lesson 5 homework 1.6 answer key

2.1 The Rectangular Coordinate Systems and Graphs

x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .

125 = 5 5 125 = 5 5

( − 5 , 5 2 ) ( − 5 , 5 2 )

2.2 Linear Equations in One Variable

x = −5 x = −5

x = −3 x = −3

x = 10 3 x = 10 3

x = 1 x = 1

x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .

x = 1 3 x = 1 3

m = − 2 3 m = − 2 3

y = 4 x −3 y = 4 x −3

x + 3 y = 2 x + 3 y = 2

Horizontal line: y = 2 y = 2

Parallel lines: equations are written in slope-intercept form.

y = 5 x + 3 y = 5 x + 3

2.3 Models and Applications

C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650

L = 37 L = 37 cm, W = 18 W = 18 cm

2.4 Complex Numbers

−24 = 0 + 2 i 6 −24 = 0 + 2 i 6

( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i

5 2 − i 5 2 − i

18 + i 18 + i

−3 −4 i −3 −4 i

2.5 Quadratic Equations

( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1

( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.

( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.

( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4

x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1

x = 4 ± 5 x = 4 ± 5

x = 3 ± 22 x = 3 ± 22

x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3

2.6 Other Types of Equations

{ −1 } { −1 }

0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2

1 ; 1 ; extraneous solution − 2 9 − 2 9

−2 ; −2 ; extraneous solution −1 −1

−1 , −1 , 3 2 3 2

−3 , 3 , − i , i −3 , 3 , − i , i

2 , 12 2 , 12

−1 , −1 , 0 0 is not a solution.

2.7 Linear Inequalities and Absolute Value Inequalities

[ −3 , 5 ] [ −3 , 5 ]

( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )

x < 1 x < 1

x ≥ −5 x ≥ −5

( 2 , ∞ ) ( 2 , ∞ )

[ − 3 14 , ∞ ) [ − 3 14 , ∞ )

6 < x ≤ 9 ​ or ( 6 , 9 ] 6 < x ≤ 9 ​ or ( 6 , 9 ]

( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )

| x −2 | ≤ 3 | x −2 | ≤ 3

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) . ( − ∞ , 1 ] ∪ [ 7 , ∞ ) .

2.1 Section Exercises

Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.

The y -intercept is the point where the graph crosses the y -axis.

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .

The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .

y = 4 − 2 x y = 4 − 2 x

y = 5 − 2 x 3 y = 5 − 2 x 3

y = 2 x − 4 5 y = 2 x − 4 5

d = 74 d = 74

d = 36 = 6 d = 36 = 6

d ≈ 62.97 d ≈ 62.97

( 3 , − 3 2 ) ( 3 , − 3 2 )

( 2 , −1 ) ( 2 , −1 )

( 0 , 0 ) ( 0 , 0 )

y = 0 y = 0

not collinear

A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )

d = 8.246 d = 8.246

d = 5 d = 5

( −3 , 4 ) ( −3 , 4 )

x = 0          y = −2 x = 0          y = −2

x = 0.75 y = 0 x = 0.75 y = 0

x = − 1.667 y = 0 x = − 1.667 y = 0

15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter

6 .0 42 6 .0 42

Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.

2.2 Section Exercises

It means they have the same slope.

The exponent of the x x variable is 1. It is called a first-degree equation.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

x = 2 x = 2

x = 2 7 x = 2 7

x = 6 x = 6

x = 3 x = 3

x = −14 x = −14

x ≠ −4 ; x ≠ −4 ; x = −3 x = −3

x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution

x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2

y = − 4 5 x + 14 5 y = − 4 5 x + 14 5

y = − 3 4 x + 2 y = − 3 4 x + 2

y = 1 2 x + 5 2 y = 1 2 x + 5 2

y = −3 x − 5 y = −3 x − 5

y = 7 y = 7

y = −4 y = −4

8 x + 5 y = 7 8 x + 5 y = 7

Perpendicular

m = − 9 7 m = − 9 7

m = 3 2 m = 3 2

m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular . m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular .

y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40

y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 ,   y max = 10 y min = −10 ,   y max = 10

y = − A B x + C B y = − A B x + C B

The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 . The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 .

Yes they are perpendicular.

