Question

Evaluate - {15/14}^-1 × {(12/15)^-1 ÷ (144/255)^-1 pls ans my question and pls write it in a notebook pls​

Answer 1

Answer:

32/15

Step-by-step explanation:

((15/4)^-1 ) * [((12/15)^-1)/((144/225)^-1)]

=(4/15)* [(15/12)/(225/144)]

=(4/15)* [(15/12)*(144/225)]

=(4/15)*[2160/2700]

=8640/40500

=32/15

Answer 2

$\underline{\underline{\sf{\maltese\:\:Solution}}}$

$\sf \dashrightarrow \: \left( \dfrac{15}{4} \right)^{ - 1} \times \left(\left( \dfrac{12}{15}\right)^{ - 1} \div \left( \dfrac{144}{225} \right)^{ - 1} \right)$

$\sf \dashrightarrow \: \dfrac{4}{15} \times \left(\dfrac{15}{12} \div \dfrac{225}{144} \right)$

$\sf \dashrightarrow \: \dfrac{4}{15} \times \left(\dfrac{ {15}}{12} \times \dfrac{144}{ {225} }\right)$

$\sf \dashrightarrow \: \dfrac{4}{15} \times \left(\dfrac{ {1}}{1} \times \dfrac{12}{ {15} }\right)$

$\sf \dashrightarrow \: \dfrac{4}{15} \times \left( \dfrac{1 \times 12}{ {1 \times 15} }\right)$

$\sf \dashrightarrow \: \dfrac{4}{15} \times \left( \dfrac{12}{ {15} }\right)$

$\sf \dashrightarrow \: \dfrac{4}{ 15} \times \dfrac{ {12}}{ {15} }$

$\sf \dashrightarrow \: \dfrac{4}{ {5} } \times \dfrac{ {4}}{ {15} }$

$\sf \dashrightarrow \: \dfrac{16}{ {75} }$

$\sf \dashrightarrow \:Answer = \underline{\boxed{ \sf \dfrac{16}{ {75} }} }$

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$\underline{\underline{\sf{\maltese\:\:Laws\:of\: Exponents :}}}$

$\sf \: 1^{st} \: Law = \bigg( \dfrac{m}{n} \bigg)^{a} \times \bigg( \dfrac{m}{n} \bigg)^{b} = \bigg( \dfrac{m}{n} \bigg)^{a + b}$

$\sf 2^{nd} \: Law =$

$\sf Case : (i) \: if \: a > b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \bigg( \dfrac{m}{n}\bigg)^{a - b}$

$\sf Case : (ii) \: if \: a < b \: then, \bigg( \dfrac{m}{n}\bigg) ^{a} \div \bigg( \dfrac{m}{n}\bigg)^{b} = \dfrac{1}{\bigg( \dfrac{m}{n}\bigg)^{b - a}}$

$\sf \: 3^{rd} \: Law = \bigg\{ \bigg( \dfrac{m}{n} \bigg)^{a} \bigg\}^{b} = \bigg( \dfrac{m}{n} \bigg)^{a \times b} =\bigg( \dfrac{m}{n} \bigg)^{ab}$

$\sf \: 4^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{ - 1} = \bigg( \dfrac{n}{m} \bigg) =\dfrac{n}{m}$

$\sf \: 5^{th} \: Law = \bigg( \dfrac{m}{n} \bigg)^{0} = 1$