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)
Subtract 12x from 84x = 72x
The better bargain is the 4 quarts
Nine and ninety-eight hundreths
Answer:
A
Step-by-step explanation:
Cosine is the ratio of "adjacent" to "hypotenuse"
With respect to the Angle E, the adjacent side is FE and the hypotenuse (always opposite side to 90 degree angle) is DE.
We are given DE = 26, but we need FE. We will solve for FE using Pythagorean Theorem, which tells us:
leg^2 + another leg^2 = hypotenuse^2
So, we will have:
10^2 + FE^2 = 26^2
Now, we solve for FE:

So,
Cos(E) = adj/hyp = 24/26
Correct answer is A