Question

Prove that (2^30+2^29+2^28)/(2^31+2^30-2^29)=7/10

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Answer 1

( 2^30 + 2^29 + 2^28 )/ ( 2^31 + 2^30 + 2^29 )

= 2^28 ( 2^2 + 2^1 + 1 ) / 2^29 ( 2^2 + 2^1 - 1 )

= 2^28 ( 4 + 2 + 1 ) / 2^29 ( 4 + 2 - 1 )

= 2^28 × 7 / 2^29 × 5

= 7 / 2^29 - 28 × 5

= 7/ 5× 2

= 7/10

Answer 2

Step-by-step explanation:

$\bf \underline{Prove\: that-}$

$\sf\dfrac{ {2}^{30} + {2}^{29} + {2}^{28} }{ {2}^{31} + {2}^{30} - {2}^{29} } = \dfrac{7}{10}$

$\bf \underline{Solution-}$

${~~~~~~:~~~\implies\sf\dfrac{ {2}^{30} + {2}^{29} + {2}^{28} }{ {2}^{31} + {2}^{30} - {2}^{29} } = \dfrac{7}{10} }$

$\textsf{Taking 2²⁸ Common }$

$\\{~~~~~~:~~~\implies\sf\dfrac{ {2}^{28}( {2}^{2} + {2}^{1} + {2}^{0} )}{ {2}^{28} ({2}^{3} + {2}^{2} - {2}^{1}) } = \dfrac{7}{10} }$

$\\{~~~~~~:~~~\implies\sf\dfrac{ \cancel{{2}^{28}}( 4 + 2 + 1 )}{ \cancel{{2}^{28} }(8 + 4 - 2) } = \dfrac{7}{10} }$

$\\{~~~~~~:~~~\implies\sf\dfrac{ 1 \times 7}{ 1(12 - 2) } = \dfrac{7}{10} }$

$\\{~~~~~~:~~~\implies\sf\dfrac{ 7}{ 10} = \dfrac{7}{10} }$

$\textsf{LHS = RHS}$

$\bf \underline{Hence\: proved.}$