Answer:
ΔNAS≅ΔSEN by SSA axiom of congruency.
Step-by-step explanation:
Consider ΔNAS and ΔSEN,
NS=SN(Common ie . Both are the same side)
SA=NE( Given in the question that SA≅ NE)
∠SNA=∠NSE( Due to corresponding angle property where SE ║ NA)
Therefore, ΔNAS ≅ΔSEN by SSA axiom of congruency.
∴ NA≅SA by congruent parts of congruent Δ. Hence, proved.
X^3 - x^2 - x + 1 = 0
Plug in x = 1:-
1 -1 - 1 + 1 = 2 - 2 = 0 so x =1 is a root.
If it has multiplicity 3 then the function will factor to (x - 1)^3:-
(x - 1)^3 = (x - 1)(x^2 - 2x + 1) = x^3 - 2x^2 + x - x^2 + 2x - 1
= x^3 - 3x^2 + 3x - 1 so it is FALSE
Step-by-step explanation:

The quadratic formula is honestly the most straightforward way of solving here.
Your other options are completing the square (which is the same thing as the quadratic formula but it's good to know that method if you have to take Integral Calculus at some point) or maybe factoring by grouping if it's appropriate. But the quadratic formula will work for you in all three equations:
1) a=1, b=0, c=8
This reduces pretty quickly into x=8i,-8i due to the negative under the radical. (Actually we didn't even really need the formula here.)
2) a=-9, b=4, c=-10
This reduces into x=(-4+i√(344))/-18, (-4-i√(344))/-18 and doesn't go any further because 344 isn't a perfect square.
3) a=1, b=8, c=17
This reduces to x=(-4+i), (-4-i)
So those are the answers for each.
Check the picture below.
the distance from 1,2 to 1,8 is simply 6 units, we can read that off the grid. Now let's see what the other distances are, and add them all up to get the perimeter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{2})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AC=\sqrt{(5-1)^2+(5-2)^2}\implies AC=\sqrt{4^2+3^2} \\\\\\ AC=\sqrt{25}\implies AC=5 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B2%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AC%3D%5Csqrt%7B%285-1%29%5E2%2B%285-2%29%5E2%7D%5Cimplies%20AC%3D%5Csqrt%7B4%5E2%2B3%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AC%3D%5Csqrt%7B25%7D%5Cimplies%20AC%3D5%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf B(\stackrel{x_1}{1}~,~\stackrel{y_1}{8})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{5}) \\\\\\ BC=\sqrt{(5-1)^2+(5-8)^2}\implies BC=\sqrt{4^2+3^2} \\\\\\ BC=\sqrt{25}\implies BC=5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{perimeter}{6+5+5\implies 16}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20B%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B8%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B%285-1%29%5E2%2B%285-8%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B4%5E2%2B3%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B25%7D%5Cimplies%20BC%3D5%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7Bperimeter%7D%7B6%2B5%2B5%5Cimplies%2016%7D~%5Chfill)