Chapter 1 standard form
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Chapter 1 standard form

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Chapter 1 standard form

Chapter 1 standard form

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Chapter 1 standard form Chapter 1 standard form Presentation Transcript

  • 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 1off 430 618 to the 700 Round nearest thousand 431 000
  • Round off 30 106 correct to the nearest hundred. 0 3 0106 0<5
  • 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
  • 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
  • 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)
  • 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)
  • 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
  • Before Getting started……  MODES Before starting a calculation, you must enter the correct mode as indicated in the table below MODE 1 MODE 2
  • MODE 3 MODE 1 MODE 2 MODE 2
  • Arithmetic Calculations Use the MODE key to enter the COMP when you want to perform basic calculations. MODE COMP 1 1
  • FIX, SCI, RND (Fix) : Number of Decimal Places 1 (Sci) 2 : Number of Significant Digits (Norm) : Exponential of significant Digits 3
  • Round off 5.316 to 1 decimal place 5x MODE Fix 1 1 Fix 0 ٨ 9 ? 1 5 . 3 = 5.3 1 6
  • 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
  • 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
  • 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
  • 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
  • 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
  • ositive numbers greater than 1 can be ounded off to a given number of significa gures
  • 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
  • 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
  • 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
  • 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
  • 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 )
  • 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 )
  • 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 )
  • 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
  • Example 1 Express 3.15 x 105 as a single number 3 . 1 5 = 315000 EXP 5
  • 3 . 1 5 EXP 5 5x = 3.15 x 105 Norm 1^2 ? MODE Norm 2 315000 3 3
  • Example 2 Express 4.23 x 10-4 as a single number 4 . 2 3 = 0.000423 EXP (-) 4
  • 4 . 2 3 EXP (-) 4 5x = 4.23 x 10-4 Norm 1^2 ? MODE Norm 2 0.000423 3 3
  • 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
  • Round off 30 106 correct to three significant figures. 0 3 0106 0<5
  • 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
  • Round off 0.05098 correct to three significant figures. 1 0 +1 0. 0 5098 8>5 00 0 . 0 51 9 8
  • 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
  •  To clear the Sci specification…… 5X Press MODE Norm 3 Norm 1 ⱱ 2 ? 3 1  To continue the Sci specification…… ON Press
  • Round off 0.0724789 correct to four significant figures. +1 8 0. 0 724789 8>5 8 0. 0 724789
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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 =
  • 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
  • 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
  • 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
  • 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 10 4 =3 . 4 0 5 0 Move the decimal point 4 places to the right =34050
  • Express 3.405 x 104 as a single number MODE COMP 1 1 3 . 4 0 = 34 050 5 EXP 4
  • 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
  • Express 7.53 x 10-4 as a single number 7 . 5 3 = 0.000753 EXP (-) 4
  • 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
  • 3.25 X 105 = 325000 -5 7.14 X 10 = 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 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
  • 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 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
  • 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 =
  • 0.0000398_ 3.98x10 −5 2.9 x10 − 0.29 x10 = (3.98 − 0.29) x10 = 3.69 x10 −5 −5 −5 −6
  • 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
  • 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
  • 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
  • 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 x10 −2 6 x10 7 .2 5 −( −2 ) = x10 6 7 = 1.2 x10 5
  • 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
  • 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 ÷
  • −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
  • −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
  • 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 6.4 x10 6 = 1.6x10 3 6.4 = x103 1.6 3 = 4 x10 m
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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