2 Answers

Jmargolisvt · Answer 1 · 2022-08-06T03:25:26+0000

$\red{➤}\:$
$\sf ({(3)/(2)})^(-1)÷ ({(-2)/(5)})^(-1)$

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Since the power is in negetive,we write the reciprocal of the number and then solve it like positive exponents-

$\begin{gathered}\\\quad\longrightarrow\quad\sf( {(3)/(2)})^(-1)÷ ({(-2)/(5)})^(-1)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf ({(2)/(3)})^(1)÷ ({(5)/(-2)})^(1)\quad (a^1=a)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (2)/(3)÷ (5)/((-2))\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (2)/(3)×((-2))/(5)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\sf (2×(-2))/(3×5)\\\end{gathered}$

$\begin{gathered}\\\quad\longrightarrow\quad\boxed{\sf{ (-4)/(15)}}\\\end{gathered}$

Laws of Exponents-

$\sf a^m×a^n = a^(m+n) \\ \sf a^m/a^n = a^(m-n) \\ \sf{(a^m)}^n = a^(mn) \\ \sf a^n/b^n = (a/b)^n \\ \sf a^0 = 1 \\ \sf a^(-m )= 1/a^m$

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