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Go Math Grade 8 Texas 1st Edition Solutions Chapter 2 Scientific Notation Exercise 2.1 Essential Questions

 

Page 33  Exercise 1  Problem 1

To Find – Explanation how we can use scientific notation to express very large quantities.

Scientific notation is a form of presenting very large numbers or very small numbers in a simpler form.

Scientific notation form is:  a × 10± b

We can assume a very large quantities to explain scientific number

Lets assuming number is ⇒ 9000000

We can adjust all zeros in power of 10 , and main number written by multiply of 10± b

So, in scientific notation – 9000000 = 9 × 10⁶

So, we can use scientific notation is the form of  a × 10^{± b} to express the very large quantities.

 

Page 33  Exercise 2  Problem 2

Given – A standard number 41,200.

We need to find the number of places to the left which we can move the decimal point to write scientific notation.

Convert the given number into the scientific notation and then the power of ten will give a number of places to the left to move the decimal point to write scientific notation.

Given value is ⇒  41,200

We can change into scientific notation

⇒  4.1 × 10⁴

The power of 10 is represents how many places to the left we moved the decimal point to write scientific notation.

So, we moved the decimal 4 places to the left to write scientific notation.

We moved the decimal 14 places to the left did to write  41,200  in scientific notation.

 

Page 33  Exercise 3  Problem 3

Given: A standard number 41,200

To Find – Exponent on 10 when we write 41,200 in scientific notation.

Convert the given number into the scientific notation and then the power of ten will give the exponent.

Given value is ⇒  41,200

We can change into scientific notation

⇒     4.1 × 10⁴

So, the exponent on 10when we write  41,200  in scientific notation is 4

 

Page 34  Exercise 4  Problem 4

Given: A standard number 6,400

To Find – Change into scientific number.

Convert the given standard number into simplest possible.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given value is ⇒  6,400

W e can change into scientific notation in the form of a  × 10± b

Now, move the decimal point to the left.

So, comparing it with standard forma is greater than one and less than ten.

So, standard notation is

⇒  6.4 × 10³

In scientific notation  6.4 × 10³

 

Page 34  Exercise 5  Problem 5

Given: A standard number 570,000,000,000.

To Find – Change into scientific number Convert the given standard number into simplest possible.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given value is ⇒  570,000,000,000

We can change into scientific notation in the form of  a × 10± b

Now, move the decimal point to the left.

So, comparing it with standard form such that a is greater than one and less than ten.

So, standard notation is ⇒  5.7 × 10¹¹

In scientific notation of   570,000,000,000  is 5.7 × 10¹¹

 

Page 34  Exercise 6  Problem 6

Given:  A standard number 9,461,000,000,000

To Find – Change into scientific number Convert the given standard number into simplest possible.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given value is ⇒  9,461,000,000,000

We can change into scientific notation in the form o f  a × 10 ^{± b}

w, move the decimal point to the left.

So, comparing it with standard form such that a is greater than one and less than ten.

So, standard notation is  ⇒  9.461 ×10¹²

In scientific notation  9.461 × 10¹²

 

Page 34  Exercise 7   Problem 7

Given: A scientific number 3.5 × 10⁶

To explain why the exponent in 3.5 × 10⁶ is 6 , while there are only 5 zeros in 3,500,000.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given scientific number ⇒  3.5 × 10⁶

Changing into standard form

⇒  3.5 × 10⁶

⇒  3.5 × 1,000,000

⇒  3,500,000

So, the this way there are only 5 zeros in 3,500,000

3.5 × 10⁶  means that decimal should be moved 6 decimals.

Moving one decimal gives 35 and the remaining five zeros are represented by placeholder zeros.

Moving one decimal gives 35 and the remaining five zeros are represented by placeholder zeros.

 

Page 35  Exercise 8  Problem 8

Given:  A scientific number 5.3

To Find – Change into scientific number Convert the given standard number into its standard form

Given scientific notation is ⇒  5.3

On the other way to write this

⇒  5.3

⇒  5.3 × 10⁰

The exponent on 10 when we write 5.3 in scientific notation is 0

The exponent on 10 when we write 5.3 in scientific notation is 0

 

Page 35  Exercise 9  Problem 9

Given: A standard number.

To Find – Change into scientific number Move the decimal to right by inspecting the exponent of ten.

Given scientific notation is ⇒  7.034 ×  10⁹

Move the decimal to right by inspecting the exponent of ten.

So, move the decimal nine place right.

Change into standard number

⇒  7.034 × 10⁹

⇒  7,034,000,000

The standard number is  7,034,000,000

 

Page 35  Exercise 10  Problem 10

Given: A scientific number 2.36 × 10⁵

To Find – Change into standard number.

The definition of the standard form of a number is representing the very large expanded number in a small number.

Given scientific notation is  ⇒    2.36 × 10⁵

Change into standard number

⇒  2.36 × 105

⇒  2.36 × 100000

⇒  236,000

The standard number is  236,000

 

Page 35 Exercise 11 Problem 11

Given: A scientific number5×106

To Find – Change into standard number.

The definition of the standard form of a number is representing the very large expanded number in a small number.

Given scientific notation is  ⇒   5 × 10⁶

Changing into standard number

⇒   5 × 10⁶

⇒   5 × 1,000,000

⇒  5,000,000

The mass of one roosting colony of Monarch butterflies in Mexico in standard notation is  5,000,000  gram

 

Page 36  Exercise 12  Problem 12

Given:  A standard number 1,304,000,000.

