Hint: Study we remaillues wTulugu3. Express the following in the form

1.	Hint: Study we remaillues wTulugu3. Express the following in the form where p and q are integers and q +0.0 0.6(ii) 0.47(ii) 0.001
Answer» (1) Let us assume that x=0.6 bar= 0.6666............------> (1) Now, we need to Multiply (1) with 10 10x = 10×( 0.666...) 10x=6.666... Subtracting (2) from (1)10x-x=6.6666.... - 0.6666...9x = 6 x= 9/6 = 2/3 (2). 0.47 bar = 0.4777777----> (1) Now we need to multiply with (1) with 10 10x = 10× (0.47777)... 10x= 4.7777.... Now we need to subtract (2) from (1) 10x-x = 4.7777...- 0.47777..... 9x = 4.3000 x= 4.3 /9 Therefore x = 43/90 3) 0.001 bar Let us assume that x= 0.001 = 0.001001...... Now let us multiply equation with 1000 1000 x = 1000× ( 0.001001...) Now we need to subtract (2) (1) 1000x - x = 1.001001...- 0.001001... 999x = 1 x = 1/999 ok thanks but explain thank u

Hint: Study we remaillues wTulugu3. Express the following in the form where p and q are integers and q +0.0 0.6(ii) 0.47(ii) 0.001

Answer»

(1) Let us assume that x=0.6 bar= 0.6666............------> (1)

Now, we need to Multiply (1) with 10

10x = 10×( 0.666...)

10x=6.666...

Subtracting (2) from (1)10x-x=6.6666.... - 0.6666...9x = 6

x= 9/6 = 2/3

(2). 0.47 bar = 0.4777777----> (1)

Now we need to multiply with (1) with 10

10x = 10× (0.47777)...

10x= 4.7777....

Now we need to subtract (2) from (1)

10x-x = 4.7777...- 0.47777.....

9x = 4.3000

x= 4.3 /9

Therefore x = 43/90

3) 0.001 bar

Let us assume that x= 0.001 = 0.001001......

Now let us multiply equation with 1000

1000 x = 1000× ( 0.001001...)

Now we need to subtract (2) (1)

1000x - x = 1.001001...- 0.001001...

999x = 1

x = 1/999

ok thanks but explain

thank u