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Dilating Lines and Angles

Angle \(ABC\) is taken by a dilation with center \(P\) and scale factor 3 to angle \(A’B’C’\) . The measure of angle \(ABC\) is \(21^\circ\) . What is the measure of angle \(A’B’C’\) ?

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Select all lines that could be the image of line \(m\) by a dilation.

Expand image

Dilate line \(f\) with a scale factor of 2. The image is line \(g\) . Which labeled point could be the center of this dilation?

Description: <p>Line f with point C to the right of the center. Line g above line f with point B to the right of the center. Point A is above line g and to the left of point B. Point D is below line f and to the left of point C.</p>

Quadrilateral \(A’B’C’E’\) is the image of quadrilateral \(ABCE\) after a dilation centered at \(F\) . What is the scale factor of this dilation?

Description: <p>Quadrilaterals A B C E and A prime B prime C prime E prime. Side lengths as follows: A B, 3. B C 6, C E, 5. E A, unlabeled. A prime B prime, 1 point 5. B prime C prime, 3. C prime E prime, 2 point 5. E prime A prime, unlabeled.<br>  </p>

A polygon has a perimeter of 18 units. It is dilated with a scale factor of \(\frac32\) . What is the perimeter of its image?

Solve the equation. 

\(\frac{4}{7}=\frac{10}{x}\)

Here are some measurements for triangle  \(ABC \)  and triangle  \(XYZ\) :

  • Angle  \(CAB\)  and angle  \(ZXY\)  are both 30 degrees
  • \(AC\) and \(XZ\) both measure 3 units
  • \(CB\) and \(ZY\) both measure 2 units

Andre thinks thinks these triangles must be congruent. Clare says she knows they might not be congruent. Construct 2 triangles with the given measurements that aren't congruent. Explain why triangles with 3 congruent parts aren't necessarily congruent.

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Answer Key 3.4

For questions 1 to 10, sketch the linear equation using the slope intercept method.

Graph with line that passes through -4,-2) (0,-3), (4,4)

For questions 11 to 20, sketch the linear equation using the [latex]x[/latex] and [latex]y[/latex] intercepts.

Line on graph passees through (-4,0), (0,-1)

For questions 21 to 28, sketch the linear equation using any method.

Line on graph passess through (-4,-5), (-2,-4), (0,3), (2,2), (4,1)

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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geometry 3.4 homework answers

( 3 , 2 ) ( 3 , 2 )

( 2 , 3 ) ( 2 , 3 )

( 3 , 4 ) ( 3 , 4 )

( 5 , −4 ) ( 5 , −4 )

no solution

infinitely many solutions

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent

( 6 , 1 ) ( 6 , 1 )

( −3 , 5 ) ( −3 , 5 )

( 2 , 3 2 ) ( 2 , 3 2 )

( − 1 2 , −2 ) ( − 1 2 , −2 )

( 2 , −1 ) ( 2 , −1 )

( −2 , 3 ) ( −2 , 3 )

( 1 , 3 ) ( 1 , 3 )

( 4 , −3 ) ( 4 , −3 )

( 6 , 2 ) ( 6 , 2 )

( 1 , −2 ) ( 1 , −2 )

ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x , using substitution will be most convenient.

ⓐ Since one equation is already solved for y , using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.

160 policies

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

The angle measures are 55 and 35.

The angle measures are 5 and 85.

The angle measures are 42 and 138.

The angle measures are 66 and 114.

The length is 60 feet and the width is 35 feet.

The length is 60 feet and the width is 38 feet.

It will take Clark 4 hours to catch Mitchell.

