math Algebra 1 Module 1 Lesson 5 Problem Set Answer Key 75$

b. Approximately when do the cars pass each other? Answer: The cars pass after about 2 $\frac{1}{2}$ hr., after 4 hr., and after about 5 $\frac{1}{2}$ hr.

c. Tell the entire story of the graph from the point of view of Car 2. (What does the driver of Car 2 see along the way and when?) Answer; The driver of Car 2 is carefully driving along at 25 mph, and he sees Car 1 pass him at 100 mph after about 2 $\frac{1}{2}$ hr. About 1 $\frac{1}{2}$ hr. later, he sees Car 1 broken down along the road. After about another 1 $\frac{1}{2}$ hr., Car 1 whizzes past again.

d. Create linear equations representing each car’s distance in terms of time (in hours). Note that you will need four equations for Car 1 and only one for Car 2. Use these equations to find the exact coordinates of when the cars meet. Answer: Using the variables, d for distance (in miles) and t for time (in hours): Equation for Car 2: d=25t+100 Equations for Car 1: d=50t, 0≤t≤2 d=100(t-2)+100=100(t-1), 2<t≤3 d=200, 3<t≤5 d=100(t-5)+200=100(t-3), 5<t

Intersection points: First: solving 100(t-1)=25t+100 gives ($\frac{200}{75}$, $\frac{(25)(200)}{75}$+100)≈(2.7,166.7), Second: solving 200=25t+100 gives (4,200), and Third: solving 100(t-3)=25t+100 gives ($\frac{400}{75}$, $\frac{(25)(400)}{75}$+100)≈(5.3,233.3).

$Eureka Math Algebra 1 Module 1 Lesson 5 Problem Set Answer Key 66$

Question 5. Generate six distinct random whole numbers between 2 and 9 inclusive, and fill in the blanks below with the numbers in the order in which they were generated. A (0 ,_______), B (_______,_______), C (10 ,_______) D (0 ,_______), E (10 ,_______).

(Link to a random number generator http://www.mathgoodies.com/calculators/random_no_custom.html)

a. On a coordinate plane, plot points A, B, and C. Draw line segments from point A to point B, and from point B to point C. b. On the same coordinate plane, plot points D and E and draw a line segment from point D to point E. c. Write a graphing story that describes what is happening in this graph. Include a title, x- and y-axis labels, and scales on your graph that correspond to your story. Answer: Answers will vary depending on the random points generated.

$Eureka Math Algebra 1 Module 1 Lesson 5 Problem Set Answer Key 61$

a. How are revenue and total cost related to the number of units of coffee mugs produced? Answer: Definition: Profit = Revenue – Cost. Revenue is the income from the sales and is directly proportional to the number of coffee mugs actually sold; it does not depend on the units of coffee mugs produced. Total cost is the sum of the fixed costs (overhead, maintaining the machines, rent, etc.) plus the production costs associated with the number of coffee mugs produced; it does not depend on the number of coffee mugs sold.

b. What is the meaning of the point (0,4000) on the total cost line? Answer: The overhead costs, the costs incurred regardless of whether 0 or 1,000 coffee mugs are made or sold, is $4,000.

c. What are the coordinates of the intersection point? What is the meaning of this point in this situation? Answer: (500,6000). The revenue, $6,000, from selling 500 coffee mugs, is equal to the total cost, $6,000, of producing 500 coffee mugs. This is known as the break-even point. When Revenue = Cost, the Profit is $0. After this point, the more coffee mugs sold, the more the positive profit; before this point, the company loses money.

d. Create linear equations for revenue and total cost in terms of units produced and sold. Verify the coordinates of the intersection point. Answer: If u is a whole number for the number of coffee mugs produced and sold, C is the total cost to produce u mugs, and R is the total revenue when u mugs are sold, then C=4000+4u, R=12u. When u=500, both C=4000+4∙500=6000 and R=12∙500=6000.

e. Profit for selling 1,000 units is equal to revenue generated by selling 1,000 units minus the total cost of making 1,000 units. What is the company’s profit if 1,000 units are produced and sold? Answer: Profit = Revenue – Total Cost. Hence, P=R-C=12∙1000-(4000+4∙1000)=12000-8000=4000 The company’s profit is $4,000.

Eureka Math Algebra 1 Module 1 Lesson 5 Exit Ticket Answer Key

Maya and Earl live at opposite ends of the hallway in their apartment building. Their doors are 50 ft. apart. Each person starts at his or her own door and walks at a steady pace toward the other. They stop walking when they meet. Suppose: → Maya walks at a constant rate of 3 ft. every second. → Earl walks at a constant rate of 4 ft. every second.

$Eureka Math Algebra 1 Module 1 Lesson 5 Exit Ticket Answer Key 10$

Question 1. Graph both people’s distance from Maya’s door versus time in seconds. Answer: Graphs should be scaled and labeled appropriately. Maya’s graph should start at (0,0) and have a slope of 3, and Earl’s graph should start at (0,50) and have a slope of -4.

Question 2. According to your graphs, approximately how far will they be from Maya’s door when they meet? Answer: The graphs intersect at approximately 7 sec. at a distance of about 21 ft. from Maya’s door.

Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 1.5 Answer Key Round Decimals.

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The Gold Frog of South America is one of the smallest frogs in the world. It is 0.386 of an inch long. What is this length rounded to the nearest hundredth of an inch?

Underline the length of the Gold Frog.
Is the frog’s length about the same as the length or the width of a large paper clip?

One Way Use a place-value chart.

Write the number in a place-value chart and circle the digit in the place value to which you want to round.
In the place-value chart, underline the digit to the right of the place to which you are rounding.
If the digit to the right is less than 5, the digit in the place value to which you are rounding stays the same.
If the digit to the right is 5 or greater, the digit in the rounding place increases by 1.

$Texas Go Math Grade 5 Lesson 1.5 Answer Key 1$

So, to the nearest hundredth of an inch, a Gold Frog is about 0.39 of an inch long.

Another Way

Use place value. The Little Grass Frog is the smallest frog in North America. It is 0.437 of an inch long

$Texas Go Math Grade 5 Lesson 1.5 Answer Key 2$

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Write the place value of the underlined digit. Round each number to the place of the underlined digit.