## CPM Educational Program

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- Core Connections Integrated I, 2013
- Core Connections Algebra 1, 2013
- Core Connections Geometry, 2013
- Core Connections Algebra 2, 2013
- Core Connections Integrated I, 2014
- Core Connections Integrated II, 2015
- Core Connections: Course 1
- Core Connections: Course 2
- Core Connections: Course 3
- Core Connections Integrated III, 2015

## 3.1 Functions and Function Notation

- ⓑ yes (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

w = f ( d ) w = f ( d )

g ( 5 ) = 1 g ( 5 ) = 1

m = 8 m = 8

y = f ( x ) = x 3 2 y = f ( x ) = x 3 2

g ( 1 ) = 8 g ( 1 ) = 8

x = 0 x = 0 or x = 2 x = 2

- ⓐ yes, because each bank account has a single balance at any given time;
- ⓑ no, because several bank account numbers may have the same balance;
- ⓒ no, because the same output may correspond to more than one input.
- ⓐ Yes, letter grade is a function of percent grade;
- ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

No, because it does not pass the horizontal line test.

## 3.2 Domain and Range

{ − 5 , 0 , 5 , 10 , 15 } { − 5 , 0 , 5 , 10 , 15 }

( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ ) ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 5 2 , ∞ ) [ − 5 2 , ∞ )

- ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3
- ⓑ { x | x ≤ − 2 or − 1 ≤ x < 3 } { x | x ≤ − 2 or − 1 ≤ x < 3 }
- ⓒ ( − ∞ , − 2 ] ∪ [ − 1 , 3 ) ( − ∞ , − 2 ] ∪ [ − 1 , 3 )

domain =[1950,2002] range = [47,000,000,89,000,000]

domain: ( − ∞ , 2 ] ; ( − ∞ , 2 ] ; range: ( − ∞ , 0 ] ( − ∞ , 0 ]

## 3.3 Rates of Change and Behavior of Graphs

$ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 $ 2.84 − $ 2.31 5 years = $ 0.53 5 years = $ 0.106 per year.

a + 7 a + 7

The local maximum appears to occur at ( − 1 , 28 ) , ( − 1 , 28 ) , and the local minimum occurs at ( 5 , − 80 ) . ( 5 , − 80 ) . The function is increasing on ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) ( − ∞ , − 1 ) ∪ ( 5 , ∞ ) and decreasing on ( − 1 , 5 ) . ( − 1 , 5 ) .

## 3.4 Composition of Functions

( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2 ( f g ) ( x ) = f ( x ) g ( x ) = ( x − 1 ) ( x 2 − 1 ) = x 3 − x 2 − x + 1 ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x − 1 ) − ( x 2 − 1 ) = x − x 2

No, the functions are not the same.

A gravitational force is still a force, so a ( G ( r ) ) a ( G ( r ) ) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G ( a ( F ) ) G ( a ( F ) ) does not make sense.

f ( g ( 1 ) ) = f ( 3 ) = 3 f ( g ( 1 ) ) = f ( 3 ) = 3 and g ( f ( 4 ) ) = g ( 1 ) = 3 g ( f ( 4 ) ) = g ( 1 ) = 3

g ( f ( 2 ) ) = g ( 5 ) = 3 g ( f ( 2 ) ) = g ( 5 ) = 3

[ − 4 , 0 ) ∪ ( 0 , ∞ ) [ − 4 , 0 ) ∪ ( 0 , ∞ )

Possible answer:

g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g g ( x ) = 4 + x 2 h ( x ) = 4 3 − x f = h ∘ g

## 3.5 Transformation of Functions

The graphs of f ( x ) f ( x ) and g ( x ) g ( x ) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

g ( x ) = 1 x - 1 + 1 g ( x ) = 1 x - 1 + 1

g ( x ) = − f ( x ) g ( x ) = − f ( x )

-2 | 0 | 2 | 4 | |

h ( x ) = f ( − x ) h ( x ) = f ( − x )

-2 | 0 | 2 | 4 | |

15 | 10 | 5 | unknown |

Notice: g ( x ) = f ( − x ) g ( x ) = f ( − x ) looks the same as f ( x ) f ( x ) .

2 | 4 | 6 | 8 | |

9 | 12 | 15 | 0 |

g ( x ) = 3 x - 2 g ( x ) = 3 x - 2

g ( x ) = f ( 1 3 x ) g ( x ) = f ( 1 3 x ) so using the square root function we get g ( x ) = 1 3 x g ( x ) = 1 3 x

## 3.6 Absolute Value Functions

using the variable p p for passing, | p − 80 | ≤ 20 | p − 80 | ≤ 20

f ( x ) = − | x + 2 | + 3 f ( x ) = − | x + 2 | + 3

x = − 1 x = − 1 or x = 2 x = 2

## 3.7 Inverse Functions

h ( 2 ) = 6 h ( 2 ) = 6

The domain of function f − 1 f − 1 is ( − ∞ , − 2 ) ( − ∞ , − 2 ) and the range of function f − 1 f − 1 is ( 1 , ∞ ) . ( 1 , ∞ ) .

- ⓐ f ( 60 ) = 50. f ( 60 ) = 50. In 60 minutes, 50 miles are traveled.
- ⓑ f − 1 ( 60 ) = 70. f − 1 ( 60 ) = 70. To travel 60 miles, it will take 70 minutes.

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

f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ] f − 1 ( x ) = ( 2 − x ) 2 ; domain of f : [ 0 , ∞ ) ; domain of f − 1 : ( − ∞ , 2 ]