It will take Sally 1 1 2 1 1 2 hours to catch up to Charlie.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

206 adults, 347 children

42 adults, 105 children

13 dimes and 29 quarters

19 quarters and 51 nickels

3 pounds peanuts and 2 pounds cashews

10 pounds of beans, 10 pounds of ground beef

120 ml of 25% solution and 30 ml of 50% solution

125 ml of 10% solution and 125 ml of 40% solution

$42,000 in the stock fund and $8000 in the savings account

$1750 at 11% and $5250 at 13%

Bank $4,000; Federal $14,000

$41,200 at 4.5%, $24,000 at 7.2%

ⓐ C ( x ) = 15 x + 25 , 500 C ( x ) = 15 x + 25 , 500

ⓑ R ( x ) = 32 x R ( x ) = 32 x

ⓓ 1,500 1,500 ; when 1,500 benches are sold, the cost and revenue will be both 48,000

ⓐ C ( x ) = 120 x + 150,000 C ( x ) = 120 x + 150,000

ⓑ R ( x ) = 170 x R ( x ) = 170 x

ⓓ 3,000 3,000 ; when 3,000 benches are sold, the revenue and costs are both $510,000

( 2 , −1 , 3 ) ( 2 , −1 , 3 )

( −2 , 3 , 4 ) ( −2 , 3 , 4 )

( −3 , 4 , −2 ) ( −3 , 4 , −2 )

( −2 , 3 , −1 ) ( −2 , 3 , −1 )

infinitely many solutions ( x , 3 , z ) ( x , 3 , z ) where x = z − 3 ; y = 3 ; z x = z − 3 ; y = 3 ; z is any real number

infinitely many solutions ( x , y , z ) ( x , y , z ) where x = 5 z − 2 ; y = 4 z − 3 ; z x = 5 z − 2 ; y = 4 z − 3 ; z is any real number

The fine arts department sold 75 adult tickets, 200 student tickets, and 75 child tickets.

The soccer team sold 200 adult tickets, 300 student tickets, and 100 child tickets.

ⓐ [ 3 8 −3 2 5 −3 ] [ 3 8 −3 2 5 −3 ] ⓑ [ 2 −5 3 8 3 −1 4 7 1 3 2 −3 ] [ 2 −5 3 8 3 −1 4 7 1 3 2 −3 ]

ⓐ [ 11 9 −5 7 5 −1 ] [ 11 9 −5 7 5 −1 ] ⓑ [ 5 −3 2 −5 2 −1 −1 4 3 −2 2 −7 ] [ 5 −3 2 −5 2 −1 −1 4 3 −2 2 −7 ]

{ x − y + 2 z = 3 2 x + y − 2 z = 1 4 x − y + 2 z = 0 { x − y + 2 z = 3 2 x + y − 2 z = 1 4 x − y + 2 z = 0

{ x + y + z = 4 2 x + 3 y − z = 8 x + y − z = 3 { x + y + z = 4 2 x + 3 y − z = 8 x + y − z = 3

ⓐ [ −2 3 0 −2 4 −1 −4 4 5 −2 −2 −2 ] [ −2 3 0 −2 4 −1 −4 4 5 −2 −2 −2 ] ⓑ [ −2 3 0 −2 4 −1 −4 4 15 −6 −6 −6 ] [ −2 3 0 −2 4 −1 −4 4 15 −6 −6 −6 ] ⓒ [ −2 3 0 −2 3 4 −13 −16 −8 15 −6 −6 −6 ] [ −2 3 0 −2 3 4 −13 −16 −8 15 −6 −6 −6 ]

ⓐ [ 4 1 −3 2 2 −3 −2 −4 5 0 4 −1 ] [ 4 1 −3 2 2 −3 −2 −4 5 0 4 −1 ] ⓑ [ 8 2 −6 4 2 −3 −2 −4 5 0 4 −1 ] [ 8 2 −6 4 2 −3 −2 −4 5 0 4 −1 ] ⓒ [ 14 −7 −12 −8 2 −3 −2 −4 5 0 4 −1 ] [ 14 −7 −12 −8 2 −3 −2 −4 5 0 4 −1 ]

[ 1 −1 2 0 −3 −4 ] [ 1 −1 2 0 −3 −4 ]

[ 1 −1 3 0 −5 8 ] [ 1 −1 3 0 −5 8 ]

The solution is ( 4 , −1 ) . ( 4 , −1 ) .

The solution is ( −2 , 0 ) . ( −2 , 0 ) .

