Topic 10 Scientific Notationx

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Topic 10 Scientific Notation
Topic 10 Scientific Notation
Definition: Scientific Notation, also known as Standard Form is a way of writing numbers
that are too large or too small to be written in the conventional way. Writing in
scientific notation allows us to eliminate zeros in front of a very small decimal or
behind a very large number. This very concise way of writing numbers is often
used by scientists, mathematicians and engineers, who find that this way of writing
numbers is more convenient than the usual decimal form of the number.
In scientific notation all numbers are written like this:
a × 10n
Where a is any real number called the coefficient and it is usually a decimal with a
value between 1 and 10 ( 1 < a < 10) , and the exponent 10n will always have a base of
10 and n is always an integer.
If a = 1 or a = 10 then we usually omit the a and write the number as 10n.
1. How to convert large numbers into Scientific Notation.
We start by looking at simple numbers that are all powers of ten; these numbers are the building
blocks of the decimal number system and are easily converted into Scientific Notation as can be
seen in the table below.
Name
Number
one
ten
hundred
thousand
ten thousand
hundred thousand
million
1
10
100
1,000
10,000
100,000
1,000,000
Scientific
Notation
0
10
1
10
2
10
3
10
4
10
5
10
6
10
For example the number ten thousand is 10,000 = 10 x 10 x 10 x 10 = 104
The number one is a little unusual as it can be written in a number of ways, for example the
number one can be written as 20 , 40 etc but we used the form 1 = 100 to be consistent with the
other numbers so that they all use a base of ten.
It is easy to convert any of these powers of ten into Scientific Notation – all you need to do is to
count the number of zeros that follow after the initial 1, so 10,000,000,000,000 = 1013
Page | 1
Topic 10 Scientific Notation
Example 1: Convert the following numbers into Scientific Notation.
(a) 100,000,000,000
(b) A hundred million
Solution(a): 100,000,000,000 =
1011
Solution(b): A hundred million
=
Solution(c): Ten trillion =
(c) Ten trillion
(1 followed by 11 zeros)
100,000,000 =
10,000,000,000,000
=
108
(1 followed by 8 zeros)
1013
(1 followed by 13 zeros)
Next we look at numbers that start with a single digit and then have trailing zeros.
Example 2: Convert the following numbers into Scientific Notation.
(a) 30,000,000,000
Solution(a): 30,000,000,000 =
=
Solution(b): 4 hundred million
(b) 4 hundred million
(c) 500,000,000,000,000
3 x 10,000,000,000
3 x 1010
=
400,000,000
=
4 x 100,000,000
=
4 x 108
Solution(c): 500,000,000,000,000 =
=
(1 followed by 10 zeros)
(1 followed by 8 zeros)
5 x 100,000,000,000,000
5 x 1014
(1 followed by 14 zeros)
Lastly we look at how to convert any large number into a x 10n its Scientific Notation.
The process involves moving the decimal point to the left until the new value of the number is
between 1 and 10 this new value will be our coefficient a, while the number of places that we
move the decimal point will be our integer power n.
Example 3: Convert the following numbers into Scientific Notation.
(a) 87,000
(b) 16,000,000
(c) 200,000,000
(d) 150,000
(e) 2,400,000
(f) 8,210,000,000
(g) 16,400
(h) 115,000
(i) 1,234
(j) 54
(k) 123.45
(l) 32 million
(m) 23.5 thousand
(n) 156 billion
(0) 23.45 million
Page | 2
Topic 10 Scientific Notation
Solution (a):
87,000 is written as 87,000.
Decimal point
We move the decimal point to its new location 8.7000 and remove the trailing
zeros so that it is now 8.7 this is the value of the coefficient a = 8.7
Also since the decimal point moved 4 places to the left the power of n = 4
So 87,000
=
8.7 × 104
107
Solution (b):
16,000,000. =
16
Solution (c):
200,000,000 =
2
Solution (d):
150,000.
=
15
105
move the decimal point 5 places to the left.
Solution (e):
2,400,000.
=
24
106
move the decimal point 6 places to the left.
Solution (f):
8,210,000,000 = 8 21
109
move the decimal point 9 places to the left.
Solution (h):
115,000.
105
move the decimal point 5 places to the left.
Solution (i):
1,234.
1 15
=
1 234
Solution (j):
54. =
Solution (k):
123.45 =
Solution (l):
Solution (m):
32 million
=
=
23.5 thousand
Solution (n):
156 billion
=
=
Solution (l):
23.45 million=
=
5.4
move the decimal point 8 places to the left.
