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

 Jun 17, 2024

Best Answer 

 #1
avatar+1252 
+1

We already have all the information we need to find \(\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}\), so let's plug them in. 

We get

\(\sqrt[3]{2(4)^2 (2)^4} + \dfrac{4 +5}{(2-5)^2}\)

 

Simplifying and multiplying everything out, we get

\(\sqrt[3]{512} + \dfrac{9}{(-3)^2}\\ 8+1\\ 9\)

 

So our final answer is 9. 

I may have made a mistake with computation. If I did, let me know. 

 

Thanks! :)

 Jun 17, 2024
 #1
avatar+1252 
+1
Best Answer

We already have all the information we need to find \(\sqrt[3]{2a^2 b^4} + \dfrac{a - c}{(b+c)^2}\), so let's plug them in. 

We get

\(\sqrt[3]{2(4)^2 (2)^4} + \dfrac{4 +5}{(2-5)^2}\)

 

Simplifying and multiplying everything out, we get

\(\sqrt[3]{512} + \dfrac{9}{(-3)^2}\\ 8+1\\ 9\)

 

So our final answer is 9. 

I may have made a mistake with computation. If I did, let me know. 

 

Thanks! :)

NotThatSmart Jun 17, 2024

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