2.3 Section Exercises

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

2 , 000 − x 2 , 000 − x

v + 10 v + 10

Ann: 23 ; 23 ; Beth: 46 46

20 + 0.05 m 20 + 0.05 m

90 + 40 P 90 + 40 P

50 , 000 − x 50 , 000 − x

She traveled for 2 h at 20 mi/h, or 40 miles.

$5,000 at 8% and $15,000 at 12%

B = 100 + .05 x B = 100 + .05 x

R = 9 R = 9

r = 4 5 r = 4 5 or 0.8

W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14

f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21

m = − 5 4 m = − 5 4

h = 2 A b 1 + b 2 h = 2 A b 1 + b 2

length = 360 ft; width = 160 ft

A = 88 in . 2 A = 88 in . 2

h = V π r 2 h = V π r 2

r = V π h r = V π h

C = 12 π C = 12 π

2.4 Section Exercises

Add the real parts together and the imaginary parts together.

Possible answer: i i times i i equals -1, which is not imaginary.

−8 + 2 i −8 + 2 i

14 + 7 i 14 + 7 i

− 23 29 + 15 29 i − 23 29 + 15 29 i

8 − i 8 − i

−11 + 4 i −11 + 4 i

2 −5 i 2 −5 i

6 + 15 i 6 + 15 i

−16 + 32 i −16 + 32 i

−4 −7 i −4 −7 i

2 − 2 3 i 2 − 2 3 i

4 − 6 i 4 − 6 i

2 5 + 11 5 i 2 5 + 11 5 i

1 + i 3 1 + i 3

( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1

5 −5 i 5 −5 i

9 2 − 9 2 i 9 2 − 9 2 i

2.5 Section Exercises

It is a second-degree equation (the highest variable exponent is 2).

We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.

One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .

x = 6 , x = 6 , x = 3 x = 3

x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3

x = 5 , x = 5 , x = −5 x = −5

x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2

x = −2 , 3 x = −2 , 3

x = 0 , x = 0 , x = − 3 7 x = − 3 7

x = −6 , x = −6 , x = 6 x = 6

x = 6 , x = 6 , x = −4 x = −4

x = 1 , x = 1 , x = −2 x = −2

x = −2 , x = −2 , x = 11 x = 11

z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2

x = 3 ± 17 4 x = 3 ± 17 4

One rational

Two real; rational

x = − 1 ± 17 2 x = − 1 ± 17 2

x = 5 ± 13 6 x = 5 ± 13 6

x = − 1 ± 17 8 x = − 1 ± 17 8

x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535

x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7

a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a

x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.

maximum at x = 70 x = 70

The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.

2.6 Section Exercises

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

x = 81 x = 81

x = 17 x = 17

x = 8 ,     x = 27 x = 8 ,     x = 27

x = −2 , 1 , −1 x = −2 , 1 , −1

y = 0 ,     3 2 ,     − 3 2 y = 0 ,     3 2 ,     − 3 2

m = 1 , −1 m = 1 , −1

x = 2 5 , ±3 i x = 2 5 , ±3 i

x = 32 x = 32

t = 44 3 t = 44 3

x = −2 x = −2

x = 4 , −4 3 x = 4 , −4 3

x = − 5 4 , 7 4 x = − 5 4 , 7 4

x = 3 , −2 x = 3 , −2

x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3

x = 2 , −2 x = 2 , −2

x = 1 , 5 x = 1 , 5

x ≥ 0 x ≥ 0

x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8

2.7 Section Exercises

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

( − ∞ , ∞ ) ( − ∞ , ∞ )

We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.

( − ∞ , 3 4 ] ( − ∞ , 3 4 ]

[ − 13 2 , ∞ ) [ − 13 2 , ∞ )

( − ∞ , 3 ) ( − ∞ , 3 )

( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]

All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )

( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )

No solution

( −5 , 11 ) ( −5 , 11 )

[ 6 , 12 ] [ 6 , 12 ]

[ −10 , 12 ] [ −10 , 12 ]

x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ )

x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ ) x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ )

( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )

[ −11 , −3 ] [ −11 , −3 ]

It is never less than zero. No solution.

Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.

( − ∞ , −3 ) ( − ∞ , −3 )

Where the blue line is above the orange line; always. All real numbers.

( − ∞ , − ∞ ) ( − ∞ , − ∞ )

( −1 , 3 ) ( −1 , 3 )

( − ∞ , 4 ) ( − ∞ , 4 )

{ x | x < 6 } { x | x < 6 }

{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }

( −2 , 1 ] ( −2 , 1 ]

( − ∞ , 4 ] ( − ∞ , 4 ]

Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .

Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .

x = 2 , − 4 5 x = 2 , − 4 5

( −7 , 5 ] ( −7 , 5 ]

80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400

[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]

Review Exercises

x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )

y = 5 3 x + 4 y = 5 3 x + 4

72 = 6 2 72 = 6 2

620.097 620.097

midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )

x = 4 x = 4

x = 12 7 x = 12 7

y = 1 6 x + 4 3 y = 1 6 x + 4 3

y = 2 3 x + 6 y = 2 3 x + 6

females 17, males 56

x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4

horizontal component −2 ; −2 ; vertical component −1 −1

7 + 11 i 7 + 11 i

−16 − 30 i −16 − 30 i

−4 − i 10 −4 − i 10

x = 7 − 3 i x = 7 − 3 i

x = −1 , −5 x = −1 , −5

x = 0 , 9 7 x = 0 , 9 7

x = 10 , −2 x = 10 , −2

x = − 1 ± 5 4 x = − 1 ± 5 4

x = 2 5 , − 1 3 x = 2 5 , − 1 3

x = 5 ± 2 7 x = 5 ± 2 7

x = 0 , 256 x = 0 , 256

x = 0 , ± 2 x = 0 , ± 2

x = 11 2 , −17 2 x = 11 2 , −17 2

[ − 10 3 , 2 ] [ − 10 3 , 2 ]

( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )

Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.

( 3.5 , ∞ ) ( 3.5 , ∞ )

Practice Test

y = 3 2 x + 2 y = 3 2 x + 2

( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )

( − ∞ , 9 ] ( − ∞ , 9 ]

x = −15 x = −15

x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1

x = 3 ± 3 2 x = 3 ± 3 2

( −4 , 1 ) ( −4 , 1 )

y = −5 9 x − 2 9 y = −5 9 x − 2 9

y = 5 2 x − 4 y = 5 2 x − 4

5 13 − 14 13 i 5 13 − 14 13 i

x = 2 , − 4 3 x = 2 , − 4 3

x = 1 2 ± 2 2 x = 1 2 ± 2 2

x = 1 2 , 2 , −2 x = 1 2 , 2 , −2

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra/pages/chapter-2

© Dec 8, 2021 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

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Eureka Math Grade 6 Module 1 Lesson 5 Answer Key

Engage ny eureka math 6th grade module 1 lesson 5 answer key, eureka math grade 6 module 1 lesson 5 example answer key.

Example 1. A County Superintendent of Highways is interested in the numbers of different types of vehicles that regularly travel within his county. In the month of August, a total of 192 registrations were purchased for passenger cars and pickup trucks at the local Department of Motor Vehicles (DMV). The DMV reported that in the month of August, for every 5 passenger cars registered, there were 7 pickup trucks registered. How many of each type of vehicle were registered in the county in the month of August? a. Using the information in the problem, write four different ratios and describe the meaning of each. Answer: The ratio of cors to trucks is 5: 7 and is a part-to-part ratio. The ratio of trucks to cors is 7:5, and that is a part-to-part ratio. The ratio of cars to total vehicles is 5 to 12, and that is a part-to-whole ratio. The ratio of trucks to total vehicles is 7 to 12, and that is a part-to-whole ratio.

c. How many equal-sized parts does the tape diagram consist of? Answer: 12

d. What total quantity does the tape diagram represent? Answer: 192 vehicles

e. What value does each individual part of the tape diagram represent? Answer: Divide the total quantity into 12 equal-sized parts: \(\frac{192}{12}\) = 16

f. How many of each type of vehicle were registered in August? Answer: 5 . 16 = 80 passenger cars 7 . 16 = 112 pickup trucks

Eureka Math Grade 6 Module 1 Lesson 5 Exercise Answer Key

Exercise 1. The ratio of the number of people who own a smartphone to the number of people who own a flip phone is 4: 3. If 500 more people own a smartphone than a flip phone, how many people own each type of phone? Answer: 2,000 people own a smartphone, and 1,500 people own a flip phone.

Exercise 2. Sammy and David were selling water bottles to raise money for new football uniforms. Sammy sold 5 water bottles for every 3 water bottles David sold. Together they sold 160 water bottles. How many did each boy sell? Answer: Sammy sold 100 Water bottles, and David sold 60 water bottles.