To Find – Change into scientific number.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given standard number is   ⇒   1,304,000,000

Move the decimal point to the left until we left with number greater than one and less than ten.

Changing into scientific notation is

⇒  1,304,000,000

⇒  1.304 × 10⁹

The scientific notation is  1.304 × 10⁹

 

Page 36  Exercise 13  Problem 13

Given:  A standard number 6,730,000.

To Find – Change into scientific number.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given standard number is  ⇒   6,730,000

Move the decimal point to the left until we left with number greater than one and less than ten.

Changing into standard number is

⇒  6,730,000

⇒  6.730 × 10⁶

The scientific notation is  6.730 × 10⁶

 

Page 36  Exercise 14  Problem 14

Given: A standard number 13,300.

To Find – Change into scientific number.

Move the decimal point to the left until we left with number greater than one and less than ten

Given standard number is  ⇒   13,300

Move the decimal point to the left until we left with number greater than one and less than ten

Changing into scientific number is

⇒  13,300

⇒  1.33 × 10⁴

The scientific number is 1.33 × 10⁴

 

Page 36 Exercise 15 Problem 15

Given: An ordinary quarter contains about 97,700,000,000,000,000,000,000 atoms.

To Write number in scientific notation.

Simplify the given number in scientific notation.

Given number is 97,700,000,000,000,000,000,000 97,700,000,000,000,000,000,000

Move decimal 22 places to left side 97,700,000,000,000,000,000,000 = 9.77 × 10²²

Scientific notation of 97,700,000,000,000,000,000,000 is  9.77 × 10²² atoms.

 

Page 36 Exercise 16 Problem 16

Given: The distance from Earth to the Moon is about 384,000 kilometers.

To Write number in scientific notation.

Simplify the given number in scientific notation.

Given number is  384,000

⇒  384,000

Move decimal 5 places to left side – 384,000 = 3.84 × 105

Scientific notation of  384,000  is  3.84 × 105  kilometers.

 

Page 36  Exercise 17  Problem 17

Given: Number is 4 × 10⁵

To Write number in standard notation.

Simplify the given number in standard notation.

Given number is  4 × 10⁵

⇒  4 × 10⁵

Move decimal 5 places to right side 4 × 10⁵

4 × 10⁵ = 400,000

Standard notation of 4 × 10⁵  is  400,000

 

Page 36  Exercise 18  Problem 18

Given: Number is 1.8499 × 10⁹

To Write number in standard notation.

Simplify the given number in standard notation.

Given number is 1.8499 × 10⁹

⇒ 1.8499 × 10⁹

Move decimal 9 places to right side  1.8499 × 10⁹

1.8499 × 10⁹ = 1,849,900,000

Standard notation of 1.8499 × 10⁹  is 1,849,900,000

 

Page 36  Exercise 19  Problem 19

Given:  Number is 6.41 × 10³

To Write number in standard notation.

Simplify the given number in standard notation.

Given number is  6.41 × 10³

⇒ 6.41 × 10³

Move decimal 3 places to right side  6.41 × 10³

6.41 × 10³ = 6,410

Standard notation of 6.41 × 10³  is 6,410

 

Page 36  Exercise 20  Problem 20

Given: Number is 8.456 × 10⁷

To Write number in standard notation.

Simplify the given number in standard notation.

Given number is 8.456 × 10⁷

⇒  8.456 × 10⁷

Move decimal 7 places to right side  8.456 × 10⁷

8.456 × 10⁷= 84,560,000

Standard notation of  8.456 × 10⁷   is 84,560,000

 

Page 36  Exercise 21  Problem 21

Given:  Number is  9 × 10¹⁰

To Write number in standard notation.

Simplify the given number in standard notation.

Given number is 9 × 10¹⁰

⇒  9 × 10¹⁰

Move decimal 10 places to right side  9 × 10¹⁰

9 × 10¹⁰ = 90,000,000,000

Standard notation of 9 × 10¹⁰  is 90,000,000,000

 

Page 36  Exercise 22  Problem 22

Given:  7.6 × 10⁶

To Find – Write this time in standard notation.

Simplify the given number in standard notation.

Move the decimal to the right in accordance with the exponent of ten.

Given number is 7.6 × 10⁶

⇒  7.6 × 10⁶

∴ 10⁶ = 1000000

∴ 7.6 = \(\frac{76}{10}\)

7.6 × 10⁶ =  \(\frac{76}{10}\) × 1000000

7.6 × 10⁶ = 7600000 cans

Standard notation of 7.6 × 10⁶  is 7600000 cans

 

Page 36 Exercise 23  Problem 23

Given:  3,482,000,000.

To Find- Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is 3,482,000,000

Move the decimal point 9 places to the left   ⇒   3.482000000

Remove extra zeroes   ⇒  3.482

Divide the original number by 3.482 ⇒ 1000000000 = 10⁹

Multiply numbers 3.482 and 10⁹  ⇒  3.482 × 10⁹

Standard notation of  3,482,000,000 is 3.482 × 10⁹

 

Page 36  Exercise 24  Problem 24

Given: The weight of Apatosaurus is 66,000 pounds.

