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# NEED HELP URGENT

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

Apr 27, 2021

#1
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Plug in the values:
$$\sqrt[3]{4a^4b^5} + \frac{a-c}{(b+c)^2}$$

$$= \sqrt[3]{4(4)^4(2)^5} + \frac{(4)-(-5)}{((2)+(-5))^2}$$

$$= \sqrt[3]{1024 \cdot 32} + \frac{9}{(-3)^2}$$

$$= \sqrt[3]{32768} + \frac{9}{9}$$

$$= 32 + 1$$

$$= 33$$

Apr 27, 2021
#2
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EZ the best answer ever THANK U

Guest Apr 27, 2021
#3
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No problemo :D

Logarhythm  Apr 27, 2021