Exercise 3. Ms. Johnson and Ms. Siple were folding report cards to send home to parents. The ratio of the number of report cards Ms. Johnson folded to the number of report cards Ms. Siple folded is 2: 3. At the end of the day, Ms. Johnson and Ms. Siple folded a total of 300 report cards. How many did each person fold? Answer: Ms. Johnson folded 120 report cards, and Ms. Siple folded 180 report cards.

Exercise 4. At a country concert, the ratio of the number of boys to the number of girls is 2: 7. If there are 250 more girls than boys, how many boys are at the concert? Answer: There are 100 boys at the country concert.

Eureka Math Grade 6 Module 1 Lesson 5 Problem Set Answer Key

Question 1. Last summer, at Camp Okey-Fun-Okey, the ratio of the number of boy campers to the number of girl campers was 8: 7. If there were a total of 195 campers, how many boy campers were there? How many girl campers? Answer: 104 boys and 91 girls are at Camp Okey-Fun-Okey.

Question 2. The student-to-faculty ratio at a small college is 17: 3. The total number of students and faculty is 740. How many faculty members are there at the college? How many students? Answer: 111 faculty members and 629 students are at the college.

Question 3. The Speedy Fast Ski Resort has started to keep track of the number of skiers and snowboarders who bought season passes. The ratio of the number of skiers who bought season passes to the number of snowboarders who bought season passes is 1: 2. If 1,250 more snowboarders bought season passes than skiers, how many snowboarders and how many skiers bought season passes? Answer: 1,250 skiers bought season passes, and 2,500 snowboarders bought season passes.

Question 4. The ratio of the number of adults to the number of students at the prom has to be 1: 10. Last year there were 477 more students than adults at the prom. If the school is expecting the same attendance this year, how many adults have to attend the prom? Answer: 53 adults have to be at the prom to keep the 1: 10 ratio.

Eureka Math Grade 6 Module 1 Lesson 5 Exit Ticket Answer Key

Question 1. When Carla looked out at the school parking lot, she noticed that for every 2 minivans, there were 5 other types of vehicles. If there are 161 vehicles in the parking lot, how many of them are not minivans? Answer: 5 out of 7 vehicles are not minivans. 7 × 23 = 161. So, 5 × 23 = 115. 115 of the vehicles are not minivans.

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Texas Go Math Grade 5 Lesson 3.6 Answer Key Decimal Multiplication

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 3.6 Answer Key Decimal Multiplication.

Investigate

Materials; color pencils

The distance from Charlene’s house to her school is 0.8 mile. Charlene rides her bike 7 tenths of the distance and walks the rest of the way. How far does Charlene ride her bike to school?

B. Using a color pencil, shade columns on the grid to represent the distance to Charlene’s school. The distance to the school is 0.8 mile. How many columns did you shade?____8______

C. Divide the square into 10 equal rows. What decimal value does each row represent? ___0.1_______

D. Using a different color, shade rows that overlap the shaded columns to represent the distance to school that Charlene rides her bike.

• What part of the distance to school does Charlene ride her bike? _____0.56_____
• How many rows of the shaded columns did you shade? _____7_____

E. Count the number of squares that you shaded twice. There are _____56_____ squares. Each square represents ____0.01______. Record the value of the squares as the product. 0.7 × 0.8 = ___0.56_______

Explanation: A. Drawn a square with 10 equal columns as shown above, The decimal value does each column represent is 0.1, B. Using a color pencil, shade columns on the grid to represent the distance to Charlene’s school, The distance to the school is 0.8 mile, So columns shaded are 8, C. Divided the square into 10 equal rows, The decimal value does each row represent is 0.1, D. Using a different color, shade rows that overlap the shaded columns to represent the distance to school that Charlene rides her bike.

• The distance to school does Charlene ride her bike 0.56,
• Number of rows of the shaded columns did he shade is 7.

Make Connections

You can use decimal squares to multiply decimals greater than 1.

Multiply. 0.3 × 1.4 STEP 1: Shade columns to represent 1.4. How many tenths are in 1.4?

STEP 2: Shade rows that overlap the shaded columns to represent 0.3. How many rows of the shaded columns did you shade? _____0.42_______

Explanation: Multiplying 0.3 × 1.4 STEP 1: Shaded columns to represent 1.4,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.3, Therefore rows of the shaded columns shaded are 0.42.

