Answer:
x=37°
y=6
Step-by-step explanation:
(ASSUMING BD is a perpendicular bisector of AC, otherwise answers maybe be wrong)
Consider the triangle ABC:
AB is the same length as BC (marked), meaning ABC is an isosceles triangle.
Since ABC is isosceles, ∠BAC=∠BCA=53
All angles in a triangle add up to 180:
180=∠ABC + ∠BAC + ∠BCA
180= ∠ABC + 53 +53
∠ABC = 180 - 106
∠ABC = 74
Assuming BD bisects ∠ABC perfectly, x is half of ∠ABC.
x=74/2=37
If BD bisects AC perfectly, AD=DC=y=6
Bread - 3 choices
Meat - 5 choices
Veggies - 7 choices
1br 1mt 1vg = 3×5×7 = 105
1br 2mt 3 vg = 3×5×4×7×6×5 =12600
7vg 1br = 3×1 = 3
Answer:
a^2+B^2=c^2
Step-by-step explanation:
4-3i is the complex conjugate
since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.
we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B-5-%28-7%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%28-5%2B7%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B2%5E2%2B2%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B2%5Ccdot%202%5E2%7D%5Cimplies%20d%3D2%5Csqrt%7B2%7D~%5Chfill%20%5Cstackrel%7B~%5Chfill%20radius%7D%7B%5Ccfrac%7B2%5Csqrt%7B2%7D%7D%7B2%7D%5Cimplies%5Cboxed%7B%20%5Csqrt%7B2%7D%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-4%7D~%2C~%5Cstackrel%7By_1%7D%7B-7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-2%7D~%2C~%5Cstackrel%7By_2%7D%7B-5%7D%29%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%20%5Ccfrac%7B-2-4%7D%7B2%7D~~%2C~~%5Ccfrac%7B-5-7%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-6%7D%7B2%7D~%2C~%5Ccfrac%7B-12%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%5Cboxed%7B%28-3%2C-6%29%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Cqquad%20center~~%28%5Cstackrel%7B-3%7D%7B%20h%7D%2C%5Cstackrel%7B-6%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%5Csqrt%7B2%7D%7D%7B%20r%7D%20%5C%5C%5B2em%5D%20%5Bx-%28-3%29%5D%5E2%2B%5By-%28-6%29%5D%5E2%3D%28%5Csqrt%7B2%7D%29%5E2%5Cimplies%20%28x%2B3%29%5E2%2B%28y%2B6%29%5E2%3D2)