## 3.1 Section Exercises

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

not a function

f ( − 3 ) = − 11 ; f ( − 3 ) = − 11 ; f ( 2 ) = − 1 ; f ( 2 ) = − 1 ; f ( − a ) = − 2 a − 5 ; f ( − a ) = − 2 a − 5 ; − f ( a ) = − 2 a + 5 ; − f ( a ) = − 2 a + 5 ; f ( a + h ) = 2 a + 2 h − 5 f ( a + h ) = 2 a + 2 h − 5

f ( − 3 ) = 5 + 5 ; f ( − 3 ) = 5 + 5 ; f ( 2 ) = 5 ; f ( 2 ) = 5 ; f ( − a ) = 2 + a + 5 ; f ( − a ) = 2 + a + 5 ; − f ( a ) = − 2 − a − 5 ; − f ( a ) = − 2 − a − 5 ; f ( a + h ) = 2 − a − h + 5 f ( a + h ) = 2 − a − h + 5

f ( − 3 ) = 2 ; f ( − 3 ) = 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( 2 ) = 1 − 3 = − 2 ; f ( − a ) = | − a − 1 | − | − a + 1 | ; f ( − a ) = | − a − 1 | − | − a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; − f ( a ) = − | a − 1 | + | a + 1 | ; f ( a + h ) = | a + h − 1 | − | a + h + 1 | f ( a + h ) = | a + h − 1 | − | a + h + 1 |

g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a g ( x ) − g ( a ) x − a = x + a + 2 , x ≠ a

a. f ( − 2 ) = 14 ; f ( − 2 ) = 14 ; b. x = 3 x = 3

a. f ( 5 ) = 10 ; f ( 5 ) = 10 ; b. x = − 1 x = − 1 or x = 4 x = 4

- ⓐ f ( t ) = 6 − 2 3 t ; f ( t ) = 6 − 2 3 t ;
- ⓑ f ( − 3 ) = 8 ; f ( − 3 ) = 8 ;
- ⓒ t = 6 t = 6
- ⓐ f ( 0 ) = 1 ; f ( 0 ) = 1 ;
- ⓑ f ( x ) = − 3 , x = − 2 f ( x ) = − 3 , x = − 2 or x = 2 x = 2

not a function so it is also not a one-to-one function

one-to- one function

function, but not one-to-one

f ( x ) = 1 , x = 2 f ( x ) = 1 , x = 2

f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2 f ( − 2 ) = 14 ; f ( − 1 ) = 11 ; f ( 0 ) = 8 ; f ( 1 ) = 5 ; f ( 2 ) = 2

f ( − 2 ) = 4 ; f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236 f ( − 2 ) = 4 ; f ( − 1 ) = 4.414 ; f ( 0 ) = 4.732 ; f ( 1 ) = 5 ; f ( 2 ) = 5.236

f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9 f ( − 2 ) = 1 9 ; f ( − 1 ) = 1 3 ; f ( 0 ) = 1 ; f ( 1 ) = 3 ; f ( 2 ) = 9

[ 0 , 100 ] [ 0 , 100 ]

[ − 0.001 , 0 .001 ] [ − 0.001 , 0 .001 ]

[ − 1 , 000 , 000 , 1,000,000 ] [ − 1 , 000 , 000 , 1,000,000 ]

[ 0 , 10 ] [ 0 , 10 ]

[ −0.1 , 0.1 ] [ −0.1 , 0.1 ]

[ − 100 , 100 ] [ − 100 , 100 ]

- ⓐ g ( 5000 ) = 50 ; g ( 5000 ) = 50 ;
- ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.
- ⓐ The height of a rocket above ground after 1 second is 200 ft.
- ⓑ The height of a rocket above ground after 2 seconds is 350 ft.

## 3.2 Section Exercises

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on x x for f ( x ) = x 3 f ( x ) = x 3 because you can take the cube root of any real number. So the domain is all real numbers, ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . When dealing with the set of real numbers, you cannot take the square root of negative numbers. So x x -values are restricted for f ( x ) = x f ( x ) = x to nonnegative numbers and the domain is [ 0 , ∞ ) . [ 0 , ∞ ) .

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the x x -axis and y y -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate − ∞ − ∞ or ∞ . ∞ . Combine the graphs to find the graph of the piecewise function.

( − ∞ , 3 ] ( − ∞ , 3 ]

( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ )

( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ ) ( − ∞ , − 11 ) ∪ ( − 11 , 2 ) ∪ ( 2 , ∞ )

( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ ) ( − ∞ , − 3 ) ∪ ( − 3 , 5 ) ∪ ( 5 , ∞ )

( − ∞ , 5 ) ( − ∞ , 5 )

[ 6 , ∞ ) [ 6 , ∞ )

( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ ) ( − ∞ , − 9 ) ∪ ( − 9 , 9 ) ∪ ( 9 , ∞ )

domain: ( 2 , 8 ] , ( 2 , 8 ] , range [ 6 , 8 ) [ 6 , 8 )

domain: [ − 4 , 4], [ − 4 , 4], range: [ 0 , 2] [ 0 , 2]

domain: [ − 5 , 3 ) , [ − 5 , 3 ) , range: [ 0 , 2 ] [ 0 , 2 ]

domain: ( − ∞ , 1 ] , ( − ∞ , 1 ] , range: [ 0 , ∞ ) [ 0 , ∞ )

domain: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] ; range: [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ] [ − 6 , − 1 6 ] ∪ [ 1 6 , 6 ]

domain: [ − 3 , ∞ ) ; [ − 3 , ∞ ) ; range: [ 0 , ∞ ) [ 0 , ∞ )

domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0 f ( − 3 ) = 1 ; f ( − 2 ) = 0 ; f ( − 1 ) = 0 ; f ( 0 ) = 0

f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34 f ( − 1 ) = − 4 ; f ( 0 ) = 6 ; f ( 2 ) = 20 ; f ( 4 ) = 34

f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16 f ( − 1 ) = − 5 ; f ( 0 ) = 3 ; f ( 2 ) = 3 ; f ( 4 ) = 16

domain: ( − ∞ , 1 ) ∪ ( 1 , ∞ ) ( − ∞ , 1 ) ∪ ( 1 , ∞ )

window: [ − 0.5 , − 0.1 ] ; [ − 0.5 , − 0.1 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

window: [ 0.1 , 0.5 ] ; [ 0.1 , 0.5 ] ; range: [ 4 , 100 ] [ 4 , 100 ]

[ 0 , 8 ] [ 0 , 8 ]

Many answers. One function is f ( x ) = 1 x − 2 . f ( x ) = 1 x − 2 .

- ⓐ The fixed cost is $500.
- ⓑ The cost of making 25 items is $750.
- ⓒ The domain is [0, 100] and the range is [500, 1500].