108
=
move the decimal point 7 places to the left.
103
101
move the decimal point 3 places to the left.
move the decimal point 1 places to the left.
1 234 5 102
move the decimal point 2 places to the left.
32,000,000.
3.2 107
move the decimal point 7 places to the left.
=
=
move the decimal point 4 places to the left.
23,500.
2.35 104
156,000,000,000.
move the decimal point 11 places to the left.
1.56 1011
23,450,000.
2.345 107
move the decimal point 7 places to the left.
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Topic 10 Scientific Notation
Exercise 1:
1.
Convert the following numbers into Scientific Notation.
For example 10,000 would be written in Scientific Notation as 104.
(a) 100,000,000,000,000
2.
4.
5.
(c) one hundred trillion
Convert the following numbers into Scientific Notation.
For example 80,000 would be written in Scientific Notation as 8 x 104.
(a) 900,000,000,000
3.
(b) one thousand million
(b) 40 thousand million
(c) 8,000,000,000,000
Write down each of the following numbers in standard form.
For example 34 000 would be written in standard form as 3.4 x 104.
(a) 324
(b) 32
(c) 3,890,000
(d) 120
(e) 356,000,000
(f) 19 000 0000 000
(g) 89 000 000 000
(h) 67,000
(i) 8,600
(j) 2 300
(k) 425 000
(l) 84
(m) 67 000 000
(n) 41 000
(o) 5 000 000 000
(p) 3 451
Write down each of the following numbers in standard form.
For example 22,100,000 would be written in standard form as 2.21 x 107.
(a) 1 ,903
(b) 346, 000,000
(c) 28 ,000,000,000,000,000
(d) 63 4
(e) 121 7
(f) 281,000
(g) 23 99
(h) 5,120,000
(i) 700 3
For each of the following numbers
(i) Write it out in figures.
(ii) Write it in scientific notation.
(a) 20 thousand
(b) 16 million
(c) 200 million
(d) 150 thousand
(e) 2 4 million
(f) 8 21 million
(g) 2.3 billion
(h) 9 thousand million
(i) 8.67 thousand
(j) sixteen thousand five hundred
(k) one million two hundred and fifty thousand
Page | 4
Topic 10 Scientific Notation
2. How to convert Scientific Notation into large numbers
When a number is written in its Scientific Notation form a x 10n where n is a positive integer
then it can be converted into a normal number.
The method that is used is to move the decimal point in the coefficient a, n places to the right.
Example 1: Convert the number 1.234 x 102 into an ordinary number.
Solution:
We take the term 1.234 x 102 and start with its coefficient 1.234 and move its
decimal point 2 places to the right.
1.234 becomes 123.4
Decimal point moved 2 points to the right
So 1.234 x 102 = 123.4
Example 2: Convert the number 1.234 x 103 into an ordinary number.
Solution:
We take the term 1.234 x 103 and start with its coefficient 1.234 and move its
decimal point 3 places to the right.
1.234 becomes 1234.
Decimal point moved 3 points to the right
So 1.234 x 103 = 1234
Example 3: Convert the number 1.234 x 107 into an ordinary number.
Solution:
We take the term 1.234 x 107 and start with its coefficient 1.234 and move its
decimal point 3 places to the right.
1.234 becomes 12,340,000.
Decimal point moved 7 points to the right
So 1.234 x 107 = 12,340,000
Page | 5
Topic 10 Scientific Notation
Example 4: Write down each of the following numbers as a normal number
(a) 4.5 x 103
(b) 3.8722 x 103
(c) 3.87 x 104
(d) 4 x 106
(e) 2.999 x 102
(f) 2.999 x 104
(g) 5 x 102
(h) 1.83 x 107
(i) 3.7 x 101
Solution (a):
4.5 x 103
=
4,500
move the decimal point 3 places to the right.
3,872.2
move the decimal point 3 places to the right.
Solution (b):
3.8722 x 103
=
Solution (c):
3.87 x 104 =
38,70 0
move the decimal point 4 places to the right.
Solution (d):
4 x 106 =
4,000,000
move the decimal point 6 places to the right.
Solution (e):
2.999 x 102
=
299.9
move the decimal point 2 places to the right.
Solution (f):
2.999 x 104
=
29,990
move the decimal point 4 places to the right.
Solution (g):
5 x 102
Solution (h):
1.83 x 107 =
Solution (i):
3.7 x 101
Exercise 2: 1.
