7.6 homework handout geometry answer key

Geometry Worksheets(pdf)

Free worksheets with answer keys.

Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

• Graphic Organizer on All Formulas
• Interior Angles of Polygons
• Exterior Angles of Polygons
• Similar Polygons
• Area of Triangle
• Interior Angles of Triangle

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

Popular pages @ mathwarehouse.com.

7.1 Solving Trigonometric Equations with Identities

sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ

This is a difference of squares formula: 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) . 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) .

7.2 Sum and Difference Identities

2 + 6 4 2 + 6 4

2 − 6 4 2 − 6 4

1 − 3 1 + 3 1 − 3 1 + 3

cos ( 5 π 14 ) cos ( 5 π 14 )

7.3 Double-Angle, Half-Angle, and Reduction Formulas

cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32

cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ ) cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ )

cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ

10 cos 4 x = 10 cos 4 x = 10 ( cos 2 x ) 2              = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x .              = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ]              = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x .              = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x )              = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x )              = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 cos 4 x = 10 ( cos 2 x ) 2              = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x .              = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ]              = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x .              = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x )              = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x )              = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

− 2 5 − 2 5

7.4 Sum-to-Product and Product-to-Sum Formulas

1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )

1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )

− 2 − 3 4 − 2 − 3 4

2 sin ( 2 θ ) cos ( θ ) 2 sin ( 2 θ ) cos ( θ )

tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ

7.5 Solving Trigonometric Equations

x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6

π 3 ± π k π 3 ± π k

θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k

cos θ = − 1 , θ = π cos θ = − 1 , θ = π

π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2

7.6 Modeling with Trigonometric Functions

The amplitude is 3 , 3 , and the period is 2 3 . 2 3 .

y = 8 sin ( π 12 t ) + 32 y = 8 sin ( π 12 t ) + 32 The temperature reaches freezing at noon and at midnight.

initial displacement =6, damping constant = -6, frequency = 2 π 2 π

y = 10 e − 0.5 t cos ( π t ) y = 10 e − 0.5 t cos ( π t )

y = 5 cos ( 6 π t ) y = 5 cos ( 6 π t )

7.1 Section Exercises

All three functions, F F , G G , and H , H , are even.

This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .

When cos t = 0 , cos t = 0 , then sec t = 1 0 , sec t = 1 0 , which is undefined.

sin x sin x

sec x sec x

csc t csc t

sec 2 x sec 2 x

sin 2 x + 1 sin 2 x + 1

1 sin x 1 sin x

1 cot x 1 cot x

tan x tan x

− 4 sec x tan x − 4 sec x tan x

± 1 cot 2 x + 1 ± 1 cot 2 x + 1

± 1 − sin 2 x sin x ± 1 − sin 2 x sin x

Answers will vary. Sample proof:

cos x − cos 3 x = cos x ( 1 − cos 2 x ) cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x = cos x sin 2 x

Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x

Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x

Proved with negative and Pythagorean Identities

True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ

7.2 Section Exercises

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) . sin x = cos ( π 2 − x ) . The same holds for the other cofunction identities. The key is that the angles are complementary.

sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.

6 − 2 4 6 − 2 4

− 2 − 3 − 2 − 3

− 2 2 sin x − 2 2 cos x − 2 2 sin x − 2 2 cos x

− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x

csc θ csc θ

cot x cot x

tan ( x 10 ) tan ( x 10 )

sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15

cot ( π 6 − x ) cot ( π 6 − x )

cot ( π 4 + x ) cot ( π 4 + x )

sin x 2 + cos x 2 sin x 2 + cos x 2

They are the same.

They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .

They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .

They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .

− 3 − 1 2 2 ,  or  − 0.2588 − 3 − 1 2 2 ,  or  − 0.2588

1 + 3 2 2 , 1 + 3 2 2 , or 0.9659

tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x

cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b

cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h

True. Note that sin ( α + β ) = sin ( π − γ ) sin ( α + β ) = sin ( π − γ ) and expand the right hand side.

7.3 Section Exercises

Use the Pythagorean identities and isolate the squared term.

