Write each of the following numbers in usual form:

Question:

Write each of the following numbers in usual form:

(i) 2.06 × 10−5

(ii) 5 × 10−7

(iii) 6.82 × 10−6

(iv) 5.673 × 10−4

(v) 1.8 × 10−2

(vi) 4.129 × 10−3

Solution:

(i) $2.06 \times 10^{-5}=\frac{206}{100} \times \frac{1}{10^{5}}=\frac{206}{10^{2} \times 10^{5}}=\frac{206}{10^{(5+2)}}=\frac{206}{10^{7}}=\frac{206}{10000000}=0.0000206$

(ii) $5 \times 10^{-7}=\frac{5}{10^{7}}=\frac{5}{10000000}=0.0000005$

(iii) $6.82 \times 10^{-6}=\frac{682}{100} \times \frac{1}{10^{6}}=\frac{682}{10^{2} \times 10^{6}}=\frac{682}{10^{(2+6)}}=\frac{682}{10^{8}}=\frac{682}{100000000}=0.00000682$

(iv) $5.673 \times 10^{-4}=\frac{5673}{1000} \times \frac{1}{10^{4}}=\frac{5673}{10^{3} \times 10^{4}}=\frac{5673}{10^{(3+4)}}=\frac{5673}{10^{7}}=\frac{5673}{10000000}=0.0005673$

(v) $1.8 \times 10^{-2}=\frac{18}{10} \times \frac{1}{10^{2}}=\frac{18}{10 \times 10^{2}}=\frac{18}{10^{(1+2)}}=\frac{18}{10^{3}}=\frac{18}{1000}=0.018$

(vi) $4.129 \times 10^{-3}=\frac{4129}{1000} \times \frac{1}{10^{3}}=\frac{4129}{10^{3} \times 10^{3}}=\frac{4129}{10^{(3+3)}}=\frac{4129}{10^{6}}=\frac{4129}{1000000}=0.004129$

 

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now