Question

please tell the answer fast​

Answer 1

Answer:

proved

Step-by-step explanation:

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Answer 2

Prove

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

Solutions

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

• Taking 2²⁸ Common

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

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

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

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

• $\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \bigstar \: \underline{\bf{\blue{~~~~~~~~~Don't~ Study~ More~~~~~~~~~~~\bigstar}}}\\ {\boxed{\begin{array}{cc} \sf \bigstar{a}^{m} \times {a}^{n} = {a}^{m + n} \\ \\ \sf \bigstar{a}^{m} \div {a}^{n} = {a}^{m - n} \\ \\ \sf{\bigstar( {a}^{m} ) ^{n} = {a}^{mn} } \\ \\ {\bigstar\sf a {}^{m} \times {n}^{m} = (ab) ^{m} } \ \\\\ \ \sf\bigstar{a}^{0} = 1 \\ \\ \sf \bigstar \: {\dfrac{ {a}^{m} }{ {b}^{m} }= \left( \dfrac{a}{b} \right) ^{m} } \\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$