Question

me ajudem, isso é para agr​

Answer 1

Resposta:

Explicação passo a passo:

$1)$

$a)\:3^{4}$

$b)\:4^{33}$

$c)\:5^{3}$

$d)\:12^{2}$

$e)\:7^{3}$

$f)\:10^{6}$

$2)$

$a)\:32=2^{5}$

$b)\:\frac{1}{128}=2^{-7}$

$c)\:\sqrt{16}=4=2^{2}$

$d)\:0,125=\frac{125}{1000}=(\frac{5}{10}) ^{3} =(\frac{1}{2}) ^{3} =2^{-3}$

$3)$

  $[3^{12}:3^{9}] ^{-2} =[3^{3}] ^{-2} =3^{-6}$

$4)$

$4^{10}=(2^{2}) ^{10} =2^{20}$

$8^{-3} =(2^{3})^{-3} =2^{-9}$

$16^{-2} =(2^{4})^{-2} =2^{-8}$

$32=2^{5}$

$y=\frac{2^{20}\:.\:2^{-9}\:.\:2^{-8} }{2^{5} } =\frac{2^{3} }{2^{5} } =2^{-2}$

$y:2=\:?$

$(2^{-2}):(2)= 2^{(-2-1)} =2^{-3}$

$Resposta:(a)$

$5)$

$a)$

$3^{2}=9$

$3^{-1}=\frac{1}{3}$

$(\frac{1}{3}) ^{-1} =(\frac{3}{1}) ^{1} =3^{1}=3$

$3^{-2}=\frac{1}{9}$

$Colocando\:em\:ordem\:crescente...$

$\{\frac{1}{9};\frac{1}{3};3;9\}$

$b)$

$4^{\frac{1}{2} }=(2^{2})^{\frac{1}{2} } =2^{1}=2$

$4^{-\frac{1}{2} }=(2^{2})^{-\frac{1}{2} } =2^{-1} =\frac{1}{2}$

$4^{2}=16$

$4^{-3} =(2^{2}) ^{-3} =2^{-6} =\frac{1}{2^{6} }=\frac{1}{64}$

$Colocando\:em\:ordem\:crescente...$

$\{\frac{1}{64};\frac{1}{2};2;16\}$