+0  
 
0
32
3
avatar

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
avatar+529 
+1

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
avatar
+1

EZ the best answer ever THANK U

Guest Apr 27, 2021
 #3
avatar+529 
0

No problemo :D

Logarhythm  Apr 27, 2021

21 Online Users

avatar
avatar