## 3.3 Section Exercises

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

4 ( b + 1 ) 4 ( b + 1 )

4 x + 2 h 4 x + 2 h

− 1 13 ( 13 + h ) − 1 13 ( 13 + h )

3 h 2 + 9 h + 9 3 h 2 + 9 h + 9

4 x + 2 h − 3 4 x + 2 h − 3

increasing on ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , ( − ∞ , − 2.5 ) ∪ ( 1 , ∞ ) , decreasing on ( − 2.5 , 1 ) ( − 2.5 , 1 )

increasing on ( − ∞ , 1 ) ∪ ( 3 , 4 ) , ( − ∞ , 1 ) ∪ ( 3 , 4 ) , decreasing on ( 1 , 3 ) ∪ ( 4 , ∞ ) ( 1 , 3 ) ∪ ( 4 , ∞ )

local maximum: ( − 3 , 60 ) , ( − 3 , 60 ) , local minimum: ( 3 , − 60 ) ( 3 , − 60 )

absolute maximum at approximately ( 7 , 150 ) , ( 7 , 150 ) , absolute minimum at approximately ( −7.5 , −220 ) ( −7.5 , −220 )

Local minimum at ( 3 , − 22 ) , ( 3 , − 22 ) , decreasing on ( − ∞ , 3 ) , ( − ∞ , 3 ) , increasing on ( 3 , ∞ ) ( 3 , ∞ )

Local minimum at ( − 2 , − 2 ) , ( − 2 , − 2 ) , decreasing on ( − 3 , − 2 ) , ( − 3 , − 2 ) , increasing on ( − 2 , ∞ ) ( − 2 , ∞ )

Local maximum at ( − 0.5 , 6 ) , ( − 0.5 , 6 ) , local minima at ( − 3.25 , − 47 ) ( − 3.25 , − 47 ) and ( 2.1 , − 32 ) , ( 2.1 , − 32 ) , decreasing on ( − ∞ , − 3.25 ) ( − ∞ , − 3.25 ) and ( − 0.5 , 2.1 ) , ( − 0.5 , 2.1 ) , increasing on ( − 3.25 , − 0.5 ) ( − 3.25 , − 0.5 ) and ( 2.1 , ∞ ) ( 2.1 , ∞ )

b = 5 b = 5

2.7 gallons per minute

approximately –0.6 milligrams per day

## 3.4 Section Exercises

Find the numbers that make the function in the denominator g g equal to zero, and check for any other domain restrictions on f f and g , g , such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let f ( x ) = x + 1 and g ( x ) = x − 1. f ( x ) = x + 1 and g ( x ) = x − 1. Then f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x f ( g ( x ) ) = f ( x − 1 ) = ( x − 1 ) + 1 = x and g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . g ( f ( x ) ) = g ( x + 1 ) = ( x + 1 ) − 1 = x . So f ∘ g = g ∘ f . f ∘ g = g ∘ f .

( f + g ) ( x ) = 2 x + 6 , ( f + g ) ( x ) = 2 x + 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f − g ) ( x ) = 2 x 2 + 2 x − 6 , ( f − g ) ( x ) = 2 x 2 + 2 x − 6 , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , ( f g ) ( x ) = − x 4 − 2 x 3 + 6 x 2 + 12 x , domain: ( − ∞ , ∞ ) ( − ∞ , ∞ )

( f g ) ( x ) = x 2 + 2 x 6 − x 2 , ( f g ) ( x ) = x 2 + 2 x 6 − x 2 , domain: ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 6 ) ∪ ( − 6 , 6 ) ∪ ( 6 , ∞ )

( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , ( f + g ) ( x ) = 4 x 3 + 8 x 2 + 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , ( f − g ) ( x ) = 4 x 3 + 8 x 2 − 1 2 x , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = x + 2 , ( f g ) ( x ) = x + 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f g ) ( x ) = 4 x 3 + 8 x 2 , ( f g ) ( x ) = 4 x 3 + 8 x 2 , domain: ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ( − ∞ , 0 ) ∪ ( 0 , ∞ )

( f + g ) ( x ) = 3 x 2 + x − 5 , ( f + g ) ( x ) = 3 x 2 + x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f − g ) ( x ) = 3 x 2 − x − 5 , ( f − g ) ( x ) = 3 x 2 − x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: [ 5 , ∞ ) [ 5 , ∞ )

( f g ) ( x ) = 3 x 2 x − 5 , ( f g ) ( x ) = 3 x 2 x − 5 , domain: ( 5 , ∞ ) ( 5 , ∞ )

- ⓑ f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1 f ( g ( x ) ) = 2 ( 3 x − 5 ) 2 + 1
- ⓒ f ( g ( x ) ) = 6 x 2 − 2 f ( g ( x ) ) = 6 x 2 − 2
- ⓓ ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20 ( g ∘ g ) ( x ) = 3 ( 3 x − 5 ) − 5 = 9 x − 20
- ⓔ ( f ∘ f ) ( − 2 ) = 163 ( f ∘ f ) ( − 2 ) = 163

f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7 f ( g ( x ) ) = x 2 + 3 + 2 , g ( f ( x ) ) = x + 4 x + 7

f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x f ( g ( x ) ) = x + 1 x 3 3 = x + 1 3 x , g ( f ( x ) ) = x 3 + 1 x

( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4 ( f ∘ g ) ( x ) = 1 2 x + 4 − 4 = x 2 , ( g ∘ f ) ( x ) = 2 x − 4

f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1 f ( g ( h ( x ) ) ) = ( 1 x + 3 ) 2 + 1

- ⓐ ( g ∘ f ) ( x ) = − 3 2 − 4 x ( g ∘ f ) ( x ) = − 3 2 − 4 x
- ⓑ ( − ∞ , 1 2 ) ( − ∞ , 1 2 )
- ⓐ ( 0 , 2 ) ∪ ( 2 , ∞ ) ; ( 0 , 2 ) ∪ ( 2 , ∞ ) ;
- ⓑ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ; ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) ;
- ⓒ ( 0 , ∞ ) ( 0 , ∞ )

