unit 4 linear equations answer key homework 5

4.1 Linear Functions

m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; decreasing because m < 0. m < 0.

y = − 7 x + 3 y = − 7 x + 3

H ( x ) = 0.5 x + 12.5 H ( x ) = 0.5 x + 12.5

Possible answers include ( − 3 , 7 ) , ( − 3 , 7 ) , ( − 6 , 9 ) , ( − 6 , 9 ) , or ( − 9 , 11 ) . ( − 9 , 11 ) .

( 16 , 0 ) ( 16 , 0 )

• ⓐ f ( x ) = 2 x ; f ( x ) = 2 x ;
• ⓑ g ( x ) = − 1 2 x g ( x ) = − 1 2 x

y = – 1 3 x + 6 y = – 1 3 x + 6

4.2 Modeling with Linear Functions

ⓐ C ( x ) = 0.25 x + 25 , 000 C ( x ) = 0.25 x + 25 , 000 ⓑ The y -intercept is ( 0 , 25 , 000 ) ( 0 , 25 , 000 ) . If the company does not produce a single doughnut, they still incur a cost of $25,000.

ⓐ 41,100 ⓑ 2020

21.57 miles

4.3 Fitting Linear Models to Data

54 ° F 54 ° F

150.871 billion gallons; extrapolation

4.1 Section Exercises

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

d ( t ) = 100 − 10 t d ( t ) = 100 − 10 t

The point of intersection is ( a , a ) . ( a , a ) . This is because for the horizontal line, all of the y y coordinates are a a and for the vertical line, all of the x x coordinates are a . a . The point of intersection is on both lines and therefore will have these two characteristics.

y = 3 5 x − 1 y = 3 5 x − 1

y = 3 x − 2 y = 3 x − 2

y = − 1 3 x + 11 3 y = − 1 3 x + 11 3

y = − 1.5 x − 3 y = − 1.5 x − 3

perpendicular

f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 ) f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 )

h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 ) h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 )

− 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 ) − 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 )

Line 1: m = –10 Line 2: m = –10 Parallel

Line 1: m = –2 Line 2: m = 1 Neither

Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel

y = 3 x − 3 y = 3 x − 3

y = − 1 3 t + 2 y = − 1 3 t + 2

y = − 5 4 x + 5 y = − 5 4 x + 5

y = 3 x − 1 y = 3 x − 1

y = − 2.5 y = − 2.5

y = 3 y = 3

x = − 3 x = − 3

Linear, g ( x ) = − 3 x + 5 g ( x ) = − 3 x + 5

Linear, f ( x ) = 5 x − 5 f ( x ) = 5 x − 5

Linear, g ( x ) = − 25 2 x + 6 g ( x ) = − 25 2 x + 6

Linear, f ( x ) = 10 x − 24 f ( x ) = 10 x − 24

f ( x ) = − 58 x + 17.3 f ( x ) = − 58 x + 17.3

• ⓐ a = 11,900 , b = 1000.1 a = 11,900 , b = 1000.1
• ⓑ q ( p ) = 1000 p – 100 q ( p ) = 1000 p – 100

y = − 16 3 y = − 16 3

x = a x = a

y = d c – a x – a d c – a y = d c – a x – a d c – a

y = 100 x – 98 y = 100 x – 98

x < 1999 201 , x > 1999 201 x < 1999 201 , x > 1999 201

$45 per training session.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

The slope is –400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

4.2 Section Exercises

Determine the independent variable. This is the variable upon which the output depends.

To determine the initial value, find the output when the input is equal to zero.

6 square units

20.01 square units

P ( t ) = 75 , 000 + 2500 t P ( t ) = 75 , 000 + 2500 t

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

Ten years after the model began

W ( t ) = 0.5 t + 7.5 W ( t ) = 0.5 t + 7.5

( − 15 , 0 ) ( − 15 , 0 ) : The x -intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. ( 0 , 7 . 5 ) ( 0 , 7 . 5 ) : The baby weighed 7.5 pounds at birth.