1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.

a) 3 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 7 31

a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3

cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2

2 sin ( π 2 ) 2 sin ( π 2 )

2 − 2 2 2 − 2 2

2 − 3 2 2 − 3 2

2 + 3 2 + 3

− 1 − 2 − 1 − 2

a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2

a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3

120 169 , – 119 169 , – 120 119 120 169 , – 119 169 , – 120 119

2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3

cos ( 74 ∘ ) cos ( 74 ∘ )

cos ( 18 x ) cos ( 18 x )

3 sin ( 10 x ) 3 sin ( 10 x )

− 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x ) − 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x )

sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = tan θ tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = tan θ tan 2 θ = tan 3 θ

1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2

3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8

2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32

3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )

1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8

3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 )

( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2

4 sin x cos x ( cos 2 x − sin 2 x ) 4 sin x cos x ( cos 2 x − sin 2 x )

2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x ) 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x )

2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )

sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x

1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1

( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )

7.4 Section Exercises

Substitute α α into cosine and β β into sine and evaluate.

Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( 3 x ) + sin x cos x = 1. sin ( 3 x ) + sin x cos x = 1. When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos x = 1 2 sin ( 2 x ) cos x cos x = 1

8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )

sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )

1 2 ( cos ( 6 x ) − cos ( 4 x ) ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) )

2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t

2 cos ( 7 x ) 2 cos ( 7 x )

2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )

1 4 ( 1 + 3 ) 1 4 ( 1 + 3 )

1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )

1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )

cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )

1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )

2 cos ( 31° ) 2 cos ( 31° )

2 cos ( 66.5 ° ) sin ( 34.5 ° ) 2 cos ( 66.5 ° ) sin ( 34.5 ° )

2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )

2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )

sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x 4 sin x cos 2 x

2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )

2 cos ( 35 ∘ ) cos ( 23 ∘ ) ,  1 .5081 2 cos ( 35 ∘ ) cos ( 23 ∘ ) ,  1 .5081

− 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078 − 2 sin ( 33 ∘ ) sin ( 11 ∘ ) , − 0.2078

1 2 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410 1 2 ( cos ( 99 ∘ ) − cos ( 71 ∘ ) ) , − 0.2410

It is and identity.

It is not an identity, but 2 cos 3 x 2 cos 3 x is.

tan ( 3 t ) tan ( 3 t )

2 cos ( 2 x ) 2 cos ( 2 x )

− sin ( 14 x ) − sin ( 14 x )

Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + cos y becomes cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β

Since x = α + β x = α + β and y = α − β , y = α − β , we can solve for α α and β β in terms of x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .

cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x

cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y

cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x

tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t

7.5 Section Exercises

There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

π 3 , 2 π 3 π 3 , 2 π 3

3 π 4 , 5 π 4 3 π 4 , 5 π 4

π 4 , 5 π 4 π 4 , 5 π 4

π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4

π 4 , 7 π 4 π 4 , 7 π 4

7 π 6 , 11 π 6 7 π 6 , 11 π 6

π 18 π 18 , 5 π 18 5 π 18 , 13 π 18 13 π 18 , 17 π 18 17 π 18 , 25 π 18 25 π 18 , 29 π 18 29 π 18

3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12

1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6

0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3

π 3 , π , 5 π 3 π 3 , π , 5 π 3

π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3

0 , π 0 , π

π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )

1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )

θ = sin - 1 2 3 , π - sin - 1 2 3 , π + sin - 1 2 3 , 2 π - sin - 1 2 3 θ = sin - 1 2 3 , π - sin - 1 2 3 , π + sin - 1 2 3 , 2 π - sin - 1 2 3

3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6

0 , π 3 , π , 4 π 3 0 , π 3 , π , 4 π 3

There are no solutions.

cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )

tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )

0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3

sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2

cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) 2 π − cos − 1 ( − 1 4 )

π 3 , cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) , 5 π 3 π 3 , cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) , 5 π 3

cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )

0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2

π 3 , cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) , 5 π 3 π 3 , cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) , 5 π 3

π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 )

2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682

π k − 0.3277 π k − 0.3277

0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703

1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360

0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2

π 2 , 3 π 2 π 2 , 3 π 2

7.2 ∘ 7.2 ∘

5.7 ∘ 5.7 ∘

82.4 ∘ 82.4 ∘

31.0 ∘ 31.0 ∘

88.7 ∘ 88.7 ∘

59.0 ∘ 59.0 ∘

36.9 ∘ 36.9 ∘

7.6 Section Exercises

Physical behavior should be periodic, or cyclical.

Since cumulative rainfall is always increasing, a sinusoidal function would not be ideal here.

y = − 3 cos ( π 6 x ) − 1 y = − 3 cos ( π 6 x ) − 1

5 sin ( 2 x ) + 2 5 sin ( 2 x ) + 2

y = 4 - 6 cos ( x π 2 ) y = 4 - 6 cos ( x π 2 )

y = tan ( x π 8 ) y = tan ( x π 8 )

tan ( x π 12 ) tan ( x π 12 )

From June 15 through November 16

From day 31 through day 58

Floods: April 16 to July 15. Drought: October 16 to January 15.

Amplitude: 8, period: 1 3 , 1 3 , frequency: 3 Hz

Amplitude: 4, period: 4 , 4 , frequency: 1 4 1 4 Hz

P ( t ) = − 19 cos ( π 6 t ) + 800 + 160 12 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 40 3 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 160 12 t P ( t ) = − 19 cos ( π 6 t ) + 800 + 40 3 t

P ( t ) = − 33 cos ( π 6 t ) + 900 + ( 1.07 ) t P ( t ) = − 33 cos ( π 6 t ) + 900 + ( 1.07 ) t

D ( t ) = 10 ( 0.85 ) t cos ( 36 π t ) D ( t ) = 10 ( 0.85 ) t cos ( 36 π t )

D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t ) D ( t ) = 17 ( 0.9145 ) t cos ( 28 π t )

15.4 seconds

Spring 2 comes to rest first after 7.3 seconds.

234.3 miles, at 72.2°

y = 6 ( 4 ) x + 5 sin ( π 2 x ) y = 6 ( 4 ) x + 5 sin ( π 2 x )

y = 4 ( – 2 ) x + 8 sin ( π 2 x ) y = 4 ( – 2 ) x + 8 sin ( π 2 x )

y = 3 ( 2 ) x cos ( π 2 x ) + 1 y = 3 ( 2 ) x cos ( π 2 x ) + 1

Review Exercises

sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 )

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )

cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x                                    = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x                                    = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x                                    = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x                                    = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x                                    = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x                                    = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x                                    = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x                                    = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x                                    = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x                                    = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x                                    = sin 2 x − 4 cos 2 x sin 2 x

tan ( 5 8 x ) tan ( 5 8 x )

− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7

2 ( 2 + 2 ) 2 ( 2 + 2 )

2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4

cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x )                      = cot x − cos x sin x ( 2 ) sin 2 x                      = − 2 sin x cos x + cot x                      = − sin ( 2 x ) + cot x cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x )                      = cot x − cos x sin x ( 2 ) sin 2 x                      = − 2 sin x cos x + cot x                      = − sin ( 2 x ) + cot x

10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )

− 2 2 − 2 2

1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )

2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )

3 π 4 , 7 π 4 3 π 4 , 7 π 4

0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π

3 π 2 3 π 2

No solution

0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124

1.3694 1.3694 , 1.9106 1.9106 , 4.3726 4.3726 , 4.9137 4.9137

3 sin ( x π 2 ) − 2 3 sin ( x π 2 ) − 2

71.6 ∘ 71.6 ∘

P ( t ) = 950 − 450 sin ( π 6 t ) P ( t ) = 950 − 450 sin ( π 6 t )

Amplitude: 3, period: 2, frequency: 1 2 1 2 Hz

C ( t ) = 20 sin ( 2 π t ) + 100 ( 1.4427 ) t C ( t ) = 20 sin ( 2 π t ) + 100 ( 1.4427 ) t

Practice Test

π 2 , 3π 2 π 2 , 3π 2

2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )

x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )

3 5 , − 4 5 , − 3 4 3 5 , − 4 5 , − 3 4

tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan – x )

sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x

Amplitude: 1 4 1 4 , period 1 60 1 60 , frequency: 60 Hz

Amplitude: 8 8 , fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz

D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 seconds

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Precalculus
• Publication date: Oct 23, 2014
• Location: Houston, Texas
• Book URL: https://openstax.org/books/precalculus/pages/1-introduction-to-functions
• Section URL: https://openstax.org/books/precalculus/pages/chapter-7

© 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.

• Texas Go Math
• Big Ideas Math
• Engageny Math
• McGraw Hill My Math
• enVision Math
• 180 Days of Math
• Math in Focus Answer Key
• Math Expressions Answer Key
• Privacy Policy

Everyday Math Grade 6 Answers Unit 7 Variables and Algebraic Relationships

Everyday mathematics 6th grade answer key unit 7 variables and algebraic relationships, everyday mathematics grade 6 home link 7.1 answers.

c. List three numbers that could be the mystery number. Check that they are in the solution sets for both inequalities. Possible numbers f could be: ____________ Answer: From part (b), We know that, When we observe the graph of both the inequalities, the mystery numbers for both lie between 12 and 13 only Hence, from the above, We can conclude that The possible numbers of f could be 12.1, 12.2,……………… 12.9

b. Write a different set of inequalities that could also represent the graph in Problem 2a. Inequality C: ____________ Inequality D: ____________ Answer: From part (a), Inequality A: x > 3 Inequality B: x < 7 So, The representation of inequality A and B in a different set of inequalities are: Inequality C: Different set inequality of Inequality A Inequality D: Different set Inequality of Inequality B So, Inequality C: x + 5 > 8 Inequality D: x – 4 < 3

Practice Evaluate. Question 3. |-4| = ________ Answer: We know that, | x| = x for x > 0 | x| = -x for x < 0 Hence, from the above, We can conclude that | -4| = 4 or -4

Question 4. |-0.5| = _____ Answer: We know that, | x| = x for x > 0 | x| = -x for x < 0 Hence, from the above, We can conclude that | -0.5| = 0.5 or -0.5

Question 5. |z| = 6; z = ______ Answer: The given expression is: | z | = 6 We know that, | x| = x for x > 0 | x| = -x for x < 0 Hence, from the above, We can conclude that z = 6 or -6

Everyday Math Grade 6 Home Link 7.2 Answer Key

Solving Problems with Inequalities Fast and Healthy sells bags of trail mix. Customers choose the ingredients to put in their trail mix. The bag is weighed at the checkout counter to determine the cost. Fast and Healthy charges $5 per pound. They also sell granola bars for $1.50 each. Question 1. Li has $9 to spend on trail mix. How many pounds of trail mix can she buy? Let y be the number of pounds of trail mix. a. Inequality for the situation: ___________ Answer: It is given that Fast and Healthy charges $5 per pound and they also sell granola bars for $1.50 each It is also given that Li has $9 to spend on trail mix Now, Let y be the number of pounds of trail mix So, The total money charged by Fast and Healthy is: $5y pounds But, Li can only buy the trail mix that is less than or equal to $9 Hence, from the above, We can conclude that the inequality for the above situation is: 5y ≤ 9

b. Solution set for y using set notation: ____________ Answer: From part (a), The inequality for the given situation is: 5y ≤ 9 y ≤ \(\frac{9}{5}\) y ≤ 1.8 We know that, The money will not be negative Hence, from the above, We can conclude that the solution set for the given inequality is: {All numbers should be greater than 0 and less than or equal to 1.8}

c. Inequalities for the values of y: _____________ Answer: From part (b), The inequality for the given situation is: y ≤ 1.8 We know that, The money will not be negative Hence, from the above, We can conclude that the inequalities for the value of y are: y > 0 and y ≤ 1.8

Question 2. The price for a plain smoothie is $2.00. Each additional ingredient costs $0.75. Li has $5. Let m be the number of ingredients. How many ingredients can Li add to a plain smoothie? a. Inequality for the situation: __________ Answer: It is given that the price for a plain smoothie is $2 and each additional ingredient costs $0.75 It is also given that Li has only $5 Now, Let m be the number of additional ingredients So, The total cost of additional ingredients is: $0.75m So, The number of ingredients that Li can add to a plain smoothie is: 2 + 0.75m But, We know that Li can only spend $5 So, 2 + 0.75m ≤ 5 Subtract with 2 on both sides 2 + 0.75m – 2 ≤ 5 – 2 0.75m ≤ 3 Hence, from the above, We can conclude that The inequality for the given situation is: 0.75m ≤ 3

b. Solution set for ‘m’ using set notation: __________ Answer: From part (a), The inequality for the given situation is: 0.75m ≤ 3 m ≤ \(\frac{3}{0.75}\) m ≤ 4 We know that, The number of ingredients will not be negative Hence, from the above, We can conclude that The solution set for the given inequality is: {All numbers greater than or equal to 0 and less than or equal to 4}

c. Inequalities for the whole number values of m: __________ Answer: From part (b), The inequality for the given situation is: m ≤ 4 We know that, The “Whole numbers” are the numbers that start from 0 Hence, from the above, We can conclude that The whole number set for the inequality in the given situation is: {0, 1, 2, 3, 4}

Question 3. Describe how the graph in Problem 2d represents the solution to the problem. Answer: We know that, The ingredients will not be the decimal numbers like 1.5, 2.5, etc So, Only the whole numbers will be the solution set for the problem 2d Hence, from the graph present in problem 2d, The solution set of the problem 2d is: {0, 1, 2, 3, 4}

