Answer:
See below
Step-by-step explanation:
When we talk about the function 
, the domain and codomain are generally defaulted to be subsets of the Real set. Once 
and 
such that 
for 
. Therefore,
![\[\sqrt{\cdot}: \mathbb R_{\geq 0} \to \mathbb R_{\geq 0} \]](https://tex.z-dn.net/?f=%5C%5B%5Csqrt%7B%5Ccdot%7D%3A%20%5Cmathbb%20R_%7B%5Cgeq%200%7D%20%5Cto%20%5Cmathbb%20R_%7B%5Cgeq%200%7D%20%5C%5D)
![\[x \mapsto \sqrt{x}\]](https://tex.z-dn.net/?f=%5C%5Bx%20%5Cmapsto%20%5Csqrt%7Bx%7D%5C%5D)
But this table just shows the perfect square solutions.
If you simplify what's in the bracket, you get 3n
=(3n)+4=32/2
=3n+4= 32/2(here we can either crossmultiply or equate)
3n+4-4=32/2-4(subtract 4 from both sides)
3n=32/2-4
3n=24/2
3n=12
3n=12(divide both sides by the coefficient of the unknown)
3n/3=12/3
n=4
Answer: The answer is (A) ∠T.
Step-by-step explanation: Given that the polygon ABCDE is congruent to the polygon TVSRK. We are to find the corresponding angle of ∠EAB.
In the two polygons, the corresponding vertices are
A ⇒ T
B ⇒ V
C ⇒ S
D ⇒ R
E ⇒ K.
Therefore, ∠EAB, which is ∠A will correspond to ∠T.
Thus, the correct option is (A) ∠T.
Answer: 10√3
Step-by-step explanation:
