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JSS2: MATHEMATICS - 1ST TERM

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  1. Properties of Whole Numbers I | Week 1
    4Topics
    |
    2 Quizzes
  2. Properties of Whole Numbers II | Week 2
    4Topics
    |
    2 Quizzes
  3. Properties of Whole Numbers III | Week 3
    4Topics
    |
    1 Quiz
  4. Indices | Week 4
    2Topics
    |
    1 Quiz
  5. Laws of Indices | Week 5
    5Topics
    |
    1 Quiz
  6. Whole Numbers & Decimal Numbers | Week 6
    4Topics
    |
    1 Quiz
  7. Standard Form | Week 7
    3Topics
    |
    1 Quiz
  8. Significant Figures (S.F) | Week 8
    4Topics
    |
    1 Quiz
  9. Fractions, Ratios, Proportions & Percentages I | Week 9
    6Topics
    |
    1 Quiz
  10. Fractions, Ratios, Proportions & Percentages II | Week 10
    4Topics
    |
    1 Quiz
  11. Fractions, Ratios, Proportions & Percentages III | Week 11
    3Topics
    |
    1 Quiz
  12. Approximation & Estimation | Week 12
    1Topic
    |
    1 Quiz
Lesson 7, Topic 1
In Progress

Standard Index Form – Numbers Greater than 1

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Standard index form is a way of expressing numbers as a product of two terms: The first term being a number between 1 and 10 and the second part is a power of 10.

In general, it is expressed as \( \scriptsize \; a \; \times \; 10^n \!\!\;\)where a is a number between 1 and 10 an n is an integer which can be positive, negative or zero.

Numbers Greater Than 1:

Every number has a decimal point usually at the back of the last digit of the number. E.g. 15, the decimal point is after 5. Also, 706 the decimal point is after 6.

Example 1

Express these numbers in standard form 

a. 25 000 000

b. 5 000

c. 7

d. 270,000

e. 15

Solution

a. 25 000 000

Step 1: Move the decimal point to the back of the first significant digit 

decimal e1606114621821

Step 2: Count the number of movements made from the initial position of the decimal point.

=   2.5 x 107

b. 5 000

decim2 e1606114763181

= 5 x 103

c. 7

There is 0 movement

therefore the power of 10, in this case, is 0.

= 7 x 100

d.270,000

dec3 e1606114972111

=  2.7 x 105

e. 15

dec4 e1606115106484

1.5 x 101

Example 2

Express the following in standard form. 

a. 18.902

b. 40 830

c. 5123.59

d. 193.25

Solution

a. 18.902

new dec 4 e1606116867473

=  1.8902 x 101

b. 40 830

new dec 3 e1606117002286

=  4.0830 x 104

c. 5123.59

Screen Shot 2021 10 13 at 12.57.01 PM

= 5.12359 x 103

d. 193.25

NEW DEC e1606117135192

= 1.9325 x 102

Example 3

Change these standard forms to ordinary numbers.

a. 6 x 105

b. 2.3×104

c. 1.05×106

d. 7003 x104

Solution

a. 6 x 10

= 6 x 10 x 10 x 10 x 10 x 10

= 6 x 100 000 

= 600 000

b. 2.3 x 10

dec point 1 e1606119820985

= 23 000 

c. 1.05 x 10  

adec e1606120059371

= 1 050 000

d. 7003 x 104

adec2 e1606120284339

= 70 030 000

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