Math, asked by monukumar14719, 9 months ago

Solve this in detail and wrote how ​

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Answered by Anonymous
35

Answer:

\sf\large a)\:\:\bigg\lgroup\dfrac{-\:3}{7}\bigg\rgroup^2\div\bigg\lgroup\dfrac{7}{3}\bigg\rgroup^{\large\frac{12}{-\:4}}\\\\\\:\implies\sf\large \dfrac{9}{49} \div \bigg\lgroup\dfrac{7}{3}\bigg\rgroup^{-3}\\\\\\:\implies\sf\large \dfrac{9}{49} \div \bigg\lgroup\dfrac{3}{7}\bigg\rgroup^{3}\\\\\\:\implies\sf\large \dfrac{9}{49} \times \bigg\lgroup\dfrac{7}{3}\bigg\rgroup^{3}\\\\\\:\implies\sf\large \dfrac{9}{49} \times \dfrac{49 \times 7}{9 \times 3}\\\\\\:\implies\sf\large \dfrac{7}{3}

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\sf\large b)\:\:(3)^5 \times \bigg\lgroup\dfrac{5}{3}\bigg\rgroup^{7}\\\\\\:\implies\sf\large (3)^5 \times \dfrac{(5)^7}{(3)^7}\\\\\\:\implies\sf\large \dfrac{(5)^7}{(3)^2}

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\sf\large c)\:\:\bigg\lgroup\dfrac{1}{4^{-3}}\bigg\rgroup^{3}\\\\\\:\implies\sf\large(4^{-(-3)})^3\\\\\\:\implies\sf \large(4^3)^3\\\\\\:\implies\sf \large(4)^{(3 \times 3)}\\\\\\:\implies\sf \large(4)^{9}

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\sf\large d)\:\:2^{-\:6} \times (-\:7)^{-\:2}\\\\\\:\implies\sf\large\dfrac{1}{(2)^6} \times \dfrac{1}{(-\:7)^2}\\\\\\:\implies\sf\large\dfrac{1}{(2)^6} \times \dfrac{1}{49}

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\sf\large e)\:\:(2^{-\:7}\div2^{-\:9}) \times 2^{-4}\\\\\\:\implies\sf \large\bigg\lgroup\dfrac{1}{2^7} \div \dfrac{1}{2^9}\bigg\rgroup \times \dfrac{1}{2^4}\\\\\\:\implies\sf \large\bigg\lgroup \dfrac{1}{2^7} \times 2^9\bigg\rgroup \times \dfrac{1}{2^4}\\\\\\:\implies\sf \large2^2\times \dfrac{1}{2^4}\\\\\\:\implies\sf\large \dfrac{1}{2^2}\\\\\\:\implies\sf \large \dfrac{1}{4}

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\sf\large f)\:\:6^{-\:3} \times (-\:5)^{-\:3}\\\\\\:\implies\sf\large\dfrac{1}{6^3} \times \dfrac{1}{(-\:5)^3}

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\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}

Answered by bhanuprakashreddy23
3

Step-by-step explanation:

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