Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

2

Share

Download to read offline

Chapter 1 standard form

Download to read offline

Chapter 1 standard form

Related Books

Free with a 30 day trial from Scribd

See all

Chapter 1 standard form

  1. 1. Created By: Mohd Said B Tegoh
  2. 2. Required Basic Mathematical Skills Rounding off whole numbers to a specified place value Round off 1 688 to the nearest hundred 1off 430 618 to the 700 Round nearest thousand 431 000
  3. 3. Round off 30 106 correct to the nearest hundred. 0 3 0106 0<5
  4. 4. Round off 14.78 to the nearest whole number +1 15. 7 8 4 Add I the decimals Drop all to digit 4 Understand !!! The first digit on the right is greater than 5
  5. 5. Required Basic Mathematical Skills Rounding off whole numbers to a specified number of decimal places Express 1.8523 to three decimal places 1.852 to Express 0.4968 two decimal places 0.50
  6. 6. Round off 5.316 to 1 decimal place 5 . 3 1 6 Do not The first digit on the change Underline digit 3 right is less than 5 digit 3 st (1 decimal place)
  7. 7. Round off 4.387 to 2 decimal places +1 9 4.38 7 Add 1 todigit on The first digit 8 Underline 8 the right is more (2 nd decimal 5 than place)
  8. 8. CALC √ sin cos ab/c x2 tan ( M+ ENG log ln RCL x -1 CONST hyp fdx DEL AC 7 0 8 . + 9 = EXP (-) ^ Ans
  9. 9. Before Getting started……  MODES Before starting a calculation, you must enter the correct mode as indicated in the table below MODE 1 MODE 2
  10. 10. MODE 3 MODE 1 MODE 2 MODE 2
  11. 11. Arithmetic Calculations Use the MODE key to enter the COMP when you want to perform basic calculations. MODE COMP 1 1
  12. 12. FIX, SCI, RND (Fix) : Number of Decimal Places 1 (Sci) 2 : Number of Significant Digits (Norm) : Exponential of significant Digits 3
  13. 13. Round off 5.316 to 1 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 1 5 . 3 = 5.3 1 6
  14. 14. Round off 5.316 to 2 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 5 . 3 = 5.32 5.32 1 6 2
  15. 15. Round off 4.387 to 2 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 4 . 3 = 4.39 4.39 8 7 2
  16. 16. Round off 4.387 to 1 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 4 . 3 = 4.4 4.4 8 7 1
  17. 17. Required Basic Mathematical Skills Law of Indices 10m x 10n = 10m + n 10 ÷ 10 = 10 m n m -n Simplify the following 103 x 10-5 10-2 102 ÷ 106 10-4
  18. 18. 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 ignificant figures refer to the relevant digits in an integer or a decimal number which has been rounded off to a given degree of accuracy
  19. 19. ositive numbers greater than 1 can be ounded off to a given number of significa gures
  20. 20. The rules for determining the number of significant figures in a number are as follows: All non-zero digits are significant figures 2.73 has 3 significant figures 1346 has 4 significant figures
  21. 21. The rules for determining the number of significant figures in a number are as follows: All zeros between non-zero are significant figures 2.03 has 3 significant figures 3008 has 4 significant figures
  22. 22. The rules for determining the number of significant figures in a number are as follows: In a decimal, all zeros after any non-zero digit are significant figures 3.60 has 3 significant figures 27.00 has 4 significant figures
