Chapter 1 standard form

Created By: Mohd Said B Tegoh

Required Basic Mathematical Skills

Rounding off whole numbers to a specified place value

Round off 1 688 to the nearest hundred

Round off 430 618 to the nearest thousand

1 700

431 000

3 10 0 60 < 5

0 0

Round off 30 106 correct to thenearest hundred.

Round off 14.78 to the nearest whole number

4. 7 8

The first digit on the right is greater than 5

+1

Add I to digit 4

15Drop all the decimalsUnderstand !!

!


Rounding off whole numbers to a specified number of decimal places

Express 1.8523 to three decimal places

Express 0.4968 to two decimal places

1.852

0.50

Round off 5.316 to 1 decimal place

5 . 3 1 6

Underline digit 3

(1st decimal place)

The first digit on the right is less than 5

Do not change digit 3

Round off 4.387 to 2 decimal places

4 . 3 8 7

Underline digit 8

(2nd decimal place)

The first digit on the right is more

than 5

Add 1 to 8

9

+1

(

+

^

EXP (-)

log ln

hyp fdx

.Ans

tan

CALC √

0

ENGM+

sin cos

ab/c x2

RCL x-1 CONST

DEL AC

7 8 9 =

Before Getting started…… MODES

Before starting a calculation, you mustenter the correct mode as indicated in the table below

MODE 1

MODE 2

MODE 3

MODE 1

MODE 2

MODE 2

Arithmetic CalculationsArithmetic Calculations

Use the key to enter the COMPwhen you want to perform basiccalculations.

MODE

MODECOMP 1 1

FIX, SCI, RNDFIX, SCI, RND

(Fix) : Number of Decimal Places(Fix) : Number of Decimal Places

(Sci) : Number of Significant (Sci) : Number of Significant DigitsDigits

(Norm) : Exponential of significant(Norm) : Exponential of significant DigitsDigits

1

2

3

1

MODE5x

Fix 1

Fix 0 ٨ 9 ?

1


5 . 3 6=

1

5.3

1MODE5x

Fix 1

Fix 0 ٨ 9 ?

2


5 . 3 6=

1

5.325.32

1MODE5x

Fix 1

Fix 0 ٨ 9 ?

2


4 . 3 7=

8

4.394.39

1MODE5x

Fix 1

Fix 0 ٨ 9 ?

1


4 . 3 7=

8

4.44.4


Law of Indices

10m x 10n = 10m + n

10m ÷ 10n = 10m - n

Simplify the following103 x 10-5

102 ÷ 106

10-2

10-4

Very large and very small numbers are conveniently rounded off to a specified number of significant figures

The concept of significant figures is another way of stating the accuracy of a measurement

Significant figures refer to the relevant digits in an integer or a decimal number which has been rounded off to a given degree of accuracy

Positive numbers greater than 1 can be rounded off to a given number of significant figures

The rules for determining the number of significant figures in a number are as follows:

All non-zero digits are significantfigures

2.73 has 3 significant figures

1346 has 4 significant figures


All zeros between non-zero aresignificant figures


3008 has 4 significant figures


In a decimal, all zeros after any non-zero digit are significant figures




In a decimal, all zeros before the first non-zero digit arenot significant




All zeros after any non-zero digit in a whole number are not significant unless stated other wise

1999 = 2000 ( one s.f )



1999 = 2000 ( two s.f )



1999 = 2000 ( three s.f )

State the number of significant figures ineach of the following

(a) 4 576

(b) 603

(c) 25 009

(d) 2.10

(a) 0.0706

(f) 0.80

4353

32

Example 1Example 1Express 3.15 x 105 as a singlenumber

3 . 1 EXP 5

=5

315000

3 . 1 EXP 5

=

5

3.15 x 105MODE5x

Norm 3 3

Norm 1^2 ? 2 315000

Example 2Example 2Express 4.23 x 10-4 as a singlenumber

4 . 2 EXP (-) 4

=3

0.000423

4 . 2 EXP (-) 4

=

3

4.23 x 10-4MODE5x

Norm 3 3

Norm 1^2 ? 2 0.000423

Method of rounding off to a specified number of significant figures

Identify the digit (x) that is to be rounded off

Is the digit after x greater than or equal to 5

Add 1 to x x remains unchanged

Do the digit after x lie before the decimal point?

Replace each digit with zero Drop the digits

Write the number according to the specified number of significant figures

YES NO

YES (BEFORE) NO (AFTER)

3 10 0 60 < 5

0 0

Round off 30 106 correct to three

significant figures.

MODESci2 2 Sci 0 ٨ 9 ?

1

3

3.01 x 1043.01 x 104

3

=

5x

0 0 6

30 10030 100



Round off 0.05098 correct to three


0 5. 0 0 988 > 5

+11 0

0 5. 0 0 981 0

Round off 0.05098 correct to three significant figures.

MODE

Sci2 2 Sci 0 ٨ 9 ?

0

3

5.10 x 10-25.10 x 10-2

0

=

5x

. 5 0 9 8

0.05100.0510

To clear the Sci specification……

MODENorm 3 3Press

5X

Norm 1 ⱱ 2 ? 1

To continue the Sci specification……

PressON

Round off 0.0724789 correct to four significant figures.

