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
Wow how do you not know how and what kind of bar are you talking about
80 square units
Divide the figure into 4 small triangles, 2 rectangles, and one big rectangle on the center.
Area of ONE small triangle:
1/2 • 2 • 2 = 2 square units
Multiply that by 4 because we have 4 small triangles: 2 • 4 = 8 square units
Area of ONE small rectangle:
2 • 6 = 12 square units
Multiply that by 2 bcos we have 2 of those rectangles: 12 • 2 = 24 square units
Area of the big rectangle on the center:
6 • 8 = 48 square units
ADD the area of the big rectangle, 4 small triangles, and 2 small rectangles:
48 + 24 + 8 = 80
FINAL ANSWER: 80 square units
BRAINLIEST WILL BE APPRECIATED IF I GOT THIS RIGHT (pls comment me back if my answer was correct)
Have a nice day -SpaceMarsh
Answer:
102
Step-by-step explanation:
triangle area = 180 degrees
straight line = 180 degrees
180 - 127 = 53
49
49 + 53 + 3rd angle = 180
3rd angle = 78
180-78 = x
x = 102
