lesson 3 homework 2.4 answer key

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Eureka Math Grade 5 Module 2 Lesson 3 Answer Key

Engage ny eureka math 5th grade module 2 lesson 3 answer key, eureka math grade 5 module 2 lesson 3 problem set answer key.

Question 1. Draw a model. Then, write the numerical expressions. a. The sum of 8 and 7, doubled

Answer: The sum of 8 and 7, doubled = 30.

b. 4 times the sum of 14 and 26

Answer: The sum of 14 and 26, 4 times = 160.

c. 3 times the difference between 37.5 and 24.5

Answer: The difference of 37.5 and 24.5, 3 times = 13.0.

d. The sum of 3 sixteens and 2 nines

Answer: The sum of 3 sixteens and 2 nines = 30.

e. The difference between 4 twenty-fives and 3 twenty-fives

Answer: The difference between 4 25 and 3 25 = 25.

f. Triple the sum of 33 and 27

Answer: The sum of 33 and 27, 3 times = 180.

Question 2. Write the numerical expressions in words. Then, solve.

Answer: 360.

Explanation: In the above-given question, given that, 12 x ( 5 + 25). 12 x (30). 360.

Answer: 550.

Explanation: In the above-given question, given that, (62 – 12) × 11 50 x 11 = 550.

Answer: 2300.

Explanation: In the above-given question, given that, (45 + 55) × 23. 100 x 23. 2300.

Answer: 76.

Explanation: In the above-given question, given that, (30 × 2) + (8 × 2) 60 + 16. 76.

Answer: 600 > 360.

Explanation: In the above-given question, given that, 24 x (20 + 5). 24 x 25. 600. (20 + 5) x 12. 30 x 12. 360. 600 > 360.

Answer: 487 > 513.

Explanation: In the above-given question, given that, 18 x 27. 487. 20 twenty-seven – 1 27. 540 – 27. 513. 487 < 513.

Answer: 171 = 171.

Question 4. Mr. Huynh wrote the sum of 7 fifteens and 38 fifteens on the board. Draw a model, and write the correct expression.

Answer: The sum of 7 fifteens and 38 fifteens = 675.

Question 5. Two students wrote the following numerical expressions. Angeline: (7 + 15) × (38 + 15) MeiLing: 15 × (7 + 38)   Are the students’ expressions equivalent to your answer in Problem 4? Explain your answer.

Answer: Yes, the student’s expressions equivalent.

Explanation: In the above-given question, given that, Angeline: (7 + 15) × (38 + 15) Meiling: 15 × (7 + 38)   105 + 570 = 675.

Question 6. A box contains 24 oranges. Mr. Lee ordered 8 boxes for his store and 12 boxes for his restaurant. a. Write an expression to show how to find the total number of oranges ordered.

Answer: The total number of oranges ordered= 480.

Explanation: In the above-given question, given that, A box contains 24 oranges. Mr. Lee ordered 8 boxes for his store and 12 boxes for his restaurant. ( 24 x 8 ) + ( 24 x 12) 192 + 288 480

b. Next week, Mr. Lee will double the number of boxes he orders. Write a new expression to represent the number of oranges in next week’s order.

Answer: The total number of oranges ordered= 960.

Explanation: In the above-given question, given that, A box contains 24 oranges. Mr. Lee ordered 8 boxes for his store and 12 boxes for his restaurant. ( 24 x 8 ) + ( 24 x 12) 192 + 288 480 + 480 = 960

c. Evaluate your expression from Part (b) to find the total number of oranges ordered in both weeks.

Answer: The total number of oranges ordered in both weeks = 1440.

Explanation: In the above-given question, given that, 960 + 480 = 1440.

Eureka Math Grade 5 Module 2 Lesson 3 Exit Ticket Answer Key

Question 1. Draw a model. Then, write the numerical expressions. a. The difference between 8 forty-sevens and 7 forty-sevens

Answer: The difference of 8 forty-sevens and 7 forty-sevens = 47.

Explanation: In the above-given question, given that, the difference of 8 forty sevens and 7 forty-sevens. 8 x 47 – 7 x 47. 376 – 329. 47

b. 6 times the sum of 12 and 8

Answer: The sum of 12 and 8, 6 times = 120.

