## 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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Lesson 4.1.1, lesson 4.1.2, lesson 4.2.1, lesson 4.2.2, lesson 4.2.3, lesson 4.2.4, lesson 4.3.1, lesson 4.3.2, lesson 4.3.3.

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## Eureka Math Grade 4 Module 3 Lesson 28 Answer Key

Engage ny eureka math 4th grade module 3 lesson 28 answer key, eureka math grade 4 module 3 lesson 28 problem set answer key.

Explanation: Given each filled 581 one-liter bottles with apple cider. He distributed the bottles to 4 stores. Each store received the same number of bottles of 581 ÷ 4 = 145 liters, Remainder 1, So, 145 liters bottles each of the stores receive, there are 1 bottle left over.

## Eureka Math Grade 4 Module 3 Lesson 28 Exit Ticket Answer Key

Question 2. A carton of milk contains 128 ounces. Sara’s son drinks 4 ounces of milk at each meal. How many 4-ounce servings will one carton of milk provide? Answer: 32 ounces of 4-ounce servings will one carton of milk provide,

Explanation: Given a carton of milk contains 128 ounces. Sara’s son drinks 4 ounces of milk at each meal, So 128 ounces ÷ 4 = 32 4|128 12 008 008 0 therefore, 32 ounces of 4-ounce servings will one carton of milk provide.

## Eureka Math Grade 4 Module 3 Lesson 28 Homework Answer Key

Question 2. Selena’s dog completed an obstacle course that was 932 meters long. There were 4 parts to the course, all equal in length. How long was 1 part of the course? Answer: 1 part of the course is 233 meters long,

Explanation: Given Selena’s dog completed an obstacle course that was 932 meters long. There were 4 parts to the course, all equal in length. Number of meters long was 1 part of the course is 932 meters ÷ 4 = 233 4|932 8 13 12 12 12 0 Therefore, 1 part of the course is 233 meters long.

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A STORY OF UNITS TEKS EDITION Lesson 2 Answer Key 4 • 3 Lesson 2 Problem Set 1. a. Width 4 ft, length 12 ft b. 32 ft 2. a. Diagram drawn; width 5 in, length 30 in b. 70 in; 150 sq in 3. a. 6 cm b. Diagram drawn; width 18 cm, length 7 cm c. 50 cm 4. a. Diagram drawn and labeled; 18 ft b. Diagram drawn and labeled; 36 ft

HomeworkHelper. G4-M3-HWH-1.3.-09.2015. 2015-16. Lesson 1 : Investigate and use the formulas for area and perimeter of rectangles. 4•3A Story of Units 3. The perimeter of this rectangle is 250 centimeters. Find the unknown side length of this rectangle . 4. The following rectangle has whole number side lengths.

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1. neither 2. corresponding angles. 3. corresponding sides 4. neither. 5. false; n IHJ ù n KHJ 6. true; SSS. 7. false; n ACE ù n BFD 8. true; SSS. 9. Stable; the figure forms triangles of fixed side lengths which cannot change shape by the SSS Congru-ence Postulate. 10.

Pete wants to make turkey sandwiches for two friends and himself. He wants each sandwich to contain 3.5 ounces of turkey. How many ounces of turkey does he need? @ 3.5 ounces 7 ounces 10.5 ounces 14 ounces Spiral Review (5.0A.1, 5.0A.2, 5.NBT.6, 5.NBT.7) 3. A group of 5 boys and 8 girls goes to the fair. Admission costs $9 per person.

Lesson 2 Homework 4-4 3. Construct each of the following using a straightedge and/or the right angle template that you created. Explain the characteristics of each by comparing the angle to a right angle. Use the words greater than, less than, or equal to in your explanations. a. b. c. acute angle right angle obtuse angle 19 A Z VOC IS -FHA N

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Lesson 4.3 Homework Answers Pg 197 - #1-25 odd, 29-33 odd, 42-45 Pg 201 - #1-10 Pg 197 because ║ lines have 1) ' PQR # ' VXW 3) RS 5) yes 7) yes 9) AAS 11) not possible 13) a) UWV b) UW ... 3rd are . Lesson 4.3 Homework Answers Pg 197 - #1-25 odd, 29-33 odd, 42-45 Pg 201 - #1-10 Pg 201 1) RS # JK ST# KL RT # JL R# J S# K T # L

