eureka math grade 5 module 4 lesson 11 homework

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5th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value and decimal fractions, unit 2: module 2: multi-digit whole number and decimal fraction operations, unit 3: module 3: addition and subtractions of fractions, unit 4: module 4: multiplication and division of fractions and decimal fractions, unit 5: module 5: addition and multiplication with volume and area, unit 6: module 6: problem solving with the coordinate plane.

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Learn, Practice, and Succeed from   Eureka Math™   offer teachers multiple ways to differentiate instruction, provide extra practice, and assess student learning. These versatile companions to   A Story of Units®   (Grades K–5) guide teachers in response to intervention (RTI), provide extra practice, and inform instruction.

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The Learn book serves as a student’s in-class companion where they show their thinking, share what they know, and watch their knowledge build every day!

Application Problems:  Problem solving in a real-world context is a daily part of   Eureka Math , building student confidence and perseverance as students apply their knowledge in new and varied ways.

Problem Sets :  A carefully sequenced Problem Set provides an in-class opportunity for independent work, with multiple entry points for differentiation.

Exit Tickets:   These exercises check student understanding, providing the teacher with immediate, valuable evidence of the efficacy of that day’s instruction and informing next steps.

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Eureka Math   fluency activities provide differentiated practice through a variety of formats—some are conducted orally, some use manipulatives, others use a personal whiteboard, or a handout and paper-and-pencil format.

Sprints:  Sprint fluency activities in  Eureka Math Practice  build speed and accuracy with already acquired skills. Used when students are nearing optimum proficiency, Sprints leverage tempo to build a low-stakes adrenaline boost that increases memory and recall. Their intentional design makes Sprints inherently differentiated – the problems build from simple to complex, with the first quadrant of problems being the simplest, and each subsequent quadrant adding complexity.

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Eureka Math Grade 4 Module 4 Lesson 11 Answer Key

Engage ny eureka math 4th grade module 4 lesson 11 answer key, eureka math grade 4 module 4 lesson 11 problem set answer key.

Write an equation, and solve for the unknown angle measurements numerically.

Answer: The value of d° is 10°.

Explanation: Given that the value of the angle acute angle is 30° and the value of the other angle is 20° and the value of another angle is d°. So the equation will be d° + 20°= 30° so the value of d° is 30° – 20° = 10°.

Answer: The value of c° is 270°.

Explanation: Here, in the above image, we can see that an arc that represents a complete rotation which means 360°, and the other angle 90°. So the equation will be c°+90°= 360° and the value of c° is c°= 360°-90° = 270°.

Answer: The value of e° is 196°.

Explanation: Here we will measure the angles using a protractor and the values of the angles will be 90° and 196°. So the equation will be 74°+90°+e°= 360° and the value of e is e° = 360°- 164° = 196°. So the value of e° is 196°.

Answer: The value of f° is 110°.

Explanation: Here we will measure the angles using a protractor and the values of the angles will be 90° and 160°. So the equation will be 90°+160°+f°= 360° and the value of f is f° = 360°- 250° = 110°. So the value of f° is 110°.

Write an equation, and solve for the unknown angles numerically.

Answer: The value of x° is 160°. The value of y° is 20°.

Explanation: In the above image, we can see that the angle COD is 180° and <DOA is 160° and <AOC is 20°. So the equation will be 160° + y°= 180° and the value of y°= 180° – 160° y°= 20°. So the value of y° is 20°. And the other equation will be 20° + x°= 180° x°= 180° – 20° = 160°. So the value of the x°is 160°.

Answer: The value of i° is 55°. The value of h° is 125°. The value of g° is 55°.

Explanation: In the above image, we can see that the angle TOV is 180° and angle SOR is 180° and <TOS is 125°. So the equation will be 125° + i°= 180° and the value of i°= 180° – 125° i°= 55°. So the value of i° is 55°. And the other equation will be 55° + h°= 180° h°= 180° – 55° = 125°. So the value of the h°is 125°. And the other equation will be 125° + g°= 180° g°= 180° – 125° = 55°. So the value of the g°is 55°.

Explanation: In the above image, we can see that the angle WOX is 180° and angle XOZ is 36°, and the value of angle UOX is 90°. So the equation will be 90°+m+ 36°= 180° and the value of m°= 180° – 126° m°= 54°. So the value of m° is 54°. And the other equation will be 54° +90°+k°= 180° k°= 180° – 144° = 36°. So the value of the k°is 36°. And the other equation will be 36° +n°= 180° n°= 180° – 36° = 144°. So the value of the n°is 144°.

Eureka Math Grade 4 Module 4 Lesson 11 Exit Ticket Answer Key

Write equations using variables to represent the unknown angle measurements. Find the unknown angle measurements numerically.

Question 1. x° = Answer: The value of x° is 24°.

Explanation: In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 90°+x+66= 180° and the value of m°= 180° – 156° x°= 24°. So the value of x° is 24°.

Question 2. y° = Answer: The value of y° is 156°.

Explanation: In the above image, we can see that the angle AEB is 180° and angle AEF is 90°. So the equation will be 24°+y= 180° and the value of y°= 180° – 24° y°= 156°. So the value of y° is 156°.

Question 3. z° = Answer: The value of z° is 24°.

Explanation: In the above image, we can see that the angle AEB is 180° and angle FEB is 90°. So the equation will be 90°+z+66= 180° and the value of m°= 180° – 156° z°= 24°. So the value of z° is 24°.

Eureka Math Grade 4 Module 4 Lesson 11 Homework Answer Key

Explanation: Given that the value of the angle is complete rotation which 360° and the value of the other angle is 320° and the value of another angle is a°. So the equation will be a° + 320°= 360° so the value of a° is 360° – 320° = 40°.

Explanation: Given that the value of the angle is complete rotation which 360° and the value of the other angle is 45° and the value of another angle is b°. So the equation will be b° + 45°= 360° so the value of b° is 360° – 45° =315°

Explanation: Given that the value of the angle is complete rotation which 360° and the value of the other angle is 115° and the value of another angle is 100° and the value of another angle is c°. So the equation will be c°+115°+100°= 360°. c° + 215°= 360° so the value of c° is 360° – 215° =145°.

Explanation: Given that the value of the angle is complete rotation which 360° and the value of the other angle is 135° and the value of another angle is 145° and the value of another angle is d°. So the equation will be d°+135°+145°= 360°. d° + 280°= 360° so the value of c° is 360° – 280° =80°.

Explanation: In the above image, we can see that the angle COD is 180° and <COB is 145° and <AOC is 35°. So the equation will be 35° + e°= 180° and the value of e°= 180° – 35° e°= 145°. So the value of e° is 145°. And the other equation will be 145° + f°= 180° x°= 180° – 145° = 35°.

Explanation: In the above image, we can see that the angle SOT is 180° and angle QOR is 180° and <QOS is 55°. So the equation will be 55° + g°= 180° and the value of g°= 180° – 55° g°= 125°. So the value of g° is 125°. And the other equation will be 55° + h°= 180° h°= 180° – 55° = 125°. So the value of the h°is 125°. And the other equation will be 125° +g°= 180° g°= 180° – 125° = 55°. So the value of the g°is 55°.

Explanation: In the above image, we can see that the angle WOX is 180° and angle UOV is 180° and <VOX is 46°. So the equation will be 46° +90°+ j°= 180° and the value of j°= 180° – 136° j°= 44°. So the value of j° is 44°. And the other equation will be 44° +90°+ k°= 180° k°= 180° – 134° = 46°. So the value of the k°is 46°. And the other equation will be 46° +m°= 180° m°= 180° – 46° = 134°. So the value of the m°is 134°.

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