Question

Помогитееееееееееееее добрыееее люди, я вас умоляююююююю

Answer 1

$6^{-2} = \frac{1}{6^2} =[tex] \frac{1}{36}$
$6^{0} = 1$
$2^{0} =1$
$(- 5)^{-2} = \frac{1}{-5^2} = \frac{1}{-25}$
$3^{0} =1$
$1^{-89} = [tex] \frac{1}{1^8^9} = 1$
$0,3^{-2} = \frac{1}{0,3^2} = \frac{1}{0,09}$ 
$0,4^{-1} = [tex] \frac{1}{0,4}$
$0,6^{-2} = \frac{1}{0,6^2} = \frac{1}{0,36}$
$( \frac{3}{5} )^-^3= (\frac{5}{3} )^3$
$( \frac{1}{2} )^-^6= (\frac{2}{1} )^6= 2^{6} =64$
$( \frac{4}{5} )^-^1= (\frac{5}{4} )^1$
$4^{-3} = \frac{1}{4^3} = \frac{1}{64}$
$(-1)^{-90} = \frac{1}{1^9^0} =1$
$(-10)^{-5} = \frac{1}{-10^5} = -100000$
$-100^{-5} = \frac{1}{100^5} = - 10000000000$
$0,9^{-2} = \frac{1}{0,9^2} = \frac{1}{0,81}$
$0,7^{-2} = \frac{1}{0,7^2} = \frac{1}{0,49}$
$(- \frac{1}{4} )^-^3= (-\frac{4}{1} )^3= -4^{3}=-64$
$(-\frac{9}{7} )^-^2= (-\frac{7}{9} )^2$
$6n^{-3} =6 \frac{1}{n^3}$
$3n^{-2} =3 \frac{1}{n^2}$
$2x^{-2} = 2 \frac{1}{x^2}$
$2x^{-2} y^{4} = 2 \frac{1}{x^2y^4}$ 
$x^{-2} = \frac{1}{x^2}$
$( 2b^{4} )^-^1= \frac{1}{(2b^4)}$

Answer 2

6^-2= 1/6; 6^0=1; 2^0=1; (-5)^-2=1/25; 3^0=1; 1^-89=1; 0.3^-2=1/0.09; 0.4^-1= 1/0.4; 0.6^-2=1/0.36; (3/5)^-3= 125/27; (1/2)^-6= 64; (4/5)^-1=5/4