+0

0
124
6

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
+283
+1

First plug in, the numbers and then simplify.

Jun 19, 2020
#2
0

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

Guest Jun 19, 2020
#6
0

4*4 ≠  ( 4*4 )4

It's not   -1   it's  +1

Guest Jun 19, 2020
#3
0

was i right???

Jun 19, 2020
#4
+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
0

(3√ 4a4 b5 ) + ( a-c ) / ( b+c )2

(3√ 45 25 ) + ( 4- ( -5) / ( 2+(-5)2

(3√ 1024 * 32) + 1 = 33

Jun 19, 2020