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.