0 7. 0 2 478 98 > 5

+18

0 7. 0 2 478 98

Round off 0.0724789 correct to four significant figures.


0

4

7.248 x 10-27.248 x 10-2

0

=

5x

. 97 2 4 7 8

0.072480.07248

Complete the following table (Round off to)

Number 3 sig. fig. 2 sig. fig. 1 sig. fig.

47 103

20 464

1 978

3.465

70.067

4.004

0.04567

0.06045

0.0007805

47100 47000 50 000

20 500 20 000 20 000

1 980 2 000 2 0003.47 3.5 3

70.1 70 70

4.00 4.0 4

0.0457 0.046 0.05

0.0605 0.060 0.06

0.000781 0.00078 0.0008

We usually use standard form for writing very large and very small numbers

A standard form is a number that is written as the product of a number A (between 1 and 10) and a power of 10

A x 10n, where 1 ≤ A < 10, and n is an integer

Positive numbers greater than or equal to 10 can be written in the standard formA x 10n , where 1 ≤ A ≤ 10 and n is the positive integer, i.e. n = 1, 2, 3,……… Example 58 000 000 = 5.8 x 107

Positive numbers less than or equal to 1 can be written in the standard formA x 10n , where 1 ≤ A ≤ 10 and n is the negative integer, i.e. n = …..,-3, -2, -1

Example 0.000073 = 7.3 x 10-5

Express 431 000 in standard form

431 000

Express the number as a product ofA (1 ≤ A < 10) and a power of 10

= 4.31

A

x 100 000

Power of 10

= 4.31 x 105

431 000 = 4 3 1 0 0 0

= 4.31 x 105

5 is the number of places, the decimal point is moved to the left

Express 431 000 in standard form


1

3

4.31 x 1054.31 x 105

4

=

5x

3 0 0 0

0.000709

Express the number as a product ofA (1 ≤ A < 10) and a power of 10

= 7.09

A

x

Power of 10

= 7.09 x 10-4

Express 0.000709 in standard form

10000

1

= 7.09 x410

1

0.000709 =


0 . 0 0 0 7 0 9

= 7.09 x 10-4

-4 is the number of places, the decimal point is moved to the right


MODE Sci

2 2 Sci 0 ٨ 9 ?

. 0 0 0

3

7.09 x 10-47.09 x 10-4

0

=

7 0 9

5x

Write the following numbers in standard form

NUMBER STANDARD FORM

8765

32154

6900000

0.7321

0.00452

0.0000376

0.0000000183

8.765 x 103

3.2154 x 104

6.9 x 106

7.321 x 10-1

4.52 x 10-3

3.76 x 10-5

1.83 x 10-8

Number in the standard form, A x 10n , can be converted to single numbers by moving the decimal point A

(a) n places to the right if n is positive

(b) n places to the left if n is negative

Express 1.205 x 104 as a single number

3.405 x 104

= 3 . 4 0 5 0

3 4 0 5 0=Move the decimal point 4 places to the right

3 . 4 EXP 4

=0

34 050

5

MODECOMP 1 1


Express 7.53x 10-4 as a single number

7.53 x 10-4

= 7 . 5 3 0 0 0 0

= 0.000753 Move the decimal point 4 places to the left

7 . 5 EXP (-) 4

=3

0.000753

Express 7.53 x 10-4 as a single number

Express the following in single numbers

STANDARD FORM NUMBER

4.863 x 103

7.2051 x 104

4.31 x 106

5.164 x 10-1

1.93 x 10-3

2.04 x 10-5

9.16 x 10-8

4863

72051

43100000.5164

0.00193

0.00002040.0000000916

3.25 X 105 = 325000

7.14 X 10-5 = 0.0000714

4537000 = 4.537 X 106

0.0000006398 = 6.398 X 10-7

325 X 105

32.5 X 106

3.25 X 107

0.325 X 108

===

431 X 10-8

43.1 X 10-7

4.31 X 10-6

0.431 X 10-5

===

Two numbers in standard form can be added or subtracted if both numbers have the same index

s M A R T

a x 10m + b x 10m =(a + b) x 10m

a x 10m - b x 10m =(a - b) x 10m

5.3 x 105 + 3.8 x 105

= (5.3 + 3.8 ) x 105

= 9.1 x 105

7.8 x 10-2 - 3.5 x 10-2

= (7.8 - 3.5 ) x 10-2

= 4.3 x 10-2

Two numbers in standard form with difference indices can only be added or subtracted if the differing indices are made equal

4.6 x 106 + 5 x 105

=

=

=

4.6 x 106 + 0.5 x 106

(4.6 + 0.5 ) x 106

5.1 x 106

6.4 x 10-4 - 8 x 10-5

=

=

=

6.4 x 10-4 - 0.8 x 10-4

(6.4 - 0.8) x 10-4

5.6 x 10-4

Calculate 3.2 x 104 – 6.7 x 103. Stating your answer in standard form.