( 6 , −1 , −3 ) ( 6 , −1 , −3 )

( 5 , 7 , 4 ) ( 5 , 7 , 4 )

infinitely many solutions ( x , y , z ) , ( x , y , z ) , where x = z − 3 ; y = 3 ; z x = z − 3 ; y = 3 ; z is any real number.

infinitely many solutions ( x , y , z ) , ( x , y , z ) , where x = 5 z − 2 ; y = 4 z − 3 ; z x = 5 z − 2 ; y = 4 z − 3 ; z is any real number.

ⓐ −14 ; −14 ; ⓑ −28 −28

ⓐ 2 ⓑ −15 −15

ⓐ 3 ⓑ 11 ⓒ 2

ⓐ −3 −3 ⓑ 2 ⓒ 3

( − 15 7 , 24 7 ) ( − 15 7 , 24 7 )

( −2 , 0 ) ( −2 , 0 )

( −9 , 3 , −1 ) ( −9 , 3 , −1 )

( −6 , 3 , −2 ) ( −6 , 3 , −2 )

infinite solutions

The solution is the grey region.

No solution.

ⓐ { 30 m + 20 p ≤ 160 2 m + 3 p ≤ 15 { 30 m + 20 p ≤ 160 2 m + 3 p ≤ 15 ⓑ

ⓐ { a ≥ p + 5 a + 2 p ≤ 400 { a ≥ p + 5 a + 2 p ≤ 400 ⓑ

ⓐ { 0.75 d + 2 e ≤ 25 360 d + 110 e ≥ 1000 { 0.75 d + 2 e ≤ 25 360 d + 110 e ≥ 1000 ⓑ

ⓐ { 140 p + 125 j ≥ 1000 1.80 p + 1.25 j ≤ 12 { 140 p + 125 j ≥ 1000 1.80 p + 1.25 j ≤ 12 ⓑ

Section 4.1 Exercises

( 0 , 2 ) ( 0 , 2 )

( 2 , 4 ) ( 2 , 4 )

( −2 , 2 ) ( −2 , 2 )

( 3 , 3 ) ( 3 , 3 )

( 6 , −4 ) ( 6 , −4 )

No solutions, inconsistent, independent

1 point, consistent and independent

infinite solutions, consistent, dependent

( 1 , −4 ) ( 1 , −4 )

( −3 , 2 ) ( −3 , 2 )

( −1 / 2 , 5 / 2 ) ( −1 / 2 , 5 / 2 )

( −5 , 4 ) ( −5 , 4 )

( 0 , 10 ) ( 0 , 10 )

( 4 , −2 ) ( 4 , −2 )

( 4 , 0 ) ( 4 , 0 )

( 4 , 5 ) ( 4 , 5 )

( 7 , 12 ) ( 7 , 12 )

( −3 , −5 ) ( −3 , −5 )

( 2 , −3 ) ( 2 , −3 )

( −11 , 2 ) ( −11 , 2 )

( 6 / −9 , 24 / 7 ) ( 6 / −9 , 24 / 7 )

infinitely many

ⓐ substitution ⓑ elimination

ⓐ elimination ⓑ substituion

Answers will vary.

Section 4.2 Exercises

−7 −7 and −19 −19

22 and −67 −67

Eighty cable packages would need to be sold to make the total pay the same.

Mitchell would need to sell 120 stoves for the companies to be equal.

8 and 40 gallons

1000 calories playing basketball and 400 calories canoeing

Oranges cost $2 per pound and bananas cost $1 per pound

Package of paper $4, stapler $7

Hot dog 150 calories, cup of cottage cheese 220 calories

Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade.

53.5 53.5 degrees and 36.5 36.5 degrees

16 degrees and 74 degrees

134 degrees and 46 degrees

37 degrees and 143 degrees

16 ° 16 ° and 74 ° 74 °

45 ° 45 ° and 45 ° 45 °

Width is 41 feet and length is 118 feet.

Width is 10 feet and length is 40 feet.

1.5 1.5 hour

Boat rate is 16 mph and current rate is 4 mph.

Boat rate is 18 mph and current rate is 2 mph.