( 1 , ∞ ) ( 1 , ∞ )

sample: f ( x ) = x 3 g ( x ) = x − 5 f ( x ) = x 3 g ( x ) = x − 5

sample: f ( x ) = 4 x g ( x ) = ( x + 2 ) 2 f ( x ) = 4 x g ( x ) = ( x + 2 ) 2

sample: f ( x ) = x 3 g ( x ) = 1 2 x − 3 f ( x ) = x 3 g ( x ) = 1 2 x − 3

sample: f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5 f ( x ) = x 4 g ( x ) = 3 x − 2 x + 5

sample: f ( x ) = x g ( x ) = 2 x + 6 f ( x ) = x g ( x ) = 2 x + 6

sample: f ( x ) = x 3 g ( x ) = ( x − 1 ) f ( x ) = x 3 g ( x ) = ( x − 1 )

sample: f ( x ) = x 3 g ( x ) = 1 x − 2 f ( x ) = x 3 g ( x ) = 1 x − 2

sample: f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 f ( x ) = x g ( x ) = 2 x − 1 3 x + 4

f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94 f ( g ( 0 ) ) = 27 , g ( f ( 0 ) ) = − 94

f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5 f ( g ( 0 ) ) = 1 5 , g ( f ( 0 ) ) = 5

18 x 2 + 60 x + 51 18 x 2 + 60 x + 51

g ∘ g ( x ) = 9 x + 20 g ∘ g ( x ) = 9 x + 20

( f ∘ g ) ( 6 ) = 6 ( f ∘ g ) ( 6 ) = 6 ; ( g ∘ f ) ( 6 ) = 6 ( g ∘ f ) ( 6 ) = 6

( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11 ( f ∘ g ) ( 11 ) = 11 , ( g ∘ f ) ( 11 ) = 11

A ( t ) = π ( 25 t + 2 ) 2 A ( t ) = π ( 25 t + 2 ) 2 and A ( 2 ) = π ( 25 4 ) 2 = 2500 π A ( 2 ) = π ( 25 4 ) 2 = 2500 π square inches

A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π A ( 5 ) = π ( 2 ( 5 ) + 1 ) 2 = 121 π square units

- ⓐ N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1 N ( T ( t ) ) = 23 ( 5 t + 1.5 ) 2 − 56 ( 5 t + 1.5 ) + 1
- ⓑ 3.38 hours

## 3.5 Section Exercises

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

For a function f , f , substitute ( − x ) ( − x ) for ( x ) ( x ) in f ( x ) . f ( x ) . Simplify. If the resulting function is the same as the original function, f ( − x ) = f ( x ) , f ( − x ) = f ( x ) , then the function is even. If the resulting function is the opposite of the original function, f ( − x ) = − f ( x ) , f ( − x ) = − f ( x ) , then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

g ( x ) = | x - 1 | − 3 g ( x ) = | x - 1 | − 3

g ( x ) = 1 ( x + 4 ) 2 + 2 g ( x ) = 1 ( x + 4 ) 2 + 2

The graph of f ( x + 43 ) f ( x + 43 ) is a horizontal shift to the left 43 units of the graph of f . f .

The graph of f ( x - 4 ) f ( x - 4 ) is a horizontal shift to the right 4 units of the graph of f . f .

The graph of f ( x ) + 8 f ( x ) + 8 is a vertical shift up 8 units of the graph of f . f .

The graph of f ( x ) − 7 f ( x ) − 7 is a vertical shift down 7 units of the graph of f . f .

The graph of f ( x + 4 ) − 1 f ( x + 4 ) − 1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f . f .

decreasing on ( − ∞ , − 3 ) ( − ∞ , − 3 ) and increasing on ( − 3 , ∞ ) ( − 3 , ∞ )

decreasing on ( 0 , ∞ ) ( 0 , ∞ )

g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1 g ( x ) = f ( x - 1 ) , h ( x ) = f ( x ) + 1

f ( x ) = | x - 3 | − 2 f ( x ) = | x - 3 | − 2

f ( x ) = x + 3 − 1 f ( x ) = x + 3 − 1

f ( x ) = ( x - 2 ) 2 f ( x ) = ( x - 2 ) 2

f ( x ) = | x + 3 | − 2 f ( x ) = | x + 3 | − 2

f ( x ) = − x f ( x ) = − x

f ( x ) = − ( x + 1 ) 2 + 2 f ( x ) = − ( x + 1 ) 2 + 2

f ( x ) = − x + 1 f ( x ) = − x + 1

The graph of g g is a vertical reflection (across the x x -axis) of the graph of f . f .

The graph of g g is a vertical stretch by a factor of 4 of the graph of f . f .

The graph of g g is a horizontal compression by a factor of 1 5 1 5 of the graph of f . f .

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f . f .

The graph of g g is a horizontal reflection across the y y -axis and a vertical stretch by a factor of 3 of the graph of f . f .

g ( x ) = | − 4 x | g ( x ) = | − 4 x |

g ( x ) = 1 3 ( x + 2 ) 2 − 3 g ( x ) = 1 3 ( x + 2 ) 2 − 3

g ( x ) = 1 2 ( x - 5 ) 2 + 1 g ( x ) = 1 2 ( x - 5 ) 2 + 1

The graph of the function f ( x ) = x 2 f ( x ) = x 2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

The graph of f ( x ) = | x | f ( x ) = | x | is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

The graph of the function f ( x ) = x 3 f ( x ) = x 3 is compressed vertically by a factor of 1 2 . 1 2 .

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

The graph of f ( x ) = x f ( x ) = x is shifted right 4 units and then reflected across the vertical line x = 4. x = 4.

## 3.6 Section Exercises

Isolate the absolute value term so that the equation is of the form | A | = B . | A | = B . Form one equation by setting the expression inside the absolute value symbol, A , A , equal to the expression on the other side of the equation, B . B . Form a second equation by setting A A equal to the opposite of the expression on the other side of the equation, − B . − B . Solve each equation for the variable.

The graph of the absolute value function does not cross the x x -axis, so the graph is either completely above or completely below the x x -axis.

The distance from x to 8 can be represented using the absolute value statement: ∣ x − 8 ∣ = 4.

∣ x − 10 ∣ ≥ 15

There are no x-intercepts.

(−4, 0) and (2, 0)

( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 ) ( 0 , − 4 ) , ( 4 , 0 ) , ( − 2 , 0 )

( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 ) ( 0 , 7 ) , ( 25 , 0 ) , ( − 7 , 0 )

range: [ – 400 , 100 ] [ – 400 , 100 ]

There is no solution for a a that will keep the function from having a y y -intercept. The absolute value function always crosses the y y -intercept when x = 0. x = 0.

| p − 0.08 | ≤ 0.015 | p − 0.08 | ≤ 0.015

| x − 5.0 | ≤ 0.01 | x − 5.0 | ≤ 0.01

## 3.7 Section Exercises

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that y y -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no y y -values repeat and the function is one-to-one.

