Answer:
someone is adding dont worrie
Step-by-step explanation:
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
solution:
we know that ,
u.v = ΙuΙ ΙvΙcosθ
here,
θ =60° (since the given triangle is equilateral triangle)
u.v = ΙuΙ ΙvΙcos60°
= 1 x 1 x 1/2
u.v = 1/2
now, u.w = ΙuΙ ΙwΙcosθ
= ΙuΙ x cos(60x2)
u.w = -1/2
Well, anything that's rotated (turned), reflected (flipped), or translated (moved) is still the same shape of the same size, so the new figures would be congruent with the originals. Dilation, however, is when the shape stays the same but the size changes, so when a figure is dilated, it's not congruent, just similar. 2 out of the 4 are dilated, so those ones would be the answer.