Jet rate is 265 mph and wind speed is 22 mph.

Jet rate is 415 mph and wind speed is 25 mph.

Section 4.3 Exercises

110 adult tickets, 190 child tickets

6 good seats, 10 cheap seats

92 adult tickets, 220 children tickets

13 nickels, 3 dimes

42 dimes, 8 quarters

17 $10 bills, 37 $20 bills

80 pounds nuts and 40 pounds raisins

9 pounds of Chicory coffee, 3 pounds of Jamaican Blue Mountain coffee

10 bags of M&M’s, 15 bags of Reese’s Pieces

7.5 7.5 liters of each solution

80 liters of the 25% solution and 40 liters of the 10% solution

240 liters of the 90% solution and 120 liters of the 75% solution

$1600 at 8%, 960 at 6%

$28,000 at 9%, $36,000 at 5.5 % 5.5 %

$8500 CD, $1500 savings account

$55,000 on loan at 6% and $30,000 on loan at 4.5 % 4.5 %

ⓐ C ( x ) = 5 x + 6500 C ( x ) = 5 x + 6500

ⓑ R ( x ) = 10 x R ( x ) = 10 x

ⓓ 1,500; when 1,500 water bottles are sold, the cost and the revenue equal $15,000

Section 4.4 Exercises

( 4 , 5 , 2 ) ( 4 , 5 , 2 )

( 7 , 12 , −2 ) ( 7 , 12 , −2 )

( −3 , −5 , 4 ) ( −3 , −5 , 4 )

( 2 , −3 , −2 ) ( 2 , −3 , −2 )

( 6 , −9 , −3 ) ( 6 , −9 , −3 )

( 3 , −4 , −2 ) ( 3 , −4 , −2 )

( −3 , 2 , 3 ) ( −3 , 2 , 3 )

( −2 , 0 , −3 ) ( −2 , 0 , −3 )

x = 203 16 ; y = –25 16 ; z = –231 16 ; x = 203 16 ; y = –25 16 ; z = –231 16 ;

( x , y , z ) ( x , y , z ) where x = 5 z + 2 ; y = −3 z + 1 ; z x = 5 z + 2 ; y = −3 z + 1 ; z is any real number

( x , y , z ) ( x , y , z ) where x = 5 z − 2 ; y = 4 z − 3 ; z x = 5 z − 2 ; y = 4 z − 3 ; z is any real number

$20, $5, $10

Section 4.5 Exercises

ⓐ [ 2 4 −5 3 −2 2 ] [ 2 4 −5 3 −2 2 ] ⓑ [ 3 −2 −1 −2 −2 1 0 5 5 4 1 −1 ] [ 3 −2 −1 −2 −2 1 0 5 5 4 1 −1 ]

ⓐ [ 2 −5 −3 4 −3 −1 ] [ 2 −5 −3 4 −3 −1 ] ⓑ [ 4 3 −2 −3 −2 1 −3 4 −1 −4 5 −2 ] [ 4 3 −2 −3 −2 1 −3 4 −1 −4 5 −2 ]

{ 2 x − 4 y = −2 3 x − 3 y = −1 { 2 x − 4 y = −2 3 x − 3 y = −1

{ 2 x − 2 y = −1 2 y − z = 2 3 x − z = −2 { 2 x − 2 y = −1 2 y − z = 2 3 x − z = −2

ⓐ [ 3 2 1 4 −6 −3 ] [ 3 2 1 4 −6 −3 ] ⓑ [ 12 8 4 4 −6 −3 ] [ 12 8 4 4 −6 −3 ] ⓒ [ 12 8 4 24 −10 −5 ] [ 12 8 4 24 −10 −5 ]

ⓐ [ 2 1 −4 5 6 −5 2 3 3 −3 1 −1 ] [ 2 1 −4 5 6 −5 2 3 3 −3 1 −1 ] ⓑ [ 2 1 −4 5 6 −5 2 3 3 −3 1 −1 ] [ 2 1 −4 5 6 −5 2 3 3 −3 1 −1 ] ⓒ [ 2 1 −4 5 6 −5 2 3 −4 7 −6 7 ] [ 2 1 −4 5 6 −5 2 3 −4 7 −6 7 ]