To Find –  Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is  ⇒   66,000

Move the decimal point 4 places to the left    ⇒   6.6000

Remove extra zeroes    ⇒   6.6

Divide the original number by 6.6  ⇒   \(\frac{66,000}{6.6}\) = 10⁴

Multiply numbers 6.6 and 10⁴

⇒  6.6 × 10⁴

Standard notation of 66,000 is 6.6 × 10⁴

 

Page 37  Exercise 25  Problem 25

Given: The weight of Argentinosaurus 220,000.

To Find – Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is   ⇒  220,000

Move the decimal point 5 places to the left  ⇒  2.20000

Remove extra zeroes   ⇒   2.2

Divide the original number by ⇒ \(\frac{220000}{2.2}\) = 10⁵

Multiply numbers 2.2 and  10⁵

⇒ 2.2 × 10⁵

The estimated weight of each dinosaur in scientific notation is  2.2 ×  10⁵

 

Page 37  Exercise 26  Problem 26

Given: The weight of Brachiosaurus100.000.

To Find –  Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is 100,000

∴ 100000 =  10⁵

100000 = 1 × 10⁵

Standard notation of  100000  is 1 × 10⁵

 

Page 37  Exercise 27  Problem 27

Given: The weight of Camarasaurus 40,000 pound.

To Find – Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is  ⇒  40,000

10⁴ = 10000

40000 = 4 × 10⁴

The estimated weight of each dinosaur in scientific notation is  4 × 10⁴.

 

Page 37  Exercise 28  Problem 28

Given: The weight of Cetiosauriscus 19,850 pound.

To Find –  Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Given number is ⇒ 19,850

Move the decimal point 4 places to the left  ⇒ 1.9850

Divide the original number by  ⇒ \(\frac{19850}{1.9850}\)= 10⁴

Multiply numbers 1.9850 and 10⁴

19850 = 1.9850 × 10⁴

The estimated weight of each dinosaur in scientific notation is 1.9850 × 10⁴.

 

Page 37  Exercise 29  Problem 29

Given: The weight of Diplodocus 50,000 pound.

To Find –  Write this in standard notation.

Simplify the given number in standard notation.

Move the decimal point to the left until we left with number greater than one and less than ten

Given number is 50,000

10000 = 10⁴

50000 = 5 × 10⁴

The estimated weight of each dinosaur in scientific notation is 5 × 10⁴

 

Page 37  Exercise 30  Problem 30

Given: A single little brown bat can eat up to 1000 mosquitoes in a single hour.

To Find – Express in scientific notation how many mosquitoes a little brown bat might eat in 10.5 hours.

Write the given number and express in scientific notation.

Move the decimal point to the left until we left with number greater than one and less than ten.

Since a little brown bat can eat up to 1000 mosquitoes in an hour it can eat 10.5 times more in 10.5 hours

10.5 × 1000 = 10500

⇒ 10500

1.0500 × 10⁴

10500 = 1.0500 × 10⁴

Standard notation of 10500 is 1.0500 × 10⁴

 

Page 37  Exercise 31  Problem 31

Given: Samuel can type nearly 40 words per minute.

To Find – Find the number of hours it would take him to type 2.6 × 10⁵  words.

To find number of hours, divide the total number of words by typing speed.

To find the number of hours N, we need to divide the total number of words by typing speed (words per minute).

We have:  N = \(\frac{2.6 \times 10^5}{40}\)

N = \(\frac{26 \times 10^5}{4 \times 10^2}\)

N = 6.5 × 10 5−2

N = 6.5 × 10³

To convert from minutes to hours, we divide the result by 60

N = \(\frac{6.5 \times 10^3}{60}\)

N = \(\frac{65 \times 10^3}{6 \times 10^2}\)

N = \(\frac{65 \times 10^{3-2}}{6}\)

N = 10.83 × 10¹

N= 1.083 × 10²

The number of hours that he would take to type 2.6 × 10⁵ word is N = 1.083 × 10².

 

Page 37 Exercise 32 Problem 32

Given: It can lift up to 1.182 × 10 ³ times its own weight.

To Find –  If you were as strong as this insect, explain how you could find how many pounds you could lift.

Solution: Number of pounds you can lift by multiplying 1.182 × 10 ³ by your weight.

Since you are as strong as the ant which can lift up to 1.182 × 10 ³ its own weight.

Since you are as strong as the ant which can lift up to 1.182 × 10 ³ its own weight.

 

Page 37  Exercise 32  Problem 33

Given: It can lift up to 1.182 × 10³ times its own weight.

We need to find how much you could lift, in pounds and Express your answer in both scientific notation and standard notation.

Write the given number and solve it.

Given number is 1.182 × 10³

​Let weight = 100 pounds

Number of pounds =100 × 1.182 × 10³

= 1.182 × 105

The scientific notation is 1.182 × 105

Now =1.182 × 10 ⁵

1.182 × 10 ⁵= 1182 × 10⁵⁻³

1.182 × 10 ⁵= 1182 × 10²

1.182 × 10 ⁵= 118200

The standard notation is 118200

The scientific notation is 1.182×105 of weight that he could lift in pounds. The standard notation is 118200 of weight that he could lift in pounds.

 

Page 37 Exercise 33 Problem 34

To Find: Which measurement would be least likely to be written in scientific notation: number of stars in a galaxy, number of grains of sand on a beach, speed of a car, or population of a country?

Explain your reasoning.

Solution: Scientific notation is used to express measurements that are extremely large or extremely small.