Practice Solve. Question 4. \(\frac{2}{3}\) × _____ = 1 Answer: Let the missing number be x So, \(\frac{2}{3}\) × x = 1 Multiply by \(\frac{3}{2}\) on both sides So, \(\frac{2}{3}\) × \(\frac{3}{2}\) × x = 1 × \(\frac{3}{2}\) x = \(\frac{3}{2}\) Hence, from the above, We can conclude that the value of x from the given expression is: \(\frac{3}{2}\)

Question 5. ______ × 5 = 1 Answer: Let the missing number be x So, x × 5 = 1 Divide by 5 on both sides So, x × \(\frac{5}{5}\) = \(\frac{1}{5}\) x = \(\frac{1}{5}\) Hence, from the above, We can conclude that the value of x from the given expression is: \(\frac{1}{5}\)

Question 6. 3 \(\frac{3}{4}\) × ______ = 1 Answer: Let the missing number be x So, 3\(\frac{3}{4}\) × x = 1 We know that, 3\(\frac{3}{4}\) = \(\frac{15}{4}\) So, \(\frac{15}{4}\) × x = 1 Multiply by \(\frac{4}{15}\) on both sides So, \(\frac{15}{4}\) × \(\frac{4}{15}\) × x = 1 × \(\frac{4}{15}\) x = \(\frac{4}{15}\) Hence, from the above, We can conclude that the value of x from the given expression is: \(\frac{4}{15}\)

Everyday Mathematics Grade 6 Home Link 7.3 Answers

Question 4. Describe two differences between the two graphs in Problem 2. Answer: In Problem 2, From the 1st graph, We can observe that the graph is a straight line that increases as x increases From the 2nd graph, We can observe that the graph is a curve that is decreasing as x decreases

Practice Find the GCF. Question 4. GCF (34, 42) = _______ Answer: We know that, “GCF” is defined as the greatest common factor of the 2 numbers So, Factors of 42 and 34 are: 42 = 7 × 2 × 3 34 = 2 × 17 Hence, from the above, We can conclude that GCF (34, 42) = 2

Question 5. GCF (49, 560) = ________ Answer: We know that, “GCF” is defined as the greatest common factor of the 2 numbers So, Factors of 49 and 560 are: 49 = 7 × 7 560 = 7 × 5 × 4 × 4 Hence, from the above, We can conclude that GCF (49, 560) = 7

Question 6. GCF (30, 75) = _______ Answer: We know that, “GCF” is defined as the greatest common factor of the 2 numbers So, Factors of 30 and 75 are: 30 = 15 × 2 75 = 15 × 5 Hence, from the above, We can conclude that GCF (30, 75) = 15

Everyday Math Grade 6 Home Link 7.4 Answer Key

Question 2. To solve Problem 1a, why might you start with an even number of pennies? Answer: To solve problem 1a, We might have to start with an even number of pennies because you will divide by 2 for the number of dimes, which only come in whole quantities

Question 3. What formula would you use to find the total value of the coins? Answer: The formula used to find the total number of coins is: The total number of coins = $0.05 (Number of nickels) + $0.10 (Number of dimes) + $0.01 (Number of pennies)

Practice Divide. Question 5. \(\frac{7}{8}\) ÷ \(\frac{1}{8}\) = ______ Answer: The given expression is: \(\frac{7}{8}\) ÷ \(\frac{1}{8}\) So, \(\frac{7}{8}\) ÷ \(\frac{1}{8}\) = \(\frac{7}{8}\) × \(\frac{8}{1}\) = \(\frac{7 × 8}{8 × 1}\) = \(\frac{7}{1}\) = 7 Hence, from the above, We can conclude that the value of the given expression is: 7

Question 6. 2 \(\frac{1}{4}\) ÷ \(\frac{7}{8}\) = ______ Answer: The given expression is: 2 \(\frac{1}{4}\) ÷ \(\frac{7}{8}\) We know that, 2\(\frac{1}{4}\) = \(\frac{9}{4}\) So, \(\frac{9}{4}\) ÷ \(\frac{7}{8}\) =  \(\frac{9}{4}\) × \(\frac{8}{7}\) = \(\frac{9 × 8}{4 × 7}\) = \(\frac{18}{7}\) Hence, from the above, We can conclude that the value of the given expression is: \(\frac{18}{7}\)