[ 1 −2 3 −4 0 5 −11 17 0 1 −10 7 ] [ 1 −2 3 −4 0 5 −11 17 0 1 −10 7 ]

( 1 , −1 ) ( 1 , −1 )

( −2 , 5 , 2 ) ( −2 , 5 , 2 )

infinitely many solutions ( x , y , z ) ( x , y , z ) where x = 1 2 z + 4 ; y = 1 2 z − 6 ; z x = 1 2 z + 4 ; y = 1 2 z − 6 ; z is any real number

infinitely many solutions ( x , y , z ) ( x , y , z ) where x = 5 z + 2 ; y = −3 z + 1 ; z x = 5 z + 2 ; y = −3 z + 1 ; z is any real number

Section 4.6 Exercises

ⓐ 6 ⓑ −14 −14 ⓒ −6 −6

ⓐ 9 ⓑ −3 −3 ⓒ 8

( 7 , 6 ) ( 7 , 6 )

( −9 , 3 ) ( −9 , 3 )

inconsistent

Section 4.7 Exercises

ⓐ false ⓑ true

ⓐ { f ≥ 0 p ≥ 0 f + p ≤ 20 2 f + 5 p ≤ 50 { f ≥ 0 p ≥ 0 f + p ≤ 20 2 f + 5 p ≤ 50 ⓑ

ⓐ { c ≥ 0 a ≥ 0 c + a ≤ 24 a ≥ 3 c { c ≥ 0 a ≥ 0 c + a ≤ 24 a ≥ 3 c ⓑ

ⓐ { w ≥ 0 b ≥ 0 27 w + 16 b > 80 3.20 w + 1.75 b ≤ 10 { w ≥ 0 b ≥ 0 27 w + 16 b > 80 3.20 w + 1.75 b ≤ 10 ⓑ

ⓐ { w ≥ 0 r ≥ 0 w + r ≥ 4 270 w + 650 r ≥ 1500 { w ≥ 0 r ≥ 0 w + r ≥ 4 270 w + 650 r ≥ 1500 ⓑ

Review Exercises

( 3 , −1 ) ( 3 , −1 )

one solution, consistent system, independent equations

( 3 , 1 ) ( 3 , 1 )

( 4 , −1 ) ( 4 , −1 )

elimination

50 irises and 150 tulips

10 calories jogging and 10 calories cycling

35 ° 35 ° and 55 ° 55 °

the length is 450 feet, the width is 264 feet

1 2 1 2 an hour

the rate of the jet is 395 mph, the rate of the wind is 7 mph

41 dimes and 11 pennies

46 2 3 46 2 3 liters of 30% solution, 23 1 3 23 1 3 liters of 60% solution

$29,000 for the federal loan, $14,000 for the private loan

( −3 , 2 , −4 ) ( −3 , 2 , −4 )

[ 4 3 0 −2 1 −2 −3 7 2 −1 2 −6 ] [ 4 3 0 −2 1 −2 −3 7 2 −1 2 −6 ]

{ x − 3 z = −1 x − 2 y = −27 − y + 2 z = 3 { x − 3 z = −1 x − 2 y = −27 − y + 2 z = 3

ⓐ [ 1 −3 −2 4 4 −2 −3 −1 2 2 −1 −3 ] [ 1 −3 −2 4 4 −2 −3 −1 2 2 −1 −3 ] ⓑ [ 2 −6 −4 8 4 −2 −3 −1 2 2 −1 −3 ] [ 2 −6 −4 8 4 −2 −3 −1 2 2 −1 −3 ] ⓒ [ 2 −6 −4 8 4 −2 −3 −1 0 −6 −1 5 ] [ 2 −6 −4 8 4 −2 −3 −1 0 −6 −1 5 ]

( −2 , 5 , −2 ) ( −2 , 5 , −2 )

ⓐ { b ≥ 0 n ≥ 0 b + n ≤ 40 12 b + 18 n ≥ 500 { b ≥ 0 n ≥ 0 b + n ≤ 40 12 b + 18 n ≥ 500 ⓑ

Practice Test

( 2 , 1 ) ( 2 , 1 )

( 2 , −2 , 1 ) ( 2 , −2 , 1 )

15 liters of 1% solution, 5 liters of 5% solution

The candy cost $20; the cookies cost $5; and the popcorn cost $10.

