FURTHER PROBLEMS ON STANDARD FORMS

FURTHER PROBLEMS ON STANDARD FORMS

PROBLEM. FIND THE VALUE OF:

(a) 7.9×10⁻² −5.4×10⁻²

(b) 8.3×10³ +5.415×10³ and

(c) 9.293×10² +1.3×10³

expressing the answers in standard form. Numbers having the same exponent can be added or subtracted by adding or subtracting the mantissa and keeping the exponent the same. Thus:

(a) 7.9×10⁻² −5.4×10⁻²

=(7.9−5.4)×10⁻² =2.5×10⁻²

(b) 8.3×10³ +5.415×10³

=(8.3+5.415)×10³ =13.715×10³

=1.3715×10⁴ in standard form

(c) Since only numbers having the same exponents can be added by straight addition of the mantissa, the Numbers are converted to this form before adding.

Thus:

9.293 × 10² + 1.3 × 10³

= 9.293 × 10² + 13 × 10²

= (9.293 + 13) × 10²

= 22.293 × 10² = 2.2293×10³ in standard form.

Alternatively, the numbers can be expressed as decimal fractions, giving:

9.293 × 102 + 1.3 × 103

= 929.3 + 1300 = 2229.3

= 2.2293×10³  in standard form as obtained previously. This method is often the ‘safest ‘way of doing this type of problem.

PROBLEM. Evaluate

(a) (3.75×10³)(6×10⁴)

and (b) 3.5×10⁵ /7×10²

expressing answers in standard form

(a) (3.75×10³)(6× 10⁴)=(3.75 × 6)(10³⁺⁴)

=22.50×10⁷

=2.25×10⁸

(b)

3.5×10⁵/7×10²

= 3.5/7×10⁵⁻²

=0.5×10³ =5×10²

NOW PRACTICES THIS

In Problems 1 to 4, find values of the expressions given, stating the answers in standard form

1. (a) 3.7×10² +9.81×10²

(b) 1.431×10⁻¹ +7.3×10⁻¹

[(a) 1.351×10³ (b) 8.731×10⁻¹]

2. (a) 4.831×10² +1.24×10³

(b) 3.24×10⁻³ −1.11×10⁻⁴

[(a) 1.7231×10³ (b) 3.129×10⁻³]

3. (a) (4.5×10⁻²)(3×10³)

(b) 2×(5.5×10⁴)

[(a) 1.35×10² (b) 1.1×10⁵]

4. (a)6 × 10⁻³/3 × 10⁻⁵

 (b)(2.4 × 10³)(3 × 10⁻²)(4.8 × 10⁴)

[(a) 2×10² (b) 1.5×10⁻³]

ALWAYS STUDY TO SHOW YOUR SELF APPROVED LEAVE A COMMENT!!!!!

Read More

NUMERIC PROCESS 2 (STANDARD FORMS)

NUMERIC PROCESS 2 (STANDARD FORMS)

A number written with one digit to the left of the decimal point and multiplied by 10 raised to some power is said to be written in Standard Form. Thus: 5837 is written as 5.837×10^3 in standard form, and 0.0415 is written as 4.15×10⁻² in standard form. When a number is written in standard form, the first factor is called the Mantissa and the second factor is called the EXPONENT. Thus the number 5.8 × 10³ has a mantissa of 5.8 and an exponent of 10³.

(i) Numbers having the same exponent can be added or subtracted in standard form by adding or subtracting the mantissa and keeping the exponent

the same.

 Thus:

2.3 × 10⁴ + 3.7 × 10⁴

= (2.3 + 3.7) × 10⁴ = 6.0 × 10⁴

and 5.9 × 10⁻² − 4.6 × 10⁻²

= (5.9 − 4.6) × 10⁻² = 1.3 × 10⁻²

When the numbers have different exponents, one way of adding or subtracting the numbers is to express one of the numbers in non-standard form, so that both numbers have the same exponent.

Thus:

2.3 × 10⁴ + 3.7 × 10³

= 2.3 × 10⁴ + 0.37 × 10⁴

= (2.3 + 0.37) × 104 = 2.67 × 104

Alternatively,

2.3 × 10⁴ + 3.7 × 10³

= 23 000 + 3700 = 26 700

= 2.67 × 10⁴

(ii) The laws of indices are used when multiplying or dividing numbers given in standard form. For example,

(2.5 × 10³) × (5 × 10²)

= (2.5 × 10⁵) × (10³⁺²)

= 12.5 × 10⁵ or 1.25 × 10⁶

Similarly,

6 × 10⁴

1.5 × 10²

=

 6

1.5× (10⁴⁻²) = 4 × 10²

2.5 Worked problems on standard

Problem  

Express in standard form:

(a) 38.71 (b) 3746 (c) 0.0124

For a number to be in standard form, it is expressed with only one digit to the left of the decimal point. Thus:

(a) 38.71 must be divided by 10 to achieve one digit to the left of the decimal point and it must also be multiplied by 10 to maintain the equality, i.e.

38.71 =

38.71    ×10 = 3.871×10 in standard form

  10       

(b) 3746=

3746

1000    ×1000=3.746×10³ in standard form

(c) 0.0124=0.0124× 100

100

1.24

100     =1.24×10⁻² in standard form

Problem 15. Express the following numbers, which are in standard form, as decimal numbers:

(a) 1.725×10−2 (b) 5.491×104 (c) 9.84×100

(a) 1.725×10⁻² =

1.725

100         =0.01725

(b) 5.491×10⁴ =5.491×10 000= 54 910

(c) 9.84×10⁰ =9.84×1=9.84 (since 10⁰ =1)

Read More