Math, asked by haashika, 1 year ago

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

Answers

Answered by dhruv0002
4

(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[7] / 2^29[5]

= 7/(2×5)

= 7/10, hence proved

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dhruv0002: please mark it as brainliest if it helps :)
dhruv0002: kindly mark the answer as brainliest, it will help me :)
Answered by Salmonpanna2022
1

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.} \\

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