ⓐ { C ≥ 0 L ≥ 0 C + 0.5 L ≤ 50 L ≥ 3 C { C ≥ 0 L ≥ 0 C + 0.5 L ≤ 50 L ≥ 3 C ⓑ

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Eureka Math Grade 3 Module 4 Lesson 10 Answer Key

Engage ny eureka math 3rd grade module 4 lesson 10 answer key, eureka math grade 3 module 4 lesson 10 problem set answer key.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 1

Answer: The area of the larger rectangle = 56 sq unit. the area of the small rectangle A = 35 sq unit. the area of the small rectangle B = 21 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 56 sq unit. area = length x breadth. area = l x b. area = 8 x 7. area = 56. the area of the rectangle A = 35 sq unit. area = length x breadth. area = l x b. area = 5 x 7. area = 35. the area of the rectangle B = 21 sq unit. area = length x breadth. area = l x b. area = 3 x 7. area = 21.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 2

Answer: The area of the larger rectangle = 48 sq unit. the area of the small rectangle A = 40 sq unit. the area of the small rectangle B = 8 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 48 sq unit. area = length x breadth. area = l x b. area = 12 x 4. area = 48. the area of the rectangle A = 40 sq unit. area = length x breadth. area = l x b. area = 10 x 4. area = 40. the area of the rectangle B = 8 sq unit. area = length x breadth. area = l x b. area = 2 x 4. area = 8.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 3

Answer: The area of the larger rectangle = 78 sq unit. the area of the small rectangle A = 60 sq unit. the area of the small rectangle B = 18 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 78 sq unit. area = length x breadth. area = l x b. area = 6 x 13. area = 78. the area of the rectangle A = 60 sq unit. area = length x breadth. area = l x b. area = 10 x 6. area = 60. the area of the rectangle B = 8 sq unit. area = length x breadth. area = l x b. area = 6 x 3. area = 18.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 4

Answer: The area of the larger rectangle = 96 sq unit. the area of the small rectangle A = 80 sq unit. the area of the small rectangle B = 16 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 96 sq unit. area = length x breadth. area = l x b. area = 8 x 12. area = 96. the area of the rectangle A = 80 sq unit. area = length x breadth. area = l x b. area = 10 x 8. area = 80. the area of the rectangle B = 16 sq unit. area = length x breadth. area = l x b. area = 8 x 2.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 5

Answer: The area of the rectangle = 72 sq cm.

Explanation: In the above-given question, given that, the area of the rectangle = l x b. where l = length and b = breadth. area = 9 x 8. area = 72 sq cm. so the area of the rectangle = 72 sq cm.

Eureka Math Grade 3 Module 4 Lesson 10 Problem Set Answer Key 6

Answer: The area of the rectangle = 40 sq cm.

Explanation: In the above-given question, given that, the area of the rectangle = l x b. where l = length and b = breadth. area = 5 x 8. area = 40 sq cm. so the area of the rectangle = 40 sq cm.

Answer: The area of the rectangle = 35 sq cm.

Eureka-Math-Grade-3-Module-4-Lesson-10-Answer Key-1

Eureka Math Grade 3 Module 4 Lesson 10 Exit Ticket Answer Key

Label the side lengths of the shaded and unshaded rectangles. Then, find the total area of the large rectangle by adding the areas of the 2 smaller rectangles.

Eureka Math Grade 3 Module 4 Lesson 10 Exit Ticket Answer Key 7

Answer: The area of the larger rectangle = 56 sq unit. the area of the small rectangle A = 40 sq unit. the area of the small rectangle B = 16 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 56 sq unit. area = length x breadth. area = l x b. area = 8 x 7. area = 56. the area of the rectangle A = 40 sq unit. area = length x breadth. area = l x b. area = 8 x 5. area = 40. the area of the rectangle B = 16 sq unit. area = length x breadth. area = l x b. area = 8 x 2. area = 16.