Math Talk Mathematical Processes

Explain why the product is less than only one of the decimal factors. Answer: We are finding a fractional amount of a quantity,

Explanation: When multiplying a number by a decimal less than one, the product will be smaller than the number being multiplied. This is because we are finding a fractional amount of a quantity. For example, 0.1 x 0.8 = 0.08, because the question is asking us to find one tenth of eight tenths.

Share and Show . Multiply. Use the decimal model.

Explanation: Multiplying 0.8 × 0.4 STEP 1: Shaded columns represent 0.8,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.4, Therefore rows of the shaded columns are 0.32.

Explanation: Multiplying 0.1 × 0.7 STEP 1: Shaded columns represents 0.1,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.7, Therefore rows of the shaded columns shaded are 0.32.

Explanation: Multiplying 0.4 × 1.6 STEP 1: Shaded columns represent 0.4,

STEP 2: Shaded rows that overlap the shaded columns to represent 1.6, Therefore rows of the shaded columns are 0.64.

Problem-Solving

Sense or Nonsense?

Explanation: Given Randy and Stacy used area models to find 0.3 of 0.5. Both Randy’s and Stacy’s models are shown above, Randy’s model makes sense which matches with 0.5, Stacy’s model is nosense as it doesnot match with 0.3. the correct answer is 0.3 X 0.5 = 0.15.

Explanation: Given a large bottle contains 1.2 liters of olive oil, A medium-sized bottle has 0.6 times the amount of olive oil as the large bottle, therefore medium-sized bottle contains 1.2 liters X 0.6 = 0.72 liters, Now more olive oil does the large bottle contain than the medium-sized bottle is 1.2 liters – 0.72 liters = 0.48 liters.

Daily Assessment Task

Fill in the bubble completely to show your answer.

Explanation: Given Farmer Green grows 0.9 tons of grain each year. Seven-tenths of the grain grown is barley, So farmer grows each year is 0.9 X 100 = 90 grains and seven-tenth is 90 X 7÷ 10 = 63, the model showing the amount of barley grown each year is 63 matches with (A).

Explanation: Given the multiplication problem modeled by the diagram as frame 1- 10 X 10 = 100 ÷ 100 = 1, and frame 2 as 1 X 10 ÷ 100 = 0.1, therefore result in matches with (C) 1.1 × 0.3 = 0.3.

Question 8. Multi-Step An outdoor swimming race is 1.6 miles long. Simon swims 8 tenths of the course and his friend Sheila swims 9 tenths of the course. How much farther did Sheila swim? (A) 0.16 mile (B) 0.72 mile (C) 0.1 mile (D) 0.2 mile Answer: (C) 0.1 mile,

Explanation: Given an outdoor swimming race is 1.6 miles long, Simon swims 8 tenths of the course means 8÷ 10 = 0.8 and his friendSheila swims 9 tenths of the course so 9 ÷ 10 = 0.9, So much farther did Sheila swims is 0.9 – 0.8 = 0.1 which matches with (C).

Texas Test Prep

Question 9. Alex is 1.3 meters tall. His brother, Raul, is 0.6 times as tall as Alex. How tall is Raul? (A) 1.9 meters (B) 0.7 meter (C) 0.78 meter (D) 7.8 meters Answer: (C) 0.78 meter,

Explanation: Given Alex is 1.3 meters tall. His brother Raul is 0.6 times as tall as Alex. Therefore Raul is 1.3 X 0.6 = 0.78 meters matches with (C).

Texas Go Math Grade 5 Lesson 3.6 Homework and Practice Answer Key

Multiply. Use the decimal model.

STEP 2: Shaded rows that overlap the shaded columns to represent 0.7, Therefore rows of the shaded columns shaded are 0.54.

Explanation: Multiplying 0.5 × 1.7 STEP 1: Shaded columns represents 0.5,

STEP 2: Shaded rows that overlap the shaded columns to represent 1.7, Therefore rows of the shaded columns shaded are 0.85.

STEP 2: Shaded rows that overlap the shaded columns to represent 0.8, Therefore rows of the shaded columns shaded are 0.16.

STEP 2: Shaded rows that overlap the shaded columns to represent 0.7, Therefore rows of the shaded columns are 0.21.

STEP 2: Shaded rows that overlap the shaded columns to represent 1.2, Therefore rows of the shaded columns shaded are 0.84.

