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avatar+36 

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

 Jun 23, 2024

Best Answer 

 #1
avatar+1230 
+1

Let's plug everything in and go from there. 

Plugging in 4, 2, and -5,

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

 

Simplfying, we get

\(\sqrt[3]{256*32*4}+\frac{9}{(-3)^2}\\ = \sqrt[3]{32768}+1\\ =32+1\\ =33\)

 

So our final answer is 33. 

 

Thanks! :)

 Jun 24, 2024
 #1
avatar+1230 
+1
Best Answer

Let's plug everything in and go from there. 

Plugging in 4, 2, and -5,

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

 

Simplfying, we get

\(\sqrt[3]{256*32*4}+\frac{9}{(-3)^2}\\ = \sqrt[3]{32768}+1\\ =32+1\\ =33\)

 

So our final answer is 33. 

 

Thanks! :)

NotThatSmart Jun 24, 2024
 #2
avatar+36 
+1

thank you!

onyulee  Jun 24, 2024
edited by onyulee  Jun 24, 2024
 #3
avatar+1230 
+1

My pleasure! 

I hope you understood! :)

 

Thanks! :)

NotThatSmart  Jun 24, 2024

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