Number of stars in a galaxy and number of grains of sand on a beach are extremely large, so we use scientific notation for those.

Comparing speed of a car and population of a country, it is clear that the speed of a car is a smaller number.

Therefore, the speed of a car is less likely to be written in scientific notation.

The speed of a car is less likely to be written in scientific notation.

Scientific notation is used to express measurements that are extremely large or extremely small.

Number of stars in a galaxy and number of grains of sand on a beach are extremely large, so we use scientific notation for those.

Comparing speed of a car and population of a country, it is clear that the speed of a car is a smaller number.

Therefore, the speed of a car is less likely to be written in scientific notation.

The speed of a car is less likely to be written in scientific notation.

The speed of a car is less likely to be written in scientific notation

 

Page 37  Exercise 34  Problem 35

Given: 4.5 × 10⁶  and 2.1 × 10⁸

We need to compare the two numbers and determine which is greater.

Convert both into standard form and then compare.

Given numbers 4.5 × 10⁶ and 2.1 × 10⁸

4.5 × 10⁶ = 4500000

2.1 × 10⁸ = 210000000

Now, comparing both, we conclude that 4500000 < 210000000

So, 4.5 × 10⁶ < 2.1 × 10⁸

Comparing the exponents we have  4.5 × 10⁶ < 2.1 × 10⁸

 

Page 37  Exercise 35  Problem 36

We have to do tests to determine whether the number is written in scientific notation or not.

Solution is:  Scientific notation is in the form of a×10n where a is a first factor and 10n is the second factor.

For a number to be written in scientific notation, it’s base a should lie between 1 and 10.

If it is a power of 10 it can be a second factor in a scientific notation.

And the multiplication of first factor and second factor should be equal to standard number given.

First factor we can apply the test : if it decimal number greater than or equal to 1 but less than 10 it can be a first factor in a scientific notation. Second factor we can apply the test : If it is a power of 10 it can be a second factor in a scientific notation.

Go Math Grade 8 Texas 1st Edition Solutions Chapter 2 Scientific Notation Exercise

 

Page 29  Exercise 1  Problem 1

We can solve real-world problems by use of scientific notation with help of scientific notation rules.

 

Page 30  Exercise 2  Problem 2

Given: Exponential expression 10².

We need to write the above exponential expression as a decimal.

Solution is -10² represent in decimal as 10 × 10 = 100.0

Exponential expression 10² as a decimal is = 100.0

 

Page 30  Exercise 3  Problem 3

Given: Exponential expression 10⁷.

We need to write the above exponential expression as a decimal.

Solution is -10⁷ represent in decimal as =10 × 10 × 10 × 10 × 10 × 10 × 10 = 10000000.0.

Exponential expression 10⁷ as a decimal =10000000.0

 

Page 30  Exercise 4  Problem 4

Given: 45.3 ×10³

To find – Product of the given expression.

Multiply the expression and shift the decimal to the right according to the exponent.

Given –  45.3 × 10³.

Product of:

45.3 × 10³ = (453 × 10⁻¹)×(1 × 10³)

45.3 × 10³ = (453 × 1)×(10⁻¹ × 10³)

45.3 × 10³ =  453 × (10 ^{− 1 + 3})

45.3 × 10³ =  453 × 10².

Product of 45.3×10³  is = 453 × 10².

 

Page 30  Exercise 5  Problem 5

Given: 7.08 ÷10²

To find – Quotient of the expression.

Move the decimal to left in accordance with the exponent of ten.

Given- 7.08 ÷ 10².

Quotient of 7.08 ÷  10² = \(\frac{7.08}{10^2}\)

7.08 ÷10² = \(\frac{708 \times 10^{-2}}{1 \times 10^2}\)

7.08 ÷ 10² = \(=\left(\frac{708}{1}\right) \times\left(\frac{10^{-2}}{10^2}\right)\)

7.08 ÷ 10²  = 708 × (10⁻²⁻² )

7.08 ÷ 10⁻⁴

= 708 × 10⁻⁴

Quotient of 7.08 ÷ 10²  is = 708×10⁻⁴.

 

Page 30  Exercise 6 Problem 6

Given: 0.00235 × 10⁶

To find –  Quotient of the expression.

Move the decimal to left in accordance with the exponent of ten

Given- 0.00235 × 10⁶.

Product of 0.00235 × 10⁶ = (235 ×10 ^{− 5} ) × (1 × 10⁶ )

0.00235 × 10⁶ = (235 × 1) × (10 ^{− 5} × 10⁶ )

0.00235 × 10⁶ = 235 × (10 ^{− 5 + 6} )

0.00235 × 10⁶ = 235 × 10¹.

Product of 0.00235  ×  106 is = 235 × 10¹.

 

Page 30 Exercise 7 Problem 7

Given: 0.5 × 10².

To find –  product or quotient of above expression . 0.5 convert in exponential form as 5 × 10^-1 and solve it.

Product of 0.5 × 10²

0.5 × 10² =  (5 × 10⁻¹) × (1 × 10² )

0.5 × 10² =  (5 × 1)×(10⁻¹ × 10² )

0.5 × 10² =  5 × (10 ⁻¹⁺² )

0.5 × 10² =  5 × 10¹.

Product of  0.5 × 10²  is = 5 × 10¹.

 

Page 30  Exercise 8  Problem 8

Given: 67.7 ÷ 10⁵.