  23. 23. The rules for determining the number of significant figures in a number are as follows: In a decimal, all zeros before the first non-zero digit are not significant 0.0032 has 2 significant figures 0.0156 has 3 significant figures
  24. 24. The rules for determining the number of significant figures in a number are as follows: All zeros after any non-zero digit in a whole number are not significant unless stated other wise 1999 = 2000 ( one s.f )
  25. 25. The rules for determining the number of significant figures in a number are as follows: All zeros after any non-zero digit in a whole number are not significant unless stated other wise 1999 = 2000 ( two s.f )
  26. 26. The rules for determining the number of significant figures in a number are as follows: All zeros after any non-zero digit in a whole number are not significant unless stated other wise 1999 = 2000 ( three s.f )
  27. 27. State the number of significant figures in each of the following (a) 4 576 (b) 603 (c) 25 009 (d) 2.10 (a) 0.0706 (f) 0.80 4 3 5 3 3 2
  28. 28. Example 1 Express 3.15 x 105 as a single number 3 . 1 5 = 315000 EXP 5
  29. 29. 3 . 1 5 EXP 5 5x = 3.15 x 105 Norm 1^2 ? MODE Norm 2 315000 3 3
  30. 30. Example 2 Express 4.23 x 10-4 as a single number 4 . 2 3 = 0.000423 EXP (-) 4
  31. 31. 4 . 2 3 EXP (-) 4 5x = 4.23 x 10-4 Norm 1^2 ? MODE Norm 2 0.000423 3 3
  32. 32. hod of rounding off to a specified number of significant figur Identify the digit (x) that is to be rounded off Is the digit after x greater than or equal to 5 YES Add 1 to x NO x remains unchanged Do the digit after x lie before the decimal point? YES (BEFORE) Replace each digit with zero NO (AFTER) Drop the digits Write the number according to the specified number of significant figures
  33. 33. Round off 30 106 correct to three significant figures. 0 3 0106 0<5
  34. 34. Round off 30 106 correct to three significant figures. 5x MODE 3 = Sci 2 0 2 1 Sci 0 ٨ 9 ? 0 3.01 x 10 44 3.01 x 10 30 100 30 100 6 3
  35. 35. Round off 0.05098 correct to three significant figures. 1 0 +1 0. 0 5098 8>5 00 0 . 0 51 9 8
  36. 36. Round off 0.05098 correct to three significant figures. 5x MODE 0 = Sci 2 . 2 0 Sci 0 ٨ 9 ? 5 5.10 x 10-2 5.10 x 10-2 0.0510 0.0510 0 3 9 8
  37. 37.  To clear the Sci specification…… 5X Press MODE Norm 3 Norm 1 ⱱ 2 ? 3 1  To continue the Sci specification…… ON Press
  38. 38. Round off 0.0724789 correct to four significant figures. +1 8 0. 0 724789 8>5 8 0. 0 724789
  39. 39. Round off 0.0724789 correct to four significant figures. 5x MODE Sci 2 2 Sci 0 ٨ 9 ? 0 . = 7.248 x 10 -2 7.248 x 10 -2 0 7 0.07248 0.07248 2 4 4 7 8 9
  40. 40. Complete the following table (Round off to) Number 3 sig. fig. 2 sig. fig. 1 sig. fig. 47 103 47100 47000 50 000 20 464 20 500 20 000 20 000 1 978 1 980 3.465 3.47 2 000 3.5 2 0003 70.067 70.1 70 70 4.004 4.00 4.0 4 0.04567 0.0457 0.046 0.05 0.06045 0.0605 0.060 0.06 0.0007805 0.000781 0.00078 0.0008
  41. 41. We usually use standard form for writing very large a 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
  42. 42. Positive numbers greater than or equal to 10 can be written in the standard form A 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
