sophieposarlie sophieposarlie
  • 20-04-2022
  • Mathematics
contestada

If $a = 4$, $b = 2$, and $c = -5$, then what is the value of $\sqrt[3]{4a^4b^5} + \dfrac{a - c}{(b+c)^2}$?

Answer is not 13 or 31.

Respuesta :

LammettHash
LammettHash LammettHash
  • 20-04-2022

[tex]\sqrt[3]{4 a^4 b^5} = \sqrt[3]{4 \times 4^4 \times 2^5} = \sqrt[3]{(4\times 2)^5} = \sqrt[3]{(2^2\times2)^5} = \sqrt[3]{(2^3)^5} = \sqrt[3]{2^{15}} = \sqrt[3]{(2^5)^3} = 2^5 = 32[/tex]

[tex]\dfrac{a-c}{(b+c)^2} = \dfrac{4-(-5)}{(2+(-5))^2} = \dfrac{4+5}{(2-5)^2} = \dfrac9{(-3)^2} = \dfrac99 = 1[/tex]

Then the expression evaluates to 32 + 1 = 33.

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