Answer:
C for sure
Step-by-step explanation:
Answer:
659_₹/good your friends and friends happily ever before 4890₹/
Answer:
Step-by-step explanation:
Given
Required
The phase
We have:
Rewrite as:
A cosine function is represented as:
Where:
Phase
By comparison:
Hence, the phase is:
Answer:
Either, (9+√69)/6 or (9-√69)/6
Step wise:
3x²-9x+1=0-------------------(i)
comparing equation onw with (ax²+bx+c=0), we get,
a=3, b=-9, c=1
now,
using Quadratic Formula,
(-b±√b²-4ac)/2a=x
{-(-9)±√(-9)²-4.3.1}/2.3=x
(9±√81-12)/6=x
(9±√69)/6=x
Taking +(ve) sign Taking -(ve) sign
(9+√69)/6=0 (9-√69)/6=0
∴(9+√69)/6=0 ∴(9-√69)/6=0
[∵They cannot be further solved]
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