Explanation: In the above-given question, given that, the sum of 12 and 8 is 120. 20 + 20 + 20 + 20 + 20 + 20 = 120. 12 + 8 = 20.

Answer: 4836 > 2028.

Eureka Math Grade 5 Module 2 Lesson 3 Homework Answer Key

Question 1. Draw a model. Then, write the numerical expressions. a. The sum of 21 and 4, doubled b. 5 times the sum of 7 and 23 c. 2 times the difference between 49.5 and 37.5 d. The sum of 3 fifteens and 4 twos e. The difference between 9 thirty-sevens and 8 thirty-sevens f. Triple the sum of 45 and 55

Answer: The sum of 21 and 4, 2 times = 50.

Answer: The sum of 7 and 23, 5 times = 150.

Answer: 2 times the difference between 49.5 and 37.5 = 24.

Answer: The sum of 3 fifteens and 4 twos = 53.

Answer: The difference between 9 thirty-sevens and 8 thirty-sevens = 37.

Explanation: In the above-given question, given that, The difference between 9 thirty-sevens and 8 thirty-sevens 9 x 37 – 8 x 37. 333 – 296. 37.

Triple the sum of 45 and 55 = 300.

Explanation: In the above-given question, given that, Triple the sum of 45 and 55 45 + 55 = 100. 100 + 100 + 100 = 300.

Answer: 160.

Explanation: In the above-given question, given that, 10 x ( 2.5 + 13.5). 10 x (16). 160.

Answer: 220.

Explanation: In the above-given question, given that, 11 x ( 98 – 78). 11 x (20). 220.

Answer: 2600.

Explanation: In the above-given question, given that, 26 x ( 71 + 29). 26 x (30). 2600.

Answer: 3000.

Explanation: In the above-given question, given that, (50 x 2) + ( 15 + 2). 100 x (30). 3000.

Answer: 3906 > 1438.

Explanation: In the above-given question, given that, 93 x ( 40 + 2). 93 x (42). 3906. (40 + 2) x 36 42 x 36. 1438. 3906 > 1438.

Answer: 1525 > 1475.

Explanation: In the above-given question, given that, 61 x ( 25). 1525. 60 twenty fives minus one twenty-five. 60  x 25 – 25. 1500 – 25. 1475. 1525 > 1475.

Question 4. Larry claims that (14 + 12) × (8 + 12) and (14 × 12) + (8 × 12) are equivalent because they have the same digits and the same operations. a. Is Larry correct? Explain your thinking.

Answer: Yes, Larry was not correct.

Explanation: In the above-given question, given that, Larry claims that (14 + 12) x (8 + 12) and (14 x 12) + (8 x 12). 42 + 12 = 26. 8 + 12 = 20. 26 x 20 = 520. (14 x 12) + (8 x 12). 168 + 96. 264. 520 is not equal to 264.

b. Which expression is greater? How much greater?

Answer: (14 + 12) x (8 + 12) is greater. 256 is greater.

Explanation: In the above-given question, given that, 42 + 12 = 26. 8 + 12 = 20. 26 x 20 = 520. (14 x 12) + (8 x 12). 168 + 96. 264. 520 is not equal to 264.

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Home > CC3 > Chapter Ch3 > Lesson 3.2.2

Lesson 3.1.1, lesson 3.1.2, lesson 3.1.3, lesson 3.1.4, lesson 3.1.5, lesson 3.1.6, lesson 3.1.7, lesson 3.2.1, lesson 3.2.2, lesson 3.2.3, lesson 3.2.4, lesson 3.2.5.

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Texas Go Math Grade 4 Lesson 2.4 Answer Key Relate Hundredths and Decimals

Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 2.4 Answer Key Relate Hundredths and Decimals.

Essential Question

How can you record hundredths as fractions and decimals? Answer: We know that, The representation of the hundredths in the place-value chart is: 0.01 So, The representation of the hundredths in the form of a fraction is: \(\frac{1}{100}\) Hence, from the above, We can conclude that The representation of the hundredths in the form of a fraction is: \(\frac{1}{100}\) The representation of the hundredths in the form of a decimal number is 0.01

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In the 2008 Summer Olympic Games, the winning time in the men’s 100-meter butterfly race was only \(\frac{1}{100}\) second faster than the second-place time. What decimal represents this fraction of a second? • Circle the numbers you need to use. Answer: It is given that In the 2008 Summer Olympic Games, the winning time in the men’s 100-meter butterfly race was only \(\frac{1}{100}\) second faster than the second-place time Hence, from the above, We can conclude that The numbers we need to use are: 1 and 100

You can write hundredths as fractions or decimals.

