eureka math grade 5 lesson 2 homework module 2

eureka math grade 5 lesson 2 homework module 2

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eureka math grade 5 module 3 lesson 9 homework

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Eureka Math Grade 8 Module 4 Lesson 16 Answer Key

Engage ny eureka math 8th grade module 4 lesson 16 answer key, eureka math grade 8 module 4 lesson 16 example answer key.

Engage NY Math 8th Grade Module 4 Lesson 16 Example Answer Key 1

→ Make a conjecture about how you could find the slope of this line.

Have students write their conjectures and share their ideas about how to find the slope of the line in this example; then, continue with the Discussion that follows.

Eureka Math Grade 8 Module 4 Lesson 16 Exercise Answer Key

Eureka Math Grade 8 Module 4 Lesson 16 Exercise Answer Key 4

b. Identify the coordinates of points P and R. Answer: Sample points are labeled on the graph.

c. Find the slope of the line using as many different points as you can. Identify your points, and show your work below. Answer: Points selected by students will vary, but the slope should always equal 2. Students could choose to use points (0,5), (-1,3), (-2,1), (-3,-1), (-4,-3), and (-5,-5).

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key

Students practice finding slope between any two points on a line. Students also see that m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) yields the same result as m=\(\frac{r_{2}-p_{2}}{r_{1}-p_{1}}\).

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 6

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\) =\(\frac{1-2}{6-2}\) =\(\frac{-1}{4}\) =-\(\frac{1}{4}\)

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 8

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\) =\(\frac{-3-5}{4-6}\) =\(\frac{-8}{-2}\) =4

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 9

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\) =\(\frac{3-1}{4-5}\) =\(\frac{2}{-1}\) =-2

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 10

m=\(\frac{q_{2}-r_{2}}{q_{1}-r_{1}}\) =\(\frac{0-1}{11-6}\) =\(\frac{-1}{5}\) =-\(\frac{1}{5}\)

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 11

b. Select two different points on the line to calculate the slope. Answer: Let the two new points be (-3,1) and (-1,4). m=\(\frac{q_{2}-s_{2}}{q_{1}-s_{1}}\) =\(\frac{1-4}{-3-(-1)}\) =\(\frac{-3}{-2}\) =\(\frac{3}{2}\)

c. What do you notice about your answers in parts (a) and (b)? Explain. Answer: The slopes are equal in parts (a) and (b). This is true because of what we know about similar triangles. The slope triangle that is drawn between the two points selected in part (a) is similar to the slope triangle that is drawn between the two points in part (b) by the AA criterion. Then, because the corresponding sides of similar triangles are equal in ratio, the slopes are equal.

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 12

Question 8. Your teacher tells you that a line goes through the points (-6, \(\frac{1}{2}\)) and (-4,3). a. Calculate the slope of this line. Answer: m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) =\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) =\(\frac{\frac{1}{2}-3}{-6-(-4)}\) =\(\frac{-\frac{5}{2}}{-2}\) = \(\frac{\frac{5}{2}}{2}\) =\(\frac{5}{2}\)÷2 =\(\frac{5}{2}\)×\(\frac{1}{2}\) = \(\frac{5}{4}\)

b. Do you think the slope will be the same if the order of the points is reversed? Verify by calculating the slope, and explain your result. The slope should be the same because we are joining the same two points. Answer: m=\(\frac{p_{2}-r_{2}}{p_{1}-r_{1}}\) = \(\frac{3-\frac{1}{2}}{-4-(-6)}\) = \(\frac{\frac{5}{2}}{2}\) = \(\frac{5}{4}\) Since the slope of a line can be computed using any two points on the same line, it makes sense that it does not matter which point we name as P and which point we name as R.

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 12.1

b. Compute the slope again, this time reversing the order of the coordinates. Answer: m=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\) =\(\frac{-3-1}{0-(-3)}\) =\(\frac{-4}{3}\) =-\(\frac{4}{3}\)

c. What do you notice about the slopes you computed in parts (a) and (b)? Answer: The slopes are equal.

d. Why do you think m=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\).? Answer: If I multiply the first fraction by \(\frac{-1}{-1}\), then I get the second fraction: \(\frac{-1}{-1}\)×\(\left(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\right)\)=\(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\). I can do the same thing to the second fraction to obtain the first: \(\frac{-1}{-1}\)×\(\left(\frac{\left(r_{2}-p_{2}\right)}{\left(r_{1}-p_{1}\right)}\right)\)=\(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\) Also, since I know that I can find the slope between any two points, it should not matter which point I pick first.

Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key 50

Question 11. A certain line has a slope of \(\frac{1}{2}\). Name two points that may be on the line. Answer: Answers will vary. Accept any answers that have a difference in y-values equal to 1 and a difference of x-values equal to 2. Points (6,4) and (4,3) may be on the line, for example.

Eureka Math Grade 8 Module 4 Lesson 16 Exit Ticket Answer Key

Eureka Math Grade 8 Module 4 Lesson 16 Exit Ticket Answer Key 5

a. Select any two points on the line to label as P and R. Name their coordinates. Answer: Answers will vary. Other points on the graph may have been chosen. P(-1,0) and R(5,3)

b. Compute the rate of change of the line. Answer: m = \(\frac{\left(p_{2}-r_{2}\right)}{\left(p_{1}-r_{1}\right)}\) m=\(\frac{0-3}{-1-5}\) =\(\frac{-3}{-6}\) =\(\frac{1}{2}\)

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    Eureka Math Grade 8 Module 4 Lesson 16 Problem Set Answer Key. Students practice finding slope between any two points on a line. Students also see that m= p2−r2 p1−r1 yields the same result as m= r2−p2 r1−p1. Question 1. Calculate the slope of the line using two different pairs of points. m= p2−r2 p1−r1. = 30−(−10) −10−30.