Yes. For example, f ( x ) = 1 x f ( x ) = 1 x is its own inverse.

Given a function y = f ( x ) , y = f ( x ) , solve for x x in terms of y . y . Interchange the x x and y . y . Solve the new equation for y . y . The expression for y y is the inverse, y = f − 1 ( x ) . y = f − 1 ( x ) .

f − 1 ( x ) = x − 3 f − 1 ( x ) = x − 3

f − 1 ( x ) = 2 − x f − 1 ( x ) = 2 − x

f − 1 ( x ) = − 2 x x − 1 f − 1 ( x ) = − 2 x x − 1

domain of f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7 f ( x ) : [ − 7 , ∞ ) ; f − 1 ( x ) = x − 7

domain of f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5 f ( x ) : [ 0 , ∞ ) ; f − 1 ( x ) = x + 5

a. f ( g ( x ) ) = x f ( g ( x ) ) = x and g ( f ( x ) ) = x . g ( f ( x ) ) = x . b. This tells us that f f and g g are inverse functions

f ( g ( x ) ) = x , g ( f ( x ) ) = x f ( g ( x ) ) = x , g ( f ( x ) ) = x

not one-to-one

[ 2 , 10 ] [ 2 , 10 ]

1 | 4 | 7 | 12 | 16 | |

3 | 6 | 9 | 13 | 14 |

f − 1 ( x ) = ( 1 + x ) 1 / 3 f − 1 ( x ) = ( 1 + x ) 1 / 3

f − 1 ( x ) = 5 9 ( x − 32 ) . f − 1 ( x ) = 5 9 ( x − 32 ) . Given the Fahrenheit temperature, x , x , this formula allows you to calculate the Celsius temperature.

t ( d ) = d 50 , t ( d ) = d 50 , t ( 180 ) = 180 50 . t ( 180 ) = 180 50 . The time for the car to travel 180 miles is 3.6 hours.

## Review Exercises

f ( − 3 ) = − 27 ; f ( − 3 ) = − 27 ; f ( 2 ) = − 2 ; f ( 2 ) = − 2 ; f ( − a ) = − 2 a 2 − 3 a ; f ( − a ) = − 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; − f ( a ) = 2 a 2 − 3 a ; f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2 f ( a + h ) = − 2 a 2 + 3 a − 4 a h + 3 h − 2 h 2

x = − 1.8 x = − 1.8 or or x = 1.8 or x = 1.8

− 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64 − 64 + 80 a − 16 a 2 − 1 + a = − 16 a + 64

( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ ) ( − ∞ , − 2 ) ∪ ( − 2 , 6 ) ∪ ( 6 , ∞ )

increasing ( 2 , ∞ ) ; ( 2 , ∞ ) ; decreasing ( − ∞ , 2 ) ( − ∞ , 2 )

increasing ( − 3 , 1 ) ; ( − 3 , 1 ) ; constant ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) ( − ∞ , − 3 ) ∪ ( 1 , ∞ )

local minimum ( − 2 , − 3 ) ; ( − 2 , − 3 ) ; local maximum ( 1 , 3 ) ( 1 , 3 )

( − 1.8 , 10 ) ( − 1.8 , 10 )

( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x ( f ∘ g ) ( x ) = 17 − 18 x ; ( g ∘ f ) ( x ) = − 7 − 18 x

( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2 ( f ∘ g ) ( x ) = 1 x + 2 ; ( g ∘ f ) ( x ) = 1 x + 2

( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4 ( f ∘ g ) ( x ) = 1 + x 1 + 4 x , x ≠ 0 , x ≠ − 1 4

( f ∘ g ) ( x ) = 1 x , x > 0 ( f ∘ g ) ( x ) = 1 x , x > 0

sample: g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x g ( x ) = 2 x − 1 3 x + 4 ; f ( x ) = x

f ( x ) = | x − 3 | f ( x ) = | x − 3 |

f ( x ) = 1 2 | x + 2 | + 1 f ( x ) = 1 2 | x + 2 | + 1

f ( x ) = − 3 | x − 3 | + 3 f ( x ) = − 3 | x − 3 | + 3

f − 1 ( x ) = x - 9 10 f − 1 ( x ) = x - 9 10

f − 1 ( x ) = x - 1 f − 1 ( x ) = x - 1

The function is one-to-one.

## Practice Test

The relation is a function.

The graph is a parabola and the graph fails the horizontal line test.

2 a 2 − a 2 a 2 − a

− 2 ( a + b ) + 1 − 2 ( a + b ) + 1

f − 1 ( x ) = x + 5 3 f − 1 ( x ) = x + 5 3

( − ∞ , − 1.1 ) and ( 1.1 , ∞ ) ( − ∞ , − 1.1 ) and ( 1.1 , ∞ )

( 1.1 , − 0.9 ) ( 1.1 , − 0.9 )

f ( 2 ) = 2 f ( 2 ) = 2

f ( x ) = { | x | if x ≤ 2 3 if x > 2 f ( x ) = { | x | if x ≤ 2 3 if x > 2

x = 2 x = 2

f − 1 ( x ) = − x − 11 2 f − 1 ( x ) = − x − 11 2

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## Unit 6 – Quadratic Functions and Their Algebra

Quadratic Function Review

LESSON/HOMEWORK

LESSON VIDEO

EDITABLE LESSON

EDITABLE KEY

Factoring Trinomials

Complete Factoring

Factoring by Grouping

The Zero Product Law

Quadratic Inequalities in One Variable

Completing the Square and Shifting Parabolas

Modeling with Quadratic Functions

Equations of Circles

The Locus Definition of a Parabola

Unit Review

Unit 6 Review – Quadratic Functions

UNIT REVIEW

EDITABLE REVIEW

Unit 6 Assessment Form A

EDITABLE ASSESSMENT

Unit 6 Assessment Form B

Unit 6 Assessment Form C

Unit 6 Assessment Form D

Unit 6 Exit Tickets

Unit 6 – Mid-Unit Quiz (Through Lesson #6) – Form A

Unit 6 – Mid-Unit Quiz (Through Lesson #6) – Form B

Unit 6 – Mid-Unit Quiz (Through Lesson #6) – Form C

Unit 6 – Mid-Unit Quiz (Through Lesson #6) – Form D

U06.AO.01 – Lesson 5.4 – Factoring Trinomials Using the AC Method

EDITABLE RESOURCE

U06.AO.02 – Lesson 5.5 – Using Structure to Factor

U06.AO.03 – Lesson 9.5 – The Vertex Form of a Parabola

U06.AO.04 – Lesson 12 – More Work with the Directrix and Focus

U06.AO.05 – Parabola Practice

U06.AO.06 – Factoring Practice

U06.AO.07 – Quadratic Systems Practice

U06.AO.08 – Quadratic Inequalities in One Variable Practice

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## Home > CC3 > Chapter 7 > Lesson 7.3.3

Lesson 7.1.1, lesson 7.1.2, lesson 7.1.3, lesson 7.2.1, lesson 7.2.2, lesson 7.2.3, lesson 7.2.4, lesson 7.2.5, lesson 7.3.1, lesson 7.3.2, lesson 7.3.3.

