• Matéria: Matemática
  • Autor: lourinho74838
  • Perguntado 4 anos atrás

Calcule a potenciação com expoente inteiro negativo:

a) 11-¹ =

b) (√7)-² =

c) (1-4) -³ =

d) 2 -⁵ =

e) 0,05-⁶ =

prfvr me ajudem

Respostas

respondido por: Helvio
4

                       Potência com expoente negativo

Quando uma potência possui expoente negativo, transforme em fração.

Formula:

                              x^{-n} ~~~ = ~~~(\dfrac{1}{y}) ^n ~~= ~~\dfrac{1}{x^n}

===

a) ~  11^{-1} \\ \\ =>  \dfrac{1}{11}

===

b) ~~(\sqrt{7})^{-2}\\\\ \\ ( \dfrac{1}{\sqrt{7}})^2\\ \\ \\ \dfrac{1^2}{\not\sqrt{7}^\not^2} \\ \\\\=>  \dfrac{1}{7}

===

c) ~~( 1 - 4)^{-3}\\ \\ \\ ( -3)^{-3}\\ \\ \\\dfrac{-1}{3^3} \\ \\ \\=> -\dfrac{1}{27}

===

d) ~~~2^{-5}\\ \\ \\ \dfrac{1}{2^5} \\ \\ \\ => \dfrac{1}{32}

===

e) ~~ (0,05) ^{-6}\\\\\\ (\dfrac{5}{100} )^{-6}\\ \\ \\ (\dfrac{1}{20}) ^{-6} \\ \\ \\ \dfrac{20}{1} ^6 \\ \\ \\=>  20^6

===

Anexos:

Imalazy: https://brainly.com.br/tarefa/42963914?answeringSource=feedPublic%2FhomePage%2F1
respondido por: JovemLendário
1

\Box \ \ \boxed{\begin{array}{l}\sf a)11^-^1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{1}{11}  \end{array}}\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf b)(\sqrt{7})^-^2 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf (\frac{1}{\sqrt{7}})^2 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{1^2}{\backslash\!\!\!\sqrt{7\backslash\!\!\!^2}} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{1}{7} \end{array}}

\Box \ \ \boxed{\begin{array}{l}\sf c)(1-4)^-^3 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf (-3)^-^3 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf -\frac{1}{3^3} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf -\frac{1}{27}  \end{array}}

\Box \ \ \boxed{\begin{array}{l}\sf d) 2^-^5 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{1}{2^5}  \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{1}{32} \end{array}}

\Box \ \ \boxed{\begin{array}{l}\sf e)0,05^-^6 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf (\frac{5}{100})^-^6 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf (\frac{1}{20})^-^6 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \frac{20^6}{1} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf 20^6 \end{array}}

\Box \ \ \boxed{\begin{array}{l}\sf 20^6=64.000.000 \end{array}}

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