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]

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