Please help me2 a,b,c
1 answer:
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The complete question in the attached figure
we know that
the diagonals of a rhombus intersect to form right angles,
so
angle ACE is ----------> (90°-64°)-----------> 26°
ACE is the angle bisector of ACD, this means that ACD is ---------> 26 x 2 = 52°
The diagonals are angle bisectors to the opposite corners
so
ACD = ACB = 52°
and
BCD = 52 x 2 = 104°
For a rhombus, opposite angles are equivalent,
so
BAD = BCD = 104°
the answer is
angle BAD=104°
we know that
the diagonals of a rhombus intersect to form right angles,
so
angle ACE is ----------> (90°-64°)-----------> 26°
ACE is the angle bisector of ACD, this means that ACD is ---------> 26 x 2 = 52°
The diagonals are angle bisectors to the opposite corners
so
ACD = ACB = 52°
and
BCD = 52 x 2 = 104°
For a rhombus, opposite angles are equivalent,
so
BAD = BCD = 104°
the answer is
angle BAD=104°
So, notice, the focus point is at -7, 5, and the directrix is at y = -11.
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-7\\ k=-3\\ p=8 \end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2 \\\\\\ 32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2 \\\\\\ y=\cfrac{1}{32}(x+7)^2-3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-7%5C%5C%0Ak%3D-3%5C%5C%0Ap%3D8%0A%5Cend%7Bcases%7D%5Cimplies%204%288%29%5By-%28-3%29%5D%3D%5Bx-%28-7%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A32%28y%2B3%29%3D%28x%2B7%29%5E2%5Cimplies%20y%2B3%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2-3)
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.