Answer is B .....hope this help
Answer:
all work pictured and shown
Check the picture below.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.

![\bf -------------------------------\\\\ d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2} \\\\\\ d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\ -------------------------------\\\\ d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\ -------------------------------\\\\ d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2} \\\\\\ d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%284-2%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%282%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B13%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%283-2%29%5E2%2B%28-2-4%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B37%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-2-3%29%5E2%2B%5B-3-%28-2%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-5%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
![\\\\ -------------------------------\\\\ d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2} \\\\\\ d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-1-%28-2%29%5D%5E2%2B%5B2-%28-3%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B2%29%5E2%2B%282%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%281%29%5E2%2B%285%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
so, those are their lengths, sum them all up, that's the polygon's perimeter.
Answer 32
Step-by-step explanation
32, Give Me Brainiest Please.
Inflection point is the point where the second derivative of a graph is zero.
y = (x+1)arctan xy' = (x+1)(arctan x)' + (1)arctan xy' = (x+1)/(x^2+1) + arctan xy'' = (x+1)(1/(1+x^2))' + 1/(1+x^2) + 1/(1+x^2)y'' = (x+1)(-1/(1+x^2)^2)(2x)+2/(1+x^2)y'' = ((x+1)(-2x)+1+x^2)/(1+x^2)^2y'' = (-2x^2-2x+2+2x^2)/(1+x^2)^2y'' = (-2x+2)/(1+x^2)^2
Solving for point of inflection: y'' = 00 = (-2x+2)/(1+x^2)^20 = -2x+2x = 1y(1) = (1+1)arctan(1) = 2 * pi/4 = pi/2
Therefore, E(1, pi/2).
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