Question

exponents
$( \frac{4}{1} ) ^{2} - 3 \times 8 \times \frac{2}{3} \times 4 ^{0} + ( \frac{16}{9} ) \frac{1}{2}$
solve this sum please​

Answer 1

Answer:

8/9 = 0.88

Step-by-step explanation:

$(\frac{4}{1} )^{2} -3*8*\frac{2}{3} *4^{0} +(\frac{16}{9} )\frac{1}{2} \\\\16-(1)*(8)*\frac{2}{1} *1+(\frac{8}{9} )\frac{1}{1} \\\\16-16+\frac{8}{9}\\\\0+\frac{8}{9}\\\\\frac{8}{9}\\\\0.88$

Answer 2

$\bf \underline{Solution-}$

$\textsf{Given that,}$

$\sf{ \bigg(\frac { 1} { 4 } \bigg ) ^ { - 2 } - 3 ( 8 ) ^ \frac { 2 }{3} \times ( {4})^{0} + \bigg( \frac {9} { 16 } \bigg ) ^{ - \frac{1}{2} } }$

$\sf{ = \bigg(\frac { 4 } { 1 } \bigg ) ^ { 2 } - 3 ( {2}^{3} ) ^ \frac { 2 }{3} \times ( 1 ) + \bigg( \frac { 16} { 9 } \bigg ) ^{ \frac{1}{2} } }$

$\sf{ = 16 - 3 \times {2}^{ \cancel3 \times \frac { 2 }{ \cancel3}} + \left \{\bigg( \frac { 4} {3} \bigg ) ^{ 2} \right \} ^{ \frac{1}{2} } }$

$\sf{ = 16 - 3 \times {2}^{2} + \bigg( \frac { 4} {3} \bigg ) ^{ \cancel2 \times \frac{1}{ \cancel2} } }$

$\sf{ = 16 - 3 \times 4 + \frac { 4} {3} }$

$\sf{ = 16 - 12+ \frac { 4} {3} }$

$\sf{ = 4+ \frac { 4} {3} }$

$\sf{ = \frac { 16} {3} }$

$\sf{ =5 \frac { 1} {3} } \: \bf{Ans}.$