Question 7. _____ = 1 \(\frac{2}{3}\) ÷ 3 Answer: The given expression is: 1 \(\frac{2}{3}\) ÷ 3 We know that, 1\(\frac{2}{3}\) = \(\frac{5}{3}\) So, \(\frac{5}{3}\) ÷ 3 =  \(\frac{5}{3}\) × \(\frac{1}{3}\) = \(\frac{5 × 1}{3 × 3}\) = \(\frac{5}{9}\) Hence, from the above, We can conclude that the value of the given expression is: \(\frac{5}{9}\)

Everyday Mathematics Grade 6 Home Link 7.5 Answers

Question 4. On Monday Edgar ran for 29 minutes and burned 270 calories. On Wednesday he biked for 25 minutes and burned 207 calories. On Friday he played soccer for 13 minutes and burned 124 calories. Which activity burns the most calories per minute? Explain how you know. Answer: It is given that On Monday Edgar ran for 29 minutes and burned 270 calories. On Wednesday he biked for 25 minutes and burned 207 calories. On Friday he played soccer for 13 minutes and burned 124 calories. So, The Calories / Minute burn by Edgar on Monday = \(\frac{Number of calories}{Number of Minutes}\) = \(\frac{270}{29}\) = 9.31 Calories / Minute The Calories / Minute burn by Edgar on Wednesday = \(\frac{Number of calories}{Number of Minutes}\) = \(\frac{207}{25}\) = 8.28 Calories / Minute The Calories / Minute burn by Edgar on Friday = \(\frac{Number of calories}{Number of Minutes}\) = \(\frac{124}{13}\) = 9.53 Calories / Minute Hence, from the above, We can conclude that soccer activity burns the most Calorie / Minute

Practice Find the LCM. Question 5. LCM (12, 48) = _______ Answer: We know that, “LCM” is the lowest common divisor of the 2 numbers that are evenly divisible by  both the numbers Now, The multiples of 12 are: 12 = 12, 24, 36, 48, 60 The multiples of 48 are: 48 = 48, 128, 144, 192, 240 Hence, from the above, We can conclude that LCM (12, 48) = 48

Question 6. LCM (14, 21) = _________ Answer: We know that, “LCM” is the lowest common divisor of the 2 numbers that are evenly divisible by  both the numbers Now, The multiples of 14 are: 14 = 14, 28, 42, 56, 70 The multiples of 21 are: 21 = 21, 42, 63, 84, 105 Hence, from the above, We can conclude that LCM (14, 21) = 42

Question 7. LCM (8, 25) = _______ Answer: We know that, “LCM” is the lowest common divisor of the 2 numbers that are evenly divisible by  both the numbers Now, The multiples of 8 are: 8 = 8, 16, 24, 32 …………. 200 The multiples of 25 are: 25 = 25, 50, 75, 100 …….. 200 Hence, from the above, We can conclude that LCM (8, 25) = 200

Everyday Math Grade 6 Home Link 7.6 Answer Key

Marathon Mathematics In 2006, Deena Kastpor set the U.S. women’s record for both the half marathon (13.1 miles) and the full marathon (26.2 miles). Her time for the half marathon was 1 hour 7 minutes 34 seconds. Her time for the full marathon was 2 hours 19 minutes 36 seconds. Question 1. Compare her rates (seconds per mile) for the two races. a. Which rate was faster? Answer: It is given that Deena Kastpor set the U.S. women’s record for both the half marathon (13.1 miles) and the full marathon (26.2 miles). Her time for the half marathon was 1 hour 7 minutes 34 seconds. Her time for the full marathon was 2 hours 19 minutes 36 seconds. Now, We have to convert the given time into seconds We know that, 1 hour = 3600 seconds 1 minute = 60 seconds So, 1 hour 7 minutes and 34 seconds = 1 (3600 seconds) + 7 (60 seconds) + 34 seconds = 3600 seconds + 420 seconds + 34 seconds = 4,054 seconds 2 hours 19 minutes and 36 seconds = 2 (3600 seconds) + 19 (60 seconds) + 36 seconds = 7,200 seconds + 1140 seconds + 36 seconds = 8,376 seconds The rate at which half marathon completed (Seconds per Mile) = \(\frac{The time taken to complete half marathon in seconds}{The distance record for half marathon in miles}\) = \(\frac{4,054}{13.1}\) = 309 miles per second The rate at which full marathon completed (Seconds per Mile) = \(\frac{The time taken to complete a full marathon in seconds}{The distance record for a full marathon in miles}\) = \(\frac{8,376}{26.2}\) = 319 miles per second Hence, from the above, We can conclude that the rate is faster for the “half marathon”

b. How much faster is her rate for that race? Answer: From part (a), The rate at which the half marathon completed is: 309 miles per second The rate at which the full marathon completed is: 319 miles per second So, The amount of faster rate = 319 – 309 = 10 seconds per mile Hence, from the above, We can conclude that she ran about 10 seconds per mile faster in the half marathon

