For midsegment theorem
And parallels lune are twice
So AB = YZ /2
AB = 10/2 =5
BUT AB = X–1 =5
X–1 = 5 = X = 1+5 =6
X =;6
..
MARK AS Brainliest
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:
can you give us more info maybe we can answer it then....?
Step-by-step explanation:
Answer:
The answer to the equation should be C
Answer:
2884.8 millimeters squared
Step-by-step explanation:
Given:
The radius of the disc is 35 millimeters, so the area of it is:
π
= 3.14*
= 3846.5
Then, we find out the area of the circular hold cut out of the bigger one, its radius is a haft of the radius of the bigger circle = 35/2 = 17.5
π
= 3.14*
=961.6
=> the area of the pendant = 3846.5 - 961.6 =2884.8 millimeters squared