Explanation: Multiplying 0.4 × 0.6 STEP 1: Shaded columns represents 0.4,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.6, Therefore rows of the shaded columns shaded are 0.24.

Explanation: Multiplying 0.8 × 0.8 STEP 1: Shaded columns represents 0.8,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.8, Therefore rows of the shaded columns shaded are 0.64.

Explanation: Multiplying 0.4 × 1.9 STEP 1: Shaded columns represent 0.4,

STEP 2: Shaded rows that overlap the shaded columns to represent 1.9, Therefore rows of the shaded columns are 0.76.

Explanation: Multiplying 0.9 × 0.7 STEP 1: Shaded columns represents 0.9,

STEP 2: Shaded rows that overlap the shaded columns to represent 0.7, Therefore rows of the shaded columns shaded are 0.63.

Problem Solving

Explanation: Given Vanesh uses 1.6 gallons of paint for his bedroom walls. For his study, he uses 0.7 times the amount of paint as he uses for his bedroom is 1.6 X 0.7 = 1.12 gallons of paint Vanesh used to paint his study.

Lesson Check

Question 12. Alex’s tablet computer weighs 1.5 pounds. The weight of his calculator is 0.6 times the weight of his tablet computer. What is the weight of Alex’s calculator? (A) 2.1 pounds (B) 1.56 pounds (C) 0.3 pound (D) 0.9 pound Answer: (D) 0.9 pound,

Explanation: Given Alex’s tablet computer weighs 1.5 pounds. The weight of his calculator is 0.6 times the weight of his tablet computer. So the weight of Alex’s calculator is 1.5 pounds X 0.6 = 0.9 pound matches with (D).

Explanation: Given model above is 0.1 X 0.1 = 0.01, So each block is 0.1 and we have 7 blocks like that 0.1 X 0.7 = 0.07 which matches with 0.07.

Question 14. Sasha’s hundredths decimal model shows the product 0.72. If she shaded 9 columns, what factor represents the shaded rows? (A) 0.6 (B) 0.9 (C) 0.7 (D) 0.8 Answer: (D) 0.8,

Explanation: Given Sasha’s hundredths decimal model shows the product 0.72. If she shaded 9 columns factor represents the shaded rowsis 0.72 ÷ 0.9 = 0.8 matches with (D).

Question 15. The postman delivers a package and an envelope to Mr. O’Connor. The package weighs 1.8 pounds. The envelope is 0.8 times the weight of the package. What is the weight of the envelope? (A) 1.44 pounds (B) 1 pound (C) 1.16 pounds (D) 1.64 pounds Answer: (A) 1.44 pounds,

Explanation: Given the postman delivers a package and an envelope to Mr. O’Connor. The package weighs 1.8 pounds. The envelope is 0.8 times the weight of the package. The weight of the envelope is 1.8 pounds X 0.8 = 1.44 pounds matches with (A).

Question 16. Multi-Step Carter solves 0.5 × 0.5. Liam solves 0.4 × 1.4. How much greater is Liam’s product than Carter’s product? (A) 0.31 (B) 0.56 (C) 0.25 (D) 0.81 Answer: (A) 0.31,

Explanation: Given Carter solved 0.5 X 0.5 as = 0.25 and Liam solved greater as 0.4 X 1.4 = 0.56, Greater is Liam’s product than Carter’s product is 0.56 – 0.25 = 0.31 matches with (A).

Question 17. Multi-Step Jessica makes punch for her party. She uses 1.6 liters of orange juice. The amount of cranberry juice she uses is 0.5 times the amount of orange juice, and the amount of apple juice is 0.5 times the amount of cranberry juice. What is the total amount of juice in Jessica’s punch? (A) 2.6 liters (B) 2.65 liters (C) 2.8 liters (D) 1.8 liters Answer: (C) 2.8 liters,

Explanation: Given Jessica makes punch for her party. She uses 1.6 liters of orange juice. The amount of cranberry juice she uses is 0.5 times the amount of orange juice, So cranberry juice Jessica makes is 1.6 liters X 0.5 = 0.8 liter + 1.6 liters = 2.4 liters and the amount of apple juice is 0.5 times the amount of cranberry juice, So the amount of apple juice is 0.8 liter X 0.5 = 0.4 liter + 2.4 liters = 2.8 liters of cranberry juice is the total amount of juice in Jessica’s punch.

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