=
=
500
move the decimal point 2 places to the right.
18,300,000
move the decimal point 7 places to the right.
37
move the decimal point 1 places to the right.
Write down each of the following numbers as a normal number
For example 3.87 x 105 would be written as 387,000.
(a) 2 4
102
(b) 3 61
101
(c) 7 003
(d) 5 8
107
(e) 6 04
103
(f) 2
(g) 2 356
103
(h) 4 254
(j) 5
106
(k) 4 13
105
104
104
100
(i) 1 24
(l) 8 423
(m) 3 45
106
(n) 1 903
103
(o) 3 4645
(p) 6 34
101
(q) 1 217
102
(r) 4 007
107
1010
108
103
Page | 6
Topic 10 Scientific Notation
3. How to convert small numbers into Scientific Notation.
We start by looking at simple numbers that are all powers of ten; these numbers are the building
blocks of the decimal number system and are easily converted into Scientific Notation as can be
seen in the table below.
Name
one
Decimal
Form
1
Fractional
Form
1
tenth
0.1
10
hundredth
0.01
10
thousandth
0.001
10
ten thousandth
0.000 1
10
hundred thousandth
0.000 01
10
millionth
0.000 001
10
For example the number ten thousandth is 0.000 1=
Scientific
Notation
0
10
-1
-2
-3
-4
-5
-6
= 10- 4
The number one is a little unusual as it can be written in a number of ways, for example the
number one can be written as 20 , 40 etc but we used the form 1 = 100 to be consistent with the
other numbers so that they all use a base of ten.
It is easy to convert any of these decimals into powers of ten, you just count the number of zeros
to the left of the first non zero number.
For example
0.0 0 0 01 has 5 zeros to the left of the number 1 so 0.00001 = 10-5
For example
0.0 0 0 2 has 4 zeros to the left of the number 2 so 0.0002 = 2 x 10-4
For example
0.0 0 324 has 3 zeros to the left of the number 3 so 0.00324 = 3.24 x 10-3
Page | 7
Topic 10 Scientific Notation
Example 1: Convert the following numbers into Scientific Notation.
(a) 0 043
(b) 0 000 357
(c) 0 19
(d) 0 009 8
(e) 0 000 070 4
(f) 0 06
(g) 0 000 000 8
(h) 0 522
(i) 0 000 020 07
(j) 0 000 000 007 8
(k) 0 001 4
(l) 0 000 45
Solution(a): 0 043
=
4.3 x 10-2
Since there are 2 zeros at the start.
Solution(b): 0 000 357
=
Since there are 4 zeros at the start.
Solution(c): 0 19
1.9 x 10-1
Since there is 1 zero at the start.
=
9.8 x 10-3
Since there are 3 zeros at the start.
7.04 x 10-5
Since there are 5 zeros at the start.
=
Solution(d): 0 009 8
Solution(e): 0 000 070 4 =
Solution(f): 0 06
=
=
Since there are 2 zeros at the start.
6 x 10-2
Solution(g): 0 000 000 8 =
Solution(h): 0 522
3.57 x 10-4
Since there are 7 zeros at the start.
8 x 10-7
Since there is 1 zero at the start.
5.22 x 10-1
Solution(i): 0 000 020 07 =
2.007 x 10-5
Solution(j): 0 000 000 007 8 =
7.8 x 10-9
Since there are 5 zeros at the start.
Since there are 9 zeros at the start.
Solution(k): 0 001 4 =
1.4 x 10-3
Since there are 3 zeros at the start.
Solution(l): 0 000 45 =
4.5 x 10-4
Since there are 4 zeros at the start.
Exercise 3: Convert the following decimal numbers into Scientific Notation.
(a) 0 00456
(b) 0 00304
(c) 0 000 000 5
(d) 0 8
(e) 0 000 000 0896
(f) 0 000000645
(g) 0.000 83
(h) 0 000 000 000 8
(i) 0 528
(j) 0 000 020 7
(k) 0 000 000 009 8
(l) 0 000 019
(m) 0 000 000 7
(n) 0 000 000 005 67
(o) 0 000 09
Page | 8
Topic 10 Scientific Notation
4. How to convert Scientific Notation into small numbers
In these situations we will be given a number written in scientific notation a x 10n where n will
be a negative integer. There are three possible strategies for doing this conversion and an
example of each is given below.
Example 1: Change 3.1 x 10- 4 into an ordinary number.