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## Eureka Math Grade 3 Module 3 Lesson 21 Answer Key

Grade 3 Eureka Math Answer Key is provided by subject experts as per the Common Core curriculum. Grade 3 Eureka Math Solutions provided educates the primary school Kids on all Math concepts of the chapter efficiently. Eureka Math Grade 3 Textbook Answers makes it easy for kids to Grasp the Concepts as well as solve any problem from chapter Tests, Practice Tests, Performance tests, cumulative practice, etc.

## Engage NY Eureka Math 3rd Grade Module 3 Lesson 21 Answer Key

Eureka Grade 3 ensures students with the learning targets and success criteria. The Diverse opportunities will develop problem-solving skills among the primary school kids practicing Bigideas Math Grade 3 Solutions. Eureka Textbook Grade 3 answers help Students Master the subject on Consistent practice.

## Eureka Math Grade 3 Module 3 Lesson 21 Sprint Answer Key

Answer: 2 x 3 = 6, 2 x 30 = 60, 20 x 3 = 60, 2 x 2 = 4, 2 x 20 = 40, 20 x 2 = 40, 4 x 2 = 8, 4 x 20 = 80, 40 x 2 = 80, 5 x 3 = 15, 50 x 3 = 150, 3 x 50 = 150, 4 x 4 = 16, 40 x 4 = 160, 4 x 40 = 160, 6 x 3 = 18, 6 x 30 = 180, 60 x 3 = 180, 7 x 5 = 35, 70 x 5 = 350, 7 x 50 = 350, 8 x 4 = 32, 8 x 40 = 320, 4 x 80 = 320, 9 x 6 = 54, 90 x 6 = 540, 2 x 5 = 10, 2 x 50 = 100, 3 x 90 = 270, 40 x 7 = 280, 5 x 40 = 200, 6 x 60 = 360, 70 x 6 = 420, 8 x 70 = 560, 80 x 6 = 480, 9 x 70 = 630, 50 x 6 = 300, 8 x 80 = 640, 9 x 80 = 720, 60 x 8 = 480, 70 x 7 = 490, 5 x 80 = 400, 60 x 9 = 540, 9 x 90 = 810.

Answer: 4 x 2 = 8, 4 x 20 = 80, 40 x 2 = 80, 3 x 3 = 9, 3 x 30 = 90, 30 x 3 = 90, 3 x 2 = 6, 3 x 20 = 60, 30 x 2 = 60, 5 x 5 = 25, 50 x 5 = 250, 5 x 50 = 250, 4 x 3 = 12, 40 x 3 = 120, 4 x 30 = 120, 7 x 3 = 21, 7 x 30 = 210, 70 x 3 = 210, 6 x 4 = 24, 60 x 4 = 240, 6 x 40 = 240, 9 x 4 = 36, 9 x 40 = 360, 90 x 4 = 360, 8 x 6 = 48, 80 x 6 = 480, 5 x 2 = 10, 5 x 20 = 100, 3 x 80 = 240, 40 x 8 = 320, 4 x 50 = 200, 8 x 80 = 640, 90 x 6 = 540, 6 x 70 = 420, 60 x 6 = 360, 7 x 70 = 490, 60 x 5 = 300, 6 x 80 = 480, 7 x 80 = 560, 80 x 6 = 480, 90 x 7 = 630, 8 x 50 = 400, 80 x 9 = 720, 7 x 90 = 630.

## Eureka Math Grade 3 Module 3 Lesson 21 Problem Set Answer Key

Use the RDW process to solve each problem. Use a letter to represent the unknown.

Question 1. There are 60 seconds in 1 minute. Use a tape diagram to find the total number of seconds in 5 minutes and 45 seconds.

Answer: The total number of seconds = 345 seconds.

Explanation: In the above-given question, given that, There are 60 seconds in 1 minute. 60 x 5 = 300. 300 + 45 = 345.

Answer: No, she does not have enough money to buy the art supplies.

Explanation: In the above-given question, given that, Lupe saves $30 each month for 4 months. 30 x 4 = 120. but the Art Supplies cost is $142. 120 is less than 142. so she does not buy.

Question 3. Brad receives 5 cents for each can or bottle he recycles. How many cents does Brad earn if he recycles 48 cans and 32 bottles?

Answer: The number of cents he recycles = 400 cents.

Explanation: In the above-given question, given that, Brad receives 5 cents for each can or bottle he recycles. 48 x 5 = 240. 32 x 5 = 150. 240 + 150 = 400.

Question 4. A box of 10 markers weighs 105 grams. If the empty box weighs 15 grams, how much does each marker weigh?

Answer: Each marker weight= 9.

Explanation: In the above-given question, given that, A box of 10 markers weighs 105 grams. If the empty box weighs 15 grams. 9 x 10 = 90. 105 – 15 = 90.

Question 5. Mr. Perez buys 3 sets of cards. Each set comes with 18 striped cards and 12 polka dot cards. He uses 49 cards. How many cards does he have left?

Answer: The number of cards he left = 51 cards.

Explanation: In the above-given question, given that, Mr. Perez buys 3 sets of cards. Each set comes with 18 striped cards and 12 polka dot cards. He uses 49 cards. 18 + 12 = 30. 30 x 3 = 90. 90 – 49 = 51. so the number of cards he left = 51.

Question 6. Ezra earns $9 an hour working at a book store. She works for 7 hours each day on Mondays and Wednesdays. How much does Ezra earn each week?

Answer: The amount does Ezra earns = $126.

