Question

" Obrigada pela atenção "

Aplicando as propriedades da potenciação escreva clara expressão em uma muito potência.a) 9⁹ × 9⁹ b) (20⁴)³ c) 10⁷ ÷ 10⁵ d) (8¹⁰)³ e) (0,7)⁴]⁵ f) [(2,5)⁴]⁵ g) (1,9)¹²÷( 1,9 )¹⁰ h) ( ⅖ ) ¹⁴÷(⅖)² i) (⅕)⁵ × (⅕)⁴ × (⅕)​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{a)\: 9^9 \times 9^9 = 9^{(9 + 9)} = 9^{18}}$

$\mathsf{b)\: (20^4)^3 = 20^{(4 \times 3)} = 20^{12}}$

$\mathsf{c)\: 10^7 \div 10^5 = 10^{(7 - 5)} = 10^2}$

$\mathsf{d)\: (8^{10})^3 = 8^{(10 \times 3)} = 8^{30}}$

$\mathsf{e)\: [(0,7)^4]^5 = (0,7)^{(4 \times 5)} = (0,7)^{20}}$

$\mathsf{f)\: [(2,5)^4]^5 = (2,5)^{(4 \times 5)} = (2,5)^{20}}$

$\mathsf{g)\: (1,9)^{12} \div (1,9)^{10} = (1,0)^{(12 - 10)} = (1,9)^2}$

$\mathsf{h)\: (2/5)^{14} \div (2/5)^{2} = (2/5)^{(14 - 2)} = (2/5)^{12}}$

$\mathsf{i)\: (1/5)^{5} \times (1/5)^{4} \times (1/5) = (1/5)^{(5 + 4 + 1)} = (1/5)^{10}}$

Answer 2

Olá!!!

Resposta...

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a) 9⁹ × 9⁹ = 9(⁹+⁹) = 9¹⁸

b) (20⁴)³ = 20(⁴ ˣ ³) = 20¹²

c) 10⁷ ÷ 10⁵ = 10(⁷⁻⁵) = 10²

d) (8¹⁰)³ = 8(¹⁰ ˣ ³) = 8³⁰

e) (0,7)⁴]⁵ = (0,7)⁽⁴ˣ⁵⁾ = (0,7)²⁰

f) [(2,5)⁴]⁵ = (2,5)⁽⁴ˣ⁵⁾ = (2,5)²⁰

g) (1,9)¹²÷( 1,9 )¹⁰ = (1,0)⁽¹²⁻¹⁰⁾ = (1,0)²

h) ( ⅖ ) ¹⁴÷(⅖)² = (⅖)⁽¹⁴⁻²⁾ = (⅖)¹²

i) (⅕)⁵ × (⅕)⁴ × (⅕)​ = (⅕)​⁽⁵⁺⁴⁺¹⁾ = (⅕)​¹⁰