At age 5.8 months

C ( t ) = 12 , 025 − 205 t C ( t ) = 12 , 025 − 205 t

( 58 . 7 , 0 ) : ( 58 . 7 , 0 ) : In roughly 59 years, the number of people inflicted with the common cold would be 0. ( 0 , 12 , 0 25 ) ( 0 , 12 , 0 25 ) Initially there were 12,025 people afflicted by the common cold.

y = − 2 t +180 y = − 2 t +180

In 2070, the company’s profit will be zero.

y = 3 0 t − 3 00 y = 3 0 t − 3 00

(10, 0) In the year 1990, the company’s profits were zero

During the year 1933

• ⓐ 696 people
• ⓒ 174 people per year
• ⓓ 305 people
• ⓔ P(t) = 305 + 174t
• ⓕ 2,219 people
• ⓐ C(x) = 0.15x + 10
• ⓑ The flat monthly fee is $10 and there is a $0.15 fee for each additional minute used

P(t) = 190t + 4,360

• ⓐ R ( t ) = − 2 . 1 t   +   16 R ( t ) = − 2 . 1 t   +   16
• ⓑ 5.5 billion cubic feet
• ⓒ During the year 2017

More than 133 minutes

More than $42,857.14 worth of jewelry

More than $66,666.67 in sales

4.3 Section Exercises

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

We predict a value outside the domain and range of the data.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

61.966 years

Interpolation. About 60 ° F . 60 ° F .

This value of r indicates a strong negative correlation or slope, so C This value of r indicates a strong negative correlation or slope, so C

This value of r indicates a weak negative correlation, so B This value of r indicates a weak negative correlation, so B

Yes, trend appears linear because r = 0. 985 r = 0. 985 and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

y = 1 . 64 0 x + 13 . 8 00 , y = 1 . 64 0 x + 13 . 8 00 , r = 0. 987 r = 0. 987

y = − 0.962 x + 26.86 ,       r = − 0.965 y = − 0.962 x + 26.86 ,       r = − 0.965

y = − 1 . 981 x + 6 0. 197; y = − 1 . 981 x + 6 0. 197; r = − 0. 998 r = − 0. 998

y = 0. 121 x − 38.841 , r = 0.998 y = 0. 121 x − 38.841 , r = 0.998

( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; ( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; Yes, the function is a good fit.

( 189 .8 , 0 ) ( 189 .8 , 0 ) If 18,980 units are sold, the company will have a profit of zero dollars.

y = 0.00587 x + 1985 .4 1 y = 0.00587 x + 1985 .4 1

y = 2 0. 25 x − 671 . 5 y = 2 0. 25 x − 671 . 5

y = − 1 0. 75 x + 742 . 5 0 y = − 1 0. 75 x + 742 . 5 0

Review Exercises

y = − 3 x + 26 y = − 3 x + 26

y = 2 x − 2 y = 2 x − 2

Not linear.

( –9 , 0 ) ; ( 0 , –7 ) ( –9 , 0 ) ; ( 0 , –7 )

Line 1: m = − 2 ; m = − 2 ; Line 2: m = − 2 ; m = − 2 ; Parallel

y = − 0.2 x + 21 y = − 0.2 x + 21

More than 250

y = − 3 00 x + 11 , 5 00 y = − 3 00 x + 11 , 5 00

• ⓑ 100 students per year
• ⓒ P ( t ) = 1 00 t + 17 00 P ( t ) = 1 00 t + 17 00

Extrapolation

y = − 1.294 x + 49.412 ;   r = − 0.974 y = − 1.294 x + 49.412 ;   r = − 0.974

Practice Test

y = −1.5x − 6

y = −2x − 1

Perpendicular

(−7, 0); (0, −2)

y = −0.25x + 12

Slope = −1 and y-intercept = 6

y = 875x + 10,625

• ⓑ dropped an average of 46.875, or about 47 people per year
• ⓒ y = −46.875t + 1250

In early 2018

y = 0.00455x + 1979.5

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Unit 4 - Linear Equations and Linear Systems

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Algebra 1 Unit 4: Linear Equations

This unit includes 73 pages of guided notes, homework assignments, three quizzes, a study guide, and a unit test that cover the topics listed in the description below.

