+0  
 
0
90
6
avatar

I really need help on this problem 

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

 

Plz help 

 Jun 19, 2020
 #1
avatar+243 
+1

First plug in, the numbers and then simplify.

 Jun 19, 2020
 #2
avatar
0

ok but in the end i got \(\sqrt[3]{16^42^5}-1\)

Guest Jun 19, 2020
 #6
avatar
0

You made 2 errors:

 

4*4 ≠  ( 4*4 )4

 

It's not   -1   it's  +1

Guest Jun 19, 2020
 #3
avatar
0

was i right???

 Jun 19, 2020
 #4
avatar+307 
0

Put in the numbers to get \( \sqrt[3]{4*4^4*2^5}+1\)

We can write the numbers inside the square root as powers of 2.

\( \sqrt[3]{2^2*2^8*2^5}+1\).

Now because they all have a common base, you can add the exponents. (To see why, take a basic example like\(2^2*2^3=2*2*2*2*2=2^5\). We are multiplying two 2's and three 2's which makes five threes.)

Adding the exponents, we get \( \sqrt[3]{2^{15}}+1\).

Simplifying, we get \(2^5+1\), which you can figure out on your own.

 Jun 19, 2020
 #5
avatar
0

(3√ 4a4 b5 ) + ( a-c ) / ( b+c )2

 

(3√ 45 25 ) + ( 4- ( -5) / ( 2+(-5)2

 

(3√ 1024 * 32) + 1 = 33   smiley

 Jun 19, 2020

23 Online Users

avatar
avatar
avatar
avatar