Mathematical Processes Explain how the size of one-tenth is related to the size one hundredth. Answer: We know that, One-hundredth is the tenth of the one-tenth Hence, from the above, We can conclude that One-hundredth = \(\frac{1}{10}\) × One-tenth

Alicia Won her 400-meter freestyle race by 4\(\frac{25}{100}\) seconds. How can you write this mixed number as a decimal?

Share and Show

Write the fraction or mixed number and the decimal shown by the model.

Mathematical Processes Are 0.5 and 0.50 equivalent? Explain. Answer: The given decimal numbers are: 0.5 and 0.50 Now, We know that, If zero is placed on the right side as the last digit in the decimal number, then that zero don’t have any value So, 0.5 = 0.50 Hence, from the above, We can conclude that 0.5 and 0.50 are equivalent

Problem Solving

Question 6. Write Math Explain how one whole, one-tenth, and one hundredth are related. Answer: We know that, “One-whole” is represented as 1 “One-tenth” is \(\frac{1}{10}\) of 1 “One-hundredth” is \(\frac{1}{10}\) of “One-tenth” and \(\frac{1}{100}\) of 1 Hence, from the above, We can conclude that One-hundredth = \(\frac{1}{10}\) of “One-tenth” = \(\frac{1}{100}\) of 1

Sense or Nonsense?

Daily Assessment Task

Fill in the bubble completely to show your answer.

TEXAS Test Prep

Texas Go Math Grade 4 Lesson 2.4 Homework and Practice Answer Key

Relate Hundredths and Decimals

Whose statement makes sense? whose statement is nonsense? Explain your reasoning. Answer: It is given that Emma said, “It is seventy-five hundredths of a mile to Waterslide Park.” Charles said, “It is seven and five-tenths miles to Waterslide Park.” Now, The representation of seventy-five hundredths in the Standard form is: 0.75 The representation of seven and five-tenths in the Standard form is: 7.5 It is given that The Waterslide park is: 0.75 miles Hence, from the above, We can conclude that Emma’s statement makes sense Charles’s statement makes not sense

Question 3. Jake said he has \(\frac{8}{100}\) of a dollar in his pocket. Margie said that the decimal for \(\frac{8}{100}\) is 0.8. What error did she make? What is the decimal form for \(\frac{8}{100}\)? Answer: It is given that Jake said he has \(\frac{8}{100}\) of a dollar in his pocket. Margie said that the decimal for \(\frac{8}{100}\) is 0.8 Now, The representation of \(\frac{8}{100}\) in the decimal form is: 0.08 Now, 0.8 represents only one digit after the decimal point But, From the given fraction, The digits after the decimal point must have 2 So, The error made by Margie is: The misinterprettion of digits after the decimal point Hence, from the above, We can conclude that The error made by Margie is: The misinterprettion of digits after the decimal point The representation of \(\frac{8}{100}\) in the decimal form is: 0.08

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2.1 The Rectangular Coordinate Systems and Graphs

x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .

125 = 5 5 125 = 5 5

( − 5 , 5 2 ) ( − 5 , 5 2 )

2.2 Linear Equations in One Variable

x = −5 x = −5

x = −3 x = −3

x = 10 3 x = 10 3

x = 1 x = 1

x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .

x = 1 3 x = 1 3

m = − 2 3 m = − 2 3

y = 4 x −3 y = 4 x −3

x + 3 y = 2 x + 3 y = 2

Horizontal line: y = 2 y = 2

Parallel lines: equations are written in slope-intercept form.

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

2.3 Models and Applications

C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650

L = 37 L = 37 cm, W = 18 W = 18 cm

2.4 Complex Numbers

−24 = 0 + 2 i 6 −24 = 0 + 2 i 6

( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i

5 2 − i 5 2 − i

18 + i 18 + i

−3 −4 i −3 −4 i

2.5 Quadratic Equations

( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1

( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.

( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.

( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4

x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1

x = 4 ± 5 x = 4 ± 5

x = 3 ± 22 x = 3 ± 22

x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3

2.6 Other Types of Equations

{ −1 } { −1 }

0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2

1 ; 1 ; extraneous solution − 2 9 − 2 9

−2 ; −2 ; extraneous solution −1 −1

−1 , −1 , 3 2 3 2

−3 , 3 , − i , i −3 , 3 , − i , i

2 , 12 2 , 12

−1 , −1 , 0 0 is not a solution.

2.7 Linear Inequalities and Absolute Value Inequalities

[ −3 , 5 ] [ −3 , 5 ]

( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )

x < 1 x < 1

x ≥ −5 x ≥ −5

( 2 , ∞ ) ( 2 , ∞ )

[ − 3 14 , ∞ ) [ − 3 14 , ∞ )

6 < x ≤ 9 ​ or ( 6 , 9 ] 6 < x ≤ 9 ​ or ( 6 , 9 ]

( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )

| x −2 | ≤ 3 | x −2 | ≤ 3

k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) . ( − ∞ , 1 ] ∪ [ 7 , ∞ ) .

2.1 Section Exercises

Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.

The y -intercept is the point where the graph crosses the y -axis.

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .

The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .

The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .

y = 4 − 2 x y = 4 − 2 x

y = 5 − 2 x 3 y = 5 − 2 x 3

y = 2 x − 4 5 y = 2 x − 4 5

d = 74 d = 74

d = 36 = 6 d = 36 = 6

d ≈ 62.97 d ≈ 62.97

( 3 , − 3 2 ) ( 3 , − 3 2 )

( 2 , −1 ) ( 2 , −1 )

( 0 , 0 ) ( 0 , 0 )

y = 0 y = 0

not collinear

A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )

d = 8.246 d = 8.246

d = 5 d = 5

( −3 , 4 ) ( −3 , 4 )

x = 0          y = −2 x = 0          y = −2

x = 0.75 y = 0 x = 0.75 y = 0

x = − 1.667 y = 0 x = − 1.667 y = 0

15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter

6 .0 42 6 .0 42

Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.

2.2 Section Exercises

It means they have the same slope.

The exponent of the x x variable is 1. It is called a first-degree equation.

If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).

x = 2 x = 2

x = 2 7 x = 2 7

x = 6 x = 6

x = 3 x = 3

x = −14 x = −14

x ≠ −4 ; x ≠ −4 ; x = −3 x = −3

x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution

x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2

y = − 4 5 x + 14 5 y = − 4 5 x + 14 5

y = − 3 4 x + 2 y = − 3 4 x + 2

y = 1 2 x + 5 2 y = 1 2 x + 5 2

y = −3 x − 5 y = −3 x − 5

y = 7 y = 7

y = −4 y = −4

8 x + 5 y = 7 8 x + 5 y = 7

Perpendicular

m = − 9 7 m = − 9 7

m = 3 2 m = 3 2

m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular . m 1 = − 1 3 ,   m 2 = 3 ;   Perpendicular .

y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40

y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 ,   y max = 10 y min = −10 ,   y max = 10

y = − A B x + C B y = − A B x + C B

The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 . The slope for  ( −1 , 1 ) to  ( 0 , 4 ) is  3. The slope for  ( −1 , 1 ) to  ( 2 , 0 ) is  − 1 3 . The slope for  ( 2 , 0 ) to  ( 3 , 3 ) is  3. The slope for  ( 0 , 4 ) to  ( 3 , 3 ) is  − 1 3 .

Yes they are perpendicular.

2.3 Section Exercises

Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.

2 , 000 − x 2 , 000 − x

v + 10 v + 10

Ann: 23 ; 23 ; Beth: 46 46

20 + 0.05 m 20 + 0.05 m

90 + 40 P 90 + 40 P

50 , 000 − x 50 , 000 − x

She traveled for 2 h at 20 mi/h, or 40 miles.

$5,000 at 8% and $15,000 at 12%

B = 100 + .05 x B = 100 + .05 x

R = 9 R = 9

r = 4 5 r = 4 5 or 0.8

W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14

f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21

m = − 5 4 m = − 5 4

h = 2 A b 1 + b 2 h = 2 A b 1 + b 2

length = 360 ft; width = 160 ft

A = 88 in . 2 A = 88 in . 2

h = V π r 2 h = V π r 2

r = V π h r = V π h

C = 12 π C = 12 π

2.4 Section Exercises

Add the real parts together and the imaginary parts together.