  43. 43. Positive numbers less than or equal to 1 can be written in the standard form A 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
  44. 44. Express 431 000 in standard form Express the number as a product of A (1 ≤ A < 10) and a power of 10 A Power of 10 431 000 = 4.31 x 100 000 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
  45. 45. Express 431 000 in standard form 5x MODE 4 = Sci 2 3 2 1 Sci 0 ٨ 9 ? 3 0 0 4.31 x 1055 4.31 x 10 0
  46. 46. Express 0.000709 in standard form Express the number as a product of A (1 ≤ A < 10) and a power of 10 Power of 10 A 1 0.000709 = 7.09 x 10000 1 = 7.09 x 4 10 7.09 x 10-4 =
  47. 47. Express 0.000709 in standard form 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
  48. 48. Express 0.000709 in standard form 5x MODE Sci 2 2 Sci 0 ٨ 9 ? 0 . = 7.09 x 10-4 7.09 x 10-4 0 0 0 3 7 0 9
  49. 49. Write the following numbers in standard form NUMBER 8765 32154 6900000 0.7321 0.00452 0.0000376 0.0000000183 STANDARD FORM 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
  50. 50. 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
  51. 51. Express 1.205 x 104 as a single number 3.405 x 10 4 =3 . 4 0 5 0 Move the decimal point 4 places to the right =34050
  52. 52. Express 3.405 x 104 as a single number MODE COMP 1 1 3 . 4 0 = 34 050 5 EXP 4
  53. 53. Express 7.53x 10-4 as a single number 7.53 x 10 -4 = 0 0 00 7.5 3 Move the decimal point 4 places to the left = 0.000753
  54. 54. Express 7.53 x 10-4 as a single number 7 . 5 3 = 0.000753 EXP (-) 4
  55. 55. Express the following in single numbers STANDARD FORM 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 NUMBER 4863 72051 4310000 0.5164 0.00193 0.0000204 0.0000000916
  56. 56. 3.25 X 105 = 325000 -5 7.14 X 10 = 0.0000714 4537000 = 4.537 X 106 0.0000006398 = 6.398 X 10-7
  57. 57. 325 X 105 32.5 X 106 = 3.25 X 107 = 0.325 X 108 =
  58. 58. 431 X 10-8 43.1 X 10-7 = 4.31 X 10-6 = 0.431 X 10-5 =
  59. 59. Two numbers in standard form can be added or subtracted if both numbers have the same index
  60. 60. s MA R T a x 10 + b x 10 m = (a + b) x 10 m m a x 10 - b x 10 m = (a - b) x 10 m m
  61. 61. 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 =
  62. 62. Two numbers in standard form with difference indices can only be added or subtracted if the differing indices are made equal
  63. 63. 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 =
  64. 64. 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 =
  65. 65. Calculate 3.2 x 10 4 – 6.7 x 10 3 . Stating your answer in standard form. 3.2 x10 − 0.67 x10 4 = (3.2 − 0.67) x10 = 2.53x10 4 4 4
  66. 66. Calculate 3.2 x 104 – 6.7 x 103. Stating your answer in standard form. 5x MODE Sci 2 3 6 . . 2 Sci 0 ٨ 9 ? 2 EXP 7 EXP 2. 53 x 1044 2. 53 x 10 4 3 3 =
  67. 67. 0.0000398_ 3.98x10 −5 2.9 x10 − 0.29 x10 = (3.98 − 0.29) x10 = 3.69 x10 −5 −5 −5 −6
  68. 68. 0.0000398_ 2.9 x10 −6 5x MODE Sci 2 2 0 . 0 0 0 0 3 9 8 - 2 . 6 = 3.69 x 10-5 3.69 x 10-5 EXP (-) Sci 0 ٨ 9 ? 3 9
  69. 69. When two numbers in standard form are multiplied or divided, the ordinary numbers are multiplied or divided with each other While their indices are added or subtracted
  70. 70. s MA R T a x 10 x b x 10 m+n = (a x b) x 10 m n a x 10 ÷ b x 10 m-n = (a ÷ b) x 10 m n
