Wait so are the sides of the octagon 9 inches each or do we have to find that?
well, looking at the picture of this vertically opening parabola, it has a vertex at 0,0 and it passes through 2,1 hmm ok
![~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = a(x-0)^2+0\qquad \stackrel{\textit{we also know that}}{x=2\qquad y = 1}\qquad \implies 1=a(2-0)^2+0 \\\\\\ 1=4a\implies \cfrac{1}{4}=a~\hspace{10em} \boxed{y=\cfrac{1}{4}x^2}](https://tex.z-dn.net/?f=~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20y%20%3D%20a%28x-0%29%5E2%2B0%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bwe%20also%20know%20that%7D%7D%7Bx%3D2%5Cqquad%20y%20%3D%201%7D%5Cqquad%20%5Cimplies%201%3Da%282-0%29%5E2%2B0%20%5C%5C%5C%5C%5C%5C%201%3D4a%5Cimplies%20%5Ccfrac%7B1%7D%7B4%7D%3Da~%5Chspace%7B10em%7D%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B4%7Dx%5E2%7D)
Answer:
Because the kind of compounfds that you try to determine using paper chromatography (organic compounds) are usually not soluble in water. Furthermore, water could react chemically with some of this compounds, because it's a very reactive molecule. You need organic solvents that are mostly inert.
Answer:
a =3
b = -5
c = 4
Step-by-step explanation:
3x^2+4=5x
Subtract 5x from each side
3x^2-5x+4=5x-5x
3x^2-5x+4=0
The equation is in the form
ax^2 +bx+c =0
a =3
b = -5
c = 4