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
20 and 2
20 - 2 = 18
......................................
Answer: f(3) = 7
Step-by-step explanation:
f(3) = 5(3) - 8
15 - 8
f(3) = 7
Answer:
C.
Step-by-step explanation:
<span> high and the </span>diameter<span> is </span>8cm using<span> the formula </span>V<span>=.π r ²h, </span>work out<span> how </span>much<span> ...the </span>height<span> h is given as </span>12 cm<span> ... Note: Generally we would </span>use<span> 3.14 as the rounded </span>value<span> of </span>pi<span>. ... when dealing </span>with<span> liquids, turn cm</span>3<span> to milliliters. ... Adding and Subtracting Polynomials Math Word Problem Solution </span>Method<span> ..</span>