Introduction to Systems of Equations and Inequalities; 7.1 Systems of Linear Equations: Two Variables; 7.2 Systems of Linear Equations: Three Variables; 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables; 7.4 Partial Fractions; 7.5 Matrices and Matrix Operations; 7.6 Solving Systems with Gaussian Elimination; 7.7 Solving Systems with Inverses; 7.8 Solving Systems with Cramer's Rule

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Find step-by-step solutions and answers to College Algebra - 9780321729682, as well as thousands of textbooks so you can move forward with confidence. ... Section 3.5: Rational Functions and Their Graphs. Section 3.6: Polynomial and Rational Inequalities. Section 3.7: Modeling Using Variation. Page 445: Review Exercises. Page 449: Chapter Test ...

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Explanation: Drawn a number bond and used the long division algorithm to record my work from Part (a) as shown above 960 ÷ 4 = 240. Question 3. a. Draw an area model to solve 774 ÷ 3. Answer: Explanation: Drawn an area model as shown above for 774 ÷ 3 =. (600 ÷ 3) + (150 ÷ 3) + (24 ÷ 3) = 200 + 50 + 8 = 258.

3. PQ −− RQ −−, PS −− RS −− 3. Given 4. QS −− −− QS 4. Reflexive Property 5. PQS RQS 5. CPCTC 6. QUILTING a. Indicate the triangles that appear to be congruent. b. Name the congruent angles and congruent sides of a pair of congruent triangles. B A I E H G F CD 4-3 "# $ 2 34 % 1 (2x + 4)° (3y-3) 80° 100° 10 12 ∠A ∠ ...

7. 608. X 9. 5,472 ,here 9 X 8 ones = 72 ones, we write 2 at ones place and. take 7 to tens places then 9 x 0 tens = 0 tens + 7 tens = 7 tens, we write 7 at tens place and 9 X 6 hundreds = 54 hundreds, now we write 4 at hundreds place and 5 at thousands place as shown above, So 9 X 608 = 5,472. b. Answer:

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The source for the homework pages is the full module PDF, available for free here:https://www.engageny.org/resource/grade-3-mathematics-module-4

rted License.3.E.25NYS COMMON CORE. ATHEMATICS CURRICULUMSolve us. rea model. The first. ne is done for you.Lesson 15:Date: ghts reserved. commoncore.orgLesson 15 Homework 43.E.26Understand and solve divisio. ay and area models.8/28/13This work is licensed under a Creative Common.

Engage NY Eureka Math 4th Grade Module 3 Lesson 32 Answer Key Eureka Math Grade 4 Module 3 Lesson 32 Problem Set Answer Key. Solve the following problems. Draw tape diagrams to help you solve. If there is a remainder, shade in a small portion of the tape diagram to represent that portion of the whole. Question 1.

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Engage NY Eureka Math 4th Grade Module 3 Lesson 16 Answer Key Eureka Math Grade 4 Module 3 Lesson 16 Problem Set Answer Key. Show the division using disks. Relate your work on the place value chart to long division. Check your quotient and remainder by using multiplication and addition. Question 1. 7 ÷ 2. quotient = ____3_____ remainder = ___1

Answer: 4 bows can be made from 3 feet of ribbon, Left over ribbon will be 4 inches, Explanation. Given If it takes 8 inches of ribbon to make a bow, So how many bows can be made from 3 feet of ribbon. (1 foot = 12 inches) are 3 X 12 = 36 inches ÷ 8 inches =. 4 quotient (8 X 4 = 32) and 4 inches remainder means|.

Eureka Math Grade 4 Module 3 Lesson 28 Problem Set Answer Key. Question 1. Divide. Check your work by multiplying. Draw disks on a place value chart as needed. multiplying 287 X 2 =574 as shown above. Drawn disks on a place value chart as needed as shown above. Drawn disks on a place value chart as needed as shown above.