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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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Home > CC3 > Chapter 6 > Lesson 6.2.3

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Eureka Math Grade 2 Answer Key | Engage NY Math 2nd Grade Answer Key Solutions

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Math Expressions Grade 5 Unit 3 Lesson 13 Answer Key Review Operations with Fractions

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Math Expressions Common Core Grade 5 Unit 3 Lesson 13 Answer Key Review Operations with Fractions

Math Expressions Grade 5 Unit 3 Lesson 13 Homework

Review Operations with Fractions Math Expressions Grade 5 Unit 3 Lesson 13 Question 1. Dan’s Ice Cream comes in cartons of two sizes. The large carton holds 4\(\frac{1}{2}\) pounds. The small carton holds 1\(\frac{3}{4}\) pounds less. How much ice cream does the small carton hold?

Answer: The small carton holds 2.25 pounds ice cream.

Explanation: In the above-given question, given that, Dan’s Ice Cream comes in cartons of two sizes. The large carton holds 4\(\frac{1}{2}\) pounds. The small carton holds 1\(\frac{3}{4}\) pounds less. 4(1/2) – 1(3/4). 9/2 – 7/4. 9/2 = 4.5. 7/4 = 1.75. 4.5 – 1.75 = 2.25. so the small carton holds 2.25 pounds of ice cream.

Question 2. Mac picked four baskets of blueberries. The weights of the berries in pounds are given below. Order the weights from lightest to heaviest. \(\frac{5}{4}\) \(\frac{9}{10}\) \(\frac{4}{5}\) \(\frac{13}{20}\)

Answer: The weights from lightest to heaviest = 13/20, 4/5, 9/10, and 5/4.

Explanation: In the above-given question, given that, Mac picked four baskets of blueberries. Order the weights from lightest to heaviest. 5/4 = 1.25. 9/10 = 0.9. 4/5 = 0.8. 13/20 = 0.65. so the weights from lightest to heaviest = 13/20, 4/5, 9/10, and 5/4.

Question 3. Four cones of Dan’s Ice Cream hold \(\frac{1}{2}\) pound. How much ice cream does each cone hold?

Answer: The ice cream does each cone holds = 3.5 pounds.

Explanation: In the above-given question, given that, Four cones of Dan’s Ice Cream hold \(\frac{1}{2}\) pound. 1/2 = 0.5. 4 – 0.5 = 3.5. so the ice cream does each cone holds = 3.5 pounds.

Question 4. If a dish of ice cream holds \(\frac{1}{4}\) pound, how many dishes can you get from a 4\(\frac{1}{2}\)-pound carton of Dan’s Ice Cream?

Answer: Dan’s carton of Ice cream = 4.25.

Explanation: In the above-given question, given that, If a dish of ice cream holds \(\frac{1}{4}\) pound. 4(1/2) = 9/2. 9/2 = 4.5. 1/4 = 0.25. 4.5 – 0.25 = 4.25.

Solve. Give your answer in simplest form.

Question 5. 3 ÷ \(\frac{1}{5}\) = ___

Answer: 3 ÷ 0.2 = 15.

Explanation: In the above-given question, given that, the fractions are 3 and 1/5. divide the fractions. 3 ÷ 1/5. 1/5 = 0.2. 3 ÷ 0.2 = 15.

Question 6. 1\(\frac{3}{4}\) + \(\frac{11}{16}\) = ___

Answer: 1(3/4) + 11/16 = 2.58.

Explanation: In the above-given question, given that, the fractions are 7/4 and 11/6. add the fractions. 7/4 = 1.75. 11/6 = 1.83. 1.75 + 1.83 = 2.58.

Question 7. \(\frac{9}{14}\) . 2\(\frac{1}{3}\) = ___

Answer: 9/14 . 2(1/3) = 1.472.

Explanation: In the above-given question, given that, the fractions are 9/14 and 2(1/3). multiply the fractions. 7/3 = 2.3. 9/14 = 0.64. 2.3 x 0.64 = 1.472.

Question 8. 2\(\frac{2}{3}\) . 6 = ___

Answer: 2(2/3) . 6 = 15.6.

Explanation: In the above-given question, given that, the fractions are 2(2/3) and 6. multiply the fractions. 2(2/3) = 8/3. 8/3 = 2.6. 2.6 . 6 = 15.6.

Question 9. \(\frac{1}{3}\) + \(\frac{2}{5}\) = ___

Answer: 1/3 + 2/5 = 0.7.

Explanation: In the above-given question, given that, the fractions are 1/3 and 2/5. add the fractions. 1/3 = 0.3. 2/5 = 0.4. 0.3 + 0.4 = 0.7.

Question 10. \(\frac{5}{6}\) + \(\frac{8}{9}\) = ___

Answer: 5/6 + 8/9 = 1.71.

Explanation: In the above-given question, given that, the fractions are 5/6 and 8/9. add the fractions. 5/6 = 0.83. 8/9 = 0.88. 0.83 + 0.88 = 1.71.

Question 11. \(\frac{1}{8}\) ÷ 4 = ___

Answer: 1/8 / 4 = 0.075.