Explanation: In the above-given question, given that, Ezra earns $9 an hour working at a book store. She works for 7 hours each day on Mondays and Wednesdays. 7 + 7 = 14. 14 x 9 = 126. so the amount Ezra earns = $126.

Eureka Math Grade 3 Module 3 Lesson 21 Exit Ticket Answer Key

Use the RDW process to solve. Use a letter to represent the unknown.

Frederick buys a can of 3 tennis balls. The empty can weighs 20 grams, and each tennis ball weighs 60 grams. What is the total weight of the can with 3 tennis balls?

Answer: The total weight of the can with 3 tennis balls = 160 grams.

Explanation: In the above-given question, given that, Frederick buys a can of 3 tennis balls. The empty can weighs 20 grams, and each ball weighs 60 grams. 60 x 3 = 180. 180 – 20 = 160.

Eureka Math Grade 3 Module 3 Lesson 21 Homework Answer Key

Use the RDW process for each problem. Use a letter to represent the unknown.

Question 1. There are 60 minutes in 1 hour. Use a tape diagram to find the total number of minutes in 6 hours and 15 minutes.

Answer: The total number of minutes in 6 hours and 15 minutes = 375 minutes.

Explanation: In the above-given question, given that, There are 60 minutes in 1 hour. 60 / n = 360. 360 / 60 = n. n = 60 + 15.

Question 2. Ms. Lemus buys 7 boxes of snacks. Each box has 12 packets of fruit snacks and 18 packets of cashews. How many snack packets does she buy altogether?

Answer: The number of snack packets = 210.

Explanation: In the above-given question, given that, Ms. Lemus buys 7 boxes of snacks. Each box has 12 packets of fruit snacks and 18 packets of cashews. 12 + 18 = 30. 30 x 7 = 210. so the number of snack packets = 210.

Question 3. Tamara wants to buy a tablet that costs $437. She saves $50 a month for 9 months. Does she have enough money to buy the tablet? Explain why or why not.

Answer: Yes, she has enough money.

Explanation: In the above-given question, given that, Tamara wants to buy a table that costs $437. she saves $50 a month for 9 months. 50 x 9 = 450. so she has enough money to buy.

Question 4. Mr. Ramirez receives 4 sets of books. Each set has 16 fiction books and 14 nonfiction books. He puts 97 books in his library and donates the rest. How many books does he donate?

Answer: The number of books he donates = 23.

Explanation: In the above-given question, given that, Mr. Ramirez receives 4 sets of books. Each set has 16 fiction books and 14 nonfiction books. He puts 97 books in his library and donates the rest. 16 + 14 = 30. 30 x 4 = 120. 120 – 97 = 23. so the number of books he donates = 23.

Question 5. Celia sells calendars for a fundraiser. Each calendar costs $9. She sells 16 calendars to her family members and 14 calendars to the people in her neighborhood. Her goal is to earn $300. Does Celia reach her goal? Explain your answer.

Answer: No, she does not reach her goal.

Explanation: In the above-given question, given that, Celia sells calendars for a fundraiser. Each calendar costs $9. she sells 16 calendars to her family members and 14 calendars to the people in her neighborhood. 16 + 14 = 30. 30 x 9 = 270. so she does not reach her goal.

Question 6. The video store sells science and history movies for $5 each. How much money does the video store make if it sells 33 science movies and 57 history movies?

Answer: The video store makes = $450.

Explanation: In the above-given question, given that, The video store sells science and history movies for $5 each. 33 + 57 = 90. 90 x 5 = 450. so the video store makes = $450.

Eureka Math Grade 3 Module 3 Lesson 21 Template Answer Key

Answer: 30 x 6 = 180, 9 x 60 = 540, 40 x 2 = 80, 10 x 6 = 60, 70 x 3 = 210, 50 x 6 = 300, 80 x 9 = 720, 20 x 5 = 100, 8 x 30 = 240, 3 x 30 = 90, 5 x 50 = 250, 4 x 40 = 160, 6 x 80 = 480, 70 x 7 = 490, 20 x 7 = 140, 10 x 7 = 70, 90 x 7 = 630, 2 x 60 = 120, 50 x 7 = 350, 80 x 5 = 400, 60 x 6 = 360, 9 x 50 = 450, 30 x 9 = 270, 4 x 80 = 320.

Explanation: In the above-given table, given that, the multiplication of 6, 7, 8, and 9 are given.