4

4

44

1053.2

10)67.02.3(

1067.0102.3

x

x

x

x

MODE Sci2 2 Sci 0 ٨ 9 ?

. 2 EXP

3

2. 53 x 1042. 53 x 104

3

6

4 -

5x

EXP7. 3 =

Calculate 3.2 x 104 – 6.7 x 103. Stating your answer in standard form.

0000398.0 6109.2 x_

5

5

55

1069.3

10)29.098.3(

1029.01098.3

x

x

xx


0

3

0

5x

. 0 0 0

3 9 8 - 2 . 9

EXP (-) 6 = 3.69 x 10-53.69 x 10-5

0000398.0 6109.2 x_

When two numbers in standard form are multiplied or divided, the ordinary numbers are multiplied or divided with each otherWhile their indices are added or subtracted

s M A R T

a x 10m x b x 10n = (a x b) x 10m + n

a x 10m ÷ b x 10n

= (a ÷ b) x 10m - n

9.5 x 103 x 2.2 x 102

= (9.5 x 2.2) x (103 x 102)

= 20.9 x 103+2

= 20.9 x 105

= 2.09 x 106

7

)2(5

2

5

102.1

106

2.7106

102.7

x

x

xx

Calculate 1.17 x 10-2 . Stating your answer in 3 x 106

standard form.

9

8

62

109.3

1039.0

103

17.1

x

x

x

Calculate 1.17 x 10-2 . Stating your answer in 3 x 106

standard form.

MODE Sci2 Sci 0 ٨

9 ?. 1 7 EXP

3

3. 90 x 10-9

1

3

(-) 2 ÷

5x

EXP 6 =

2

2

23

106

)1024.9(

Calculate , expressing the answer

in standard form.

3

4

)2(6

2

232

1042.1

102.14

10

106

10)24.9(

x

x

x6

85.4x

x x

2

23

106

)1024.9(


. 2 EXP

3

1. 42 x 10-31. 42 x 10-3

9

)

(-) 3

5x

EXP6x2

=

(

4

÷ ( (-)

2 )

Calculate , expressing the answer

in standard form.

1 km2 = (1000 x 1000) m2

= (103 x 103) m2

= 106 m2

The area of a piece of rectangular land is 6.4 km2. If the width of the land is 1600 m, calculate the length, in m, of the land

Length of the land = Area Width

m3

3

6

104

106.1

4.6

104.6

x

x

1.6x10x

3

Round off 0.05098 correct to three


A 0.051B 0.0500C 0.0509D 0.0510

0 5. 0 0 988 > 5

+11 0

0 5. 0 0 981 0


MODE

Sci2 2 Sci 0 ٨ 9 ?

0

3

5.10 x 10-25.10 x 10-2

0

=

5x

. 5 0 9 8

0.05100.0510


A 0.083B 0.084C 0.0830D 0.0831

0 8. 0 3 055 = 5

+11

0 8. 0 3 051


MODE

Sci2 2 Sci 0 ٨ 9 ?

0

3

8.31 x 10-28.31 x 10-2

0

=

5x

. 8 3 0 5

0.08310.0831

3 10 0 60 < 5

0 0



A 30 000B 30 100C 30 110D 30 200



A 30 000B 30 100C 30 110D 30 200


1

3

3.01 x 1043.01 x 104

3

=

5x

0 0 6

30 10030 100


A 1 205B 12 050C 1 205 000D 12 050 000

1 . 2 EXP 4

=0

12 050

5

1 02 50

MODECOMP 1 1

Express 4.23 x 10-4 as a single number

A 0. 423B 0. 0423C 0. 00423D 0. 000423

4 . 2 EXP (-) 4

=3

0.000423

Express 52 700 in standard form.

A 5.27 102

B 5.27 104

C 5.27 102

D 5.27 x 10-4

MODE

Sci2 2 Sci 0 ٨ 9 ?

7

3

5.27 x 1045.27 x 104

5

=

5x

2 0 0

5 72 0 0

6900102.3 4x

ABCD

41089.3 x81089.3 x41001.1 x81001.1 x

4

4

44

4

1089.3

10)69.02.3(

1069.0102.3

6900102.3

x

x

xx

x


2

3

3

5x

. EXP 4 +

6 9 0 0 = 3.89 x 1043.89 x 104

6900102.3 4x

76 108.11015.8 xx

ABCD

71035.6 x61035.6 x71097.7 x61097.7 x

6

6

66

1097.7

10)18.015.8(

1018.01015.8

x

x

xx


. 1 EXP

3

7. 97 x 10-67. 97 x 10-6

8

1

(-) 6

5x

EXP8. (-)

=

76 108.11015.8 xx

5

- 7

24

3

)104(

1096.2

xx

ABCD

41024.7 x51024.7 x41085.1 x51085.1 x

4

5

8

3

1085.1

10185.0

16

96.2101

1096.2

x

x

x10

6x

x

(-8)-3-


. 9 EXP

3

1. 85 x 1041. 85 x 104

2 (-) 3

5x

EXP4

=

6

÷ ( (-)

x2

)

24

3

)104(

1096.2

xx

4

Education

Chapter 1 standard form