Step-by-step explanation:
Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
Answer:
Round to the nearest thousand.
5,568
= 5,000
8h/3+19
Move all terms to the left
8-(h/3+19)=0
Get rid of parentheses
-h/3-19+8=0
Multiply all terms by denominator
-h-19*3+8*3=0
Add all numbers and variables together
-1h-33=0
Move all terms containing h to the left all other terms to the right
-h=33
h=33/-1
h=-33
<h3>
Answer: A) parabola</h3>
Some degenerate parabola cases form a single straight line, while other cases form one pair of parallel lines.
A degenerate hyperbola forms two lines that intersect at the vertex of the cone. We can rule out choice B.
A degenerate circle is a single point, so we can rule out choice C.
A degenerate ellipse is also a single point. Any circle is an ellipse (but not the other way around). We can rule out choice D.