Solution:
Our first strategy is to convert the power 10n to a decimal and then multiply the
coefficient a and the converted power together to obtain the result.
3.1 x 10- 4
=
=
3.1 x 0.0001
0.00031
This method is seldom used as it is not concise and can easily lead to very messy
calculations especially if n is a large integer.
In our second strategy we start with the coefficient a and move the decimal point n
places to the left filling in with zeros when necessary.So in this situation we take
the coefficient a = 3.1 and move the decimal point 4 places to the left to get the
result 0.00031
So
3.1 x 10- 4
= 0.00031
In our third strategy we remove the decimal point from the coefficient a = 3.1 to get
31 and then adding 4 zeros to the left of a to get the result 0.00031
This is a very concise method as it involves a few steps that usually can be
performed at once to quickly give the result.
Example 2: Write each of the following as an ordinary number :
(a) 4 3
(d) 7 65
10-3
Solution(a): 4 3
Solution(b): 2
(b) 2 1
10-3
10-5
(e) 3 11
10-7
10-5
10-1
(c) 6 41
10-2
(f) 1 253
10-8
=
0.0043
3 zeros to the left of 43
=
0.0000
5 zeros to the left of 2
Solution(c): 6 41
10-2
=
0.00641
2 zeros to the left of 641
Solution(d): 7 65
10-7
=
0.000 000 765
7 zeros to the left of 765
Solution(e): 3 11
10-1
=
0.311
1 zeros to the left of 311
0.000 000 012 53
8 zeros to the left of 1253
Solution(f): 1 253
10-8 =
Page | 9
Topic 10 Scientific Notation
Exercise 4: Write each of the following as an ordinary number.
For example 7 12 10-3 would be written as 0.007 12
1.
(a) 2 6
10-3
(d) 5 55
(g) 8
10-7
(b) 2
(e) 7 06
(h) 9
10-6
10-2
(k) 6 53
(m) 9 167
10-2
(n) 9
10-10
10-3
10-1
(f) 5 0007
(l) 3
10-6
10-2
10-5
10-10
(o) 9 569
10-6
(q) 1.566
10-2
(i) 7 02
10-5
(j) 8 545
(p) 3 23
(c) 2 4
10-6
(r) 9.96
10-1
10-4
Page | 10
Topic 10 Scientific Notation
5. Problems involving Scientific Notation
There are many situations that involve using numbers written in Scientific Notation in
calculations. This can happen when we use very large or very small numbers in formula such as
those used by physicist, chemists as well as engineers.
The main strategy for performing multiplications that involve Scientific Notation is to follow
these 4 steps.
Step 1: Separate the terms into coefficients and powers of 10
Step 2: Multiply the coefficients together and the powers of ten together as separate
calculations.
Step 3: Convert the product of the coefficients into Scientific Notation if needed.
Step 4: Combine the individual parts to obtain the final result.
Here are some examples of this process in action.
3
Solution:
5
(7.2 x 103 ) x (2.5 x 105 ) =
=
=
=
( 7.2 x 2.5) x ( 103 x 105 )
18 x 108
1.8 x 101 x 108
1.8 x 109
Step 1
Step 2
Step 3
Step 4
Example 2: Calculate (7.51x 1020 ) x (9 x 10-26 ) giving your answer in Scientific Notation.
Solution:
(7.51 x 1020 ) x (9 x 10-26 ) =
=
=
=
-2
Solution:
( 7.51 x 9) x ( 1020 x 10-26 )
67.59 x 10-6
6.759 x 101 x 10-6
6.759 x 10-5
Step 1
Step 2
Step 3
Step 4
-13
(6.32 x 10-2) x ( 3.2 x 10-13)=
=
=
=
( 7.51 x 3.2) x ( 10-2 x 10-13 )
24.032 x 10-15
2.4032 x 101 x 10-15
2.4032 x 10-14
Step 1
Step 2
Step 3
Step 4
Example 4: Calculate (1.5 x 108) x ( 2.6 x 10-6) giving your answer in Scientific Notation.
Solution:
(1.5 x 108) x ( 2.6 x 10-6)
=
=
( 1.5 x 2.6) x ( 108 x 10-6 )
3.45 x 102
Step 1
Step 2
Page | 11
Topic 10 Scientific Notation
The main strategy for performing a division that involve Scientific Notation is to follow these 4
steps.
Step 1: Separate the terms into coefficients and powers of 10
Step 2: Divide the coefficients together and the powers of ten together as separate calculations.