To find – Product or quotient of above expression .

67.7convert in exponential form as 677 × 10⁻¹ and solve it.

Quotient of 67.7 ÷ 10⁵ = \(\frac{677 \times 10^{-1}}{10^5}\)

67.7 ÷ 10⁵ =  \(677 \times\left(\frac{10^{-1}}{10^5}\right)\)

67.7 ÷ 10⁵ =  677 × (10 ⁻¹⁻⁵ )

67.7 ÷ 10⁵ =  677 × (10 ⁻⁶).

Quotient of 67.7 ÷ 105 is = 677 × 10 ⁻⁶.

 

Page 30  Exercise 9  Problem 9

Given: 0.0057 × 10⁴.

To find – Product or quotient of above expression .

0.0057 convert in exponential form as 57 × 10⁻⁴ and solve it.

Product of  0.0057 × 10⁴ 

0.0057 × 10⁴  = (57 × 10⁻⁴ ) × (1 × 10⁴ )

0.0057 × 10⁴  =   (57 × 1) × (10⁻⁴ × 10⁴ )

0.0057 × 10⁴  =  57 × ( 10⁻⁴⁺⁴ )

0.0057 × 10⁴  =  57.0

Product of 0.0057 × 10⁴ is = 57.0

 

Page 30 Exercise 10 Problem 10

Given: 195 ÷10⁶.

To find-product or quotient of above expression .

195 convert in exponential form as 195 × 10⁰  and solve it.

Quotient of 195 ÷10⁶ = \(\frac{195 \times 10^0}{10^6}\)

195 ÷10⁶= \(\left(\frac{10^0}{10^6}\right)\)

195 ÷10⁶=  195 × 10⁰−6

195 ÷10⁶= 195 × 10−6

Quotient of 195 ÷106 is = 195 × 10−6.

 

Page 31  Exercise 11  Problem 11

Given:

To find –  Complete the Venn diagram .

Given expression 10 compare with exponential expression b^a and solve it.

10² represent the exponential expression where 10 is base and 2 is exponent.

So in box 1 : _____________ 10 is base .

In box 2 :______________  2 is exponent.

Venn diagram 

 

Page 31  Exercise 12  Problem 12

Given:

To find – Complete the sentences

Solution is – A number produced by raising a base to an exponent is a power.

Complete sentence is – A number produced by raising a base to an exponent is a power .

 

Page 31  Exercise 13  Problem 13

Given:

To find – Complete the sentences.

Solution is – Scientific notation is a method of writing very large or very small numbers by using powers of 10 .

Complete sentence is – Scientific notation is a method of writing very large or very small numbers by using powers of 10 .

 

Page 31  Exercise 14  Problem 14

Given: A  __________  is any number that can be expressed as a ratio of two integers.

To Complete the sentence.

Scientific notation is a form of presenting very large numbers or very small numbers in a simpler form.

A scientific notation is any number that can be expressed as a ratio of two integers.

A scientific notation is any number that can be expressed as a ratio of two integers.

Go Math Grade 8 Texas 1st Edition Solutions Chapter 1 Real Numbers

 

Page 7   Exercise 1   Problem 1

We divide the numerator by the denominator to convert a rational number to a decimal.

We simply convert a rational number to a decimal by converting it to the form of a fraction.

The numerator is then divided by the denominator, yielding the division’s exact value.

Because a/b is a non-terminating, non-repeating decimal, it cannot be used to represent irrational values.

In order to approximate the value of irrational numbers, students should know the perfect squares (1 to 15). , as well as square roots of numbers less than 225, are examples of irrational numbers.

In order to change a rational number to a decimal, we divide the numerator with the denominator or can be converted to a decimal by the division method. 

 

Page 8 Exercise 2 Problem 2

Given: \(\frac{1}{8}\).

To convert fractions into decimals.

Method – We use the division method to convert fractions into decimals that means dividing the numerator by denominator.

It is given \(\frac{1}{8}\).

We have to convert fractions into a decimal.

Divide 1 by 8.

We will get

=  \(\frac{1}{8}\)

\(\frac{1}{8}\) =  0.125.

The fraction \(\frac{1}{8}\) will be 0.125 in decimal.

 

Page 8  Exercise 3  Problem 3

Given: 2\(\frac{1}{3}\)

To convert fractions into decimals.

Method- Convert mixed fraction into an improper fraction.

It is given,2\(\frac{1}{3}\)

We have to convert fractions into a decimal.

First, we convert the mixed number to an improper fraction:

2\(\frac{1}{3}\) = 2 + \(\frac{1}{3}\)

= \(\frac{6}{3}\)+\(\frac{1}{3}\)

2\(\frac{1}{3}\) = \(\frac{7}{3}\)

To write 7/3 as a decimal, we divide the numerator by the denominator until the remainder is zero or until the digits in the quotient begin to repeat.

We add as many zeros after the decimal point in the dividend as needed.
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When a decimal has one or more digits that repeat indefinitely, we write the decimal with a bar over the repeating digit(s). In our case, 3 repeats indefinitely.

The decimal form of the given fraction is 2\(\frac{1}{3}\) = \(2 . \overline{3}\) or 2.3333 

 

Page 8  Exercise 4  Problem 4

A positive number has two square roots because a positive number multiplied by itself is positive and a negative number multiplied by itself is also positive.

The principal square root is the non negative number that when multiplied by itself equals a.