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• Additional Information
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This unit contains the following topics:

• Slope from a Graph • Slope from Ordered Pairs (The Slope Formula) • Linear Equations: Slope Intercept Form vs. Standard Form • Graphing by Slope Intercept Form • Writing Linear Equations Given a Graph • Graphing by Intercepts • Vertical vs. Horizontal Lines • Writing Linear Equations given Point and Slope • Writing Linear Equations given Two Points • Linear Equation Word Problems • Parallel vs. Perpendicular Lines • Scatter Plots & Line of Best Fit • Linear Regression

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2.4: Graphing Linear Equations- Answers to the Homework Exercises

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• Page ID 45036

• Darlene Diaz
• Santiago Canyon College via ASCCC Open Educational Resources Initiative

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Graphing and Slope

• \(\frac{1}{3}\)
• \(\frac{4}{3}\)
• \(\frac{1}{2}\)
• \(-\frac{1}{3}\)
• \(\frac{16}{7}\)
• \(-\frac{7}{17}\)
• \(\frac{1}{16}\)
• \(\frac{24}{11}\)
• \(x=\frac{23}{6}\)
• \(y=-\frac{29}{6}\)

Equations of Lines

• \(y=-\frac{3}{4}x-1\)
• \(y = −6x + 4\)
• \(y = − \frac{1}{4} x + 3\)
• \(y = \frac{1}{3} x + 3\)
• \(y = −3x + 5\)
• \(y = − \frac{1}{10} x − \frac{37}{10}\)
• \(y = \frac{7x}{3} − 8\)
• \(y = −4x + 3\)
• \(y = \frac{1}{10} x − \frac{3}{10}\)
• \(y = − \frac{4}{7} x + 4\)
• \(y=\frac{5}{2}x\)

• \(y − (−5) = 9(x − (−1))\)
• \(y − (−2) = −3(x − 0)\)
• \(y − (−3) = \frac{1}{5} (x − (−5))\)
• \(y − 2 = 0(x − 1)\)
• \(y − (−2) = −2(x − 2)\)
• \(y − 1 = 4(x − (−1))\)
• \(y − (−4) = − \frac{2}{3} (x − (−1))\)
• \(y = − \frac{3}{5} x + 2\)
• \(y = − \frac{3}{2} x + 4\)
• \(y = x − 4\)
• \(y = − \frac{1}{2} x\)
• \(y = − \frac{2}{3} x − \frac{10}{3}\)
• \(y = − \frac{5}{2} x − 5\)
• \(y = −3\)
• \(y − 3 = −2(x + 4)\)
• \(y + 2 = \frac{3}{2} (x + 4)\)
• \(y + 3 = − \frac{8}{7} (x − 3)\)
• \(y − 5 = − \frac{1}{8} (x + 4)\)
• \(y + 4 = −(x + 1)\)
• \(y = − \frac{8}{7} x − \frac{5}{7}\)
• \(y = −x + 2\)
• \(y = − \frac{1}{10} x − \frac{3}{2}\)
• \(y=\frac{1}{3}x+1\)

Parallel and Perpendicular Lines

• \(m_{||} = 2\)
• \(m_{||} = 1\)
• \(m_{||} = − \frac{2}{3}\)
• \(m_{||} = \frac{6}{5}\)
• \(m_{⊥} = 0\)
• \(m_{⊥} = −3\)
• \(m_{⊥} = 2\)
• \(m_{⊥} = − \frac{1}{3}\)
• \(y − 4 = \frac{9}{2} (x − 3)\)
• \(y − 3 = \frac{7}{5} (x − 2)\)
• \(y + 5 = −(x − 1)\)
• \(y − 2 = \frac{1}{5} (x − 5)\)
• \(y − 2 = − \frac{1}{4} (x − 4)\)
• \(y + 2 = −3(x − 2)\)
• \(y = −2x + 5\)
• \(y = − \frac{4}{3} x − 3\)
• \(y = − \frac{1}{2} x − 3\)
• \(y = − \frac{1}{2} x − 2\)
• \(y = x − 1\)
• \(y=-2x+5\)

Linear Equations (Algebra 1 Curriculum - Unit 4) | All Things Algebra®

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This Linear Equations Unit Bundle contains guided notes, homework assignments, three quizzes, study guide and a unit test that cover the following topics:

• Slope from a Graph

• Slope from Ordered Pairs (The Slope Formula)

• Linear Equations: Slope Intercept Form vs. Standard Form

• Graphing by Slope Intercept Form

• Writing Linear Equations Given a Graph

• Graphing by Intercepts

• Vertical vs. Horizontal Lines

• Writing Linear Equations given Point and Slope

• Writing Linear Equations given Two Points

• Linear Equation Word Problems

• Parallel vs. Perpendicular Lines

• Scatter Plots & Line of Best Fit

• Linear Regression

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Unit 7 – Polynomials & Factoring

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Unit 9 – Linear, Quadratic, and Exponential Functions

Unit 10 – Radical Expressions & Equations

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Unit 12 – Statistics

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Unit 4 Linear Equations Homework 2 Answer Key

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