Class 9th Math

class 9th math 2.5 solution

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Question#01: Evaluate \((i)\) \[{{i}^{7}}\] Solution: \[\begin{align}  & ={{i}^{7}} \\ & ={{i}^{6}}\times i \\ & ={{\left( {{i}^{2}} \right)}^{3}}\times i \\ & ={{\left( -1 \right)}^{3}}\times i \\ & =-1\times i \\ & =-i \\\end{align}\] \((ii)\) \[{{i}^{50}}\] Solution: \[\begin{align}  & ={{i}^{50}} \\ & ={{\left( {{i}^{2}} \right)}^{25}} \\ & ={{\left( -1 \right)}^{25}} \\ & =-1 \\  \end{align}\] \((iii)\) \[{{i}^{12}}\] Solution: \[\begin{align}  & ={{i}^{12}} \\ & ={{\left( {{i}^{2}} \right)}^{6}} … Read more

class 9th math 2.4 solution

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Question#1: Use the law of exponents to simplify. \((i)\) \[\frac{{{\left( 243 \right)}^{-\frac{2}{3}}}{{\left( 32 \right)}^{\frac{1}{5}}}}{\sqrt{{{\left( 196 \right)}^{-1}}}}\] Solution: \[\begin{aligned}&=\frac{{{\left( 3^5 \right)}^{-\frac{2}{3}}}{{\left( 2^5 \right)}^{-\frac{1}{5}}}}{\sqrt{{{\left( 14^2 \right)}^{-1}}}}\\&=\frac{{{\left( 3 \right)}^{-\frac{5\times 2}{3}}}\times{{\left( 2^\cancel{5} \right)}^{-\frac{1}{\cancel{5}}}}}{\sqrt{{{\left( 14^{-1} \right)}^{2}}}}\\&=\frac{{{\left( 3 \right)}^{-\frac{10}{3}}}\times{{\left( 2 \right)}^{-1}}}{\sqrt{{{\left( 14^{-1} \right)}^{\cancel{2}}}}}\\&=\frac{{{\left( 3 \right)}^{-\frac{10}{3}}}\times{{ 2}^{-1}}}{{{ 14^{-1}}}}\\&=\frac{14}{{{3}^{\frac{10}{3}}}\times 2}\\&=\frac{\cancelto{7}{14}}{3^{\frac{9}{3}}\times {{3}^{\frac{1}{3}}}\times\cancel{2} }\\&=\frac{7}{3^{\frac{\cancelto{3}{9}}{\cancel{3}}}\times {{3}^{\frac{1}{3}}}}\\&=\frac{7}{{{3}^{3}}\times \sqrt[3]{3}}\\&=\frac{7}{27 \sqrt[3]{3}}\end{aligned}\] \((ii)\) \[\left( 2{{x}^{5}}{{y}^{-4}} \right)\left( -8{{x}^{-3}}{{y}^{2}} \right)\] Solution: \[\begin{aligned}&=\left( … Read more