We can rewrite the expression under the radical as
then taking the fourth root, we get
![\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4%7D%3D%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C)
Why the absolute value? It's for the same reason that
since both 
and 
return the same number , and 
captures both possibilities. From here, we have
The absolute values disappear on all but the 
term because all of 
, 
and 
are positive, while 
could potentially be negative. So we end up with
Answer:
x=12 or x= -12 (answer A)
Step-by-step explanation:
|x| – 5 = 7
|x| = 12
x=12 or x= -12 (answer A)
Answer:
Step-by-step explanation:
Answer:
= 17 v^2 + 7v + 11
Step-by-step explanation:
Given that:
= (8v^2+6v+5)+(9v^2+v+6)
by removing brackets
= 8v^2 + 6v + 5 + 9v^2 + v + 6
Addin like terms:
= (8v^2 + 9v^2) + (6v + v)+ (5 + 6)
= 17 v^2 + 7v + 11
I hope it will help you!