Step 3: Convert the division of the coefficients into Scientific Notation if needed.
Step 4: Combine the individual parts to obtain the final result.
Here are some examples of this process in action.
giving your answer in Scientific Notation.
Example 5: Calculate
=
Solution:
Step 1
=
0.125 x 10-12
Step 2
=
1.25 x 10-1 x 10-12
Step 3
=
1.25 x 10-13
Step 4
giving your answer in Scientific Notation.
Example 6: Calculate
=
Solution:
Step 1
=
0.25 x 10-3-(-7)
=
0.25 x 104
Step 2
=
2.5 x 10-1 x 104
Step 3
=
2.5 x 10-3
Step 4
Example 7: Calculate (7.2 x 103 )
Solution:
(7.2 x 103 )
(2.5 x 105 ) giving your answer in Scientific Notation.
(2.5 x 105 ) =
=
=
Step 1
2.88 x 10-2
Step 2
Page | 12
Topic 10 Scientific Notation
Another common situation is to use Scientific Notation in formula and in multiple calculations.
Example 8: There are approximately 5.9 x 1012 miles in one light year. How many miles are
there in 2,500 light years? Write your answer in scientific notation.
Solution:
Number of miles =
=
=
=
=
=
( 2,500 ) x (5.9 x 1012 )
( 2.5 x 103 ) x (5.9 x 1012 )
( 2.5 x 5.9 ) x (103 x 1012 )
14.75 x 1015
1.475 x 101 x 1015
1.475 x 1016
Step 1
Step 2
Step 3
Step 4
Example 9: The number of atoms in 12 grams of carbon is 6.12 x 1023. Find the number of
atoms in 1 gram of carbon. Write your answer in scientific notation.
Solution:
Number of atoms =
=
=
=
=
5.1 x 10-1 x
5.1 x
Example 10: The amount of energy that is equivalent to a mass of m kg is given by the equation
E = mc2 where m is the mass in kg and c = 3 x 109 m/sec is the speed of light.
(a) How much energy is contained in a single proton that weighs
m = 0.000 000 000 000 000 000 000 000 001 672 kg?
(b) 1 gram of hydrogen contains 6.02 x 1023 protons, how much energy is there in
1 gram of hydrogen?
Solution(a): We express the mass in Scientific notation.
m = 0.000 000 000 000 000 000 000 000 001 672 = 1.672 x 10-27
E
Solution(b): E
=
=
=
=
=
=
=
mc2
(1.672 x 10-27 ) x ( 3 x 109 )2
(1.672 x 10-27 ) x ( 3 x 109 ) x ( 3 x 109 )
(1.672 x 3 x 3 ) x (10-27 x 109 x 109)
15.048 x 10-9
1.5048 x 101 x 10-9
1.5048 x 10-8 Joules
=
=
=
=
( 6.02 x 1023 ) x ( 1.5048 x 10-8 )
( 6.02 x 1.5048 ) x (1023 x 10-8 )
9.058896 x 1015 Joules
9,058,889,600,000,000 Joules
Step 1
Step 2
Step 3
Step 4
Step 1
Step 2
Page | 13
Topic 10 Scientific Notation
Exercise 5:
1.
(a) The distance between the planet Earth and the sun is a approximately 90 000 000 miles.
Write this distance in Scientific Notation.
(b) Jupiter's closest satellite is called Amalthea and is approximately 110 000 miles from
the centre of the planet. Write this distance in Scientific Notation.
2.
What is 3.015 103 written in standard form?
A 0.0003015
B 0.003015
C 3,015
D 30,150
3.
An electrons mass is about 0.00000000000000000000000000000091 kg.
Write this number in Scientific Notation.
4.
The Earths mass is about 5973600000000000000000000 kg
Write this number in Scientific Notation.
5.
Calculate the following, giving your answer in standard form.
(a) 42 million times 56 million.
(c) ( 1.2 x 105 ) x ( 2.5 x 109 )
(e) ( 2.9 x 103 ) x ( 5.9 x 103 )
(g)
6.
(b) 56 000 x 98 000 000
(d) ( 4 x 10-3 ) ( 8 x 108 )
(f) 1 0.0000 000 4
(h)
There are an estimated 1022 stars in the visible universe. It has been suggested by a scientist
that 0.03% of these stars will have earth like planets around them.
Which of the following calculations will give you the number of earth like planets in the
visible universe?
A. 3 1022
B. 3
1020
C. 3
1018
D. 3
1026
7.
