Question

*Find the value of:
1
$( {3}^{0} + {4}^{ - 1} ) \times {2}^{2}$
2
$( {2}^{ - 1} \times {4}^{ - 1} ) \div {2}^{ - 2}$
3
$( \frac{1}{ 2} )^{ - 2} + (\frac{1}{3} )^{ - 2} + ( \frac{1}{4} )^{ - 2}$
4
$( {3}^{ - 1} + {4}^{ - 1} + {5}^{ - 1} ) ^{0}$
5
$(( \frac{ - 2}{3} )^{ - 2} ) ^{2}$
Find value
​

Answer 1

Step-by-step explanation:

Answer 1:-

$\[ \begin{array}{l} \bf \left(3^{0}+4^{-1}\right) \times 2^{2} \\ \\ \sf =\left(1+\dfrac{1}{4}\right) \times 4 \\ \\ \sf =\left(\dfrac{4+1}{4}\right) \times 4 \\ \\ \sf =\dfrac{5}{4} \times 4 \\ \\ \boxed{ \red{=5 }}\end{array} \]$

Answer 2:-

$\[ \begin{array}{l} \tt \left(2^{-1} \times 4^{-1}\right) \div 2^{-2} \\ \\ \tt =\left\{2^{-1} \times\left(2^{2}\right)^{-1}\right\} \div 2^{-2} \\ \\ \tt=\left\{2^{-1} \times 2^{2} \times(-1)\right\} \div 2^{-2} \\ \\ \tt =\left(2^{-1} \times 2^{-2}\right) \div 2^{-2} \\ \\ \tt =2^{(-1)} \times 2^{(-2)} \div 2^{-2} \\ \\ \tt =2^{-3} \div 2^{-2} \\ \\ \tt =\dfrac{2^{-3}}{2^{-2}} \\ \\ \tt =\dfrac{1}{2^{(-2)-(-3)}} \\ \\ \tt =\dfrac{1}{2^{-2+3}} \\ \\ \tt =\dfrac{1}{2^{1}} \\ \\ \boxed{\red{ \tt=\dfrac{1}{2}}} \end{array} \]$

Answer 3:--

$\begin{aligned}\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} &=(2)^{2}+(3)^{2}+(4)^{2} \\ \\ & \sf=4+9+16 \\ \\ & \pink{ \pmb{ \boxed{\bf=29 }}}\end{aligned}$

Answer 4:-

We know that any number(except 0 ) with a power of zero is equal to one. And,

$\displaystyle\sf a^{-1}=\frac{1}{a}, a \neq 0$

So,

$\[ \begin{array}{l} \displaystyle \rm \left(3^{-1}+4^{-1}+5^{-1}\right)^{0} \\ \\ \displaystyle \rm=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)^{0} \\ \\ \boxed{\orange{ =1 }}\end{array} \]$

Answer 5:-

$\pmb{\begin{aligned}\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^{2} &=\left\{\left(-\frac{3}{2}\right)^{2}\right\}^{2} \\ \\ & \sf=\left(\frac{9}{4}\right)^{2}\\ \\ & \boxed{\green{ \bf=\frac{81}{16}}} \end{aligned}}$