Eureka Math Grade 3 Module 4 Lesson 10 Exit Ticket Answer Key 8

Answer: The area of the larger rectangle = 117 sq unit. the area of the small rectangle A = 90 sq unit. the area of the small rectangle B = 27 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 117 sq unit. area = length x breadth. area = l x b. area = 9 x 13. area = 117. the area of the rectangle A = 90 sq unit. area = length x breadth. area = l x b. area = 9 x 10. area = 90. the area of the rectangle B = 27 sq unit. area = length x breadth. area = l x b. area = 9 x 3. area = 27.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 9

Answer: The area of the larger rectangle = 72 sq unit. the area of the small rectangle A = 40 sq unit. the area of the small rectangle B = 32 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 78 sq unit. area = length x breadth. area = l x b. area = 9 x 8. area = 72. the area of the rectangle A = 40 sq unit. area = length x breadth. area = l x b. area = 5 x 8. area = 40. the area of the rectangle B = 32 sq unit. area = length x breadth. area = l x b. area = 4 x 8. area = 32.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 10

Answer: The area of the larger rectangle = 60 sq unit. the area of the small rectangle A = 50 sq unit. the area of the small rectangle B = 10 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 60 sq unit. area = length x breadth. area = l x b. area = 12 x 5. area = 60. the area of the rectangle A = 50 sq unit. area = length x breadth. area = l x b. area = 10 x 5. area = 50. the area of the rectangle B = 10 sq unit. area = length x breadth. area = l x b. area = 2 x 5. area = 10.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 11

Answer: The area of the larger rectangle = 91 sq unit. the area of the small rectangle A = 70 sq unit. the area of the small rectangle B = 21 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 91 sq unit. area = length x breadth. area = l x b. area = 7 x 13. area = 91. the area of the rectangle A = 70 sq unit. area = length x breadth. area = l x b. area = 7 x 10. area = 70. the area of the rectangle B = 21 sq unit. area = length x breadth. area = l x b. area = 7 x 3. area = 21.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 12

Answer: The area of the larger rectangle = 108 sq unit. the area of the small rectangle A = 90 sq unit. the area of the small rectangle B = 18 sq unit.

Explanation: In the above-given question, given that, the area of the larger rectangle = 108 sq unit. area = length x breadth. area = l x b. area = 9 x 12. area = 108. the area of the rectangle A = 90 sq unit. area = length x breadth. area = l x b. area = 9 x 10. area = 90. the area of the rectangle B = 18 sq unit. area = length x breadth. area = l x b. area = 9 x 2. area = 18.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 13

Answer: The area of the rectangle = 81 sq cm.

Explanation: In the above-given question, given that, the area of the rectangle = l x b. where l = length and b = breadth. area = 9 x 9. area = 81 sq cm. so the area of the rectangle = 81 sq cm.

Eureka Math Grade 3 Module 4 Lesson 10 Homework Answer Key 14

Explanation: In the above-given question, given that, the area of the rectangle = l x b. where l = length and b = breadth. area = 10 x 4. area = 40 sq cm. so the area of the rectangle = 40 sq cm.

Answer: The area of the rectangle = 24 sq cm.

Eureka-Math-Grade-3-Module-4-Lesson-10-Answer Key-2

Eureka Math Grade 3 Module 4 Lesson 10 Template Answer Key

Eureka Math Grade 3 Module 4 Lesson 10 Template Answer Key 15

Explanation: In the above-given question, given that, the area of the rectangle = l x b. where l = length and b = breadth. area = 4 x 4. area = 16 sq cm. so the area of the rectangle = 16 sq cm.

Answer: The area of the rectangle = 8 sq cm.

Eureka-Math-Grade-3-Module-4-Lesson-10-Answer Key-3

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College Algebra - Section 3.4

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