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)
I wrote the domain and range in interval notation, not sure if that’s how you’re asked to do it. This is how I was taught to state the domain/range of a graph.
You add the like terms. In this case, the like terms are -6w,+7w & 5,-4.
-6w + 7w = 1w.
5 - 4 = -1.
The coefficients (terms with variables - letters) comes firm, then the terms (numbers)
.
so the final answer is: 1w -1 :)
Answer:
11.31 units
Step-by-step explanation:
a^2 + b^2 = c^2
8^2 + 8^2 = c^2
64 + 64 = c^2
128 = c^2
Opposite of exponents is square roots, and we need to get rid of that exponent in c^2, meaning we find the square root.
= c
c = 11.31 units
Parent function = √x
On stretching vertically by factor of 6
New function G(x) = 6√x