Question

1 – Marque a segunda coluna de acordo com a primeira:
( A) Todo numero elevado a zero é ( ) Nove
(B) Todo numero elevado a um é ( ) Um
(C) No caso de potencia de potencia ( ) vinte e sete
(D) Três elevado a dois é: ( ) Multiplica-se os
expoentes
(E) 3³ é ( ) È igual a ele mesmo

2 – Transforme os produtos em potencia:
a) 5.5.5.5.5.5.5.5.5.5.=
b) 10.10.10.10.10 =
c) 2.2.2.2.2.2.2.2.2.2.2.2.2.2.2=
d) 3.3.3.3 =
e) (0,5).(0,5).(0,5).(0,5) =
f)).(1/2).(1/2).(1/2) =

3 – Utilizando as propriedades das potencias resolva as operações:
a) 3 3 . 3 7 : 3 6 =
b) 10 8 : 10 -4 .10 -10 =
c) 5 -10 : 5 12 : 5 22 =
d) 2 7 . 2 9 : 2 2 =
e) (1/6) 12 . (1/2) -8 : (1/2) 0 =
f) 5 4 . (1/5 3 ) : 5 1 =
g) 9 23 : 9 -20 : 9 -47 =

Answer 1

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$\boxed{\begin{array}{l}\underline{\tt1)~ Marque~a~segunda~coluna~de~acordo~com~a~primeira\!:}\\\sf (A)~todo~n\acute umero~elevado~a~zero~\acute e~~~~~~~(D)~nove\\\sf(B)~Todo~n\acute umero~elevado~a~um~\acute e~~~~~~~~(A)~Um\\\sf(C)~No~caso~de~pot\hat encia~de~pot\hat encia~~~(E)~vinte~e~sete\\\sf(D)~Tr\hat es~elevado~a~dois~\acute e:~~~~~~~~~~~~~(C) ~Multiplica~os~expoentes\\\sf(E)3^3~\acute e~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)~\acute E~igual~a~ele~mesmo\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 2)~Transforme~os~produtos~em~pot\hat encia:\\\sf a)5\cdot5\cdot5\cdot5\cdot5\cdot5\cdot5\cdot5\cdot5\cdot5=5^{10}\\\sf b)10\cdot10\cdot10\cdot10\cdot10\cdot10=10^5\\\sf c)2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2=2^{15}\\\sf d)3\cdot3\cdot3\cdot3=3^4\\\sf e)(0,5)\cdot(0,5)\cdot(0,5)\cdot(0,5)=(0,5)^4\\\sf f)\bigg(\dfrac{1}{2}\bigg)\cdot\bigg(\dfrac{1}{2}\bigg)\cdot\bigg(\dfrac{1}{2}\bigg)=\bigg(\dfrac{1}{2}\bigg)^3\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm 3)~ Utilizando~as~propriedades~das~pot\hat encias}\\\underline{\rm~resolva~as~operac_{\!\!,}\tilde oes:}\\\sf a)3^3\cdot3^7\div3^6=3^{3+7-6}=3^4=81\\\sf b)10^8\div10^{-4}\cdot10^{-10}=10^{8-(-4)+(-10)}\\\sf=10^{8+4-10}=3^2=9\\\sf c)5^{-10}\div5^{12}\div5^{22}=5^{-10-12-22}=5^{-44}\\\sf d)2^7\cdot2^9\div2^2=2^{7+9-2}=2^{14}\end{array}}$$\Large\boxed{\begin{array}{l}\sf e)\bigg(\dfrac{1}{2}\bigg)^{12}\cdot\bigg(\dfrac{1}{2}\bigg)^{-8}\div\bigg(\dfrac{1}{2}\bigg)^0=\bigg(\dfrac{1}{2}\bigg)^{12+8}=\bigg(\dfrac{1}{2}\bigg)^{20}\\\sf f) 5^4\cdot\bigg(\dfrac{1}{5^3}\bigg)\div5^1=5^4\cdot5^{-3}\div5^1\\\sf=5^{4-3-1}=5^0=1\\\sf g)9^{23}\div9^{-20}\div9^{-47}=9^{23-[-20]-[-47]}\\\sf9^{23+20+47}=9^{90}\end{array}}$