The square root obtained using a calculator is the principal square root. The principal square root of a is written as √​a​​​.

The answer to the equation x² = b is the square root of a number b.

It’s a number that equals b when multiplied by itself.

Every positive number b has two square roots, which are indicated by the letters √b and −√b.

The positive square root of b denoted b, is the major square root.

 

Page 8  Exercise 5  Problem 5

As, the number √2 be irrational because it is not an integer (2 is not a perfect square).

Any square root of any natural number that is not the square of a natural number is irrational.

Squares are integers obtained by multiplying one number by itself.

When you multiply a whole number by itself, the outcome is always another whole number, which is known as a perfect square.

As a result, perfect squares’ square roots are always whole numbers.

 

Page 9  Exercise 6  Problem 6

Given number: 64

To find out the two number roots of the 64

Method −  For finding the square root prime factorization method.

It is given that,64

We have to find the two square roots of each number.

The positive square root and the negative square root are the two square roots of any positive number.

Therefore, 8 × 8 = 64

The positive square root of 64 is 8

While the negative square root is−8

⇒ (−8) × (−8) = 64

The two square roots of 64 is 8 and −8.

 

Page 9  Exercise 7  Problem 7

Given number: 100

To find out the two number roots of the 100

Method −  For finding the square root prime factorization method.

It is given that, 100

We have to find the two square roots of each number.

The positive square root and the negative square root are the two square roots of any positive number.

Therefore, 10 × 10 = 100

The positive square root of 100 is 10

While the negative square root is −10

⇒ (−10) × (−10) = 100

The two square roots of 100 is 10 and −10.

 

Page 9  Exercise 8  Problem 8

Given: \(\frac{1}{9}\)

To find the two square root of \(\frac{1}{9}\).

Method – There must be two square roots of a positive real number, one is positive and another is the negative square root.

It is given,\(\frac{1}{9}\).

We have to find two square roots of a number.

For finding two square roots, we will take square root on both sides.

We will get

x² = \(\frac{1}{9}\)

⇒  x = \(\sqrt{\frac{1}{9}}\)

⇒  x = ± \(\frac{1}{3}\)

= + \(\frac{1}{3}\), –\(\frac{1}{3}\).

The two square roots of \(\sqrt{\frac{1}{9}}\) will be + \(\frac{1}{3}\), –\(\frac{1}{3}\).

 

Page 9  Exercise 9  Problem 9

Given: Area of a square garden is 144 ft².

To find the length of each side.

Method – By using the area formula of the square.
img

It is given, area of a square garden is 144 ft².

To find the length of each side of garden.

we will use the area of the square.

We will get

⇒  a² = 144

⇒  a = \(\sqrt{144}\)

⇒  a = 12ft.

Each side of the square garden will be 12ft.

 

Page 10  Exercise 10  Problem 10

Given: \(\sqrt{2}\)

To find an estimation of   \(\sqrt{2}\)

Method – Square root method

The square root of 2 or root 2 is written as√2 with a value of 1.414.

It is represented by the square root symbol.

The square root of 2 is the number which when multiplied with itself gives the result as 2. It is generally represented as √2 or 2​\(\frac{1}{2}\).

The numerical value of square root 2 up to 50 decimal places is as follows:  

\(\sqrt{2}\) =  1.41421356237309504880168872420969807856967187537694…

We can choose numbers with two decimal points instead of one and see in between which lies the number  \(\sqrt{2}\)

 

Page 11 Exercise 11 Problem 11

An irrational number cannot be expressed as a ratio between two numbers and it cannot be written as a simple fraction because there is not a finite number of numbers when written as a decimal.

Instead, the numbers in the decimal would go on forever, without repeating.

The decimal numeral system is the standard system for denoting integer and non-integer numbers.

Since π is irrational, it means that its decimal representation goes on forever. It cannot be expressed as the ratio of two integers.

 

Page 11 Exercise 12 Problem 12

Given: The figure is img

Plot π on the number line.

Method – The number line method

The value of pi in decimal notation is about 3.14.

However, pi is an irrational number, which means that its decimal form does not terminate (such as \(\frac{1}{4}\) = 0.25) or become repetitious (such as \(\frac{1}{6}\) = 0.166666…).

As we have to plot the π on the given number line.

So, the value of π = 3.14
img

Hence, the answer is
img

Go Math Grade 8 Texas 1st Edition Solutions Chapter 1 Real Numbers Exercise

 

Page 2  Exercise 1  Problem 1

The puzzle is given as

To find a puzzle solution.

By using number concepts, preview key vocabulary.

The given puzzle is

NOLRATAI

RUNMEB

The given statement is

Any number that can be written as a ratio of two integers.

The above statement give hint as the number related to rational value.

So, the solution is “Rational Number”.

The key vocabulary from this unit is “RATIONAL NUMBER”.

 

Page 2  Exercise 2  Problem 2

The puzzle is given as

To find a puzzle solution.

Method: To preview key vocabulary by using number concepts.

The given puzzle is

PERTIANEG

MALCEDI

The given statement is

A decimal in which one or more digits repeat infinitely.

The infinitely repeat of number is known as repeating.

The number can be integer and non-integer so the second one is decimal.

So, the answer we get is “Repeating Decimal”.

The key vocabulary from this unit is “REPEATING DECIMAL”.

 

Page 2  Exercise 3  Problem 3

The given statement is the set of rational and the set of irrational numbers.