There are approximately 5.9 x 1012 miles in one light year.
How many miles are there in 3,500 light years? Write your answer in scientific notation.
8.
The mass of one proton is approximately 1.7 x 10-24 gram.
Find the mass of 4 million protons. Write your answer in scientific notation.
9.
A certain computer does one calculation in 6 x 10-7 second. How long does this
computer take to do 8 million calculations?
10. The number of atoms in 12 grams of carbon is 6.12 x 1023. Find the number of atoms in 18
grams of carbon. Write your answer in scientific notation.
11. The amount of energy that is equivalent to a mass of m kg is given by the equation
E = mc2 where m is the mass in kg and c = 3 x 109 m/sec is the speed of light.
How much energy is contained in an object that weighs a million Kg
Page | 14
Topic 10 Scientific Notation
Solutions
Exercise 1: 1.(a) 1014
(b) 109
(c) 1014
11
10
2.(a) 9 x 10
(b) 4 x 10
(c) 8 x 1012
3.(a) 3.24 x 102
(b) 3.2 x 101
(c) 3.89 x 106
(d) 1.2 x 102
8
10
10
3.(e) 3.56 x 10
(f) 1.9 x 10
(g) 8.9 x 10
(h) 6.7 x 104
3.(i) 8.6 x 103
(j) 2.3 x 103
(k) 4.25 x 105
(h) 8.4 x 101
7
4
9
3.(m) 6.7 x 10
(j) 4.1 x 10
(k) 5 x 10
(h) 3.451 x 103
4.(a) 1.901 x 103
(b) 3.46 x 108
(c) 2.8 x 1016
(d) 6.34 x 101
2
5
1
4.(e) 1.217 x 10
(f) 2.81 x 10
(g) 2.399 x 10
(h) 5.12 x 106
4.(i) 7.003 x 102
5.(a) 20,000 = 2 x 104
(b) 16,000,000 = 1.6 x 107
8
5.(c) 200,000,000 = 2 x 10
(d) 150,000 = 1.5 x 105
5.(e) 2,400,000 = 2.4 x 106
(f) 8,210,000 = 8.21 x 106
9
5.(g) 2,300,000,000 = 2.3 x 10
(h) 9,000,000,000 = 9 x 109
5.(i) 8,670 = 6.67 x 103
(j) 16,500= 1.65 x 104
6
5.(k) 1,250,000 = 1.25 x 10
Exercise 2: 1.(a) 240
1.(e) 6,040
1.(i) 12,400,000
1.(m) 3,450,000
1.(q) 121.7
(b) 36.1
(f) 2
(j) 5,000,000
(n) 1,903
(r) 4,007
(c) 70,030
(g) 2356
(k) 41,300
(o) 346,450,000
(d) 58,000,000
(h) 2,356
(l) 84,320,000,000
(p) 63.4
3.04 x 10-3
6.45 x 10-7
2.07 x 10-5
5.67 x 10-9
(c) 5 x 10-7
(g) 8.3 x 10-4
(k) 9.8 x 10-9
(k) 9 x 10-5
(d) 8 x 10-1
(h) 8 x 10-10
(l) 1.9 x 10-5
Exercise 4: 1.(a) 0.0026
1.(e) 0.00706
1.(i) 0.000 070 2
1.(m) 0.091 67
1.(q) 0.000 001 566
(b) 0.000002
(f) 0.0050007
(j) 0.085 45
(n) 0.000 009
(r) 0.000 996
(c) 0.024
(g) 0.000 008
(k) 0.653
(o) 0.956 9
(d) 0.000 000 555
(h) 0.000 009
(l) 0.000 000 000 3
(p) 0.000 000 000 323
Exercise 5: 1.(a) 9 x 107
2. C
3. 9.1 x 10-31
4. 5.9736 x 1024
5.(a) 2.352 x 1015
5.(e) 1.711 x 107
6.C
7. 2.065 x 1016
9. 4.8 x 100 sec
11. 9 x 1024
1.(b) 1.1 x 105
Exercise 3: (a) 4.56 x 10-3
(e) 8.96 x 10-8
(i) 5.28 x 10-1
(m) 7 x 10-7
(b)
(f)
(j)
(n)
(b) 5.488 x 1012
(f) 2.5 x 10-8
(c) 3 x 1014
(g) 2 x 108
(d) 5 x 10-12
(h) 9 x 10-7
8. 6.8 x 10-18
10. 9.18 x 1023
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