Question

Find the valur of (1/4)-2-3(8)2/3(4)0+(9/16)-1/2

Answer 1

Question:

(1/4)^-2 - 3(8)^⅔ (4)^0 + (9/16)^-1/2

Solution :

Given :

(1/4)^-2 - 3(8)^⅔ (4)^0 + (9/16)^-1/2

To find :

• Simplify the expression =?

Step-by-step explanation :

(1/4)^-2 - 3(8)^⅔ (4)^0 + (9/16)^-1/2

4^2 - 3(2^3)^⅔ × 1 + (3/4)^2 × (-1/2)

16 - 3 × 2^2 + (3/4)^-1

16 - 12 + 4/3

4 + 4/3

12 + 3/3

16/3.

Therefore, (1/4)^-2 - 3(8)^⅔ (4)^0 + (9/16)^-1/2 = 16/3

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\textsf{Given that,}$

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

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

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

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

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

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

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

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

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

Hope this helps!