(64)−23×(14)−2is equal to ?(64)−23×(14)−2is equal to ?\eqalign{ & {\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & {\text{is equal to ?}} \cr} " /> (64)−23×(14)−2is equal to ?(64)−23×(14)−2is equal to ?\eqalign{ & {\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & {\text{is equal to ?}} \cr} " /> (64)−23×(14)−2is equal to ?(64)−23×(14)−2is equal to ?\eqalign{ & {\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & {\text{is equal to ?}} \cr} " />

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$\begin{aligned} {(64)}^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ is equal to ? \end{aligned}$ $\eqalign{ & {\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & {\text{is equal to ?}} \cr}$

11
22
3¹/₂
4¹/₁₆

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 6 4^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = {(4^{3})}^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = 4^{- 2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2 - 2} \\ = {(\frac{1}{4})}^{0} \\ = 1 \end{aligned}

$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr}$

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", "text": "

\begin{aligned} {(64)}^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ is equal to ? \end{aligned}

$\eqalign{ & {\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & {\text{is equal to ?}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 6 4^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = {(4^{3})}^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = 4^{- 2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2 - 2} \\ = {(\frac{1}{4})}^{0} \\ = 1 \end{aligned}

$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 6 4^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = {(4^{3})}^{- \frac{2}{3}} \times {(\frac{1}{4})}^{- 2} \\ = 4^{- 2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2} \times {(\frac{1}{4})}^{- 2} \\ = {(\frac{1}{4})}^{2 - 2} \\ = {(\frac{1}{4})}^{0} \\ = 1 \end{aligned}

$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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