Question 2. If Deena could run a full marathon at her half-marathon pace, how long would it take her to run the full marathon? Answer:

Question 3. Which record do you think would be easier to break: half marathon or full marathon? Explain. Answer: The half marathon record would be easier to break because it is more likely that you could increase speed for a short distance

Practice Find the value of x that makes each number sentence true. Question 4. 6x = 54 _______ Answer: The given expression is: 6x = 54 Divide by 6 on both sides So, \(\frac{6x}{6}\) = \(\frac{54}{6}\) x = 9 Hence, from the above, We can conclude that the value of x for the given expresion is: 9

Question 5. x – 14 = 152 _____ Answer: The given expression is: x – 14 = 152 Add with 14 on both sides So, x – 14 + 14 = 152 + 14 x = 166 Hence, from the above, We can conclude that the value of x for the given expression is: 166

Question 6. 300 = x + 199 ________ Answer: The given expression is: 300 = x + 199 Subtract with 199 on both sides So, 300 – 199 = x + 199 – 199 x = 101 Hnece, from the above, We can conclude that the value of x for the given expression is: 101

Everyday Mathematics Grade 6 Home Link 7.7 Answers

Doing the Dishes Question 1. Ronald’s family washes dishes by hand. Hand washing the dinner dishes takes about 10 minutes, and the faucet is running the whole time. The kitchen faucet runs at about 2.2 gallons per minute. a. In one evening, about how much water does Ronald’s family use to wash dinner dishes? Answer: It is given that Ronald’s family washes dishes by hand. Hand washing the dinner dishes takes about 10 minutes, and the faucet is running the whole time. The kitchen faucet runs at about 2.2 gallons per minute. So, In one evening, The dinner dishes washing only takes about 10 minutes So, The amount of water Ronald’s family used to wash dishes in one evening = (The time taken for hand washing the dishes by Ronald’s family) × ( The rate at which the kitchen faucet runs) = 10 minutes × 2.2 gallons per minute = 22 gallons Hence, from the above, We can conclude that The amount of water used by Ronald’s family to wash dishes in one evening is: 22 gallons

b. In seven evenings, how much water does the family use to wash dishes? Answer: From part (a), We know that, The amount of water used by Ronald’s family to wash dishes in 1 evening is: 22 gallons So, The amount of water used by Ronald’s family to wash dishes in seven evenings) = ( The amount of water used by Ronald’s family to wash dishes in 1 evening) × 7 = 22 × 7 = 154 gallons Hence, from the above, We can conclude that The amount of water used by Ronald’s family to wash dishes in seven evenings is: 154 gallons

Question 2. A high-efficiency faucet runs at about 1.5 gallons per minute. a. About how much water would the family save each time they wash their dinner dishes if they replace their old faucet with a high-efficiency faucet? Answer: The amount of water Ronald’s family used to wash dishes in one evening by using a high-efficiency faucet = (The time taken for hand washing the dishes by Ronald’s family) × ( The rate at which the  high-efficiency kitchen faucet runs) = 10 minutes × 1.5 gallons per minute = 15 gallons Now, The amount of water Ronald’s family used to wash dishes in one evening = (The time taken for hand washing the dishes by Ronald’s family) × ( The rate at which the kitchen faucet runs) = 10 minutes × 2.2 gallons per minute = 22 gallons So, The amount of water the family would save each time they wash their dinner dishes = 22 – 15 = 7 gallons Hence, from the above, We can conclude that The amount of water the family would save each time they wash their dishes is: 7 gallons

b. About how much water would they save washing dinner dishes in a year (365 days)? Answer: From part (a), We know that, The amount of water the family would save each time they wash their dishes is: 7 gallons So, The amount of water saved by the family they wash their dishes in 365 days = 365 × 7 = 2.555 gallons Hence, from the above, We can conclude that The amount of water saved by the family they wash their dishes in 365 days is: 2,555 gallons