To find a puzzle solution.

By using number concepts, preview key vocabulary.

The given puzzle is

LAER

SEBMNUR

To find a puzzle solution.

By using number concepts, preview key vocabulary

From the definition of real numbers

Real numbers are numbers that include both rational and irrational numbers.

So, the definition of a real number matches the given statement.

Also, rearrange the alphabets in the puzzle.

LAER which is unscrambled to REAL.

The next word is SEBMNUR which is unscrambled to NUMBERS.

So, the answer is “Real Numbers”

The key vocabulary from this unit is “REAL NUMBERS”.

 

Page 2  Exercise 4  Problem 4

Given: A method of writing very large or very small numbers by using powers of 10.

To find a puzzle solution.

By using number concepts, preview key vocabulary.

The given puzzle is

NIISICFTCE

OITANTON

As we have statement

A method of writing very large or very small numbers by using powers of 10 .

The statement talk about some notation concept as we know.

Scientific notation is a way to express numbers in a form that makes numbers that are too small or too large more convenient to write.

So, the answer is “Scientific Notation”

The key vocabulary from this unit is “SCIENTIFIC NOTATION”.

 

Page 3  Exercise 5  Problem 5

Use real numbers to solve real-world problems.

We have to finding real world problems.

The set of real numbers is the combination of rational and irrational numbers.

The entire number line from ±∞ represents the set of all real numbers.

Real numbers are used in a multitude of real-world scenarios.

For example, they are used to describe distances, weights, area, volume, and price.

Real numbers are used in a multitude of real-world scenarios. They are used to describe distances, weights, area, volume, and price.

 

Page 4  Exercise 6  Problem 6

Given number:  7

To find out the square of the 7

Method –  For finding square multiplying the number by itself.

It is given that,7

We have to find the square of 7

To get a square of a given number multiplies the number by itself, we get

So, Multiplying 7 by itself, we will get

Square of the 7 ​= 7 × 7

= 49

​The square of a given number 7 will be 49.

 

Page 4  Exercise 7  Problem 7

Given: 21.

To find the square of 21.

Method – For finding square multiplying the number by itself.

It is given, 21

We have to find the square of 21

To get a square of a given number multiplies the number by itself.

By itself, we will get

So, Multiplying 21 by itself, we will get

Square of 21 = 21 × 21

= 441
​
The square of a given number 21 will be 441.

 

Page 4  Exercise 8  Problem 8

Given: −3.

To find the square of −3

Method – For finding the square of the given number multiply the number by itself.

It is given,−3.

We have to find the square of −3.

To get a square of a given number multiplies the number by itself.

So, Multiplying −3 by itself, we will get

Square of −3 ​= −3 × −3

= 9
​
The square of a given number −3 will be 9.

 

Page 4  Exercise 9  Problem 9

Given: 2.7.

To find the square of 2.7.

Method – For finding the square of a given number multiply the number by itself.

It is given,2.7.

We have to find the square of 2.7.

To get a square of a given number multiplies the number by itself.

So, multiplying by 2.7 itself, we will get.

Square of 2.7. ​= 2.7 × 2.7

= 7.29

The square of given numbers 2.7 will be 7.29.

 

Page 4  Exercise 10  Problem 10

Given: \(\frac{−1}{4}\).

To find the square of \(\frac{−1}{4}\).

Method – for finding the square of a given number multiply the number by itself.

It is given \(\frac{−1}{4}\).

We have to find the square of \(\frac{−1}{4}\).

To get a square of a given number multiplies the number by itself.

So, multiplying by \(\frac{−1}{4}\) itself, we will get.

The square of \(\frac{−1}{4}\) = \(\frac{−1}{4}\) × \(\frac{−1}{4}\)

= \(\frac{1}{16}\).

The square of given numbers\(\frac{−1}{4}\) will be \(\frac{1}{16}\).

 

Page 4  Exercise 11  Problem 11

Given:−5.7.

To find the square of −5.7.

Method – for finding the square of a given number multiply the number by itself.

It is given,−5.7

We have to find the square of−5.7

To get a square of a given number multiplies the number by itself.

So, Multiplying by 5.7 itself, we will get

The square of −5.7= −5.7 × −5.7

= 32.49.

The square of given numbers 5.7 will be 32.49.

 

Page 4  Exercise 12  Problem 12

Given: 1\(\frac{1}{2}\)

To find the square of 1\(\frac{1}{2}\).

Method – For finding the square of a given number multiply the number by itself.

It is given 1\(\frac{1}{2}\)

We have to find the square of 1\(\frac{1}{2}\).

To get a square of a given number multiplies the number by itself.

But first, we have to convert mixed fractions into fractions.

​= 1\(\frac{1}{2}\)

= \(\frac{5×1+2}{5}\)

= \(\frac{7}{5}\)

Further, we have to find the square of \(\frac{7}{5}\).

So, Multiplying by \(\frac{7}{5}\) itself, we will get

The square of 1\(\frac{1}{2}\) = \(\frac{7}{5}\) × \(\frac{7}{5}\)

= \(\frac{49}{25}\).

The square of a given number \(\frac{7}{5}\) will be \(\frac{49}{25}\).

 

Page 4  Exercise 13  Problem 13

Given: 9².

To find the square of 9².

Method – For finding the square of a given number multiply the number by itself.

It is given, 9².

