Question

Evaluate: (5/9)-²×(3/5)-³×(3/5)°​

Answer 1

QUESTION:-

Evaluate: $\sf\therefore \bigg( \dfrac{9}{5}^{-2} \bigg) \times \bigg ( \dfrac{5}{3}^{-3} \bigg) \times \bigg( \frac{3}{5} \bigg)^0$

ANSWER✔

$\sf\therefore \bigg( \dfrac{9}{5}^{-2} \bigg) \times \bigg( \dfrac{5}{3}^{-3} \bigg) \times \bigg( \frac{3}{5} \bigg)^0$

$\Large\bold{now,}$

$\sf\underline{\sf{using\:law\:of\:exponents,}}$

$\sf\dashrightarrow negative\:exponent\:= a^{-x} = \dfrac{1}{a^x}$

$\sf\implies \bigg( \dfrac{5}{9} \bigg)^{-2} = \bigg( \dfrac{9}{5} \bigg)^{2} \:........from\: negative\:exponent\:law$

$\sf\implies \bigg( \dfrac{3}{5} \bigg)^{-3} = \bigg( \dfrac{5}{3} \bigg)^{3} \: ........from\: negative\:exponent\:law$

$\sf\dashrightarrow zero\:exponent\:= a^{0}= 1$

$\sf\implies \bigg( \frac{3}{5} \bigg)^0=1\:.....from\:zero\:exponent\:law$

$\Large\bold{solution,}$

$\sf\implies \bigg[ \bigg( \dfrac{9}{5} \bigg)^2 \times \bigg( \dfrac{5}{3} \bigg)^3 \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{81}{25} \bigg) \times \bigg( \dfrac{125}{27} \bigg) \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{81}{25} \bigg) \times \bigg( \dfrac{125}{27}\bigg) \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{\cancel {81}}{\cancel {25}} \bigg) \times \bigg( \dfrac{\cancel {125}}{\cancel {27}}\bigg) \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{\cancel {27}}{\cancel {5}} \bigg) \times \bigg( \dfrac{\cancel {25}}{\cancel {9}} \bigg) \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{9}{1} \bigg) \times \bigg( \dfrac{5}{3}\bigg) \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{\cancel {9}}{1} \bigg) \times \bigg( \dfrac{5}{\cancel {3}}\bigg) \bigg] \times 1$

$\sf\implies 3 \times 5 \times 1$

$\large{\fbox{ =15}}$

_________________________

Answer 2

$\rule{300}3$

UR QUESTION:-

$\huge {\mathbf {Q.}}$Evaluate: $\sf\therefore \bigg( \dfrac{9}{5}^{-2} \bigg) \times \bigg ( \dfrac{5}{3}^{-3} \bigg) \times \bigg( \frac{3}{5} \bigg)^0$

$\rule{300}3$

UR ANSWER✓

$\sf\therefore \bigg( \dfrac{9}{5}^{-2} \bigg) \times \bigg( \dfrac{5}{3}^{-3} \bigg) \times \bigg( \frac{3}{5} \bigg)^0$

$\Large\bold{now,}$

$\underline{\sf{using\:law\:of\:exponents,}}$

$\sf\dashrightarrow negative\:exponent\:= a^{-x} = \dfrac{1}{a^x}$

$\sf\implies \bigg( \dfrac{5}{9} \bigg)^{-2} = \bigg( \dfrac{9}{5} \bigg)^{2} \:........from\: negative\:exponent\:law$

$\sf\implies \bigg( \dfrac{3}{5} \bigg)^{-3} = \bigg( \dfrac{5}{3} \bigg)^{3} \: ........from\: negative\:exponent\:law$

$\sf\dashrightarrow zero\:exponent\:= a^{0}= 1$

$\sf\implies \bigg( \frac{3}{5} \bigg)^0=1\:...............from\:zero\:exponent\:law$

$\Large\bold{solution,}$

$\sf\implies \bigg[ \bigg( \dfrac{9}{5} \bigg) }^2 \times \bigg( \dfrac{5}{3} \bigg) }^3 \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{81}{25} \bigg) } \times \bigg( \dfrac{125}{27} \bigg) } \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{81}{25} \bigg) } \times \bigg( \dfrac{125}{27}\bigg) } \bigg] \times 1$

$\sf\implies \bigg[ \cancel \bigg( \dfrac{81}{25} \bigg) } \times \cancel\bigg( \dfrac{125}{27}\bigg) } \bigg] \times 1$

$\sf\implies \bigg[ \cancel \bigg( \dfrac{27}{5} \bigg) } \times \cancel \bigg( \dfrac{25}{9}\bigg) } \bigg] \times 1$

$\sf\implies \bigg[ \bigg( \dfrac{9}{1} \bigg) } \times \bigg( \dfrac{5}{3}\bigg) } \bigg] \times 1$

$\sf\implies \bigg[ \cancel \bigg( \dfrac{9}{1} \bigg) } \times \cancel \bigg( \dfrac{5}{3}\bigg) \bigg] \times 1$

$\sf\implies 3 \times 5 \times 1$

$\huge{\fbox{ =15}}$

$\rule{300}3$