Question

pls answer the following questions​

Answer 1

Answer:

$\underline{\boxed{\mathsf{ (1)\dfrac{217}{99}\:,(2) \dfrac{13}{990} and (3) \dfrac{2724}{900}}}}$

$\mathsf{Solution}$

$\mathsf{Here\: we \: are\:asked\:to\:covert\:the\:following\:term\:in \dfrac{p}{q}\: form}$

$\underline{\mathbf{\mathsf{ 1)\: 2.\overline{19}}}}$

$\underline{\mathsf{Therefore \implies }}$

$\mathsf{Let,\: }$

$\mathsf{x \: = \:2.19191919....}$ ------- eq (1)

$\implies \textsf{Multiplying eq (1) by 100 we have,}$

$\mathsf{100x \:=\: 219.1919191....}$ --------- eq (2)

$\implies \textsf{Subtracting eq(1) from eq(2) we have,}$

100x = 219.191919...

- x = 2.191919...

$\rule{90}{1}$

= 99x = 217.00

$\rule{90}{1}$

$\underline{\mathsf{that\:is}}$

$\boxed{\mathsf{x = \dfrac{217}{99}}}$

_______________________________________________

$\underline{\mathbf{2) \: 0.0 \overline{13}}}$

$\mathsf{Let,\: }$

$\mathsf{x \: = \:0.013131313....}$ ------- eq (1)

$\implies \textsf{Multiplying eq (1) by 10 we have,}$

$\mathsf{10x \:=\: 0.13131313....}$ --------- eq (2)

$\implies \textsf{Multiplying eq (2) by 100 we have,}$

$\mathsf{1000x \:=\: 13.131313....}$ --------- eq (3)

$\implies \textsf{Subtracting eq(2) from eq(3) we have,}$

1000x= 13.131313...

- 10x = 0.131313...

$\rule{90}{1}$

= 990x = 13.00

$\rule{90}{1}$

$\underline{\mathsf{That\:is}}$

$\boxed{\mathsf{x = \dfrac{13}{990}}}$

_______________________________________________

$\underline{\mathbf{2) \: 3.02 \overline{6}}}$

$\mathsf{Let,\: }$

$\mathsf{x \: = \:3.0266666666...}$ ------- eq (1)

$\implies \textsf{Multiplying eq (1) by 100 we have,}$

$\mathsf{100x \:=\: 302.66666666...}$ --------- eq (2)

$\implies \textsf{Multiplying eq (2) by 10 we have,}$

$\mathsf{1000x \:=\: 3026.6666666...}$ --------- eq (3)

$\implies \textsf{Subtracting eq(2) from eq(3) we have,}$

1000x = 3026.6666...

- 100x = 302.6666...

$\rule{90}{1}$

=900x = 2724.00

$\rule{90}{1}$

$\underline{\mathsf{That\:is}}$

$\boxed{\mathsf{x = \dfrac{2724}{900}}}$