  71. 71. 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 =
  72. 72. 7.2 x10 −2 6 x10 7 .2 5 −( −2 ) = x10 6 7 = 1.2 x10 5
  73. 73. Calculate 1.17 x 10-2 . Stating your answer in 3 x 106 standard form. 1.17 −2 −6 x10 3 −8 = 0.39 x10 = 3.9 x10 −9
  74. 74. Calculate 1.17 x 10-2 . Stating your answer in 3 x 106 standard form. 5x MODE Sci 2 1 . 3 EXP 2 3 Sci 0 ٨ 9? 1 7 6 = EXP (-) 2 3. 90 x 10 -9 ÷
  75. 75. −3 2 Calculate (9.24 ×10 ) , expressing the answer 6 ×10 −2 in standard form. (9.24) 2 x10 −3 x 2 6 x10 −2 85.4 = x10 −6−( −2 ) 6 = 14.2 x10 −4 = 1.42 x10 −3
  76. 76. −3 2 Calculate (9.24 ×10 ) , expressing the answer 6 ×10 −2 in standard form. 5x MODE Sci 2 2 Sci 0 ٨ 9 ? 3 ( 9 . 2 4 EXP (-) 3 ) x2 ÷ ( 6 EXP (-) 2 ) = 1. 42 x 10-3 1. 42 x 10-3
  77. 77. 1 km2 = (1000 x 1000) m2 = (103 x 103) m2 = 106 m2
  78. 78. 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 6.4 x10 6 = 1.6x10 3 6.4 = x103 1.6 3 = 4 x10 m
  79. 79. Round off 0.05098 correct to three significant figures. +1 1 0 A B C D 0.051 0.0500 0.0509 0.0510 0. 0 5098 8>5 00 0 . 0 51 9 8
  80. 80. Round off 0.05098 correct to three significant figures. 5x MODE 0 = Sci 2 . 2 0 Sci 0 ٨ 9 ? 5 5.10 x 10 -2 5.10 x 10 -2 0.0510 0.0510 0 3 9 8
  81. 81. Round off 0.08305 correct to three significant figures. A B C D 0.083 0.084 0.0830 0.0831 1 +1 0. 08305 5=5 0 0 . 0 8 31 5
  82. 82. Round off 0.08305 correct to three significant figures. 5x MODE 0 = Sci 2 . 2 0 Sci 0 ٨ 9 ? 8 8.31 x 10-2 8.31 x 10-2 0.0831 0.0831 3 3 0 5
  83. 83. Round off 30 106 correct to three significant figures. A B C D 30 000 30 100 30 110 30 200 0 3 0106 0<5
  84. 84. Round off 30 106 correct to three significant figures. A B C D 30 000 30 100 30 110 30 200 5x MODE 3 = Sci 2 0 2 1 Sci 0 ٨ 9 ? 0 3.01 x 1044 3.01 x 10 30 100 30 100 6 3
  85. 85. Express 1.205 x 104 as a single number A B C D 1 205 12 050 1 205 000 12 050 000 MODE COMP 1 1 2 0 50 1 1 . 2 0 = 12 050 5 EXP 4
  86. 86. Express 4.23 x 10-4 as a single number A B C D 0. 423 0. 0423 0. 00423 0. 000423 4 . 2 3 = 0.000423 EXP (-) 4
  87. 87. Express 52 700 in standard form. A B C D 5.27 × 102 5.27 × 104 5.27 × 10−2 5.27 x 10-4 52 700 5x MODE 5 = Sci 2 2 2 7 Sci 0 ٨ 9 ? 0 5.27 x 1044 5.27 x 10 0 3
  88. 88. 3.2 x10 + 6900 = 4 A B C D 3.89 x10 8 3.89 x10 4 1.01x10 8 1.01x10 4 3.2 x10 + 6900 4 = 3.2 x10 + 0.69 x10 4 = (3.2 + 0.69) x10 = 3.89 x10 4 4 4
  89. 89. 3.2 x10 + 6900 = 4 5x MODE Sci 2 2 3 . 2 EXP 4 6 9 0 Sci 0 ٨ 9 ? 0 = 3 + 3.89 x 1044 3.89 x 10
  90. 90. 8.15x10 A B C D −6 −1.8x10 −7 = 6.35x10 −6 6.35x10 −7 7.97 x10 −6 7.97 x10 8.15x10 −7 −6 − 0.18x10 = (8.15 − 0.18) x10 = 7.97 x10 −6 −6 −6
  91. 91. 8.15x10 −6 −1.8x10 −7 = 5x MODE Sci 2 8 . 2 1 . Sci 0 ٨ 9 ? 5 EXP 8 EXP - 1 = 7. 97 x 10-6 7. 97 x 10-6 3 (-) (-) 6 7
  92. 92. 2.96 x10 = −4 2 (4 x10 ) −3 A B C D 7.24 x10 5 7.24 x10 4 1.85x10 5 1.85x10 4 2.96 x10 −3 16x10 2.96 -3(-8) = x10 16 5 = 0.185x10 −8 = 1.85x10 4
  93. 93. 2.96 x10 = −4 2 (4 x10 ) −3 5x MODE Sci 2 2 Sci 0 ٨ 9 ? 3 2 . 9 6 EXP (-) 3 ÷ ( 4 EXP (-) 4 ) x2 = 1. 85 x 1044 1. 85 x 10
  • cikwani2

    Feb. 17, 2017
  • beautyshoppe1

    Apr. 1, 2015

Chapter 1 standard form

Views

Total views

4,984

On Slideshare

0

From embeds

0

Number of embeds

25

Actions

Downloads

116

Shares

0

Comments

0

Likes

2

×