Explanation: In the above-given question, given that, the fractions are 1/8 and 4. divide the fractions. 1/8 = 0.3. 0.3 / 4 = 0.075.

Question 12. \(\frac{2}{5}\) – \(\frac{1}{10}\) = ___

Answer: 2/5 – 1/10 = 0.3.

Explanation: In the above-given question, given that, the fractions are 2/5 and 1/10. subtract the fractions. 1/10 = 0.1. 2/5 = 0.4. 0.4 – 0.1 = 0.3.

Question 13. 3\(\frac{5}{7}\) – 1\(\frac{1}{2}\) = ____

Answer: 3(5/7) – 1(1/2) = 2.2.

Explanation: In the above-given question, given that, the fractions are 3(5/7) and 1(1/2). subtract the fractions. 26/7 = 3.7. 3/2 = 1.5. 3.7 – 1.5 = 2.2.

Question 14. \(\frac{7}{8}\) ∙ \(\frac{2}{7}\) = ___

Answer: 7/8 . 2/7 = 0.245.

Explanation: In the above-given question, given that, the fractions are 7/8 and 2/7. multiply the fractions. 7/8 = 0.875. 2/7 = 0.28. 0.875 . 0.28 = 0.245.

Math Expressions Grade 5 Unit 3 Lesson 13 Remembering

Use benchmarks of 0, \(\frac{1}{2}\), and 1 to estimate the sum or difference. Then find the actual sum or difference.

Question 1. \(\frac{5}{10}\) + \(\frac{4}{9}\) Estimate: ____ Sum: ____

Answer: 5/10 + 4/9 = 0.9.

Explanation: In the above-given question, given that, the fractions are 5/10 and 4/9. add the fractions. 5/10 = 0.5. 4/9 = 0.4. 0.5 + 0.4 = 0.9.

Question 2. \(\frac{13}{14}\) – \(\frac{3}{7}\) Estimate: ____ Difference: ____

Answer: 13/14 – 3/7 = 0.5.

Explanation: In the above-given question, given that, the fractions are 13/14 and 3/7. subtract the fractions. 13/14 = 0.92. 3/7 = 0.42. 0.92 – 0.42 = 0.5.

Question 3. \(\frac{8}{9}\) – \(\frac{7}{8}\) Estimate: ____ Difference: ____

Answer: 8/9 – 7/8 = 0.01.

Explanation: In the above-given question, given that, the fractions are 8/9 and 7/8. subtract the fractions. 8/9 = 0.88. 7/8 = 0.87. 0.88 – 0.87 = 0.01.

Question 4. \(\frac{13}{14}\) + \(\frac{3}{4}\) Estimate: ____ Sum: ____

Answer: 13/14 + 3/4 = 1.67.

Explanation: In the above-given question, given that, the fractions are 13/14 and 3/4. add the fractions. 13/14 = 0.92. 3/4 = 0.75. 0.92 + 0.75 = 1.67.

Write an equation. Then solve. Show your Work.

Question 5. A rectangle has an area of 20 square feet and a length of 6 feet. What is its width?

Answer: The width of the rectangle = 3.3 feet.

Explanation: In the above-given question, given that, A rectangle has an area of 20 square feet and a length of 6 feet. area of the rectangle = l x b. 20 = 6 x w. 20/6 = w. w = 3.3 feet.

Question 6. Bailey attends gymnastics practice for 8 hours each week. This is \(\frac{1}{4}\) the number of hours that the gym is open for practice. How many hours is the gym open for practice?

Answer: The number of hours is the gym open for practice = 2.

Explanation: In the above-given question, given that, Bailey attends gymnastics practice for 8 hours each week. This is \(\frac{1}{4}\) the number of hours that the gym is open for practice. 8 x 1/4. 2 x 1 = 2. so the number of hours is the gym open for practice = 2.

Question 7. \(\frac{1}{4}\) ÷ 3 = ___

Answer: 1/4 ÷ 3 = 0.08.

Explanation: In the above-given question, given that, the fractions are 1/4 and 3. divide the fractions. 1/4 = 0.25. 0.25 / 3 = 0.08.

Question 8. \(\frac{1}{4}\) . 3 = ___

Answer: 1/4 . 3 = 0.75.

Explanation: In the above-given question, given that, the fractions are 1/4 and 3. multiply the fractions. 1/4 = 0.25. 0.25 . 3 = 0.75.

Question 9. 14 . \(\frac{1}{6}\) = ___

Answer: 14 . 1/6 = 0.75.

Explanation: In the above-given question, given that, the fractions are 14 and 1/6. multiply the fractions. 1/6 = 0.25. 0.25 . 3 = 0.75.

Question 10. Stretch Your Thinking How is solving \(\frac{1}{8}\) ÷ 5 different from solving \(\frac{1}{8}\) . 5?

Answer: Yes, both of them are different.

Explanation: In the above-given question, given that, the fractions are 1/8 and 5. multiply and divide the fractions. 1/8 = 0.125. 0.125 x 5 = 0.625. 0.125 / 5 = 0.025. so both of them are different.

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