Question 3. A high-efficiency dishwasher uses about 4 gallons of water per load. The family would run the dishwasher 4 times per week to do their dinner dishes. Should they install a high-efficiency faucet (see Problem 2) or use the dishwasher to save water? Explain. Answer: The amount of water Ronald’s family used to wash dishes in 7 days by using a high-efficiency faucet = (The time taken for hand washing the dishes by Ronald’s family) × ( The rate at which the  high-efficiency kitchen faucet runs) = 10 minutes × 1.5 gallons per minute × 7 = 15 gallons × 7 = 105 gallons The amount of water used by a high-efficiency dishwasher = (The amount of water used per load by a high-efficiency dishwasher) × ( The number of times the family would run dishwasher per week) = 4 × 4 = 16 gallons Hence, from the above, We can conclude that we will use the dishwasher to save the water

Try This Question 4. A typical circular pool that is 18 feet across and 4 feet deep requires about 3,800 gallons of water. Ronald’s parents agree to get this pool if they cut their water usage enough to fill the pool. If they use the dishwasher, can Ronald’s family save enough water during the year to justify getting the pool? Explain. Answer:

Practice Write whether each number sentence is true or false. Question 5. 4 × 7 > 6 × 3 + 4 ______ Answer: The given sentence is: 4 × 7 > 6 × 3 + 4 28 > 18 + 4 28 > 22 Hence, from the above, We can conclude that the given sentence is: True

Question 6. 15 + 9 ≤ 6 × 4 _______ Answer: The given sentence is: 15 + 9 ≤ 6 × 4 24 ≤ 24 Hence, from the above, We can conclude that the given sentence is: True

Everyday Math Grade 6 Home Link 7.8 Answer Key

Practice Evaluate. Question 4. 15% of 80 = ______ Answer: The given expression is: 15% of 80 So, 15% of 80 = \(\frac{15}{100}\) × 80 = 0.15 × 80 = 12 Hence, from the above, We can conclude that the value of the given expression is: 12

Question 5. 45% of 200 = _______ Answer: The given expression is: 45% of 200 So, 45% of 200 = \(\frac{45}{100}\) × 200 = 90 Hence, from the above, We can conclude that the value of the given expression is: 90

Question 6. 85% of 2,200 = ________ Answer: The given expression is: 85% of 2,200 So, 85% of 2,200 = \(\frac{85}{100}\) × 2,200 = 0.85 × 2,200 = 1,870 Hence, from the above, We can conclude that the value of the given expression is: 1,870

Everyday Mathematics Grade 6 Home Link 7.9 Answers

Question 3. Explain how you know which variable is independent and which is dependent. Answer: From Question 1, The equation that represents the rule for calculating your ideal maximum heart rate is: y = 220 – x So, From the equation, We can observe that the maximum ideal heart rate is dependent on age and the age is independent Hence, from the above, We can conclude that Independent variable: Age (x) Dependent variable: Maximum ideal heart rate (y)

Practice Evaluate. Question 5. -(4) = _______ Answer: We know that, – * + = – + * + = + + * – = – – * – = + So, -(4) = – (+4) = -4

Question 6. -(-9) = ______ Answer: We know that, – * + = – + * + = + + * – = – – * – = + So, – (-9) = +9 = 9

Question 7. -(-1.5) = ______ Answer: We know that, – * + = – + * + = + + * – = – – * – = + So, – (-1.5) = +1.5 = 1.5

Everyday Math Grade 6 Home Link 7.10 Answer Key

Question 3. Explain why the graphs look different. Answer: The area grows faster when compared to the perimeter The change in perimeter remains constant whereas the change in the area changes every time Hence, The graphs of the perimeter and the area look different

Practice Find each number based on the given percents. Question 4. 10% of n is 4; n = ________ Answer: It is given that 10% of n is 4 So, 10% of n = 4 We know that, 1% = 0.01 So, 10% = 0.10 So, 0.10 of n = 4 0.10 × n = 4 Divide by 0.10 on both sides So, \(\frac{0.10}{0.10}\) × n = \(\frac{4}{0.10}\) n = 4 ×10 n = 40 Hence, from the above, We can conclude that the value of n is: 40

Question 5. 30% of n is 18; n = __________ Answer: It is given that 30% of n is 18 So, 30% of n = 18 We know that, 1% = 0.01 So, 30% = 0.30 So, 0.30 of n = 18 0.30 × n = 18 Divide by 0.30 on both sides So, \(\frac{0.30}{0.30}\) × n = \(\frac{18}{0.30}\) n = 6 ×10 n = 60 Hence, from the above, We can conclude that the value of n is: 60

Everyday Mathematics Grade 6 Home Link 7.11 Answers

Leave a Comment Cancel Reply

You must be logged in to post a comment.