Answer:
x = 38, y = 4
Step-by-step explanation:
Since AB = BC then the triangle is isosceles and the base angles are congruent, that is
∠ DAB = ∠ DCB = 52°
Subtract the sum of the base angle from 180° for ∠ ABC
∠ ABC = 180° - (52 + 52)° = 180° - 104° = 76°
Note that ∠ ABD = ∠ CBD, thus
x = 76 ÷ 2 = 38
BD bisects the side AC, thus DC = AD = 4
Thus y = 4
End fraction is a lot of math lol but try the 16=6 x
Answer:
5x+4
Step-by-step explanation:
Check the picture below, so the parabola looks more or less like so.
the vertex is always half-way between the focus point and the directrix, and since the parabola is opening downwards, the "p" distance is negative.
![\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