Possible answer: i i times i i equals -1, which is not imaginary.

−8 + 2 i −8 + 2 i

14 + 7 i 14 + 7 i

− 23 29 + 15 29 i − 23 29 + 15 29 i

8 − i 8 − i

−11 + 4 i −11 + 4 i

2 −5 i 2 −5 i

6 + 15 i 6 + 15 i

−16 + 32 i −16 + 32 i

−4 −7 i −4 −7 i

2 − 2 3 i 2 − 2 3 i

4 − 6 i 4 − 6 i

2 5 + 11 5 i 2 5 + 11 5 i

1 + i 3 1 + i 3

( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1

5 −5 i 5 −5 i

9 2 − 9 2 i 9 2 − 9 2 i

2.5 Section Exercises

It is a second-degree equation (the highest variable exponent is 2).

We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.

One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .

x = 6 , x = 6 , x = 3 x = 3

x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3

x = 5 , x = 5 , x = −5 x = −5

x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2

x = −2 , 3 x = −2 , 3

x = 0 , x = 0 , x = − 3 7 x = − 3 7

x = −6 , x = −6 , x = 6 x = 6

x = 6 , x = 6 , x = −4 x = −4

x = 1 , x = 1 , x = −2 x = −2

x = −2 , x = −2 , x = 11 x = 11

z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2

x = 3 ± 17 4 x = 3 ± 17 4

One rational

Two real; rational

x = − 1 ± 17 2 x = − 1 ± 17 2

x = 5 ± 13 6 x = 5 ± 13 6

x = − 1 ± 17 8 x = − 1 ± 17 8

x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535

x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7

a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a

x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.

maximum at x = 70 x = 70

The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.

2.6 Section Exercises

This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.

He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.

A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.

x = 81 x = 81

x = 17 x = 17

x = 8 ,     x = 27 x = 8 ,     x = 27

x = −2 , 1 , −1 x = −2 , 1 , −1

y = 0 ,     3 2 ,     − 3 2 y = 0 ,     3 2 ,     − 3 2

m = 1 , −1 m = 1 , −1

x = 2 5 , ±3 i x = 2 5 , ±3 i

x = 32 x = 32

t = 44 3 t = 44 3

x = −2 x = −2

x = 4 , −4 3 x = 4 , −4 3

x = − 5 4 , 7 4 x = − 5 4 , 7 4

x = 3 , −2 x = 3 , −2

x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3

x = 2 , −2 x = 2 , −2

x = 1 , 5 x = 1 , 5

x ≥ 0 x ≥ 0

x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8

2.7 Section Exercises

When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.

( − ∞ , ∞ ) ( − ∞ , ∞ )

We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.

( − ∞ , 3 4 ] ( − ∞ , 3 4 ]

[ − 13 2 , ∞ ) [ − 13 2 , ∞ )

( − ∞ , 3 ) ( − ∞ , 3 )

( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]

All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )

( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )

( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )

No solution

( −5 , 11 ) ( −5 , 11 )

[ 6 , 12 ] [ 6 , 12 ]

[ −10 , 12 ] [ −10 , 12 ]

x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 ,   ( − 2 , + ∞ )

x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ ) x < − 3   or   x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ ​ ​ [ 1 , ∞ )

( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )

[ −11 , −3 ] [ −11 , −3 ]

It is never less than zero. No solution.

Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.

( − ∞ , −3 ) ( − ∞ , −3 )

Where the blue line is above the orange line; always. All real numbers.

( − ∞ , − ∞ ) ( − ∞ , − ∞ )

( −1 , 3 ) ( −1 , 3 )

( − ∞ , 4 ) ( − ∞ , 4 )

{ x | x < 6 } { x | x < 6 }

{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }

( −2 , 1 ] ( −2 , 1 ]

( − ∞ , 4 ] ( − ∞ , 4 ]

Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .

Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .

x = 2 , − 4 5 x = 2 , − 4 5

( −7 , 5 ] ( −7 , 5 ]

80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400

[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]

Review Exercises

x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )

y = 5 3 x + 4 y = 5 3 x + 4

72 = 6 2 72 = 6 2

620.097 620.097

midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )

x = 4 x = 4

x = 12 7 x = 12 7

y = 1 6 x + 4 3 y = 1 6 x + 4 3

y = 2 3 x + 6 y = 2 3 x + 6

females 17, males 56

x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4

horizontal component −2 ; −2 ; vertical component −1 −1

7 + 11 i 7 + 11 i

−16 − 30 i −16 − 30 i

−4 − i 10 −4 − i 10

x = 7 − 3 i x = 7 − 3 i

x = −1 , −5 x = −1 , −5

x = 0 , 9 7 x = 0 , 9 7

x = 10 , −2 x = 10 , −2

x = − 1 ± 5 4 x = − 1 ± 5 4

x = 2 5 , − 1 3 x = 2 5 , − 1 3

x = 5 ± 2 7 x = 5 ± 2 7

x = 0 , 256 x = 0 , 256

x = 0 , ± 2 x = 0 , ± 2

x = 11 2 , −17 2 x = 11 2 , −17 2

[ − 10 3 , 2 ] [ − 10 3 , 2 ]

( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )

Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.

( 3.5 , ∞ ) ( 3.5 , ∞ )

Practice Test

y = 3 2 x + 2 y = 3 2 x + 2

( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )

( − ∞ , 9 ] ( − ∞ , 9 ]

x = −15 x = −15

x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1

x = 3 ± 3 2 x = 3 ± 3 2

( −4 , 1 ) ( −4 , 1 )

y = −5 9 x − 2 9 y = −5 9 x − 2 9

y = 5 2 x − 4 y = 5 2 x − 4

5 13 − 14 13 i 5 13 − 14 13 i

x = 2 , − 4 3 x = 2 , − 4 3

x = 1 2 ± 2 2 x = 1 2 ± 2 2

x = 1 2 , 2 , −2 x = 1 2 , 2 , −2

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Access for free at https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra/pages/chapter-2

© 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.

Alg2.4 Exponential Functions and Equations

In this unit, students build on their understanding of exponential functions from an earlier course. Previously, they saw functions whose domain is the integers. Here, they write, interpret, and evaluate exponential functions whose domain is the real numbers. In the second half of the unit, students learn about logarithms in base 2 and 10 as a way to express the exponent that makes an exponential equation true. They then use logarithms to solve exponential equations and to answer questions about exponential functions. During this time, students encounter the constant \(e\) and learn that it is used to model situations with continuous growth rates, leading to working with the natural logarithm. The unit ends with an exposure to logarithmic functions.

Growing and Shrinking

• 1 Growing and Shrinking
• 2 Representations of Growth and Decay

Understanding Non-Integer Inputs

• 3 Understanding Rational Inputs
• 4 Representing Functions at Rational Inputs
• 5 Changes Over Rational Intervals
• 6 Writing Equations for Exponential Functions
• 7 Interpreting and Using Exponential Functions

Missing Exponents

• 8 Unknown Exponents
• 9 What is a Logarithm?
• 10 Interpreting and Writing Logarithmic Equations
• 11 Evaluating Logarithmic Expressions

The Constant $e$

• 12 The Number $e$
• 13 Exponential Functions with Base $e$
• 14 Solving Exponential Equations

Logarithmic Functions and Graphs

• 15 Using Graphs and Logarithms to Solve Problems (Part 1)
• 16 Using Graphs and Logarithms to Solve Problems (Part 2)
• 17 Logarithmic Functions
• 18 Applications of Logarithmic Functions

• Precalculus
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About This Course

Welcome to the Math Medic Algebra 2 course! Here you will find a ready-to-be-taught lesson for every day of the school year, along with expert tips and questioning techniques to help the lesson be successful. Each lesson is designed to be taught in an Experience First, Formalize Later (EFFL) approach, in which students work in small groups on an engaging activity before the teacher formalizes the learning.

Our Algebra 2 course focuses on several function families and develops the graphical, analytical, verbal, and numerical analysis skills needed to flexibly work with these functions, with an emphasis on quadratics. The concept of function transformations is a big theme and woven throughout the course. This course was created using the Common Core State Standards as a guide. The standards taught in each Math Medic Algebra 2 lesson can be found here . The unit overviews and learning targets for the Math Medic Algebra 2 course can be found here .

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