We have to find the square of 9².

To get a square of a given number multiplies the number by itself.

So, multiplying by 9 itself, we will get

Square of 9² = 9 × 9

= 81.

The square of a given numbers 9² will be 81.

 

Page 4  Exercise 14  Problem 14

Given: 2⁴

To simplify exponential expression of 2⁴.

Method – For finding the value of a given number multiply 4 times the number by itself.

It is given, 2⁴.

We have to simplify the exponential expression of 2⁴.

To simplify the exponential expression of a given number multiplies four times itself.

So, multiplying, we will get

2⁴ = 2 × 2 × 2 × 2

= 16

Simplified value of the exponential expression 2⁴ will be 16.

 

Page 4  Exercise 15  Problem 15

Given: (\(\frac{1}{3}\))²

To simplify the exponential expression of (\(\frac{1}{3}\))².

Method – For finding the square of a given number multiply the number by itself.

It is given, (\(\frac{1}{3}\))²

We have to simplify exponential expression of (\(\frac{1}{3}\))².

To get a square of a given number multiplies the number by itself.

So, multiplying by \(\frac{1}{3}\) itself, we will get

(\(\frac{1}{3}\))² = \(\frac{1}{3}\) × \(\frac{1}{3}\)

= \(\frac{1}{9}\)

The simplified exponential expression of (\(\frac{1}{3}\))² will be \(\frac{1}{9}\).

 

Page 4  Exercise 16  Problem 16

Given: (−7)².

To simplify the exponential expression of (−7)².

Method – For finding the square of a given number multiply the number by itself.

It is given, (−7)².

We have to simplify the exponential expression of(−7)².

To get a square of a given number multiplies the number by itself.

So, multiplying by −7 itself, we will get

(−7)². = −7 × −7

= 49.

The simplified exponential expression of (−7)² will be 49.

 

Page 4  Exercise 17  Problem 17

Given: 4³.

To simplify the exponential expression of 4³.

Method -For finding the cube of a number, multiply that number itself, then multiply the product obtained with the original number again.

It is given,4³.

Four multiply by itself three times.

We will get

4³ = 4 × 4 × 4

= 64.

The simplified exponential expression of 4³ will be 64.

 

Page 4  Exercise 18  Problem 18

Given: 10⁵.

To simplify the exponential expression of 10⁵.

Method- To simplify the exponential expression, we will use exponent rules.

It is given,10⁵.

We have to simplify the exponential expression of 10⁵.

To get a simplified expression of a given number multiplies the number five times by itself.

So, multiply 10 to itself at five times.

We will get

10⁵ = 10 × 10 × 10 × 10 × 10

= 100000

The simplified exponential expression of 10⁵ will be 100000.

 

Page 4  Exercise 19  Problem 19

Given: 3\(\frac{1}{3}\).

To convert a mixed fraction into an improper fraction.

Method- To convert a mixed fraction into an improper fraction, we multiply the denominator of the fraction by the whole number and then add the product to the numerator.

Like, for converting the mixed fraction x \(\frac{a}{b}\) we will write the same as \(\frac{(b×x)+a}{b}\).

It is given 3\(\frac{1}{3}\).

We have to convert mixed fractions into improper fractions.

Multiply 3 by 3, the product will be 9.

= 3 × 3

= 9

Then, We will add the product with the numerator We will get

= 9 + 1

= 10

After that Writing 10 over 3.

We will get improper fraction as \(\frac{10}{3}\).

The mixed fraction 3\(\frac{1}{3}\) will be \(\frac{10}{3}\) in improper fraction.

 

Page 4 Exercise 20 Problem 20

Given: 5 \(\frac{5}{6}\).

To convert a mixed fraction into an improper fraction.

Method- To convert a mixed fraction into an improper fraction, we multiply the denominator of the fraction by the whole number and then add the product to the numerator.Like, for converting the mixed fraction x \(\frac{a}{b}\) we will write the same as \(\frac{(b×x)+a}{b}\).

It is given 5 \(\frac{5}{6}\).

We have to convert mixed fractions into improper fractions.

Multiply 6 by 5, the product will be 30.

=  6 × 5

=  30

Then, We will add the product with the numerator We will get

= 30 + 5

= 35.

After that Writing 35 over 6.

We will get an improper fraction as \(\frac{35}{6}\).

The mixed fraction 5 \(\frac{5}{6}\) will be \(\frac{35}{6}\)  in improper fraction.

 

Page 4  Exercise 21  Problem 21

The whole number and positive number is 0,10,200.

The whole number and positive number is 21, 44, 308.

The integer and a negative number are -21,-78, -93.

Hence , the answer is

 

Page 5  Exercise 22  Problem 22

Prime factors that, themselves, have only factors of 1 and themselves. Non-prime numbers, like 10, have prime factors.

Note that 1 is a factor of all integers.

If the two factors are equal (the same), like 4 × 4 = 16 , then each of them is called a “square root.“

One of the two equal factors of a number is a square root.

Hence, One of the two equal factors of a number is a square root.

 

Page 5  Exercise 23  Problem 23

The square root of a non-negative number, is a non-negative number that when multiplied by itself results in the original number.

The square root of a negative number does not exist in the real numbers.

Example: Since 25 is a non-negative number, there is a non-negative number 5, such that 52 = 25.

The real number is the non negative square root of a number.

Hence, the real number is the non negative square root of a number.