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## Answer Key 11.3

- [latex](g\circ f)(x)=-(\sqrt[5]{-x-3})^5-3[/latex] [latex]\begin{array}[t]{rrl} \phantom{(g\circ f)x}&=&-(-x-3)-3 \\ &=&x+3-3 \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}[/latex]
- [latex](g\circ f)(x)=4-\left(\dfrac{4}{x}\right)[/latex] [latex]\phantom{(g\circ f)(x)}=4-\dfrac{4}{x}\hspace{0.5in}\text{Not inverse}[/latex]
- [latex](g\circ f)(x)=-10\left(\dfrac{x-5}{10}\right)+5[/latex] [latex]\begin{array}[t]{rrl} \phantom{(g\circ f)x}&=& -x+5+5\\ \\ \phantom{(g\circ f)x}&=&-x+10\hspace{0.75in}\text{Not inverse} \end{array}[/latex]
- [latex](f\circ g)(x)=\dfrac{(10x+5)-5}{10}[/latex] [latex]\begin{array}[t]{rrl} \phantom{(f\circ g)x}&=&\dfrac{10x+5-5}{10} \\ \\ &=&\dfrac{10x}{10} \\ \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}[/latex]
- [latex](f\circ g)(x)=\dfrac{-2}{\dfrac{3x+2}{x+2}+3}[/latex] [latex]\begin{array}[t]{rrl} \phantom{(f\circ g)x}&=& \dfrac{-2(x+2)}{3x+2+3(x+2)}\\ \\ &=& \dfrac{-2x-4}{3x+2+3x+6}\\ \\ &=& \dfrac{-2x-4}{6x+8}\\ \\ &=& \dfrac{-x-2}{3x+4}\hspace{0.75in}\text{Not inverse} \end{array}[/latex]
- [latex](f\circ g)=\dfrac{-\left(\dfrac{-2x+1}{-x-1}\right)-1}{\dfrac{-2x+1}{-x-1}-2}[/latex] [latex]\begin{array}[t]{rrl} \phantom{(f\circ g)}&=&\dfrac{-(-2x+1)-1(-x-1)}{-2x+1-2(-x-1)} \\ \\ &=&\dfrac{2x-1+x+1}{-2x+1+2x+2} \\ \\ &=&\dfrac{3x}{3} \\ \\ &=&x\hspace{0.75in}\text{Inverse} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrcrr} y&=&(x-2)^5&+&3 \\ x&=&(y-2)^5&+&3 \\ -3&&&-&3 \\ \hline x-3&=&(y-2)^5&& \\ \sqrt[5]{x-3}&=&y-2&& \\ \\ y&=&\sqrt[5]{x-3}&+&2 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrcrr} y&=&\sqrt[3]{x+1}&+&2 \\ x&=&\sqrt[3]{y+1}&+&2 \\ -2&&&-&2 \\ \hline x-2&=&\sqrt[3]{y+1}&& \\ (x-2)^3&=&y+1&& \\ \\ y&=&(x-2)^3&-&1 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{4}{x+2} \\ \\ x&=&\dfrac{4}{y+2} \\ \\ y+2&=&\dfrac{4}{x} \\ \\ y&=&\dfrac{4}{x}-2 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{3}{x-3} \\ \\ x&=&\dfrac{3}{y-3} \\ \\ y-3&=&\dfrac{3}{x} \\ \\ y&=&\dfrac{3}{x}+3 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{-2x-2}{x+2} \\ \\ x&=&\dfrac{-2y-2}{y+2} \\ \\ x(y+2)&=&-2y-2 \\ xy+2x&=&-2y-2 \\ -xy+2&&-xy+2 \\ \hline 2x+2&=&-2y-xy \\ 2x+2&=&y(-2-x) \\ \\ y&=&\dfrac{2x+2}{-2-x} \\ \\ y&=&-\dfrac{2x+2}{2+x} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{9+x}{3} \\ \\ x&=&\dfrac{9+y}{3} \\ \\ 3x&=&9+y \\ y&=&3x-9 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{10-x}{5} \\ \\ x&=&\dfrac{10-y}{5} \\ \\ 5x&=&10-y \\ y&=&10-5x \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{5x-15}{2} \\ \\ x&=&\dfrac{5y-15}{2} \\ \\ 5y-15&=&2x \\ \\ 5y&=&2x+15 \\ \\ y&=&\dfrac{2x+15}{5} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&-(x-1)^3 \\ x&=&-(y-1)^3 \\ \sqrt[3]{x}&=&-(y-1) \\ \sqrt[3]{x}&=&-y+1 \\ -y&=&\sqrt[3]{x}-1 \\ \\ y&=&1-\sqrt[3]{x} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{12-3x}{4} \\ \\ x&=&\dfrac{12-3y}{4} \\ \\ 4x&=&12-3y \\ \\ 3y&=&12-4x \\ \\ y&=&\dfrac{12-4x}{3} \\ \\ y&=&4-\dfrac{4}{3}x \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&(x-3)^3 \\ x&=&(y-3)^3 \\ \sqrt[3]{x}&=&y-3 \\ \\ y&=&\sqrt[3]{x}+3 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\sqrt[5]{-x}+2 \\ x&=&\sqrt[5]{-y}+2 \\ x-2&=&\sqrt[5]{-y} \\ (x-2)^5&=&-y \\ \\ y&=&-(x-2)^5 \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{x}{x-1} \\ \\ x&=&\dfrac{y}{y-1} \\ \\ x(y-1)&=&y \\ \\ xy-x&=&y \\ \\ y-xy&=&-x \\ \\ y(1-x)&=&-x \\ \\ y&=&\dfrac{-x}{1-x} \\ \\ y&=&\dfrac{x}{x-1} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{-3-2x}{x+3} \\ \\ x&=&\dfrac{-3-2y}{y+3} \\ \\ x(y+3)&=&-3-2y \\ xy+3x&=&-3-2y \\ +2y-3x&&-3x+2y \\ \hline xy+2y&=&-3-3x \\ y(x+2)&=&-3-3x \\ \\ y&=&\dfrac{-3-3x}{x+2} \\ \\ y&=&-\dfrac{3x+3}{x+2} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{x-1}{x+1} \\ \\ x&=&\dfrac{y-1}{y+1} \\ \\ x(y+1)&=&y-1 \\ xy+x&=&y-1 \\ xy-y&=&-x-1 \\ y(x-1)&=&-x-1 \\ \\ y&=&\dfrac{-x-1}{x-1} \\ \\ y&=&-\dfrac{x+1}{x-1} \end{array}[/latex]
- [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrl} y&=&\dfrac{x}{x+2} \\ \\ x&=&\dfrac{y}{y+2} \\ \\ x(y+2)&=&y \\ xy+2x&=&y \\ 2x&=&y-xy \\ 2x&=&y(1-x) \\ \\ y&=&\dfrac{2x}{1-x} \end{array}[/latex]

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PowerPoint Presentation. New York State Common Core KEY Mathematics Curriculum GRADE 3. MODULE 3 Multiplication and Division with Units of O, 1, 6—9, and Multiples of 10 Module Overview „ Topic A: The Properties of Multiplication and Division MODULE 3 ..3.A.1 3.B.1 .3.c.1 .3.D.1 3.E.1 3.F.1 L 2 a. A tricyde 3 N of Tota I of b.

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Multiplication and Division with Units of 0, 1, 6─9, and Multiples of 10 Module 3: Date:

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Add-on U06.AO.04 - Lesson 12 - More Work with the Directrix and Focus RESOURCE ANSWER KEY EDITABLE RESOURCE EDITABLE KEY

Name LESSON 3.3 For use with pages 161—169 Is there enough information to prove t at In parallel? If so, state the postula e or eorem you would use. 250 1220 Find the value of x that makes m Il n.

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Lesson 3.3 Practice Level A yes; Corresponding Angles Converse yes; Alternate Interior Angles Converse yes; Alternate Exterior Angles Converse

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Answer Key 11.3 (g∘f)(x) = −( 5√−x−3)5 −3 ( g ∘ f) ( x) = − ( − x − 3 5) 5 − 3 (g ∘f)x = −(−x−3)− 3 = x+ 3−3 = x Inverse ( g ∘ f ...