Question

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

Answer 1

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

On LHS

( 2^30 + 2^29 + 2^28 ) ÷ ( 2^31 + 2^30 - 2^29 )

= 2^28 ( 2^2 + 2^1 + 2^0 ) ÷ 2^29 ( 2^2 + 2^1 + 2^0 )

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

= 7 ÷ 2^1 ( 5 )

= 7/ 2× 5

= 7/10

LHS = RHS

Hence proved

✌

Answer 2

Step-by-step explanation:

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

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

$\bf \underline{Solution-}$

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

$\textsf{Taking 2²⁸ Common }$

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

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

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

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

$\textsf{LHS = RHS}$

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